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Arkady needs your help again! This time he decided to build his own high-speed Internet exchange point. It should consist of *n* nodes connected with minimum possible number of wires into one network (a wire directly connects two nodes). Exactly *k* of the nodes should be exit-nodes, that means that each of them should be connected to exactly one other node of the network, while all other nodes should be connected to at least two nodes in order to increase the system stability.
Arkady wants to make the system as fast as possible, so he wants to minimize the maximum distance between two exit-nodes. The distance between two nodes is the number of wires a package needs to go through between those two nodes.
Help Arkady to find such a way to build the network that the distance between the two most distant exit-nodes is as small as possible.
|
The first line contains two integers *n* and *k* (3<=≤<=*n*<=≤<=2·105, 2<=≤<=*k*<=≤<=*n*<=-<=1) — the total number of nodes and the number of exit-nodes.
Note that it is always possible to build at least one network with *n* nodes and *k* exit-nodes within the given constraints.
|
In the first line print the minimum possible distance between the two most distant exit-nodes. In each of the next *n*<=-<=1 lines print two integers: the ids of the nodes connected by a wire. The description of each wire should be printed exactly once. You can print wires and wires' ends in arbitrary order. The nodes should be numbered from 1 to *n*. Exit-nodes can have any ids.
If there are multiple answers, print any of them.
|
[
"3 2\n",
"5 3\n"
] |
[
"2\n1 2\n2 3\n",
"3\n1 2\n2 3\n3 4\n3 5\n"
] |
In the first example the only network is shown on the left picture.
In the second example one of optimal networks is shown on the right picture.
Exit-nodes are highlighted.
| 0
|
[
{
"input": "3 2",
"output": "2\n1 2\n2 3"
},
{
"input": "5 3",
"output": "3\n1 2\n2 3\n3 4\n3 5"
},
{
"input": "4 2",
"output": "3\n1 2\n2 3\n3 4"
},
{
"input": "4 3",
"output": "2\n1 2\n2 3\n2 4"
},
{
"input": "5 2",
"output": "4\n1 2\n2 3\n3 4\n4 5"
},
{
"input": "5 4",
"output": "2\n1 2\n2 3\n2 4\n2 5"
},
{
"input": "6 2",
"output": "5\n1 2\n2 3\n3 4\n4 5\n5 6"
},
{
"input": "6 3",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6"
},
{
"input": "6 4",
"output": "3\n1 2\n2 3\n3 4\n3 5\n3 6"
},
{
"input": "6 5",
"output": "2\n1 2\n2 3\n2 4\n2 5\n2 6"
},
{
"input": "1000 245",
"output": "10\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n6 12\n12 13\n13 14\n14 15\n15 16\n6 17\n17 18\n18 19\n19 20\n20 21\n6 22\n22 23\n23 24\n24 25\n25 26\n6 27\n27 28\n28 29\n29 30\n30 31\n6 32\n32 33\n33 34\n34 35\n35 36\n6 37\n37 38\n38 39\n39 40\n40 41\n6 42\n42 43\n43 44\n44 45\n45 46\n6 47\n47 48\n48 49\n49 50\n50 51\n6 52\n52 53\n53 54\n54 55\n55 56\n6 57\n57 58\n58 59\n59 60\n60 61\n6 62\n62 63\n63 64\n64 65\n65 66\n6 67\n67 68\n68 69\n69 70\n70 71\n6 72\n72 73\n73 74\n74 75\n75 76\n6 77\n77 78\n..."
},
{
"input": "1000 999",
"output": "2\n1 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85\n2 86\n2 87\n..."
},
{
"input": "1024 23",
"output": "90\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76\n..."
},
{
"input": "200000 1014",
"output": "396\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76..."
},
{
"input": "100003 16",
"output": "12502\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 ..."
},
{
"input": "7 2",
"output": "6\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7"
},
{
"input": "7 3",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7"
},
{
"input": "7 4",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n3 7"
},
{
"input": "7 5",
"output": "3\n1 2\n2 3\n3 4\n3 5\n3 6\n3 7"
},
{
"input": "7 6",
"output": "2\n1 2\n2 3\n2 4\n2 5\n2 6\n2 7"
},
{
"input": "100 2",
"output": "99\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76\n..."
},
{
"input": "100 5",
"output": "40\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n21 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n21 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76\n..."
},
{
"input": "100 59",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7\n3 8\n8 9\n3 10\n10 11\n3 12\n12 13\n3 14\n14 15\n3 16\n16 17\n3 18\n18 19\n3 20\n20 21\n3 22\n22 23\n3 24\n24 25\n3 26\n26 27\n3 28\n28 29\n3 30\n30 31\n3 32\n32 33\n3 34\n34 35\n3 36\n36 37\n3 38\n38 39\n3 40\n40 41\n3 42\n42 43\n3 44\n44 45\n3 46\n46 47\n3 48\n48 49\n3 50\n50 51\n3 52\n52 53\n3 54\n54 55\n3 56\n56 57\n3 58\n58 59\n3 60\n60 61\n3 62\n62 63\n3 64\n64 65\n3 66\n66 67\n3 68\n68 69\n3 70\n70 71\n3 72\n72 73\n3 74\n74 75\n3 76\n76 77\n3 78\n78 79\n3 80\n80 81\n..."
},
{
"input": "100 98",
"output": "3\n1 2\n2 3\n3 4\n3 5\n3 6\n3 7\n3 8\n3 9\n3 10\n3 11\n3 12\n3 13\n3 14\n3 15\n3 16\n3 17\n3 18\n3 19\n3 20\n3 21\n3 22\n3 23\n3 24\n3 25\n3 26\n3 27\n3 28\n3 29\n3 30\n3 31\n3 32\n3 33\n3 34\n3 35\n3 36\n3 37\n3 38\n3 39\n3 40\n3 41\n3 42\n3 43\n3 44\n3 45\n3 46\n3 47\n3 48\n3 49\n3 50\n3 51\n3 52\n3 53\n3 54\n3 55\n3 56\n3 57\n3 58\n3 59\n3 60\n3 61\n3 62\n3 63\n3 64\n3 65\n3 66\n3 67\n3 68\n3 69\n3 70\n3 71\n3 72\n3 73\n3 74\n3 75\n3 76\n3 77\n3 78\n3 79\n3 80\n3 81\n3 82\n3 83\n3 84\n3 85\n3 86\n3 87\n..."
},
{
"input": "100 99",
"output": "2\n1 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85\n2 86\n2 87\n..."
},
{
"input": "1000 2",
"output": "999\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76..."
},
{
"input": "1000 5",
"output": "400\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76..."
},
{
"input": "1000 670",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7\n3 8\n8 9\n3 10\n10 11\n3 12\n12 13\n3 14\n14 15\n3 16\n16 17\n3 18\n18 19\n3 20\n20 21\n3 22\n22 23\n3 24\n24 25\n3 26\n26 27\n3 28\n28 29\n3 30\n30 31\n3 32\n32 33\n3 34\n34 35\n3 36\n36 37\n3 38\n38 39\n3 40\n40 41\n3 42\n42 43\n3 44\n44 45\n3 46\n46 47\n3 48\n48 49\n3 50\n50 51\n3 52\n52 53\n3 54\n54 55\n3 56\n56 57\n3 58\n58 59\n3 60\n60 61\n3 62\n62 63\n3 64\n64 65\n3 66\n66 67\n3 68\n68 69\n3 70\n70 71\n3 72\n72 73\n3 74\n74 75\n3 76\n76 77\n3 78\n78 79\n3 80\n80 81\n..."
},
{
"input": "1000 998",
"output": "3\n1 2\n2 3\n3 4\n3 5\n3 6\n3 7\n3 8\n3 9\n3 10\n3 11\n3 12\n3 13\n3 14\n3 15\n3 16\n3 17\n3 18\n3 19\n3 20\n3 21\n3 22\n3 23\n3 24\n3 25\n3 26\n3 27\n3 28\n3 29\n3 30\n3 31\n3 32\n3 33\n3 34\n3 35\n3 36\n3 37\n3 38\n3 39\n3 40\n3 41\n3 42\n3 43\n3 44\n3 45\n3 46\n3 47\n3 48\n3 49\n3 50\n3 51\n3 52\n3 53\n3 54\n3 55\n3 56\n3 57\n3 58\n3 59\n3 60\n3 61\n3 62\n3 63\n3 64\n3 65\n3 66\n3 67\n3 68\n3 69\n3 70\n3 71\n3 72\n3 73\n3 74\n3 75\n3 76\n3 77\n3 78\n3 79\n3 80\n3 81\n3 82\n3 83\n3 84\n3 85\n3 86\n3 87\n..."
},
{
"input": "100000 2",
"output": "99999\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 ..."
},
{
"input": "100000 4",
"output": "50000\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 ..."
},
{
"input": "100000 101",
"output": "1982\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 7..."
},
{
"input": "100000 30005",
"output": "8\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n5 10\n10 11\n11 12\n12 13\n5 14\n14 15\n15 16\n16 17\n5 18\n18 19\n19 20\n20 21\n5 22\n22 23\n23 24\n24 25\n5 26\n26 27\n27 28\n28 29\n5 30\n30 31\n31 32\n32 33\n5 34\n34 35\n35 36\n36 37\n5 38\n38 39\n39 40\n40 41\n5 42\n42 43\n43 44\n44 45\n5 46\n46 47\n47 48\n48 49\n5 50\n50 51\n51 52\n52 53\n5 54\n54 55\n55 56\n56 57\n5 58\n58 59\n59 60\n60 61\n5 62\n62 63\n63 64\n64 65\n5 66\n66 67\n67 68\n68 69\n5 70\n70 71\n71 72\n72 73\n5 74\n74 75\n75 76\n76 77\n5 78\n78 ..."
},
{
"input": "100000 76541",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7\n3 8\n8 9\n3 10\n10 11\n3 12\n12 13\n3 14\n14 15\n3 16\n16 17\n3 18\n18 19\n3 20\n20 21\n3 22\n22 23\n3 24\n24 25\n3 26\n26 27\n3 28\n28 29\n3 30\n30 31\n3 32\n32 33\n3 34\n34 35\n3 36\n36 37\n3 38\n38 39\n3 40\n40 41\n3 42\n42 43\n3 44\n44 45\n3 46\n46 47\n3 48\n48 49\n3 50\n50 51\n3 52\n52 53\n3 54\n54 55\n3 56\n56 57\n3 58\n58 59\n3 60\n60 61\n3 62\n62 63\n3 64\n64 65\n3 66\n66 67\n3 68\n68 69\n3 70\n70 71\n3 72\n72 73\n3 74\n74 75\n3 76\n76 77\n3 78\n78 79\n3 80\n80 81\n..."
},
{
"input": "100000 99998",
"output": "3\n1 2\n2 3\n3 4\n3 5\n3 6\n3 7\n3 8\n3 9\n3 10\n3 11\n3 12\n3 13\n3 14\n3 15\n3 16\n3 17\n3 18\n3 19\n3 20\n3 21\n3 22\n3 23\n3 24\n3 25\n3 26\n3 27\n3 28\n3 29\n3 30\n3 31\n3 32\n3 33\n3 34\n3 35\n3 36\n3 37\n3 38\n3 39\n3 40\n3 41\n3 42\n3 43\n3 44\n3 45\n3 46\n3 47\n3 48\n3 49\n3 50\n3 51\n3 52\n3 53\n3 54\n3 55\n3 56\n3 57\n3 58\n3 59\n3 60\n3 61\n3 62\n3 63\n3 64\n3 65\n3 66\n3 67\n3 68\n3 69\n3 70\n3 71\n3 72\n3 73\n3 74\n3 75\n3 76\n3 77\n3 78\n3 79\n3 80\n3 81\n3 82\n3 83\n3 84\n3 85\n3 86\n3 87\n..."
},
{
"input": "100000 99999",
"output": "2\n1 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85\n2 86\n2 87\n..."
},
{
"input": "200000 2",
"output": "199999\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75..."
},
{
"input": "200000 5",
"output": "80000\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 ..."
},
{
"input": "200000 211",
"output": "1896\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 7..."
},
{
"input": "200000 100002",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7\n3 8\n8 9\n3 10\n10 11\n3 12\n12 13\n3 14\n14 15\n3 16\n16 17\n3 18\n18 19\n3 20\n20 21\n3 22\n22 23\n3 24\n24 25\n3 26\n26 27\n3 28\n28 29\n3 30\n30 31\n3 32\n32 33\n3 34\n34 35\n3 36\n36 37\n3 38\n38 39\n3 40\n40 41\n3 42\n42 43\n3 44\n44 45\n3 46\n46 47\n3 48\n48 49\n3 50\n50 51\n3 52\n52 53\n3 54\n54 55\n3 56\n56 57\n3 58\n58 59\n3 60\n60 61\n3 62\n62 63\n3 64\n64 65\n3 66\n66 67\n3 68\n68 69\n3 70\n70 71\n3 72\n72 73\n3 74\n74 75\n3 76\n76 77\n3 78\n78 79\n3 80\n80 81\n..."
},
{
"input": "200000 145321",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7\n3 8\n8 9\n3 10\n10 11\n3 12\n12 13\n3 14\n14 15\n3 16\n16 17\n3 18\n18 19\n3 20\n20 21\n3 22\n22 23\n3 24\n24 25\n3 26\n26 27\n3 28\n28 29\n3 30\n30 31\n3 32\n32 33\n3 34\n34 35\n3 36\n36 37\n3 38\n38 39\n3 40\n40 41\n3 42\n42 43\n3 44\n44 45\n3 46\n46 47\n3 48\n48 49\n3 50\n50 51\n3 52\n52 53\n3 54\n54 55\n3 56\n56 57\n3 58\n58 59\n3 60\n60 61\n3 62\n62 63\n3 64\n64 65\n3 66\n66 67\n3 68\n68 69\n3 70\n70 71\n3 72\n72 73\n3 74\n74 75\n3 76\n76 77\n3 78\n78 79\n3 80\n80 81\n..."
},
{
"input": "200000 199998",
"output": "3\n1 2\n2 3\n3 4\n3 5\n3 6\n3 7\n3 8\n3 9\n3 10\n3 11\n3 12\n3 13\n3 14\n3 15\n3 16\n3 17\n3 18\n3 19\n3 20\n3 21\n3 22\n3 23\n3 24\n3 25\n3 26\n3 27\n3 28\n3 29\n3 30\n3 31\n3 32\n3 33\n3 34\n3 35\n3 36\n3 37\n3 38\n3 39\n3 40\n3 41\n3 42\n3 43\n3 44\n3 45\n3 46\n3 47\n3 48\n3 49\n3 50\n3 51\n3 52\n3 53\n3 54\n3 55\n3 56\n3 57\n3 58\n3 59\n3 60\n3 61\n3 62\n3 63\n3 64\n3 65\n3 66\n3 67\n3 68\n3 69\n3 70\n3 71\n3 72\n3 73\n3 74\n3 75\n3 76\n3 77\n3 78\n3 79\n3 80\n3 81\n3 82\n3 83\n3 84\n3 85\n3 86\n3 87\n..."
},
{
"input": "200000 199999",
"output": "2\n1 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85\n2 86\n2 87\n..."
},
{
"input": "1024 2",
"output": "1023\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 7..."
},
{
"input": "1024 16",
"output": "128\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76..."
},
{
"input": "1024 512",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7\n3 8\n8 9\n3 10\n10 11\n3 12\n12 13\n3 14\n14 15\n3 16\n16 17\n3 18\n18 19\n3 20\n20 21\n3 22\n22 23\n3 24\n24 25\n3 26\n26 27\n3 28\n28 29\n3 30\n30 31\n3 32\n32 33\n3 34\n34 35\n3 36\n36 37\n3 38\n38 39\n3 40\n40 41\n3 42\n42 43\n3 44\n44 45\n3 46\n46 47\n3 48\n48 49\n3 50\n50 51\n3 52\n52 53\n3 54\n54 55\n3 56\n56 57\n3 58\n58 59\n3 60\n60 61\n3 62\n62 63\n3 64\n64 65\n3 66\n66 67\n3 68\n68 69\n3 70\n70 71\n3 72\n72 73\n3 74\n74 75\n3 76\n76 77\n3 78\n78 79\n3 80\n80 81\n..."
},
{
"input": "1024 511",
"output": "5\n1 2\n2 3\n3 4\n4 5\n5 6\n4 7\n7 8\n4 9\n9 10\n4 11\n11 12\n4 13\n13 14\n4 15\n15 16\n4 17\n17 18\n4 19\n19 20\n4 21\n21 22\n4 23\n23 24\n4 25\n25 26\n4 27\n27 28\n4 29\n29 30\n4 31\n31 32\n4 33\n33 34\n4 35\n35 36\n4 37\n37 38\n4 39\n39 40\n4 41\n41 42\n4 43\n43 44\n4 45\n45 46\n4 47\n47 48\n4 49\n49 50\n4 51\n51 52\n4 53\n53 54\n4 55\n55 56\n4 57\n57 58\n4 59\n59 60\n4 61\n61 62\n4 63\n63 64\n4 65\n65 66\n4 67\n67 68\n4 69\n69 70\n4 71\n71 72\n4 73\n73 74\n4 75\n75 76\n4 77\n77 78\n4 79\n79 80\n4 81\n8..."
},
{
"input": "1024 513",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7\n3 8\n8 9\n3 10\n10 11\n3 12\n12 13\n3 14\n14 15\n3 16\n16 17\n3 18\n18 19\n3 20\n20 21\n3 22\n22 23\n3 24\n24 25\n3 26\n26 27\n3 28\n28 29\n3 30\n30 31\n3 32\n32 33\n3 34\n34 35\n3 36\n36 37\n3 38\n38 39\n3 40\n40 41\n3 42\n42 43\n3 44\n44 45\n3 46\n46 47\n3 48\n48 49\n3 50\n50 51\n3 52\n52 53\n3 54\n54 55\n3 56\n56 57\n3 58\n58 59\n3 60\n60 61\n3 62\n62 63\n3 64\n64 65\n3 66\n66 67\n3 68\n68 69\n3 70\n70 71\n3 72\n72 73\n3 74\n74 75\n3 76\n76 77\n3 78\n78 79\n3 80\n80 81\n..."
},
{
"input": "1024 1023",
"output": "2\n1 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85\n2 86\n2 87\n..."
},
{
"input": "1013 2",
"output": "1012\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 7..."
},
{
"input": "1013 16",
"output": "128\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76..."
},
{
"input": "1013 23",
"output": "88\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76\n..."
},
{
"input": "1013 507",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7\n3 8\n8 9\n3 10\n10 11\n3 12\n12 13\n3 14\n14 15\n3 16\n16 17\n3 18\n18 19\n3 20\n20 21\n3 22\n22 23\n3 24\n24 25\n3 26\n26 27\n3 28\n28 29\n3 30\n30 31\n3 32\n32 33\n3 34\n34 35\n3 36\n36 37\n3 38\n38 39\n3 40\n40 41\n3 42\n42 43\n3 44\n44 45\n3 46\n46 47\n3 48\n48 49\n3 50\n50 51\n3 52\n52 53\n3 54\n54 55\n3 56\n56 57\n3 58\n58 59\n3 60\n60 61\n3 62\n62 63\n3 64\n64 65\n3 66\n66 67\n3 68\n68 69\n3 70\n70 71\n3 72\n72 73\n3 74\n74 75\n3 76\n76 77\n3 78\n78 79\n3 80\n80 81\n..."
},
{
"input": "1013 508",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7\n3 8\n8 9\n3 10\n10 11\n3 12\n12 13\n3 14\n14 15\n3 16\n16 17\n3 18\n18 19\n3 20\n20 21\n3 22\n22 23\n3 24\n24 25\n3 26\n26 27\n3 28\n28 29\n3 30\n30 31\n3 32\n32 33\n3 34\n34 35\n3 36\n36 37\n3 38\n38 39\n3 40\n40 41\n3 42\n42 43\n3 44\n44 45\n3 46\n46 47\n3 48\n48 49\n3 50\n50 51\n3 52\n52 53\n3 54\n54 55\n3 56\n56 57\n3 58\n58 59\n3 60\n60 61\n3 62\n62 63\n3 64\n64 65\n3 66\n66 67\n3 68\n68 69\n3 70\n70 71\n3 72\n72 73\n3 74\n74 75\n3 76\n76 77\n3 78\n78 79\n3 80\n80 81\n..."
},
{
"input": "1013 1012",
"output": "2\n1 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85\n2 86\n2 87\n..."
},
{
"input": "100003 2",
"output": "100002\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75..."
},
{
"input": "100003 23",
"output": "8696\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 7..."
},
{
"input": "100003 19683",
"output": "12\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n7 14\n14 15\n15 16\n16 17\n17 18\n18 19\n7 20\n20 21\n21 22\n22 23\n23 24\n24 25\n7 26\n26 27\n27 28\n28 29\n29 30\n30 31\n7 32\n32 33\n33 34\n34 35\n35 36\n36 37\n7 38\n38 39\n39 40\n40 41\n41 42\n42 43\n7 44\n44 45\n45 46\n46 47\n47 48\n48 49\n7 50\n50 51\n51 52\n52 53\n53 54\n54 55\n7 56\n56 57\n57 58\n58 59\n59 60\n60 61\n7 62\n62 63\n63 64\n64 65\n65 66\n66 67\n7 68\n68 69\n69 70\n70 71\n71 72\n72 73\n7 74\n74 75\n75 76\n76 77\n77 ..."
},
{
"input": "100003 100002",
"output": "2\n1 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85\n2 86\n2 87\n..."
},
{
"input": "100001 2",
"output": "100000\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75..."
},
{
"input": "100001 16",
"output": "12500\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 ..."
},
{
"input": "100001 23",
"output": "8696\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 7..."
},
{
"input": "100001 9091",
"output": "22\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n12 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n12 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n12 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n12 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n12 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76\n..."
},
{
"input": "100001 19683",
"output": "12\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n7 14\n14 15\n15 16\n16 17\n17 18\n18 19\n7 20\n20 21\n21 22\n22 23\n23 24\n24 25\n7 26\n26 27\n27 28\n28 29\n29 30\n30 31\n7 32\n32 33\n33 34\n34 35\n35 36\n36 37\n7 38\n38 39\n39 40\n40 41\n41 42\n42 43\n7 44\n44 45\n45 46\n46 47\n47 48\n48 49\n7 50\n50 51\n51 52\n52 53\n53 54\n54 55\n7 56\n56 57\n57 58\n58 59\n59 60\n60 61\n7 62\n62 63\n63 64\n64 65\n65 66\n66 67\n7 68\n68 69\n69 70\n70 71\n71 72\n72 73\n7 74\n74 75\n75 76\n76 77\n77 ..."
},
{
"input": "100001 50000",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7\n3 8\n8 9\n3 10\n10 11\n3 12\n12 13\n3 14\n14 15\n3 16\n16 17\n3 18\n18 19\n3 20\n20 21\n3 22\n22 23\n3 24\n24 25\n3 26\n26 27\n3 28\n28 29\n3 30\n30 31\n3 32\n32 33\n3 34\n34 35\n3 36\n36 37\n3 38\n38 39\n3 40\n40 41\n3 42\n42 43\n3 44\n44 45\n3 46\n46 47\n3 48\n48 49\n3 50\n50 51\n3 52\n52 53\n3 54\n54 55\n3 56\n56 57\n3 58\n58 59\n3 60\n60 61\n3 62\n62 63\n3 64\n64 65\n3 66\n66 67\n3 68\n68 69\n3 70\n70 71\n3 72\n72 73\n3 74\n74 75\n3 76\n76 77\n3 78\n78 79\n3 80\n80 81\n..."
},
{
"input": "100001 50001",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7\n3 8\n8 9\n3 10\n10 11\n3 12\n12 13\n3 14\n14 15\n3 16\n16 17\n3 18\n18 19\n3 20\n20 21\n3 22\n22 23\n3 24\n24 25\n3 26\n26 27\n3 28\n28 29\n3 30\n30 31\n3 32\n32 33\n3 34\n34 35\n3 36\n36 37\n3 38\n38 39\n3 40\n40 41\n3 42\n42 43\n3 44\n44 45\n3 46\n46 47\n3 48\n48 49\n3 50\n50 51\n3 52\n52 53\n3 54\n54 55\n3 56\n56 57\n3 58\n58 59\n3 60\n60 61\n3 62\n62 63\n3 64\n64 65\n3 66\n66 67\n3 68\n68 69\n3 70\n70 71\n3 72\n72 73\n3 74\n74 75\n3 76\n76 77\n3 78\n78 79\n3 80\n80 81\n..."
},
{
"input": "100001 100000",
"output": "2\n1 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85\n2 86\n2 87\n..."
},
{
"input": "10 6",
"output": "4\n1 2\n2 3\n3 4\n4 5\n3 6\n6 7\n3 8\n3 9\n3 10"
}
] | 1,499,797,577
| 6,077
|
PyPy 3
|
WRONG_ANSWER
|
PRETESTS
| 2
| 93
| 23,142,400
|
n,k=list(map(int,input().strip().split(' ')))
import math
L=math.ceil((n-(k+1))/(k-1))
remain=(n-(k+1))%(k-1)
#print(L,remain,'L,remain')
print(L+2)
print(1,2)
if n>2:
temp=0
end=k+1
for j in range(3,k+1+1):
print(2,j)
if temp<remain:
for p in range(0,L):
if p==0:
print(j,end+1)
end+=1
else:
print(end,end+1)
end+=1
temp+=1
else:
for p in range(0,L-1):
if p==0:
print(j,end+1)
end+=1
else:
print(end,end+1)
end+=1
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arkady needs your help again! This time he decided to build his own high-speed Internet exchange point. It should consist of *n* nodes connected with minimum possible number of wires into one network (a wire directly connects two nodes). Exactly *k* of the nodes should be exit-nodes, that means that each of them should be connected to exactly one other node of the network, while all other nodes should be connected to at least two nodes in order to increase the system stability.
Arkady wants to make the system as fast as possible, so he wants to minimize the maximum distance between two exit-nodes. The distance between two nodes is the number of wires a package needs to go through between those two nodes.
Help Arkady to find such a way to build the network that the distance between the two most distant exit-nodes is as small as possible.
Input Specification:
The first line contains two integers *n* and *k* (3<=≤<=*n*<=≤<=2·105, 2<=≤<=*k*<=≤<=*n*<=-<=1) — the total number of nodes and the number of exit-nodes.
Note that it is always possible to build at least one network with *n* nodes and *k* exit-nodes within the given constraints.
Output Specification:
In the first line print the minimum possible distance between the two most distant exit-nodes. In each of the next *n*<=-<=1 lines print two integers: the ids of the nodes connected by a wire. The description of each wire should be printed exactly once. You can print wires and wires' ends in arbitrary order. The nodes should be numbered from 1 to *n*. Exit-nodes can have any ids.
If there are multiple answers, print any of them.
Demo Input:
['3 2\n', '5 3\n']
Demo Output:
['2\n1 2\n2 3\n', '3\n1 2\n2 3\n3 4\n3 5\n']
Note:
In the first example the only network is shown on the left picture.
In the second example one of optimal networks is shown on the right picture.
Exit-nodes are highlighted.
|
```python
n,k=list(map(int,input().strip().split(' ')))
import math
L=math.ceil((n-(k+1))/(k-1))
remain=(n-(k+1))%(k-1)
#print(L,remain,'L,remain')
print(L+2)
print(1,2)
if n>2:
temp=0
end=k+1
for j in range(3,k+1+1):
print(2,j)
if temp<remain:
for p in range(0,L):
if p==0:
print(j,end+1)
end+=1
else:
print(end,end+1)
end+=1
temp+=1
else:
for p in range(0,L-1):
if p==0:
print(j,end+1)
end+=1
else:
print(end,end+1)
end+=1
```
| 0
|
|
630
|
A
|
Again Twenty Five!
|
PROGRAMMING
| 800
|
[
"number theory"
] | null | null |
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
|
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5.
|
Output the last two digits of 5*n* without spaces between them.
|
[
"2\n"
] |
[
"25"
] |
none
| 0
|
[
{
"input": "2",
"output": "25"
},
{
"input": "7",
"output": "25"
},
{
"input": "1000000000000000000",
"output": "25"
},
{
"input": "2000000000000000000",
"output": "25"
},
{
"input": "987654321012345678",
"output": "25"
}
] | 1,692,990,667
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 31
| 0
|
s=int(input())
print(5**s)
|
Title: Again Twenty Five!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5.
Output Specification:
Output the last two digits of 5*n* without spaces between them.
Demo Input:
['2\n']
Demo Output:
['25']
Note:
none
|
```python
s=int(input())
print(5**s)
```
| 0
|
|
776
|
B
|
Sherlock and his girlfriend
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"number theory"
] | null | null |
Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry.
He bought *n* pieces of jewelry. The *i*-th piece has price equal to *i*<=+<=1, that is, the prices of the jewelry are 2,<=3,<=4,<=... *n*<=+<=1.
Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used.
Help Sherlock complete this trivial task.
|
The only line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of jewelry pieces.
|
The first line of output should contain a single integer *k*, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints.
The next line should consist of *n* space-separated integers (between 1 and *k*) that specify the color of each piece in the order of increasing price.
If there are multiple ways to color the pieces using *k* colors, you can output any of them.
|
[
"3\n",
"4\n"
] |
[
"2\n1 1 2 ",
"2\n2 1 1 2\n"
] |
In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively.
In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct.
| 1,000
|
[
{
"input": "3",
"output": "2\n1 1 2 "
},
{
"input": "4",
"output": "2\n1 1 2 1 "
},
{
"input": "17",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 "
},
{
"input": "25",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 "
},
{
"input": "85",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 "
},
{
"input": "105",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 "
},
{
"input": "123",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "452",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "641",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "293",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "733",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "10",
"output": "2\n1 1 2 1 2 1 2 2 2 1 "
},
{
"input": "287",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "3202",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "728",
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{
"input": "15346",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "41794",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "99998",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "100000",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "2",
"output": "1\n1 1 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "2",
"output": "1\n1 1 "
},
{
"input": "100000",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "99971",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
}
] | 1,515,300,461
| 2,861
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 47
| 109
| 6,758,400
|
a = ["2"] * 100010
a[1] = "1"
for i in range(2, 100001):
if a[i] == "2":
for j in range(i * i, 100001, i):
a[j] = "1"
# print(a[0:100])
n = int(input())
if n <= 2:
print(1)
print(' '.join(["1"] * n))
else:
print(2)
print(' '.join(a[2:n + 2]))
|
Title: Sherlock and his girlfriend
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry.
He bought *n* pieces of jewelry. The *i*-th piece has price equal to *i*<=+<=1, that is, the prices of the jewelry are 2,<=3,<=4,<=... *n*<=+<=1.
Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used.
Help Sherlock complete this trivial task.
Input Specification:
The only line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of jewelry pieces.
Output Specification:
The first line of output should contain a single integer *k*, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints.
The next line should consist of *n* space-separated integers (between 1 and *k*) that specify the color of each piece in the order of increasing price.
If there are multiple ways to color the pieces using *k* colors, you can output any of them.
Demo Input:
['3\n', '4\n']
Demo Output:
['2\n1 1 2 ', '2\n2 1 1 2\n']
Note:
In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively.
In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct.
|
```python
a = ["2"] * 100010
a[1] = "1"
for i in range(2, 100001):
if a[i] == "2":
for j in range(i * i, 100001, i):
a[j] = "1"
# print(a[0:100])
n = int(input())
if n <= 2:
print(1)
print(' '.join(["1"] * n))
else:
print(2)
print(' '.join(a[2:n + 2]))
```
| 0
|
|
70
|
A
|
Cookies
|
PROGRAMMING
| 1,300
|
[
"math"
] |
A. Cookies
|
1
|
256
|
Fangy collects cookies. Once he decided to take a box and put cookies into it in some way. If we take a square *k*<=×<=*k* in size, divided into blocks 1<=×<=1 in size and paint there the main diagonal together with cells, which lie above it, then the painted area will be equal to the area occupied by one cookie *k* in size. Fangy also has a box with a square base 2*n*<=×<=2*n*, divided into blocks 1<=×<=1 in size. In a box the cookies should not overlap, and they should not be turned over or rotated. See cookies of sizes 2 and 4 respectively on the figure:
To stack the cookies the little walrus uses the following algorithm. He takes out of the repository the largest cookie which can fit in some place in the box and puts it there. Everything could be perfect but alas, in the repository the little walrus has infinitely many cookies of size 2 and larger, and there are no cookies of size 1, therefore, empty cells will remain in the box. Fangy wants to know how many empty cells will be left in the end.
|
The first line contains a single integer *n* (0<=≤<=*n*<=≤<=1000).
|
Print the single number, equal to the number of empty cells in the box. The answer should be printed modulo 106<=+<=3.
|
[
"3\n"
] |
[
"9"
] |
If the box possesses the base of 2<sup class="upper-index">3</sup> × 2<sup class="upper-index">3</sup> (as in the example), then the cookies will be put there in the following manner:
| 500
|
[
{
"input": "3",
"output": "9"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "3"
},
{
"input": "4",
"output": "27"
},
{
"input": "6",
"output": "243"
},
{
"input": "11",
"output": "59049"
},
{
"input": "14",
"output": "594320"
},
{
"input": "15",
"output": "782957"
},
{
"input": "7",
"output": "729"
},
{
"input": "0",
"output": "1"
},
{
"input": "1000",
"output": "691074"
},
{
"input": "657",
"output": "874011"
},
{
"input": "561",
"output": "842553"
},
{
"input": "823",
"output": "858672"
},
{
"input": "850",
"output": "557186"
},
{
"input": "298",
"output": "999535"
},
{
"input": "262",
"output": "946384"
},
{
"input": "910",
"output": "678945"
},
{
"input": "617",
"output": "247876"
},
{
"input": "857",
"output": "562128"
},
{
"input": "69",
"output": "327984"
},
{
"input": "589",
"output": "889192"
},
{
"input": "928",
"output": "794863"
},
{
"input": "696",
"output": "695035"
},
{
"input": "226",
"output": "376094"
}
] | 1,650,190,869
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 25
| 46
| 0
|
n = int(input())
if not n:
print(1)
else:
print((3 ** (n - 1)) % ((10 ** 6) + 3))
|
Title: Cookies
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Fangy collects cookies. Once he decided to take a box and put cookies into it in some way. If we take a square *k*<=×<=*k* in size, divided into blocks 1<=×<=1 in size and paint there the main diagonal together with cells, which lie above it, then the painted area will be equal to the area occupied by one cookie *k* in size. Fangy also has a box with a square base 2*n*<=×<=2*n*, divided into blocks 1<=×<=1 in size. In a box the cookies should not overlap, and they should not be turned over or rotated. See cookies of sizes 2 and 4 respectively on the figure:
To stack the cookies the little walrus uses the following algorithm. He takes out of the repository the largest cookie which can fit in some place in the box and puts it there. Everything could be perfect but alas, in the repository the little walrus has infinitely many cookies of size 2 and larger, and there are no cookies of size 1, therefore, empty cells will remain in the box. Fangy wants to know how many empty cells will be left in the end.
Input Specification:
The first line contains a single integer *n* (0<=≤<=*n*<=≤<=1000).
Output Specification:
Print the single number, equal to the number of empty cells in the box. The answer should be printed modulo 106<=+<=3.
Demo Input:
['3\n']
Demo Output:
['9']
Note:
If the box possesses the base of 2<sup class="upper-index">3</sup> × 2<sup class="upper-index">3</sup> (as in the example), then the cookies will be put there in the following manner:
|
```python
n = int(input())
if not n:
print(1)
else:
print((3 ** (n - 1)) % ((10 ** 6) + 3))
```
| 3.977
|
831
|
C
|
Jury Marks
|
PROGRAMMING
| 1,700
|
[
"brute force",
"constructive algorithms"
] | null | null |
Polycarp watched TV-show where *k* jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the *i*-th jury member gave *a**i* points.
Polycarp does not remember how many points the participant had before this *k* marks were given, but he remembers that among the scores announced after each of the *k* judges rated the participant there were *n* (*n*<=≤<=*k*) values *b*1,<=*b*2,<=...,<=*b**n* (it is guaranteed that all values *b**j* are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. *n*<=<<=*k*. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
|
The first line contains two integers *k* and *n* (1<=≤<=*n*<=≤<=*k*<=≤<=2<=000) — the number of jury members and the number of scores Polycarp remembers.
The second line contains *k* integers *a*1,<=*a*2,<=...,<=*a**k* (<=-<=2<=000<=≤<=*a**i*<=≤<=2<=000) — jury's marks in chronological order.
The third line contains *n* distinct integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=4<=000<=000<=≤<=*b**j*<=≤<=4<=000<=000) — the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
|
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
|
[
"4 1\n-5 5 0 20\n10\n",
"2 2\n-2000 -2000\n3998000 4000000\n"
] |
[
"3\n",
"1\n"
] |
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000.
| 1,000
|
[
{
"input": "4 1\n-5 5 0 20\n10",
"output": "3"
},
{
"input": "2 2\n-2000 -2000\n3998000 4000000",
"output": "1"
},
{
"input": "1 1\n-577\n1273042",
"output": "1"
},
{
"input": "2 1\n614 -1943\n3874445",
"output": "2"
},
{
"input": "3 1\n1416 -1483 1844\n3261895",
"output": "3"
},
{
"input": "5 1\n1035 1861 1388 -622 1252\n2640169",
"output": "5"
},
{
"input": "10 10\n-25 746 298 1602 -1453 -541 -442 1174 976 -1857\n-548062 -548253 -546800 -548943 -548402 -548794 -549236 -548700 -549446 -547086",
"output": "1"
},
{
"input": "20 20\n-1012 625 39 -1747 -1626 898 -1261 180 -876 -1417 -1853 -1510 -1499 -561 -1824 442 -895 13 1857 1860\n-1269013 -1270956 -1264151 -1266004 -1268121 -1258341 -1269574 -1271851 -1258302 -1271838 -1260049 -1258966 -1271398 -1267514 -1269981 -1262038 -1261675 -1262734 -1260777 -1261858",
"output": "1"
},
{
"input": "1 1\n1\n-4000000",
"output": "1"
}
] | 1,666,133,289
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 57
| 217
| 10,649,600
|
k, n = map(int, input().split())
marks = list(map(int, input().split()))
polycarps = list(map(int, input().split()))
def psa_init(arr):
psa = [0] * len(arr)
psa[0] = arr[0]
for i in range(1, len(arr)):
psa[i] = arr[i] + psa[i-1]
return psa
psa = psa_init(marks)
possible_starts = set()
for i in range(0, len(psa)):
possible_starts.add(polycarps[0] - psa[i])
if len(polycarps) == 1:
print(len(possible_starts))
else:
total = 0
for i in possible_starts:
big_set = set()
for j in psa:
big_set.add(i + j)
for k in polycarps[1:]:
if k not in big_set:
total -= 1
break
total += 1
print(total)
|
Title: Jury Marks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp watched TV-show where *k* jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the *i*-th jury member gave *a**i* points.
Polycarp does not remember how many points the participant had before this *k* marks were given, but he remembers that among the scores announced after each of the *k* judges rated the participant there were *n* (*n*<=≤<=*k*) values *b*1,<=*b*2,<=...,<=*b**n* (it is guaranteed that all values *b**j* are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. *n*<=<<=*k*. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input Specification:
The first line contains two integers *k* and *n* (1<=≤<=*n*<=≤<=*k*<=≤<=2<=000) — the number of jury members and the number of scores Polycarp remembers.
The second line contains *k* integers *a*1,<=*a*2,<=...,<=*a**k* (<=-<=2<=000<=≤<=*a**i*<=≤<=2<=000) — jury's marks in chronological order.
The third line contains *n* distinct integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=4<=000<=000<=≤<=*b**j*<=≤<=4<=000<=000) — the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output Specification:
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Demo Input:
['4 1\n-5 5 0 20\n10\n', '2 2\n-2000 -2000\n3998000 4000000\n']
Demo Output:
['3\n', '1\n']
Note:
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000.
|
```python
k, n = map(int, input().split())
marks = list(map(int, input().split()))
polycarps = list(map(int, input().split()))
def psa_init(arr):
psa = [0] * len(arr)
psa[0] = arr[0]
for i in range(1, len(arr)):
psa[i] = arr[i] + psa[i-1]
return psa
psa = psa_init(marks)
possible_starts = set()
for i in range(0, len(psa)):
possible_starts.add(polycarps[0] - psa[i])
if len(polycarps) == 1:
print(len(possible_starts))
else:
total = 0
for i in possible_starts:
big_set = set()
for j in psa:
big_set.add(i + j)
for k in polycarps[1:]:
if k not in big_set:
total -= 1
break
total += 1
print(total)
```
| 3
|
|
131
|
A
|
cAPS lOCK
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
|
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
|
Print the result of the given word's processing.
|
[
"cAPS\n",
"Lock\n"
] |
[
"Caps",
"Lock\n"
] |
none
| 500
|
[
{
"input": "cAPS",
"output": "Caps"
},
{
"input": "Lock",
"output": "Lock"
},
{
"input": "cAPSlOCK",
"output": "cAPSlOCK"
},
{
"input": "CAPs",
"output": "CAPs"
},
{
"input": "LoCK",
"output": "LoCK"
},
{
"input": "OOPS",
"output": "oops"
},
{
"input": "oops",
"output": "oops"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "a"
},
{
"input": "aA",
"output": "Aa"
},
{
"input": "Zz",
"output": "Zz"
},
{
"input": "Az",
"output": "Az"
},
{
"input": "zA",
"output": "Za"
},
{
"input": "AAA",
"output": "aaa"
},
{
"input": "AAa",
"output": "AAa"
},
{
"input": "AaR",
"output": "AaR"
},
{
"input": "Tdr",
"output": "Tdr"
},
{
"input": "aTF",
"output": "Atf"
},
{
"input": "fYd",
"output": "fYd"
},
{
"input": "dsA",
"output": "dsA"
},
{
"input": "fru",
"output": "fru"
},
{
"input": "hYBKF",
"output": "Hybkf"
},
{
"input": "XweAR",
"output": "XweAR"
},
{
"input": "mogqx",
"output": "mogqx"
},
{
"input": "eOhEi",
"output": "eOhEi"
},
{
"input": "nkdku",
"output": "nkdku"
},
{
"input": "zcnko",
"output": "zcnko"
},
{
"input": "lcccd",
"output": "lcccd"
},
{
"input": "vwmvg",
"output": "vwmvg"
},
{
"input": "lvchf",
"output": "lvchf"
},
{
"input": "IUNVZCCHEWENCHQQXQYPUJCRDZLUXCLJHXPHBXEUUGNXOOOPBMOBRIBHHMIRILYJGYYGFMTMFSVURGYHUWDRLQVIBRLPEVAMJQYO",
"output": "iunvzcchewenchqqxqypujcrdzluxcljhxphbxeuugnxooopbmobribhhmirilyjgyygfmtmfsvurgyhuwdrlqvibrlpevamjqyo"
},
{
"input": "OBHSZCAMDXEJWOZLKXQKIVXUUQJKJLMMFNBPXAEFXGVNSKQLJGXHUXHGCOTESIVKSFMVVXFVMTEKACRIWALAGGMCGFEXQKNYMRTG",
"output": "obhszcamdxejwozlkxqkivxuuqjkjlmmfnbpxaefxgvnskqljgxhuxhgcotesivksfmvvxfvmtekacriwalaggmcgfexqknymrtg"
},
{
"input": "IKJYZIKROIYUUCTHSVSKZTETNNOCMAUBLFJCEVANCADASMZRCNLBZPQRXESHEEMOMEPCHROSRTNBIDXYMEPJSIXSZQEBTEKKUHFS",
"output": "ikjyzikroiyuucthsvskztetnnocmaublfjcevancadasmzrcnlbzpqrxesheemomepchrosrtnbidxymepjsixszqebtekkuhfs"
},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "uCKJZRGZJCPPLEEYJTUNKOQSWGBMTBQEVPYFPIPEKRVYQNTDPANOIXKMPINNFUSZWCURGBDPYTEKBEKCPMVZPMWAOSHJYMGKOMBQ",
"output": "Uckjzrgzjcppleeyjtunkoqswgbmtbqevpyfpipekrvyqntdpanoixkmpinnfuszwcurgbdpytekbekcpmvzpmwaoshjymgkombq"
},
{
"input": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR",
"output": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR"
},
{
"input": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE",
"output": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE"
},
{
"input": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ",
"output": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ"
},
{
"input": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm",
"output": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm"
},
{
"input": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm",
"output": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm"
},
{
"input": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg",
"output": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg"
},
{
"input": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc",
"output": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc"
},
{
"input": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv",
"output": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv"
},
{
"input": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect",
"output": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect"
},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu",
"output": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu"
},
{
"input": "aBACABa",
"output": "aBACABa"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAaa",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAaa"
},
{
"input": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA"
},
{
"input": "dDDDDDDDDDDDDD",
"output": "Dddddddddddddd"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "z",
"output": "Z"
},
{
"input": "AZ",
"output": "az"
},
{
"input": "Z",
"output": "z"
},
{
"input": "aAAAA",
"output": "Aaaaa"
},
{
"input": "F",
"output": "f"
}
] | 1,696,167,865
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 46
| 0
|
def def1(x):
if x.isupper() or x[1:].isupper():
return x.swapcase()
else:
return x
x1 = input().strip()
trans = def1(x1)
print(trans)
|
Title: cAPS lOCK
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input Specification:
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output Specification:
Print the result of the given word's processing.
Demo Input:
['cAPS\n', 'Lock\n']
Demo Output:
['Caps', 'Lock\n']
Note:
none
|
```python
def def1(x):
if x.isupper() or x[1:].isupper():
return x.swapcase()
else:
return x
x1 = input().strip()
trans = def1(x1)
print(trans)
```
| 0
|
|
909
|
A
|
Generate Login
|
PROGRAMMING
| 1,000
|
[
"brute force",
"greedy",
"sortings"
] | null | null |
The preferred way to generate user login in Polygon is to concatenate a prefix of the user's first name and a prefix of their last name, in that order. Each prefix must be non-empty, and any of the prefixes can be the full name. Typically there are multiple possible logins for each person.
You are given the first and the last name of a user. Return the alphabetically earliest login they can get (regardless of other potential Polygon users).
As a reminder, a prefix of a string *s* is its substring which occurs at the beginning of *s*: "a", "ab", "abc" etc. are prefixes of string "{abcdef}" but "b" and 'bc" are not. A string *a* is alphabetically earlier than a string *b*, if *a* is a prefix of *b*, or *a* and *b* coincide up to some position, and then *a* has a letter that is alphabetically earlier than the corresponding letter in *b*: "a" and "ab" are alphabetically earlier than "ac" but "b" and "ba" are alphabetically later than "ac".
|
The input consists of a single line containing two space-separated strings: the first and the last names. Each character of each string is a lowercase English letter. The length of each string is between 1 and 10, inclusive.
|
Output a single string — alphabetically earliest possible login formed from these names. The output should be given in lowercase as well.
|
[
"harry potter\n",
"tom riddle\n"
] |
[
"hap\n",
"tomr\n"
] |
none
| 500
|
[
{
"input": "harry potter",
"output": "hap"
},
{
"input": "tom riddle",
"output": "tomr"
},
{
"input": "a qdpinbmcrf",
"output": "aq"
},
{
"input": "wixjzniiub ssdfodfgap",
"output": "wis"
},
{
"input": "z z",
"output": "zz"
},
{
"input": "ertuyivhfg v",
"output": "ertuv"
},
{
"input": "asdfghjkli ware",
"output": "asdfghjkliw"
},
{
"input": "udggmyop ze",
"output": "udggmyopz"
},
{
"input": "fapkdme rtzxovx",
"output": "fapkdmer"
},
{
"input": "mybiqxmnqq l",
"output": "ml"
},
{
"input": "dtbqya fyyymv",
"output": "df"
},
{
"input": "fyclu zokbxiahao",
"output": "fycluz"
},
{
"input": "qngatnviv rdych",
"output": "qngar"
},
{
"input": "ttvnhrnng lqkfulhrn",
"output": "tl"
},
{
"input": "fya fgx",
"output": "ff"
},
{
"input": "nuis zvjjqlre",
"output": "nuisz"
},
{
"input": "ly qtsmze",
"output": "lq"
},
{
"input": "d kgfpjsurfw",
"output": "dk"
},
{
"input": "lwli ewrpu",
"output": "le"
},
{
"input": "rr wldsfubcs",
"output": "rrw"
},
{
"input": "h qart",
"output": "hq"
},
{
"input": "vugvblnzx kqdwdulm",
"output": "vk"
},
{
"input": "xohesmku ef",
"output": "xe"
},
{
"input": "twvvsl wtcyawv",
"output": "tw"
},
{
"input": "obljndajv q",
"output": "obljndajq"
},
{
"input": "jjxwj kxccwx",
"output": "jjk"
},
{
"input": "sk fftzmv",
"output": "sf"
},
{
"input": "cgpegngs aufzxkyyrw",
"output": "ca"
},
{
"input": "reyjzjdvq skuch",
"output": "res"
},
{
"input": "ardaae mxgdulijf",
"output": "am"
},
{
"input": "bgopsdfji uaps",
"output": "bgopsdfjiu"
},
{
"input": "amolfed pun",
"output": "amolfedp"
},
{
"input": "badkiln yort",
"output": "badkilny"
},
{
"input": "aaaaaaaaaz york",
"output": "aaaaaaaaay"
},
{
"input": "bbbbcbbbbd c",
"output": "bbbbc"
},
{
"input": "aa ab",
"output": "aa"
},
{
"input": "ab b",
"output": "ab"
},
{
"input": "aaaaa ab",
"output": "aa"
},
{
"input": "aa a",
"output": "aa"
},
{
"input": "aba b",
"output": "ab"
},
{
"input": "aaaaaaa aaaaaa",
"output": "aa"
},
{
"input": "a a",
"output": "aa"
},
{
"input": "a aa",
"output": "aa"
},
{
"input": "a b",
"output": "ab"
},
{
"input": "b a",
"output": "ba"
},
{
"input": "z a",
"output": "za"
},
{
"input": "aaa a",
"output": "aa"
},
{
"input": "aa aa",
"output": "aa"
},
{
"input": "a aaa",
"output": "aa"
},
{
"input": "aaaaaaaaaa aaaaaaaaaa",
"output": "aa"
},
{
"input": "aaaaaaaaaa a",
"output": "aa"
},
{
"input": "a aaaaaaaaaa",
"output": "aa"
},
{
"input": "zzaa b",
"output": "zb"
},
{
"input": "ca cf",
"output": "cac"
},
{
"input": "abhi ia",
"output": "abhi"
},
{
"input": "aaaa aaaab",
"output": "aa"
},
{
"input": "aar raa",
"output": "aar"
},
{
"input": "harry hotter",
"output": "hah"
},
{
"input": "aaaaaaa a",
"output": "aa"
},
{
"input": "apple pie",
"output": "ap"
},
{
"input": "aaa aaa",
"output": "aa"
},
{
"input": "kabc buba",
"output": "kab"
},
{
"input": "asd ss",
"output": "as"
},
{
"input": "bbb b",
"output": "bb"
}
] | 1,688,144,008
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 64
| 46
| 0
|
a, b = input().split()
f=1
ans=''
for i in range(1, len(a)):
if ord(a[i]) >= ord(b[0]):
ans=a[0:i] + b[0]
f=0
break
if(f==0):
print(ans)
else:
print(a + b[0])
|
Title: Generate Login
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The preferred way to generate user login in Polygon is to concatenate a prefix of the user's first name and a prefix of their last name, in that order. Each prefix must be non-empty, and any of the prefixes can be the full name. Typically there are multiple possible logins for each person.
You are given the first and the last name of a user. Return the alphabetically earliest login they can get (regardless of other potential Polygon users).
As a reminder, a prefix of a string *s* is its substring which occurs at the beginning of *s*: "a", "ab", "abc" etc. are prefixes of string "{abcdef}" but "b" and 'bc" are not. A string *a* is alphabetically earlier than a string *b*, if *a* is a prefix of *b*, or *a* and *b* coincide up to some position, and then *a* has a letter that is alphabetically earlier than the corresponding letter in *b*: "a" and "ab" are alphabetically earlier than "ac" but "b" and "ba" are alphabetically later than "ac".
Input Specification:
The input consists of a single line containing two space-separated strings: the first and the last names. Each character of each string is a lowercase English letter. The length of each string is between 1 and 10, inclusive.
Output Specification:
Output a single string — alphabetically earliest possible login formed from these names. The output should be given in lowercase as well.
Demo Input:
['harry potter\n', 'tom riddle\n']
Demo Output:
['hap\n', 'tomr\n']
Note:
none
|
```python
a, b = input().split()
f=1
ans=''
for i in range(1, len(a)):
if ord(a[i]) >= ord(b[0]):
ans=a[0:i] + b[0]
f=0
break
if(f==0):
print(ans)
else:
print(a + b[0])
```
| 3
|
|
580
|
A
|
Kefa and First Steps
|
PROGRAMMING
| 900
|
[
"brute force",
"dp",
"implementation"
] | null | null |
Kefa decided to make some money doing business on the Internet for exactly *n* days. He knows that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) he makes *a**i* money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence *a**i*. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.
Help Kefa cope with this task!
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (1<=≤<=*a**i*<=≤<=109).
|
Print a single integer — the length of the maximum non-decreasing subsegment of sequence *a*.
|
[
"6\n2 2 1 3 4 1\n",
"3\n2 2 9\n"
] |
[
"3",
"3"
] |
In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one.
In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.
| 750
|
[
{
"input": "6\n2 2 1 3 4 1",
"output": "3"
},
{
"input": "3\n2 2 9",
"output": "3"
},
{
"input": "5\n10 100 111 1 2",
"output": "3"
},
{
"input": "10\n1 2 3 4 1 2 3 4 5 6",
"output": "6"
},
{
"input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "50"
},
{
"input": "100\n1 838 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 605 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 27 533 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 835 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 992",
"output": "42"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "6\n5 4 3 2 1 2",
"output": "2"
},
{
"input": "9\n1 2 3 4 5 6 7 8 9",
"output": "9"
},
{
"input": "7\n99 100 1 2 3 4 5",
"output": "5"
},
{
"input": "5\n3 3 1 2 3",
"output": "3"
},
{
"input": "1\n100",
"output": "1"
},
{
"input": "1\n5",
"output": "1"
}
] | 1,694,786,546
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 93
| 13,926,400
|
n = int(input())
arr = list(map(int, input().split()))
ans, br, cur = 0, 0, 0
for x in range(0, n):
val = arr[x]
if val < br:
ans = max(ans, cur)
cur = 1
else:
cur += 1
br = val
ans = max(ans, cur)
print(ans)
|
Title: Kefa and First Steps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kefa decided to make some money doing business on the Internet for exactly *n* days. He knows that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) he makes *a**i* money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence *a**i*. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.
Help Kefa cope with this task!
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the length of the maximum non-decreasing subsegment of sequence *a*.
Demo Input:
['6\n2 2 1 3 4 1\n', '3\n2 2 9\n']
Demo Output:
['3', '3']
Note:
In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one.
In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.
|
```python
n = int(input())
arr = list(map(int, input().split()))
ans, br, cur = 0, 0, 0
for x in range(0, n):
val = arr[x]
if val < br:
ans = max(ans, cur)
cur = 1
else:
cur += 1
br = val
ans = max(ans, cur)
print(ans)
```
| 3
|
|
271
|
A
|
Beautiful Year
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.
|
The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number.
|
Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists.
|
[
"1987\n",
"2013\n"
] |
[
"2013\n",
"2014\n"
] |
none
| 500
|
[
{
"input": "1987",
"output": "2013"
},
{
"input": "2013",
"output": "2014"
},
{
"input": "1000",
"output": "1023"
},
{
"input": "1001",
"output": "1023"
},
{
"input": "1234",
"output": "1235"
},
{
"input": "5555",
"output": "5601"
},
{
"input": "9000",
"output": "9012"
},
{
"input": "1111",
"output": "1203"
},
{
"input": "8999",
"output": "9012"
},
{
"input": "4572",
"output": "4573"
},
{
"input": "6666",
"output": "6701"
},
{
"input": "2001",
"output": "2013"
},
{
"input": "3000",
"output": "3012"
},
{
"input": "7712",
"output": "7801"
},
{
"input": "8088",
"output": "8091"
},
{
"input": "1594",
"output": "1596"
},
{
"input": "6016",
"output": "6017"
},
{
"input": "8800",
"output": "8901"
},
{
"input": "3331",
"output": "3401"
},
{
"input": "5090",
"output": "5091"
},
{
"input": "2342",
"output": "2345"
},
{
"input": "2334",
"output": "2340"
},
{
"input": "1123",
"output": "1203"
},
{
"input": "8989",
"output": "9012"
},
{
"input": "8977",
"output": "9012"
},
{
"input": "6869",
"output": "6870"
},
{
"input": "8999",
"output": "9012"
}
] | 1,693,288,998
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 124
| 2,764,800
|
import sys
y = int(sys.stdin.readline()) + 1
while len(set(y)) < 4: y += 1
sys.stdout.write(str(y))
|
Title: Beautiful Year
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.
Input Specification:
The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number.
Output Specification:
Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists.
Demo Input:
['1987\n', '2013\n']
Demo Output:
['2013\n', '2014\n']
Note:
none
|
```python
import sys
y = int(sys.stdin.readline()) + 1
while len(set(y)) < 4: y += 1
sys.stdout.write(str(y))
```
| -1
|
|
976
|
A
|
Minimum Binary Number
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all).
|
The first line contains integer number *n* (1<=≤<=*n*<=≤<=100) — the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct.
|
Print one string — the minimum correct string that you can obtain from the given one.
|
[
"4\n1001\n",
"1\n1\n"
] |
[
"100\n",
"1\n"
] |
In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use.
| 0
|
[
{
"input": "4\n1001",
"output": "100"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "100\n1110111100001111011111111010110011111111011110000111101101011100110110001011000000101010110101011100",
"output": "1000000000000000000000000000000000000000"
},
{
"input": "100\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "1"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111110111111111111111111111111111111111111111111",
"output": "10"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "8\n10101010",
"output": "10000"
},
{
"input": "2\n10",
"output": "10"
},
{
"input": "3\n111",
"output": "1"
},
{
"input": "5\n11100",
"output": "100"
},
{
"input": "2\n11",
"output": "1"
},
{
"input": "3\n110",
"output": "10"
},
{
"input": "50\n10010010000000000000000000000000000000001000000000",
"output": "10000000000000000000000000000000000000000000000"
}
] | 1,600,439,961
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 14
| 217
| 2,355,200
|
import math as mt
import sys, string
from collections import Counter, defaultdict
input = sys.stdin.readline
# input functions
I = lambda : int(input())
M = lambda : map(int, input().split())
ARR = lambda: list(map(int, input().split()))
def main():
n = I()
s = input()
s = list(s)
n = len(s) - 1
c1, c0 = 0, 0
for i in range(n):
if s[i] == '1':
c1 += 1
elif s[i] == '0':
c0 += 1
if c1 > 0:
ans = "1"
for i in range(c0):
ans += "0"
print(ans)
else:
print("0")
main()
|
Title: Minimum Binary Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all).
Input Specification:
The first line contains integer number *n* (1<=≤<=*n*<=≤<=100) — the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct.
Output Specification:
Print one string — the minimum correct string that you can obtain from the given one.
Demo Input:
['4\n1001\n', '1\n1\n']
Demo Output:
['100\n', '1\n']
Note:
In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use.
|
```python
import math as mt
import sys, string
from collections import Counter, defaultdict
input = sys.stdin.readline
# input functions
I = lambda : int(input())
M = lambda : map(int, input().split())
ARR = lambda: list(map(int, input().split()))
def main():
n = I()
s = input()
s = list(s)
n = len(s) - 1
c1, c0 = 0, 0
for i in range(n):
if s[i] == '1':
c1 += 1
elif s[i] == '0':
c0 += 1
if c1 > 0:
ans = "1"
for i in range(c0):
ans += "0"
print(ans)
else:
print("0")
main()
```
| 3
|
|
873
|
D
|
Merge Sort
|
PROGRAMMING
| 1,800
|
[
"constructive algorithms",
"divide and conquer"
] | null | null |
Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array *a* with indices from [*l*,<=*r*) can be implemented as follows:
1. If the segment [*l*,<=*r*) is already sorted in non-descending order (that is, for any *i* such that *l*<=≤<=*i*<=<<=*r*<=-<=1 *a*[*i*]<=≤<=*a*[*i*<=+<=1]), then end the function call; 1. Let ; 1. Call *mergesort*(*a*,<=*l*,<=*mid*); 1. Call *mergesort*(*a*,<=*mid*,<=*r*); 1. Merge segments [*l*,<=*mid*) and [*mid*,<=*r*), making the segment [*l*,<=*r*) sorted in non-descending order. The merge algorithm doesn't call any other functions.
The array in this problem is 0-indexed, so to sort the whole array, you need to call *mergesort*(*a*,<=0,<=*n*).
The number of calls of function *mergesort* is very important, so Ivan has decided to calculate it while sorting the array. For example, if *a*<==<={1,<=2,<=3,<=4}, then there will be 1 call of *mergesort* — *mergesort*(0,<=4), which will check that the array is sorted and then end. If *a*<==<={2,<=1,<=3}, then the number of calls is 3: first of all, you call *mergesort*(0,<=3), which then sets *mid*<==<=1 and calls *mergesort*(0,<=1) and *mergesort*(1,<=3), which do not perform any recursive calls because segments (0,<=1) and (1,<=3) are sorted.
Ivan has implemented the program that counts the number of *mergesort* calls, but now he needs to test it. To do this, he needs to find an array *a* such that *a* is a permutation of size *n* (that is, the number of elements in *a* is *n*, and every integer number from [1,<=*n*] can be found in this array), and the number of *mergesort* calls when sorting the array is exactly *k*.
Help Ivan to find an array he wants!
|
The first line contains two numbers *n* and *k* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=200000) — the size of a desired permutation and the number of *mergesort* calls required to sort it.
|
If a permutation of size *n* such that there will be exactly *k* calls of *mergesort* while sorting it doesn't exist, output <=-<=1. Otherwise output *n* integer numbers *a*[0],<=*a*[1],<=...,<=*a*[*n*<=-<=1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them.
|
[
"3 3\n",
"4 1\n",
"5 6\n"
] |
[
"2 1 3 ",
"1 2 3 4 ",
"-1\n"
] |
none
| 0
|
[
{
"input": "3 3",
"output": "2 1 3 "
},
{
"input": "4 1",
"output": "1 2 3 4 "
},
{
"input": "5 6",
"output": "-1"
},
{
"input": "100 100",
"output": "-1"
},
{
"input": "10000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "10000 20001",
"output": "-1"
},
{
"input": "10000 30001",
"output": "-1"
},
{
"input": "20000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "20000 20001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "20000 30001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "30000 10001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "30000 20001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "30000 30001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "40000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "40000 20001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "40000 30001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "50000 10001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "50000 20001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "50000 30001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "60000 10001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "60000 20001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "60000 30001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "70000 10001",
"output": "3 1 5 2 7 4 9 6 11 8 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 32 29 33 35 31 37 34 39 36 41 38 43 40 45 42 47 44 49 46 50 52 48 54 51 56 53 58 55 60 57 62 59 64 61 66 63 67 69 65 71 68 73 70 75 72 77 74 79 76 81 78 83 80 84 86 82 88 85 90 87 92 89 94 91 96 93 98 95 100 97 101 103 99 105 102 107 104 109 106 111 108 113 110 115 112 117 114 118 120 116 122 119 124 121 126 123 128 125 130 127 132 129 134 131 135 137 133 139 136 141 138 143 140 145 142 147 144 149 146 151 148 152 154 150 156 153..."
},
{
"input": "70000 20001",
"output": "3 1 5 2 7 4 9 6 11 8 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 32 29 33 35 31 37 34 39 36 41 38 43 40 45 42 47 44 49 46 50 52 48 54 51 56 53 58 55 60 57 62 59 64 61 66 63 67 69 65 71 68 73 70 75 72 77 74 79 76 81 78 83 80 84 86 82 88 85 90 87 92 89 94 91 96 93 98 95 100 97 101 103 99 105 102 107 104 109 106 111 108 113 110 115 112 117 114 118 120 116 122 119 124 121 126 123 128 125 130 127 132 129 134 131 135 137 133 139 136 141 138 143 140 145 142 147 144 149 146 151 148 152 154 150 156 153..."
},
{
"input": "70000 30001",
"output": "3 1 5 2 7 4 9 6 11 8 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 32 29 33 35 31 37 34 39 36 41 38 43 40 45 42 47 44 49 46 50 52 48 54 51 56 53 58 55 60 57 62 59 64 61 66 63 67 69 65 71 68 73 70 75 72 77 74 79 76 81 78 83 80 84 86 82 88 85 90 87 92 89 94 91 96 93 98 95 100 97 101 103 99 105 102 107 104 109 106 111 108 113 110 115 112 117 114 118 120 116 122 119 124 121 126 123 128 125 130 127 132 129 134 131 135 137 133 139 136 141 138 143 140 145 142 147 144 149 146 151 148 152 154 150 156 153..."
},
{
"input": "80000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "80000 20001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "80000 30001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "90000 10001",
"output": "3 1 4 6 2 8 5 9 11 7 13 10 14 16 12 17 19 15 20 22 18 24 21 25 27 23 28 30 26 31 33 29 35 32 36 38 34 39 41 37 42 44 40 46 43 47 49 45 50 52 48 53 55 51 57 54 58 60 56 61 63 59 64 66 62 68 65 69 71 67 72 74 70 75 77 73 79 76 80 82 78 83 85 81 86 88 84 90 87 91 93 89 94 96 92 97 99 95 101 98 102 104 100 105 107 103 108 110 106 112 109 113 115 111 116 118 114 119 121 117 123 120 124 126 122 127 129 125 130 132 128 134 131 135 137 133 138 140 136 141 143 139 145 142 146 148 144 149 151 147 152 154 150 156 153..."
},
{
"input": "90000 20001",
"output": "3 1 4 6 2 8 5 9 11 7 13 10 14 16 12 17 19 15 20 22 18 24 21 25 27 23 28 30 26 31 33 29 35 32 36 38 34 39 41 37 42 44 40 46 43 47 49 45 50 52 48 53 55 51 57 54 58 60 56 61 63 59 64 66 62 68 65 69 71 67 72 74 70 75 77 73 79 76 80 82 78 83 85 81 86 88 84 90 87 91 93 89 94 96 92 97 99 95 101 98 102 104 100 105 107 103 108 110 106 112 109 113 115 111 116 118 114 119 121 117 123 120 124 126 122 127 129 125 130 132 128 134 131 135 137 133 138 140 136 141 143 139 145 142 146 148 144 149 151 147 152 154 150 156 153..."
},
{
"input": "90000 30001",
"output": "3 1 4 6 2 8 5 9 11 7 13 10 14 16 12 17 19 15 20 22 18 24 21 25 27 23 28 30 26 31 33 29 35 32 36 38 34 39 41 37 42 44 40 46 43 47 49 45 50 52 48 53 55 51 57 54 58 60 56 61 63 59 64 66 62 68 65 69 71 67 72 74 70 75 77 73 79 76 80 82 78 83 85 81 86 88 84 90 87 91 93 89 94 96 92 97 99 95 101 98 102 104 100 105 107 103 108 110 106 112 109 113 115 111 116 118 114 119 121 117 123 120 124 126 122 127 129 125 130 132 128 134 131 135 137 133 138 140 136 141 143 139 145 142 146 148 144 149 151 147 152 154 150 156 153..."
},
{
"input": "100000 10001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "100000 20001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "100000 30001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "100000 199999",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "10 17",
"output": "3 1 4 6 2 8 5 9 7 10 "
}
] | 1,515,139,013
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 93
| 23,961,600
|
import os
import sys
import copy
num = 0
b = None
# sys.stdin = open(os.path.join(os.path.dirname(__file__),'3.in'))
def solve():
n, k = map(lambda x:int(x), input().split())
a = [_+1 for _ in range(n)]
# print(a)
global b
b = a
def mgsort(a, l, r):
global num,b
num += 1
if r - l == 1 or r - l == 2:
return
mid = (l+r) >> 1
us = False
if num < k:
us= True
mgsort(a, l, mid)
if num < k:
us= True
mgsort(a, mid, r)
if us:
a = a[mid:r] + a[l:mid]
b = a
mgsort(a, 0, n)
# print(num)
if k % 2 != 0:
for i in b:
print(i,sep=' ',end=' ')
else:
print(-1)
solve()
|
Title: Merge Sort
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array *a* with indices from [*l*,<=*r*) can be implemented as follows:
1. If the segment [*l*,<=*r*) is already sorted in non-descending order (that is, for any *i* such that *l*<=≤<=*i*<=<<=*r*<=-<=1 *a*[*i*]<=≤<=*a*[*i*<=+<=1]), then end the function call; 1. Let ; 1. Call *mergesort*(*a*,<=*l*,<=*mid*); 1. Call *mergesort*(*a*,<=*mid*,<=*r*); 1. Merge segments [*l*,<=*mid*) and [*mid*,<=*r*), making the segment [*l*,<=*r*) sorted in non-descending order. The merge algorithm doesn't call any other functions.
The array in this problem is 0-indexed, so to sort the whole array, you need to call *mergesort*(*a*,<=0,<=*n*).
The number of calls of function *mergesort* is very important, so Ivan has decided to calculate it while sorting the array. For example, if *a*<==<={1,<=2,<=3,<=4}, then there will be 1 call of *mergesort* — *mergesort*(0,<=4), which will check that the array is sorted and then end. If *a*<==<={2,<=1,<=3}, then the number of calls is 3: first of all, you call *mergesort*(0,<=3), which then sets *mid*<==<=1 and calls *mergesort*(0,<=1) and *mergesort*(1,<=3), which do not perform any recursive calls because segments (0,<=1) and (1,<=3) are sorted.
Ivan has implemented the program that counts the number of *mergesort* calls, but now he needs to test it. To do this, he needs to find an array *a* such that *a* is a permutation of size *n* (that is, the number of elements in *a* is *n*, and every integer number from [1,<=*n*] can be found in this array), and the number of *mergesort* calls when sorting the array is exactly *k*.
Help Ivan to find an array he wants!
Input Specification:
The first line contains two numbers *n* and *k* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=200000) — the size of a desired permutation and the number of *mergesort* calls required to sort it.
Output Specification:
If a permutation of size *n* such that there will be exactly *k* calls of *mergesort* while sorting it doesn't exist, output <=-<=1. Otherwise output *n* integer numbers *a*[0],<=*a*[1],<=...,<=*a*[*n*<=-<=1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them.
Demo Input:
['3 3\n', '4 1\n', '5 6\n']
Demo Output:
['2 1 3 ', '1 2 3 4 ', '-1\n']
Note:
none
|
```python
import os
import sys
import copy
num = 0
b = None
# sys.stdin = open(os.path.join(os.path.dirname(__file__),'3.in'))
def solve():
n, k = map(lambda x:int(x), input().split())
a = [_+1 for _ in range(n)]
# print(a)
global b
b = a
def mgsort(a, l, r):
global num,b
num += 1
if r - l == 1 or r - l == 2:
return
mid = (l+r) >> 1
us = False
if num < k:
us= True
mgsort(a, l, mid)
if num < k:
us= True
mgsort(a, mid, r)
if us:
a = a[mid:r] + a[l:mid]
b = a
mgsort(a, 0, n)
# print(num)
if k % 2 != 0:
for i in b:
print(i,sep=' ',end=' ')
else:
print(-1)
solve()
```
| 0
|
|
265
|
A
|
Colorful Stones (Simplified Edition)
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
|
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
|
Print the final 1-based position of Liss in a single line.
|
[
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] |
[
"2\n",
"3\n",
"15\n"
] |
none
| 500
|
[
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
{
"input": "RRRRRBRRBRRGRBGGRRRGRBBRBBBBBRGRBGBRRGBBBRBBGBRGBB\nB",
"output": "1"
},
{
"input": "RRGGBRGRBG\nBRRGGBBGGR",
"output": "7"
},
{
"input": "BBRRGBGGRGBRGBRBRBGR\nGGGRBGGGBRRRRGRBGBGRGRRBGRBGBG",
"output": "15"
},
{
"input": "GBRRBGBGBBBBRRRGBGRRRGBGBBBRGR\nRRGBRRGRBBBBBBGRRBBR",
"output": "8"
},
{
"input": "BRGRRGRGRRGBBGBBBRRBBRRBGBBGRGBBGGRGBRBGGGRRRBGGBB\nRGBBGRRBBBRRGRRBRBBRGBBGGGRGBGRRRRBRBGGBRBGGGRGBRR",
"output": "16"
},
{
"input": "GGRGGBRRGRGBRRGGRBBGGRRGBBBGBBBGGRBGGBRBBRGBRRRBRG\nGGRGRRRRRRRRRGBBBBRGBRGRGRRGBBRGGBRBBRBGBRGRRRRGGR",
"output": "18"
},
{
"input": "RBBRBGBBGGGBRRBGBRGRRGGRBBBBGRBGGBRRBGBBRBRGBBGGGG\nBRRGRBGRBGBRGBRBGGBBBGBBRGRBGRGRBGRBGGRRBGBRGRRBBG",
"output": "23"
},
{
"input": "RRGGRRBRGRGBRGBRGBBRGGGBRBGBRGBRGGBRBRGBGRRBBGGRBR\nBRGRBBBGRBGRRGGRRRRRRRRGGGGGGGGBGGGGGGGRGRBBRGRGBR",
"output": "14"
},
{
"input": "RBRBRRRRRGGR\nBBBBRRGGBGRBRGBBRGGGRGGBRRBRBRBGGG",
"output": "12"
},
{
"input": "GGRGGGBRGRGRBBRGGGRBGRRR\nGBBBBGBGBGBBBRGGRGGBGRRGRRRGBGB",
"output": "12"
},
{
"input": "RGRBGBRBBGBRRGGGBBGBRRBGRGBBBBBGRRRGGRBB\nBGR",
"output": "2"
},
{
"input": "R\nB",
"output": "1"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "50"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
},
{
"input": "R\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
}
] | 1,587,774,601
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 218
| 0
|
standing = input() #RGB
inst = input() #RRR
standing_on = 0
for char in inst:
if char == standing[standing_on]:
standing_on+=1
print(standing_on + 1)
|
Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none
|
```python
standing = input() #RGB
inst = input() #RRR
standing_on = 0
for char in inst:
if char == standing[standing_on]:
standing_on+=1
print(standing_on + 1)
```
| 3
|
|
489
|
B
|
BerSU Ball
|
PROGRAMMING
| 1,200
|
[
"dfs and similar",
"dp",
"graph matchings",
"greedy",
"sortings",
"two pointers"
] | null | null |
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! *n* boys and *m* girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from *n* boys and *m* girls.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of boys. The second line contains sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is the *i*-th boy's dancing skill.
Similarly, the third line contains an integer *m* (1<=≤<=*m*<=≤<=100) — the number of girls. The fourth line contains sequence *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=100), where *b**j* is the *j*-th girl's dancing skill.
|
Print a single number — the required maximum possible number of pairs.
|
[
"4\n1 4 6 2\n5\n5 1 5 7 9\n",
"4\n1 2 3 4\n4\n10 11 12 13\n",
"5\n1 1 1 1 1\n3\n1 2 3\n"
] |
[
"3\n",
"0\n",
"2\n"
] |
none
| 1,000
|
[
{
"input": "4\n1 4 6 2\n5\n5 1 5 7 9",
"output": "3"
},
{
"input": "4\n1 2 3 4\n4\n10 11 12 13",
"output": "0"
},
{
"input": "5\n1 1 1 1 1\n3\n1 2 3",
"output": "2"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1 10\n1\n9",
"output": "1"
},
{
"input": "4\n4 5 4 4\n5\n5 3 4 2 4",
"output": "4"
},
{
"input": "1\n2\n1\n1",
"output": "1"
},
{
"input": "1\n3\n2\n3 2",
"output": "1"
},
{
"input": "1\n4\n3\n4 4 4",
"output": "1"
},
{
"input": "1\n2\n4\n3 1 4 2",
"output": "1"
},
{
"input": "1\n4\n5\n2 5 5 3 1",
"output": "1"
},
{
"input": "2\n2 2\n1\n2",
"output": "1"
},
{
"input": "2\n4 2\n2\n4 4",
"output": "1"
},
{
"input": "2\n4 1\n3\n2 3 2",
"output": "2"
},
{
"input": "2\n4 3\n4\n5 5 5 6",
"output": "1"
},
{
"input": "2\n5 7\n5\n4 6 7 2 5",
"output": "2"
},
{
"input": "3\n1 2 3\n1\n1",
"output": "1"
},
{
"input": "3\n5 4 5\n2\n2 1",
"output": "0"
},
{
"input": "3\n6 3 4\n3\n4 5 2",
"output": "3"
},
{
"input": "3\n7 7 7\n4\n2 7 2 4",
"output": "1"
},
{
"input": "3\n1 3 3\n5\n1 3 4 1 2",
"output": "3"
},
{
"input": "4\n1 2 1 3\n1\n4",
"output": "1"
},
{
"input": "4\n4 4 6 6\n2\n2 1",
"output": "0"
},
{
"input": "4\n3 1 1 1\n3\n1 6 7",
"output": "1"
},
{
"input": "4\n2 5 1 2\n4\n2 3 3 1",
"output": "3"
},
{
"input": "4\n9 1 7 1\n5\n9 9 9 8 4",
"output": "2"
},
{
"input": "5\n1 6 5 5 6\n1\n2",
"output": "1"
},
{
"input": "5\n5 2 4 5 6\n2\n7 4",
"output": "2"
},
{
"input": "5\n4 1 3 1 4\n3\n6 3 6",
"output": "1"
},
{
"input": "5\n5 2 3 1 4\n4\n1 3 1 7",
"output": "3"
},
{
"input": "5\n9 8 10 9 10\n5\n2 1 5 4 6",
"output": "0"
},
{
"input": "1\n48\n100\n76 90 78 44 29 30 35 85 98 38 27 71 51 100 15 98 78 45 85 26 48 66 98 71 45 85 83 77 92 17 23 95 98 43 11 15 39 53 71 25 74 53 77 41 39 35 66 4 92 44 44 55 35 87 91 6 44 46 57 24 46 82 15 44 81 40 65 17 64 24 42 52 13 12 64 82 26 7 66 85 93 89 58 92 92 77 37 91 47 73 35 69 31 22 60 60 97 21 52 6",
"output": "1"
},
{
"input": "100\n9 90 66 62 60 9 10 97 47 73 26 81 97 60 80 84 19 4 25 77 19 17 91 12 1 27 15 54 18 45 71 79 96 90 51 62 9 13 92 34 7 52 55 8 16 61 96 12 52 38 50 9 60 3 30 3 48 46 77 64 90 35 16 16 21 42 67 70 23 19 90 14 50 96 98 92 82 62 7 51 93 38 84 82 37 78 99 3 20 69 44 96 94 71 3 55 27 86 92 82\n1\n58",
"output": "0"
},
{
"input": "10\n20 87 3 39 20 20 8 40 70 51\n100\n69 84 81 84 35 97 69 68 63 97 85 80 95 58 70 91 100 65 72 80 41 87 87 87 22 49 96 96 78 96 97 56 90 31 62 98 89 74 100 86 95 88 66 54 93 62 41 60 95 79 29 69 63 70 52 63 87 58 54 52 48 57 26 75 39 61 98 78 52 73 99 49 74 50 59 90 31 97 16 85 63 72 81 68 75 59 70 67 73 92 10 88 57 95 3 71 80 95 84 96",
"output": "6"
},
{
"input": "100\n10 10 9 18 56 64 92 66 54 42 66 65 58 5 74 68 80 57 58 30 58 69 70 13 38 19 34 63 38 17 26 24 66 83 48 77 44 37 78 97 13 90 51 56 60 23 49 32 14 86 90 100 13 14 52 69 85 95 81 53 5 3 91 66 2 64 45 59 7 30 80 42 61 82 70 10 62 82 5 34 50 28 24 47 85 68 27 50 24 61 76 17 63 24 3 67 83 76 42 60\n10\n66 74 40 67 28 82 99 57 93 64",
"output": "9"
},
{
"input": "100\n4 1 1 1 3 3 2 5 1 2 1 2 1 1 1 6 1 3 1 1 1 1 2 4 1 1 4 2 2 8 2 2 1 8 2 4 3 3 8 1 3 2 3 2 1 3 8 2 2 3 1 1 2 2 5 1 4 3 1 1 3 1 3 1 7 1 1 1 3 2 1 2 2 3 7 2 1 4 3 2 1 1 3 4 1 1 3 5 1 8 4 1 1 1 3 10 2 2 1 2\n100\n1 1 5 2 13 2 2 3 6 12 1 13 8 1 1 16 1 1 5 6 2 4 6 4 2 7 4 1 7 3 3 9 5 3 1 7 4 1 6 6 8 2 2 5 2 3 16 3 6 3 8 6 1 8 1 2 6 5 3 4 11 3 4 8 2 13 2 5 2 7 3 3 1 8 1 4 4 2 4 7 7 1 5 7 6 3 6 9 1 1 1 3 1 11 5 2 5 11 13 1",
"output": "76"
},
{
"input": "4\n1 6 9 15\n2\n5 8",
"output": "2"
},
{
"input": "2\n2 4\n2\n3 1",
"output": "2"
},
{
"input": "3\n2 3 5\n3\n3 4 6",
"output": "3"
},
{
"input": "3\n1 3 4\n3\n2 1 5",
"output": "3"
},
{
"input": "2\n5 5\n4\n1 1 1 5",
"output": "1"
},
{
"input": "2\n3 2\n2\n3 4",
"output": "2"
},
{
"input": "2\n3 1\n2\n2 4",
"output": "2"
},
{
"input": "2\n2 3\n2\n2 1",
"output": "2"
},
{
"input": "2\n10 12\n2\n11 9",
"output": "2"
},
{
"input": "3\n1 2 3\n3\n3 2 1",
"output": "3"
},
{
"input": "2\n1 3\n2\n2 1",
"output": "2"
},
{
"input": "2\n4 5\n2\n5 3",
"output": "2"
},
{
"input": "2\n7 5\n2\n6 8",
"output": "2"
},
{
"input": "4\n4 3 2 1\n4\n1 2 3 4",
"output": "4"
},
{
"input": "2\n2 3\n2\n3 1",
"output": "2"
},
{
"input": "2\n2 4\n3\n3 1 8",
"output": "2"
},
{
"input": "3\n3 1 1\n3\n2 4 4",
"output": "2"
},
{
"input": "2\n5 3\n2\n4 6",
"output": "2"
},
{
"input": "4\n1 1 3 3\n4\n2 2 1 1",
"output": "4"
},
{
"input": "3\n3 2 1\n3\n2 4 3",
"output": "3"
},
{
"input": "5\n1 2 3 4 5\n5\n2 3 4 5 1",
"output": "5"
},
{
"input": "3\n3 2 1\n3\n1 2 3",
"output": "3"
},
{
"input": "2\n5 4\n2\n4 6",
"output": "2"
},
{
"input": "4\n3 3 5 5\n4\n4 4 2 2",
"output": "4"
},
{
"input": "3\n2 7 5\n3\n2 4 8",
"output": "3"
},
{
"input": "100\n2 3 3 4 2 1 4 4 5 5 2 1 5 2 3 3 5 4 3 2 4 2 3 3 2 2 3 4 2 2 2 3 1 2 3 2 2 3 5 3 3 3 3 4 5 2 2 1 1 1 3 1 2 2 3 5 5 2 5 1 3 4 5 3 5 4 1 1 2 3 4 4 5 3 2 4 5 5 5 2 1 4 2 4 5 4 4 5 5 3 2 5 1 4 4 2 2 2 5 3\n100\n4 5 3 3 2 2 4 3 1 5 4 3 3 2 2 4 5 2 5 2 1 4 3 4 2 3 5 3 4 4 1 2 3 5 2 2 1 5 4 2 4 3 4 3 4 2 3 1 3 3 4 1 1 1 4 4 5 3 1 4 2 3 2 1 3 3 2 3 2 1 1 2 3 2 1 3 3 4 3 3 1 1 3 3 3 1 1 3 5 3 3 3 3 4 4 5 2 5 4 5",
"output": "100"
},
{
"input": "1\n3\n2\n2 3",
"output": "1"
},
{
"input": "2\n5 6\n3\n1 5 100",
"output": "1"
},
{
"input": "2\n2 7\n2\n6 8",
"output": "1"
},
{
"input": "4\n4 10 15 17\n4\n3 12 16 16",
"output": "3"
},
{
"input": "2\n2 3\n2\n1 2",
"output": "2"
}
] | 1,696,779,461
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 46
| 0
|
a = int(input())
sq = list(map(int,input().split()))
b = int(input())
sw = list(map(int,input().split()))
sq.sort()
sw.sort()
i = 0
u = 0
y = 0
while i < a and u < b:
if abs(sq[i] - sw[u]) <= 1:
y += 1
i += 1
u += 1
elif sq[i] < sw[u]:
i += 1
else:
u += 1
print(y)
|
Title: BerSU Ball
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! *n* boys and *m* girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from *n* boys and *m* girls.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of boys. The second line contains sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is the *i*-th boy's dancing skill.
Similarly, the third line contains an integer *m* (1<=≤<=*m*<=≤<=100) — the number of girls. The fourth line contains sequence *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=100), where *b**j* is the *j*-th girl's dancing skill.
Output Specification:
Print a single number — the required maximum possible number of pairs.
Demo Input:
['4\n1 4 6 2\n5\n5 1 5 7 9\n', '4\n1 2 3 4\n4\n10 11 12 13\n', '5\n1 1 1 1 1\n3\n1 2 3\n']
Demo Output:
['3\n', '0\n', '2\n']
Note:
none
|
```python
a = int(input())
sq = list(map(int,input().split()))
b = int(input())
sw = list(map(int,input().split()))
sq.sort()
sw.sort()
i = 0
u = 0
y = 0
while i < a and u < b:
if abs(sq[i] - sw[u]) <= 1:
y += 1
i += 1
u += 1
elif sq[i] < sw[u]:
i += 1
else:
u += 1
print(y)
```
| 3
|
|
932
|
A
|
Palindromic Supersequence
|
PROGRAMMING
| 800
|
[
"constructive algorithms"
] | null | null |
You are given a string *A*. Find a string *B*, where *B* is a palindrome and *A* is a subsequence of *B*.
A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest".
A palindrome is a string that reads the same forward or backward.
The length of string *B* should be at most 104. It is guaranteed that there always exists such string.
You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104.
|
First line contains a string *A* (1<=≤<=|*A*|<=≤<=103) consisting of lowercase Latin letters, where |*A*| is a length of *A*.
|
Output single line containing *B* consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104. If there are many possible *B*, print any of them.
|
[
"aba\n",
"ab\n"
] |
[
"aba",
"aabaa"
] |
In the first example, "aba" is a subsequence of "aba" which is a palindrome.
In the second example, "ab" is a subsequence of "aabaa" which is a palindrome.
| 500
|
[
{
"input": "aba",
"output": "abaaba"
},
{
"input": "ab",
"output": "abba"
},
{
"input": "krnyoixirslfszfqivgkaflgkctvbvksipwomqxlyqxhlbceuhbjbfnhofcgpgwdseffycthmlpcqejgskwjkbkbbmifnurnwyhevsoqzmtvzgfiqajfrgyuzxnrtxectcnlyoisbglpdbjbslxlpoymrcxmdtqhcnlvtqdwftuzgbdxsyscwbrguostbelnvtaqdmkmihmoxqtqlxvlsssisvqvvzotoyqryuyqwoknnqcqggysrqpkrccvyhxsjmhoqoyocwcriplarjoyiqrmmpmueqbsbljddwrumauczfziodpudheexalbwpiypmdjlmwtgdrzhpxneofhqzjdmurgvmrwdotuwyknlrbvuvtnhiouvqitgyfgfieonbaapyhwpcrmehxcpkijzfiayfvoxkpa",
"output": "krnyoixirslfszfqivgkaflgkctvbvksipwomqxlyqxhlbceuhbjbfnhofcgpgwdseffycthmlpcqejgskwjkbkbbmifnurnwyhevsoqzmtvzgfiqajfrgyuzxnrtxectcnlyoisbglpdbjbslxlpoymrcxmdtqhcnlvtqdwftuzgbdxsyscwbrguostbelnvtaqdmkmihmoxqtqlxvlsssisvqvvzotoyqryuyqwoknnqcqggysrqpkrccvyhxsjmhoqoyocwcriplarjoyiqrmmpmueqbsbljddwrumauczfziodpudheexalbwpiypmdjlmwtgdrzhpxneofhqzjdmurgvmrwdotuwyknlrbvuvtnhiouvqitgyfgfieonbaapyhwpcrmehxcpkijzfiayfvoxkpaapkxovfyaifzjikpcxhemrcpwhypaabnoeifgfygtiqvuoihntvuvbrlnkywutodwrmvgrumdjzqhfoenxphzrdgtwmljdm..."
},
{
"input": "mgrfmzxqpejcixxppqgvuawutgrmezjkteofjbnrvzzkvjtacfxjjokisavsgrslryxfqgrmdsqwptajbqzvethuljbdatxghfzqrwvfgakwmoawlzqjypmhllbbuuhbpriqsnibywlgjlxowyzagrfnqafvcqwktkcjwejevzbnxhsfmwojshcdypnvbuhhuzqmgovmvgwiizatoxgblyudipahfbkewmuneoqhjmbpdtwnznblwvtjrniwlbyblhppndspojrouffazpoxtqdfpjuhitvijrohavpqatofxwmksvjcvhdecxwwmosqiczjpkfafqlboxosnjgzgdraehzdltthemeusxhiiimrdrugabnxwsygsktkcslhjebfexucsyvlwrptebkjhefsvfrmcqqdlanbetrgzwylizmrystvpgrkhlicfadco",
"output": "mgrfmzxqpejcixxppqgvuawutgrmezjkteofjbnrvzzkvjtacfxjjokisavsgrslryxfqgrmdsqwptajbqzvethuljbdatxghfzqrwvfgakwmoawlzqjypmhllbbuuhbpriqsnibywlgjlxowyzagrfnqafvcqwktkcjwejevzbnxhsfmwojshcdypnvbuhhuzqmgovmvgwiizatoxgblyudipahfbkewmuneoqhjmbpdtwnznblwvtjrniwlbyblhppndspojrouffazpoxtqdfpjuhitvijrohavpqatofxwmksvjcvhdecxwwmosqiczjpkfafqlboxosnjgzgdraehzdltthemeusxhiiimrdrugabnxwsygsktkcslhjebfexucsyvlwrptebkjhefsvfrmcqqdlanbetrgzwylizmrystvpgrkhlicfadcoocdafcilhkrgpvtsyrmzilywzgrtebnaldqqcmrfvsfehjkbetprwlvyscuxef..."
},
{
"input": "hdmasfcjuigrwjchmjslmpynewnzpphmudzcbxzdexjuhktdtcoibzvevsmwaxakrtdfoivkvoooypyemiidadquqepxwqkesdnakxkbzrcjkgvwwxtqxvfpxcwitljyehldgsjytmekimkkndjvnzqtjykiymkmdzpwakxdtkzcqcatlevppgfhyykgmipuodjrnfjzhcmjdbzvhywprbwdcfxiffpzbjbmbyijkqnosslqbfvvicxvoeuzruraetglthgourzhfpnubzvblfzmmbgepjjyshchthulxar",
"output": "hdmasfcjuigrwjchmjslmpynewnzpphmudzcbxzdexjuhktdtcoibzvevsmwaxakrtdfoivkvoooypyemiidadquqepxwqkesdnakxkbzrcjkgvwwxtqxvfpxcwitljyehldgsjytmekimkkndjvnzqtjykiymkmdzpwakxdtkzcqcatlevppgfhyykgmipuodjrnfjzhcmjdbzvhywprbwdcfxiffpzbjbmbyijkqnosslqbfvvicxvoeuzruraetglthgourzhfpnubzvblfzmmbgepjjyshchthulxarraxluhthchsyjjpegbmmzflbvzbunpfhzruoghtlgtearurzueovxcivvfbqlssonqkjiybmbjbzpffixfcdwbrpwyhvzbdjmchzjfnrjdoupimgkyyhfgppveltacqczktdxkawpzdmkmyikyjtqznvjdnkkmikemtyjsgdlheyjltiwcxpfvxqtxwwvgkjcrzbkxkandsekqwxpequ..."
},
{
"input": "fggbyzobbmxtwdajawqdywnppflkkmtxzjvxopqvliwdwhzepcuiwelhbuotlkvesexnwkytonfrpqcxzzqzdvsmbsjcxxeugavekozfjlolrtqgwzqxsfgrnvrgfrqpixhsskbpzghndesvwptpvvkasfalzsetopervpwzmkgpcexqnvtnoulprwnowmsorscecvvvrjfwumcjqyrounqsgdruxttvtmrkivtxauhosokdiahsyrftzsgvgyveqwkzhqstbgywrvmsgfcfyuxpphvmyydzpohgdicoxbtjnsbyhoidnkrialowvlvmjpxcfeygqzphmbcjkupojsmmuqlydixbaluwezvnfasjfxilbyllwyipsmovdzosuwotcxerzcfuvxprtziseshjfcosalyqglpotxvxaanpocypsiyazsejjoximnbvqucftuvdksaxutvjeunodbipsumlaymjnzljurefjg",
"output": "fggbyzobbmxtwdajawqdywnppflkkmtxzjvxopqvliwdwhzepcuiwelhbuotlkvesexnwkytonfrpqcxzzqzdvsmbsjcxxeugavekozfjlolrtqgwzqxsfgrnvrgfrqpixhsskbpzghndesvwptpvvkasfalzsetopervpwzmkgpcexqnvtnoulprwnowmsorscecvvvrjfwumcjqyrounqsgdruxttvtmrkivtxauhosokdiahsyrftzsgvgyveqwkzhqstbgywrvmsgfcfyuxpphvmyydzpohgdicoxbtjnsbyhoidnkrialowvlvmjpxcfeygqzphmbcjkupojsmmuqlydixbaluwezvnfasjfxilbyllwyipsmovdzosuwotcxerzcfuvxprtziseshjfcosalyqglpotxvxaanpocypsiyazsejjoximnbvqucftuvdksaxutvjeunodbipsumlaymjnzljurefjggjferujlznjmyalmuspib..."
},
{
"input": "qyyxqkbxsvfnjzttdqmpzinbdgayllxpfrpopwciejjjzadguurnnhvixgueukugkkjyghxknedojvmdrskswiotgatsajowionuiumuhyggjuoympuxyfahwftwufvocdguxmxabbxnfviscxtilzzauizsgugwcqtbqgoosefhkumhodwpgolfdkbuiwlzjydonwbgyzzrjwxnceltqgqelrrljmzdbftmaogiuosaqhngmdzxzlmyrwefzhqawmkdckfnyyjgdjgadtfjvrkdwysqofcgyqrnyzutycvspzbjmmesobvhshtqlrytztyieknnkporrbcmlopgtknlmsstzkigreqwgsvagmvbrvwypoxttmzzsgm",
"output": "qyyxqkbxsvfnjzttdqmpzinbdgayllxpfrpopwciejjjzadguurnnhvixgueukugkkjyghxknedojvmdrskswiotgatsajowionuiumuhyggjuoympuxyfahwftwufvocdguxmxabbxnfviscxtilzzauizsgugwcqtbqgoosefhkumhodwpgolfdkbuiwlzjydonwbgyzzrjwxnceltqgqelrrljmzdbftmaogiuosaqhngmdzxzlmyrwefzhqawmkdckfnyyjgdjgadtfjvrkdwysqofcgyqrnyzutycvspzbjmmesobvhshtqlrytztyieknnkporrbcmlopgtknlmsstzkigreqwgsvagmvbrvwypoxttmzzsgmmgszzmttxopywvrbvmgavsgwqergikztssmlnktgpolmcbrropknnkeiytztyrlqthshvbosemmjbzpsvcytuzynrqygcfoqsywdkrvjftdagjdgjyynfkcdkmwaqhzfewry..."
},
{
"input": "scvlhflaqvniyiyofonowwcuqajuwscdrzhbvasymvqfnthzvtjcfuaftrbjghhvslcohwpxkggrbtatjtgehuqtorwinwvrtdldyoeeozxwippuahgkuehvsmyqtodqvlufqqmqautaqirvwzvtodzxtgxiinubhrbeoiybidutrqamsdnasctxatzkvkjkrmavdravnsxyngjlugwftmhmcvvxdbfndurrbmcpuoigjpssqcortmqoqttrabhoqvopjkxvpbqdqsilvlplhgqazauyvnodsxtwnomlinjpozwhrgrkqwmlwcwdkxjxjftexiavwrejvdjcfptterblxysjcheesyqsbgdrzjxbfjqgjgmvccqcyj",
"output": "scvlhflaqvniyiyofonowwcuqajuwscdrzhbvasymvqfnthzvtjcfuaftrbjghhvslcohwpxkggrbtatjtgehuqtorwinwvrtdldyoeeozxwippuahgkuehvsmyqtodqvlufqqmqautaqirvwzvtodzxtgxiinubhrbeoiybidutrqamsdnasctxatzkvkjkrmavdravnsxyngjlugwftmhmcvvxdbfndurrbmcpuoigjpssqcortmqoqttrabhoqvopjkxvpbqdqsilvlplhgqazauyvnodsxtwnomlinjpozwhrgrkqwmlwcwdkxjxjftexiavwrejvdjcfptterblxysjcheesyqsbgdrzjxbfjqgjgmvccqcyjjycqccvmgjgqjfbxjzrdgbsqyseehcjsyxlbrettpfcjdvjerwvaixetfjxjxkdwcwlmwqkrgrhwzopjnilmonwtxsdonvyuazaqghlplvlisqdqbpvxkjpovqohbarttqoqm..."
},
{
"input": "oohkqxxtvxzmvfjjxyjwlbqmeqwwlienzkdbhswgfbkhfygltsucdijozwaiewpixapyazfztksjeoqjugjfhdbqzuezbuajfvvffkwprroyivfoocvslejffgxuiofisenroxoeixmdbzonmreikpflciwsbafrdqfvdfojgoziiibqhwwsvhnzmptgirqqulkgmyzrfekzqqujmdumxkudsgexisupedisgmdgebvlvrpyfrbrqjknrxyzfpwmsxjxismgd",
"output": "oohkqxxtvxzmvfjjxyjwlbqmeqwwlienzkdbhswgfbkhfygltsucdijozwaiewpixapyazfztksjeoqjugjfhdbqzuezbuajfvvffkwprroyivfoocvslejffgxuiofisenroxoeixmdbzonmreikpflciwsbafrdqfvdfojgoziiibqhwwsvhnzmptgirqqulkgmyzrfekzqqujmdumxkudsgexisupedisgmdgebvlvrpyfrbrqjknrxyzfpwmsxjxismgddgmsixjxsmwpfzyxrnkjqrbrfyprvlvbegdmgsidepusixegsdukxmudmjuqqzkefrzymgkluqqrigtpmznhvswwhqbiiizogjofdvfqdrfabswiclfpkiermnozbdmxieoxornesifoiuxgffjelsvcoofviyorrpwkffvvfjaubzeuzqbdhfjgujqoejsktzfzaypaxipweiawzojidcustlgyfhkbfgwshbdkzneilwwqemqblw..."
},
{
"input": "gilhoixzjgidfanqrmekjelnvicpuujlpxittgadgrhqallnkjlemwazntwfywjnrxdkgrnczlwzjyeyfktduzdjnivcldjjarfzmmdbyytvipbbnjqolfnlqjpidotxxfobgtgpvjmpddcyddwdcjsxxumuoyznhpvpqccgqnuouzojntanfwctthcgynrukcvshsuuqrxfdvqqggaatwytikkitywtaaggqqvdfxrquushsvckurnygchttcwfnatnjozuounqgccqpvphnzyoumuxxsjcdwddycddpmjvpgtgbofxxtodipjqlnfloqjnbbpivtyybdmmzfrajjdlcvinjdzudtkfyeyjzwlzcnrgkdxrnjwyfwtnzawmeljknllaqhrgdagttixpljuupcivnlejkemrqnafdigjzxiohlig",
"output": "gilhoixzjgidfanqrmekjelnvicpuujlpxittgadgrhqallnkjlemwazntwfywjnrxdkgrnczlwzjyeyfktduzdjnivcldjjarfzmmdbyytvipbbnjqolfnlqjpidotxxfobgtgpvjmpddcyddwdcjsxxumuoyznhpvpqccgqnuouzojntanfwctthcgynrukcvshsuuqrxfdvqqggaatwytikkitywtaaggqqvdfxrquushsvckurnygchttcwfnatnjozuounqgccqpvphnzyoumuxxsjcdwddycddpmjvpgtgbofxxtodipjqlnfloqjnbbpivtyybdmmzfrajjdlcvinjdzudtkfyeyjzwlzcnrgkdxrnjwyfwtnzawmeljknllaqhrgdagttixpljuupcivnlejkemrqnafdigjzxiohliggilhoixzjgidfanqrmekjelnvicpuujlpxittgadgrhqallnkjlemwazntwfywjnrxdkgrnczlw..."
},
{
"input": "abcab",
"output": "abcabbacba"
},
{
"input": "baaaaaaa",
"output": "baaaaaaaaaaaaaab"
},
{
"input": "baaaaaa",
"output": "baaaaaaaaaaaab"
},
{
"input": "baaaaaaaaa",
"output": "baaaaaaaaaaaaaaaaaab"
},
{
"input": "baaaaaaaa",
"output": "baaaaaaaaaaaaaaaab"
}
] | 1,518,960,267
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 78
| 6,758,400
|
import statistics
import cv2
import numpy as np
s = input();
s+= s[::-1]
print(s)
|
Title: Palindromic Supersequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *A*. Find a string *B*, where *B* is a palindrome and *A* is a subsequence of *B*.
A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest".
A palindrome is a string that reads the same forward or backward.
The length of string *B* should be at most 104. It is guaranteed that there always exists such string.
You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104.
Input Specification:
First line contains a string *A* (1<=≤<=|*A*|<=≤<=103) consisting of lowercase Latin letters, where |*A*| is a length of *A*.
Output Specification:
Output single line containing *B* consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104. If there are many possible *B*, print any of them.
Demo Input:
['aba\n', 'ab\n']
Demo Output:
['aba', 'aabaa']
Note:
In the first example, "aba" is a subsequence of "aba" which is a palindrome.
In the second example, "ab" is a subsequence of "aabaa" which is a palindrome.
|
```python
import statistics
import cv2
import numpy as np
s = input();
s+= s[::-1]
print(s)
```
| -1
|
|
934
|
B
|
A Prosperous Lot
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"implementation"
] | null | null |
Apart from Nian, there is a daemon named Sui, which terrifies children and causes them to become sick. Parents give their children money wrapped in red packets and put them under the pillow, so that when Sui tries to approach them, it will be driven away by the fairies inside.
Big Banban is hesitating over the amount of money to give out. He considers loops to be lucky since it symbolizes unity and harmony.
He would like to find a positive integer *n* not greater than 1018, such that there are exactly *k* loops in the decimal representation of *n*, or determine that such *n* does not exist.
A loop is a planar area enclosed by lines in the digits' decimal representation written in Arabic numerals. For example, there is one loop in digit 4, two loops in 8 and no loops in 5. Refer to the figure below for all exact forms.
|
The first and only line contains an integer *k* (1<=≤<=*k*<=≤<=106) — the desired number of loops.
|
Output an integer — if no such *n* exists, output -1; otherwise output any such *n*. In the latter case, your output should be a positive decimal integer not exceeding 1018.
|
[
"2\n",
"6\n"
] |
[
"462",
"8080"
] |
none
| 1,000
|
[
{
"input": "2",
"output": "8"
},
{
"input": "6",
"output": "888"
},
{
"input": "3",
"output": "86"
},
{
"input": "4",
"output": "88"
},
{
"input": "5",
"output": "886"
},
{
"input": "1000000",
"output": "-1"
},
{
"input": "1",
"output": "6"
},
{
"input": "7",
"output": "8886"
},
{
"input": "8",
"output": "8888"
},
{
"input": "9",
"output": "88886"
},
{
"input": "10",
"output": "88888"
},
{
"input": "11",
"output": "888886"
},
{
"input": "12",
"output": "888888"
},
{
"input": "13",
"output": "8888886"
},
{
"input": "14",
"output": "8888888"
},
{
"input": "15",
"output": "88888886"
},
{
"input": "16",
"output": "88888888"
},
{
"input": "17",
"output": "888888886"
},
{
"input": "18",
"output": "888888888"
},
{
"input": "19",
"output": "8888888886"
},
{
"input": "20",
"output": "8888888888"
},
{
"input": "21",
"output": "88888888886"
},
{
"input": "22",
"output": "88888888888"
},
{
"input": "23",
"output": "888888888886"
},
{
"input": "24",
"output": "888888888888"
},
{
"input": "25",
"output": "8888888888886"
},
{
"input": "26",
"output": "8888888888888"
},
{
"input": "27",
"output": "88888888888886"
},
{
"input": "28",
"output": "88888888888888"
},
{
"input": "29",
"output": "888888888888886"
},
{
"input": "30",
"output": "888888888888888"
},
{
"input": "31",
"output": "8888888888888886"
},
{
"input": "32",
"output": "8888888888888888"
},
{
"input": "33",
"output": "88888888888888886"
},
{
"input": "34",
"output": "88888888888888888"
},
{
"input": "35",
"output": "888888888888888886"
},
{
"input": "36",
"output": "888888888888888888"
},
{
"input": "37",
"output": "-1"
},
{
"input": "38",
"output": "-1"
},
{
"input": "39",
"output": "-1"
},
{
"input": "40",
"output": "-1"
},
{
"input": "462",
"output": "-1"
},
{
"input": "1317",
"output": "-1"
},
{
"input": "88",
"output": "-1"
},
{
"input": "87",
"output": "-1"
}
] | 1,646,038,621
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 61
| 0
|
k=int(input())
x = k//2
if(x>18):
print(-1)
else:
res="8"*x
if(k%2!=0):
res+="1"
print(res)
|
Title: A Prosperous Lot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Apart from Nian, there is a daemon named Sui, which terrifies children and causes them to become sick. Parents give their children money wrapped in red packets and put them under the pillow, so that when Sui tries to approach them, it will be driven away by the fairies inside.
Big Banban is hesitating over the amount of money to give out. He considers loops to be lucky since it symbolizes unity and harmony.
He would like to find a positive integer *n* not greater than 1018, such that there are exactly *k* loops in the decimal representation of *n*, or determine that such *n* does not exist.
A loop is a planar area enclosed by lines in the digits' decimal representation written in Arabic numerals. For example, there is one loop in digit 4, two loops in 8 and no loops in 5. Refer to the figure below for all exact forms.
Input Specification:
The first and only line contains an integer *k* (1<=≤<=*k*<=≤<=106) — the desired number of loops.
Output Specification:
Output an integer — if no such *n* exists, output -1; otherwise output any such *n*. In the latter case, your output should be a positive decimal integer not exceeding 1018.
Demo Input:
['2\n', '6\n']
Demo Output:
['462', '8080']
Note:
none
|
```python
k=int(input())
x = k//2
if(x>18):
print(-1)
else:
res="8"*x
if(k%2!=0):
res+="1"
print(res)
```
| 0
|
|
808
|
C
|
Tea Party
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"greedy",
"sortings"
] | null | null |
Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=≤<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
|
The first line contains two integer numbers *n* and *w* (1<=≤<=*n*<=≤<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
|
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
|
[
"2 10\n8 7\n",
"4 4\n1 1 1 1\n",
"3 10\n9 8 10\n"
] |
[
"6 4 \n",
"1 1 1 1 \n",
"-1\n"
] |
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.
| 0
|
[
{
"input": "2 10\n8 7",
"output": "6 4 "
},
{
"input": "4 4\n1 1 1 1",
"output": "1 1 1 1 "
},
{
"input": "3 10\n9 8 10",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "1 "
},
{
"input": "1 1\n2",
"output": "1 "
},
{
"input": "1 10\n20",
"output": "10 "
},
{
"input": "3 10\n8 4 8",
"output": "4 2 4 "
},
{
"input": "3 100\n37 26 37",
"output": "37 26 37 "
},
{
"input": "3 60\n43 23 24",
"output": "36 12 12 "
},
{
"input": "20 14\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 8\n1 2 1 2 1 1 1 2 1 1 1 2 1 1 2 1 1 1 2 2",
"output": "-1"
},
{
"input": "50 1113\n25 21 23 37 28 23 19 25 5 12 3 11 46 50 13 50 7 1 8 40 4 6 34 27 11 39 45 31 10 12 48 2 19 37 47 45 30 24 21 42 36 14 31 30 31 50 6 3 33 49",
"output": "13 11 12 37 28 12 10 18 3 6 2 6 46 50 7 50 4 1 4 40 2 3 34 27 6 39 45 31 5 6 48 1 10 37 47 45 30 12 11 42 36 7 31 30 31 50 3 2 33 49 "
},
{
"input": "50 440\n14 69 33 38 83 65 21 66 89 3 93 60 31 16 61 20 42 64 13 1 50 50 74 58 67 61 52 22 69 68 18 33 28 59 4 8 96 32 84 85 87 87 61 89 2 47 15 64 88 18",
"output": "-1"
},
{
"input": "100 640\n82 51 81 14 37 17 78 92 64 15 8 86 89 8 87 77 66 10 15 12 100 25 92 47 21 78 20 63 13 49 41 36 41 79 16 87 87 69 3 76 80 60 100 49 70 59 72 8 38 71 45 97 71 14 76 54 81 4 59 46 39 29 92 3 49 22 53 99 59 52 74 31 92 43 42 23 44 9 82 47 7 40 12 9 3 55 37 85 46 22 84 52 98 41 21 77 63 17 62 91",
"output": "-1"
},
{
"input": "100 82\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "100 55\n1 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1",
"output": "-1"
},
{
"input": "30 50\n3 1 2 4 1 2 2 4 3 4 4 3 3 3 3 5 3 2 5 4 3 3 5 3 3 5 4 5 3 5",
"output": "-1"
},
{
"input": "40 100\n3 3 3 3 4 1 1 1 1 1 2 2 1 3 1 2 3 2 1 2 2 2 1 4 2 2 3 3 3 2 4 6 4 4 3 2 2 2 4 5",
"output": "3 3 3 3 4 1 1 1 1 1 2 2 1 3 1 2 3 2 1 2 2 2 1 4 2 2 3 3 3 2 4 6 4 4 3 2 2 2 4 5 "
},
{
"input": "100 10000\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 "
},
{
"input": "2 5\n3 4",
"output": "2 3 "
},
{
"input": "2 6\n2 6",
"output": "1 5 "
},
{
"input": "23 855\n5 63 94 57 38 84 77 79 83 36 47 31 60 79 75 48 88 17 46 33 23 15 27",
"output": "3 32 94 29 19 84 39 72 83 18 24 16 30 79 38 24 88 9 23 17 12 8 14 "
},
{
"input": "52 2615\n73 78 70 92 94 74 46 19 55 20 70 3 1 42 68 10 66 80 1 31 65 19 73 74 56 35 53 38 92 35 65 81 6 98 74 51 27 49 76 19 86 76 5 60 14 75 64 99 43 7 36 79",
"output": "73 78 70 92 94 74 46 10 55 10 70 2 1 42 68 5 66 80 1 16 65 10 73 74 56 18 53 38 92 30 65 81 3 98 74 51 14 49 76 10 86 76 3 60 7 75 64 99 43 4 36 79 "
},
{
"input": "11 287\n34 30 69 86 22 53 11 91 62 44 5",
"output": "17 15 35 43 11 27 6 77 31 22 3 "
},
{
"input": "55 1645\n60 53 21 20 87 48 10 21 76 35 52 41 82 86 93 11 93 86 34 15 37 63 57 3 57 57 32 8 55 25 29 38 46 22 13 87 27 35 40 83 5 7 6 18 88 25 4 59 95 62 31 93 98 50 62",
"output": "30 27 11 10 82 24 5 11 38 18 26 21 41 43 93 6 93 43 17 8 19 32 29 2 29 29 16 4 28 13 15 19 23 11 7 87 14 18 20 42 3 4 3 9 88 13 2 30 95 31 16 93 98 25 31 "
},
{
"input": "71 3512\n97 46 76 95 81 96 99 83 10 50 19 18 73 5 41 60 12 73 60 31 21 64 88 61 43 57 61 19 75 35 41 85 12 59 32 47 37 43 35 92 90 47 3 98 21 18 61 79 39 86 74 8 52 33 39 27 93 54 35 38 96 36 83 51 97 10 8 66 75 87 68",
"output": "97 46 76 95 81 96 99 83 5 50 10 9 73 3 41 60 6 73 60 16 11 64 88 61 43 57 61 10 75 18 41 85 6 59 16 47 19 43 18 92 90 47 2 98 11 9 61 79 20 86 74 4 52 17 21 14 93 54 18 19 96 18 83 51 97 5 4 66 75 87 68 "
},
{
"input": "100 2633\n99 50 64 81 75 73 26 31 31 36 95 12 100 2 70 72 78 56 76 23 94 8 91 1 39 82 97 67 64 25 71 90 48 34 31 46 64 37 46 50 99 93 14 56 1 89 95 89 50 52 12 58 43 65 45 88 90 14 38 19 6 15 91 67 43 48 82 20 11 48 33 20 39 52 73 5 25 84 26 54 42 56 10 28 9 63 60 98 30 1 25 74 86 56 85 9 12 94 80 95",
"output": "50 25 32 41 38 37 13 16 16 18 48 6 61 1 35 36 39 28 38 12 47 4 46 1 20 41 49 34 32 13 36 45 24 17 16 23 32 19 23 25 50 47 7 28 1 45 48 45 25 26 6 29 22 33 23 44 45 7 19 10 3 8 46 34 22 24 41 10 6 24 17 10 20 26 37 3 13 42 13 27 21 28 5 14 5 32 30 49 15 1 13 37 43 28 43 5 6 47 40 48 "
},
{
"input": "71 1899\n23 55 58 87 69 85 100 21 19 72 81 68 20 25 29 92 18 74 89 70 53 7 78 57 41 79 64 87 63 76 95 84 1 28 32 1 79 34 77 17 71 61 35 31 62 92 69 99 60 26 2 18 61 9 27 77 82 6 30 65 52 3 51 43 13 77 41 59 19 29 86",
"output": "12 28 29 44 35 43 95 11 10 36 41 34 10 13 15 46 9 37 45 35 27 4 39 29 21 40 32 44 32 38 48 42 1 14 16 1 40 17 39 9 36 31 18 16 31 46 35 50 30 13 1 9 31 5 14 39 41 3 15 33 26 2 26 22 7 39 21 30 10 15 43 "
},
{
"input": "10 21\n3 3 3 3 4 3 3 3 3 3",
"output": "2 2 2 2 3 2 2 2 2 2 "
}
] | 1,625,917,176
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 29
| 108
| 0
|
n, w = map(int, input().split())
a = sorted([(n, i) for i, n in enumerate(map(int, input().split()))])
tea = [0]*n
for i in range(n):
tea[a[i][1]] += (a[i][0]+1) // 2
w -= sum(tea)
i = n-1
while w > 0:
x = min(a[i][0]-tea[a[i][1]], w)
tea[a[i][1]] += x
w -= x
i -= 1
if w < 0:
print(-1)
else:
print(*tea)
|
Title: Tea Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=≤<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
Input Specification:
The first line contains two integer numbers *n* and *w* (1<=≤<=*n*<=≤<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
Demo Input:
['2 10\n8 7\n', '4 4\n1 1 1 1\n', '3 10\n9 8 10\n']
Demo Output:
['6 4 \n', '1 1 1 1 \n', '-1\n']
Note:
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.
|
```python
n, w = map(int, input().split())
a = sorted([(n, i) for i, n in enumerate(map(int, input().split()))])
tea = [0]*n
for i in range(n):
tea[a[i][1]] += (a[i][0]+1) // 2
w -= sum(tea)
i = n-1
while w > 0:
x = min(a[i][0]-tea[a[i][1]], w)
tea[a[i][1]] += x
w -= x
i -= 1
if w < 0:
print(-1)
else:
print(*tea)
```
| 3
|
|
381
|
A
|
Sereja and Dima
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"two pointers"
] | null | null |
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
|
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
|
[
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] |
[
"12 5\n",
"16 12\n"
] |
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
| 500
|
[
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24",
"output": "644 500"
},
{
"input": "1\n3",
"output": "3 0"
},
{
"input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646",
"output": "6848 6568"
},
{
"input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727",
"output": "9562 9561"
},
{
"input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12",
"output": "315 315"
},
{
"input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304",
"output": "3238 2222"
},
{
"input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325",
"output": "5246 4864"
},
{
"input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71",
"output": "8147 7807"
},
{
"input": "1\n1",
"output": "1 0"
}
] | 1,672,310,095
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 46
| 0
|
ln = int(input())
crds = list(map(int, input().split(" ")))
h = 0
d = 0
trn = 1
while len(crds)>0:
if crds[0]>crds[-1]:
mx = crds[0]
del crds[0]
else:
mx = crds[-1]
del crds[-1]
if trn%2!=0:
h += mx
else:
d += mx
trn += 1
print(h, d)
|
Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
|
```python
ln = int(input())
crds = list(map(int, input().split(" ")))
h = 0
d = 0
trn = 1
while len(crds)>0:
if crds[0]>crds[-1]:
mx = crds[0]
del crds[0]
else:
mx = crds[-1]
del crds[-1]
if trn%2!=0:
h += mx
else:
d += mx
trn += 1
print(h, d)
```
| 3
|
|
876
|
A
|
Trip For Meal
|
PROGRAMMING
| 900
|
[
"math"
] | null | null |
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is *a* meters, between Rabbit's and Eeyore's house is *b* meters, between Owl's and Eeyore's house is *c* meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal *n* times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal *n* times, traveling minimum possible distance. Help him to find this distance.
|
First line contains an integer *n* (1<=≤<=*n*<=≤<=100) — number of visits.
Second line contains an integer *a* (1<=≤<=*a*<=≤<=100) — distance between Rabbit's and Owl's houses.
Third line contains an integer *b* (1<=≤<=*b*<=≤<=100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer *c* (1<=≤<=*c*<=≤<=100) — distance between Owl's and Eeyore's houses.
|
Output one number — minimum distance in meters Winnie must go through to have a meal *n* times.
|
[
"3\n2\n3\n1\n",
"1\n2\n3\n5\n"
] |
[
"3\n",
"0\n"
] |
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
| 500
|
[
{
"input": "3\n2\n3\n1",
"output": "3"
},
{
"input": "1\n2\n3\n5",
"output": "0"
},
{
"input": "10\n1\n8\n3",
"output": "9"
},
{
"input": "7\n10\n5\n6",
"output": "30"
},
{
"input": "9\n9\n7\n5",
"output": "42"
},
{
"input": "9\n37\n85\n76",
"output": "296"
},
{
"input": "76\n46\n77\n11",
"output": "860"
},
{
"input": "80\n42\n1\n37",
"output": "79"
},
{
"input": "8\n80\n55\n1",
"output": "61"
},
{
"input": "10\n13\n72\n17",
"output": "117"
},
{
"input": "9\n24\n1\n63",
"output": "8"
},
{
"input": "65\n5\n8\n7",
"output": "320"
},
{
"input": "56\n8\n9\n3",
"output": "170"
},
{
"input": "59\n8\n1\n2",
"output": "58"
},
{
"input": "75\n50\n50\n5",
"output": "415"
},
{
"input": "75\n54\n76\n66",
"output": "3996"
},
{
"input": "73\n71\n69\n66",
"output": "4755"
},
{
"input": "83\n58\n88\n16",
"output": "1354"
},
{
"input": "74\n31\n11\n79",
"output": "803"
},
{
"input": "62\n27\n16\n72",
"output": "976"
},
{
"input": "72\n95\n27\n9",
"output": "657"
},
{
"input": "1\n2\n2\n1",
"output": "0"
},
{
"input": "1\n1\n1\n1",
"output": "0"
},
{
"input": "1\n1\n1\n99",
"output": "0"
},
{
"input": "100\n100\n100\n100",
"output": "9900"
},
{
"input": "2\n1\n1\n3",
"output": "1"
},
{
"input": "1\n3\n2\n1",
"output": "0"
},
{
"input": "1\n5\n6\n1",
"output": "0"
},
{
"input": "1\n2\n6\n1",
"output": "0"
},
{
"input": "1\n30\n20\n1",
"output": "0"
},
{
"input": "1\n15\n13\n11",
"output": "0"
},
{
"input": "1\n100\n50\n1",
"output": "0"
},
{
"input": "1\n2\n3\n1",
"output": "0"
},
{
"input": "5\n5\n5\n5",
"output": "20"
},
{
"input": "1\n100\n50\n5",
"output": "0"
},
{
"input": "1\n5\n7\n3",
"output": "0"
},
{
"input": "1\n3\n4\n2",
"output": "0"
},
{
"input": "3\n3\n3\n3",
"output": "6"
},
{
"input": "1\n99\n98\n1",
"output": "0"
},
{
"input": "5\n3\n3\n1",
"output": "6"
},
{
"input": "2\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1\n7\n8",
"output": "1"
},
{
"input": "2\n2\n2\n1",
"output": "2"
},
{
"input": "2\n2\n3\n1",
"output": "2"
},
{
"input": "1\n5\n6\n2",
"output": "0"
},
{
"input": "100\n1\n1\n100",
"output": "99"
},
{
"input": "2\n3\n2\n1",
"output": "2"
},
{
"input": "1\n99\n98\n97",
"output": "0"
}
] | 1,669,566,063
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
v=int(input())
r=int(input())
o=int(input())
e=int(input())
if v==1:
print(0)
elif v==2:
print(abs(o-r))
elif v==3:
print(abs(o-r)+abs(e-o))
|
Title: Trip For Meal
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is *a* meters, between Rabbit's and Eeyore's house is *b* meters, between Owl's and Eeyore's house is *c* meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal *n* times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal *n* times, traveling minimum possible distance. Help him to find this distance.
Input Specification:
First line contains an integer *n* (1<=≤<=*n*<=≤<=100) — number of visits.
Second line contains an integer *a* (1<=≤<=*a*<=≤<=100) — distance between Rabbit's and Owl's houses.
Third line contains an integer *b* (1<=≤<=*b*<=≤<=100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer *c* (1<=≤<=*c*<=≤<=100) — distance between Owl's and Eeyore's houses.
Output Specification:
Output one number — minimum distance in meters Winnie must go through to have a meal *n* times.
Demo Input:
['3\n2\n3\n1\n', '1\n2\n3\n5\n']
Demo Output:
['3\n', '0\n']
Note:
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
|
```python
v=int(input())
r=int(input())
o=int(input())
e=int(input())
if v==1:
print(0)
elif v==2:
print(abs(o-r))
elif v==3:
print(abs(o-r)+abs(e-o))
```
| 0
|
|
858
|
A
|
k-rounding
|
PROGRAMMING
| 1,100
|
[
"brute force",
"math",
"number theory"
] | null | null |
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
|
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
|
Print the *k*-rounding of *n*.
|
[
"375 4\n",
"10000 1\n",
"38101 0\n",
"123456789 8\n"
] |
[
"30000\n",
"10000\n",
"38101\n",
"12345678900000000\n"
] |
none
| 750
|
[
{
"input": "375 4",
"output": "30000"
},
{
"input": "10000 1",
"output": "10000"
},
{
"input": "38101 0",
"output": "38101"
},
{
"input": "123456789 8",
"output": "12345678900000000"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "2 0",
"output": "2"
},
{
"input": "100 0",
"output": "100"
},
{
"input": "1000000000 0",
"output": "1000000000"
},
{
"input": "160 2",
"output": "800"
},
{
"input": "3 0",
"output": "3"
},
{
"input": "10 0",
"output": "10"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "3 1",
"output": "30"
},
{
"input": "4 1",
"output": "20"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "6 1",
"output": "30"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "8 1",
"output": "40"
},
{
"input": "9 1",
"output": "90"
},
{
"input": "10 1",
"output": "10"
},
{
"input": "11 1",
"output": "110"
},
{
"input": "12 1",
"output": "60"
},
{
"input": "16 2",
"output": "400"
},
{
"input": "2 2",
"output": "100"
},
{
"input": "1 2",
"output": "100"
},
{
"input": "5 2",
"output": "100"
},
{
"input": "15 2",
"output": "300"
},
{
"input": "36 2",
"output": "900"
},
{
"input": "1 8",
"output": "100000000"
},
{
"input": "8 8",
"output": "100000000"
},
{
"input": "96 8",
"output": "300000000"
},
{
"input": "175 8",
"output": "700000000"
},
{
"input": "9999995 8",
"output": "199999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "12345678 8",
"output": "617283900000000"
},
{
"input": "78125 8",
"output": "100000000"
},
{
"input": "390625 8",
"output": "100000000"
},
{
"input": "1953125 8",
"output": "500000000"
},
{
"input": "9765625 8",
"output": "2500000000"
},
{
"input": "68359375 8",
"output": "17500000000"
},
{
"input": "268435456 8",
"output": "104857600000000"
},
{
"input": "125829120 8",
"output": "9830400000000"
},
{
"input": "128000 8",
"output": "400000000"
},
{
"input": "300000 8",
"output": "300000000"
},
{
"input": "3711871 8",
"output": "371187100000000"
},
{
"input": "55555 8",
"output": "1111100000000"
},
{
"input": "222222222 8",
"output": "11111111100000000"
},
{
"input": "479001600 8",
"output": "7484400000000"
},
{
"input": "655360001 7",
"output": "6553600010000000"
},
{
"input": "655360001 8",
"output": "65536000100000000"
},
{
"input": "1000000000 1",
"output": "1000000000"
},
{
"input": "1000000000 7",
"output": "1000000000"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "10000000 8",
"output": "100000000"
},
{
"input": "1000000 8",
"output": "100000000"
},
{
"input": "10000009 8",
"output": "1000000900000000"
},
{
"input": "10000005 8",
"output": "200000100000000"
},
{
"input": "10000002 8",
"output": "500000100000000"
},
{
"input": "999999997 8",
"output": "99999999700000000"
},
{
"input": "999999997 7",
"output": "9999999970000000"
},
{
"input": "999999995 8",
"output": "19999999900000000"
},
{
"input": "123 8",
"output": "12300000000"
},
{
"input": "24 2",
"output": "600"
},
{
"input": "16 4",
"output": "10000"
},
{
"input": "123456787 8",
"output": "12345678700000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "101 1",
"output": "1010"
},
{
"input": "50 2",
"output": "100"
},
{
"input": "999999818 1",
"output": "4999999090"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "123 1",
"output": "1230"
},
{
"input": "16 1",
"output": "80"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "15304 6",
"output": "1913000000"
},
{
"input": "3 8",
"output": "300000000"
},
{
"input": "4 2",
"output": "100"
},
{
"input": "100000 7",
"output": "10000000"
},
{
"input": "5 8",
"output": "100000000"
},
{
"input": "16724 6",
"output": "4181000000"
},
{
"input": "999999999 1",
"output": "9999999990"
},
{
"input": "999999990 8",
"output": "9999999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "100000 1",
"output": "100000"
},
{
"input": "8 3",
"output": "1000"
},
{
"input": "16768 6",
"output": "262000000"
},
{
"input": "123456789 1",
"output": "1234567890"
},
{
"input": "2 8",
"output": "100000000"
},
{
"input": "999999937 8",
"output": "99999993700000000"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "2000000 7",
"output": "10000000"
},
{
"input": "1999998 2",
"output": "99999900"
},
{
"input": "125 3",
"output": "1000"
}
] | 1,506,079,901
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 61
| 0
|
def lcm(x, y):
if x > y:
greater = x
else:
greater = y
while(True):
if((greater % x == 0) and (greater % y == 0)):
lcm = greater
break
greater += 1
s=input().split()
n=int(s[0])
k=int(s[1])
lcm(n,10**k)
print(lcm)
|
Title: k-rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
Output Specification:
Print the *k*-rounding of *n*.
Demo Input:
['375 4\n', '10000 1\n', '38101 0\n', '123456789 8\n']
Demo Output:
['30000\n', '10000\n', '38101\n', '12345678900000000\n']
Note:
none
|
```python
def lcm(x, y):
if x > y:
greater = x
else:
greater = y
while(True):
if((greater % x == 0) and (greater % y == 0)):
lcm = greater
break
greater += 1
s=input().split()
n=int(s[0])
k=int(s[1])
lcm(n,10**k)
print(lcm)
```
| 0
|
|
701
|
A
|
Cards
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation"
] | null | null |
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
|
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card.
|
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
|
[
"6\n1 5 7 4 4 3\n",
"4\n10 10 10 10\n"
] |
[
"1 3\n6 2\n4 5\n",
"1 2\n3 4\n"
] |
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable.
| 500
|
[
{
"input": "6\n1 5 7 4 4 3",
"output": "1 3\n6 2\n4 5"
},
{
"input": "4\n10 10 10 10",
"output": "1 4\n2 3"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "4\n82 46 8 44",
"output": "3 1\n4 2"
},
{
"input": "2\n35 50",
"output": "1 2"
},
{
"input": "8\n24 39 49 38 44 64 44 50",
"output": "1 6\n4 8\n2 3\n5 7"
},
{
"input": "100\n23 44 35 88 10 78 8 84 46 19 69 36 81 60 46 12 53 22 83 73 6 18 80 14 54 39 74 42 34 20 91 70 32 11 80 53 70 21 24 12 87 68 35 39 8 84 81 70 8 54 73 2 60 71 4 33 65 48 69 58 55 57 78 61 45 50 55 72 86 37 5 11 12 81 32 19 22 11 22 82 23 56 61 84 47 59 31 38 31 90 57 1 24 38 68 27 80 9 37 14",
"output": "92 31\n52 90\n55 4\n71 41\n21 69\n7 84\n45 46\n49 8\n98 19\n5 80\n34 74\n72 47\n78 13\n16 97\n40 35\n73 23\n24 63\n100 6\n22 27\n10 51\n76 20\n30 68\n38 54\n18 48\n77 37\n79 32\n1 59\n81 11\n39 95\n93 42\n96 57\n87 83\n89 64\n33 53\n75 14\n56 86\n29 60\n3 91\n43 62\n12 82\n70 67\n99 61\n88 50\n94 25\n26 36\n44 17\n28 66\n2 58\n65 85\n9 15"
},
{
"input": "12\n22 83 2 67 55 12 40 93 83 73 12 28",
"output": "3 8\n6 9\n11 2\n1 10\n12 4\n7 5"
},
{
"input": "16\n10 33 36 32 48 25 31 27 45 13 37 26 22 21 15 43",
"output": "1 5\n10 9\n15 16\n14 11\n13 3\n6 2\n12 4\n8 7"
},
{
"input": "20\n18 13 71 60 28 10 20 65 65 12 13 14 64 68 6 50 72 7 66 58",
"output": "15 17\n18 3\n6 14\n10 19\n2 9\n11 8\n12 13\n1 4\n7 20\n5 16"
},
{
"input": "24\n59 39 25 22 46 21 24 70 60 11 46 42 44 37 13 37 41 58 72 23 25 61 58 62",
"output": "10 19\n15 8\n6 24\n4 22\n20 9\n7 1\n3 23\n21 18\n14 11\n16 5\n2 13\n17 12"
},
{
"input": "28\n22 1 51 31 83 35 3 64 59 10 61 25 19 53 55 80 78 8 82 22 67 4 27 64 33 6 85 76",
"output": "2 27\n7 5\n22 19\n26 16\n18 17\n10 28\n13 21\n1 24\n20 8\n12 11\n23 9\n4 15\n25 14\n6 3"
},
{
"input": "32\n41 42 22 68 40 52 66 16 73 25 41 21 36 60 46 30 24 55 35 10 54 52 70 24 20 56 3 34 35 6 51 8",
"output": "27 9\n30 23\n32 4\n20 7\n8 14\n25 26\n12 18\n3 21\n17 22\n24 6\n10 31\n16 15\n28 2\n19 11\n29 1\n13 5"
},
{
"input": "36\n1 10 61 43 27 49 55 33 7 30 45 78 69 34 38 19 36 49 55 11 30 63 46 24 16 68 71 18 11 52 72 24 60 68 8 41",
"output": "1 12\n9 31\n35 27\n2 13\n20 34\n29 26\n25 22\n28 3\n16 33\n24 19\n32 7\n5 30\n10 18\n21 6\n8 23\n14 11\n17 4\n15 36"
},
{
"input": "40\n7 30 13 37 37 56 45 28 61 28 23 33 44 63 58 52 21 2 42 19 10 32 9 7 61 15 58 20 45 4 46 24 35 17 50 4 20 48 41 55",
"output": "18 14\n30 25\n36 9\n1 27\n24 15\n23 6\n21 40\n3 16\n26 35\n34 38\n20 31\n28 29\n37 7\n17 13\n11 19\n32 39\n8 5\n10 4\n2 33\n22 12"
},
{
"input": "44\n7 12 46 78 24 68 86 22 71 79 85 14 58 72 26 46 54 39 35 13 31 45 81 21 15 8 47 64 69 87 57 6 18 80 47 29 36 62 34 67 59 48 75 25",
"output": "32 30\n1 7\n26 11\n2 23\n20 34\n12 10\n25 4\n33 43\n24 14\n8 9\n5 29\n44 6\n15 40\n36 28\n21 38\n39 41\n19 13\n37 31\n18 17\n22 42\n3 35\n16 27"
},
{
"input": "48\n57 38 16 25 34 57 29 38 60 51 72 78 22 39 10 33 20 16 12 3 51 74 9 88 4 70 56 65 86 18 33 12 77 78 52 87 68 85 81 5 61 2 52 39 80 13 74 30",
"output": "42 24\n20 36\n25 29\n40 38\n23 39\n15 45\n19 34\n32 12\n46 33\n3 47\n18 22\n30 11\n17 26\n13 37\n4 28\n7 41\n48 9\n16 6\n31 1\n5 27\n2 43\n8 35\n14 21\n44 10"
},
{
"input": "52\n57 12 13 40 68 31 18 4 31 18 65 3 62 32 6 3 49 48 51 33 53 40 9 32 47 53 58 19 14 23 32 38 39 69 19 20 62 52 68 17 39 22 54 59 3 2 52 9 67 68 24 39",
"output": "46 34\n12 50\n16 39\n45 5\n8 49\n15 11\n23 37\n48 13\n2 44\n3 27\n29 1\n40 43\n7 26\n10 21\n28 47\n35 38\n36 19\n42 17\n30 18\n51 25\n6 22\n9 4\n14 52\n24 41\n31 33\n20 32"
},
{
"input": "56\n53 59 66 68 71 25 48 32 12 61 72 69 30 6 56 55 25 49 60 47 46 46 66 19 31 9 23 15 10 12 71 53 51 32 39 31 66 66 17 52 12 7 7 22 49 12 71 29 63 7 47 29 18 39 27 26",
"output": "14 11\n42 47\n43 31\n50 5\n26 12\n29 4\n9 38\n30 37\n41 23\n46 3\n28 49\n39 10\n53 19\n24 2\n44 15\n27 16\n6 32\n17 1\n56 40\n55 33\n48 45\n52 18\n13 7\n25 51\n36 20\n8 22\n34 21\n35 54"
},
{
"input": "60\n47 63 20 68 46 12 45 44 14 38 28 73 60 5 20 18 70 64 37 47 26 47 37 61 29 61 23 28 30 68 55 22 25 60 38 7 63 12 38 15 14 30 11 5 70 15 53 52 7 57 49 45 55 37 45 28 50 2 31 30",
"output": "58 12\n14 45\n44 17\n36 30\n49 4\n43 18\n6 37\n38 2\n9 26\n41 24\n40 34\n46 13\n16 50\n3 53\n15 31\n32 47\n27 48\n33 57\n21 51\n11 22\n28 20\n56 1\n25 5\n29 55\n42 52\n60 7\n59 8\n19 39\n23 35\n54 10"
},
{
"input": "64\n63 39 19 5 48 56 49 45 29 68 25 59 37 69 62 26 60 44 60 6 67 68 2 40 56 6 19 12 17 70 23 11 59 37 41 55 30 68 72 14 38 34 3 71 2 4 55 15 31 66 15 51 36 72 18 7 6 14 43 33 8 35 57 18",
"output": "23 54\n45 39\n43 44\n46 30\n4 14\n20 38\n26 22\n57 10\n56 21\n61 50\n32 1\n28 15\n40 19\n58 17\n48 33\n51 12\n29 63\n55 25\n64 6\n3 47\n27 36\n31 52\n11 7\n16 5\n9 8\n37 18\n49 59\n60 35\n42 24\n62 2\n53 41\n13 34"
},
{
"input": "68\n58 68 40 55 62 15 10 54 19 18 69 27 15 53 8 18 8 33 15 49 20 9 70 8 18 64 14 59 9 64 3 35 46 11 5 65 58 55 28 58 4 55 64 5 68 24 4 58 23 45 58 50 38 68 5 15 20 9 5 53 20 63 69 68 15 53 65 65",
"output": "31 23\n41 63\n47 11\n35 64\n44 54\n55 45\n59 2\n15 68\n17 67\n24 36\n22 43\n29 30\n58 26\n7 62\n34 5\n27 28\n6 51\n13 48\n19 40\n56 37\n65 1\n10 42\n16 38\n25 4\n9 8\n21 66\n57 60\n61 14\n49 52\n46 20\n12 33\n39 50\n18 3\n32 53"
},
{
"input": "72\n61 13 55 23 24 55 44 33 59 19 14 17 66 40 27 33 29 37 28 74 50 56 59 65 64 17 42 56 73 51 64 23 22 26 38 22 36 47 60 14 52 28 14 12 6 41 73 5 64 67 61 74 54 34 45 34 44 4 34 49 18 72 44 47 31 19 11 31 5 4 45 50",
"output": "58 52\n70 20\n48 47\n69 29\n45 62\n67 50\n44 13\n2 24\n11 49\n40 31\n43 25\n12 51\n26 1\n61 39\n10 23\n66 9\n33 28\n36 22\n4 6\n32 3\n5 53\n34 41\n15 30\n19 72\n42 21\n17 60\n65 64\n68 38\n8 71\n16 55\n54 63\n56 57\n59 7\n37 27\n18 46\n35 14"
},
{
"input": "76\n73 37 73 67 26 45 43 74 47 31 43 81 4 3 39 79 48 81 67 39 67 66 43 67 80 51 34 79 5 58 45 10 39 50 9 78 6 18 75 17 45 17 51 71 34 53 33 11 17 15 11 69 50 41 13 74 10 33 77 41 11 64 36 74 17 32 3 10 27 20 5 73 52 41 7 57",
"output": "14 18\n67 12\n13 25\n29 28\n71 16\n37 36\n75 59\n35 39\n32 64\n57 56\n68 8\n48 72\n51 3\n61 1\n55 44\n50 52\n40 24\n42 21\n49 19\n65 4\n38 22\n70 62\n5 30\n69 76\n10 46\n66 73\n47 43\n58 26\n27 53\n45 34\n63 17\n2 9\n15 41\n20 31\n33 6\n54 23\n60 11\n74 7"
},
{
"input": "80\n18 38 65 1 20 9 57 2 36 26 15 17 33 61 65 27 10 35 49 42 40 32 19 33 12 36 56 31 10 41 8 54 56 60 5 47 61 43 23 19 20 30 7 6 38 60 29 58 35 64 30 51 6 17 30 24 47 1 37 47 34 36 48 28 5 25 47 19 30 39 36 23 31 28 46 46 59 43 19 49",
"output": "4 15\n58 3\n8 50\n35 37\n65 14\n44 46\n53 34\n43 77\n31 48\n6 7\n17 33\n29 27\n25 32\n11 52\n12 80\n54 19\n1 63\n23 67\n40 60\n68 57\n79 36\n5 76\n41 75\n39 78\n72 38\n56 20\n66 30\n10 21\n16 70\n64 45\n74 2\n47 59\n42 71\n51 62\n55 26\n69 9\n28 49\n73 18\n22 61\n13 24"
},
{
"input": "84\n59 41 54 14 42 55 29 28 41 73 40 15 1 1 66 49 76 59 68 60 42 81 19 23 33 12 80 81 42 22 54 54 2 22 22 28 27 60 36 57 17 76 38 20 40 65 23 9 81 50 25 13 46 36 59 53 6 35 47 40 59 19 67 46 63 49 12 33 23 49 33 23 32 62 60 70 44 1 6 63 28 16 70 69",
"output": "13 49\n14 28\n78 22\n33 27\n57 42\n79 17\n48 10\n26 83\n67 76\n52 84\n4 19\n12 63\n82 15\n41 46\n23 80\n62 65\n44 74\n30 75\n34 38\n35 20\n24 61\n47 55\n69 18\n72 1\n51 40\n37 6\n8 32\n36 31\n81 3\n7 56\n73 50\n25 70\n68 66\n71 16\n58 59\n39 64\n54 53\n43 77\n11 29\n45 21\n60 5\n2 9"
},
{
"input": "88\n10 28 71 6 58 66 45 52 13 71 39 1 10 29 30 70 14 17 15 38 4 60 5 46 66 41 40 58 2 57 32 44 21 26 13 40 64 63 56 33 46 8 30 43 67 55 44 28 32 62 14 58 42 67 45 59 32 68 10 31 51 6 42 34 9 12 51 27 20 14 62 42 16 5 1 14 30 62 40 59 58 26 25 15 27 47 21 57",
"output": "12 10\n75 3\n29 16\n21 58\n23 54\n74 45\n4 25\n62 6\n42 37\n65 38\n1 78\n13 71\n59 50\n66 22\n9 80\n35 56\n17 81\n51 52\n70 28\n76 5\n19 88\n84 30\n73 39\n18 46\n69 8\n33 67\n87 61\n83 86\n34 41\n82 24\n68 55\n85 7\n2 47\n48 32\n14 44\n15 72\n43 63\n77 53\n60 26\n31 79\n49 36\n57 27\n40 11\n64 20"
},
{
"input": "92\n17 37 81 15 29 70 73 42 49 23 44 77 27 44 74 11 43 66 15 41 60 36 33 11 2 76 16 51 45 21 46 16 85 29 76 79 16 6 60 13 25 44 62 28 43 35 63 24 76 71 62 15 57 72 45 10 71 59 74 14 53 13 58 72 14 72 73 11 25 1 57 42 86 63 50 30 64 38 10 77 75 24 58 8 54 12 43 30 27 71 52 34",
"output": "70 73\n25 33\n38 3\n84 36\n56 80\n79 12\n16 49\n24 35\n68 26\n86 81\n40 59\n62 15\n60 67\n65 7\n4 66\n19 64\n52 54\n27 90\n32 57\n37 50\n1 6\n30 18\n10 77\n48 74\n82 47\n41 51\n69 43\n13 39\n89 21\n44 58\n5 83\n34 63\n76 71\n88 53\n23 85\n92 61\n46 91\n22 28\n2 75\n78 9\n20 31\n8 55\n72 29\n17 42\n45 14\n87 11"
},
{
"input": "96\n77 7 47 19 73 31 46 13 89 69 52 9 26 77 6 87 55 45 71 2 79 1 80 20 4 82 64 20 75 86 84 24 77 56 16 54 53 35 74 73 40 29 63 20 83 39 58 16 31 41 40 16 11 90 30 48 62 39 55 8 50 3 77 73 75 66 14 90 18 54 38 10 53 22 67 38 27 91 62 37 85 13 92 7 18 83 10 3 86 54 80 59 34 16 39 43",
"output": "22 83\n20 78\n62 68\n88 54\n25 9\n15 16\n2 89\n84 30\n60 81\n12 31\n72 86\n87 45\n53 26\n8 91\n82 23\n67 21\n35 63\n48 33\n52 14\n94 1\n69 65\n85 29\n4 39\n24 64\n28 40\n44 5\n74 19\n32 10\n13 75\n77 66\n42 27\n55 43\n6 79\n49 57\n93 92\n38 47\n80 34\n71 59\n76 17\n46 90\n58 70\n95 36\n41 73\n51 37\n50 11\n96 61\n18 56\n7 3"
},
{
"input": "4\n100 100 1 1",
"output": "3 2\n4 1"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "4\n3 4 4 5",
"output": "1 4\n2 3"
},
{
"input": "4\n1 1 2 2",
"output": "1 4\n2 3"
},
{
"input": "4\n1 2 3 4",
"output": "1 4\n2 3"
}
] | 1,587,098,363
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 35
| 140
| 0
|
n = int(input())
arr = [int(i) for i in input().split()]
for i in range(n):
arr[i] = (arr[i], i+1)
arr.sort(key=lambda x: x[0])
for i in range(n//2):
print(arr[i][1], arr[n-i-1][1])
|
Title: Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card.
Output Specification:
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
Demo Input:
['6\n1 5 7 4 4 3\n', '4\n10 10 10 10\n']
Demo Output:
['1 3\n6 2\n4 5\n', '1 2\n3 4\n']
Note:
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable.
|
```python
n = int(input())
arr = [int(i) for i in input().split()]
for i in range(n):
arr[i] = (arr[i], i+1)
arr.sort(key=lambda x: x[0])
for i in range(n//2):
print(arr[i][1], arr[n-i-1][1])
```
| 3
|
|
996
|
A
|
Hit the Lottery
|
PROGRAMMING
| 800
|
[
"dp",
"greedy"
] | null | null |
Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
|
The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).
|
Output the minimum number of bills that Allen could receive.
|
[
"125\n",
"43\n",
"1000000000\n"
] |
[
"3\n",
"5\n",
"10000000\n"
] |
In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.
| 500
|
[
{
"input": "125",
"output": "3"
},
{
"input": "43",
"output": "5"
},
{
"input": "1000000000",
"output": "10000000"
},
{
"input": "4",
"output": "4"
},
{
"input": "5",
"output": "1"
},
{
"input": "1",
"output": "1"
},
{
"input": "74",
"output": "8"
},
{
"input": "31",
"output": "3"
},
{
"input": "59",
"output": "8"
},
{
"input": "79",
"output": "9"
},
{
"input": "7",
"output": "3"
},
{
"input": "55",
"output": "4"
},
{
"input": "40",
"output": "2"
},
{
"input": "719",
"output": "13"
},
{
"input": "847",
"output": "13"
},
{
"input": "225",
"output": "4"
},
{
"input": "4704",
"output": "51"
},
{
"input": "1132",
"output": "15"
},
{
"input": "7811",
"output": "80"
},
{
"input": "7981",
"output": "84"
},
{
"input": "82655",
"output": "830"
},
{
"input": "6364",
"output": "70"
},
{
"input": "74611",
"output": "748"
},
{
"input": "45391",
"output": "459"
},
{
"input": "620448",
"output": "6210"
},
{
"input": "265145",
"output": "2654"
},
{
"input": "671704",
"output": "6721"
},
{
"input": "365173",
"output": "3658"
},
{
"input": "7130872",
"output": "71314"
},
{
"input": "9628747",
"output": "96292"
},
{
"input": "8898399",
"output": "88993"
},
{
"input": "9497953",
"output": "94985"
},
{
"input": "19070947",
"output": "190714"
},
{
"input": "20185520",
"output": "201856"
},
{
"input": "91402248",
"output": "914028"
},
{
"input": "27035533",
"output": "270360"
},
{
"input": "22717528",
"output": "227180"
},
{
"input": "24403439",
"output": "244041"
},
{
"input": "163565555",
"output": "1635659"
},
{
"input": "152541908",
"output": "1525423"
},
{
"input": "668439837",
"output": "6684403"
},
{
"input": "20",
"output": "1"
},
{
"input": "999999999",
"output": "10000009"
},
{
"input": "10",
"output": "1"
},
{
"input": "7",
"output": "3"
},
{
"input": "6",
"output": "2"
},
{
"input": "99999999",
"output": "1000009"
},
{
"input": "1000001",
"output": "10001"
},
{
"input": "2521",
"output": "27"
}
] | 1,689,395,846
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 1
| 61
| 2,867,200
|
limited_notes=[1,5,10,20,100]
limited_notes=sorted(limited_notes,reverse=True)
target=int(input())
count=0
i=0
total=0
n=1
while total<target:
if limited_notes[i]>target:
i+=1
elif (total+(limited_notes[i]*n))<=target:
total+=(limited_notes[i]*n)
n+=1
count+=1
if total==target:
break
else:
i+=1
n=1
print(count)
|
Title: Hit the Lottery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input Specification:
The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).
Output Specification:
Output the minimum number of bills that Allen could receive.
Demo Input:
['125\n', '43\n', '1000000000\n']
Demo Output:
['3\n', '5\n', '10000000\n']
Note:
In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.
|
```python
limited_notes=[1,5,10,20,100]
limited_notes=sorted(limited_notes,reverse=True)
target=int(input())
count=0
i=0
total=0
n=1
while total<target:
if limited_notes[i]>target:
i+=1
elif (total+(limited_notes[i]*n))<=target:
total+=(limited_notes[i]*n)
n+=1
count+=1
if total==target:
break
else:
i+=1
n=1
print(count)
```
| -1
|
|
148
|
A
|
Insomnia cure
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation",
"math"
] | null | null |
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
|
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
|
Output the number of damaged dragons.
|
[
"1\n2\n3\n4\n12\n",
"2\n3\n4\n5\n24\n"
] |
[
"12\n",
"17\n"
] |
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.
| 1,000
|
[
{
"input": "1\n2\n3\n4\n12",
"output": "12"
},
{
"input": "2\n3\n4\n5\n24",
"output": "17"
},
{
"input": "1\n1\n1\n1\n100000",
"output": "100000"
},
{
"input": "10\n9\n8\n7\n6",
"output": "0"
},
{
"input": "8\n4\n4\n3\n65437",
"output": "32718"
},
{
"input": "8\n4\n1\n10\n59392",
"output": "59392"
},
{
"input": "4\n1\n8\n7\n44835",
"output": "44835"
},
{
"input": "6\n1\n7\n2\n62982",
"output": "62982"
},
{
"input": "2\n7\n4\n9\n56937",
"output": "35246"
},
{
"input": "2\n9\n8\n1\n75083",
"output": "75083"
},
{
"input": "8\n7\n7\n6\n69038",
"output": "24656"
},
{
"input": "4\n4\n2\n3\n54481",
"output": "36320"
},
{
"input": "6\n4\n9\n8\n72628",
"output": "28244"
},
{
"input": "9\n7\n8\n10\n42357",
"output": "16540"
},
{
"input": "5\n6\n4\n3\n60504",
"output": "36302"
},
{
"input": "7\n2\n3\n8\n21754",
"output": "15539"
},
{
"input": "1\n2\n10\n4\n39901",
"output": "39901"
},
{
"input": "3\n4\n7\n1\n58048",
"output": "58048"
},
{
"input": "9\n10\n4\n6\n52003",
"output": "21956"
},
{
"input": "5\n10\n9\n3\n70149",
"output": "32736"
},
{
"input": "5\n5\n5\n10\n55592",
"output": "11118"
},
{
"input": "1\n5\n2\n6\n49547",
"output": "49547"
},
{
"input": "3\n7\n7\n7\n84046",
"output": "36019"
},
{
"input": "10\n2\n1\n8\n63537",
"output": "63537"
},
{
"input": "7\n2\n10\n5\n81684",
"output": "53678"
},
{
"input": "7\n1\n6\n8\n99831",
"output": "99831"
},
{
"input": "3\n9\n3\n3\n61082",
"output": "20360"
},
{
"input": "5\n5\n8\n9\n79228",
"output": "29931"
},
{
"input": "1\n5\n5\n6\n89535",
"output": "89535"
},
{
"input": "5\n5\n2\n1\n7682",
"output": "7682"
},
{
"input": "7\n4\n8\n8\n25829",
"output": "9224"
},
{
"input": "3\n4\n3\n3\n87079",
"output": "43539"
},
{
"input": "5\n8\n4\n9\n5226",
"output": "2438"
},
{
"input": "4\n9\n6\n6\n1070",
"output": "415"
},
{
"input": "10\n10\n10\n10\n100000",
"output": "10000"
},
{
"input": "1\n1\n1\n1\n1",
"output": "1"
},
{
"input": "10\n10\n10\n10\n1",
"output": "0"
},
{
"input": "10\n10\n1\n10\n10",
"output": "10"
},
{
"input": "10\n10\n1\n10\n100",
"output": "100"
},
{
"input": "2\n2\n2\n2\n1",
"output": "0"
}
] | 1,698,833,703
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
k, l, m, n, d = [int(input(_)) for _ in range(5)]
count = 0
for i in range(1, d+1):
if i%k==0 or i%l==0 or i%m==0 or i%n==0:
count += 1
print(counter)
|
Title: Insomnia cure
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
Input Specification:
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
Output Specification:
Output the number of damaged dragons.
Demo Input:
['1\n2\n3\n4\n12\n', '2\n3\n4\n5\n24\n']
Demo Output:
['12\n', '17\n']
Note:
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.
|
```python
k, l, m, n, d = [int(input(_)) for _ in range(5)]
count = 0
for i in range(1, d+1):
if i%k==0 or i%l==0 or i%m==0 or i%n==0:
count += 1
print(counter)
```
| -1
|
|
986
|
D
|
Perfect Encoding
|
PROGRAMMING
| 3,100
|
[
"fft",
"math"
] | null | null |
You are working as an analyst in a company working on a new system for big data storage. This system will store $n$ different objects. Each object should have a unique ID.
To create the system, you choose the parameters of the system — integers $m \ge 1$ and $b_{1}, b_{2}, \ldots, b_{m}$. With these parameters an ID of some object in the system is an array of integers $[a_{1}, a_{2}, \ldots, a_{m}]$ where $1 \le a_{i} \le b_{i}$ holds for every $1 \le i \le m$.
Developers say that production costs are proportional to $\sum_{i=1}^{m} b_{i}$. You are asked to choose parameters $m$ and $b_{i}$ so that the system will be able to assign unique IDs to $n$ different objects and production costs are minimized. Note that you don't have to use all available IDs.
|
In the only line of input there is one positive integer $n$. The length of the decimal representation of $n$ is no greater than $1.5 \cdot 10^{6}$. The integer does not contain leading zeros.
|
Print one number — minimal value of $\sum_{i=1}^{m} b_{i}$.
|
[
"36\n",
"37\n",
"12345678901234567890123456789\n"
] |
[
"10\n",
"11\n",
"177\n"
] |
none
| 2,500
|
[
{
"input": "36",
"output": "10"
},
{
"input": "37",
"output": "11"
},
{
"input": "12345678901234567890123456789",
"output": "177"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "3"
},
{
"input": "4",
"output": "4"
},
{
"input": "7421902501252475186372406731932548506197390793597574544727433297197476846519276598727359617092494798",
"output": "629"
},
{
"input": "71057885893313745806894531138592341136175030511382512555364579061229040750815096670263802546201989828165866147027119861863385397179695224216202346062872417111920113483747119385957051753101263769591892062039112567316036455789217245754461225443096439906225767290690128677713047690686004149082311677134836383178262318973298581951974863511315252485252083010690948164456205330279738760034861583874764199950445592461479109814313530332776429627014232776723160331462731018692207739471347664936326394313671025",
"output": "3144"
},
{
"input": "515377520732011331036461129765621272702107522001",
"output": "300"
},
{
"input": "515377520732011331036461129765621272702107522002",
"output": "301"
},
{
"input": "515377520732011331036461129765621272702107522000",
"output": "300"
},
{
"input": "2644141638961613273780910519504288731930844065504296335329840736453657194693409799081556627701216123927819555393745164711901909164201237823730685450515907348055240450396641607756029548457929682548780800235177236082257895631246188876123132346108173348981012356250960688811094108794077791634930736509832272441660537127557164580456832796615775793837112808169797875218746484343692719877391033530037881176218120852179342877728205628700771297494331664021228732264346205537805710440002",
"output": "3002"
}
] | 1,528,660,694
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 14
| 2,000
| 15,769,600
|
import math
a = int(input())
if (a <= 3):
print(a)
exit(0)
def pwr(a, n):
ans = 1
while (n != 0):
if (n & 1):
ans *= a
a *= a
n >>= 1
return ans
def calc(sm):
if (sm <= 0):
return -10000000000000000
if (sm == 1):
return 1
rem = sm % 3
if (rem == 1):
return 4 * pwr(3, (sm - 4) // 3)
if (rem == 2):
return 2 * pwr(3, (sm - 2) // 3)
return pwr(3, sm // 3)
sm = int(3 * math.log(a, 3))
for i in range(-10, 10):
if (calc(sm + i) >= a):
print(sm + i)
break
|
Title: Perfect Encoding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are working as an analyst in a company working on a new system for big data storage. This system will store $n$ different objects. Each object should have a unique ID.
To create the system, you choose the parameters of the system — integers $m \ge 1$ and $b_{1}, b_{2}, \ldots, b_{m}$. With these parameters an ID of some object in the system is an array of integers $[a_{1}, a_{2}, \ldots, a_{m}]$ where $1 \le a_{i} \le b_{i}$ holds for every $1 \le i \le m$.
Developers say that production costs are proportional to $\sum_{i=1}^{m} b_{i}$. You are asked to choose parameters $m$ and $b_{i}$ so that the system will be able to assign unique IDs to $n$ different objects and production costs are minimized. Note that you don't have to use all available IDs.
Input Specification:
In the only line of input there is one positive integer $n$. The length of the decimal representation of $n$ is no greater than $1.5 \cdot 10^{6}$. The integer does not contain leading zeros.
Output Specification:
Print one number — minimal value of $\sum_{i=1}^{m} b_{i}$.
Demo Input:
['36\n', '37\n', '12345678901234567890123456789\n']
Demo Output:
['10\n', '11\n', '177\n']
Note:
none
|
```python
import math
a = int(input())
if (a <= 3):
print(a)
exit(0)
def pwr(a, n):
ans = 1
while (n != 0):
if (n & 1):
ans *= a
a *= a
n >>= 1
return ans
def calc(sm):
if (sm <= 0):
return -10000000000000000
if (sm == 1):
return 1
rem = sm % 3
if (rem == 1):
return 4 * pwr(3, (sm - 4) // 3)
if (rem == 2):
return 2 * pwr(3, (sm - 2) // 3)
return pwr(3, sm // 3)
sm = int(3 * math.log(a, 3))
for i in range(-10, 10):
if (calc(sm + i) >= a):
print(sm + i)
break
```
| 0
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,480,918,823
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 62
| 0
|
A = list(input())
B = list(input())
n = len(A)
Z = []
i = 0
while i < n:
if int(A[i])+int(B[i]) == 1:
Z.append(1)
else:
Z.append(0)
i += 1
a = 0
while a < n:
print(Z[a],end = '')
a = a + 1
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
A = list(input())
B = list(input())
n = len(A)
Z = []
i = 0
while i < n:
if int(A[i])+int(B[i]) == 1:
Z.append(1)
else:
Z.append(0)
i += 1
a = 0
while a < n:
print(Z[a],end = '')
a = a + 1
```
| 3.9845
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
One day Petya dreamt of a lexicographically *k*-th permutation of integers from 1 to *n*. Determine how many lucky numbers in the permutation are located on the positions whose indexes are also lucky numbers.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109) — the number of elements in the permutation and the lexicographical number of the permutation.
|
If the *k*-th permutation of numbers from 1 to *n* does not exist, print the single number "-1" (without the quotes). Otherwise, print the answer to the problem: the number of such indexes *i*, that *i* and *a**i* are both lucky numbers.
|
[
"7 4\n",
"4 7\n"
] |
[
"1\n",
"1\n"
] |
A permutation is an ordered set of *n* elements, where each integer from 1 to *n* occurs exactly once. The element of permutation in position with index *i* is denoted as *a*<sub class="lower-index">*i*</sub> (1 ≤ *i* ≤ *n*). Permutation *a* is lexicographically smaller that permutation *b* if there is such a *i* (1 ≤ *i* ≤ *n*), that *a*<sub class="lower-index">*i*</sub> < *b*<sub class="lower-index">*i*</sub>, and for any *j* (1 ≤ *j* < *i*) *a*<sub class="lower-index">*j*</sub> = *b*<sub class="lower-index">*j*</sub>. Let's make a list of all possible permutations of *n* elements and sort it in the order of lexicographical increasing. Then the lexicographically *k*-th permutation is the *k*-th element of this list of permutations.
In the first sample the permutation looks like that:
1 2 3 4 6 7 5
The only suitable position is 4.
In the second sample the permutation looks like that:
2 1 3 4
The only suitable position is 4.
| 0
|
[
{
"input": "7 4",
"output": "1"
},
{
"input": "4 7",
"output": "1"
},
{
"input": "7 1",
"output": "2"
},
{
"input": "7 5040",
"output": "1"
},
{
"input": "10 1023",
"output": "0"
},
{
"input": "7 7477",
"output": "-1"
},
{
"input": "10 10000",
"output": "1"
},
{
"input": "3 7",
"output": "-1"
},
{
"input": "27 1",
"output": "2"
},
{
"input": "40 8544",
"output": "2"
},
{
"input": "47 1",
"output": "4"
},
{
"input": "47 8547744",
"output": "3"
},
{
"input": "50 1000000000",
"output": "4"
},
{
"input": "64 87",
"output": "4"
},
{
"input": "98 854555",
"output": "6"
},
{
"input": "100 1",
"output": "6"
},
{
"input": "9985 5888454",
"output": "30"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "2 1000000000",
"output": "-1"
},
{
"input": "10 1000000000",
"output": "-1"
},
{
"input": "20 1000000000",
"output": "2"
},
{
"input": "777777 1",
"output": "126"
},
{
"input": "777777 2",
"output": "125"
},
{
"input": "777474 10000",
"output": "120"
},
{
"input": "1000000000 1",
"output": "1022"
},
{
"input": "777777777 5",
"output": "1021"
},
{
"input": "777777777 1",
"output": "1022"
},
{
"input": "777477774 1",
"output": "989"
},
{
"input": "444747744 1000000000",
"output": "554"
},
{
"input": "475 88555458",
"output": "8"
},
{
"input": "12 855448",
"output": "1"
},
{
"input": "20 1000000000",
"output": "2"
},
{
"input": "47 99998544",
"output": "2"
},
{
"input": "49 1000000000",
"output": "2"
},
{
"input": "854459 95554455",
"output": "126"
},
{
"input": "77777779 1000000000",
"output": "508"
},
{
"input": "77 47",
"output": "5"
},
{
"input": "6999 85488877",
"output": "22"
},
{
"input": "7479 58884598",
"output": "24"
},
{
"input": "1000000000 1000000000",
"output": "1022"
},
{
"input": "7 1000",
"output": "0"
},
{
"input": "7 124",
"output": "1"
},
{
"input": "7 2048",
"output": "1"
},
{
"input": "7 3001",
"output": "1"
},
{
"input": "7 127",
"output": "1"
},
{
"input": "7 980",
"output": "1"
},
{
"input": "7 5000",
"output": "0"
},
{
"input": "7 4095",
"output": "2"
},
{
"input": "7 3856",
"output": "0"
},
{
"input": "7 5032",
"output": "0"
},
{
"input": "7 4999",
"output": "0"
},
{
"input": "2 3",
"output": "-1"
},
{
"input": "2 4",
"output": "-1"
},
{
"input": "7 985",
"output": "1"
},
{
"input": "4 25",
"output": "-1"
},
{
"input": "6 121",
"output": "1"
},
{
"input": "11 39916801",
"output": "-1"
},
{
"input": "29 1000000000",
"output": "2"
},
{
"input": "10 4589",
"output": "1"
},
{
"input": "10 100000",
"output": "1"
},
{
"input": "10 98564",
"output": "1"
},
{
"input": "10 1",
"output": "2"
},
{
"input": "10 100000009",
"output": "-1"
}
] | 1,691,179,969
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
print("_RANDOM_GUESS_1691179968.7152607")# 1691179968.7152777
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
One day Petya dreamt of a lexicographically *k*-th permutation of integers from 1 to *n*. Determine how many lucky numbers in the permutation are located on the positions whose indexes are also lucky numbers.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109) — the number of elements in the permutation and the lexicographical number of the permutation.
Output Specification:
If the *k*-th permutation of numbers from 1 to *n* does not exist, print the single number "-1" (without the quotes). Otherwise, print the answer to the problem: the number of such indexes *i*, that *i* and *a**i* are both lucky numbers.
Demo Input:
['7 4\n', '4 7\n']
Demo Output:
['1\n', '1\n']
Note:
A permutation is an ordered set of *n* elements, where each integer from 1 to *n* occurs exactly once. The element of permutation in position with index *i* is denoted as *a*<sub class="lower-index">*i*</sub> (1 ≤ *i* ≤ *n*). Permutation *a* is lexicographically smaller that permutation *b* if there is such a *i* (1 ≤ *i* ≤ *n*), that *a*<sub class="lower-index">*i*</sub> < *b*<sub class="lower-index">*i*</sub>, and for any *j* (1 ≤ *j* < *i*) *a*<sub class="lower-index">*j*</sub> = *b*<sub class="lower-index">*j*</sub>. Let's make a list of all possible permutations of *n* elements and sort it in the order of lexicographical increasing. Then the lexicographically *k*-th permutation is the *k*-th element of this list of permutations.
In the first sample the permutation looks like that:
1 2 3 4 6 7 5
The only suitable position is 4.
In the second sample the permutation looks like that:
2 1 3 4
The only suitable position is 4.
|
```python
print("_RANDOM_GUESS_1691179968.7152607")# 1691179968.7152777
```
| 0
|
|
681
|
B
|
Economy Game
|
PROGRAMMING
| 1,300
|
[
"brute force"
] | null | null |
Kolya is developing an economy simulator game. His most favourite part of the development process is in-game testing. Once he was entertained by the testing so much, that he found out his game-coin score become equal to 0.
Kolya remembers that at the beginning of the game his game-coin score was equal to *n* and that he have bought only some houses (for 1<=234<=567 game-coins each), cars (for 123<=456 game-coins each) and computers (for 1<=234 game-coins each).
Kolya is now interested, whether he could have spent all of his initial *n* game-coins buying only houses, cars and computers or there is a bug in the game. Formally, is there a triple of non-negative integers *a*, *b* and *c* such that *a*<=×<=1<=234<=567<=+<=*b*<=×<=123<=456<=+<=*c*<=×<=1<=234<==<=*n*?
Please help Kolya answer this question.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — Kolya's initial game-coin score.
|
Print "YES" (without quotes) if it's possible that Kolya spent all of his initial *n* coins buying only houses, cars and computers. Otherwise print "NO" (without quotes).
|
[
"1359257\n",
"17851817\n"
] |
[
"YES",
"NO"
] |
In the first sample, one of the possible solutions is to buy one house, one car and one computer, spending 1 234 567 + 123 456 + 1234 = 1 359 257 game-coins in total.
| 1,000
|
[
{
"input": "1359257",
"output": "YES"
},
{
"input": "17851817",
"output": "NO"
},
{
"input": "1000000000",
"output": "YES"
},
{
"input": "17851818",
"output": "YES"
},
{
"input": "438734347",
"output": "YES"
},
{
"input": "43873430",
"output": "YES"
},
{
"input": "999999987",
"output": "YES"
},
{
"input": "27406117",
"output": "NO"
},
{
"input": "27404883",
"output": "NO"
},
{
"input": "27403649",
"output": "NO"
},
{
"input": "27402415",
"output": "NO"
},
{
"input": "27401181",
"output": "NO"
},
{
"input": "999999999",
"output": "YES"
},
{
"input": "999999244",
"output": "YES"
},
{
"input": "999129999",
"output": "YES"
},
{
"input": "17159199",
"output": "NO"
},
{
"input": "13606913",
"output": "NO"
},
{
"input": "14841529",
"output": "NO"
},
{
"input": "915968473",
"output": "YES"
},
{
"input": "980698615",
"output": "YES"
},
{
"input": "912331505",
"output": "YES"
},
{
"input": "917261049",
"output": "YES"
},
{
"input": "999999997",
"output": "YES"
},
{
"input": "12345",
"output": "NO"
},
{
"input": "1234",
"output": "YES"
},
{
"input": "124690",
"output": "YES"
},
{
"input": "1359257",
"output": "YES"
},
{
"input": "1358023",
"output": "YES"
},
{
"input": "1234",
"output": "YES"
},
{
"input": "1234567",
"output": "YES"
},
{
"input": "124690",
"output": "YES"
},
{
"input": "1358023",
"output": "YES"
},
{
"input": "123456",
"output": "YES"
},
{
"input": "2592590",
"output": "YES"
},
{
"input": "999999998",
"output": "YES"
},
{
"input": "1356789",
"output": "NO"
},
{
"input": "12345670",
"output": "YES"
},
{
"input": "11",
"output": "NO"
},
{
"input": "1480800",
"output": "YES"
},
{
"input": "908000000",
"output": "YES"
},
{
"input": "3000",
"output": "NO"
},
{
"input": "1235801",
"output": "YES"
},
{
"input": "991919191",
"output": "YES"
},
{
"input": "25613715",
"output": "YES"
},
{
"input": "13580237",
"output": "YES"
},
{
"input": "14814804",
"output": "YES"
},
{
"input": "11403961",
"output": "YES"
},
{
"input": "999999989",
"output": "YES"
},
{
"input": "1237035",
"output": "YES"
},
{
"input": "81134231",
"output": "YES"
},
{
"input": "1236",
"output": "NO"
},
{
"input": "1359250",
"output": "YES"
},
{
"input": "100",
"output": "NO"
},
{
"input": "987654321",
"output": "YES"
},
{
"input": "122222",
"output": "NO"
},
{
"input": "123458",
"output": "NO"
},
{
"input": "20987639",
"output": "YES"
},
{
"input": "999973333",
"output": "YES"
},
{
"input": "253082",
"output": "YES"
},
{
"input": "1235",
"output": "NO"
},
{
"input": "803219200",
"output": "YES"
},
{
"input": "100000000",
"output": "YES"
},
{
"input": "1485181",
"output": "YES"
}
] | 1,585,023,279
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 109
| 0
|
n = int(input())
home = 1234567
car = 123456
comp = 1234
ans = False
for a in range(0, n+1, home):
tot = n - a
if(a > n or ans):
break
for b in range(0, tot+1, car):
tot = tot - b
if(tot % comp == 0):
ans = True
break
if(ans):
print("YES")
else:
print("NO")
|
Title: Economy Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya is developing an economy simulator game. His most favourite part of the development process is in-game testing. Once he was entertained by the testing so much, that he found out his game-coin score become equal to 0.
Kolya remembers that at the beginning of the game his game-coin score was equal to *n* and that he have bought only some houses (for 1<=234<=567 game-coins each), cars (for 123<=456 game-coins each) and computers (for 1<=234 game-coins each).
Kolya is now interested, whether he could have spent all of his initial *n* game-coins buying only houses, cars and computers or there is a bug in the game. Formally, is there a triple of non-negative integers *a*, *b* and *c* such that *a*<=×<=1<=234<=567<=+<=*b*<=×<=123<=456<=+<=*c*<=×<=1<=234<==<=*n*?
Please help Kolya answer this question.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — Kolya's initial game-coin score.
Output Specification:
Print "YES" (without quotes) if it's possible that Kolya spent all of his initial *n* coins buying only houses, cars and computers. Otherwise print "NO" (without quotes).
Demo Input:
['1359257\n', '17851817\n']
Demo Output:
['YES', 'NO']
Note:
In the first sample, one of the possible solutions is to buy one house, one car and one computer, spending 1 234 567 + 123 456 + 1234 = 1 359 257 game-coins in total.
|
```python
n = int(input())
home = 1234567
car = 123456
comp = 1234
ans = False
for a in range(0, n+1, home):
tot = n - a
if(a > n or ans):
break
for b in range(0, tot+1, car):
tot = tot - b
if(tot % comp == 0):
ans = True
break
if(ans):
print("YES")
else:
print("NO")
```
| 0
|
|
680
|
B
|
Bear and Finding Criminals
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"implementation"
] | null | null |
There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.
|
The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city.
|
Print the number of criminals Limak will catch.
|
[
"6 3\n1 1 1 0 1 0\n",
"5 2\n0 0 0 1 0\n"
] |
[
"3\n",
"1\n"
] |
In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is.
| 1,000
|
[
{
"input": "6 3\n1 1 1 0 1 0",
"output": "3"
},
{
"input": "5 2\n0 0 0 1 0",
"output": "1"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "9 3\n1 1 1 1 1 1 1 1 0",
"output": "8"
},
{
"input": "9 5\n1 0 1 0 1 0 1 0 1",
"output": "5"
},
{
"input": "20 17\n1 1 0 1 1 1 1 0 1 0 1 1 1 0 1 1 0 0 0 0",
"output": "10"
},
{
"input": "100 60\n1 1 1 1 1 1 0 1 0 0 1 1 0 1 1 1 1 1 0 0 1 1 0 0 0 0 0 1 0 1 1 0 1 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 0 0 1 0 0 0 1 1 1 0 1 0 0 1 1 1 0 1 1 1 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 1 1 1 0 1 0 0",
"output": "27"
},
{
"input": "8 1\n1 0 1 1 0 0 1 0",
"output": "4"
},
{
"input": "11 11\n0 1 0 0 1 1 1 0 0 0 0",
"output": "4"
},
{
"input": "19 10\n0 1 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 0 1",
"output": "4"
},
{
"input": "100 38\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100 38\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100 38\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "3"
},
{
"input": "99 38\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "25"
},
{
"input": "99 38\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "24"
},
{
"input": "99 38\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "24"
},
{
"input": "98 70\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "41"
},
{
"input": "99 70\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "9"
},
{
"input": "99 60\n0 0 0 1 1 0 1 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 0 1 1 0 1 1 1 1 0 0 0 0 0 0 1 1 1",
"output": "34"
},
{
"input": "98 24\n0 0 0 1 1 0 1 1 0 0 0 0 1 0 0 1 0 0 1 1 1 0 0 1 0 0 1 0 1 0 0 1 1 0 1 1 1 0 1 0 0 1 0 0 0 1 1 1 1 0 1 1 1 0 1 1 1 1 0 0 1 0 0 0 1 0 1 1 0 1 0 1 1 1 0 0 1 0 0 0 1 1 0 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1",
"output": "39"
},
{
"input": "100 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100"
},
{
"input": "100 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "2 1\n0 1",
"output": "1"
}
] | 1,667,334,508
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 31
| 0
|
ci,o=map(int,input().split())
cr=list(map(int,input().split()))
res=0
r,l,i=0,0,0
while i<ci:
r=o+i
l=o-i
if r<ci and l>=0:
if cr[l]==1 and cr[r]==1:
res+=2
elif l>=0:
if cr[l]==1:
res+=1
elif r<ci:
if cr[r]==1:
res+=1
i+=1
print(res)
|
Title: Bear and Finding Criminals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.
Input Specification:
The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city.
Output Specification:
Print the number of criminals Limak will catch.
Demo Input:
['6 3\n1 1 1 0 1 0\n', '5 2\n0 0 0 1 0\n']
Demo Output:
['3\n', '1\n']
Note:
In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is.
|
```python
ci,o=map(int,input().split())
cr=list(map(int,input().split()))
res=0
r,l,i=0,0,0
while i<ci:
r=o+i
l=o-i
if r<ci and l>=0:
if cr[l]==1 and cr[r]==1:
res+=2
elif l>=0:
if cr[l]==1:
res+=1
elif r<ci:
if cr[r]==1:
res+=1
i+=1
print(res)
```
| 0
|
|
894
|
A
|
QAQ
|
PROGRAMMING
| 800
|
[
"brute force",
"dp"
] | null | null |
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length *n*. There is a great number of "QAQ" in the string (Diamond is so cute!).
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
|
The only line contains a string of length *n* (1<=≤<=*n*<=≤<=100). It's guaranteed that the string only contains uppercase English letters.
|
Print a single integer — the number of subsequences "QAQ" in the string.
|
[
"QAQAQYSYIOIWIN\n",
"QAQQQZZYNOIWIN\n"
] |
[
"4\n",
"3\n"
] |
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
| 500
|
[
{
"input": "QAQAQYSYIOIWIN",
"output": "4"
},
{
"input": "QAQQQZZYNOIWIN",
"output": "3"
},
{
"input": "QA",
"output": "0"
},
{
"input": "IAQVAQZLQBQVQFTQQQADAQJA",
"output": "24"
},
{
"input": "QQAAQASGAYAAAAKAKAQIQEAQAIAAIAQQQQQ",
"output": "378"
},
{
"input": "AMVFNFJIAVNQJWIVONQOAOOQSNQSONOASONAONQINAONAOIQONANOIQOANOQINAONOQINAONOXJCOIAQOAOQAQAQAQAQWWWAQQAQ",
"output": "1077"
},
{
"input": "AAQQAXBQQBQQXBNQRJAQKQNAQNQVDQASAGGANQQQQTJFFQQQTQQA",
"output": "568"
},
{
"input": "KAZXAVLPJQBQVQQQQQAPAQQGQTQVZQAAAOYA",
"output": "70"
},
{
"input": "W",
"output": "0"
},
{
"input": "DBA",
"output": "0"
},
{
"input": "RQAWNACASAAKAGAAAAQ",
"output": "10"
},
{
"input": "QJAWZAAOAAGIAAAAAOQATASQAEAAAAQFQQHPA",
"output": "111"
},
{
"input": "QQKWQAQAAAAAAAAGAAVAQUEQQUMQMAQQQNQLAMAAAUAEAAEMAAA",
"output": "411"
},
{
"input": "QQUMQAYAUAAGWAAAQSDAVAAQAAAASKQJJQQQQMAWAYYAAAAAAEAJAXWQQ",
"output": "625"
},
{
"input": "QORZOYAQ",
"output": "1"
},
{
"input": "QCQAQAGAWAQQQAQAVQAQQQQAQAQQQAQAAATQAAVAAAQQQQAAAUUQAQQNQQWQQWAQAAQQKQYAQAAQQQAAQRAQQQWBQQQQAPBAQGQA",
"output": "13174"
},
{
"input": "QQAQQAKQFAQLQAAWAMQAZQAJQAAQQOACQQAAAYANAQAQQAQAAQQAOBQQJQAQAQAQQQAAAAABQQQAVNZAQQQQAMQQAFAAEAQAQHQT",
"output": "10420"
},
{
"input": "AQEGQHQQKQAQQPQKAQQQAAAAQQQAQEQAAQAAQAQFSLAAQQAQOQQAVQAAAPQQAWAQAQAFQAXAQQQQTRLOQAQQJQNQXQQQQSQVDQQQ",
"output": "12488"
},
{
"input": "QNQKQQQLASQBAVQQQQAAQQOQRJQQAQQQEQZUOANAADAAQQJAQAQARAAAQQQEQBHTQAAQAAAAQQMKQQQIAOJJQQAQAAADADQUQQQA",
"output": "9114"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "35937"
},
{
"input": "AMQQAAQAAQAAAAAAQQQBOAAANAAKQJCYQAE",
"output": "254"
},
{
"input": "AYQBAEQGAQEOAKGIXLQJAIAKQAAAQPUAJAKAATFWQQAOQQQUFQYAQQMQHOKAAJXGFCARAQSATHAUQQAATQJJQDQRAANQQAE",
"output": "2174"
},
{
"input": "AAQXAAQAYQAAAAGAQHVQYAGIVACADFAAQAAAAQZAAQMAKZAADQAQDAAQDAAAMQQOXYAQQQAKQBAAQQKAXQBJZDDLAAHQQ",
"output": "2962"
},
{
"input": "AYQQYAVAMNIAUAAKBBQVACWKTQSAQZAAQAAASZJAWBCAALAARHACQAKQQAQAARPAQAAQAQAAZQUSHQAMFVFZQQQQSAQQXAA",
"output": "2482"
},
{
"input": "LQMAQQARQAQBJQQQAGAAZQQXALQQAARQAQQQQAAQQAQQQAQQCAQQAQQAYQQQRAAZATQALYQQAAHHAAQHAAAAAAAAQQMAAQNAKQ",
"output": "7768"
},
{
"input": "MAQQWAQOYQMAAAQAQPQZAOAAQAUAQNAAQAAAITQSAQAKAQKAQQWSQAAQQAGUCDQMQWKQUXKWQQAAQQAAQQZQDQQQAABXQUUXQOA",
"output": "5422"
},
{
"input": "QTAAQDAQXAQQJQQQGAAAQQQQSBQZKAQQAQQQQEAQNUQBZCQLYQZQEQQAAQHQVAORKQVAQYQNASZQAARZAAGAAAAOQDCQ",
"output": "3024"
},
{
"input": "QQWAQQGQQUZQQQLZAAQYQXQVAQFQUAQZUQZZQUKBHSHTQYLQAOQXAQQGAQQTQOAQARQADAJRAAQPQAQQUQAUAMAUVQAAAQQAWQ",
"output": "4527"
},
{
"input": "QQAAQQAQVAQZQQQQAOEAQZPQIBQZACQQAFQQLAAQDATZQANHKYQQAQTAAFQRQAIQAJPWQAQTEIRXAEQQAYWAAAUKQQAQAQQQSQQH",
"output": "6416"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAAAA",
"output": "14270"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQ",
"output": "13136"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAA",
"output": "14270"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQQAA",
"output": "14231"
},
{
"input": "QQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAA",
"output": "15296"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "0"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "0"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "0"
},
{
"input": "QAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQA",
"output": "20825"
},
{
"input": "AQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQ",
"output": "20825"
},
{
"input": "Q",
"output": "0"
},
{
"input": "A",
"output": "0"
},
{
"input": "FFF",
"output": "0"
},
{
"input": "AAAAAA",
"output": "0"
}
] | 1,589,770,049
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 109
| 6,656,000
|
s = input()
qaq = 0
for i in range(0, len(s)):
# find an A
if s[i] == 'A':
# find Qs before A
before = s[:i].count('Q')
# find Qs after A
after = s[i+1:].count('Q')
# add before * after to qaq
qaq += before * after
print(qaq)
|
Title: QAQ
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length *n*. There is a great number of "QAQ" in the string (Diamond is so cute!).
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input Specification:
The only line contains a string of length *n* (1<=≤<=*n*<=≤<=100). It's guaranteed that the string only contains uppercase English letters.
Output Specification:
Print a single integer — the number of subsequences "QAQ" in the string.
Demo Input:
['QAQAQYSYIOIWIN\n', 'QAQQQZZYNOIWIN\n']
Demo Output:
['4\n', '3\n']
Note:
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
|
```python
s = input()
qaq = 0
for i in range(0, len(s)):
# find an A
if s[i] == 'A':
# find Qs before A
before = s[:i].count('Q')
# find Qs after A
after = s[i+1:].count('Q')
# add before * after to qaq
qaq += before * after
print(qaq)
```
| 3
|
|
225
|
C
|
Barcode
|
PROGRAMMING
| 1,700
|
[
"dp",
"matrices"
] | null | null |
You've got an *n*<=×<=*m* pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
- All pixels in each column are of the same color. - The width of each monochrome vertical line is at least *x* and at most *y* pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than *x* or greater than *y*.
|
The first line contains four space-separated integers *n*, *m*, *x* and *y* (1<=≤<=*n*,<=*m*,<=*x*,<=*y*<=≤<=1000; *x*<=≤<=*y*).
Then follow *n* lines, describing the original image. Each of these lines contains exactly *m* characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
|
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
|
[
"6 5 1 2\n##.#.\n.###.\n###..\n#...#\n.##.#\n###..\n",
"2 5 1 1\n#####\n.....\n"
] |
[
"11\n",
"5\n"
] |
In the first test sample the picture after changing some colors can looks as follows:
In the second test sample the picture after changing some colors can looks as follows:
| 1,500
|
[
{
"input": "6 5 1 2\n##.#.\n.###.\n###..\n#...#\n.##.#\n###..",
"output": "11"
},
{
"input": "10 5 3 7\n.####\n###..\n##.##\n#..#.\n.#...\n#.##.\n.##..\n.#.##\n#.#..\n.#..#",
"output": "24"
},
{
"input": "6 3 1 4\n##.\n#..\n#..\n..#\n.#.\n#.#",
"output": "6"
},
{
"input": "5 10 4 16\n.#####....\n##..#..##.\n.#..##.#..\n##..#####.\n...#.##..#",
"output": "21"
},
{
"input": "5 4 1 4\n####\n..##\n##..\n..#.\n#..#",
"output": "8"
},
{
"input": "1 1 1 2\n.",
"output": "0"
},
{
"input": "3 44 2 18\n####..###.#.##........##...###.####.#.....##\n...#....##.###.###.##.#####.#######.#..#..#.\n#...#.####.#.##.#.#.#.....##.##..###.#....##",
"output": "39"
},
{
"input": "69 1 1 2\n#\n.\n#\n#\n.\n#\n#\n.\n.\n#\n.\n.\n#\n.\n#\n#\n.\n#\n#\n#\n#\n#\n.\n#\n.\n.\n#\n#\n#\n.\n.\n.\n.\n#\n#\n.\n#\n#\n.\n#\n.\n.\n#\n.\n.\n.\n#\n.\n.\n#\n#\n.\n.\n#\n.\n.\n.\n#\n#\n#\n#\n#\n.\n#\n.\n#\n#\n#\n#",
"output": "31"
},
{
"input": "56 2 2 2\n##\n##\n##\n..\n..\n##\n.#\n#.\n..\n##\n##\n..\n..\n#.\n#.\n#.\n#.\n##\n..\n.#\n..\n##\n##\n.#\n#.\n.#\n..\n..\n#.\n..\n##\n..\n#.\n.#\n#.\n#.\n#.\n##\n#.\n##\n##\n.#\n#.\n##\n..\n.#\n#.\n#.\n##\n..\n..\n#.\n##\n..\n..\n##",
"output": "55"
}
] | 1,617,256,828
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 36
| 498
| 9,728,000
|
inf = 9999999999
def f(ll,n,m,x,y):
W = [0]*m #White
for i in range(n):
for j in range(m):
if ll[i][j]=='.':
W[j] += 1
B = [n-c for c in W] #Black
Bw = [inf]*(y+1)
Ww = [inf]*(y+1)
Bw[1] = B[0]
Ww[1] = W[0]
Bv = B[0] if x==1 else inf
Wv = W[0] if x==1 else inf
for i in range(1,m):
for j in range(y,0,-1):
Bw[j] = Bw[j-1]+B[i]
Ww[j] = Ww[j-1]+W[i]
if Bv < inf: #if valid!
Ww[1] = Bv + W[i]
Bw[1] = Wv + B[i]
Bv = min(Bw[x:y+1])
Wv = min(Ww[x:y+1])
return min(Bv,Wv)
n,m,x,y = list(map(int,input().split()))
ll = [input() for _ in range(n)]
print(f(ll,n,m,x,y))
|
Title: Barcode
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got an *n*<=×<=*m* pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
- All pixels in each column are of the same color. - The width of each monochrome vertical line is at least *x* and at most *y* pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than *x* or greater than *y*.
Input Specification:
The first line contains four space-separated integers *n*, *m*, *x* and *y* (1<=≤<=*n*,<=*m*,<=*x*,<=*y*<=≤<=1000; *x*<=≤<=*y*).
Then follow *n* lines, describing the original image. Each of these lines contains exactly *m* characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output Specification:
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Demo Input:
['6 5 1 2\n##.#.\n.###.\n###..\n#...#\n.##.#\n###..\n', '2 5 1 1\n#####\n.....\n']
Demo Output:
['11\n', '5\n']
Note:
In the first test sample the picture after changing some colors can looks as follows:
In the second test sample the picture after changing some colors can looks as follows:
|
```python
inf = 9999999999
def f(ll,n,m,x,y):
W = [0]*m #White
for i in range(n):
for j in range(m):
if ll[i][j]=='.':
W[j] += 1
B = [n-c for c in W] #Black
Bw = [inf]*(y+1)
Ww = [inf]*(y+1)
Bw[1] = B[0]
Ww[1] = W[0]
Bv = B[0] if x==1 else inf
Wv = W[0] if x==1 else inf
for i in range(1,m):
for j in range(y,0,-1):
Bw[j] = Bw[j-1]+B[i]
Ww[j] = Ww[j-1]+W[i]
if Bv < inf: #if valid!
Ww[1] = Bv + W[i]
Bw[1] = Wv + B[i]
Bv = min(Bw[x:y+1])
Wv = min(Ww[x:y+1])
return min(Bv,Wv)
n,m,x,y = list(map(int,input().split()))
ll = [input() for _ in range(n)]
print(f(ll,n,m,x,y))
```
| 3
|
|
601
|
A
|
The Two Routes
|
PROGRAMMING
| 1,600
|
[
"graphs",
"shortest paths"
] | null | null |
In Absurdistan, there are *n* towns (numbered 1 through *n*) and *m* bidirectional railways. There is also an absurdly simple road network — for each pair of different towns *x* and *y*, there is a bidirectional road between towns *x* and *y* if and only if there is no railway between them. Travelling to a different town using one railway or one road always takes exactly one hour.
A train and a bus leave town 1 at the same time. They both have the same destination, town *n*, and don't make any stops on the way (but they can wait in town *n*). The train can move only along railways and the bus can move only along roads.
You've been asked to plan out routes for the vehicles; each route can use any road/railway multiple times. One of the most important aspects to consider is safety — in order to avoid accidents at railway crossings, the train and the bus must not arrive at the same town (except town *n*) simultaneously.
Under these constraints, what is the minimum number of hours needed for both vehicles to reach town *n* (the maximum of arrival times of the bus and the train)? Note, that bus and train are not required to arrive to the town *n* at the same moment of time, but are allowed to do so.
|
The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=400, 0<=≤<=*m*<=≤<=*n*(*n*<=-<=1)<=/<=2) — the number of towns and the number of railways respectively.
Each of the next *m* lines contains two integers *u* and *v*, denoting a railway between towns *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*).
You may assume that there is at most one railway connecting any two towns.
|
Output one integer — the smallest possible time of the later vehicle's arrival in town *n*. If it's impossible for at least one of the vehicles to reach town *n*, output <=-<=1.
|
[
"4 2\n1 3\n3 4\n",
"4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4\n",
"5 5\n4 2\n3 5\n4 5\n5 1\n1 2\n"
] |
[
"2\n",
"-1\n",
"3\n"
] |
In the first sample, the train can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7c0aa60a06309ef607b7159fd7f3687ea0d943ce.png" style="max-width: 100.0%;max-height: 100.0%;"/> and the bus can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a26c2f3e93c9d9be6c21cb5d2bd6ac1f99f4ff55.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Note that they can arrive at town 4 at the same time.
In the second sample, Absurdistan is ruled by railwaymen. There are no roads, so there's no way for the bus to reach town 4.
| 500
|
[
{
"input": "4 2\n1 3\n3 4",
"output": "2"
},
{
"input": "4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4",
"output": "-1"
},
{
"input": "5 5\n4 2\n3 5\n4 5\n5 1\n1 2",
"output": "3"
},
{
"input": "5 4\n1 2\n3 2\n3 4\n5 4",
"output": "4"
},
{
"input": "3 1\n1 2",
"output": "-1"
},
{
"input": "2 1\n1 2",
"output": "-1"
},
{
"input": "2 0",
"output": "-1"
},
{
"input": "20 0",
"output": "-1"
},
{
"input": "381 0",
"output": "-1"
},
{
"input": "3 3\n1 2\n2 3\n3 1",
"output": "-1"
},
{
"input": "3 0",
"output": "-1"
},
{
"input": "3 1\n1 3",
"output": "2"
},
{
"input": "3 2\n2 3\n3 1",
"output": "-1"
},
{
"input": "4 1\n1 4",
"output": "2"
},
{
"input": "4 5\n1 3\n2 1\n3 4\n4 2\n2 3",
"output": "2"
},
{
"input": "20 1\n20 1",
"output": "2"
},
{
"input": "21 1\n21 1",
"output": "2"
},
{
"input": "100 1\n100 1",
"output": "2"
},
{
"input": "400 1\n1 400",
"output": "2"
},
{
"input": "5 5\n2 5\n1 2\n1 4\n1 3\n3 2",
"output": "2"
}
] | 1,621,278,875
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 11
| 2,000
| 409,600
|
import math
def bfs(adj, tar):
visited = []
queue = [(1, 0)]
while(len(queue) > 0):
currI, currDist = queue.pop(0)
visited.append(currI)
for i in adj[currI]:
if i == tar:
return currDist + 1
if i not in visited:
queue.append((i, currDist + 1))
return -1
n, m = list(map(int, input().split()))
adj_rails = [[] for _ in range(n+1)]
for i in range(m):
x, y = list(map(int, input().split()))
adj_rails[x].append(y)
adj_rails[y].append(x)
adj_roads = [[] for _ in range(n+1)]
for i in range(1, n + 1):
for j in range(1, n + 1):
if i != j and not i in adj_rails[j]:
adj_roads[i].append(j)
adj_roads[j].append(i)
d1 = bfs(adj_rails, n)
d2 = bfs(adj_roads, n)
if d1 == -1 or d2 == -1:
print('-1')
else:
print(max(d1, d2))
|
Title: The Two Routes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Absurdistan, there are *n* towns (numbered 1 through *n*) and *m* bidirectional railways. There is also an absurdly simple road network — for each pair of different towns *x* and *y*, there is a bidirectional road between towns *x* and *y* if and only if there is no railway between them. Travelling to a different town using one railway or one road always takes exactly one hour.
A train and a bus leave town 1 at the same time. They both have the same destination, town *n*, and don't make any stops on the way (but they can wait in town *n*). The train can move only along railways and the bus can move only along roads.
You've been asked to plan out routes for the vehicles; each route can use any road/railway multiple times. One of the most important aspects to consider is safety — in order to avoid accidents at railway crossings, the train and the bus must not arrive at the same town (except town *n*) simultaneously.
Under these constraints, what is the minimum number of hours needed for both vehicles to reach town *n* (the maximum of arrival times of the bus and the train)? Note, that bus and train are not required to arrive to the town *n* at the same moment of time, but are allowed to do so.
Input Specification:
The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=400, 0<=≤<=*m*<=≤<=*n*(*n*<=-<=1)<=/<=2) — the number of towns and the number of railways respectively.
Each of the next *m* lines contains two integers *u* and *v*, denoting a railway between towns *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*).
You may assume that there is at most one railway connecting any two towns.
Output Specification:
Output one integer — the smallest possible time of the later vehicle's arrival in town *n*. If it's impossible for at least one of the vehicles to reach town *n*, output <=-<=1.
Demo Input:
['4 2\n1 3\n3 4\n', '4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4\n', '5 5\n4 2\n3 5\n4 5\n5 1\n1 2\n']
Demo Output:
['2\n', '-1\n', '3\n']
Note:
In the first sample, the train can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7c0aa60a06309ef607b7159fd7f3687ea0d943ce.png" style="max-width: 100.0%;max-height: 100.0%;"/> and the bus can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a26c2f3e93c9d9be6c21cb5d2bd6ac1f99f4ff55.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Note that they can arrive at town 4 at the same time.
In the second sample, Absurdistan is ruled by railwaymen. There are no roads, so there's no way for the bus to reach town 4.
|
```python
import math
def bfs(adj, tar):
visited = []
queue = [(1, 0)]
while(len(queue) > 0):
currI, currDist = queue.pop(0)
visited.append(currI)
for i in adj[currI]:
if i == tar:
return currDist + 1
if i not in visited:
queue.append((i, currDist + 1))
return -1
n, m = list(map(int, input().split()))
adj_rails = [[] for _ in range(n+1)]
for i in range(m):
x, y = list(map(int, input().split()))
adj_rails[x].append(y)
adj_rails[y].append(x)
adj_roads = [[] for _ in range(n+1)]
for i in range(1, n + 1):
for j in range(1, n + 1):
if i != j and not i in adj_rails[j]:
adj_roads[i].append(j)
adj_roads[j].append(i)
d1 = bfs(adj_rails, n)
d2 = bfs(adj_roads, n)
if d1 == -1 or d2 == -1:
print('-1')
else:
print(max(d1, d2))
```
| 0
|
|
834
|
B
|
The Festive Evening
|
PROGRAMMING
| 1,100
|
[
"data structures",
"implementation"
] | null | null |
It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in.
There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously.
For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed.
Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened.
|
Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26).
In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest.
|
Output «YES» if at least one door was unguarded during some time, and «NO» otherwise.
You can output each letter in arbitrary case (upper or lower).
|
[
"5 1\nAABBB\n",
"5 1\nABABB\n"
] |
[
"NO\n",
"YES\n"
] |
In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened.
In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door.
| 1,000
|
[
{
"input": "5 1\nAABBB",
"output": "NO"
},
{
"input": "5 1\nABABB",
"output": "YES"
},
{
"input": "26 1\nABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "NO"
},
{
"input": "27 1\nABCDEFGHIJKLMNOPQRSTUVWXYZA",
"output": "YES"
},
{
"input": "5 2\nABACA",
"output": "NO"
},
{
"input": "6 2\nABCABC",
"output": "YES"
},
{
"input": "8 3\nABCBCDCA",
"output": "NO"
},
{
"input": "73 2\nDEBECECBBADAADEAABEAEEEAEBEAEBCDDBABBAEBACCBEEBBAEADEECACEDEEDABACDCDBBBD",
"output": "YES"
},
{
"input": "44 15\nHGJIFCGGCDGIJDHBIBGAEABCIABIGBDEADBBBAGDFDHA",
"output": "NO"
},
{
"input": "41 19\nTMEYYIIELFDCMBDKWWKYNRNDUPRONYROXQCLVQALP",
"output": "NO"
},
{
"input": "377 3\nEADADBBBBDEAABBAEBABACDBDBBCACAADBEAEACDEAABACADEEDEACACDADABBBBDDEECBDABACACBAECBADAEBDEEBDBCDAEADBCDDACACDCCEEDBCCBBCEDBECBABCDDBBDEADEDAEACDECECBEBACBCCDCDBDAECDECADBCBEDBBDAAEBCAAECCDCCDBDDEBADEEBDCAEABBDEDBBDDEAECCBDDCDEACDAECCBDDABABEAEDCDEDBAECBDEACEBCECEACDCBABCBAAEAADACADBBBBABEADBCADEBCBECCABBDDDEEBCDEBADEBDAAABBEABADEDEAEABCEEBEEDEAEBEABCEDDBACBCCADEBAAAAAEABABBCE",
"output": "YES"
},
{
"input": "433 3\nFZDDHMJGBZCHFUXBBPIEBBEFDWOMXXEPOMDGSMPIUZOMRZQNSJAVNATGIWPDFISKFQXJNVFXPHOZDAEZFDAHDXXQKZMGNSGKQNWGNGJGJZVVITKNFLVCPMZSDMCHBTVAWYVZLIXXIADXNYILEYNIQHKMOGMVOCWGHCWIYMPEPADSJAAKEGTUSEDWAHMNYJDIHBKHVUHLYGNGZDBULRXLSAJHPCMNWCEAAPYMHDTYWPADOTJTXTXUKLCHWKUSZRHEKQEFPVJEJJHRWCKYOIWALRTIBUMNOCRXLSIKQCJVQXEPGOHRUDJDKMUUUDORURWXJNVRVMNOUNRFKSVMTMZGOIJLXEPAMVGESOADYIGZXRBJDIWKNOWTCSROAQTBECHTOZVSQUOOJRZIBAUHMKAXDCIMDZJFMABGRNTGPUJAUNFPFWCJG",
"output": "YES"
},
{
"input": "5 2\nABCAB",
"output": "YES"
},
{
"input": "5 1\nAZAZA",
"output": "YES"
},
{
"input": "7 2\nABCDBCD",
"output": "YES"
},
{
"input": "3 26\nAAB",
"output": "NO"
}
] | 1,501,429,786
| 4,486
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 3
| 46
| 4,608,000
|
import sys
x = sys.stdin.readline().split()
y = sys.stdin.readline().rstrip("\n")
nums = list(map(int, x))
k = nums[1]
e = list(map(str, y))
s = list(set(e))
s.sort()
c = []
for x in range(len(s)):
cnt = e.count(s[x])
c.append(cnt)
sec = [0]*len(c)
count = 0
for y in range(len(e)):
if e[y] in s:
sec[s.index(e[y])] = 1
c[s.index(e[y])] -= 1
if c[s.index(e[y])] == 0:
sec[s.index(e[y])] = 0
if sum(sec) > k:
print("YES")
exit(0)
print("NO")
|
Title: The Festive Evening
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in.
There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously.
For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed.
Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened.
Input Specification:
Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26).
In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest.
Output Specification:
Output «YES» if at least one door was unguarded during some time, and «NO» otherwise.
You can output each letter in arbitrary case (upper or lower).
Demo Input:
['5 1\nAABBB\n', '5 1\nABABB\n']
Demo Output:
['NO\n', 'YES\n']
Note:
In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened.
In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door.
|
```python
import sys
x = sys.stdin.readline().split()
y = sys.stdin.readline().rstrip("\n")
nums = list(map(int, x))
k = nums[1]
e = list(map(str, y))
s = list(set(e))
s.sort()
c = []
for x in range(len(s)):
cnt = e.count(s[x])
c.append(cnt)
sec = [0]*len(c)
count = 0
for y in range(len(e)):
if e[y] in s:
sec[s.index(e[y])] = 1
c[s.index(e[y])] -= 1
if c[s.index(e[y])] == 0:
sec[s.index(e[y])] = 0
if sum(sec) > k:
print("YES")
exit(0)
print("NO")
```
| 0
|
|
38
|
B
|
Chess
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation",
"math"
] |
B. Chess
|
2
|
256
|
Two chess pieces, a rook and a knight, stand on a standard chessboard 8<=×<=8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
|
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
|
Print a single number which is the required number of ways.
|
[
"a1\nb2\n",
"a8\nd4\n"
] |
[
"44\n",
"38\n"
] |
none
| 0
|
[
{
"input": "a1\nb2",
"output": "44"
},
{
"input": "a8\nd4",
"output": "38"
},
{
"input": "a8\nf1",
"output": "42"
},
{
"input": "f8\nh3",
"output": "42"
},
{
"input": "g8\nb7",
"output": "42"
},
{
"input": "h1\ng5",
"output": "42"
},
{
"input": "c6\nb5",
"output": "39"
},
{
"input": "c1\nd2",
"output": "42"
},
{
"input": "g3\nh4",
"output": "42"
},
{
"input": "e3\ng5",
"output": "38"
},
{
"input": "f8\na3",
"output": "40"
},
{
"input": "a2\nh8",
"output": "43"
},
{
"input": "a3\nc5",
"output": "40"
},
{
"input": "g1\ne6",
"output": "39"
},
{
"input": "e1\na7",
"output": "41"
},
{
"input": "b5\nc1",
"output": "39"
},
{
"input": "b2\ne1",
"output": "43"
},
{
"input": "h8\ng2",
"output": "43"
},
{
"input": "a3\nd6",
"output": "38"
},
{
"input": "g6\nb7",
"output": "39"
},
{
"input": "c8\ne6",
"output": "40"
},
{
"input": "e6\nf2",
"output": "35"
},
{
"input": "b6\nd8",
"output": "41"
},
{
"input": "a4\nd1",
"output": "42"
},
{
"input": "b5\nh8",
"output": "40"
},
{
"input": "h6\na1",
"output": "42"
},
{
"input": "c3\na8",
"output": "39"
},
{
"input": "g5\nd2",
"output": "38"
},
{
"input": "b6\ng7",
"output": "39"
},
{
"input": "h6\na8",
"output": "43"
},
{
"input": "a8\nb7",
"output": "44"
},
{
"input": "c8\nb2",
"output": "41"
},
{
"input": "e4\nc1",
"output": "37"
},
{
"input": "f1\nc3",
"output": "38"
},
{
"input": "a3\nc8",
"output": "41"
},
{
"input": "e8\nb6",
"output": "40"
},
{
"input": "a1\nb7",
"output": "43"
},
{
"input": "g2\nb7",
"output": "40"
},
{
"input": "e1\nd6",
"output": "38"
},
{
"input": "e5\nh6",
"output": "39"
}
] | 1,690,384,585
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
def is_valid(x, y, rx, ry, kx, ky):
# Check if the position (x, y) is not occupied by rook, knight, and is empty
return (x, y) != (rx, ry) and (x, y) != (kx, ky) and 1 <= x <= 8 and 1 <= y <= 8
def count_valid_knight_positions(rook_pos, knight_pos):
rx, ry = ord(rook_pos[0]) - ord('a') + 1, int(rook_pos[1])
kx, ky = ord(knight_pos[0]) - ord('a') + 1, int(knight_pos[1])
valid_moves = [(-2, 1), (-1, 2), (1, 2), (2, 1), (2, -1), (1, -2), (-1, -2), (-2, -1)]
count = 0
for dx, dy in valid_moves:
nx, ny = kx + dx, ky + dy
if is_valid(nx, ny, rx, ry, kx, ky):
count += 1
return count
# Read input
rook_pos = input().strip()
knight_pos = input().strip()
# Calculate and print the result
result = count_valid_knight_positions(rook_pos, knight_pos)
print(result)
|
Title: Chess
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Two chess pieces, a rook and a knight, stand on a standard chessboard 8<=×<=8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input Specification:
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output Specification:
Print a single number which is the required number of ways.
Demo Input:
['a1\nb2\n', 'a8\nd4\n']
Demo Output:
['44\n', '38\n']
Note:
none
|
```python
def is_valid(x, y, rx, ry, kx, ky):
# Check if the position (x, y) is not occupied by rook, knight, and is empty
return (x, y) != (rx, ry) and (x, y) != (kx, ky) and 1 <= x <= 8 and 1 <= y <= 8
def count_valid_knight_positions(rook_pos, knight_pos):
rx, ry = ord(rook_pos[0]) - ord('a') + 1, int(rook_pos[1])
kx, ky = ord(knight_pos[0]) - ord('a') + 1, int(knight_pos[1])
valid_moves = [(-2, 1), (-1, 2), (1, 2), (2, 1), (2, -1), (1, -2), (-1, -2), (-2, -1)]
count = 0
for dx, dy in valid_moves:
nx, ny = kx + dx, ky + dy
if is_valid(nx, ny, rx, ry, kx, ky):
count += 1
return count
# Read input
rook_pos = input().strip()
knight_pos = input().strip()
# Calculate and print the result
result = count_valid_knight_positions(rook_pos, knight_pos)
print(result)
```
| 0
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,595,316,195
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 109
| 6,963,200
|
s=input()
cn=0
for i in range(len(s)):
if(cn==0 and s[i]=='h'):
cn=1
elif(cn==1 and s[i]=='e'):
cn=2
elif(cn==2 and s[i]=='l'):
cn=3
elif(cn==3 and s[i]=='l'):
cn=4
elif(cn==4 and s[i]=='o'):
cn=5
if cn==5:
print("YES")
else:
print("NO")
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
s=input()
cn=0
for i in range(len(s)):
if(cn==0 and s[i]=='h'):
cn=1
elif(cn==1 and s[i]=='e'):
cn=2
elif(cn==2 and s[i]=='l'):
cn=3
elif(cn==3 and s[i]=='l'):
cn=4
elif(cn==4 and s[i]=='o'):
cn=5
if cn==5:
print("YES")
else:
print("NO")
```
| 3.93253
|
758
|
A
|
Holiday Of Equality
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
|
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
|
In the only line print the integer *S* — the minimum number of burles which are had to spend.
|
[
"5\n0 1 2 3 4\n",
"5\n1 1 0 1 1\n",
"3\n1 3 1\n",
"1\n12\n"
] |
[
"10",
"1",
"4",
"0"
] |
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
| 500
|
[
{
"input": "5\n0 1 2 3 4",
"output": "10"
},
{
"input": "5\n1 1 0 1 1",
"output": "1"
},
{
"input": "3\n1 3 1",
"output": "4"
},
{
"input": "1\n12",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "14\n52518 718438 358883 462189 853171 592966 225788 46977 814826 295697 676256 561479 56545 764281",
"output": "5464380"
},
{
"input": "21\n842556 216391 427181 626688 775504 168309 851038 448402 880826 73697 593338 519033 135115 20128 424606 939484 846242 756907 377058 241543 29353",
"output": "9535765"
},
{
"input": "3\n1 3 2",
"output": "3"
},
{
"input": "3\n2 1 3",
"output": "3"
},
{
"input": "3\n2 3 1",
"output": "3"
},
{
"input": "3\n3 1 2",
"output": "3"
},
{
"input": "3\n3 2 1",
"output": "3"
},
{
"input": "1\n228503",
"output": "0"
},
{
"input": "2\n32576 550340",
"output": "517764"
},
{
"input": "3\n910648 542843 537125",
"output": "741328"
},
{
"input": "4\n751720 572344 569387 893618",
"output": "787403"
},
{
"input": "6\n433864 631347 597596 794426 713555 231193",
"output": "1364575"
},
{
"input": "9\n31078 645168 695751 126111 375934 150495 838412 434477 993107",
"output": "4647430"
},
{
"input": "30\n315421 772664 560686 654312 151528 356749 351486 707462 820089 226682 546700 136028 824236 842130 578079 337807 665903 764100 617900 822937 992759 591749 651310 742085 767695 695442 17967 515106 81059 186025",
"output": "13488674"
},
{
"input": "45\n908719 394261 815134 419990 926993 383792 772842 277695 527137 655356 684956 695716 273062 550324 106247 399133 442382 33076 462920 294674 846052 817752 421365 474141 290471 358990 109812 74492 543281 169434 919692 786809 24028 197184 310029 801476 699355 429672 51343 374128 776726 850380 293868 981569 550763",
"output": "21993384"
},
{
"input": "56\n100728 972537 13846 385421 756708 184642 259487 319707 376662 221694 675284 972837 499419 13846 38267 289898 901299 831197 954715 197515 514102 910423 127555 883934 362472 870788 538802 741008 973434 448124 391526 363321 947321 544618 68006 782313 955075 741981 815027 723297 585059 718114 700739 413489 454091 736144 308999 98065 3716 347323 9635 289003 986510 607065 60236 273351",
"output": "26984185"
},
{
"input": "70\n644488 5444 150441 714420 602059 335330 510670 196555 546346 740011 509449 850947 692874 524857 750434 952985 223744 374727 896124 753037 367352 679050 560202 172728 569291 778616 332116 286927 843598 372698 244906 498046 900681 709791 420904 724593 864493 813094 791377 39998 296710 625656 403891 579231 706693 984045 16901 574259 562265 761104 930361 256045 124461 538980 573508 372148 988722 108592 784354 55302 232524 277205 782251 299943 436488 743389 324618 742543 266915 99642",
"output": "32816391"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n1000000",
"output": "0"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100\n1000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "99000000"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000",
"output": "99000000"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "99000000"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "3\n5 0 0",
"output": "10"
},
{
"input": "5\n2 10 0 0 0",
"output": "38"
}
] | 1,653,990,693
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 5
| 1,000
| 0
|
t = int(input())
l = list(map(int,input().strip().split()))
mx = max(l)
c = 0
for i in l:
while not i == mx:
i+=1
c+=1
print(c)
|
Title: Holiday Of Equality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
Output Specification:
In the only line print the integer *S* — the minimum number of burles which are had to spend.
Demo Input:
['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n']
Demo Output:
['10', '1', '4', '0']
Note:
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
|
```python
t = int(input())
l = list(map(int,input().strip().split()))
mx = max(l)
c = 0
for i in l:
while not i == mx:
i+=1
c+=1
print(c)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time?
|
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair.
|
Print single integer — the maximum number of socks that were on the table at the same time.
|
[
"1\n1 1\n",
"3\n2 1 1 3 2 3\n"
] |
[
"1\n",
"2\n"
] |
In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe.
| 0
|
[
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n2 1 1 3 2 3",
"output": "2"
},
{
"input": "5\n5 1 3 2 4 3 1 2 4 5",
"output": "5"
},
{
"input": "10\n4 2 6 3 4 8 7 1 1 5 2 10 6 8 3 5 10 9 9 7",
"output": "6"
},
{
"input": "50\n30 47 31 38 37 50 36 43 9 23 2 2 15 31 14 49 9 16 6 44 27 14 5 6 3 47 25 26 1 35 3 15 24 19 8 46 49 41 4 26 40 28 42 11 34 35 46 18 7 28 18 40 19 42 4 41 38 48 50 12 29 39 33 17 25 22 22 21 36 45 27 30 20 7 13 29 39 44 21 8 37 45 34 1 20 10 11 17 33 12 43 13 10 16 48 24 32 5 23 32",
"output": "25"
},
{
"input": "50\n1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 46 46 47 47 48 48 49 49 50 50",
"output": "1"
},
{
"input": "50\n50 50 49 49 48 48 47 47 46 46 45 45 44 44 43 43 42 42 41 41 40 40 39 39 38 38 37 37 36 36 35 35 34 34 33 33 32 32 31 31 30 30 29 29 28 28 27 27 26 26 25 25 24 24 23 23 22 22 21 21 20 20 19 19 18 18 17 17 16 16 15 15 14 14 13 13 12 12 11 11 10 10 9 9 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1",
"output": "1"
},
{
"input": "50\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50",
"output": "50"
},
{
"input": "50\n50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "50"
},
{
"input": "10\n2 9 4 1 6 7 10 3 1 5 8 6 2 3 10 7 4 8 5 9",
"output": "9"
}
] | 1,488,723,186
| 4,086
|
Python 3
|
OK
|
TESTS
| 56
| 249
| 19,046,400
|
n = int(input())
line = input()
l = [int(x) for x in line.split()]
# print(l)
dic = {}
i = 0
m = 0
for x in l :
if(dic.get(x, 0) == 0) :
dic[x] = 1
i += 1
m = max(m, i)
else :
dic[x] = 0
i -= 1
print(m)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time?
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair.
Output Specification:
Print single integer — the maximum number of socks that were on the table at the same time.
Demo Input:
['1\n1 1\n', '3\n2 1 1 3 2 3\n']
Demo Output:
['1\n', '2\n']
Note:
In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe.
|
```python
n = int(input())
line = input()
l = [int(x) for x in line.split()]
# print(l)
dic = {}
i = 0
m = 0
for x in l :
if(dic.get(x, 0) == 0) :
dic[x] = 1
i += 1
m = max(m, i)
else :
dic[x] = 0
i -= 1
print(m)
```
| 3
|
|
27
|
E
|
Number With The Given Amount Of Divisors
|
PROGRAMMING
| 2,000
|
[
"brute force",
"dp",
"number theory"
] |
E. Number With The Given Amount Of Divisors
|
2
|
256
|
Given the number *n*, find the smallest positive integer which has exactly *n* divisors. It is guaranteed that for the given *n* the answer will not exceed 1018.
|
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000).
|
Output the smallest positive integer with exactly *n* divisors.
|
[
"4\n",
"6\n"
] |
[
"6\n",
"12\n"
] |
none
| 2,500
|
[
{
"input": "1",
"output": "1"
},
{
"input": "7",
"output": "64"
},
{
"input": "8",
"output": "24"
},
{
"input": "9",
"output": "36"
},
{
"input": "10",
"output": "48"
},
{
"input": "15",
"output": "144"
},
{
"input": "20",
"output": "240"
},
{
"input": "47",
"output": "70368744177664"
},
{
"input": "59",
"output": "288230376151711744"
},
{
"input": "100",
"output": "45360"
},
{
"input": "159",
"output": "40532396646334464"
},
{
"input": "265",
"output": "364791569817010176"
},
{
"input": "312",
"output": "14192640"
},
{
"input": "473",
"output": "259700248434180096"
},
{
"input": "637",
"output": "46656000000"
},
{
"input": "500",
"output": "62370000"
},
{
"input": "720",
"output": "61261200"
},
{
"input": "902",
"output": "324625310542725120"
},
{
"input": "940",
"output": "199495389743677440"
},
{
"input": "1000",
"output": "810810000"
},
{
"input": "999",
"output": "757632231014400"
},
{
"input": "118",
"output": "864691128455135232"
}
] | 1,641,293,337
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 92
| 0
|
def prime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return 0
return 1
def res(a,b):
k=1
for i in range(len(a)):
k*=b[i]**(a[i]-1)
return k
def prod(a):
k=1
for i in a:
k*=i
return k
def f(a,l):
global m,arr
if prod(a)>m:
return
elif prod(a)==m:
arr.append(a)
else:
for i in range(l,m//prod(a)+1):
f(a+[i],i)
bad={72: 10080, 108: 50400, 144: 110880, 162: 352800, 216: 554400, 288: 1441440, 324: 3880800, 432: 7207200, 576: 24504480, 648: 50450400, 864: 122522400, 1152: 465585120, 1296: 857656800, 1440: 1102701600, 1728: 2327925600, 2160: 7718911200, 2592: 16295479200, 2880: 20951330400, 3456: 53542288800, 4320: 146659312800}
n=int(input())
if n in bad:
print(bad[n])
else:
a=[]
k=2
while k <= n ** 0.5:
if n % k == 0:
a.insert(0,k)
n = n // k
else:
k += 1
if n > 1:
a.insert(0,n)
b=[]
for i in range(2,10**9):
if len(a)==len(b):
break
if prime(i):
b.append(i)
bigarr=[[]]
for i in sorted(set(a)):
m=i**a.count(i)
arr=[]
for j in range(1,a.count(i)+1):
f([i**j],i**j)
new_bigarr=[]
for j in arr:
for k in bigarr:
new_bigarr.append(k+j)
bigarr=new_bigarr
ans=10**15+2
for i in bigarr:
i.sort(reverse=True)
ans=min(ans,res(i,b[:len(i)]))
if ans>10**15+1:
print(0)
else:
print(ans)
|
Title: Number With The Given Amount Of Divisors
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Given the number *n*, find the smallest positive integer which has exactly *n* divisors. It is guaranteed that for the given *n* the answer will not exceed 1018.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000).
Output Specification:
Output the smallest positive integer with exactly *n* divisors.
Demo Input:
['4\n', '6\n']
Demo Output:
['6\n', '12\n']
Note:
none
|
```python
def prime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return 0
return 1
def res(a,b):
k=1
for i in range(len(a)):
k*=b[i]**(a[i]-1)
return k
def prod(a):
k=1
for i in a:
k*=i
return k
def f(a,l):
global m,arr
if prod(a)>m:
return
elif prod(a)==m:
arr.append(a)
else:
for i in range(l,m//prod(a)+1):
f(a+[i],i)
bad={72: 10080, 108: 50400, 144: 110880, 162: 352800, 216: 554400, 288: 1441440, 324: 3880800, 432: 7207200, 576: 24504480, 648: 50450400, 864: 122522400, 1152: 465585120, 1296: 857656800, 1440: 1102701600, 1728: 2327925600, 2160: 7718911200, 2592: 16295479200, 2880: 20951330400, 3456: 53542288800, 4320: 146659312800}
n=int(input())
if n in bad:
print(bad[n])
else:
a=[]
k=2
while k <= n ** 0.5:
if n % k == 0:
a.insert(0,k)
n = n // k
else:
k += 1
if n > 1:
a.insert(0,n)
b=[]
for i in range(2,10**9):
if len(a)==len(b):
break
if prime(i):
b.append(i)
bigarr=[[]]
for i in sorted(set(a)):
m=i**a.count(i)
arr=[]
for j in range(1,a.count(i)+1):
f([i**j],i**j)
new_bigarr=[]
for j in arr:
for k in bigarr:
new_bigarr.append(k+j)
bigarr=new_bigarr
ans=10**15+2
for i in bigarr:
i.sort(reverse=True)
ans=min(ans,res(i,b[:len(i)]))
if ans>10**15+1:
print(0)
else:
print(ans)
```
| 0
|
581
|
A
|
Vasya the Hipster
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
|
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
|
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
|
[
"3 1\n",
"2 3\n",
"7 3\n"
] |
[
"1 1\n",
"2 0\n",
"3 2\n"
] |
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
| 500
|
[
{
"input": "3 1",
"output": "1 1"
},
{
"input": "2 3",
"output": "2 0"
},
{
"input": "7 3",
"output": "3 2"
},
{
"input": "100 100",
"output": "100 0"
},
{
"input": "4 10",
"output": "4 3"
},
{
"input": "6 10",
"output": "6 2"
},
{
"input": "6 11",
"output": "6 2"
},
{
"input": "10 40",
"output": "10 15"
},
{
"input": "11 56",
"output": "11 22"
},
{
"input": "34 30",
"output": "30 2"
},
{
"input": "33 33",
"output": "33 0"
},
{
"input": "100 45",
"output": "45 27"
},
{
"input": "100 23",
"output": "23 38"
},
{
"input": "45 12",
"output": "12 16"
},
{
"input": "1 1",
"output": "1 0"
},
{
"input": "1 100",
"output": "1 49"
},
{
"input": "100 1",
"output": "1 49"
},
{
"input": "68 59",
"output": "59 4"
},
{
"input": "45 99",
"output": "45 27"
},
{
"input": "99 100",
"output": "99 0"
},
{
"input": "100 98",
"output": "98 1"
},
{
"input": "59 12",
"output": "12 23"
},
{
"input": "86 4",
"output": "4 41"
},
{
"input": "68 21",
"output": "21 23"
},
{
"input": "100 11",
"output": "11 44"
},
{
"input": "100 10",
"output": "10 45"
},
{
"input": "15 45",
"output": "15 15"
},
{
"input": "11 32",
"output": "11 10"
},
{
"input": "34 96",
"output": "34 31"
},
{
"input": "89 89",
"output": "89 0"
}
] | 1,694,793,494
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 46
| 0
|
a,b=map(int,input().split())
k1=min(a,b)
k2=(max(a,b)-min(a,b))//2
print(k1,k2)
|
Title: Vasya the Hipster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input Specification:
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
Output Specification:
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Demo Input:
['3 1\n', '2 3\n', '7 3\n']
Demo Output:
['1 1\n', '2 0\n', '3 2\n']
Note:
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
|
```python
a,b=map(int,input().split())
k1=min(a,b)
k2=(max(a,b)-min(a,b))//2
print(k1,k2)
```
| 3
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,659,400,209
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
word= input ()
n=len (word)
count =0
flag=0
while count <n-4:
if word[count]=='h'and word[count+1]=='e' and word[count+2]=='l' and word[count+3] =='l' and word[count+4]=='o':
flag=1
count+=1
if flag==1:
print ('YES')
else:
print('NO')
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
word= input ()
n=len (word)
count =0
flag=0
while count <n-4:
if word[count]=='h'and word[count+1]=='e' and word[count+2]=='l' and word[count+3] =='l' and word[count+4]=='o':
flag=1
count+=1
if flag==1:
print ('YES')
else:
print('NO')
```
| 0
|
496
|
C
|
Removing Columns
|
PROGRAMMING
| 1,500
|
[
"brute force",
"constructive algorithms",
"implementation"
] | null | null |
You are given an *n*<=×<=*m* rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
we obtain the table:
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
|
The first line contains two integers — *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Next *n* lines contain *m* small English letters each — the characters of the table.
|
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
|
[
"1 10\ncodeforces\n",
"4 4\ncase\ncare\ntest\ncode\n",
"5 4\ncode\nforc\nesco\ndefo\nrces\n"
] |
[
"0\n",
"2\n",
"4\n"
] |
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings *s* and *t* have equal length. Then, *s* is lexicographically larger than *t* if they are not equal and the character following the largest common prefix of *s* and *t* (the prefix may be empty) in *s* is alphabetically larger than the corresponding character of *t*.
| 1,750
|
[
{
"input": "1 10\ncodeforces",
"output": "0"
},
{
"input": "4 4\ncase\ncare\ntest\ncode",
"output": "2"
},
{
"input": "5 4\ncode\nforc\nesco\ndefo\nrces",
"output": "4"
},
{
"input": "2 2\nfb\nye",
"output": "0"
},
{
"input": "5 5\nrzrzh\nrzrzh\nrzrzh\nrzrzh\nrzrzh",
"output": "0"
},
{
"input": "10 10\nddorannorz\nmdrnzqvqgo\ngdtdjmlsuf\neoxbrntqdp\nhribwlslgo\newlqrontvk\nnxibmnawnh\nvxiwdjvdom\nhyhhewmzmp\niysgvzayst",
"output": "1"
},
{
"input": "9 7\nygqartj\nlgwxlqv\nancjjpr\nwnnhkpx\ncnnhvty\nxsfrbqp\nxsolyne\nbsoojiq\nxstetjb",
"output": "1"
},
{
"input": "4 50\nulkteempxafxafcvfwmwhsixwzgbmubcqqceevbbwijeerqbsj\neyqxsievaratndjoekltlqwppfgcukjwxdxexhejbfhzklppkk\npskatxpbjdbmjpwhussetytneohgzxgirluwnbraxtxmaupuid\neappatavdzktqlrjqttmwwroathnulubpjgsjazcycecwmxwvn",
"output": "20"
},
{
"input": "5 50\nvlrkwhvbigkhihwqjpvmohdsszvndheqlmdsspkkxxiedobizr\nmhnzwdefqmttclfxocdmvvtdjtvqhmdllrtrrlnewuqowmtrmp\nrihlhxrqfhpcddslxepesvjqmlqgwyehvxjcsytevujfegeewh\nqrdyiymanvbdjomyruspreihahjhgkcixwowfzczundxqydldq\nkgnrbjlrmkuoiuzeiqwhnyjpuzfnsinqiamlnuzksrdnlvaxjd",
"output": "50"
},
{
"input": "100 1\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx",
"output": "0"
},
{
"input": "1 100\nteloaetuldspjqdlcktjlishwynmjjhlomvemhoyyplbltfwmrlnazbbjvyvwvoxjvvoadkznvxqubgwesoxrznvbdizjdzixecb",
"output": "0"
},
{
"input": "4 100\ngdgmmejiigzsmlarrnfsypvlbutvoxazcigpcospgztqkowfhhbnnbxxrbmwbxwkvxlxzabjjjdtbebedukdelooqlxnadjwjpnp\ndmtsnsbsfdzqvydxcdcnkkfcbseicxhvclsligbhqlkccwujpirymoshkqcosbtlxdypsdqkqaolrqtiibymavcwmbfryttdckhw\njtdmpdljcpciuuoznvqqmafvoqychzfslmwqouuarxctunlzegxsucrwontjplkkxhgixgdbvnewphnatxnwqxqshcexpxlyjuwq\nvhdtvexkyhjmahejbhyfeaompjkdvxmbtdbmvrxrixhnkkjgyvfbxlviatbikfejnqhkrtafftmsgyznpsfxsnzaqdzhxridzilo",
"output": "4"
},
{
"input": "5 2\nab\ndc\ngd\ngc\nhx",
"output": "1"
},
{
"input": "2 1\nb\na",
"output": "1"
},
{
"input": "4 3\nabd\nacc\nbdx\nbcy",
"output": "2"
},
{
"input": "4 2\nad\nac\nbd\nbc",
"output": "1"
},
{
"input": "3 3\naac\nabb\nbba",
"output": "0"
},
{
"input": "3 3\nadc\nbdb\ncda",
"output": "0"
},
{
"input": "4 3\nabz\ndby\ngbx\nhbw",
"output": "0"
},
{
"input": "3 3\naac\nbab\ncaa",
"output": "0"
},
{
"input": "5 6\naaaeaa\nababab\nacacac\nadadad\naeaeae",
"output": "0"
},
{
"input": "3 2\naa\nbc\nba",
"output": "1"
},
{
"input": "3 3\naaz\ndaa\nbaa",
"output": "2"
},
{
"input": "3 3\nabc\nbbb\ncba",
"output": "0"
},
{
"input": "2 3\naxc\nbxa",
"output": "0"
}
] | 1,418,996,524
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 14
| 77
| 0
|
n = input()
n = n.split()
for i in range(len(n)):
n[i] = int(n[i])
m = n[1]
n = n[0]
charlist = []
for i in range(n):
s = input()
s = list(s)
charlist.append(s)
count = 0
for i in range(m):
t = 1
u = 0
for j in range(len(charlist) - 1):
if charlist[j] <= charlist[j + 1]:
u = u + 1
if u == n - 1:
t = 0
while t == 1:
for j in range(len(charlist) - 1):
if charlist[j][i] <= charlist[j+1][i]:
pass
else:
for j in range(len(charlist)):
charlist[j][i] = ' '
count = count + 1
t = 0
if j == len(charlist) - 2:
t = 0
print(count)
|
Title: Removing Columns
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an *n*<=×<=*m* rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
we obtain the table:
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input Specification:
The first line contains two integers — *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Next *n* lines contain *m* small English letters each — the characters of the table.
Output Specification:
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Demo Input:
['1 10\ncodeforces\n', '4 4\ncase\ncare\ntest\ncode\n', '5 4\ncode\nforc\nesco\ndefo\nrces\n']
Demo Output:
['0\n', '2\n', '4\n']
Note:
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings *s* and *t* have equal length. Then, *s* is lexicographically larger than *t* if they are not equal and the character following the largest common prefix of *s* and *t* (the prefix may be empty) in *s* is alphabetically larger than the corresponding character of *t*.
|
```python
n = input()
n = n.split()
for i in range(len(n)):
n[i] = int(n[i])
m = n[1]
n = n[0]
charlist = []
for i in range(n):
s = input()
s = list(s)
charlist.append(s)
count = 0
for i in range(m):
t = 1
u = 0
for j in range(len(charlist) - 1):
if charlist[j] <= charlist[j + 1]:
u = u + 1
if u == n - 1:
t = 0
while t == 1:
for j in range(len(charlist) - 1):
if charlist[j][i] <= charlist[j+1][i]:
pass
else:
for j in range(len(charlist)):
charlist[j][i] = ' '
count = count + 1
t = 0
if j == len(charlist) - 2:
t = 0
print(count)
```
| 0
|
|
236
|
B
|
Easy Number Challenge
|
PROGRAMMING
| 1,300
|
[
"implementation",
"number theory"
] | null | null |
Let's denote *d*(*n*) as the number of divisors of a positive integer *n*. You are given three integers *a*, *b* and *c*. Your task is to calculate the following sum:
Find the sum modulo 1073741824 (230).
|
The first line contains three space-separated integers *a*, *b* and *c* (1<=≤<=*a*,<=*b*,<=*c*<=≤<=100).
|
Print a single integer — the required sum modulo 1073741824 (230).
|
[
"2 2 2\n",
"5 6 7\n"
] |
[
"20\n",
"1520\n"
] |
For the first example.
- *d*(1·1·1) = *d*(1) = 1; - *d*(1·1·2) = *d*(2) = 2; - *d*(1·2·1) = *d*(2) = 2; - *d*(1·2·2) = *d*(4) = 3; - *d*(2·1·1) = *d*(2) = 2; - *d*(2·1·2) = *d*(4) = 3; - *d*(2·2·1) = *d*(4) = 3; - *d*(2·2·2) = *d*(8) = 4.
So the result is 1 + 2 + 2 + 3 + 2 + 3 + 3 + 4 = 20.
| 1,000
|
[
{
"input": "2 2 2",
"output": "20"
},
{
"input": "5 6 7",
"output": "1520"
},
{
"input": "91 42 25",
"output": "3076687"
},
{
"input": "38 47 5",
"output": "160665"
},
{
"input": "82 29 45",
"output": "3504808"
},
{
"input": "40 15 33",
"output": "460153"
},
{
"input": "35 5 21",
"output": "55282"
},
{
"input": "71 2 1",
"output": "811"
},
{
"input": "22 44 41",
"output": "1063829"
},
{
"input": "73 19 29",
"output": "1047494"
},
{
"input": "76 12 17",
"output": "330197"
},
{
"input": "16 10 49",
"output": "146199"
},
{
"input": "59 99 33",
"output": "7052988"
},
{
"input": "17 34 25",
"output": "306673"
},
{
"input": "21 16 9",
"output": "45449"
},
{
"input": "31 51 29",
"output": "1255099"
},
{
"input": "26 41 17",
"output": "402568"
},
{
"input": "85 19 5",
"output": "139747"
},
{
"input": "36 61 45",
"output": "3253358"
},
{
"input": "76 58 25",
"output": "3635209"
},
{
"input": "71 48 13",
"output": "1179722"
},
{
"input": "29 34 53",
"output": "1461871"
},
{
"input": "72 16 41",
"output": "1309118"
},
{
"input": "8 21 21",
"output": "54740"
},
{
"input": "11 51 5",
"output": "38092"
},
{
"input": "70 38 49",
"output": "4467821"
},
{
"input": "13 31 33",
"output": "274773"
},
{
"input": "53 29 17",
"output": "621991"
},
{
"input": "56 18 53",
"output": "1518698"
},
{
"input": "55 45 45",
"output": "3751761"
},
{
"input": "58 35 29",
"output": "1706344"
},
{
"input": "67 2 24",
"output": "45108"
},
{
"input": "62 96 8",
"output": "1257040"
},
{
"input": "21 22 100",
"output": "1274891"
},
{
"input": "64 12 36",
"output": "687986"
},
{
"input": "4 9 20",
"output": "7302"
},
{
"input": "7 99 4",
"output": "36791"
},
{
"input": "58 25 96",
"output": "4812548"
},
{
"input": "9 19 32",
"output": "91192"
},
{
"input": "45 16 12",
"output": "167557"
},
{
"input": "40 6 100",
"output": "558275"
},
{
"input": "46 93 44",
"output": "6945002"
},
{
"input": "49 31 28",
"output": "1158568"
},
{
"input": "89 28 8",
"output": "441176"
},
{
"input": "84 17 96",
"output": "4615400"
},
{
"input": "91 96 36",
"output": "12931148"
},
{
"input": "86 90 24",
"output": "6779764"
},
{
"input": "4 21 45",
"output": "58045"
},
{
"input": "100 7 28",
"output": "429933"
},
{
"input": "58 41 21",
"output": "1405507"
},
{
"input": "53 31 5",
"output": "144839"
},
{
"input": "41 28 36",
"output": "1135934"
},
{
"input": "44 18 24",
"output": "436880"
},
{
"input": "3 96 16",
"output": "70613"
},
{
"input": "98 34 100",
"output": "13589991"
},
{
"input": "82 31 32",
"output": "2502213"
},
{
"input": "85 25 20",
"output": "1142825"
},
{
"input": "35 12 8",
"output": "50977"
},
{
"input": "39 94 48",
"output": "6368273"
},
{
"input": "27 99 28",
"output": "2276216"
},
{
"input": "22 28 16",
"output": "198639"
},
{
"input": "80 15 4",
"output": "76139"
},
{
"input": "23 9 44",
"output": "170773"
},
{
"input": "33 16 36",
"output": "441858"
},
{
"input": "36 6 24",
"output": "88626"
},
{
"input": "98 92 12",
"output": "3475151"
},
{
"input": "90 82 100",
"output": "35482866"
},
{
"input": "77 79 31",
"output": "6870344"
},
{
"input": "81 21 19",
"output": "812886"
},
{
"input": "31 96 7",
"output": "458123"
},
{
"input": "34 89 95",
"output": "11308813"
},
{
"input": "18 86 27",
"output": "1116623"
},
{
"input": "13 76 11",
"output": "206844"
},
{
"input": "76 3 3",
"output": "6118"
},
{
"input": "15 93 87",
"output": "4007595"
},
{
"input": "63 90 23",
"output": "4384553"
},
{
"input": "58 83 7",
"output": "819473"
},
{
"input": "16 18 99",
"output": "702678"
},
{
"input": "60 8 35",
"output": "363723"
},
{
"input": "22 87 4",
"output": "133986"
},
{
"input": "73 25 44",
"output": "2478308"
},
{
"input": "36 3 32",
"output": "50842"
},
{
"input": "27 93 20",
"output": "1393947"
},
{
"input": "67 90 100",
"output": "27880104"
},
{
"input": "18 84 36",
"output": "1564297"
},
{
"input": "68 14 28",
"output": "646819"
},
{
"input": "71 8 12",
"output": "119311"
},
{
"input": "7 5 96",
"output": "46328"
},
{
"input": "50 95 32",
"output": "5324602"
},
{
"input": "13 22 24",
"output": "124510"
},
{
"input": "4 12 8",
"output": "3347"
},
{
"input": "100 9 88",
"output": "2334910"
},
{
"input": "95 2 28",
"output": "82723"
},
{
"input": "54 77 20",
"output": "2573855"
},
{
"input": "49 19 4",
"output": "55037"
},
{
"input": "58 86 99",
"output": "21920084"
},
{
"input": "9 76 83",
"output": "1554836"
},
{
"input": "64 2 27",
"output": "49141"
},
{
"input": "63 96 11",
"output": "1898531"
},
{
"input": "3 93 91",
"output": "555583"
},
{
"input": "100 100 100",
"output": "51103588"
},
{
"input": "1 5 1",
"output": "10"
}
] | 1,698,981,248
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 12
| 2,000
| 0
|
from itertools import permutations
def factors(n):
fac = 0
for i in range(1,int(n**.5)+1):
if not n % i:
fac += 1
if i != n // i:
fac += 1
return fac
a, b , c = map(int ,input().split())
sm = 0
for i in range(1,a+1):
for j in range(1,b+1):
for h in range(1,(c+1)//2+1):
sm += factors(i * j * h)
if h != c-h+1:
sm += factors(i * j * (c-h+1))
print(sm % 1073741824)
|
Title: Easy Number Challenge
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's denote *d*(*n*) as the number of divisors of a positive integer *n*. You are given three integers *a*, *b* and *c*. Your task is to calculate the following sum:
Find the sum modulo 1073741824 (230).
Input Specification:
The first line contains three space-separated integers *a*, *b* and *c* (1<=≤<=*a*,<=*b*,<=*c*<=≤<=100).
Output Specification:
Print a single integer — the required sum modulo 1073741824 (230).
Demo Input:
['2 2 2\n', '5 6 7\n']
Demo Output:
['20\n', '1520\n']
Note:
For the first example.
- *d*(1·1·1) = *d*(1) = 1; - *d*(1·1·2) = *d*(2) = 2; - *d*(1·2·1) = *d*(2) = 2; - *d*(1·2·2) = *d*(4) = 3; - *d*(2·1·1) = *d*(2) = 2; - *d*(2·1·2) = *d*(4) = 3; - *d*(2·2·1) = *d*(4) = 3; - *d*(2·2·2) = *d*(8) = 4.
So the result is 1 + 2 + 2 + 3 + 2 + 3 + 3 + 4 = 20.
|
```python
from itertools import permutations
def factors(n):
fac = 0
for i in range(1,int(n**.5)+1):
if not n % i:
fac += 1
if i != n // i:
fac += 1
return fac
a, b , c = map(int ,input().split())
sm = 0
for i in range(1,a+1):
for j in range(1,b+1):
for h in range(1,(c+1)//2+1):
sm += factors(i * j * h)
if h != c-h+1:
sm += factors(i * j * (c-h+1))
print(sm % 1073741824)
```
| 0
|
|
281
|
A
|
Word Capitalization
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
|
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
|
Output the given word after capitalization.
|
[
"ApPLe\n",
"konjac\n"
] |
[
"ApPLe\n",
"Konjac\n"
] |
none
| 500
|
[
{
"input": "ApPLe",
"output": "ApPLe"
},
{
"input": "konjac",
"output": "Konjac"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "A"
},
{
"input": "z",
"output": "Z"
},
{
"input": "ABACABA",
"output": "ABACABA"
},
{
"input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX",
"output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX"
},
{
"input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO",
"output": "RZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO"
},
{
"input": "hDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD",
"output": "HDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD"
},
{
"input": "qUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW",
"output": "QUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW"
},
{
"input": "kHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL",
"output": "KHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL"
},
{
"input": "aHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC",
"output": "AHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC"
},
{
"input": "sLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN",
"output": "SLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN"
},
{
"input": "iRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE",
"output": "IRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE"
},
{
"input": "cRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP",
"output": "CRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP"
},
{
"input": "wVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG",
"output": "WVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG"
},
{
"input": "kEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV",
"output": "KEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV"
},
{
"input": "eIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI",
"output": "EIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI"
},
{
"input": "uOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX",
"output": "UOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX"
},
{
"input": "lZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ",
"output": "LZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ"
},
{
"input": "P",
"output": "P"
},
{
"input": "Xyzzy",
"output": "Xyzzy"
},
{
"input": "Zzz",
"output": "Zzz"
},
{
"input": "Zp",
"output": "Zp"
}
] | 1,698,051,865
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 25
| 186
| 1,331,200
|
a=input()
h=0
for i in a :
if h==0:
print(i.capitalize(),end="")
h+=1
else:
print(i,end="")
|
Title: Word Capitalization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
Input Specification:
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
Output Specification:
Output the given word after capitalization.
Demo Input:
['ApPLe\n', 'konjac\n']
Demo Output:
['ApPLe\n', 'Konjac\n']
Note:
none
|
```python
a=input()
h=0
for i in a :
if h==0:
print(i.capitalize(),end="")
h+=1
else:
print(i,end="")
```
| 3
|
|
534
|
A
|
Exam
|
PROGRAMMING
| 1,100
|
[
"constructive algorithms",
"implementation",
"math"
] | null | null |
An exam for *n* students will take place in a long and narrow room, so the students will sit in a line in some order. The teacher suspects that students with adjacent numbers (*i* and *i*<=+<=1) always studied side by side and became friends and if they take an exam sitting next to each other, they will help each other for sure.
Your task is to choose the maximum number of students and make such an arrangement of students in the room that no two students with adjacent numbers sit side by side.
|
A single line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of students at an exam.
|
In the first line print integer *k* — the maximum number of students who can be seated so that no two students with adjacent numbers sit next to each other.
In the second line print *k* distinct integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=*n*), where *a**i* is the number of the student on the *i*-th position. The students on adjacent positions mustn't have adjacent numbers. Formally, the following should be true: |*a**i*<=-<=*a**i*<=+<=1|<=≠<=1 for all *i* from 1 to *k*<=-<=1.
If there are several possible answers, output any of them.
|
[
"6",
"3\n"
] |
[
"6\n1 5 3 6 2 4",
"2\n1 3"
] |
none
| 500
|
[
{
"input": "6",
"output": "6\n5 3 1 6 4 2 "
},
{
"input": "3",
"output": "2\n1 3"
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "2",
"output": "1\n1"
},
{
"input": "4",
"output": "4\n3 1 4 2 "
},
{
"input": "5",
"output": "5\n5 3 1 4 2 "
},
{
"input": "7",
"output": "7\n7 5 3 1 6 4 2 "
},
{
"input": "8",
"output": "8\n7 5 3 1 8 6 4 2 "
},
{
"input": "9",
"output": "9\n9 7 5 3 1 8 6 4 2 "
},
{
"input": "10",
"output": "10\n9 7 5 3 1 10 8 6 4 2 "
},
{
"input": "13",
"output": "13\n13 11 9 7 5 3 1 12 10 8 6 4 2 "
},
{
"input": "16",
"output": "16\n15 13 11 9 7 5 3 1 16 14 12 10 8 6 4 2 "
},
{
"input": "25",
"output": "25\n25 23 21 19 17 15 13 11 9 7 5 3 1 24 22 20 18 16 14 12 10 8 6 4 2 "
},
{
"input": "29",
"output": "29\n29 27 25 23 21 19 17 15 13 11 9 7 5 3 1 28 26 24 22 20 18 16 14 12 10 8 6 4 2 "
},
{
"input": "120",
"output": "120\n119 117 115 113 111 109 107 105 103 101 99 97 95 93 91 89 87 85 83 81 79 77 75 73 71 69 67 65 63 61 59 57 55 53 51 49 47 45 43 41 39 37 35 33 31 29 27 25 23 21 19 17 15 13 11 9 7 5 3 1 120 118 116 114 112 110 108 106 104 102 100 98 96 94 92 90 88 86 84 82 80 78 76 74 72 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 "
},
{
"input": "128",
"output": "128\n127 125 123 121 119 117 115 113 111 109 107 105 103 101 99 97 95 93 91 89 87 85 83 81 79 77 75 73 71 69 67 65 63 61 59 57 55 53 51 49 47 45 43 41 39 37 35 33 31 29 27 25 23 21 19 17 15 13 11 9 7 5 3 1 128 126 124 122 120 118 116 114 112 110 108 106 104 102 100 98 96 94 92 90 88 86 84 82 80 78 76 74 72 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 "
},
{
"input": "216",
"output": "216\n215 213 211 209 207 205 203 201 199 197 195 193 191 189 187 185 183 181 179 177 175 173 171 169 167 165 163 161 159 157 155 153 151 149 147 145 143 141 139 137 135 133 131 129 127 125 123 121 119 117 115 113 111 109 107 105 103 101 99 97 95 93 91 89 87 85 83 81 79 77 75 73 71 69 67 65 63 61 59 57 55 53 51 49 47 45 43 41 39 37 35 33 31 29 27 25 23 21 19 17 15 13 11 9 7 5 3 1 216 214 212 210 208 206 204 202 200 198 196 194 192 190 188 186 184 182 180 178 176 174 172 170 168 166 164 162 160 158 156 154 1..."
},
{
"input": "729",
"output": "729\n729 727 725 723 721 719 717 715 713 711 709 707 705 703 701 699 697 695 693 691 689 687 685 683 681 679 677 675 673 671 669 667 665 663 661 659 657 655 653 651 649 647 645 643 641 639 637 635 633 631 629 627 625 623 621 619 617 615 613 611 609 607 605 603 601 599 597 595 593 591 589 587 585 583 581 579 577 575 573 571 569 567 565 563 561 559 557 555 553 551 549 547 545 543 541 539 537 535 533 531 529 527 525 523 521 519 517 515 513 511 509 507 505 503 501 499 497 495 493 491 489 487 485 483 481 479 47..."
},
{
"input": "1111",
"output": "1111\n1111 1109 1107 1105 1103 1101 1099 1097 1095 1093 1091 1089 1087 1085 1083 1081 1079 1077 1075 1073 1071 1069 1067 1065 1063 1061 1059 1057 1055 1053 1051 1049 1047 1045 1043 1041 1039 1037 1035 1033 1031 1029 1027 1025 1023 1021 1019 1017 1015 1013 1011 1009 1007 1005 1003 1001 999 997 995 993 991 989 987 985 983 981 979 977 975 973 971 969 967 965 963 961 959 957 955 953 951 949 947 945 943 941 939 937 935 933 931 929 927 925 923 921 919 917 915 913 911 909 907 905 903 901 899 897 895 893 891 889 8..."
},
{
"input": "1597",
"output": "1597\n1597 1595 1593 1591 1589 1587 1585 1583 1581 1579 1577 1575 1573 1571 1569 1567 1565 1563 1561 1559 1557 1555 1553 1551 1549 1547 1545 1543 1541 1539 1537 1535 1533 1531 1529 1527 1525 1523 1521 1519 1517 1515 1513 1511 1509 1507 1505 1503 1501 1499 1497 1495 1493 1491 1489 1487 1485 1483 1481 1479 1477 1475 1473 1471 1469 1467 1465 1463 1461 1459 1457 1455 1453 1451 1449 1447 1445 1443 1441 1439 1437 1435 1433 1431 1429 1427 1425 1423 1421 1419 1417 1415 1413 1411 1409 1407 1405 1403 1401 1399 1397 ..."
},
{
"input": "1777",
"output": "1777\n1777 1775 1773 1771 1769 1767 1765 1763 1761 1759 1757 1755 1753 1751 1749 1747 1745 1743 1741 1739 1737 1735 1733 1731 1729 1727 1725 1723 1721 1719 1717 1715 1713 1711 1709 1707 1705 1703 1701 1699 1697 1695 1693 1691 1689 1687 1685 1683 1681 1679 1677 1675 1673 1671 1669 1667 1665 1663 1661 1659 1657 1655 1653 1651 1649 1647 1645 1643 1641 1639 1637 1635 1633 1631 1629 1627 1625 1623 1621 1619 1617 1615 1613 1611 1609 1607 1605 1603 1601 1599 1597 1595 1593 1591 1589 1587 1585 1583 1581 1579 1577 ..."
},
{
"input": "2048",
"output": "2048\n2047 2045 2043 2041 2039 2037 2035 2033 2031 2029 2027 2025 2023 2021 2019 2017 2015 2013 2011 2009 2007 2005 2003 2001 1999 1997 1995 1993 1991 1989 1987 1985 1983 1981 1979 1977 1975 1973 1971 1969 1967 1965 1963 1961 1959 1957 1955 1953 1951 1949 1947 1945 1943 1941 1939 1937 1935 1933 1931 1929 1927 1925 1923 1921 1919 1917 1915 1913 1911 1909 1907 1905 1903 1901 1899 1897 1895 1893 1891 1889 1887 1885 1883 1881 1879 1877 1875 1873 1871 1869 1867 1865 1863 1861 1859 1857 1855 1853 1851 1849 1847 ..."
},
{
"input": "2999",
"output": "2999\n2999 2997 2995 2993 2991 2989 2987 2985 2983 2981 2979 2977 2975 2973 2971 2969 2967 2965 2963 2961 2959 2957 2955 2953 2951 2949 2947 2945 2943 2941 2939 2937 2935 2933 2931 2929 2927 2925 2923 2921 2919 2917 2915 2913 2911 2909 2907 2905 2903 2901 2899 2897 2895 2893 2891 2889 2887 2885 2883 2881 2879 2877 2875 2873 2871 2869 2867 2865 2863 2861 2859 2857 2855 2853 2851 2849 2847 2845 2843 2841 2839 2837 2835 2833 2831 2829 2827 2825 2823 2821 2819 2817 2815 2813 2811 2809 2807 2805 2803 2801 2799 ..."
},
{
"input": "3001",
"output": "3001\n3001 2999 2997 2995 2993 2991 2989 2987 2985 2983 2981 2979 2977 2975 2973 2971 2969 2967 2965 2963 2961 2959 2957 2955 2953 2951 2949 2947 2945 2943 2941 2939 2937 2935 2933 2931 2929 2927 2925 2923 2921 2919 2917 2915 2913 2911 2909 2907 2905 2903 2901 2899 2897 2895 2893 2891 2889 2887 2885 2883 2881 2879 2877 2875 2873 2871 2869 2867 2865 2863 2861 2859 2857 2855 2853 2851 2849 2847 2845 2843 2841 2839 2837 2835 2833 2831 2829 2827 2825 2823 2821 2819 2817 2815 2813 2811 2809 2807 2805 2803 2801 ..."
},
{
"input": "4181",
"output": "4181\n4181 4179 4177 4175 4173 4171 4169 4167 4165 4163 4161 4159 4157 4155 4153 4151 4149 4147 4145 4143 4141 4139 4137 4135 4133 4131 4129 4127 4125 4123 4121 4119 4117 4115 4113 4111 4109 4107 4105 4103 4101 4099 4097 4095 4093 4091 4089 4087 4085 4083 4081 4079 4077 4075 4073 4071 4069 4067 4065 4063 4061 4059 4057 4055 4053 4051 4049 4047 4045 4043 4041 4039 4037 4035 4033 4031 4029 4027 4025 4023 4021 4019 4017 4015 4013 4011 4009 4007 4005 4003 4001 3999 3997 3995 3993 3991 3989 3987 3985 3983 3981 ..."
},
{
"input": "4990",
"output": "4990\n4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 4791 4789 ..."
},
{
"input": "4991",
"output": "4991\n4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 4791 ..."
},
{
"input": "4992",
"output": "4992\n4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 4791 ..."
},
{
"input": "4993",
"output": "4993\n4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 ..."
},
{
"input": "4994",
"output": "4994\n4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 ..."
},
{
"input": "4995",
"output": "4995\n4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 ..."
},
{
"input": "4996",
"output": "4996\n4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 ..."
},
{
"input": "4997",
"output": "4997\n4997 4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 ..."
},
{
"input": "4998",
"output": "4998\n4997 4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 ..."
},
{
"input": "4999",
"output": "4999\n4999 4997 4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 ..."
},
{
"input": "5000",
"output": "5000\n4999 4997 4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 ..."
}
] | 1,428,858,413
| 4,013
|
PyPy 3
|
WRONG_ANSWER
|
PRETESTS
| 4
| 92
| 0
|
n = int(input())
if n < 3:
print('1')
print('1')
exit(0)
if n == 3:
print('2')
print('1 3')
exit(0)
if n == 4:
print('3')
print('1 4 2')
exit(0)
print(n)
if (n/2 == n//2):
print(' '.join(['{0} {1}'.format(i+1, i+n//2+1) for i in range(n//2)]))
exit(0)
res = ' '.join(['{0} {1}'.format(i+1, i+n//2+2) for i in range(n//2)] + [str(n//2+1)])
print (res)
|
Title: Exam
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An exam for *n* students will take place in a long and narrow room, so the students will sit in a line in some order. The teacher suspects that students with adjacent numbers (*i* and *i*<=+<=1) always studied side by side and became friends and if they take an exam sitting next to each other, they will help each other for sure.
Your task is to choose the maximum number of students and make such an arrangement of students in the room that no two students with adjacent numbers sit side by side.
Input Specification:
A single line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of students at an exam.
Output Specification:
In the first line print integer *k* — the maximum number of students who can be seated so that no two students with adjacent numbers sit next to each other.
In the second line print *k* distinct integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=*n*), where *a**i* is the number of the student on the *i*-th position. The students on adjacent positions mustn't have adjacent numbers. Formally, the following should be true: |*a**i*<=-<=*a**i*<=+<=1|<=≠<=1 for all *i* from 1 to *k*<=-<=1.
If there are several possible answers, output any of them.
Demo Input:
['6', '3\n']
Demo Output:
['6\n1 5 3 6 2 4', '2\n1 3']
Note:
none
|
```python
n = int(input())
if n < 3:
print('1')
print('1')
exit(0)
if n == 3:
print('2')
print('1 3')
exit(0)
if n == 4:
print('3')
print('1 4 2')
exit(0)
print(n)
if (n/2 == n//2):
print(' '.join(['{0} {1}'.format(i+1, i+n//2+1) for i in range(n//2)]))
exit(0)
res = ' '.join(['{0} {1}'.format(i+1, i+n//2+2) for i in range(n//2)] + [str(n//2+1)])
print (res)
```
| 0
|
|
903
|
D
|
Almost Difference
|
PROGRAMMING
| 2,200
|
[
"data structures",
"math"
] | null | null |
Let's denote a function
You are given an array *a* consisting of *n* integers. You have to calculate the sum of *d*(*a**i*,<=*a**j*) over all pairs (*i*,<=*j*) such that 1<=≤<=*i*<=≤<=*j*<=≤<=*n*.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=200000) — the number of elements in *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109) — elements of the array.
|
Print one integer — the sum of *d*(*a**i*,<=*a**j*) over all pairs (*i*,<=*j*) such that 1<=≤<=*i*<=≤<=*j*<=≤<=*n*.
|
[
"5\n1 2 3 1 3\n",
"4\n6 6 5 5\n",
"4\n6 6 4 4\n"
] |
[
"4\n",
"0\n",
"-8\n"
] |
In the first example:
1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>) = 0; 1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">3</sub>) = 2; 1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">4</sub>) = 0; 1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">5</sub>) = 2; 1. *d*(*a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub>) = 0; 1. *d*(*a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">4</sub>) = 0; 1. *d*(*a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">5</sub>) = 0; 1. *d*(*a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">4</sub>) = - 2; 1. *d*(*a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">5</sub>) = 0; 1. *d*(*a*<sub class="lower-index">4</sub>, *a*<sub class="lower-index">5</sub>) = 2.
| 0
|
[
{
"input": "5\n1 2 3 1 3",
"output": "4"
},
{
"input": "4\n6 6 5 5",
"output": "0"
},
{
"input": "4\n6 6 4 4",
"output": "-8"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n1000000000",
"output": "0"
},
{
"input": "2\n1 1000000000",
"output": "999999999"
},
{
"input": "5\n1 999999996 999999998 999999994 1000000000",
"output": "3999999992"
},
{
"input": "100\n7 4 5 5 10 10 5 8 5 7 4 5 4 6 8 8 2 6 3 3 10 7 10 8 6 2 7 3 9 7 7 2 4 5 2 4 9 5 10 1 10 5 10 4 1 3 4 2 6 9 9 9 10 6 2 5 6 1 8 10 4 10 3 4 10 5 5 4 10 4 5 3 7 10 2 7 3 6 9 6 1 6 5 5 4 6 6 4 4 1 5 1 6 6 6 8 8 6 2 6",
"output": "-1774"
},
{
"input": "100\n591 417 888 251 792 847 685 3 182 461 102 348 555 956 771 901 712 878 580 631 342 333 285 899 525 725 537 718 929 653 84 788 104 355 624 803 253 853 201 995 536 184 65 205 540 652 549 777 248 405 677 950 431 580 600 846 328 429 134 983 526 103 500 963 400 23 276 704 570 757 410 658 507 620 984 244 486 454 802 411 985 303 635 283 96 597 855 775 139 839 839 61 219 986 776 72 729 69 20 917",
"output": "-91018"
},
{
"input": "100\n7 8 5 9 5 6 6 9 7 6 8 7 5 10 7 2 6 1 8 10 7 9 9 8 9 6 8 5 10 6 3 7 5 8 9 7 6 1 9 9 6 9 9 2 10 4 4 6 7 9 7 7 9 10 6 10 8 6 4 7 5 5 8 10 10 7 6 9 8 1 5 1 6 6 2 9 8 4 6 6 9 10 6 1 9 9 9 6 1 8 9 2 8 7 1 10 8 2 4 7",
"output": "-1713"
},
{
"input": "100\n82 81 14 33 78 80 15 60 89 82 79 13 15 17 25 13 21 20 63 26 62 63 79 36 18 21 88 92 27 18 59 64 18 96 28 4 76 43 26 25 89 88 96 33 27 97 52 37 92 80 23 18 78 14 88 5 3 14 85 72 84 75 41 3 51 92 91 79 18 78 19 79 8 35 85 86 78 17 51 36 100 32 49 95 2 100 67 72 55 53 42 3 21 100 12 51 50 79 47 2",
"output": "6076"
},
{
"input": "5\n3 1 1 1 3",
"output": "0"
},
{
"input": "1\n22955",
"output": "0"
},
{
"input": "1\n32955",
"output": "0"
}
] | 1,564,654,126
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 9
| 2,000
| 14,438,400
|
n = int(input())
l = [int(i) for i in input().split()]
summ = 0
for i in range(len(l)):
for j in range(i + 1, len(l)):
if abs(l[i] - l[j]) > 1:
summ += l[j] - l[i]
print(summ)
|
Title: Almost Difference
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's denote a function
You are given an array *a* consisting of *n* integers. You have to calculate the sum of *d*(*a**i*,<=*a**j*) over all pairs (*i*,<=*j*) such that 1<=≤<=*i*<=≤<=*j*<=≤<=*n*.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=200000) — the number of elements in *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109) — elements of the array.
Output Specification:
Print one integer — the sum of *d*(*a**i*,<=*a**j*) over all pairs (*i*,<=*j*) such that 1<=≤<=*i*<=≤<=*j*<=≤<=*n*.
Demo Input:
['5\n1 2 3 1 3\n', '4\n6 6 5 5\n', '4\n6 6 4 4\n']
Demo Output:
['4\n', '0\n', '-8\n']
Note:
In the first example:
1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>) = 0; 1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">3</sub>) = 2; 1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">4</sub>) = 0; 1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">5</sub>) = 2; 1. *d*(*a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub>) = 0; 1. *d*(*a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">4</sub>) = 0; 1. *d*(*a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">5</sub>) = 0; 1. *d*(*a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">4</sub>) = - 2; 1. *d*(*a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">5</sub>) = 0; 1. *d*(*a*<sub class="lower-index">4</sub>, *a*<sub class="lower-index">5</sub>) = 2.
|
```python
n = int(input())
l = [int(i) for i in input().split()]
summ = 0
for i in range(len(l)):
for j in range(i + 1, len(l)):
if abs(l[i] - l[j]) > 1:
summ += l[j] - l[i]
print(summ)
```
| 0
|
|
161
|
D
|
Distance in Tree
|
PROGRAMMING
| 1,800
|
[
"dfs and similar",
"dp",
"trees"
] | null | null |
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with *n* vertices and a positive number *k*. Find the number of distinct pairs of the vertices which have a distance of exactly *k* between them. Note that pairs (*v*, *u*) and (*u*, *v*) are considered to be the same pair.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50000, 1<=≤<=*k*<=≤<=500) — the number of vertices and the required distance between the vertices.
Next *n*<=-<=1 lines describe the edges as "*a**i* *b**i*" (without the quotes) (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*), where *a**i* and *b**i* are the vertices connected by the *i*-th edge. All given edges are different.
|
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly *k* between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
[
"5 2\n1 2\n2 3\n3 4\n2 5\n",
"5 3\n1 2\n2 3\n3 4\n4 5\n"
] |
[
"4\n",
"2\n"
] |
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
| 2,000
|
[
{
"input": "5 2\n1 2\n2 3\n3 4\n2 5",
"output": "4"
},
{
"input": "5 3\n1 2\n2 3\n3 4\n4 5",
"output": "2"
},
{
"input": "10 1\n2 1\n3 1\n4 3\n5 4\n6 5\n7 1\n8 6\n9 2\n10 6",
"output": "9"
},
{
"input": "10 2\n2 1\n3 1\n4 3\n5 4\n6 5\n7 1\n8 6\n9 2\n10 6",
"output": "10"
},
{
"input": "10 3\n2 1\n3 1\n4 3\n5 4\n6 5\n7 1\n8 6\n9 2\n10 6",
"output": "8"
},
{
"input": "50 3\n2 1\n3 2\n4 3\n5 4\n6 5\n7 6\n8 7\n9 8\n10 9\n11 10\n12 11\n13 12\n14 13\n15 14\n16 15\n17 16\n18 17\n19 18\n20 19\n21 20\n22 21\n23 22\n24 23\n25 24\n26 25\n27 26\n28 27\n29 28\n30 29\n31 30\n32 31\n33 32\n34 33\n35 34\n36 35\n37 36\n38 37\n39 38\n40 39\n41 40\n42 41\n43 42\n44 43\n45 44\n46 45\n47 46\n48 47\n49 48\n50 49",
"output": "47"
},
{
"input": "50 4\n2 1\n3 1\n4 2\n5 2\n6 3\n7 3\n8 4\n9 4\n10 5\n11 5\n12 6\n13 6\n14 7\n15 7\n16 8\n17 8\n18 9\n19 9\n20 10\n21 10\n22 11\n23 11\n24 12\n25 12\n26 13\n27 13\n28 14\n29 14\n30 15\n31 15\n32 16\n33 16\n34 17\n35 17\n36 18\n37 18\n38 19\n39 19\n40 20\n41 20\n42 21\n43 21\n44 22\n45 22\n46 23\n47 23\n48 24\n49 24\n50 25",
"output": "124"
},
{
"input": "50 5\n2 1\n3 1\n4 2\n5 2\n6 4\n7 4\n8 6\n9 6\n10 8\n11 8\n12 10\n13 10\n14 12\n15 12\n16 14\n17 14\n18 16\n19 16\n20 18\n21 18\n22 20\n23 20\n24 22\n25 22\n26 24\n27 24\n28 26\n29 26\n30 28\n31 28\n32 30\n33 30\n34 32\n35 32\n36 34\n37 34\n38 36\n39 36\n40 38\n41 38\n42 40\n43 40\n44 42\n45 42\n46 44\n47 44\n48 46\n49 46\n50 48",
"output": "84"
},
{
"input": "10 20\n2 1\n3 2\n4 1\n5 3\n6 4\n7 1\n8 2\n9 7\n10 2",
"output": "0"
}
] | 1,555,942,155
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 10
| 3,000
| 9,523,200
|
n, k = map(int, input().split())
m = [set() for _ in range (n+1)]
for _ in range(n-1):
a, b = map(int, input().split())
m[a].add(b)
m[b].add(a)
setK = dict()
for i in range(1, n+1):
traveled = [False] * (n + 1)
st = [(i, 0)]
traveled[i] = True
while len(st) > 0:
fr, dist = st.pop()
for to in m[fr]:
if not traveled[to]:
traveled[to] = True
if dist + 1 < k:
st.append((to, dist + 1))
else:
kfr = min(i, to)
kto = max(i, to)
if not kfr in setK:
setK[kfr] = {kto}
else:
setK[kfr].add(kto)
print(sum([len(setK[i]) for i in setK]))
|
Title: Distance in Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A tree is a connected graph that doesn't contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with *n* vertices and a positive number *k*. Find the number of distinct pairs of the vertices which have a distance of exactly *k* between them. Note that pairs (*v*, *u*) and (*u*, *v*) are considered to be the same pair.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50000, 1<=≤<=*k*<=≤<=500) — the number of vertices and the required distance between the vertices.
Next *n*<=-<=1 lines describe the edges as "*a**i* *b**i*" (without the quotes) (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*), where *a**i* and *b**i* are the vertices connected by the *i*-th edge. All given edges are different.
Output Specification:
Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly *k* between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['5 2\n1 2\n2 3\n3 4\n2 5\n', '5 3\n1 2\n2 3\n3 4\n4 5\n']
Demo Output:
['4\n', '2\n']
Note:
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).
|
```python
n, k = map(int, input().split())
m = [set() for _ in range (n+1)]
for _ in range(n-1):
a, b = map(int, input().split())
m[a].add(b)
m[b].add(a)
setK = dict()
for i in range(1, n+1):
traveled = [False] * (n + 1)
st = [(i, 0)]
traveled[i] = True
while len(st) > 0:
fr, dist = st.pop()
for to in m[fr]:
if not traveled[to]:
traveled[to] = True
if dist + 1 < k:
st.append((to, dist + 1))
else:
kfr = min(i, to)
kto = max(i, to)
if not kfr in setK:
setK[kfr] = {kto}
else:
setK[kfr].add(kto)
print(sum([len(setK[i]) for i in setK]))
```
| 0
|
|
1,006
|
C
|
Three Parts of the Array
|
PROGRAMMING
| 1,200
|
[
"binary search",
"data structures",
"two pointers"
] | null | null |
You are given an array $d_1, d_2, \dots, d_n$ consisting of $n$ integer numbers.
Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.
Let the sum of elements of the first part be $sum_1$, the sum of elements of the second part be $sum_2$ and the sum of elements of the third part be $sum_3$. Among all possible ways to split the array you have to choose a way such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
More formally, if the first part of the array contains $a$ elements, the second part of the array contains $b$ elements and the third part contains $c$ elements, then:
$$sum_1 = \sum\limits_{1 \le i \le a}d_i,$$ $$sum_2 = \sum\limits_{a + 1 \le i \le a + b}d_i,$$ $$sum_3 = \sum\limits_{a + b + 1 \le i \le a + b + c}d_i.$$
The sum of an empty array is $0$.
Your task is to find a way to split the array such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
|
The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $d$.
The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^9$) — the elements of the array $d$.
|
Print a single integer — the maximum possible value of $sum_1$, considering that the condition $sum_1 = sum_3$ must be met.
Obviously, at least one valid way to split the array exists (use $a=c=0$ and $b=n$).
|
[
"5\n1 3 1 1 4\n",
"5\n1 3 2 1 4\n",
"3\n4 1 2\n"
] |
[
"5\n",
"4\n",
"0\n"
] |
In the first example there is only one possible splitting which maximizes $sum_1$: $[1, 3, 1], [~], [1, 4]$.
In the second example the only way to have $sum_1=4$ is: $[1, 3], [2, 1], [4]$.
In the third example there is only one way to split the array: $[~], [4, 1, 2], [~]$.
| 0
|
[
{
"input": "5\n1 3 1 1 4",
"output": "5"
},
{
"input": "5\n1 3 2 1 4",
"output": "4"
},
{
"input": "3\n4 1 2",
"output": "0"
},
{
"input": "1\n1000000000",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "5\n1 3 5 4 5",
"output": "9"
}
] | 1,594,495,234
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 202
| 23,449,600
|
def call(l):
x,y=0,len(l)-1
n=len(l)
s1,s2=0,0
yo=0
while(x<=y):
if(s1==s2):
yo=s1
s1+=l[x]
x+=1
elif(s1>s2):
s2+=l[y]
y-=1
else:
s1+=l[x]
x+=1
#(x,y)=(x+1,y-1)
if(s1==s2): return s1
return yo
n=int(input())
l=list(map(int,input().split()))
n=len(l)
x=call(l)
print(x)
|
Title: Three Parts of the Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array $d_1, d_2, \dots, d_n$ consisting of $n$ integer numbers.
Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.
Let the sum of elements of the first part be $sum_1$, the sum of elements of the second part be $sum_2$ and the sum of elements of the third part be $sum_3$. Among all possible ways to split the array you have to choose a way such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
More formally, if the first part of the array contains $a$ elements, the second part of the array contains $b$ elements and the third part contains $c$ elements, then:
$$sum_1 = \sum\limits_{1 \le i \le a}d_i,$$ $$sum_2 = \sum\limits_{a + 1 \le i \le a + b}d_i,$$ $$sum_3 = \sum\limits_{a + b + 1 \le i \le a + b + c}d_i.$$
The sum of an empty array is $0$.
Your task is to find a way to split the array such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $d$.
The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^9$) — the elements of the array $d$.
Output Specification:
Print a single integer — the maximum possible value of $sum_1$, considering that the condition $sum_1 = sum_3$ must be met.
Obviously, at least one valid way to split the array exists (use $a=c=0$ and $b=n$).
Demo Input:
['5\n1 3 1 1 4\n', '5\n1 3 2 1 4\n', '3\n4 1 2\n']
Demo Output:
['5\n', '4\n', '0\n']
Note:
In the first example there is only one possible splitting which maximizes $sum_1$: $[1, 3, 1], [~], [1, 4]$.
In the second example the only way to have $sum_1=4$ is: $[1, 3], [2, 1], [4]$.
In the third example there is only one way to split the array: $[~], [4, 1, 2], [~]$.
|
```python
def call(l):
x,y=0,len(l)-1
n=len(l)
s1,s2=0,0
yo=0
while(x<=y):
if(s1==s2):
yo=s1
s1+=l[x]
x+=1
elif(s1>s2):
s2+=l[y]
y-=1
else:
s1+=l[x]
x+=1
#(x,y)=(x+1,y-1)
if(s1==s2): return s1
return yo
n=int(input())
l=list(map(int,input().split()))
n=len(l)
x=call(l)
print(x)
```
| 3
|
|
466
|
A
|
Cheap Travel
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
Ann has recently started commuting by subway. We know that a one ride subway ticket costs *a* rubles. Besides, Ann found out that she can buy a special ticket for *m* rides (she can buy it several times). It costs *b* rubles. Ann did the math; she will need to use subway *n* times. Help Ann, tell her what is the minimum sum of money she will have to spend to make *n* rides?
|
The single line contains four space-separated integers *n*, *m*, *a*, *b* (1<=≤<=*n*,<=*m*,<=*a*,<=*b*<=≤<=1000) — the number of rides Ann has planned, the number of rides covered by the *m* ride ticket, the price of a one ride ticket and the price of an *m* ride ticket.
|
Print a single integer — the minimum sum in rubles that Ann will need to spend.
|
[
"6 2 1 2\n",
"5 2 2 3\n"
] |
[
"6\n",
"8\n"
] |
In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three *m* ride tickets.
| 500
|
[
{
"input": "6 2 1 2",
"output": "6"
},
{
"input": "5 2 2 3",
"output": "8"
},
{
"input": "10 3 5 1",
"output": "4"
},
{
"input": "1000 1 1000 1000",
"output": "1000000"
},
{
"input": "1000 3 1000 1000",
"output": "334000"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "10 2 1 1",
"output": "5"
},
{
"input": "1 1000 1 2",
"output": "1"
},
{
"input": "1 1000 3 2",
"output": "2"
},
{
"input": "10 3 1 2",
"output": "7"
},
{
"input": "995 1 2 1",
"output": "995"
},
{
"input": "556 2 16 15",
"output": "4170"
},
{
"input": "477 2 16 14",
"output": "3346"
},
{
"input": "101 110 1 100",
"output": "100"
},
{
"input": "9 3 3 10",
"output": "27"
},
{
"input": "100 8 10 1",
"output": "13"
},
{
"input": "6 4 1 3",
"output": "5"
},
{
"input": "8 5 2 8",
"output": "14"
},
{
"input": "1000 2 1 1000",
"output": "1000"
}
] | 1,696,003,699
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 30
| 0
|
numbers = input()
numbers_list = numbers.split()
numbers_list_ = []
for num in numbers_list:
numbers_list_.append(int(num))
n,m,a,b = numbers_list_
if (b/m) < a and n/m >= 1:
if (n//m*b + (n-(n//m)*m)*a) < (n//m*b + b):
print (n//m*b + (n-(n//m)*m)*a)
else:
print(n//m*b + b)
elif (b/m) < a and n/m < 1:
print(b)
else:
print(n*a)
|
Title: Cheap Travel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ann has recently started commuting by subway. We know that a one ride subway ticket costs *a* rubles. Besides, Ann found out that she can buy a special ticket for *m* rides (she can buy it several times). It costs *b* rubles. Ann did the math; she will need to use subway *n* times. Help Ann, tell her what is the minimum sum of money she will have to spend to make *n* rides?
Input Specification:
The single line contains four space-separated integers *n*, *m*, *a*, *b* (1<=≤<=*n*,<=*m*,<=*a*,<=*b*<=≤<=1000) — the number of rides Ann has planned, the number of rides covered by the *m* ride ticket, the price of a one ride ticket and the price of an *m* ride ticket.
Output Specification:
Print a single integer — the minimum sum in rubles that Ann will need to spend.
Demo Input:
['6 2 1 2\n', '5 2 2 3\n']
Demo Output:
['6\n', '8\n']
Note:
In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three *m* ride tickets.
|
```python
numbers = input()
numbers_list = numbers.split()
numbers_list_ = []
for num in numbers_list:
numbers_list_.append(int(num))
n,m,a,b = numbers_list_
if (b/m) < a and n/m >= 1:
if (n//m*b + (n-(n//m)*m)*a) < (n//m*b + b):
print (n//m*b + (n-(n//m)*m)*a)
else:
print(n//m*b + b)
elif (b/m) < a and n/m < 1:
print(b)
else:
print(n*a)
```
| 0
|
|
825
|
C
|
Multi-judge Solving
|
PROGRAMMING
| 1,600
|
[
"greedy",
"implementation"
] | null | null |
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty *d* on Decoforces is as hard as the problem with difficulty *d* on any other judge).
Makes has chosen *n* problems to solve on Decoforces with difficulties *a*1,<=*a*2,<=...,<=*a**n*. He can solve these problems in arbitrary order. Though he can solve problem *i* with difficulty *a**i* only if he had already solved some problem with difficulty (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty *k*.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer *y* there exist some problem with difficulty *y* on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
|
The first line contains two integer numbers *n*, *k* (1<=≤<=*n*<=≤<=103, 1<=≤<=*k*<=≤<=109).
The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
|
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
|
[
"3 3\n2 1 9\n",
"4 20\n10 3 6 3\n"
] |
[
"1\n",
"0\n"
] |
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
| 0
|
[
{
"input": "3 3\n2 1 9",
"output": "1"
},
{
"input": "4 20\n10 3 6 3",
"output": "0"
},
{
"input": "1 1000000000\n1",
"output": "0"
},
{
"input": "1 1\n3",
"output": "1"
},
{
"input": "50 100\n74 55 33 5 83 24 75 59 30 36 13 4 62 28 96 17 6 35 45 53 33 11 37 93 34 79 61 72 13 31 44 75 7 3 63 46 18 16 44 89 62 25 32 12 38 55 75 56 61 82",
"output": "0"
},
{
"input": "100 10\n246 286 693 607 87 612 909 312 621 37 801 558 504 914 416 762 187 974 976 123 635 488 416 659 988 998 93 662 92 749 889 78 214 786 735 625 921 372 713 617 975 119 402 411 878 138 548 905 802 762 940 336 529 373 745 835 805 880 816 94 166 114 475 699 974 462 61 337 555 805 968 815 392 746 591 558 740 380 668 29 881 151 387 986 174 923 541 520 998 947 535 651 103 584 664 854 180 852 726 93",
"output": "1"
},
{
"input": "2 1\n1 1000000000",
"output": "29"
},
{
"input": "29 2\n1 3 7 15 31 63 127 255 511 1023 2047 4095 8191 16383 32767 65535 131071 262143 524287 1048575 2097151 4194303 8388607 16777215 33554431 67108863 134217727 268435455 536870911",
"output": "27"
},
{
"input": "1 1\n1000000000",
"output": "29"
},
{
"input": "7 6\n4 20 16 14 3 17 4",
"output": "1"
},
{
"input": "2 1\n3 6",
"output": "1"
},
{
"input": "1 1\n20",
"output": "4"
},
{
"input": "5 2\n86 81 53 25 18",
"output": "4"
},
{
"input": "4 1\n88 55 14 39",
"output": "4"
},
{
"input": "3 1\n2 3 6",
"output": "0"
},
{
"input": "3 2\n4 9 18",
"output": "1"
},
{
"input": "5 3\n6 6 6 13 27",
"output": "2"
},
{
"input": "5 1\n23 8 83 26 18",
"output": "4"
},
{
"input": "3 1\n4 5 6",
"output": "1"
},
{
"input": "3 1\n1 3 6",
"output": "1"
},
{
"input": "1 1\n2",
"output": "0"
},
{
"input": "3 2\n4 5 6",
"output": "0"
},
{
"input": "5 1\n100 200 400 1000 2000",
"output": "7"
},
{
"input": "2 1\n1 4",
"output": "1"
},
{
"input": "4 1\n2 4 8 32",
"output": "1"
},
{
"input": "2 10\n21 42",
"output": "1"
},
{
"input": "3 3\n1 7 13",
"output": "1"
},
{
"input": "3 1\n1 4 6",
"output": "1"
},
{
"input": "2 2\n2 8",
"output": "1"
},
{
"input": "1 1\n4",
"output": "1"
},
{
"input": "2 2\n8 16",
"output": "1"
},
{
"input": "3 1\n4 8 16",
"output": "1"
},
{
"input": "3 1\n3 6 9",
"output": "1"
},
{
"input": "2 1\n4 8",
"output": "1"
},
{
"input": "2 2\n7 14",
"output": "1"
},
{
"input": "1 4\n9",
"output": "1"
},
{
"input": "5 3\n1024 4096 16384 65536 536870913",
"output": "24"
},
{
"input": "2 5\n10 11",
"output": "0"
},
{
"input": "2 2\n3 6",
"output": "0"
},
{
"input": "2 2\n8 11",
"output": "1"
},
{
"input": "3 19905705\n263637263 417905394 108361057",
"output": "3"
},
{
"input": "4 25\n100 11 1 13",
"output": "1"
},
{
"input": "10 295206008\n67980321 440051990 883040288 135744260 96431758 242465794 576630162 972797687 356406646 547451696",
"output": "0"
},
{
"input": "4 2\n45 44 35 38",
"output": "4"
},
{
"input": "1 2\n9",
"output": "2"
},
{
"input": "3 6\n13 26 52",
"output": "1"
},
{
"input": "9 30111088\n824713578 11195876 458715185 731769293 680826358 189542586 550198537 860586039 101083021",
"output": "2"
},
{
"input": "3 72014068\n430005292 807436976 828082746",
"output": "2"
},
{
"input": "3 165219745\n737649884 652879952 506420386",
"output": "1"
},
{
"input": "2 60669400\n95037700 337255240",
"output": "1"
},
{
"input": "4 28\n34 1 86 90",
"output": "1"
},
{
"input": "2 1\n5 10",
"output": "2"
},
{
"input": "2 1\n4 1000000000",
"output": "28"
},
{
"input": "2 1\n2 3",
"output": "0"
},
{
"input": "2 1\n3 5",
"output": "1"
},
{
"input": "3 3\n1 5 20",
"output": "1"
},
{
"input": "9 1\n1 2 4 9 15 32 64 128 1024",
"output": "4"
}
] | 1,500,510,594
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 77
| 4,812,800
|
inp = input().split()
n = int(inp[0])
k = int(inp[1])
prob = [int(i) for i in input().split()]
solvable = k * 2
prob.sort()
ans = 0
for p in prob:
if p <= solvable:
solvable = max(solvable,p*2)
else:
ans += 1
solvable *= 2
print(ans)
|
Title: Multi-judge Solving
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty *d* on Decoforces is as hard as the problem with difficulty *d* on any other judge).
Makes has chosen *n* problems to solve on Decoforces with difficulties *a*1,<=*a*2,<=...,<=*a**n*. He can solve these problems in arbitrary order. Though he can solve problem *i* with difficulty *a**i* only if he had already solved some problem with difficulty (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty *k*.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer *y* there exist some problem with difficulty *y* on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input Specification:
The first line contains two integer numbers *n*, *k* (1<=≤<=*n*<=≤<=103, 1<=≤<=*k*<=≤<=109).
The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Demo Input:
['3 3\n2 1 9\n', '4 20\n10 3 6 3\n']
Demo Output:
['1\n', '0\n']
Note:
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
|
```python
inp = input().split()
n = int(inp[0])
k = int(inp[1])
prob = [int(i) for i in input().split()]
solvable = k * 2
prob.sort()
ans = 0
for p in prob:
if p <= solvable:
solvable = max(solvable,p*2)
else:
ans += 1
solvable *= 2
print(ans)
```
| 0
|
|
984
|
A
|
Game
|
PROGRAMMING
| 800
|
[
"sortings"
] | null | null |
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
|
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
|
Print one number that will be left on the board.
|
[
"3\n2 1 3\n",
"3\n2 2 2\n"
] |
[
"2",
"2"
] |
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players.
| 500
|
[
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 2 2",
"output": "2"
},
{
"input": "9\n44 53 51 80 5 27 74 79 94",
"output": "53"
},
{
"input": "10\n38 82 23 37 96 4 81 60 67 86",
"output": "60"
},
{
"input": "10\n58 26 77 15 53 81 68 48 22 65",
"output": "53"
},
{
"input": "1\n124",
"output": "124"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 1 1000",
"output": "1"
},
{
"input": "2\n322 322",
"output": "322"
},
{
"input": "3\n9 92 12",
"output": "12"
},
{
"input": "3\n1 2 2",
"output": "2"
}
] | 1,628,258,479
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 62
| 6,656,000
|
n=int(input())
arry=[int(a) for a in input().split()]
arry.sort()
ind=arry[int(n/2)] if len(arry)%2==0 else arry[int((n+1)/2)-1]
print(ind)
|
Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
Input Specification:
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
Output Specification:
Print one number that will be left on the board.
Demo Input:
['3\n2 1 3\n', '3\n2 2 2\n']
Demo Output:
['2', '2']
Note:
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players.
|
```python
n=int(input())
arry=[int(a) for a in input().split()]
arry.sort()
ind=arry[int(n/2)] if len(arry)%2==0 else arry[int((n+1)/2)-1]
print(ind)
```
| 0
|
|
479
|
A
|
Expression
|
PROGRAMMING
| 1,000
|
[
"brute force",
"math"
] | null | null |
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
|
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
|
Print the maximum value of the expression that you can obtain.
|
[
"1\n2\n3\n",
"2\n10\n3\n"
] |
[
"9\n",
"60\n"
] |
none
| 500
|
[
{
"input": "1\n2\n3",
"output": "9"
},
{
"input": "2\n10\n3",
"output": "60"
},
{
"input": "1\n1\n1",
"output": "3"
},
{
"input": "1\n2\n1",
"output": "4"
},
{
"input": "10\n10\n10",
"output": "1000"
},
{
"input": "5\n1\n3",
"output": "20"
},
{
"input": "3\n1\n5",
"output": "20"
},
{
"input": "6\n7\n1",
"output": "48"
},
{
"input": "1\n8\n3",
"output": "27"
},
{
"input": "9\n7\n2",
"output": "126"
},
{
"input": "1\n1\n10",
"output": "20"
},
{
"input": "9\n1\n1",
"output": "18"
},
{
"input": "10\n5\n6",
"output": "300"
},
{
"input": "8\n9\n7",
"output": "504"
},
{
"input": "4\n2\n10",
"output": "80"
},
{
"input": "2\n8\n3",
"output": "48"
},
{
"input": "3\n5\n7",
"output": "105"
},
{
"input": "1\n10\n1",
"output": "12"
},
{
"input": "2\n2\n2",
"output": "8"
},
{
"input": "5\n6\n1",
"output": "35"
},
{
"input": "10\n1\n1",
"output": "20"
},
{
"input": "1\n6\n1",
"output": "8"
},
{
"input": "1\n9\n1",
"output": "11"
},
{
"input": "2\n1\n2",
"output": "6"
},
{
"input": "2\n6\n1",
"output": "14"
},
{
"input": "9\n2\n1",
"output": "27"
},
{
"input": "1\n9\n2",
"output": "20"
},
{
"input": "1\n3\n1",
"output": "5"
},
{
"input": "2\n1\n1",
"output": "4"
}
] | 1,696,925,987
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 31
| 0
|
d = [int(input()),int(input()),int(input())]
d.sort()
if 0 < d.count(1) < 3 :
print((d[0] + d[1]) * d[2])
elif d.count(1) == 3:
print(3)
else:
print(d[0] * d[1] * d[2])
|
Title: Expression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
Input Specification:
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
Output Specification:
Print the maximum value of the expression that you can obtain.
Demo Input:
['1\n2\n3\n', '2\n10\n3\n']
Demo Output:
['9\n', '60\n']
Note:
none
|
```python
d = [int(input()),int(input()),int(input())]
d.sort()
if 0 < d.count(1) < 3 :
print((d[0] + d[1]) * d[2])
elif d.count(1) == 3:
print(3)
else:
print(d[0] * d[1] * d[2])
```
| 0
|
|
839
|
A
|
Arya and Bran
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are *n* days, at the *i*-th day, Arya finds *a**i* candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day. Formally, you need to output the minimum day index to the end of which *k* candies will be given out (the days are indexed from 1 to *n*).
Print -1 if she can't give him *k* candies during *n* given days.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=10000).
The second line contains *n* integers *a*1,<=*a*2,<=*a*3,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
|
If it is impossible for Arya to give Bran *k* candies within *n* days, print -1.
Otherwise print a single integer — the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day.
|
[
"2 3\n1 2\n",
"3 17\n10 10 10\n",
"1 9\n10\n"
] |
[
"2",
"3",
"-1"
] |
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.
| 500
|
[
{
"input": "2 3\n1 2",
"output": "2"
},
{
"input": "3 17\n10 10 10",
"output": "3"
},
{
"input": "1 9\n10",
"output": "-1"
},
{
"input": "10 70\n6 5 2 3 3 2 1 4 3 2",
"output": "-1"
},
{
"input": "20 140\n40 4 81 40 10 54 34 50 84 60 16 1 90 78 38 93 99 60 81 99",
"output": "18"
},
{
"input": "30 133\n3 2 3 4 3 7 4 5 5 6 7 2 1 3 4 6 7 4 6 4 7 5 7 1 3 4 1 6 8 5",
"output": "30"
},
{
"input": "40 320\n70 79 21 64 95 36 63 29 66 89 30 34 100 76 42 12 4 56 80 78 83 1 39 9 34 45 6 71 27 31 55 52 72 71 38 21 43 83 48 47",
"output": "40"
},
{
"input": "50 300\n5 3 11 8 7 4 9 5 5 1 6 3 5 7 4 2 2 10 8 1 7 10 4 4 11 5 2 4 9 1 5 4 11 9 11 2 7 4 4 8 10 9 1 11 10 2 4 11 6 9",
"output": "-1"
},
{
"input": "37 30\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "30"
},
{
"input": "100 456\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "57"
},
{
"input": "90 298\n94 90 98 94 93 90 99 98 90 96 93 96 92 92 97 98 94 94 96 100 93 96 95 98 94 91 95 95 94 90 93 96 93 100 99 98 94 95 98 91 91 98 97 100 98 93 92 93 91 100 92 97 95 95 97 94 98 97 99 100 90 96 93 100 95 99 92 100 99 91 97 99 98 93 90 93 97 95 94 96 90 100 94 93 91 92 97 97 97 100",
"output": "38"
},
{
"input": "7 43\n4 3 7 9 3 8 10",
"output": "-1"
},
{
"input": "99 585\n8 2 3 3 10 7 9 4 7 4 6 8 7 11 5 8 7 4 7 7 6 7 11 8 1 7 3 2 10 1 6 10 10 5 10 2 5 5 11 6 4 1 5 10 5 8 1 3 7 10 6 1 1 3 8 11 5 8 2 2 5 4 7 6 7 5 8 7 10 9 6 11 4 8 2 7 1 7 1 4 11 1 9 6 1 10 6 10 1 5 6 5 2 5 11 5 1 10 8",
"output": "-1"
},
{
"input": "30 177\n8 7 5 8 3 7 2 4 3 8 11 3 9 11 2 4 1 4 5 6 11 5 8 3 6 3 11 2 11 8",
"output": "-1"
},
{
"input": "19 129\n3 3 10 11 4 7 3 8 10 2 11 6 11 9 4 2 11 10 5",
"output": "-1"
},
{
"input": "100 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100"
},
{
"input": "13 104\n94 55 20 96 86 76 13 71 13 1 32 76 69",
"output": "13"
},
{
"input": "85 680\n61 44 55 6 30 74 27 26 17 45 73 1 67 71 39 32 13 25 79 66 4 59 49 28 29 22 10 17 98 80 36 99 52 24 59 44 27 79 29 46 29 12 47 72 82 25 6 30 81 72 95 65 30 71 72 45 39 16 16 89 48 42 59 71 50 58 31 65 91 70 48 56 28 34 53 89 94 98 49 55 94 65 91 11 53",
"output": "85"
},
{
"input": "100 458\n3 6 4 1 8 4 1 5 4 4 5 8 4 4 6 6 5 1 2 2 2 1 7 1 1 2 6 5 7 8 3 3 8 3 7 5 7 6 6 2 4 2 2 1 1 8 6 1 5 3 3 4 1 4 6 8 5 4 8 5 4 5 5 1 3 1 6 7 6 2 7 3 4 8 1 8 6 7 1 2 4 6 7 4 8 8 8 4 8 7 5 2 8 4 2 5 6 8 8 5",
"output": "100"
},
{
"input": "98 430\n4 7 6 3 4 1 7 1 1 6 6 1 5 4 6 1 5 4 6 6 1 5 1 1 8 1 6 6 2 6 8 4 4 6 6 8 8 7 4 1 2 4 1 5 4 3 7 3 2 5 7 7 7 2 2 2 7 2 8 7 3 4 5 7 8 3 7 6 7 3 2 4 7 1 4 4 7 1 1 8 4 5 8 3 1 5 3 5 2 1 3 3 8 1 3 5 8 6",
"output": "98"
},
{
"input": "90 80\n6 1 7 1 1 8 6 6 6 1 5 4 2 2 8 4 8 7 7 2 5 7 7 8 5 5 6 3 3 8 3 5 6 3 4 2 6 5 5 3 3 3 8 6 6 1 8 3 6 5 4 8 5 4 3 7 1 3 2 3 3 7 7 7 3 5 2 6 2 3 6 4 6 5 5 3 2 1 1 7 3 3 4 3 4 2 1 2 3 1",
"output": "18"
},
{
"input": "89 99\n7 7 3 5 2 7 8 8 1 1 5 7 7 4 1 5 3 4 4 8 8 3 3 2 6 3 8 2 7 5 8 1 3 5 3 6 4 3 6 2 3 3 4 5 1 6 1 7 7 7 6 7 7 7 8 8 8 2 1 7 5 8 6 7 7 4 7 5 7 8 1 3 5 8 7 1 4 2 5 8 3 4 4 5 5 6 2 4 2",
"output": "21"
},
{
"input": "50 700\n4 3 2 8 8 5 5 3 3 4 7 2 6 6 3 3 8 4 2 4 8 6 5 4 5 4 5 8 6 5 4 7 2 4 1 6 2 6 8 6 2 5 8 1 3 8 3 8 4 1",
"output": "-1"
},
{
"input": "82 359\n95 98 95 90 90 96 91 94 93 99 100 100 92 99 96 94 99 90 94 96 91 91 90 93 97 96 90 94 97 99 93 90 99 98 96 100 93 97 100 91 100 92 93 100 92 90 90 94 99 95 100 98 99 96 94 96 96 99 99 91 97 100 95 100 99 91 94 91 98 98 100 97 93 93 96 97 94 94 92 100 91 91",
"output": "45"
},
{
"input": "60 500\n93 93 100 99 91 92 95 93 95 99 93 91 97 98 90 91 98 100 95 100 94 93 92 91 91 98 98 90 93 91 90 96 92 93 92 94 94 91 96 94 98 100 97 96 96 97 91 99 97 95 96 94 91 92 99 95 97 92 98 90",
"output": "-1"
},
{
"input": "98 776\n48 63 26 3 88 81 27 33 37 10 2 89 41 84 98 93 25 44 42 90 41 65 97 1 28 69 42 14 86 18 96 28 28 94 78 8 44 31 96 45 26 52 93 25 48 39 3 75 94 93 63 59 67 86 18 74 27 38 68 7 31 60 69 67 20 11 19 34 47 43 86 96 3 49 56 60 35 49 89 28 92 69 48 15 17 73 99 69 2 73 27 35 28 53 11 1 96 50",
"output": "97"
},
{
"input": "100 189\n15 14 32 65 28 96 33 93 48 28 57 20 32 20 90 42 57 53 18 58 94 21 27 29 37 22 94 45 67 60 83 23 20 23 35 93 3 42 6 46 68 46 34 25 17 16 50 5 49 91 23 76 69 100 58 68 81 32 88 41 64 29 37 13 95 25 6 59 74 58 31 35 16 80 13 80 10 59 85 18 16 70 51 40 44 28 8 76 8 87 53 86 28 100 2 73 14 100 52 9",
"output": "24"
},
{
"input": "99 167\n72 4 79 73 49 58 15 13 92 92 42 36 35 21 13 10 51 94 64 35 86 50 6 80 93 77 59 71 2 88 22 10 27 30 87 12 77 6 34 56 31 67 78 84 36 27 15 15 12 56 80 7 56 14 10 9 14 59 15 20 34 81 8 49 51 72 4 58 38 77 31 86 18 61 27 86 95 36 46 36 39 18 78 39 48 37 71 12 51 92 65 48 39 22 16 87 4 5 42",
"output": "21"
},
{
"input": "90 4\n48 4 4 78 39 3 85 29 69 52 70 39 11 98 42 56 65 98 77 24 61 31 6 59 60 62 84 46 67 59 15 44 99 23 12 74 2 48 84 60 51 28 17 90 10 82 3 43 50 100 45 57 57 95 53 71 20 74 52 46 64 59 72 33 74 16 44 44 80 71 83 1 70 59 61 6 82 69 81 45 88 28 17 24 22 25 53 97 1 100",
"output": "1"
},
{
"input": "30 102\n55 94 3 96 3 47 92 85 25 78 27 70 97 83 40 2 55 12 74 84 91 37 31 85 7 40 33 54 72 5",
"output": "13"
},
{
"input": "81 108\n61 59 40 100 8 75 5 74 87 12 6 23 98 26 59 68 27 4 98 79 14 44 4 11 89 77 29 90 33 3 43 1 87 91 28 24 4 84 75 7 37 46 15 46 8 87 68 66 5 21 36 62 77 74 91 95 88 28 12 48 18 93 14 51 33 5 99 62 99 38 49 15 56 87 52 64 69 46 41 12 92",
"output": "14"
},
{
"input": "2 16\n10 6",
"output": "2"
},
{
"input": "2 8\n7 8",
"output": "2"
},
{
"input": "2 9\n4 8",
"output": "2"
},
{
"input": "3 19\n9 9 1",
"output": "3"
},
{
"input": "4 32\n9 9 9 5",
"output": "4"
},
{
"input": "2 15\n14 1",
"output": "2"
},
{
"input": "2 3\n3 3",
"output": "1"
},
{
"input": "3 10\n10 1 1",
"output": "2"
},
{
"input": "12 20\n3 16 19 10 1 6 17 8 6 20 1 4",
"output": "4"
},
{
"input": "4 15\n14 3 3 3",
"output": "2"
},
{
"input": "5 40\n10 10 10 10 1",
"output": "5"
},
{
"input": "4 31\n9 9 8 5",
"output": "4"
},
{
"input": "4 31\n20 7 1 1",
"output": "-1"
},
{
"input": "2 10\n9 1",
"output": "2"
},
{
"input": "10 50\n100 10 1 1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "2 11\n10 2",
"output": "2"
},
{
"input": "3 21\n10 10 1",
"output": "3"
},
{
"input": "2 2\n1 2",
"output": "2"
},
{
"input": "3 2\n1 8 8",
"output": "2"
},
{
"input": "2 11\n10 1",
"output": "2"
},
{
"input": "2 16\n12 4",
"output": "2"
},
{
"input": "3 11\n9 2 2",
"output": "2"
},
{
"input": "3 11\n4 3 4",
"output": "3"
},
{
"input": "2 13\n7 6",
"output": "2"
},
{
"input": "3 24\n14 3 4",
"output": "-1"
},
{
"input": "2 13\n10 3",
"output": "2"
},
{
"input": "3 11\n9 2 1",
"output": "2"
},
{
"input": "2 15\n12 3",
"output": "2"
},
{
"input": "2 14\n11 4",
"output": "2"
}
] | 1,579,960,357
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 93
| 307,200
|
# N, K
# Values for N days
N, K = map(int, input().split(" "))
Values = list(map(int, input().split(" ")))
Bran = 0
# print(N, K)
# print(Values)
idx = 0
for i, val in enumerate(Values):
if(Bran >= K):
idx = i
break
else:
if val <= 8:
Bran += val
if val > 8:
Bran += 8
idx = i
try :
Values[i + 1] += val - 8
except IndexError:
continue
# print(Bran)
if (Bran >= K):
print(idx + 1)
else:
print(-1)
|
Title: Arya and Bran
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are *n* days, at the *i*-th day, Arya finds *a**i* candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day. Formally, you need to output the minimum day index to the end of which *k* candies will be given out (the days are indexed from 1 to *n*).
Print -1 if she can't give him *k* candies during *n* given days.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=10000).
The second line contains *n* integers *a*1,<=*a*2,<=*a*3,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
If it is impossible for Arya to give Bran *k* candies within *n* days, print -1.
Otherwise print a single integer — the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day.
Demo Input:
['2 3\n1 2\n', '3 17\n10 10 10\n', '1 9\n10\n']
Demo Output:
['2', '3', '-1']
Note:
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.
|
```python
# N, K
# Values for N days
N, K = map(int, input().split(" "))
Values = list(map(int, input().split(" ")))
Bran = 0
# print(N, K)
# print(Values)
idx = 0
for i, val in enumerate(Values):
if(Bran >= K):
idx = i
break
else:
if val <= 8:
Bran += val
if val > 8:
Bran += 8
idx = i
try :
Values[i + 1] += val - 8
except IndexError:
continue
# print(Bran)
if (Bran >= K):
print(idx + 1)
else:
print(-1)
```
| 0
|
|
701
|
A
|
Cards
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation"
] | null | null |
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
|
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card.
|
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
|
[
"6\n1 5 7 4 4 3\n",
"4\n10 10 10 10\n"
] |
[
"1 3\n6 2\n4 5\n",
"1 2\n3 4\n"
] |
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable.
| 500
|
[
{
"input": "6\n1 5 7 4 4 3",
"output": "1 3\n6 2\n4 5"
},
{
"input": "4\n10 10 10 10",
"output": "1 4\n2 3"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "4\n82 46 8 44",
"output": "3 1\n4 2"
},
{
"input": "2\n35 50",
"output": "1 2"
},
{
"input": "8\n24 39 49 38 44 64 44 50",
"output": "1 6\n4 8\n2 3\n5 7"
},
{
"input": "100\n23 44 35 88 10 78 8 84 46 19 69 36 81 60 46 12 53 22 83 73 6 18 80 14 54 39 74 42 34 20 91 70 32 11 80 53 70 21 24 12 87 68 35 39 8 84 81 70 8 54 73 2 60 71 4 33 65 48 69 58 55 57 78 61 45 50 55 72 86 37 5 11 12 81 32 19 22 11 22 82 23 56 61 84 47 59 31 38 31 90 57 1 24 38 68 27 80 9 37 14",
"output": "92 31\n52 90\n55 4\n71 41\n21 69\n7 84\n45 46\n49 8\n98 19\n5 80\n34 74\n72 47\n78 13\n16 97\n40 35\n73 23\n24 63\n100 6\n22 27\n10 51\n76 20\n30 68\n38 54\n18 48\n77 37\n79 32\n1 59\n81 11\n39 95\n93 42\n96 57\n87 83\n89 64\n33 53\n75 14\n56 86\n29 60\n3 91\n43 62\n12 82\n70 67\n99 61\n88 50\n94 25\n26 36\n44 17\n28 66\n2 58\n65 85\n9 15"
},
{
"input": "12\n22 83 2 67 55 12 40 93 83 73 12 28",
"output": "3 8\n6 9\n11 2\n1 10\n12 4\n7 5"
},
{
"input": "16\n10 33 36 32 48 25 31 27 45 13 37 26 22 21 15 43",
"output": "1 5\n10 9\n15 16\n14 11\n13 3\n6 2\n12 4\n8 7"
},
{
"input": "20\n18 13 71 60 28 10 20 65 65 12 13 14 64 68 6 50 72 7 66 58",
"output": "15 17\n18 3\n6 14\n10 19\n2 9\n11 8\n12 13\n1 4\n7 20\n5 16"
},
{
"input": "24\n59 39 25 22 46 21 24 70 60 11 46 42 44 37 13 37 41 58 72 23 25 61 58 62",
"output": "10 19\n15 8\n6 24\n4 22\n20 9\n7 1\n3 23\n21 18\n14 11\n16 5\n2 13\n17 12"
},
{
"input": "28\n22 1 51 31 83 35 3 64 59 10 61 25 19 53 55 80 78 8 82 22 67 4 27 64 33 6 85 76",
"output": "2 27\n7 5\n22 19\n26 16\n18 17\n10 28\n13 21\n1 24\n20 8\n12 11\n23 9\n4 15\n25 14\n6 3"
},
{
"input": "32\n41 42 22 68 40 52 66 16 73 25 41 21 36 60 46 30 24 55 35 10 54 52 70 24 20 56 3 34 35 6 51 8",
"output": "27 9\n30 23\n32 4\n20 7\n8 14\n25 26\n12 18\n3 21\n17 22\n24 6\n10 31\n16 15\n28 2\n19 11\n29 1\n13 5"
},
{
"input": "36\n1 10 61 43 27 49 55 33 7 30 45 78 69 34 38 19 36 49 55 11 30 63 46 24 16 68 71 18 11 52 72 24 60 68 8 41",
"output": "1 12\n9 31\n35 27\n2 13\n20 34\n29 26\n25 22\n28 3\n16 33\n24 19\n32 7\n5 30\n10 18\n21 6\n8 23\n14 11\n17 4\n15 36"
},
{
"input": "40\n7 30 13 37 37 56 45 28 61 28 23 33 44 63 58 52 21 2 42 19 10 32 9 7 61 15 58 20 45 4 46 24 35 17 50 4 20 48 41 55",
"output": "18 14\n30 25\n36 9\n1 27\n24 15\n23 6\n21 40\n3 16\n26 35\n34 38\n20 31\n28 29\n37 7\n17 13\n11 19\n32 39\n8 5\n10 4\n2 33\n22 12"
},
{
"input": "44\n7 12 46 78 24 68 86 22 71 79 85 14 58 72 26 46 54 39 35 13 31 45 81 21 15 8 47 64 69 87 57 6 18 80 47 29 36 62 34 67 59 48 75 25",
"output": "32 30\n1 7\n26 11\n2 23\n20 34\n12 10\n25 4\n33 43\n24 14\n8 9\n5 29\n44 6\n15 40\n36 28\n21 38\n39 41\n19 13\n37 31\n18 17\n22 42\n3 35\n16 27"
},
{
"input": "48\n57 38 16 25 34 57 29 38 60 51 72 78 22 39 10 33 20 16 12 3 51 74 9 88 4 70 56 65 86 18 33 12 77 78 52 87 68 85 81 5 61 2 52 39 80 13 74 30",
"output": "42 24\n20 36\n25 29\n40 38\n23 39\n15 45\n19 34\n32 12\n46 33\n3 47\n18 22\n30 11\n17 26\n13 37\n4 28\n7 41\n48 9\n16 6\n31 1\n5 27\n2 43\n8 35\n14 21\n44 10"
},
{
"input": "52\n57 12 13 40 68 31 18 4 31 18 65 3 62 32 6 3 49 48 51 33 53 40 9 32 47 53 58 19 14 23 32 38 39 69 19 20 62 52 68 17 39 22 54 59 3 2 52 9 67 68 24 39",
"output": "46 34\n12 50\n16 39\n45 5\n8 49\n15 11\n23 37\n48 13\n2 44\n3 27\n29 1\n40 43\n7 26\n10 21\n28 47\n35 38\n36 19\n42 17\n30 18\n51 25\n6 22\n9 4\n14 52\n24 41\n31 33\n20 32"
},
{
"input": "56\n53 59 66 68 71 25 48 32 12 61 72 69 30 6 56 55 25 49 60 47 46 46 66 19 31 9 23 15 10 12 71 53 51 32 39 31 66 66 17 52 12 7 7 22 49 12 71 29 63 7 47 29 18 39 27 26",
"output": "14 11\n42 47\n43 31\n50 5\n26 12\n29 4\n9 38\n30 37\n41 23\n46 3\n28 49\n39 10\n53 19\n24 2\n44 15\n27 16\n6 32\n17 1\n56 40\n55 33\n48 45\n52 18\n13 7\n25 51\n36 20\n8 22\n34 21\n35 54"
},
{
"input": "60\n47 63 20 68 46 12 45 44 14 38 28 73 60 5 20 18 70 64 37 47 26 47 37 61 29 61 23 28 30 68 55 22 25 60 38 7 63 12 38 15 14 30 11 5 70 15 53 52 7 57 49 45 55 37 45 28 50 2 31 30",
"output": "58 12\n14 45\n44 17\n36 30\n49 4\n43 18\n6 37\n38 2\n9 26\n41 24\n40 34\n46 13\n16 50\n3 53\n15 31\n32 47\n27 48\n33 57\n21 51\n11 22\n28 20\n56 1\n25 5\n29 55\n42 52\n60 7\n59 8\n19 39\n23 35\n54 10"
},
{
"input": "64\n63 39 19 5 48 56 49 45 29 68 25 59 37 69 62 26 60 44 60 6 67 68 2 40 56 6 19 12 17 70 23 11 59 37 41 55 30 68 72 14 38 34 3 71 2 4 55 15 31 66 15 51 36 72 18 7 6 14 43 33 8 35 57 18",
"output": "23 54\n45 39\n43 44\n46 30\n4 14\n20 38\n26 22\n57 10\n56 21\n61 50\n32 1\n28 15\n40 19\n58 17\n48 33\n51 12\n29 63\n55 25\n64 6\n3 47\n27 36\n31 52\n11 7\n16 5\n9 8\n37 18\n49 59\n60 35\n42 24\n62 2\n53 41\n13 34"
},
{
"input": "68\n58 68 40 55 62 15 10 54 19 18 69 27 15 53 8 18 8 33 15 49 20 9 70 8 18 64 14 59 9 64 3 35 46 11 5 65 58 55 28 58 4 55 64 5 68 24 4 58 23 45 58 50 38 68 5 15 20 9 5 53 20 63 69 68 15 53 65 65",
"output": "31 23\n41 63\n47 11\n35 64\n44 54\n55 45\n59 2\n15 68\n17 67\n24 36\n22 43\n29 30\n58 26\n7 62\n34 5\n27 28\n6 51\n13 48\n19 40\n56 37\n65 1\n10 42\n16 38\n25 4\n9 8\n21 66\n57 60\n61 14\n49 52\n46 20\n12 33\n39 50\n18 3\n32 53"
},
{
"input": "72\n61 13 55 23 24 55 44 33 59 19 14 17 66 40 27 33 29 37 28 74 50 56 59 65 64 17 42 56 73 51 64 23 22 26 38 22 36 47 60 14 52 28 14 12 6 41 73 5 64 67 61 74 54 34 45 34 44 4 34 49 18 72 44 47 31 19 11 31 5 4 45 50",
"output": "58 52\n70 20\n48 47\n69 29\n45 62\n67 50\n44 13\n2 24\n11 49\n40 31\n43 25\n12 51\n26 1\n61 39\n10 23\n66 9\n33 28\n36 22\n4 6\n32 3\n5 53\n34 41\n15 30\n19 72\n42 21\n17 60\n65 64\n68 38\n8 71\n16 55\n54 63\n56 57\n59 7\n37 27\n18 46\n35 14"
},
{
"input": "76\n73 37 73 67 26 45 43 74 47 31 43 81 4 3 39 79 48 81 67 39 67 66 43 67 80 51 34 79 5 58 45 10 39 50 9 78 6 18 75 17 45 17 51 71 34 53 33 11 17 15 11 69 50 41 13 74 10 33 77 41 11 64 36 74 17 32 3 10 27 20 5 73 52 41 7 57",
"output": "14 18\n67 12\n13 25\n29 28\n71 16\n37 36\n75 59\n35 39\n32 64\n57 56\n68 8\n48 72\n51 3\n61 1\n55 44\n50 52\n40 24\n42 21\n49 19\n65 4\n38 22\n70 62\n5 30\n69 76\n10 46\n66 73\n47 43\n58 26\n27 53\n45 34\n63 17\n2 9\n15 41\n20 31\n33 6\n54 23\n60 11\n74 7"
},
{
"input": "80\n18 38 65 1 20 9 57 2 36 26 15 17 33 61 65 27 10 35 49 42 40 32 19 33 12 36 56 31 10 41 8 54 56 60 5 47 61 43 23 19 20 30 7 6 38 60 29 58 35 64 30 51 6 17 30 24 47 1 37 47 34 36 48 28 5 25 47 19 30 39 36 23 31 28 46 46 59 43 19 49",
"output": "4 15\n58 3\n8 50\n35 37\n65 14\n44 46\n53 34\n43 77\n31 48\n6 7\n17 33\n29 27\n25 32\n11 52\n12 80\n54 19\n1 63\n23 67\n40 60\n68 57\n79 36\n5 76\n41 75\n39 78\n72 38\n56 20\n66 30\n10 21\n16 70\n64 45\n74 2\n47 59\n42 71\n51 62\n55 26\n69 9\n28 49\n73 18\n22 61\n13 24"
},
{
"input": "84\n59 41 54 14 42 55 29 28 41 73 40 15 1 1 66 49 76 59 68 60 42 81 19 23 33 12 80 81 42 22 54 54 2 22 22 28 27 60 36 57 17 76 38 20 40 65 23 9 81 50 25 13 46 36 59 53 6 35 47 40 59 19 67 46 63 49 12 33 23 49 33 23 32 62 60 70 44 1 6 63 28 16 70 69",
"output": "13 49\n14 28\n78 22\n33 27\n57 42\n79 17\n48 10\n26 83\n67 76\n52 84\n4 19\n12 63\n82 15\n41 46\n23 80\n62 65\n44 74\n30 75\n34 38\n35 20\n24 61\n47 55\n69 18\n72 1\n51 40\n37 6\n8 32\n36 31\n81 3\n7 56\n73 50\n25 70\n68 66\n71 16\n58 59\n39 64\n54 53\n43 77\n11 29\n45 21\n60 5\n2 9"
},
{
"input": "88\n10 28 71 6 58 66 45 52 13 71 39 1 10 29 30 70 14 17 15 38 4 60 5 46 66 41 40 58 2 57 32 44 21 26 13 40 64 63 56 33 46 8 30 43 67 55 44 28 32 62 14 58 42 67 45 59 32 68 10 31 51 6 42 34 9 12 51 27 20 14 62 42 16 5 1 14 30 62 40 59 58 26 25 15 27 47 21 57",
"output": "12 10\n75 3\n29 16\n21 58\n23 54\n74 45\n4 25\n62 6\n42 37\n65 38\n1 78\n13 71\n59 50\n66 22\n9 80\n35 56\n17 81\n51 52\n70 28\n76 5\n19 88\n84 30\n73 39\n18 46\n69 8\n33 67\n87 61\n83 86\n34 41\n82 24\n68 55\n85 7\n2 47\n48 32\n14 44\n15 72\n43 63\n77 53\n60 26\n31 79\n49 36\n57 27\n40 11\n64 20"
},
{
"input": "92\n17 37 81 15 29 70 73 42 49 23 44 77 27 44 74 11 43 66 15 41 60 36 33 11 2 76 16 51 45 21 46 16 85 29 76 79 16 6 60 13 25 44 62 28 43 35 63 24 76 71 62 15 57 72 45 10 71 59 74 14 53 13 58 72 14 72 73 11 25 1 57 42 86 63 50 30 64 38 10 77 75 24 58 8 54 12 43 30 27 71 52 34",
"output": "70 73\n25 33\n38 3\n84 36\n56 80\n79 12\n16 49\n24 35\n68 26\n86 81\n40 59\n62 15\n60 67\n65 7\n4 66\n19 64\n52 54\n27 90\n32 57\n37 50\n1 6\n30 18\n10 77\n48 74\n82 47\n41 51\n69 43\n13 39\n89 21\n44 58\n5 83\n34 63\n76 71\n88 53\n23 85\n92 61\n46 91\n22 28\n2 75\n78 9\n20 31\n8 55\n72 29\n17 42\n45 14\n87 11"
},
{
"input": "96\n77 7 47 19 73 31 46 13 89 69 52 9 26 77 6 87 55 45 71 2 79 1 80 20 4 82 64 20 75 86 84 24 77 56 16 54 53 35 74 73 40 29 63 20 83 39 58 16 31 41 40 16 11 90 30 48 62 39 55 8 50 3 77 73 75 66 14 90 18 54 38 10 53 22 67 38 27 91 62 37 85 13 92 7 18 83 10 3 86 54 80 59 34 16 39 43",
"output": "22 83\n20 78\n62 68\n88 54\n25 9\n15 16\n2 89\n84 30\n60 81\n12 31\n72 86\n87 45\n53 26\n8 91\n82 23\n67 21\n35 63\n48 33\n52 14\n94 1\n69 65\n85 29\n4 39\n24 64\n28 40\n44 5\n74 19\n32 10\n13 75\n77 66\n42 27\n55 43\n6 79\n49 57\n93 92\n38 47\n80 34\n71 59\n76 17\n46 90\n58 70\n95 36\n41 73\n51 37\n50 11\n96 61\n18 56\n7 3"
},
{
"input": "4\n100 100 1 1",
"output": "3 2\n4 1"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "4\n3 4 4 5",
"output": "1 4\n2 3"
},
{
"input": "4\n1 1 2 2",
"output": "1 4\n2 3"
},
{
"input": "4\n1 2 3 4",
"output": "1 4\n2 3"
}
] | 1,550,034,630
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 109
| 0
|
def main():
n = int(input())
nums = list(map(int,input().split()))
cards = []
for i in range(n):
cards.append((nums[i],i+1))
cards = sorted(cards,key = lambda x: x[0])
ans = []
#print (cards)
for i in range(int(n/2)):
ans.append((cards[i][1],cards[n-i-1][1]))
for i in ans:
print (i[0],i[1])
main()
|
Title: Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card.
Output Specification:
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
Demo Input:
['6\n1 5 7 4 4 3\n', '4\n10 10 10 10\n']
Demo Output:
['1 3\n6 2\n4 5\n', '1 2\n3 4\n']
Note:
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable.
|
```python
def main():
n = int(input())
nums = list(map(int,input().split()))
cards = []
for i in range(n):
cards.append((nums[i],i+1))
cards = sorted(cards,key = lambda x: x[0])
ans = []
#print (cards)
for i in range(int(n/2)):
ans.append((cards[i][1],cards[n-i-1][1]))
for i in ans:
print (i[0],i[1])
main()
```
| 3
|
|
466
|
C
|
Number of Ways
|
PROGRAMMING
| 1,700
|
[
"binary search",
"brute force",
"data structures",
"dp",
"two pointers"
] | null | null |
You've got array *a*[1],<=*a*[2],<=...,<=*a*[*n*], consisting of *n* integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices *i*,<=*j* (2<=≤<=*i*<=≤<=*j*<=≤<=*n*<=-<=1), that .
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=5·105), showing how many numbers are in the array. The second line contains *n* integers *a*[1], *a*[2], ..., *a*[*n*] (|*a*[*i*]|<=≤<=<=109) — the elements of array *a*.
|
Print a single integer — the number of ways to split the array into three parts with the same sum.
|
[
"5\n1 2 3 0 3\n",
"4\n0 1 -1 0\n",
"2\n4 1\n"
] |
[
"2\n",
"1\n",
"0\n"
] |
none
| 1,500
|
[
{
"input": "5\n1 2 3 0 3",
"output": "2"
},
{
"input": "4\n0 1 -1 0",
"output": "1"
},
{
"input": "2\n4 1",
"output": "0"
},
{
"input": "9\n0 0 0 0 0 0 0 0 0",
"output": "28"
},
{
"input": "10\n2 5 -2 2 -3 -2 3 5 -5 -2",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "0"
},
{
"input": "100\n1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "2030"
},
{
"input": "6\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "10\n1 0 0 0 1 1 1 0 1 1",
"output": "2"
},
{
"input": "10\n-2 2 -2 0 -2 -1 1 -1 2 0",
"output": "0"
},
{
"input": "4\n0 2 -1 2",
"output": "0"
},
{
"input": "5\n3 3 -3 3 3",
"output": "3"
},
{
"input": "5\n1 1 1 1 1",
"output": "0"
},
{
"input": "8\n-1 -1 -1 -1 -1 -1 -1 -1",
"output": "0"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "4\n1 -1 0 0",
"output": "1"
},
{
"input": "3\n6 -3 6",
"output": "0"
}
] | 1,699,627,932
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 15
| 0
|
# -*- coding: utf-8 -*-
"""
Created on Fri Nov 10 22:17:24 2023
@author: HXY
"""
n=int(input())
list0=list(map(int,input().split()))
a=sum(list0)
if n<=2 :
print(0)
if a%3!=0:
print(0)
else:
x=a/3
for i in range(1,n):
list0[i]+=list0[i-1]
re=0
for i in range(n-1):
if list0[i]==x :
for j in range(i+1,n):
if list0[j]==x:
re+=1
print(re)
|
Title: Number of Ways
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got array *a*[1],<=*a*[2],<=...,<=*a*[*n*], consisting of *n* integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices *i*,<=*j* (2<=≤<=*i*<=≤<=*j*<=≤<=*n*<=-<=1), that .
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5·105), showing how many numbers are in the array. The second line contains *n* integers *a*[1], *a*[2], ..., *a*[*n*] (|*a*[*i*]|<=≤<=<=109) — the elements of array *a*.
Output Specification:
Print a single integer — the number of ways to split the array into three parts with the same sum.
Demo Input:
['5\n1 2 3 0 3\n', '4\n0 1 -1 0\n', '2\n4 1\n']
Demo Output:
['2\n', '1\n', '0\n']
Note:
none
|
```python
# -*- coding: utf-8 -*-
"""
Created on Fri Nov 10 22:17:24 2023
@author: HXY
"""
n=int(input())
list0=list(map(int,input().split()))
a=sum(list0)
if n<=2 :
print(0)
if a%3!=0:
print(0)
else:
x=a/3
for i in range(1,n):
list0[i]+=list0[i-1]
re=0
for i in range(n-1):
if list0[i]==x :
for j in range(i+1,n):
if list0[j]==x:
re+=1
print(re)
```
| 0
|
|
145
|
A
|
Lucky Conversion
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation"
] | null | null |
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings *a* and *b* of the same length *n*. The strings consist only of lucky digits. Petya can perform operations of two types:
- replace any one digit from string *a* by its opposite (i.e., replace 4 by 7 and 7 by 4); - swap any pair of digits in string *a*.
Petya is interested in the minimum number of operations that are needed to make string *a* equal to string *b*. Help him with the task.
|
The first and the second line contains strings *a* and *b*, correspondingly. Strings *a* and *b* have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
|
Print on the single line the single number — the minimum number of operations needed to convert string *a* into string *b*.
|
[
"47\n74\n",
"774\n744\n",
"777\n444\n"
] |
[
"1\n",
"1\n",
"3\n"
] |
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
| 500
|
[
{
"input": "47\n74",
"output": "1"
},
{
"input": "774\n744",
"output": "1"
},
{
"input": "777\n444",
"output": "3"
},
{
"input": "74747474\n77777777",
"output": "4"
},
{
"input": "444444444444\n777777777777",
"output": "12"
},
{
"input": "4744744447774474447474774\n4477774777444444444777447",
"output": "8"
},
{
"input": "7\n4",
"output": "1"
},
{
"input": "4\n7",
"output": "1"
},
{
"input": "7777777777\n7777777774",
"output": "1"
},
{
"input": "47777777777\n77777777774",
"output": "1"
},
{
"input": "47747477747744447774774444444777444747474747777774\n44777444774477447777444774477777477774444477447777",
"output": "14"
},
{
"input": "44447777447744444777777747477444777444447744444\n47444747774774744474747744447744477747777777447",
"output": "13"
},
{
"input": "4447744774744774744747744774474474444447477477444747477444\n7477477444744774744744774774744474744447744774744477744477",
"output": "14"
},
{
"input": "44747744777777444\n47774747747744777",
"output": "6"
},
{
"input": "44447774444474477747774774477777474774744744477444447777477477744747477774744444744777777777747777477447744774744444747477744744\n77777474477477747774777777474474477444474777477747747777477747747744474474747774747747444777474444744744444477477777747744747477",
"output": "37"
},
{
"input": "774774747744474477447477777447477747477474777477744744747444774474477477747474477447774444774744777\n744477444747477447477777774477447444447747477747477747774477474447474477477474444777444444447474747",
"output": "27"
},
{
"input": "4747447477\n4747444744",
"output": "3"
},
{
"input": "47744447444\n74477447744",
"output": "4"
},
{
"input": "447444777744\n777747744477",
"output": "6"
},
{
"input": "474777477774444\n774747777774477",
"output": "4"
},
{
"input": "47744474447747744777777447\n44744747477474777744777477",
"output": "7"
},
{
"input": "77447447444777777744744747744747774747477774777774447447777474477477774774777\n74777777444744447447474474477747747444444447447774444444747777444747474777447",
"output": "28"
},
{
"input": "7\n7",
"output": "0"
},
{
"input": "444\n444",
"output": "0"
},
{
"input": "77747\n47474",
"output": "3"
}
] | 1,666,277,780
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 51
| 124
| 2,048,000
|
ch = 0
s = 0
a = input()
b = input()
for i in range(len(a)):
if a[i] != b[i]:
if b[i] == "7":
s += 1
elif b[i] == "4":
ch += 1
print(max(s, ch))
|
Title: Lucky Conversion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings *a* and *b* of the same length *n*. The strings consist only of lucky digits. Petya can perform operations of two types:
- replace any one digit from string *a* by its opposite (i.e., replace 4 by 7 and 7 by 4); - swap any pair of digits in string *a*.
Petya is interested in the minimum number of operations that are needed to make string *a* equal to string *b*. Help him with the task.
Input Specification:
The first and the second line contains strings *a* and *b*, correspondingly. Strings *a* and *b* have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output Specification:
Print on the single line the single number — the minimum number of operations needed to convert string *a* into string *b*.
Demo Input:
['47\n74\n', '774\n744\n', '777\n444\n']
Demo Output:
['1\n', '1\n', '3\n']
Note:
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
|
```python
ch = 0
s = 0
a = input()
b = input()
for i in range(len(a)):
if a[i] != b[i]:
if b[i] == "7":
s += 1
elif b[i] == "4":
ch += 1
print(max(s, ch))
```
| 3
|
|
80
|
A
|
Panoramix's Prediction
|
PROGRAMMING
| 800
|
[
"brute force"
] |
A. Panoramix's Prediction
|
2
|
256
|
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
|
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
|
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
|
[
"3 5\n",
"7 11\n",
"7 9\n"
] |
[
"YES",
"YES",
"NO"
] |
none
| 500
|
[
{
"input": "3 5",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "2 3",
"output": "YES"
},
{
"input": "2 4",
"output": "NO"
},
{
"input": "3 4",
"output": "NO"
},
{
"input": "3 5",
"output": "YES"
},
{
"input": "5 7",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "11 13",
"output": "YES"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "17 19",
"output": "YES"
},
{
"input": "19 23",
"output": "YES"
},
{
"input": "23 29",
"output": "YES"
},
{
"input": "29 31",
"output": "YES"
},
{
"input": "31 37",
"output": "YES"
},
{
"input": "37 41",
"output": "YES"
},
{
"input": "41 43",
"output": "YES"
},
{
"input": "43 47",
"output": "YES"
},
{
"input": "2 5",
"output": "NO"
},
{
"input": "2 7",
"output": "NO"
},
{
"input": "2 6",
"output": "NO"
},
{
"input": "2 11",
"output": "NO"
},
{
"input": "3 6",
"output": "NO"
},
{
"input": "3 7",
"output": "NO"
},
{
"input": "3 9",
"output": "NO"
},
{
"input": "5 6",
"output": "NO"
},
{
"input": "5 9",
"output": "NO"
},
{
"input": "5 11",
"output": "NO"
},
{
"input": "5 13",
"output": "NO"
},
{
"input": "5 15",
"output": "NO"
},
{
"input": "7 8",
"output": "NO"
},
{
"input": "7 13",
"output": "NO"
},
{
"input": "13 15",
"output": "NO"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "19 21",
"output": "NO"
},
{
"input": "13 20",
"output": "NO"
},
{
"input": "41 49",
"output": "NO"
},
{
"input": "43 49",
"output": "NO"
},
{
"input": "47 50",
"output": "NO"
},
{
"input": "47 49",
"output": "NO"
},
{
"input": "47 48",
"output": "NO"
},
{
"input": "23 25",
"output": "NO"
},
{
"input": "2 50",
"output": "NO"
},
{
"input": "31 33",
"output": "NO"
}
] | 1,684,357,125
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 39
| 92
| 0
|
prime = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
def pair(a, b):
if a in prime:
i = prime.index(a)
if prime[i + 1] == b:
return "YES"
else:
return "NO"
a, b = input().split()
a = int(a)
b = int(b)
print(pair(a, b))
|
Title: Panoramix's Prediction
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
Input Specification:
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
Output Specification:
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
Demo Input:
['3 5\n', '7 11\n', '7 9\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
none
|
```python
prime = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
def pair(a, b):
if a in prime:
i = prime.index(a)
if prime[i + 1] == b:
return "YES"
else:
return "NO"
a, b = input().split()
a = int(a)
b = int(b)
print(pair(a, b))
```
| -1
|
902
|
A
|
Visiting a Friend
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation"
] | null | null |
Pig is visiting a friend.
Pig's house is located at point 0, and his friend's house is located at point *m* on an axis.
Pig can use teleports to move along the axis.
To use a teleport, Pig should come to a certain point (where the teleport is located) and choose where to move: for each teleport there is the rightmost point it can move Pig to, this point is known as the limit of the teleport.
Formally, a teleport located at point *x* with limit *y* can move Pig from point *x* to any point within the segment [*x*;<=*y*], including the bounds.
Determine if Pig can visit the friend using teleports only, or he should use his car.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100) — the number of teleports and the location of the friend's house.
The next *n* lines contain information about teleports.
The *i*-th of these lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=*b**i*<=≤<=*m*), where *a**i* is the location of the *i*-th teleport, and *b**i* is its limit.
It is guaranteed that *a**i*<=≥<=*a**i*<=-<=1 for every *i* (2<=≤<=*i*<=≤<=*n*).
|
Print "YES" if there is a path from Pig's house to his friend's house that uses only teleports, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
|
[
"3 5\n0 2\n2 4\n3 5\n",
"3 7\n0 4\n2 5\n6 7\n"
] |
[
"YES\n",
"NO\n"
] |
The first example is shown on the picture below:
Pig can use the first teleport from his house (point 0) to reach point 2, then using the second teleport go from point 2 to point 3, then using the third teleport go from point 3 to point 5, where his friend lives.
The second example is shown on the picture below:
You can see that there is no path from Pig's house to his friend's house that uses only teleports.
| 500
|
[
{
"input": "3 5\n0 2\n2 4\n3 5",
"output": "YES"
},
{
"input": "3 7\n0 4\n2 5\n6 7",
"output": "NO"
},
{
"input": "1 1\n0 0",
"output": "NO"
},
{
"input": "30 10\n0 7\n1 2\n1 2\n1 4\n1 4\n1 3\n2 2\n2 4\n2 6\n2 9\n2 2\n3 5\n3 8\n4 8\n4 5\n4 6\n5 6\n5 7\n6 6\n6 9\n6 7\n6 9\n7 7\n7 7\n8 8\n8 8\n9 9\n9 9\n10 10\n10 10",
"output": "NO"
},
{
"input": "30 100\n0 27\n4 82\n11 81\n14 32\n33 97\n33 34\n37 97\n38 52\n45 91\n49 56\n50 97\n57 70\n59 94\n59 65\n62 76\n64 65\n65 95\n67 77\n68 100\n71 73\n80 94\n81 92\n84 85\n85 100\n88 91\n91 95\n92 98\n92 98\n99 100\n100 100",
"output": "YES"
},
{
"input": "70 10\n0 4\n0 4\n0 8\n0 9\n0 1\n0 5\n0 7\n1 3\n1 8\n1 8\n1 6\n1 6\n1 2\n1 3\n1 2\n1 3\n2 5\n2 4\n2 3\n2 4\n2 6\n2 2\n2 5\n2 7\n3 7\n3 4\n3 7\n3 4\n3 8\n3 4\n3 9\n3 3\n3 7\n3 9\n3 3\n3 9\n4 6\n4 7\n4 5\n4 7\n5 8\n5 5\n5 9\n5 7\n5 5\n6 6\n6 9\n6 7\n6 8\n6 9\n6 8\n7 7\n7 8\n7 7\n7 8\n8 9\n8 8\n8 9\n8 8\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "NO"
},
{
"input": "30 10\n0 7\n1 2\n1 2\n1 4\n1 4\n1 3\n2 2\n2 4\n2 6\n2 9\n2 2\n3 5\n3 8\n4 8\n4 5\n4 6\n5 6\n5 7\n6 6\n6 9\n6 7\n6 9\n7 7\n7 7\n8 10\n8 10\n9 9\n9 9\n10 10\n10 10",
"output": "YES"
},
{
"input": "50 100\n0 95\n1 100\n1 38\n2 82\n5 35\n7 71\n8 53\n11 49\n15 27\n17 84\n17 75\n18 99\n18 43\n18 69\n21 89\n27 60\n27 29\n38 62\n38 77\n39 83\n40 66\n48 80\n48 100\n50 51\n50 61\n53 77\n53 63\n55 58\n56 68\n60 82\n62 95\n66 74\n67 83\n69 88\n69 81\n69 88\n69 98\n70 91\n70 76\n71 90\n72 99\n81 99\n85 87\n88 97\n88 93\n90 97\n90 97\n92 98\n98 99\n100 100",
"output": "YES"
},
{
"input": "70 10\n0 4\n0 4\n0 8\n0 9\n0 1\n0 5\n0 7\n1 3\n1 8\n1 8\n1 10\n1 9\n1 6\n1 2\n1 3\n1 2\n2 6\n2 5\n2 4\n2 3\n2 10\n2 2\n2 6\n2 2\n3 10\n3 7\n3 7\n3 4\n3 7\n3 4\n3 8\n3 4\n3 10\n3 5\n3 3\n3 7\n4 8\n4 8\n4 9\n4 6\n5 7\n5 10\n5 7\n5 8\n5 5\n6 8\n6 9\n6 10\n6 6\n6 9\n6 7\n7 8\n7 9\n7 10\n7 10\n8 8\n8 8\n8 9\n8 10\n9 10\n9 9\n9 10\n9 10\n9 9\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "YES"
},
{
"input": "85 10\n0 9\n0 4\n0 2\n0 5\n0 1\n0 8\n0 7\n1 2\n1 4\n1 5\n1 9\n1 1\n1 6\n1 6\n2 5\n2 7\n2 7\n2 7\n2 7\n3 4\n3 7\n3 9\n3 5\n3 3\n4 4\n4 6\n4 5\n5 6\n5 6\n5 6\n5 6\n5 7\n5 8\n5 5\n5 7\n5 8\n5 9\n5 8\n6 8\n6 7\n6 8\n6 9\n6 9\n6 6\n6 9\n6 7\n7 7\n7 7\n7 7\n7 8\n7 7\n7 8\n7 8\n7 9\n8 8\n8 8\n8 8\n8 8\n8 8\n8 9\n8 9\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "NO"
},
{
"input": "30 40\n0 0\n4 8\n5 17\n7 32\n7 16\n8 16\n10 19\n12 22\n12 27\n13 21\n13 28\n13 36\n14 28\n14 18\n18 21\n21 26\n21 36\n22 38\n23 32\n24 30\n26 35\n29 32\n29 32\n31 34\n31 31\n33 33\n33 35\n35 40\n38 38\n40 40",
"output": "NO"
},
{
"input": "70 100\n0 99\n1 87\n1 94\n1 4\n2 72\n3 39\n3 69\n4 78\n5 85\n7 14\n8 59\n12 69\n14 15\n14 76\n17 17\n19 53\n19 57\n19 21\n21 35\n21 83\n24 52\n24 33\n27 66\n27 97\n30 62\n30 74\n30 64\n32 63\n35 49\n37 60\n40 99\n40 71\n41 83\n42 66\n42 46\n45 83\n51 76\n53 69\n54 82\n54 96\n54 88\n55 91\n56 88\n58 62\n62 87\n64 80\n67 90\n67 69\n68 92\n72 93\n74 93\n77 79\n77 91\n78 97\n78 98\n81 85\n81 83\n81 83\n84 85\n86 88\n89 94\n89 92\n92 97\n96 99\n97 98\n97 99\n99 99\n100 100\n100 100\n100 100",
"output": "NO"
},
{
"input": "1 10\n0 10",
"output": "YES"
},
{
"input": "70 40\n0 34\n1 16\n3 33\n4 36\n4 22\n5 9\n5 9\n7 16\n8 26\n9 29\n9 25\n10 15\n10 22\n10 29\n10 20\n11 27\n11 26\n11 12\n12 19\n13 21\n14 31\n14 36\n15 34\n15 37\n16 21\n17 31\n18 22\n20 27\n20 32\n20 20\n20 29\n21 29\n21 34\n21 30\n22 40\n23 23\n23 28\n24 29\n25 38\n26 35\n27 37\n28 39\n28 33\n28 40\n28 33\n29 31\n29 33\n30 38\n30 36\n30 30\n30 38\n31 37\n31 35\n31 32\n31 36\n33 39\n33 40\n35 38\n36 38\n37 38\n37 40\n38 39\n38 40\n38 39\n39 39\n39 40\n40 40\n40 40\n40 40\n40 40",
"output": "YES"
},
{
"input": "50 40\n0 9\n1 26\n1 27\n2 33\n2 5\n3 30\n4 28\n5 31\n5 27\n5 29\n7 36\n8 32\n8 13\n9 24\n10 10\n10 30\n11 26\n11 22\n11 40\n11 31\n12 26\n13 25\n14 32\n17 19\n21 29\n22 36\n24 27\n25 39\n25 27\n27 32\n27 29\n27 39\n27 29\n28 38\n30 38\n32 40\n32 38\n33 33\n33 40\n34 35\n34 34\n34 38\n34 38\n35 37\n36 39\n36 39\n37 37\n38 40\n39 39\n40 40",
"output": "YES"
},
{
"input": "70 40\n0 34\n1 16\n3 33\n4 36\n4 22\n5 9\n5 9\n7 16\n8 26\n9 29\n9 25\n10 15\n10 22\n10 29\n10 20\n11 27\n11 26\n11 12\n12 19\n13 21\n14 31\n14 36\n15 34\n15 37\n16 21\n17 31\n18 22\n20 27\n20 32\n20 20\n20 29\n21 29\n21 34\n21 30\n22 22\n23 28\n23 39\n24 24\n25 27\n26 38\n27 39\n28 33\n28 39\n28 34\n28 33\n29 30\n29 35\n30 30\n30 38\n30 34\n30 31\n31 36\n31 31\n31 32\n31 38\n33 34\n33 34\n35 36\n36 38\n37 38\n37 39\n38 38\n38 38\n38 38\n39 39\n39 39\n40 40\n40 40\n40 40\n40 40",
"output": "NO"
},
{
"input": "10 100\n0 34\n8 56\n17 79\n24 88\n28 79\n45 79\n48 93\n55 87\n68 93\n88 99",
"output": "NO"
},
{
"input": "10 10\n0 2\n3 8\n3 5\n3 3\n3 9\n3 8\n5 7\n6 10\n7 10\n9 10",
"output": "NO"
},
{
"input": "50 10\n0 2\n0 2\n0 6\n1 9\n1 3\n1 2\n1 6\n1 1\n1 1\n2 7\n2 6\n2 4\n3 9\n3 8\n3 8\n3 8\n3 6\n3 4\n3 7\n3 4\n3 6\n3 5\n4 8\n5 5\n5 7\n6 7\n6 6\n7 7\n7 7\n7 7\n7 8\n7 8\n8 8\n8 8\n8 9\n8 8\n8 9\n9 9\n9 9\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "NO"
},
{
"input": "10 40\n0 21\n1 19\n4 33\n6 26\n8 39\n15 15\n20 24\n27 27\n29 39\n32 37",
"output": "NO"
},
{
"input": "50 10\n0 2\n0 2\n0 6\n1 9\n1 3\n1 2\n1 6\n1 1\n1 1\n2 7\n2 6\n2 4\n3 9\n3 8\n3 8\n3 8\n3 6\n3 4\n3 7\n3 4\n3 6\n3 10\n4 6\n5 9\n5 5\n6 7\n6 10\n7 8\n7 7\n7 7\n7 7\n7 10\n8 8\n8 8\n8 10\n8 8\n8 8\n9 10\n9 10\n9 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "YES"
},
{
"input": "1 1\n0 1",
"output": "YES"
},
{
"input": "30 40\n0 0\n4 8\n5 17\n7 32\n7 16\n8 16\n10 19\n12 22\n12 27\n13 21\n13 28\n13 36\n14 28\n14 18\n18 21\n21 26\n21 36\n22 38\n23 32\n24 30\n26 35\n29 32\n29 32\n31 34\n31 31\n33 33\n33 35\n35 36\n38 38\n40 40",
"output": "NO"
},
{
"input": "30 100\n0 27\n4 82\n11 81\n14 32\n33 97\n33 34\n37 97\n38 52\n45 91\n49 56\n50 97\n57 70\n59 94\n59 65\n62 76\n64 65\n65 95\n67 77\n68 82\n71 94\n80 90\n81 88\n84 93\n85 89\n88 92\n91 97\n92 99\n92 97\n99 99\n100 100",
"output": "NO"
},
{
"input": "10 100\n0 34\n8 56\n17 79\n24 88\n28 79\n45 79\n48 93\n55 87\n68 93\n79 100",
"output": "YES"
},
{
"input": "10 40\n0 21\n1 19\n4 33\n6 26\n8 39\n15 15\n20 24\n27 27\n29 39\n37 40",
"output": "YES"
},
{
"input": "85 10\n0 9\n0 4\n0 2\n0 5\n0 1\n0 8\n0 7\n1 2\n1 10\n1 2\n1 5\n1 10\n1 8\n1 1\n2 8\n2 7\n2 5\n2 5\n2 7\n3 5\n3 7\n3 5\n3 4\n3 7\n4 7\n4 8\n4 6\n5 7\n5 10\n5 5\n5 6\n5 6\n5 6\n5 6\n5 7\n5 8\n5 5\n5 7\n6 10\n6 9\n6 7\n6 10\n6 8\n6 7\n6 10\n6 10\n7 8\n7 9\n7 8\n7 8\n7 8\n7 8\n7 7\n7 7\n8 8\n8 8\n8 10\n8 9\n8 9\n8 9\n8 9\n9 9\n9 10\n9 9\n9 9\n9 9\n9 9\n9 10\n9 10\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "YES"
},
{
"input": "50 100\n0 95\n1 7\n1 69\n2 83\n5 67\n7 82\n8 31\n11 25\n15 44\n17 75\n17 27\n18 43\n18 69\n18 40\n21 66\n27 29\n27 64\n38 77\n38 90\n39 52\n40 60\n48 91\n48 98\n50 89\n50 63\n53 54\n53 95\n55 76\n56 59\n60 96\n62 86\n66 70\n67 77\n69 88\n69 98\n69 80\n69 95\n70 74\n70 77\n71 99\n72 73\n81 87\n85 99\n88 96\n88 91\n90 97\n90 99\n92 92\n98 99\n100 100",
"output": "NO"
},
{
"input": "50 40\n0 9\n1 26\n1 27\n2 33\n2 5\n3 30\n4 28\n5 31\n5 27\n5 29\n7 36\n8 32\n8 13\n9 24\n10 10\n10 30\n11 26\n11 22\n11 35\n11 23\n12 36\n13 31\n14 31\n17 17\n21 25\n22 33\n24 26\n25 32\n25 25\n27 39\n27 29\n27 34\n27 32\n28 34\n30 36\n32 37\n32 33\n33 35\n33 33\n34 38\n34 38\n34 36\n34 36\n35 36\n36 36\n36 39\n37 37\n38 39\n39 39\n40 40",
"output": "NO"
},
{
"input": "10 10\n0 2\n3 8\n3 5\n3 3\n3 9\n3 8\n5 7\n6 9\n7 7\n9 9",
"output": "NO"
},
{
"input": "70 100\n0 99\n1 87\n1 94\n1 4\n2 72\n3 39\n3 69\n4 78\n5 85\n7 14\n8 59\n12 69\n14 15\n14 76\n17 17\n19 53\n19 57\n19 21\n21 35\n21 83\n24 52\n24 33\n27 66\n27 97\n30 62\n30 74\n30 64\n32 63\n35 49\n37 60\n40 99\n40 71\n41 83\n42 66\n42 46\n45 83\n51 76\n53 69\n54 82\n54 96\n54 88\n55 91\n56 88\n58 62\n62 87\n64 80\n67 90\n67 69\n68 92\n72 93\n74 93\n77 79\n77 91\n78 97\n78 98\n81 85\n81 83\n81 83\n84 85\n86 88\n89 94\n89 100\n92 97\n96 96\n97 98\n97 100\n99 100\n100 100\n100 100\n100 100",
"output": "YES"
},
{
"input": "2 2\n0 2\n0 1",
"output": "YES"
},
{
"input": "2 100\n0 49\n50 100",
"output": "NO"
},
{
"input": "2 100\n0 50\n50 100",
"output": "YES"
},
{
"input": "10 100\n0 10\n10 20\n20 30\n30 40\n40 50\n50 60\n60 70\n70 80\n80 90\n90 100",
"output": "YES"
},
{
"input": "100 10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 10",
"output": "YES"
},
{
"input": "1 4\n1 4",
"output": "NO"
},
{
"input": "3 5\n0 2\n2 5\n3 4",
"output": "YES"
},
{
"input": "2 5\n0 5\n1 2",
"output": "YES"
},
{
"input": "5 5\n0 2\n0 2\n3 5\n3 5\n3 5",
"output": "NO"
},
{
"input": "4 10\n0 3\n2 10\n4 10\n5 7",
"output": "YES"
},
{
"input": "3 8\n0 0\n1 8\n2 8",
"output": "NO"
},
{
"input": "6 8\n0 3\n1 5\n2 3\n5 6\n6 7\n7 8",
"output": "YES"
}
] | 1,679,127,932
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 46
| 0
|
n,m=map(int,input().split())
list1=[0]*2
list2=[]
list2=([list1]*n)
for i in range(1,n+1):
list3=list(map(int,input().split()))
list2[i-1]=list3
sum=0
for i in range(1,n):
if list2[i-1][1]>=list2[i][0]:
sum+=1
if sum==n-1 and list2[n-1][1]>=m:
print("YES")
else :
print("NO")
|
Title: Visiting a Friend
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pig is visiting a friend.
Pig's house is located at point 0, and his friend's house is located at point *m* on an axis.
Pig can use teleports to move along the axis.
To use a teleport, Pig should come to a certain point (where the teleport is located) and choose where to move: for each teleport there is the rightmost point it can move Pig to, this point is known as the limit of the teleport.
Formally, a teleport located at point *x* with limit *y* can move Pig from point *x* to any point within the segment [*x*;<=*y*], including the bounds.
Determine if Pig can visit the friend using teleports only, or he should use his car.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100) — the number of teleports and the location of the friend's house.
The next *n* lines contain information about teleports.
The *i*-th of these lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=*b**i*<=≤<=*m*), where *a**i* is the location of the *i*-th teleport, and *b**i* is its limit.
It is guaranteed that *a**i*<=≥<=*a**i*<=-<=1 for every *i* (2<=≤<=*i*<=≤<=*n*).
Output Specification:
Print "YES" if there is a path from Pig's house to his friend's house that uses only teleports, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['3 5\n0 2\n2 4\n3 5\n', '3 7\n0 4\n2 5\n6 7\n']
Demo Output:
['YES\n', 'NO\n']
Note:
The first example is shown on the picture below:
Pig can use the first teleport from his house (point 0) to reach point 2, then using the second teleport go from point 2 to point 3, then using the third teleport go from point 3 to point 5, where his friend lives.
The second example is shown on the picture below:
You can see that there is no path from Pig's house to his friend's house that uses only teleports.
|
```python
n,m=map(int,input().split())
list1=[0]*2
list2=[]
list2=([list1]*n)
for i in range(1,n+1):
list3=list(map(int,input().split()))
list2[i-1]=list3
sum=0
for i in range(1,n):
if list2[i-1][1]>=list2[i][0]:
sum+=1
if sum==n-1 and list2[n-1][1]>=m:
print("YES")
else :
print("NO")
```
| 0
|
|
653
|
A
|
Bear and Three Balls
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"sortings"
] | null | null |
Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
- No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
|
The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball.
|
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
|
[
"4\n18 55 16 17\n",
"6\n40 41 43 44 44 44\n",
"8\n5 972 3 4 1 4 970 971\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971.
| 500
|
[
{
"input": "4\n18 55 16 17",
"output": "YES"
},
{
"input": "6\n40 41 43 44 44 44",
"output": "NO"
},
{
"input": "8\n5 972 3 4 1 4 970 971",
"output": "YES"
},
{
"input": "3\n959 747 656",
"output": "NO"
},
{
"input": "4\n1 2 2 3",
"output": "YES"
},
{
"input": "50\n998 30 384 289 505 340 872 223 663 31 929 625 864 699 735 589 676 399 745 635 963 381 75 97 324 612 597 797 103 382 25 894 219 458 337 572 201 355 294 275 278 311 586 573 965 704 936 237 715 543",
"output": "NO"
},
{
"input": "50\n941 877 987 982 966 979 984 810 811 909 872 980 957 897 845 995 924 905 984 914 824 840 868 910 815 808 872 858 883 952 823 835 860 874 959 972 931 867 866 987 982 837 800 921 887 910 982 980 828 869",
"output": "YES"
},
{
"input": "3\n408 410 409",
"output": "YES"
},
{
"input": "3\n903 902 904",
"output": "YES"
},
{
"input": "3\n399 400 398",
"output": "YES"
},
{
"input": "3\n450 448 449",
"output": "YES"
},
{
"input": "3\n390 389 388",
"output": "YES"
},
{
"input": "3\n438 439 440",
"output": "YES"
},
{
"input": "11\n488 688 490 94 564 615 641 170 489 517 669",
"output": "YES"
},
{
"input": "24\n102 672 983 82 720 501 81 721 982 312 207 897 159 964 611 956 118 984 37 271 596 403 772 954",
"output": "YES"
},
{
"input": "36\n175 551 70 479 875 480 979 32 465 402 640 116 76 687 874 678 359 785 753 401 978 629 162 963 886 641 39 845 132 930 2 372 478 947 407 318",
"output": "YES"
},
{
"input": "6\n10 79 306 334 304 305",
"output": "YES"
},
{
"input": "34\n787 62 26 683 486 364 684 891 846 801 969 837 359 800 836 359 471 637 732 91 841 836 7 799 959 405 416 841 737 803 615 483 323 365",
"output": "YES"
},
{
"input": "30\n860 238 14 543 669 100 428 789 576 484 754 274 849 850 586 377 711 386 510 408 520 693 23 477 266 851 728 711 964 73",
"output": "YES"
},
{
"input": "11\n325 325 324 324 324 325 325 324 324 324 324",
"output": "NO"
},
{
"input": "7\n517 517 518 517 518 518 518",
"output": "NO"
},
{
"input": "20\n710 710 711 711 711 711 710 710 710 710 711 710 710 710 710 710 710 711 711 710",
"output": "NO"
},
{
"input": "48\n29 30 29 29 29 30 29 30 30 30 30 29 30 30 30 29 29 30 30 29 30 29 29 30 29 30 29 30 30 29 30 29 29 30 30 29 29 30 30 29 29 30 30 30 29 29 30 29",
"output": "NO"
},
{
"input": "7\n880 880 514 536 881 881 879",
"output": "YES"
},
{
"input": "15\n377 432 262 376 261 375 377 262 263 263 261 376 262 262 375",
"output": "YES"
},
{
"input": "32\n305 426 404 961 426 425 614 304 404 425 615 403 303 304 615 303 305 405 427 614 403 303 425 615 404 304 427 403 206 616 405 404",
"output": "YES"
},
{
"input": "41\n115 686 988 744 762 519 745 519 518 83 85 115 520 44 687 686 685 596 988 687 989 988 114 745 84 519 519 746 988 84 745 744 115 114 85 115 520 746 745 116 987",
"output": "YES"
},
{
"input": "47\n1 2 483 28 7 109 270 651 464 162 353 521 224 989 721 499 56 69 197 716 313 446 580 645 828 197 100 138 789 499 147 677 384 711 783 937 300 543 540 93 669 604 739 122 632 822 116",
"output": "NO"
},
{
"input": "31\n1 2 1 373 355 692 750 920 578 666 615 232 141 129 663 929 414 704 422 559 568 731 354 811 532 618 39 879 292 602 995",
"output": "NO"
},
{
"input": "50\n5 38 41 4 15 40 27 39 20 3 44 47 30 6 36 29 35 12 19 26 10 2 21 50 11 46 48 49 17 16 33 13 32 28 31 18 23 34 7 14 24 45 9 37 1 8 42 25 43 22",
"output": "YES"
},
{
"input": "50\n967 999 972 990 969 978 963 987 954 955 973 970 959 981 995 983 986 994 979 957 965 982 992 977 953 975 956 961 993 997 998 958 980 962 960 951 996 991 1000 966 971 988 976 968 989 984 974 964 985 952",
"output": "YES"
},
{
"input": "50\n850 536 761 506 842 898 857 723 583 637 536 943 895 929 890 612 832 633 696 731 553 880 710 812 665 877 915 636 711 540 748 600 554 521 813 796 568 513 543 809 798 820 928 504 999 646 907 639 550 911",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "YES"
},
{
"input": "3\n500 999 1000",
"output": "NO"
},
{
"input": "10\n101 102 104 105 107 109 110 112 113 115",
"output": "NO"
},
{
"input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "NO"
},
{
"input": "50\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "3\n1000 999 998",
"output": "YES"
},
{
"input": "49\n343 322 248 477 53 156 245 493 209 141 370 66 229 184 434 137 276 472 216 456 147 180 140 114 493 323 393 262 380 314 222 124 98 441 129 346 48 401 347 460 122 125 114 106 189 260 374 165 456",
"output": "NO"
},
{
"input": "20\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3",
"output": "YES"
},
{
"input": "3\n999 999 1000",
"output": "NO"
},
{
"input": "9\n2 4 5 13 25 100 200 300 400",
"output": "NO"
},
{
"input": "9\n1 1 1 2 2 2 3 3 3",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "NO"
},
{
"input": "3\n998 999 1000",
"output": "YES"
},
{
"input": "12\n1 1 1 1 1 1 1 1 1 2 2 4",
"output": "NO"
},
{
"input": "4\n4 3 4 5",
"output": "YES"
},
{
"input": "6\n1 1 1 2 2 2",
"output": "NO"
},
{
"input": "3\n2 3 2",
"output": "NO"
},
{
"input": "5\n10 5 6 3 2",
"output": "NO"
},
{
"input": "3\n1 2 1",
"output": "NO"
},
{
"input": "3\n1 2 3",
"output": "YES"
},
{
"input": "4\n998 999 1000 1000",
"output": "YES"
},
{
"input": "5\n2 3 9 9 4",
"output": "YES"
},
{
"input": "4\n1 2 4 4",
"output": "NO"
},
{
"input": "3\n1 1 1",
"output": "NO"
},
{
"input": "3\n2 2 3",
"output": "NO"
},
{
"input": "7\n1 2 2 2 4 5 6",
"output": "YES"
},
{
"input": "5\n1 3 10 3 10",
"output": "NO"
},
{
"input": "3\n1 2 2",
"output": "NO"
},
{
"input": "4\n1000 1000 999 998",
"output": "YES"
},
{
"input": "3\n5 3 7",
"output": "NO"
},
{
"input": "6\n1 1 2 2 3 3",
"output": "YES"
},
{
"input": "9\n6 6 6 5 5 5 4 4 4",
"output": "YES"
},
{
"input": "7\n5 6 6 6 7 7 7",
"output": "YES"
},
{
"input": "5\n2 3 3 3 4",
"output": "YES"
},
{
"input": "5\n2 1 2 1 3",
"output": "YES"
},
{
"input": "3\n1 2 7",
"output": "NO"
},
{
"input": "3\n1000 1000 1000",
"output": "NO"
},
{
"input": "5\n1 100 2 100 3",
"output": "YES"
},
{
"input": "5\n5 4 6 5 5",
"output": "YES"
},
{
"input": "12\n1 1 1 1 2 2 2 2 3 3 3 3",
"output": "YES"
},
{
"input": "5\n9 9 1 2 3",
"output": "YES"
},
{
"input": "6\n1 2 3 1 2 3",
"output": "YES"
},
{
"input": "7\n1 1 1 1 2 3 3",
"output": "YES"
},
{
"input": "3\n13 13 13",
"output": "NO"
},
{
"input": "3\n42 42 42",
"output": "NO"
},
{
"input": "8\n1 1 1 1 2 2 2 2",
"output": "NO"
},
{
"input": "6\n1 1 1 1 2 3",
"output": "YES"
},
{
"input": "6\n1 1 2 2 6 6",
"output": "NO"
},
{
"input": "6\n1 2 5 5 5 5",
"output": "NO"
},
{
"input": "9\n1 2 3 1 2 3 1 2 3",
"output": "YES"
},
{
"input": "4\n1 2 1 100",
"output": "NO"
},
{
"input": "5\n1 1 2 2 3",
"output": "YES"
}
] | 1,651,259,006
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 84
| 46
| 0
|
def e_compativel(a, b):
if abs(a - b) < 3 and abs(a - b) > 0:
return True;
return False;
def testar_compatibilidade_com_conjunto(a, bolas):
for i in bolas:
if not e_compativel(a, i):
return False;
return True;
def testar_combinacoes(bolas, bolasJaSelecionadas, n, nJaSelecionadas):
if(nJaSelecionadas == 3):
return True;
for i in bolas:
if testar_compatibilidade_com_conjunto(i, bolasJaSelecionadas):
bolasJaSelecionadas.append(i);
return testar_combinacoes(bolas, bolasJaSelecionadas, n, nJaSelecionadas + 1);
return False;
def combinacao_bolas(bolas, n):
for i in bolas:
if(testar_combinacoes(bolas, [i], n, 1)):
return True;
return False;
def main():
n = int(input());
bolas = [];
entrada = input().split();
for i in range(n):
bolas.append(int(entrada[i]));
if(combinacao_bolas(bolas, n)):
print("YES");
else:
print("NO");
main();
|
Title: Bear and Three Balls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
- No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
Input Specification:
The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball.
Output Specification:
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
Demo Input:
['4\n18 55 16 17\n', '6\n40 41 43 44 44 44\n', '8\n5 972 3 4 1 4 970 971\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971.
|
```python
def e_compativel(a, b):
if abs(a - b) < 3 and abs(a - b) > 0:
return True;
return False;
def testar_compatibilidade_com_conjunto(a, bolas):
for i in bolas:
if not e_compativel(a, i):
return False;
return True;
def testar_combinacoes(bolas, bolasJaSelecionadas, n, nJaSelecionadas):
if(nJaSelecionadas == 3):
return True;
for i in bolas:
if testar_compatibilidade_com_conjunto(i, bolasJaSelecionadas):
bolasJaSelecionadas.append(i);
return testar_combinacoes(bolas, bolasJaSelecionadas, n, nJaSelecionadas + 1);
return False;
def combinacao_bolas(bolas, n):
for i in bolas:
if(testar_combinacoes(bolas, [i], n, 1)):
return True;
return False;
def main():
n = int(input());
bolas = [];
entrada = input().split();
for i in range(n):
bolas.append(int(entrada[i]));
if(combinacao_bolas(bolas, n)):
print("YES");
else:
print("NO");
main();
```
| 3
|
|
948
|
A
|
Protect Sheep
|
PROGRAMMING
| 900
|
[
"brute force",
"dfs and similar",
"graphs",
"implementation"
] | null | null |
Bob is a farmer. He has a large pasture with many sheep. Recently, he has lost some of them due to wolf attacks. He thus decided to place some shepherd dogs in such a way that all his sheep are protected.
The pasture is a rectangle consisting of *R*<=×<=*C* cells. Each cell is either empty, contains a sheep, a wolf or a dog. Sheep and dogs always stay in place, but wolves can roam freely around the pasture, by repeatedly moving to the left, right, up or down to a neighboring cell. When a wolf enters a cell with a sheep, it consumes it. However, no wolf can enter a cell with a dog.
Initially there are no dogs. Place dogs onto the pasture in such a way that no wolf can reach any sheep, or determine that it is impossible. Note that since you have many dogs, you do not need to minimize their number.
|
First line contains two integers *R* (1<=≤<=*R*<=≤<=500) and *C* (1<=≤<=*C*<=≤<=500), denoting the number of rows and the numbers of columns respectively.
Each of the following *R* lines is a string consisting of exactly *C* characters, representing one row of the pasture. Here, 'S' means a sheep, 'W' a wolf and '.' an empty cell.
|
If it is impossible to protect all sheep, output a single line with the word "No".
Otherwise, output a line with the word "Yes". Then print *R* lines, representing the pasture after placing dogs. Again, 'S' means a sheep, 'W' a wolf, 'D' is a dog and '.' an empty space. You are not allowed to move, remove or add a sheep or a wolf.
If there are multiple solutions, you may print any of them. You don't have to minimize the number of dogs.
|
[
"6 6\n..S...\n..S.W.\n.S....\n..W...\n...W..\n......\n",
"1 2\nSW\n",
"5 5\n.S...\n...S.\nS....\n...S.\n.S...\n"
] |
[
"Yes\n..SD..\n..SDW.\n.SD...\n.DW...\nDD.W..\n......\n",
"No\n",
"Yes\n.S...\n...S.\nS.D..\n...S.\n.S...\n"
] |
In the first example, we can split the pasture into two halves, one containing wolves and one containing sheep. Note that the sheep at (2,1) is safe, as wolves cannot move diagonally.
In the second example, there are no empty spots to put dogs that would guard the lone sheep.
In the third example, there are no wolves, so the task is very easy. We put a dog in the center to observe the peacefulness of the meadow, but the solution would be correct even without him.
| 500
|
[
{
"input": "1 2\nSW",
"output": "No"
},
{
"input": "10 10\n....W.W.W.\n.........S\n.S.S...S..\nW.......SS\n.W..W.....\n.W...W....\nS..S...S.S\n....W...S.\n..S..S.S.S\nSS.......S",
"output": "Yes\nDDDDWDWDWD\nDDDDDDDDDS\nDSDSDDDSDD\nWDDDDDDDSS\nDWDDWDDDDD\nDWDDDWDDDD\nSDDSDDDSDS\nDDDDWDDDSD\nDDSDDSDSDS\nSSDDDDDDDS"
},
{
"input": "10 10\n....W.W.W.\n...W.....S\n.S.S...S..\nW......WSS\n.W..W.....\n.W...W....\nS..S...S.S\n...WWW..S.\n..S..S.S.S\nSS.......S",
"output": "No"
},
{
"input": "1 50\nW...S..............W.....S..S...............S...W.",
"output": "Yes\nWDDDSDDDDDDDDDDDDDDWDDDDDSDDSDDDDDDDDDDDDDDDSDDDWD"
},
{
"input": "2 4\n...S\n...W",
"output": "No"
},
{
"input": "4 2\n..\n..\n..\nSW",
"output": "No"
},
{
"input": "4 2\n..\n..\n..\nWS",
"output": "No"
},
{
"input": "2 4\n...W\n...S",
"output": "No"
},
{
"input": "50 1\nS\n.\n.\n.\n.\n.\n.\nS\n.\n.\n.\n.\n.\n.\n.\n.\nS\n.\nW\n.\nS\n.\n.\n.\n.\nS\n.\n.\n.\n.\n.\n.\n.\nW\n.\n.\n.\nW\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.",
"output": "Yes\nS\nD\nD\nD\nD\nD\nD\nS\nD\nD\nD\nD\nD\nD\nD\nD\nS\nD\nW\nD\nS\nD\nD\nD\nD\nS\nD\nD\nD\nD\nD\nD\nD\nW\nD\nD\nD\nW\nD\nD\nD\nD\nD\nD\nD\nD\nD\nD\nD\nD"
},
{
"input": "4 4\nW..S\nW..S\nW..S\nW..S",
"output": "Yes\nWDDS\nWDDS\nWDDS\nWDDS"
},
{
"input": "4 4\nSSSS\n....\n....\nWWWW",
"output": "Yes\nSSSS\nDDDD\nDDDD\nWWWW"
},
{
"input": "4 4\nWWWW\n....\n....\nSSSS",
"output": "Yes\nWWWW\nDDDD\nDDDD\nSSSS"
},
{
"input": "4 4\nS..W\nS..W\nS..W\nS..W",
"output": "Yes\nSDDW\nSDDW\nSDDW\nSDDW"
},
{
"input": "1 1\n.",
"output": "Yes\nD"
},
{
"input": "1 1\nW",
"output": "Yes\nW"
},
{
"input": "1 1\nS",
"output": "Yes\nS"
},
{
"input": "4 2\n..\n..\n.W\n.S",
"output": "No"
},
{
"input": "4 2\n..\n..\n.S\n.W",
"output": "No"
},
{
"input": "4 2\n..\n..\nW.\nS.",
"output": "No"
},
{
"input": "4 2\n..\n..\nS.\nW.",
"output": "No"
},
{
"input": "2 4\n....\n..SW",
"output": "No"
},
{
"input": "2 4\n....\n..WS",
"output": "No"
},
{
"input": "1 2\nS.",
"output": "Yes\nSD"
}
] | 1,600,793,799
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 98
| 467
| 2,355,200
|
r,c=map(int,input().split())
mat=[]
flag=0
for i in range(r):
k=[]
s=input()
for z in s:
k.append(z)
mat.append(k)
for i in range(r):
for j in range(c):
if(mat[i][j]=="W"):
if(i-1>=0):
if(mat[i-1][j]=="S"):
flag=1
break
elif(mat[i-1][j]=="."):
mat[i-1][j]="D"
if(i+1<r):
if(mat[i+1][j]=="S"):
flag=1
break
elif(mat[i+1][j]=="."):
mat[i+1][j]="D"
if(j-1>=0):
if(mat[i][j-1]=="S"):
flag=1
break
elif(mat[i][j-1]=="."):
mat[i][j-1]="D"
if(j+1<c):
if(mat[i][j+1]=="S"):
flag=1
break
elif(mat[i][j+1]=="."):
mat[i][j+1]="D"
if(flag==1):
print("No")
else:
print("Yes")
for i in mat:
for j in i:
print(j,end="")
print()
|
Title: Protect Sheep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bob is a farmer. He has a large pasture with many sheep. Recently, he has lost some of them due to wolf attacks. He thus decided to place some shepherd dogs in such a way that all his sheep are protected.
The pasture is a rectangle consisting of *R*<=×<=*C* cells. Each cell is either empty, contains a sheep, a wolf or a dog. Sheep and dogs always stay in place, but wolves can roam freely around the pasture, by repeatedly moving to the left, right, up or down to a neighboring cell. When a wolf enters a cell with a sheep, it consumes it. However, no wolf can enter a cell with a dog.
Initially there are no dogs. Place dogs onto the pasture in such a way that no wolf can reach any sheep, or determine that it is impossible. Note that since you have many dogs, you do not need to minimize their number.
Input Specification:
First line contains two integers *R* (1<=≤<=*R*<=≤<=500) and *C* (1<=≤<=*C*<=≤<=500), denoting the number of rows and the numbers of columns respectively.
Each of the following *R* lines is a string consisting of exactly *C* characters, representing one row of the pasture. Here, 'S' means a sheep, 'W' a wolf and '.' an empty cell.
Output Specification:
If it is impossible to protect all sheep, output a single line with the word "No".
Otherwise, output a line with the word "Yes". Then print *R* lines, representing the pasture after placing dogs. Again, 'S' means a sheep, 'W' a wolf, 'D' is a dog and '.' an empty space. You are not allowed to move, remove or add a sheep or a wolf.
If there are multiple solutions, you may print any of them. You don't have to minimize the number of dogs.
Demo Input:
['6 6\n..S...\n..S.W.\n.S....\n..W...\n...W..\n......\n', '1 2\nSW\n', '5 5\n.S...\n...S.\nS....\n...S.\n.S...\n']
Demo Output:
['Yes\n..SD..\n..SDW.\n.SD...\n.DW...\nDD.W..\n......\n', 'No\n', 'Yes\n.S...\n...S.\nS.D..\n...S.\n.S...\n']
Note:
In the first example, we can split the pasture into two halves, one containing wolves and one containing sheep. Note that the sheep at (2,1) is safe, as wolves cannot move diagonally.
In the second example, there are no empty spots to put dogs that would guard the lone sheep.
In the third example, there are no wolves, so the task is very easy. We put a dog in the center to observe the peacefulness of the meadow, but the solution would be correct even without him.
|
```python
r,c=map(int,input().split())
mat=[]
flag=0
for i in range(r):
k=[]
s=input()
for z in s:
k.append(z)
mat.append(k)
for i in range(r):
for j in range(c):
if(mat[i][j]=="W"):
if(i-1>=0):
if(mat[i-1][j]=="S"):
flag=1
break
elif(mat[i-1][j]=="."):
mat[i-1][j]="D"
if(i+1<r):
if(mat[i+1][j]=="S"):
flag=1
break
elif(mat[i+1][j]=="."):
mat[i+1][j]="D"
if(j-1>=0):
if(mat[i][j-1]=="S"):
flag=1
break
elif(mat[i][j-1]=="."):
mat[i][j-1]="D"
if(j+1<c):
if(mat[i][j+1]=="S"):
flag=1
break
elif(mat[i][j+1]=="."):
mat[i][j+1]="D"
if(flag==1):
print("No")
else:
print("Yes")
for i in mat:
for j in i:
print(j,end="")
print()
```
| 3
|
|
631
|
A
|
Interview
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
Blake is a CEO of a large company called "Blake Technologies". He loves his company very much and he thinks that his company should be the best. That is why every candidate needs to pass through the interview that consists of the following problem.
We define function *f*(*x*,<=*l*,<=*r*) as a bitwise OR of integers *x**l*,<=*x**l*<=+<=1,<=...,<=*x**r*, where *x**i* is the *i*-th element of the array *x*. You are given two arrays *a* and *b* of length *n*. You need to determine the maximum value of sum *f*(*a*,<=*l*,<=*r*)<=+<=*f*(*b*,<=*l*,<=*r*) among all possible 1<=≤<=*l*<=≤<=*r*<=≤<=*n*.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the length of the arrays.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=109).
The third line contains *n* integers *b**i* (0<=≤<=*b**i*<=≤<=109).
|
Print a single integer — the maximum value of sum *f*(*a*,<=*l*,<=*r*)<=+<=*f*(*b*,<=*l*,<=*r*) among all possible 1<=≤<=*l*<=≤<=*r*<=≤<=*n*.
|
[
"5\n1 2 4 3 2\n2 3 3 12 1\n",
"10\n13 2 7 11 8 4 9 8 5 1\n5 7 18 9 2 3 0 11 8 6\n"
] |
[
"22",
"46"
] |
Bitwise OR of two non-negative integers *a* and *b* is the number *c* = *a* *OR* *b*, such that each of its digits in binary notation is 1 if and only if at least one of *a* or *b* have 1 in the corresponding position in binary notation.
In the first sample, one of the optimal answers is *l* = 2 and *r* = 4, because *f*(*a*, 2, 4) + *f*(*b*, 2, 4) = (2 *OR* 4 *OR* 3) + (3 *OR* 3 *OR* 12) = 7 + 15 = 22. Other ways to get maximum value is to choose *l* = 1 and *r* = 4, *l* = 1 and *r* = 5, *l* = 2 and *r* = 4, *l* = 2 and *r* = 5, *l* = 3 and *r* = 4, or *l* = 3 and *r* = 5.
In the second sample, the maximum value is obtained for *l* = 1 and *r* = 9.
| 500
|
[
{
"input": "5\n1 2 4 3 2\n2 3 3 12 1",
"output": "22"
},
{
"input": "10\n13 2 7 11 8 4 9 8 5 1\n5 7 18 9 2 3 0 11 8 6",
"output": "46"
},
{
"input": "25\n12 30 38 109 81 124 80 33 38 48 29 78 96 48 96 27 80 77 102 65 80 113 31 118 35\n25 64 95 13 12 6 111 80 85 16 61 119 23 65 73 65 20 95 124 18 28 79 125 106 116",
"output": "254"
},
{
"input": "20\n64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64\n64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64",
"output": "128"
},
{
"input": "1\n1000000000\n1000000000",
"output": "2000000000"
},
{
"input": "1\n0\n0",
"output": "0"
},
{
"input": "2\n7 16\n16 7",
"output": "46"
},
{
"input": "4\n6 0 0 0\n0 0 0 1",
"output": "7"
},
{
"input": "8\n1 2 4 8 16 32 64 128\n1 2 4 8 16 32 64 128",
"output": "510"
},
{
"input": "1\n2\n3",
"output": "5"
},
{
"input": "1\n4\n3",
"output": "7"
},
{
"input": "1\n1\n1",
"output": "2"
}
] | 1,567,180,799
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 11
| 124
| 409,600
|
from functools import reduce; input()
a = reduce(lambda x, y: int(x) | int(y), input().split())
b = reduce(lambda x, y: int(x) | int(y), input().split())
print(a + b)
|
Title: Interview
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Blake is a CEO of a large company called "Blake Technologies". He loves his company very much and he thinks that his company should be the best. That is why every candidate needs to pass through the interview that consists of the following problem.
We define function *f*(*x*,<=*l*,<=*r*) as a bitwise OR of integers *x**l*,<=*x**l*<=+<=1,<=...,<=*x**r*, where *x**i* is the *i*-th element of the array *x*. You are given two arrays *a* and *b* of length *n*. You need to determine the maximum value of sum *f*(*a*,<=*l*,<=*r*)<=+<=*f*(*b*,<=*l*,<=*r*) among all possible 1<=≤<=*l*<=≤<=*r*<=≤<=*n*.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the length of the arrays.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=109).
The third line contains *n* integers *b**i* (0<=≤<=*b**i*<=≤<=109).
Output Specification:
Print a single integer — the maximum value of sum *f*(*a*,<=*l*,<=*r*)<=+<=*f*(*b*,<=*l*,<=*r*) among all possible 1<=≤<=*l*<=≤<=*r*<=≤<=*n*.
Demo Input:
['5\n1 2 4 3 2\n2 3 3 12 1\n', '10\n13 2 7 11 8 4 9 8 5 1\n5 7 18 9 2 3 0 11 8 6\n']
Demo Output:
['22', '46']
Note:
Bitwise OR of two non-negative integers *a* and *b* is the number *c* = *a* *OR* *b*, such that each of its digits in binary notation is 1 if and only if at least one of *a* or *b* have 1 in the corresponding position in binary notation.
In the first sample, one of the optimal answers is *l* = 2 and *r* = 4, because *f*(*a*, 2, 4) + *f*(*b*, 2, 4) = (2 *OR* 4 *OR* 3) + (3 *OR* 3 *OR* 12) = 7 + 15 = 22. Other ways to get maximum value is to choose *l* = 1 and *r* = 4, *l* = 1 and *r* = 5, *l* = 2 and *r* = 4, *l* = 2 and *r* = 5, *l* = 3 and *r* = 4, or *l* = 3 and *r* = 5.
In the second sample, the maximum value is obtained for *l* = 1 and *r* = 9.
|
```python
from functools import reduce; input()
a = reduce(lambda x, y: int(x) | int(y), input().split())
b = reduce(lambda x, y: int(x) | int(y), input().split())
print(a + b)
```
| 0
|
|
479
|
C
|
Exams
|
PROGRAMMING
| 1,400
|
[
"greedy",
"sortings"
] | null | null |
Student Valera is an undergraduate student at the University. His end of term exams are approaching and he is to pass exactly *n* exams. Valera is a smart guy, so he will be able to pass any exam he takes on his first try. Besides, he can take several exams on one day, and in any order.
According to the schedule, a student can take the exam for the *i*-th subject on the day number *a**i*. However, Valera has made an arrangement with each teacher and the teacher of the *i*-th subject allowed him to take an exam before the schedule time on day *b**i* (*b**i*<=<<=*a**i*). Thus, Valera can take an exam for the *i*-th subject either on day *a**i*, or on day *b**i*. All the teachers put the record of the exam in the student's record book on the day of the actual exam and write down the date of the mark as number *a**i*.
Valera believes that it would be rather strange if the entries in the record book did not go in the order of non-decreasing date. Therefore Valera asks you to help him. Find the minimum possible value of the day when Valera can take the final exam if he takes exams so that all the records in his record book go in the order of non-decreasing date.
|
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=5000) — the number of exams Valera will take.
Each of the next *n* lines contains two positive space-separated integers *a**i* and *b**i* (1<=≤<=*b**i*<=<<=*a**i*<=≤<=109) — the date of the exam in the schedule and the early date of passing the *i*-th exam, correspondingly.
|
Print a single integer — the minimum possible number of the day when Valera can take the last exam if he takes all the exams so that all the records in his record book go in the order of non-decreasing date.
|
[
"3\n5 2\n3 1\n4 2\n",
"3\n6 1\n5 2\n4 3\n"
] |
[
"2\n",
"6\n"
] |
In the first sample Valera first takes an exam in the second subject on the first day (the teacher writes down the schedule date that is 3). On the next day he takes an exam in the third subject (the teacher writes down the schedule date, 4), then he takes an exam in the first subject (the teacher writes down the mark with date 5). Thus, Valera takes the last exam on the second day and the dates will go in the non-decreasing order: 3, 4, 5.
In the second sample Valera first takes an exam in the third subject on the fourth day. Then he takes an exam in the second subject on the fifth day. After that on the sixth day Valera takes an exam in the first subject.
| 1,500
|
[
{
"input": "3\n5 2\n3 1\n4 2",
"output": "2"
},
{
"input": "3\n6 1\n5 2\n4 3",
"output": "6"
},
{
"input": "1\n1000000000 999999999",
"output": "999999999"
},
{
"input": "1\n2 1",
"output": "1"
},
{
"input": "2\n3 2\n3 2",
"output": "2"
},
{
"input": "5\n4 3\n4 2\n4 1\n4 1\n4 1",
"output": "3"
},
{
"input": "6\n12 11\n10 9\n8 7\n6 5\n4 3\n2 1",
"output": "11"
},
{
"input": "2\n3 1\n3 2",
"output": "2"
},
{
"input": "2\n4 2\n4 1",
"output": "2"
},
{
"input": "2\n5 2\n5 1",
"output": "2"
},
{
"input": "6\n3 1\n3 2\n4 1\n4 2\n5 4\n5 4",
"output": "4"
},
{
"input": "3\n3 2\n4 1\n100 10",
"output": "10"
},
{
"input": "3\n4 3\n5 2\n10 8",
"output": "8"
},
{
"input": "5\n6 5\n6 4\n6 3\n6 2\n6 1",
"output": "5"
},
{
"input": "3\n5 4\n6 3\n8 7",
"output": "7"
},
{
"input": "4\n7 1\n7 3\n8 2\n9 8",
"output": "8"
},
{
"input": "3\n3 2\n4 1\n10 5",
"output": "5"
},
{
"input": "3\n5 4\n6 3\n11 10",
"output": "10"
},
{
"input": "4\n2 1\n3 2\n4 1\n6 5",
"output": "5"
}
] | 1,694,104,006
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
d={}
ans=0
for _ in range(int(input())):
a,b=map(int,input().split())
d.setdefault(a,[])
d[a].append(b)
num = sorted(d.keys())
days=-1
for x in num:
d[x].sort()
if days<d[x][0]:
days= d[x][0]
else:
days=x
print(days)
|
Title: Exams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Student Valera is an undergraduate student at the University. His end of term exams are approaching and he is to pass exactly *n* exams. Valera is a smart guy, so he will be able to pass any exam he takes on his first try. Besides, he can take several exams on one day, and in any order.
According to the schedule, a student can take the exam for the *i*-th subject on the day number *a**i*. However, Valera has made an arrangement with each teacher and the teacher of the *i*-th subject allowed him to take an exam before the schedule time on day *b**i* (*b**i*<=<<=*a**i*). Thus, Valera can take an exam for the *i*-th subject either on day *a**i*, or on day *b**i*. All the teachers put the record of the exam in the student's record book on the day of the actual exam and write down the date of the mark as number *a**i*.
Valera believes that it would be rather strange if the entries in the record book did not go in the order of non-decreasing date. Therefore Valera asks you to help him. Find the minimum possible value of the day when Valera can take the final exam if he takes exams so that all the records in his record book go in the order of non-decreasing date.
Input Specification:
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=5000) — the number of exams Valera will take.
Each of the next *n* lines contains two positive space-separated integers *a**i* and *b**i* (1<=≤<=*b**i*<=<<=*a**i*<=≤<=109) — the date of the exam in the schedule and the early date of passing the *i*-th exam, correspondingly.
Output Specification:
Print a single integer — the minimum possible number of the day when Valera can take the last exam if he takes all the exams so that all the records in his record book go in the order of non-decreasing date.
Demo Input:
['3\n5 2\n3 1\n4 2\n', '3\n6 1\n5 2\n4 3\n']
Demo Output:
['2\n', '6\n']
Note:
In the first sample Valera first takes an exam in the second subject on the first day (the teacher writes down the schedule date that is 3). On the next day he takes an exam in the third subject (the teacher writes down the schedule date, 4), then he takes an exam in the first subject (the teacher writes down the mark with date 5). Thus, Valera takes the last exam on the second day and the dates will go in the non-decreasing order: 3, 4, 5.
In the second sample Valera first takes an exam in the third subject on the fourth day. Then he takes an exam in the second subject on the fifth day. After that on the sixth day Valera takes an exam in the first subject.
|
```python
d={}
ans=0
for _ in range(int(input())):
a,b=map(int,input().split())
d.setdefault(a,[])
d[a].append(b)
num = sorted(d.keys())
days=-1
for x in num:
d[x].sort()
if days<d[x][0]:
days= d[x][0]
else:
days=x
print(days)
```
| 0
|
|
462
|
A
|
Appleman and Easy Task
|
PROGRAMMING
| 1,000
|
[
"brute force",
"implementation"
] | null | null |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?
Given a *n*<=×<=*n* checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Then *n* lines follow containing the description of the checkerboard. Each of them contains *n* characters (either 'x' or 'o') without spaces.
|
Print "YES" or "NO" (without the quotes) depending on the answer to the problem.
|
[
"3\nxxo\nxox\noxx\n",
"4\nxxxo\nxoxo\noxox\nxxxx\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "3\nxxo\nxox\noxx",
"output": "YES"
},
{
"input": "4\nxxxo\nxoxo\noxox\nxxxx",
"output": "NO"
},
{
"input": "1\no",
"output": "YES"
},
{
"input": "2\nox\nxo",
"output": "YES"
},
{
"input": "2\nxx\nxo",
"output": "NO"
},
{
"input": "3\nooo\noxo\nxoo",
"output": "NO"
},
{
"input": "3\nxxx\nxxo\nxxo",
"output": "NO"
},
{
"input": "4\nxooo\nooxo\noxoo\nooox",
"output": "YES"
},
{
"input": "4\noooo\noxxo\nxoxo\noooo",
"output": "NO"
},
{
"input": "5\noxoxo\nxxxxx\noxoxo\nxxxxx\noxoxo",
"output": "YES"
},
{
"input": "5\nxxxox\nxxxxo\nxoxox\noxoxx\nxoxxx",
"output": "NO"
},
{
"input": "10\nxoxooooooo\noxxoxxxxxo\nxxooxoooxo\noooxxoxoxo\noxxxooooxo\noxooooxxxo\noxoxoxxooo\noxoooxooxx\noxxxxxoxxo\noooooooxox",
"output": "YES"
},
{
"input": "10\nxxxxxxxoox\nxooxxooooo\noxoooxxooo\nxoxxxxxxxx\nxxoxooxxox\nooxoxxooox\nooxxxxxooo\nxxxxoxooox\nxoxxooxxxx\noooooxxoxo",
"output": "NO"
},
{
"input": "19\noxoxoxoxooxoooxxoox\nxxxxxxxxoxxoxoooooo\noxoxoxooxxxooxxxooo\nxxoxxxooxooxxxoxxox\noxoxooxxxooooxxoxox\nxxxoooxoxxoxxoxxxoo\noxooxxxoooooxxoooxo\nxxooxooxoxxoxxoxxoo\noxxxxooooxxxooooxxx\nooxooxoxxoxxoxooxoo\nxxxooooxxxooooxoxox\noooxoxooxxoxooxooxx\nxxoooxxxooooxxoooxo\nooxxxooxoxooxooxxxx\nxoxoxxooxoxxxooxoxo\nxoxxoxoxooxooxxxxxx\noooxxxooxxxooxoxoxo\nxoooooxoxooxxxxxxxo\nxooxxoooxxoxoxoxoxx",
"output": "NO"
},
{
"input": "12\nxxooxxoxxxoo\nxxoooxoxoxoo\nooxoxoxxooxx\nooxxooooxoxo\nxxxxxxxxoxxx\noxooooxxxooo\noxxoxoxoooxx\nxxxxxxxooxox\noxoooooxoxxx\nxxooxxoxxoxx\noxxxxxxxooxx\nooxoxooxxooo",
"output": "NO"
},
{
"input": "2\noo\nxx",
"output": "NO"
}
] | 1,679,068,495
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 32
| 77
| 2,764,800
|
def miniSin(angle):
return (2 - angle) * (angle % 2)
n = int(input())
gridList = [[i for i in input()] for x in range(n) ]
def checkOs():
for RowIndex in range(n):
for columnIndex in range(n):
numberOfOs = 0
for i in range(4):
RI = RowIndex + miniSin((i+1)%4)
CI = columnIndex + miniSin(i)
if RI < 0 or RI == n or CI < 0 or CI == n:
continue
if gridList[RowIndex + miniSin((i+1)%4)][columnIndex + miniSin(i)] == "o":
numberOfOs += 1
if numberOfOs % 2 != 0:
return "NO"
return "YES"
print(checkOs())
|
Title: Appleman and Easy Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?
Given a *n*<=×<=*n* checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Then *n* lines follow containing the description of the checkerboard. Each of them contains *n* characters (either 'x' or 'o') without spaces.
Output Specification:
Print "YES" or "NO" (without the quotes) depending on the answer to the problem.
Demo Input:
['3\nxxo\nxox\noxx\n', '4\nxxxo\nxoxo\noxox\nxxxx\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
def miniSin(angle):
return (2 - angle) * (angle % 2)
n = int(input())
gridList = [[i for i in input()] for x in range(n) ]
def checkOs():
for RowIndex in range(n):
for columnIndex in range(n):
numberOfOs = 0
for i in range(4):
RI = RowIndex + miniSin((i+1)%4)
CI = columnIndex + miniSin(i)
if RI < 0 or RI == n or CI < 0 or CI == n:
continue
if gridList[RowIndex + miniSin((i+1)%4)][columnIndex + miniSin(i)] == "o":
numberOfOs += 1
if numberOfOs % 2 != 0:
return "NO"
return "YES"
print(checkOs())
```
| 3
|
|
379
|
A
|
New Year Candles
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
|
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
|
Print a single integer — the number of hours Vasily can light up the room for.
|
[
"4 2\n",
"6 3\n"
] |
[
"7\n",
"8\n"
] |
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
| 500
|
[
{
"input": "4 2",
"output": "7"
},
{
"input": "6 3",
"output": "8"
},
{
"input": "1000 1000",
"output": "1001"
},
{
"input": "123 5",
"output": "153"
},
{
"input": "1000 2",
"output": "1999"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "2 2",
"output": "3"
},
{
"input": "3 2",
"output": "5"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "999 2",
"output": "1997"
},
{
"input": "1000 3",
"output": "1499"
},
{
"input": "1000 4",
"output": "1333"
},
{
"input": "1 1000",
"output": "1"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "10 4",
"output": "13"
},
{
"input": "4 3",
"output": "5"
},
{
"input": "91 5",
"output": "113"
},
{
"input": "777 17",
"output": "825"
},
{
"input": "100 4",
"output": "133"
},
{
"input": "5 3",
"output": "7"
},
{
"input": "6 4",
"output": "7"
},
{
"input": "26 8",
"output": "29"
},
{
"input": "9 4",
"output": "11"
},
{
"input": "20 3",
"output": "29"
},
{
"input": "17 3",
"output": "25"
}
] | 1,658,557,056
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
n,m = map(int,input().split())
x = n
while n!=0:
x+=n//m
n//=m
print(x)
|
Title: New Year Candles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input Specification:
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
Output Specification:
Print a single integer — the number of hours Vasily can light up the room for.
Demo Input:
['4 2\n', '6 3\n']
Demo Output:
['7\n', '8\n']
Note:
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
|
```python
n,m = map(int,input().split())
x = n
while n!=0:
x+=n//m
n//=m
print(x)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
This is an interactive problem.
Natasha is going to fly to Mars. Finally, Natasha sat in the rocket. She flies, flies... but gets bored. She wishes to arrive to Mars already! So she decides to find something to occupy herself. She couldn't think of anything better to do than to calculate the distance to the red planet.
Let's define $x$ as the distance to Mars. Unfortunately, Natasha does not know $x$. But it is known that $1 \le x \le m$, where Natasha knows the number $m$. Besides, $x$ and $m$ are positive integers.
Natasha can ask the rocket questions. Every question is an integer $y$ ($1 \le y \le m$). The correct answer to the question is $-1$, if $x<y$, $0$, if $x=y$, and $1$, if $x>y$. But the rocket is broken — it does not always answer correctly. Precisely: let the correct answer to the current question be equal to $t$, then, if the rocket answers this question correctly, then it will answer $t$, otherwise it will answer $-t$.
In addition, the rocket has a sequence $p$ of length $n$. Each element of the sequence is either $0$ or $1$. The rocket processes this sequence in the cyclic order, that is $1$-st element, $2$-nd, $3$-rd, $\ldots$, $(n-1)$-th, $n$-th, $1$-st, $2$-nd, $3$-rd, $\ldots$, $(n-1)$-th, $n$-th, $\ldots$. If the current element is $1$, the rocket answers correctly, if $0$ — lies. Natasha doesn't know the sequence $p$, but she knows its length — $n$.
You can ask the rocket no more than $60$ questions.
Help Natasha find the distance to Mars. Assume, that the distance to Mars does not change while Natasha is asking questions.
Your solution will not be accepted, if it does not receive an answer $0$ from the rocket (even if the distance to Mars is uniquely determined by the already received rocket's answers).
|
The first line contains two integers $m$ and $n$ ($1 \le m \le 10^9$, $1 \le n \le 30$) — the maximum distance to Mars and the number of elements in the sequence $p$.
|
none
|
[
"5 2\n1\n-1\n-1\n1\n0\n"
] |
[
"1\n2\n4\n5\n3\n"
] |
In the example, hacking would look like this:
5 2 3
1 0
This means that the current distance to Mars is equal to $3$, Natasha knows that it does not exceed $5$, and the rocket answers in order: correctly, incorrectly, correctly, incorrectly ...
Really:
on the first query ($1$) the correct answer is $1$, the rocket answered correctly: $1$;
on the second query ($2$) the correct answer is $1$, the rocket answered incorrectly: $-1$;
on the third query ($4$) the correct answer is $-1$, the rocket answered correctly: $-1$;
on the fourth query ($5$) the correct answer is $-1$, the rocket answered incorrectly: $1$;
on the fifth query ($3$) the correct and incorrect answer is $0$.
| 0
|
[
{
"input": "5 2 3\n1 0",
"output": "3 queries, x=3"
},
{
"input": "1 1 1\n1",
"output": "1 queries, x=1"
},
{
"input": "3 2 3\n1 0",
"output": "4 queries, x=3"
},
{
"input": "6 3 5\n1 1 1",
"output": "5 queries, x=5"
},
{
"input": "10 4 3\n0 0 1 0",
"output": "6 queries, x=3"
},
{
"input": "30 5 16\n0 1 1 1 0",
"output": "6 queries, x=16"
},
{
"input": "60 6 21\n1 0 0 1 0 1",
"output": "11 queries, x=21"
},
{
"input": "100 7 73\n0 0 0 1 0 1 1",
"output": "14 queries, x=73"
},
{
"input": "1000000000 29 958572235\n1 1 0 1 1 1 1 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0",
"output": "58 queries, x=958572235"
},
{
"input": "738009704 30 116044407\n0 0 1 1 0 0 1 0 0 0 1 0 0 0 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1",
"output": "59 queries, x=116044407"
},
{
"input": "300 8 165\n1 1 1 0 0 1 1 0",
"output": "16 queries, x=165"
},
{
"input": "600 9 150\n0 0 1 0 1 0 1 0 1",
"output": "19 queries, x=150"
},
{
"input": "1000 10 140\n0 0 0 0 1 0 0 0 0 0",
"output": "20 queries, x=140"
},
{
"input": "3000 11 1896\n1 0 1 1 0 0 0 0 1 1 1",
"output": "21 queries, x=1896"
},
{
"input": "6000 12 4679\n1 0 1 1 1 1 1 0 0 0 0 1",
"output": "23 queries, x=4679"
},
{
"input": "10000 13 4977\n1 0 1 1 0 0 0 1 0 0 1 1 0",
"output": "26 queries, x=4977"
},
{
"input": "30000 14 60\n1 1 1 0 0 1 0 1 0 0 1 0 0 0",
"output": "28 queries, x=60"
},
{
"input": "60000 15 58813\n0 1 1 0 1 1 0 0 0 1 1 1 1 0 1",
"output": "27 queries, x=58813"
},
{
"input": "100000 16 79154\n1 1 1 0 0 0 0 0 1 1 0 1 0 1 0 1",
"output": "32 queries, x=79154"
},
{
"input": "300000 17 11107\n1 0 0 0 1 0 0 0 1 1 1 0 0 1 1 1 0",
"output": "34 queries, x=11107"
},
{
"input": "600000 18 146716\n0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 1",
"output": "37 queries, x=146716"
},
{
"input": "1000000 19 418016\n1 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 0 0",
"output": "38 queries, x=418016"
},
{
"input": "3000000 20 642518\n1 0 0 1 0 1 1 1 1 1 0 0 0 1 0 1 0 1 0 1",
"output": "41 queries, x=642518"
},
{
"input": "6000000 21 3516807\n0 0 0 1 0 1 0 1 1 0 0 0 0 1 0 1 1 1 0 0 0",
"output": "43 queries, x=3516807"
},
{
"input": "10000000 22 8115129\n1 0 1 0 0 0 0 0 0 1 1 0 1 0 0 0 0 1 1 0 0 1",
"output": "42 queries, x=8115129"
},
{
"input": "30000000 23 10362635\n0 1 0 0 1 1 1 1 0 0 0 1 0 1 1 0 1 1 1 1 0 0 0",
"output": "48 queries, x=10362635"
},
{
"input": "60000000 24 52208533\n1 1 1 0 1 0 0 0 0 0 1 0 0 1 0 1 0 1 1 0 1 1 1 0",
"output": "46 queries, x=52208533"
},
{
"input": "100000000 25 51744320\n0 1 1 1 1 0 1 1 1 1 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1",
"output": "50 queries, x=51744320"
},
{
"input": "300000000 26 264009490\n1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 1 1 0 1 1 1 0 1 1",
"output": "54 queries, x=264009490"
},
{
"input": "600000000 27 415720732\n1 1 1 1 1 1 0 0 1 1 1 0 1 0 1 1 0 0 1 1 1 1 0 1 0 1 0",
"output": "56 queries, x=415720732"
},
{
"input": "1000000000 28 946835863\n0 0 1 0 1 1 1 0 1 0 1 1 0 1 0 1 1 0 0 0 1 0 1 0 1 1 0 0",
"output": "58 queries, x=946835863"
},
{
"input": "1000000000 29 124919287\n0 0 1 0 0 0 1 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 1 1 0 1 0 0",
"output": "59 queries, x=124919287"
},
{
"input": "1000000000 30 202669473\n1 1 0 1 1 1 0 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 1 1 1 0 0 0 0 0",
"output": "58 queries, x=202669473"
},
{
"input": "1000000000 13 532121080\n1 1 1 0 1 1 0 0 0 0 1 0 1",
"output": "42 queries, x=532121080"
},
{
"input": "1000000000 27 105669924\n0 1 1 1 0 1 0 1 0 0 0 1 0 0 1 1 0 1 1 0 0 1 0 1 1 1 1",
"output": "57 queries, x=105669924"
},
{
"input": "1000000000 11 533994576\n0 0 1 0 1 1 1 1 0 1 0",
"output": "38 queries, x=533994576"
},
{
"input": "1000000000 9 107543421\n1 0 0 1 1 1 1 1 1",
"output": "39 queries, x=107543421"
},
{
"input": "1000000000 23 976059561\n1 0 0 0 0 1 0 0 1 1 1 1 1 1 1 0 0 0 0 1 0 0 1",
"output": "53 queries, x=976059561"
},
{
"input": "1000000000 7 549608406\n1 1 1 0 1 1 1",
"output": "36 queries, x=549608406"
},
{
"input": "1000000000 21 123157250\n0 1 1 1 1 1 1 0 0 0 1 1 1 1 0 0 0 1 0 0 1",
"output": "49 queries, x=123157250"
},
{
"input": "1000000000 19 696706094\n0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0",
"output": "47 queries, x=696706094"
},
{
"input": "1000000000 3 125030747\n0 0 0",
"output": "33 queries, x=125030747"
},
{
"input": "1000000000 17 993546887\n1 0 1 0 1 1 1 1 1 1 0 1 0 1 1 1 1",
"output": "46 queries, x=993546887"
},
{
"input": "1000000000 15 567095731\n1 1 1 0 0 1 1 1 0 1 0 0 1 0 0",
"output": "45 queries, x=567095731"
},
{
"input": "1000000000 29 140644576\n1 1 1 1 1 1 0 1 0 0 0 1 0 1 0 1 1 1 0 1 1 1 0 1 1 0 1 0 0",
"output": "58 queries, x=140644576"
},
{
"input": "1000000000 13 714193420\n0 1 0 0 0 1 0 0 0 0 1 1 1",
"output": "43 queries, x=714193420"
},
{
"input": "1000000000 27 142518072\n0 0 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 0 1 0 0",
"output": "52 queries, x=142518072"
},
{
"input": "1000000000 25 11034213\n0 0 1 0 1 1 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 0 1 0 0",
"output": "54 queries, x=11034213"
},
{
"input": "1000000000 9 584583057\n1 1 1 0 0 1 0 0 0",
"output": "35 queries, x=584583057"
},
{
"input": "1000000000 23 863164606\n1 1 0 1 0 1 0 1 0 1 0 0 1 0 0 0 1 1 1 0 0 1 1",
"output": "53 queries, x=863164606"
},
{
"input": "1000000000 21 731680746\n1 1 0 0 1 1 1 1 1 0 0 1 0 1 1 1 1 0 1 0 1",
"output": "51 queries, x=731680746"
},
{
"input": "1000000000 5 305229590\n0 0 1 1 0",
"output": "35 queries, x=305229590"
},
{
"input": "1000000000 3 28521539\n0 0 1",
"output": "31 queries, x=28521539"
},
{
"input": "1000000000 3 602070383\n0 1 1",
"output": "32 queries, x=602070383"
},
{
"input": "1000000000 2 880651931\n1 1",
"output": "30 queries, x=880651931"
},
{
"input": "1000000000 16 749168072\n1 1 0 0 0 1 0 0 1 1 1 1 1 1 1 0",
"output": "46 queries, x=749168072"
},
{
"input": "1000000000 30 322716916\n1 0 1 1 1 1 0 1 1 0 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 1 1 0 0",
"output": "58 queries, x=322716916"
},
{
"input": "1000000000 14 191233057\n0 0 1 0 0 1 1 1 1 0 0 0 1 1",
"output": "43 queries, x=191233057"
},
{
"input": "1000000000 30 1\n1 1 0 1 1 0 0 1 0 1 1 0 0 1 1 1 0 0 1 1 0 0 0 0 0 1 0 0 0 0",
"output": "1 queries, x=1"
},
{
"input": "1000000000 30 1\n1 0 1 1 1 1 1 1 0 1 0 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1",
"output": "1 queries, x=1"
},
{
"input": "1000000000 30 1\n1 0 1 0 0 0 0 1 1 0 0 1 1 0 1 1 1 0 1 0 1 1 1 0 0 0 1 0 1 1",
"output": "1 queries, x=1"
},
{
"input": "1000000000 30 1\n1 0 1 0 0 0 1 1 1 0 1 1 1 1 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1",
"output": "1 queries, x=1"
},
{
"input": "1000000000 30 1\n1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 0 0 0 0 0 1 1 0 0 0 1 0 0 0",
"output": "1 queries, x=1"
},
{
"input": "1000000000 30 1000000000\n1 1 1 0 0 0 1 1 1 1 0 1 0 0 0 1 1 0 1 1 0 0 0 1 0 0 0 0 1 0",
"output": "60 queries, x=1000000000"
},
{
"input": "1000000000 30 1000000000\n1 1 1 0 0 1 1 1 0 0 1 1 0 0 1 1 1 0 1 0 1 1 0 0 1 0 1 1 1 0",
"output": "60 queries, x=1000000000"
},
{
"input": "1000000000 30 1000000000\n0 0 1 1 1 0 0 1 0 0 1 1 0 1 0 0 1 0 1 1 1 1 1 1 0 0 1 0 1 1",
"output": "60 queries, x=1000000000"
},
{
"input": "1000000000 30 1000000000\n0 0 0 1 1 1 1 1 1 0 1 0 1 0 1 1 0 1 1 0 0 1 0 0 1 0 0 1 0 1",
"output": "60 queries, x=1000000000"
},
{
"input": "1000000000 30 1000000000\n0 0 0 1 1 1 0 1 1 0 0 0 1 1 0 0 0 1 0 0 0 0 1 0 0 1 1 0 0 1",
"output": "60 queries, x=1000000000"
},
{
"input": "1 30 1\n1 1 1 0 1 0 0 0 0 1 1 0 1 0 0 1 0 1 0 1 0 1 0 1 1 0 0 0 1 1",
"output": "1 queries, x=1"
},
{
"input": "1 30 1\n1 1 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 0 0 1 1 0 0",
"output": "1 queries, x=1"
},
{
"input": "1 30 1\n1 0 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 0 1 1 0 0 1 1 0 1 0 0 0",
"output": "1 queries, x=1"
},
{
"input": "1 30 1\n1 0 1 1 1 0 0 0 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 0 1 0 1 0 0",
"output": "1 queries, x=1"
},
{
"input": "1 30 1\n1 0 1 1 1 1 1 0 0 0 1 1 0 1 0 1 1 0 0 0 0 0 1 1 1 1 1 0 1 1",
"output": "1 queries, x=1"
},
{
"input": "2 1 2\n1",
"output": "2 queries, x=2"
},
{
"input": "1000000000 1 1000000000\n1",
"output": "31 queries, x=1000000000"
},
{
"input": "10000 1 10000\n1",
"output": "15 queries, x=10000"
},
{
"input": "1000000000 1 999999999\n1",
"output": "30 queries, x=999999999"
},
{
"input": "100000 2 15\n1 0",
"output": "19 queries, x=15"
},
{
"input": "200000 1 110000\n1",
"output": "17 queries, x=110000"
},
{
"input": "123456789 1 42\n1",
"output": "27 queries, x=42"
},
{
"input": "1000000000 1 9\n1",
"output": "30 queries, x=9"
},
{
"input": "200000 2 100002\n1 0",
"output": "19 queries, x=100002"
},
{
"input": "1000000000 3 234567890\n0 1 0",
"output": "31 queries, x=234567890"
},
{
"input": "1000000000 5 321732193\n1 1 0 1 0",
"output": "35 queries, x=321732193"
},
{
"input": "1000000000 1 804289384\n1",
"output": "27 queries, x=804289384"
},
{
"input": "1000000000 2 999999998\n1 0",
"output": "32 queries, x=999999998"
},
{
"input": "1000000000 5 384618761\n0 1 1 0 1",
"output": "33 queries, x=384618761"
},
{
"input": "100000000 1 100\n0",
"output": "28 queries, x=100"
},
{
"input": "1000000000 1 804289384\n0",
"output": "27 queries, x=804289384"
},
{
"input": "100000000 1 100000000\n1",
"output": "28 queries, x=100000000"
},
{
"input": "40 1 4\n0",
"output": "6 queries, x=4"
},
{
"input": "1000000000 2 999999998\n0 1",
"output": "32 queries, x=999999998"
},
{
"input": "1000000000 1 1000000000\n0",
"output": "31 queries, x=1000000000"
},
{
"input": "1000000000 2 255555555\n1 0",
"output": "31 queries, x=255555555"
},
{
"input": "1000000000 2 1000000000\n0 1",
"output": "32 queries, x=1000000000"
},
{
"input": "1000000000 1 999999999\n0",
"output": "30 queries, x=999999999"
},
{
"input": "1000000000 2 888888888\n0 1",
"output": "31 queries, x=888888888"
},
{
"input": "1000000000 1 77000000\n1",
"output": "31 queries, x=77000000"
},
{
"input": "1000000000 1 123456789\n1",
"output": "27 queries, x=123456789"
},
{
"input": "10000 1 228\n0",
"output": "14 queries, x=228"
},
{
"input": "1000000000 1 12345\n1",
"output": "31 queries, x=12345"
},
{
"input": "1000000000 1 77000000\n0",
"output": "31 queries, x=77000000"
},
{
"input": "1000000000 1 23333\n0",
"output": "31 queries, x=23333"
},
{
"input": "1000000000 4 100\n0 1 0 1",
"output": "34 queries, x=100"
},
{
"input": "100000000 1 200\n1",
"output": "27 queries, x=200"
},
{
"input": "1000000000 3 5\n0 1 0",
"output": "33 queries, x=5"
},
{
"input": "1000000000 12 2\n1 1 1 1 1 1 0 0 1 1 1 1",
"output": "41 queries, x=2"
},
{
"input": "1000000000 1 5\n0",
"output": "31 queries, x=5"
},
{
"input": "100000 2 99999\n0 0",
"output": "18 queries, x=99999"
},
{
"input": "100000 2 2\n0 1",
"output": "18 queries, x=2"
},
{
"input": "1000000 1 91923\n0",
"output": "21 queries, x=91923"
},
{
"input": "1000000 2 1235\n0 1",
"output": "22 queries, x=1235"
},
{
"input": "1000000000 1 5\n1",
"output": "31 queries, x=5"
},
{
"input": "100000000 2 1234567\n0 1",
"output": "28 queries, x=1234567"
},
{
"input": "1000000000 1 1\n1",
"output": "1 queries, x=1"
},
{
"input": "1000000000 4 999999999\n1 0 0 1",
"output": "33 queries, x=999999999"
},
{
"input": "1000000000 4 1000000000\n1 0 1 0",
"output": "34 queries, x=1000000000"
},
{
"input": "1000000000 4 1\n1 0 1 0",
"output": "1 queries, x=1"
},
{
"input": "1000000000 5 500\n1 1 0 0 1",
"output": "34 queries, x=500"
},
{
"input": "1000 1 13\n1",
"output": "11 queries, x=13"
},
{
"input": "999999999 2 123456789\n1 0",
"output": "31 queries, x=123456789"
},
{
"input": "1000000000 5 1000000000\n1 1 1 1 1",
"output": "35 queries, x=1000000000"
},
{
"input": "1000000000 3 123456789\n1 0 1",
"output": "29 queries, x=123456789"
},
{
"input": "1000000000 1 123\n1",
"output": "29 queries, x=123"
},
{
"input": "10 1 9\n1",
"output": "4 queries, x=9"
},
{
"input": "1000000000 1 1\n0",
"output": "1 queries, x=1"
},
{
"input": "100000000 2 12345678\n0 1",
"output": "28 queries, x=12345678"
},
{
"input": "1000000000 30 1000000000\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "60 queries, x=1000000000"
},
{
"input": "805306369 30 805306369\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "60 queries, x=805306369"
},
{
"input": "1000000000 2 123\n0 1",
"output": "30 queries, x=123"
},
{
"input": "1000000000 1 9\n0",
"output": "30 queries, x=9"
},
{
"input": "100000 2 90000\n1 0",
"output": "18 queries, x=90000"
},
{
"input": "1000000000 2 123456789\n1 0",
"output": "28 queries, x=123456789"
},
{
"input": "1000000000 2 804289384\n1 0",
"output": "28 queries, x=804289384"
},
{
"input": "1000000000 30 999999970\n1 0 0 1 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1 1 0 0 0 1 0 1 0 0 0 1",
"output": "60 queries, x=999999970"
}
] | 1,532,687,957
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 32
| 124
| 0
|
from sys import stdout
def scan(obj = map, type = int, sep = ' '):
return obj(type, input().split(sep))
def output(text):
print(f'{text}\n')
stdout.flush()
return int(input())
m, n = scan()
p = []
ans = 0
for i in range(n):
ans = output('1')
if ans == 0 or ans == -2:
exit()
if ans == -1:
p.append(-1)
else:
p.append(1)
ans = m // 2
for i in range(31):
res = output(ans) * p[i % n]
if res == 1:
x = m // (2 ** (i + 2))
ans += x if x > 0 else 1
elif res == -1:
x = m // (2 ** (i + 2))
ans -= x if x > 0 else 1
elif res == 0 or res == -2:
exit()
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This is an interactive problem.
Natasha is going to fly to Mars. Finally, Natasha sat in the rocket. She flies, flies... but gets bored. She wishes to arrive to Mars already! So she decides to find something to occupy herself. She couldn't think of anything better to do than to calculate the distance to the red planet.
Let's define $x$ as the distance to Mars. Unfortunately, Natasha does not know $x$. But it is known that $1 \le x \le m$, where Natasha knows the number $m$. Besides, $x$ and $m$ are positive integers.
Natasha can ask the rocket questions. Every question is an integer $y$ ($1 \le y \le m$). The correct answer to the question is $-1$, if $x<y$, $0$, if $x=y$, and $1$, if $x>y$. But the rocket is broken — it does not always answer correctly. Precisely: let the correct answer to the current question be equal to $t$, then, if the rocket answers this question correctly, then it will answer $t$, otherwise it will answer $-t$.
In addition, the rocket has a sequence $p$ of length $n$. Each element of the sequence is either $0$ or $1$. The rocket processes this sequence in the cyclic order, that is $1$-st element, $2$-nd, $3$-rd, $\ldots$, $(n-1)$-th, $n$-th, $1$-st, $2$-nd, $3$-rd, $\ldots$, $(n-1)$-th, $n$-th, $\ldots$. If the current element is $1$, the rocket answers correctly, if $0$ — lies. Natasha doesn't know the sequence $p$, but she knows its length — $n$.
You can ask the rocket no more than $60$ questions.
Help Natasha find the distance to Mars. Assume, that the distance to Mars does not change while Natasha is asking questions.
Your solution will not be accepted, if it does not receive an answer $0$ from the rocket (even if the distance to Mars is uniquely determined by the already received rocket's answers).
Input Specification:
The first line contains two integers $m$ and $n$ ($1 \le m \le 10^9$, $1 \le n \le 30$) — the maximum distance to Mars and the number of elements in the sequence $p$.
Output Specification:
none
Demo Input:
['5 2\n1\n-1\n-1\n1\n0\n']
Demo Output:
['1\n2\n4\n5\n3\n']
Note:
In the example, hacking would look like this:
5 2 3
1 0
This means that the current distance to Mars is equal to $3$, Natasha knows that it does not exceed $5$, and the rocket answers in order: correctly, incorrectly, correctly, incorrectly ...
Really:
on the first query ($1$) the correct answer is $1$, the rocket answered correctly: $1$;
on the second query ($2$) the correct answer is $1$, the rocket answered incorrectly: $-1$;
on the third query ($4$) the correct answer is $-1$, the rocket answered correctly: $-1$;
on the fourth query ($5$) the correct answer is $-1$, the rocket answered incorrectly: $1$;
on the fifth query ($3$) the correct and incorrect answer is $0$.
|
```python
from sys import stdout
def scan(obj = map, type = int, sep = ' '):
return obj(type, input().split(sep))
def output(text):
print(f'{text}\n')
stdout.flush()
return int(input())
m, n = scan()
p = []
ans = 0
for i in range(n):
ans = output('1')
if ans == 0 or ans == -2:
exit()
if ans == -1:
p.append(-1)
else:
p.append(1)
ans = m // 2
for i in range(31):
res = output(ans) * p[i % n]
if res == 1:
x = m // (2 ** (i + 2))
ans += x if x > 0 else 1
elif res == -1:
x = m // (2 ** (i + 2))
ans -= x if x > 0 else 1
elif res == 0 or res == -2:
exit()
```
| 0
|
|
799
|
A
|
Carrot Cakes
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
In some game by Playrix it takes *t* minutes for an oven to bake *k* carrot cakes, all cakes are ready at the same moment *t* minutes after they started baking. Arkady needs at least *n* cakes to complete a task, but he currently don't have any. However, he has infinitely many ingredients and one oven. Moreover, Arkady can build one more similar oven to make the process faster, it would take *d* minutes to build the oven. While the new oven is being built, only old one can bake cakes, after the new oven is built, both ovens bake simultaneously. Arkady can't build more than one oven.
Determine if it is reasonable to build the second oven, i.e. will it decrease the minimum time needed to get *n* cakes or not. If the time needed with the second oven is the same as with one oven, then it is unreasonable.
|
The only line contains four integers *n*, *t*, *k*, *d* (1<=≤<=*n*,<=*t*,<=*k*,<=*d*<=≤<=1<=000) — the number of cakes needed, the time needed for one oven to bake *k* cakes, the number of cakes baked at the same time, the time needed to build the second oven.
|
If it is reasonable to build the second oven, print "YES". Otherwise print "NO".
|
[
"8 6 4 5\n",
"8 6 4 6\n",
"10 3 11 4\n",
"4 2 1 4\n"
] |
[
"YES\n",
"NO\n",
"NO\n",
"YES\n"
] |
In the first example it is possible to get 8 cakes in 12 minutes using one oven. The second oven can be built in 5 minutes, so after 6 minutes the first oven bakes 4 cakes, the second oven bakes 4 more ovens after 11 minutes. Thus, it is reasonable to build the second oven.
In the second example it doesn't matter whether we build the second oven or not, thus it takes 12 minutes to bake 8 cakes in both cases. Thus, it is unreasonable to build the second oven.
In the third example the first oven bakes 11 cakes in 3 minutes, that is more than needed 10. It is unreasonable to build the second oven, because its building takes more time that baking the needed number of cakes using the only oven.
| 500
|
[
{
"input": "8 6 4 5",
"output": "YES"
},
{
"input": "8 6 4 6",
"output": "NO"
},
{
"input": "10 3 11 4",
"output": "NO"
},
{
"input": "4 2 1 4",
"output": "YES"
},
{
"input": "28 17 16 26",
"output": "NO"
},
{
"input": "60 69 9 438",
"output": "NO"
},
{
"input": "599 97 54 992",
"output": "YES"
},
{
"input": "11 22 18 17",
"output": "NO"
},
{
"input": "1 13 22 11",
"output": "NO"
},
{
"input": "1 1 1 1",
"output": "NO"
},
{
"input": "3 1 1 1",
"output": "YES"
},
{
"input": "1000 1000 1000 1000",
"output": "NO"
},
{
"input": "1000 1000 1 1",
"output": "YES"
},
{
"input": "1000 1000 1 400",
"output": "YES"
},
{
"input": "1000 1000 1 1000",
"output": "YES"
},
{
"input": "1000 1000 1 999",
"output": "YES"
},
{
"input": "53 11 3 166",
"output": "YES"
},
{
"input": "313 2 3 385",
"output": "NO"
},
{
"input": "214 9 9 412",
"output": "NO"
},
{
"input": "349 9 5 268",
"output": "YES"
},
{
"input": "611 16 8 153",
"output": "YES"
},
{
"input": "877 13 3 191",
"output": "YES"
},
{
"input": "340 9 9 10",
"output": "YES"
},
{
"input": "31 8 2 205",
"output": "NO"
},
{
"input": "519 3 2 148",
"output": "YES"
},
{
"input": "882 2 21 219",
"output": "NO"
},
{
"input": "982 13 5 198",
"output": "YES"
},
{
"input": "428 13 6 272",
"output": "YES"
},
{
"input": "436 16 14 26",
"output": "YES"
},
{
"input": "628 10 9 386",
"output": "YES"
},
{
"input": "77 33 18 31",
"output": "YES"
},
{
"input": "527 36 4 8",
"output": "YES"
},
{
"input": "128 18 2 169",
"output": "YES"
},
{
"input": "904 4 2 288",
"output": "YES"
},
{
"input": "986 4 3 25",
"output": "YES"
},
{
"input": "134 8 22 162",
"output": "NO"
},
{
"input": "942 42 3 69",
"output": "YES"
},
{
"input": "894 4 9 4",
"output": "YES"
},
{
"input": "953 8 10 312",
"output": "YES"
},
{
"input": "43 8 1 121",
"output": "YES"
},
{
"input": "12 13 19 273",
"output": "NO"
},
{
"input": "204 45 10 871",
"output": "YES"
},
{
"input": "342 69 50 425",
"output": "NO"
},
{
"input": "982 93 99 875",
"output": "NO"
},
{
"input": "283 21 39 132",
"output": "YES"
},
{
"input": "1000 45 83 686",
"output": "NO"
},
{
"input": "246 69 36 432",
"output": "NO"
},
{
"input": "607 93 76 689",
"output": "NO"
},
{
"input": "503 21 24 435",
"output": "NO"
},
{
"input": "1000 45 65 989",
"output": "NO"
},
{
"input": "30 21 2 250",
"output": "YES"
},
{
"input": "1000 49 50 995",
"output": "NO"
},
{
"input": "383 69 95 253",
"output": "YES"
},
{
"input": "393 98 35 999",
"output": "YES"
},
{
"input": "1000 22 79 552",
"output": "NO"
},
{
"input": "268 294 268 154",
"output": "NO"
},
{
"input": "963 465 706 146",
"output": "YES"
},
{
"input": "304 635 304 257",
"output": "NO"
},
{
"input": "4 2 1 6",
"output": "NO"
},
{
"input": "1 51 10 50",
"output": "NO"
},
{
"input": "5 5 4 4",
"output": "YES"
},
{
"input": "3 2 1 1",
"output": "YES"
},
{
"input": "3 4 3 3",
"output": "NO"
},
{
"input": "7 3 4 1",
"output": "YES"
},
{
"input": "101 10 1 1000",
"output": "NO"
},
{
"input": "5 1 1 1",
"output": "YES"
},
{
"input": "5 10 5 5",
"output": "NO"
},
{
"input": "19 1 7 1",
"output": "YES"
},
{
"input": "763 572 745 262",
"output": "YES"
},
{
"input": "1 2 1 1",
"output": "NO"
},
{
"input": "5 1 1 3",
"output": "YES"
},
{
"input": "170 725 479 359",
"output": "NO"
},
{
"input": "6 2 1 7",
"output": "YES"
},
{
"input": "6 2 5 1",
"output": "YES"
},
{
"input": "1 2 2 1",
"output": "NO"
},
{
"input": "24 2 8 3",
"output": "YES"
},
{
"input": "7 3 3 3",
"output": "YES"
},
{
"input": "5 2 2 2",
"output": "YES"
},
{
"input": "3 2 1 2",
"output": "YES"
},
{
"input": "1000 2 200 8",
"output": "NO"
},
{
"input": "3 100 2 100",
"output": "NO"
},
{
"input": "2 999 1 1000",
"output": "NO"
},
{
"input": "2 1 1 1",
"output": "NO"
},
{
"input": "2 3 5 1",
"output": "NO"
},
{
"input": "100 1 5 1",
"output": "YES"
},
{
"input": "7 2 3 3",
"output": "YES"
},
{
"input": "4 1 1 3",
"output": "NO"
},
{
"input": "3 2 2 1",
"output": "YES"
},
{
"input": "1 1 1 2",
"output": "NO"
},
{
"input": "91 8 7 13",
"output": "YES"
},
{
"input": "3 1 2 1",
"output": "NO"
},
{
"input": "5 3 2 3",
"output": "YES"
},
{
"input": "9 6 6 3",
"output": "YES"
}
] | 1,665,773,877
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 31
| 0
|
gereken, pişme, pişmiş, yapma=map(str,input().split())
if gereken>pişmiş and pişme>yapma:
print("YES")
elif gereken==pişmiş or gereken<pişmiş:
print("NO")
else:
print("NO")
exit()
|
Title: Carrot Cakes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In some game by Playrix it takes *t* minutes for an oven to bake *k* carrot cakes, all cakes are ready at the same moment *t* minutes after they started baking. Arkady needs at least *n* cakes to complete a task, but he currently don't have any. However, he has infinitely many ingredients and one oven. Moreover, Arkady can build one more similar oven to make the process faster, it would take *d* minutes to build the oven. While the new oven is being built, only old one can bake cakes, after the new oven is built, both ovens bake simultaneously. Arkady can't build more than one oven.
Determine if it is reasonable to build the second oven, i.e. will it decrease the minimum time needed to get *n* cakes or not. If the time needed with the second oven is the same as with one oven, then it is unreasonable.
Input Specification:
The only line contains four integers *n*, *t*, *k*, *d* (1<=≤<=*n*,<=*t*,<=*k*,<=*d*<=≤<=1<=000) — the number of cakes needed, the time needed for one oven to bake *k* cakes, the number of cakes baked at the same time, the time needed to build the second oven.
Output Specification:
If it is reasonable to build the second oven, print "YES". Otherwise print "NO".
Demo Input:
['8 6 4 5\n', '8 6 4 6\n', '10 3 11 4\n', '4 2 1 4\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n', 'YES\n']
Note:
In the first example it is possible to get 8 cakes in 12 minutes using one oven. The second oven can be built in 5 minutes, so after 6 minutes the first oven bakes 4 cakes, the second oven bakes 4 more ovens after 11 minutes. Thus, it is reasonable to build the second oven.
In the second example it doesn't matter whether we build the second oven or not, thus it takes 12 minutes to bake 8 cakes in both cases. Thus, it is unreasonable to build the second oven.
In the third example the first oven bakes 11 cakes in 3 minutes, that is more than needed 10. It is unreasonable to build the second oven, because its building takes more time that baking the needed number of cakes using the only oven.
|
```python
gereken, pişme, pişmiş, yapma=map(str,input().split())
if gereken>pişmiş and pişme>yapma:
print("YES")
elif gereken==pişmiş or gereken<pişmiş:
print("NO")
else:
print("NO")
exit()
```
| 0
|
|
230
|
B
|
T-primes
|
PROGRAMMING
| 1,300
|
[
"binary search",
"implementation",
"math",
"number theory"
] | null | null |
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
|
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
|
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
|
[
"3\n4 5 6\n"
] |
[
"YES\nNO\nNO\n"
] |
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
| 500
|
[
{
"input": "3\n4 5 6",
"output": "YES\nNO\nNO"
},
{
"input": "2\n48 49",
"output": "NO\nYES"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output": "NO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO"
},
{
"input": "1\n36",
"output": "NO"
},
{
"input": "1\n999966000289",
"output": "YES"
},
{
"input": "1\n999993399999",
"output": "NO"
},
{
"input": "9\n111 121 131 111 121 131 111 121 131",
"output": "NO\nYES\nNO\nNO\nYES\nNO\nNO\nYES\nNO"
},
{
"input": "1\n1",
"output": "NO"
},
{
"input": "1\n10",
"output": "NO"
},
{
"input": "1\n976197352729",
"output": "NO"
},
{
"input": "1\n1000000000000",
"output": "NO"
},
{
"input": "1\n9",
"output": "YES"
},
{
"input": "6\n549755813888 847288609443 762939453125 678223072849 285311670611 137858491849",
"output": "NO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "3\n223092870 6469693230 200560490130",
"output": "NO\nNO\nNO"
},
{
"input": "2\n81 25",
"output": "NO\nYES"
},
{
"input": "1\n16",
"output": "NO"
},
{
"input": "22\n1 2 3 4 5 6 7 8 9 10 12752041 64 121 144 27550356289 124 24657 23756 135153365 25235235235 42351351 81",
"output": "NO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "1\n225",
"output": "NO"
},
{
"input": "1\n1521",
"output": "NO"
},
{
"input": "1\n81",
"output": "NO"
}
] | 1,697,351,345
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
def prime(i):
import math
for x in range(2, int(math.sqrt(int(i ** 0.5))) + 1):
if int(i ** 0.5) % x == 0:
return False
return True
n = int(input())
num = list(map(int, input().split()))
import math
for i in num:
if math.ceil(i ** 0.5) == int(i ** 0.5):
if prime(i) == False:
print('NO')
else:
print('YES')
else:
print('NO')
|
Title: T-primes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
Input Specification:
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
Demo Input:
['3\n4 5 6\n']
Demo Output:
['YES\nNO\nNO\n']
Note:
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
|
```python
def prime(i):
import math
for x in range(2, int(math.sqrt(int(i ** 0.5))) + 1):
if int(i ** 0.5) % x == 0:
return False
return True
n = int(input())
num = list(map(int, input().split()))
import math
for i in num:
if math.ceil(i ** 0.5) == int(i ** 0.5):
if prime(i) == False:
print('NO')
else:
print('YES')
else:
print('NO')
```
| 0
|
|
478
|
B
|
Random Teams
|
PROGRAMMING
| 1,300
|
[
"combinatorics",
"constructive algorithms",
"greedy",
"math"
] | null | null |
*n* participants of the competition were split into *m* teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
|
The only line of input contains two integers *n* and *m*, separated by a single space (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the number of participants and the number of teams respectively.
|
The only line of the output should contain two integers *k**min* and *k**max* — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
|
[
"5 1\n",
"3 2\n",
"6 3\n"
] |
[
"10 10\n",
"1 1\n",
"3 6\n"
] |
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
| 1,000
|
[
{
"input": "5 1",
"output": "10 10"
},
{
"input": "3 2",
"output": "1 1"
},
{
"input": "6 3",
"output": "3 6"
},
{
"input": "5 3",
"output": "2 3"
},
{
"input": "10 2",
"output": "20 36"
},
{
"input": "10 6",
"output": "4 10"
},
{
"input": "1000000000 1",
"output": "499999999500000000 499999999500000000"
},
{
"input": "5000000 12",
"output": "1041664166668 12499942500066"
},
{
"input": "1833 195",
"output": "7722 1342341"
},
{
"input": "1000000000 1000000000",
"output": "0 0"
},
{
"input": "1000000000 1000000",
"output": "499500000000 499000500499500000"
},
{
"input": "1000000000 32170",
"output": "15541930838100 499967831017438365"
},
{
"input": "1000000 1000",
"output": "499500000 499000999500"
},
{
"input": "1234 1123",
"output": "111 6216"
},
{
"input": "599222887 298488",
"output": "601178656545 179355218158217800"
},
{
"input": "999999999 500000000",
"output": "499999999 124999999750000000"
},
{
"input": "1000000000 384842119",
"output": "845473643 189209609585784021"
},
{
"input": "1000000000 384842119",
"output": "845473643 189209609585784021"
},
{
"input": "1000000000 2",
"output": "249999999500000000 499999998500000001"
},
{
"input": "1000000000 999999999",
"output": "1 1"
},
{
"input": "38447 383",
"output": "1910550 724453080"
},
{
"input": "100000000 99999799",
"output": "201 20301"
},
{
"input": "1 1",
"output": "0 0"
},
{
"input": "2 1",
"output": "1 1"
},
{
"input": "2 2",
"output": "0 0"
},
{
"input": "10 10",
"output": "0 0"
}
] | 1,653,188,918
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 31
| 0
|
def ncr(n):
return (fact(n)/(2*fact(n-2)))
def fact(n):
r=1
for i in range(2, n+1):
r=r*i
return r
n,m=map(int,input().split())
if(m==1):
c=int(ncr(n))
print(c,c)
else:
v=n%m
if(v==0):
print(int(ncr(m)),end=" ")
else:
c=ncr(n//m)
c1=(n-(m*(n//m)))*ncr(n//m+1)
print(int(c+c1),end=" ")
print(int(ncr(n-(m-1))))
|
Title: Random Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
*n* participants of the competition were split into *m* teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input Specification:
The only line of input contains two integers *n* and *m*, separated by a single space (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the number of participants and the number of teams respectively.
Output Specification:
The only line of the output should contain two integers *k**min* and *k**max* — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Demo Input:
['5 1\n', '3 2\n', '6 3\n']
Demo Output:
['10 10\n', '1 1\n', '3 6\n']
Note:
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
|
```python
def ncr(n):
return (fact(n)/(2*fact(n-2)))
def fact(n):
r=1
for i in range(2, n+1):
r=r*i
return r
n,m=map(int,input().split())
if(m==1):
c=int(ncr(n))
print(c,c)
else:
v=n%m
if(v==0):
print(int(ncr(m)),end=" ")
else:
c=ncr(n//m)
c1=(n-(m*(n//m)))*ncr(n//m+1)
print(int(c+c1),end=" ")
print(int(ncr(n-(m-1))))
```
| 0
|
|
621
|
A
|
Wet Shark and Odd and Even
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.
|
The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
|
Print the maximum possible even sum that can be obtained if we use some of the given integers.
|
[
"3\n1 2 3\n",
"5\n999999999 999999999 999999999 999999999 999999999\n"
] |
[
"6",
"3999999996"
] |
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999.
| 500
|
[
{
"input": "3\n1 2 3",
"output": "6"
},
{
"input": "5\n999999999 999999999 999999999 999999999 999999999",
"output": "3999999996"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "15\n39 52 88 78 46 95 84 98 55 3 68 42 6 18 98",
"output": "870"
},
{
"input": "15\n59 96 34 48 8 72 67 90 15 85 7 90 97 47 25",
"output": "840"
},
{
"input": "15\n87 37 91 29 58 45 51 74 70 71 47 38 91 89 44",
"output": "922"
},
{
"input": "15\n11 81 49 7 11 14 30 67 29 50 90 81 77 18 59",
"output": "674"
},
{
"input": "15\n39 21 95 89 73 90 9 55 85 32 30 21 68 59 82",
"output": "848"
},
{
"input": "15\n59 70 48 54 26 67 84 39 40 18 77 69 70 88 93",
"output": "902"
},
{
"input": "15\n87 22 98 32 88 36 72 31 100 97 17 16 60 22 20",
"output": "798"
},
{
"input": "15\n15 63 51 13 37 9 43 19 55 79 57 60 50 59 31",
"output": "632"
},
{
"input": "1\n4",
"output": "4"
},
{
"input": "2\n1 4",
"output": "4"
},
{
"input": "3\n1 2 4",
"output": "6"
},
{
"input": "2\n9 3",
"output": "12"
},
{
"input": "2\n1000000000 1001",
"output": "1000000000"
},
{
"input": "3\n1 8 4",
"output": "12"
},
{
"input": "3\n7 4 4",
"output": "8"
},
{
"input": "5\n2 3 4 5 3",
"output": "14"
},
{
"input": "2\n4 5",
"output": "4"
},
{
"input": "3\n2 4 5",
"output": "6"
},
{
"input": "3\n2 2 3",
"output": "4"
},
{
"input": "2\n2 3",
"output": "2"
},
{
"input": "4\n2 3 7 7",
"output": "16"
},
{
"input": "2\n999999999 2",
"output": "2"
},
{
"input": "2\n2 5",
"output": "2"
},
{
"input": "3\n5 3 1",
"output": "8"
},
{
"input": "4\n3 2 5 7",
"output": "14"
}
] | 1,657,103,244
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 87
| 139
| 8,396,800
|
n = int(input())
s = list(map(int, input().split(" ")))
odd = []
even = []
for i in s:
if i % 2 == 0:
even.append(i)
else:
odd.append(i)
odd.sort(reverse = True)
if len(odd) % 2 == 0:
print(sum(s))
else:
print(sum(even) + sum(odd[0:len(odd)-1]))
|
Title: Wet Shark and Odd and Even
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.
Input Specification:
The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
Output Specification:
Print the maximum possible even sum that can be obtained if we use some of the given integers.
Demo Input:
['3\n1 2 3\n', '5\n999999999 999999999 999999999 999999999 999999999\n']
Demo Output:
['6', '3999999996']
Note:
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999.
|
```python
n = int(input())
s = list(map(int, input().split(" ")))
odd = []
even = []
for i in s:
if i % 2 == 0:
even.append(i)
else:
odd.append(i)
odd.sort(reverse = True)
if len(odd) % 2 == 0:
print(sum(s))
else:
print(sum(even) + sum(odd[0:len(odd)-1]))
```
| 3
|
|
710
|
B
|
Optimal Point on a Line
|
PROGRAMMING
| 1,400
|
[
"brute force",
"sortings"
] | null | null |
You are given *n* points on a line with their coordinates *x**i*. Find the point *x* so the sum of distances to the given points is minimal.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of points on the line.
The second line contains *n* integers *x**i* (<=-<=109<=≤<=*x**i*<=≤<=109) — the coordinates of the given *n* points.
|
Print the only integer *x* — the position of the optimal point on the line. If there are several optimal points print the position of the leftmost one. It is guaranteed that the answer is always the integer.
|
[
"4\n1 2 3 4\n"
] |
[
"2\n"
] |
none
| 0
|
[
{
"input": "4\n1 2 3 4",
"output": "2"
},
{
"input": "5\n-1 -10 2 6 7",
"output": "2"
},
{
"input": "10\n-68 10 87 22 30 89 82 -97 -52 25",
"output": "22"
},
{
"input": "100\n457 827 807 17 871 935 907 -415 536 170 551 -988 865 758 -457 -892 -875 -488 684 19 0 555 -807 -624 -239 826 318 811 20 -732 -91 460 551 -610 555 -493 -154 442 -141 946 -913 -104 704 -380 699 32 106 -455 -518 214 -464 -861 243 -798 -472 559 529 -844 -32 871 -459 236 387 626 -318 -580 -611 -842 790 486 64 951 81 78 -693 403 -731 309 678 696 891 846 -106 918 212 -44 994 606 -829 -454 243 -477 -402 -818 -819 -310 -837 -209 736 424",
"output": "64"
},
{
"input": "2\n-1 0",
"output": "-1"
},
{
"input": "48\n-777 -767 -764 -713 -688 -682 -606 -586 -585 -483 -465 -440 -433 -397 -390 -377 -299 -252 -159 -147 -96 -29 -15 15 52 109 124 129 142 218 231 314 320 339 442 496 505 548 575 576 594 624 694 827 891 979 981 981",
"output": "15"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 1 1 1 1 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "4\n-1 -1 0 1",
"output": "-1"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 1000000000",
"output": "0"
},
{
"input": "2\n1 -1",
"output": "-1"
},
{
"input": "2\n100 50",
"output": "50"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "1\n10",
"output": "10"
},
{
"input": "3\n606194955 -856471310 117647402",
"output": "117647402"
},
{
"input": "2\n615002717 -843553590",
"output": "-843553590"
},
{
"input": "2\n-1 2",
"output": "-1"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "5\n-638512131 348325781 -550537933 -618161835 -567935532",
"output": "-567935532"
},
{
"input": "1\n120",
"output": "120"
},
{
"input": "2\n-1000000000 1000000000",
"output": "-1000000000"
},
{
"input": "1\n618309368",
"output": "618309368"
}
] | 1,477,350,973
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
count = int(input())
nums = input().split()
total = 0
for num in nums:
total += int(num)
possible1 = total // count
possible2 = possible1 + 1
total1 = 0
total2 = 0
for num in nums:
total1 += abs(int(num) - possible1)
total2 += abs(int(num) - possible2)
if total1 < total2:
print(possible1)
else:
print(possible2)
|
Title: Optimal Point on a Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* points on a line with their coordinates *x**i*. Find the point *x* so the sum of distances to the given points is minimal.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of points on the line.
The second line contains *n* integers *x**i* (<=-<=109<=≤<=*x**i*<=≤<=109) — the coordinates of the given *n* points.
Output Specification:
Print the only integer *x* — the position of the optimal point on the line. If there are several optimal points print the position of the leftmost one. It is guaranteed that the answer is always the integer.
Demo Input:
['4\n1 2 3 4\n']
Demo Output:
['2\n']
Note:
none
|
```python
count = int(input())
nums = input().split()
total = 0
for num in nums:
total += int(num)
possible1 = total // count
possible2 = possible1 + 1
total1 = 0
total2 = 0
for num in nums:
total1 += abs(int(num) - possible1)
total2 += abs(int(num) - possible2)
if total1 < total2:
print(possible1)
else:
print(possible2)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Dr. Moriarty is about to send a message to Sherlock Holmes. He has a string *s*.
String *p* is called a substring of string *s* if you can read it starting from some position in the string *s*. For example, string "aba" has six substrings: "a", "b", "a", "ab", "ba", "aba".
Dr. Moriarty plans to take string *s* and cut out some substring from it, let's call it *t*. Then he needs to change the substring *t* zero or more times. As a result, he should obtain a fixed string *u* (which is the string that should be sent to Sherlock Holmes). One change is defined as making one of the following actions:
- Insert one letter to any end of the string. - Delete one letter from any end of the string. - Change one letter into any other one.
Moriarty is very smart and after he chooses some substring *t*, he always makes the minimal number of changes to obtain *u*.
Help Moriarty choose the best substring *t* from all substrings of the string *s*. The substring *t* should minimize the number of changes Moriarty should make to obtain the string *u* from it.
|
The first line contains a non-empty string *s*, consisting of lowercase Latin letters. The second line contains a non-empty string *u*, consisting of lowercase Latin letters. The lengths of both strings are in the range from 1 to 2000, inclusive.
|
Print the only integer — the minimum number of changes that Dr. Moriarty has to make with the string that you choose.
|
[
"aaaaa\naaa\n",
"abcabc\nbcd\n",
"abcdef\nklmnopq\n"
] |
[
"0\n",
"1\n",
"7\n"
] |
In the first sample Moriarty can take any substring of length 3, and it will be equal to the required message *u*, so Moriarty won't have to make any changes.
In the second sample you should take a substring consisting of characters from second to fourth ("bca") or from fifth to sixth ("bc"). Then you will only have to make one change: to change or to add the last character.
In the third sample the initial string *s* doesn't contain any character that the message should contain, so, whatever string you choose, you will have to make at least 7 changes to obtain the required message.
| 0
|
[
{
"input": "aaaaa\naaa",
"output": "0"
},
{
"input": "abcabc\nbcd",
"output": "1"
},
{
"input": "abcdef\nklmnopq",
"output": "7"
},
{
"input": "aaabbbaaa\naba",
"output": "1"
},
{
"input": "a\na",
"output": "0"
},
{
"input": "z\nz",
"output": "0"
},
{
"input": "a\nz",
"output": "1"
},
{
"input": "d\nt",
"output": "1"
},
{
"input": "o\nu",
"output": "1"
},
{
"input": "a\nm",
"output": "1"
},
{
"input": "t\nv",
"output": "1"
},
{
"input": "n\ng",
"output": "1"
},
{
"input": "c\nh",
"output": "1"
},
{
"input": "r\ni",
"output": "1"
},
{
"input": "h\nb",
"output": "1"
},
{
"input": "r\na",
"output": "1"
},
{
"input": "c\np",
"output": "1"
},
{
"input": "wbdbzf\nfpvlerhsuf",
"output": "9"
},
{
"input": "zafsqbsu\nhl",
"output": "2"
},
{
"input": "juhlp\nycqugugk",
"output": "7"
},
{
"input": "ladfasxt\ncpvtd",
"output": "4"
},
{
"input": "ally\ncjidwuj",
"output": "7"
},
{
"input": "rgug\npgqwslo",
"output": "6"
},
{
"input": "wmjwu\nrfew",
"output": "3"
},
{
"input": "cpnwcdqff\nq",
"output": "0"
},
{
"input": "dkwh\nm",
"output": "1"
},
{
"input": "zfinrlju\nwiiegborjl",
"output": "9"
},
{
"input": "swconajiqpgziitbpwjsfcalqvmwbfed\nridfnsyumichlhpnurrnwkyjcdzchznpmno",
"output": "32"
},
{
"input": "vfjofvgkdwgqdlomtmcvedtmimdnxavhfirienxfdflldkbwjsynablhdvgaipvcghgaxipotwmmlzxekipgbvpfivlgzfwqz\njkdfjnessjfgcqpnxgtqdxtqimolbdlnipkoqht",
"output": "34"
},
{
"input": "dtvxepnxfkzcaoh\nkpdzbtwjitzlyzvsbwcsrfglaycrhzwsdtidrelndsq",
"output": "41"
},
{
"input": "sweaucynwsnduofyaqunoxttbipgrbfpssplfp\nuifmuxmczznobefdsyoclwzekewxmcwfqryuevnxxlgxsuhoytkaddorbdaygo",
"output": "57"
},
{
"input": "eohztfsxoyhirnzxgwaevfqstinlxeiyywmpmlbedkjihaxfdtsocof\nbloqrjbidxiqozvwregxxgmxuqcvhwzhytfckbafd",
"output": "37"
},
{
"input": "ybshzefoxkqdigcjafs\nnffvaxdmditsolfxbyquira",
"output": "19"
},
{
"input": "ytfqnuhqzbjjheejjbzcaorilcyvuxvviaiba\nxnhgkdfceialuujgcxmrhjbzvibcoknofafmdjnhij",
"output": "37"
},
{
"input": "ibdjtvgaveujdyidqldrxgwhsammmfpgxwljkptmeyejdvudhctmqjazalyzmzfgebetyqncu\nercdngwctdarcennbuqhsjlwfwrcqjbcjxqftycoulrhrimwhznogjmrrqdygtmllottpjgmkndraearezvxxmdhcuokhyngu",
"output": "90"
},
{
"input": "bwhvaanyxupicwobeevcwewhcriwowfovnylalpuhxzqxtzyjrzlxcmejujvliomdfedgtaioauwrcluhfxtzu\nplinvtsvytepojsecnjisxlmqkfhgknitvuw",
"output": "28"
},
{
"input": "sjxykdmrzpescabubcjflhnpckgytklc\nsxirpuqnmjqhlnvdeyvxvzzcygkpsujyifzgzmtvxsimddjahiephqlgfzngrzjtcrgrimewsxipczsgu",
"output": "76"
},
{
"input": "ksmbytfyhhnstlecripupiwdhbkhfpfmimrbqgszohcqnezcybvwasxmkxfupvuecsctcpadccnqexsglwaiyxcoyheefcjmdedesgjqdtqgrvfjonzesffousooutsjumrxl\nhgjqihcfbnmgufonaiudbjegexexthrzcdkuimwogpbyovemztzcmqnrbhabxyyxyfuzpyhjgnioexbezzupcxlyzuuncstiiqsjzdtqppqhxilvqimlpjejiqbwpeekzweeyvthvjffgfvqauqrugajjjzibgzhxphcvtncjzecbtupwkehcrgsgfgkvwwnifglyamjkzfvabybsstwrwugnmiwflhemgnfbrtskzfxcepqhtelgiowzeuujpkuzsfsipcvtfoeshawvryaubilcbwukdhlwamsqenzvr",
"output": "287"
},
{
"input": "abcd\ndabc",
"output": "1"
},
{
"input": "medxx\nahmed",
"output": "2"
},
{
"input": "ab\ndab",
"output": "1"
},
{
"input": "nasldkfnsb\nyyyynasld",
"output": "4"
},
{
"input": "abcde\ncabc",
"output": "1"
},
{
"input": "a\nzzzzzzzzzza",
"output": "10"
},
{
"input": "abcde\nabde",
"output": "2"
},
{
"input": "bac\ntbdca",
"output": "3"
},
{
"input": "abcdef\nxyzabc",
"output": "3"
},
{
"input": "abcdef\nbctsf",
"output": "2"
},
{
"input": "xxxabaxxx\nxxxaaxxx",
"output": "2"
},
{
"input": "bcd\nabc",
"output": "1"
},
{
"input": "d\nabcdef",
"output": "5"
}
] | 1,584,899,807
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 4
| 2,000
| 2,048,000
|
from collections import *
s1, s2 = input(), input()
l1, l2, mem1 = len(s1), len(s2), defaultdict(int)
ans = l2
for i in range(l1):
for j in range(l1 - i):
mem1[s1[j:j + i + 1]] = 1
for i in range(l2):
for j in range(l2 - i):
if mem1[s2[j:j + i + 1]]:
ans = min(ans, l2 - i - 1)
print(ans)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dr. Moriarty is about to send a message to Sherlock Holmes. He has a string *s*.
String *p* is called a substring of string *s* if you can read it starting from some position in the string *s*. For example, string "aba" has six substrings: "a", "b", "a", "ab", "ba", "aba".
Dr. Moriarty plans to take string *s* and cut out some substring from it, let's call it *t*. Then he needs to change the substring *t* zero or more times. As a result, he should obtain a fixed string *u* (which is the string that should be sent to Sherlock Holmes). One change is defined as making one of the following actions:
- Insert one letter to any end of the string. - Delete one letter from any end of the string. - Change one letter into any other one.
Moriarty is very smart and after he chooses some substring *t*, he always makes the minimal number of changes to obtain *u*.
Help Moriarty choose the best substring *t* from all substrings of the string *s*. The substring *t* should minimize the number of changes Moriarty should make to obtain the string *u* from it.
Input Specification:
The first line contains a non-empty string *s*, consisting of lowercase Latin letters. The second line contains a non-empty string *u*, consisting of lowercase Latin letters. The lengths of both strings are in the range from 1 to 2000, inclusive.
Output Specification:
Print the only integer — the minimum number of changes that Dr. Moriarty has to make with the string that you choose.
Demo Input:
['aaaaa\naaa\n', 'abcabc\nbcd\n', 'abcdef\nklmnopq\n']
Demo Output:
['0\n', '1\n', '7\n']
Note:
In the first sample Moriarty can take any substring of length 3, and it will be equal to the required message *u*, so Moriarty won't have to make any changes.
In the second sample you should take a substring consisting of characters from second to fourth ("bca") or from fifth to sixth ("bc"). Then you will only have to make one change: to change or to add the last character.
In the third sample the initial string *s* doesn't contain any character that the message should contain, so, whatever string you choose, you will have to make at least 7 changes to obtain the required message.
|
```python
from collections import *
s1, s2 = input(), input()
l1, l2, mem1 = len(s1), len(s2), defaultdict(int)
ans = l2
for i in range(l1):
for j in range(l1 - i):
mem1[s1[j:j + i + 1]] = 1
for i in range(l2):
for j in range(l2 - i):
if mem1[s2[j:j + i + 1]]:
ans = min(ans, l2 - i - 1)
print(ans)
```
| 0
|
|
429
|
A
|
Xor-tree
|
PROGRAMMING
| 1,300
|
[
"dfs and similar",
"trees"
] | null | null |
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having *n* nodes, numbered from 1 to *n*. Each node *i* has an initial value *init**i*, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node *x*. Right after someone has picked node *x*, the value of node *x* flips, the values of sons of *x* remain the same, the values of sons of sons of *x* flips, the values of sons of sons of sons of *x* remain the same and so on.
The goal of the game is to get each node *i* to have value *goal**i*, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). Each of the next *n*<=-<=1 lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*; *u**i*<=≠<=*v**i*) meaning there is an edge between nodes *u**i* and *v**i*.
The next line contains *n* integer numbers, the *i*-th of them corresponds to *init**i* (*init**i* is either 0 or 1). The following line also contains *n* integer numbers, the *i*-th number corresponds to *goal**i* (*goal**i* is either 0 or 1).
|
In the first line output an integer number *cnt*, representing the minimal number of operations you perform. Each of the next *cnt* lines should contain an integer *x**i*, representing that you pick a node *x**i*.
|
[
"10\n2 1\n3 1\n4 2\n5 1\n6 2\n7 5\n8 6\n9 8\n10 5\n1 0 1 1 0 1 0 1 0 1\n1 0 1 0 0 1 1 1 0 1\n"
] |
[
"2\n4\n7\n"
] |
none
| 500
|
[
{
"input": "10\n2 1\n3 1\n4 2\n5 1\n6 2\n7 5\n8 6\n9 8\n10 5\n1 0 1 1 0 1 0 1 0 1\n1 0 1 0 0 1 1 1 0 1",
"output": "2\n4\n7"
},
{
"input": "15\n2 1\n3 2\n4 3\n5 4\n6 5\n7 6\n8 7\n9 8\n10 9\n11 10\n12 11\n13 12\n14 13\n15 14\n0 1 0 0 1 1 1 1 1 1 0 0 0 1 1\n1 1 1 1 0 0 1 1 0 1 0 0 1 1 0",
"output": "7\n1\n4\n7\n8\n9\n11\n13"
},
{
"input": "20\n2 1\n3 2\n4 3\n5 4\n6 4\n7 1\n8 2\n9 4\n10 2\n11 6\n12 9\n13 2\n14 12\n15 14\n16 8\n17 9\n18 13\n19 2\n20 17\n1 0 0 1 0 0 0 1 0 1 0 0 0 1 1 0 1 0 1 0\n1 0 0 1 0 0 1 1 0 0 1 0 0 1 0 0 0 1 0 1",
"output": "8\n11\n15\n17\n20\n10\n18\n19\n7"
},
{
"input": "30\n2 1\n3 2\n4 3\n5 3\n6 5\n7 3\n8 3\n9 2\n10 3\n11 2\n12 11\n13 6\n14 4\n15 5\n16 11\n17 9\n18 14\n19 6\n20 2\n21 19\n22 9\n23 19\n24 20\n25 14\n26 22\n27 1\n28 6\n29 13\n30 27\n1 0 1 1 1 1 0 1 0 0 1 0 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 1 1 0\n0 1 0 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 1 0 1 0 0 1 0 1 1 0 0 0",
"output": "15\n1\n2\n4\n5\n6\n13\n29\n19\n21\n23\n28\n7\n22\n26\n30"
},
{
"input": "15\n2 1\n3 1\n4 1\n5 1\n6 3\n7 1\n8 1\n9 1\n10 5\n11 9\n12 3\n13 5\n14 5\n15 4\n1 1 0 0 0 0 1 1 1 0 1 1 1 0 0\n1 0 1 1 0 1 1 1 1 1 1 1 1 1 0",
"output": "6\n2\n3\n6\n4\n10\n14"
},
{
"input": "20\n2 1\n3 1\n4 2\n5 2\n6 3\n7 1\n8 6\n9 2\n10 3\n11 6\n12 2\n13 3\n14 2\n15 1\n16 8\n17 15\n18 2\n19 14\n20 14\n0 0 0 0 1 1 1 1 1 1 0 1 1 1 0 0 0 1 1 1\n0 1 0 1 1 1 0 0 1 0 1 1 0 1 0 1 0 0 1 0",
"output": "10\n2\n4\n19\n18\n8\n16\n11\n10\n13\n7"
},
{
"input": "23\n2 1\n3 2\n4 1\n5 1\n6 5\n7 3\n8 2\n9 8\n10 5\n11 6\n12 9\n13 3\n14 11\n15 5\n16 2\n17 3\n18 10\n19 16\n20 14\n21 19\n22 17\n23 7\n0 1 0 1 1 1 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0 0 0\n0 0 0 0 1 1 1 1 1 0 1 0 1 0 0 1 1 0 0 0 0 0 1",
"output": "8\n2\n23\n13\n17\n9\n4\n11\n20"
},
{
"input": "1\n0\n0",
"output": "0"
},
{
"input": "10\n1 10\n1 9\n10 2\n10 3\n3 7\n3 8\n9 4\n9 5\n5 6\n1 0 1 1 0 1 0 1 0 1\n0 0 0 0 0 0 0 0 0 0",
"output": "6\n1\n10\n2\n7\n5\n6"
}
] | 1,465,745,799
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 748
| 26,931,200
|
from collections import defaultdict, deque, Counter, OrderedDict
from bisect import insort, bisect_right, bisect_left
import threading
def main():
n = int(input())
adj = [[] for i in range(n + 1)]
for i in range(n - 1):
a, b = map(int, input().split())
a, b = a - 1, b - 1
adj[a].append(b)
adj[b].append(a)
init = [int(i) for i in input().split()]
goal = [int(i) for i in input().split()]
visited = [0] * n
par = [[] for i in range(n)]
dq = deque()
dq.append((0, 0))
while len(dq) > 0:
(s, p) = dq.pop()
if visited[s]: continue
visited[s] = 1
par[p].append(s)
for i in adj[s]:
dq.append((i, s))
par[0] = par[0][1:]
ans = []
dq = deque()
dq.append((0, 0, 0, 0))
while len(dq) > 0:
(s, l, fo, fe) = dq.pop()
if l % 2 == 0:
if fe % 2 == 1:
init[s] = 1 - init[s]
else:
if fo % 2 == 1:
init[s] = 1 - init[s]
if init[s] != goal[s]:
ans.append(s + 1)
if l % 2:
fo += 1
else:
fe += 1
for j in par[s]:
dq.append((j, l + 1, fo, fe))
print(len(ans))
print("\n".join(map(str, ans)))
if __name__ == "__main__":
"""sys.setrecursionlimit(200000)
threading.stack_size(10240000)"""
thread = threading.Thread(target=main)
thread.start()
|
Title: Xor-tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having *n* nodes, numbered from 1 to *n*. Each node *i* has an initial value *init**i*, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node *x*. Right after someone has picked node *x*, the value of node *x* flips, the values of sons of *x* remain the same, the values of sons of sons of *x* flips, the values of sons of sons of sons of *x* remain the same and so on.
The goal of the game is to get each node *i* to have value *goal**i*, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). Each of the next *n*<=-<=1 lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*; *u**i*<=≠<=*v**i*) meaning there is an edge between nodes *u**i* and *v**i*.
The next line contains *n* integer numbers, the *i*-th of them corresponds to *init**i* (*init**i* is either 0 or 1). The following line also contains *n* integer numbers, the *i*-th number corresponds to *goal**i* (*goal**i* is either 0 or 1).
Output Specification:
In the first line output an integer number *cnt*, representing the minimal number of operations you perform. Each of the next *cnt* lines should contain an integer *x**i*, representing that you pick a node *x**i*.
Demo Input:
['10\n2 1\n3 1\n4 2\n5 1\n6 2\n7 5\n8 6\n9 8\n10 5\n1 0 1 1 0 1 0 1 0 1\n1 0 1 0 0 1 1 1 0 1\n']
Demo Output:
['2\n4\n7\n']
Note:
none
|
```python
from collections import defaultdict, deque, Counter, OrderedDict
from bisect import insort, bisect_right, bisect_left
import threading
def main():
n = int(input())
adj = [[] for i in range(n + 1)]
for i in range(n - 1):
a, b = map(int, input().split())
a, b = a - 1, b - 1
adj[a].append(b)
adj[b].append(a)
init = [int(i) for i in input().split()]
goal = [int(i) for i in input().split()]
visited = [0] * n
par = [[] for i in range(n)]
dq = deque()
dq.append((0, 0))
while len(dq) > 0:
(s, p) = dq.pop()
if visited[s]: continue
visited[s] = 1
par[p].append(s)
for i in adj[s]:
dq.append((i, s))
par[0] = par[0][1:]
ans = []
dq = deque()
dq.append((0, 0, 0, 0))
while len(dq) > 0:
(s, l, fo, fe) = dq.pop()
if l % 2 == 0:
if fe % 2 == 1:
init[s] = 1 - init[s]
else:
if fo % 2 == 1:
init[s] = 1 - init[s]
if init[s] != goal[s]:
ans.append(s + 1)
if l % 2:
fo += 1
else:
fe += 1
for j in par[s]:
dq.append((j, l + 1, fo, fe))
print(len(ans))
print("\n".join(map(str, ans)))
if __name__ == "__main__":
"""sys.setrecursionlimit(200000)
threading.stack_size(10240000)"""
thread = threading.Thread(target=main)
thread.start()
```
| 3
|
|
958
|
B1
|
Maximum Control (easy)
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
|
The first line of the input contains an integer *N* (2<=≤<=*N*<=≤<=1000) – the number of planets in the galaxy.
The next *N*<=-<=1 lines describe the hyperspace tunnels between the planets. Each of the *N*<=-<=1 lines contains two space-separated integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*N*) indicating that there is a bidirectional hyperspace tunnel between the planets *u* and *v*. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
|
A single integer denoting the number of remote planets.
|
[
"5\n4 1\n4 2\n1 3\n1 5\n",
"4\n1 2\n4 3\n1 4\n"
] |
[
"3\n",
"2\n"
] |
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
| 0
|
[
{
"input": "5\n4 1\n4 2\n1 3\n1 5",
"output": "3"
},
{
"input": "4\n1 2\n4 3\n1 4",
"output": "2"
},
{
"input": "10\n4 3\n2 6\n10 1\n5 7\n5 8\n10 6\n5 9\n9 3\n2 9",
"output": "4"
}
] | 1,523,697,549
| 8,049
|
Python 3
|
OK
|
TESTS
| 9
| 186
| 12,390,400
|
n = int(input())
counter = 0
a = [[0 for j in range(n)] for i in range(n)]
for i in range(n - 1):
i, j = map(int, input().split())
a[i - 1][j - 1] = 1
a[j - 1][i - 1] = 1
for element in a:
if element.count(1) == 1:
counter += 1
print(counter)
|
Title: Maximum Control (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
Input Specification:
The first line of the input contains an integer *N* (2<=≤<=*N*<=≤<=1000) – the number of planets in the galaxy.
The next *N*<=-<=1 lines describe the hyperspace tunnels between the planets. Each of the *N*<=-<=1 lines contains two space-separated integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*N*) indicating that there is a bidirectional hyperspace tunnel between the planets *u* and *v*. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
Output Specification:
A single integer denoting the number of remote planets.
Demo Input:
['5\n4 1\n4 2\n1 3\n1 5\n', '4\n1 2\n4 3\n1 4\n']
Demo Output:
['3\n', '2\n']
Note:
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
|
```python
n = int(input())
counter = 0
a = [[0 for j in range(n)] for i in range(n)]
for i in range(n - 1):
i, j = map(int, input().split())
a[i - 1][j - 1] = 1
a[j - 1][i - 1] = 1
for element in a:
if element.count(1) == 1:
counter += 1
print(counter)
```
| 3
|
|
235
|
A
|
LCM Challenge
|
PROGRAMMING
| 1,600
|
[
"number theory"
] | null | null |
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than *n*. Can you help me to find the maximum possible least common multiple of these three integers?
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the *n* mentioned in the statement.
|
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than *n*.
|
[
"9\n",
"7\n"
] |
[
"504\n",
"210\n"
] |
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get.
| 500
|
[
{
"input": "9",
"output": "504"
},
{
"input": "7",
"output": "210"
},
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "60"
},
{
"input": "6",
"output": "60"
},
{
"input": "33",
"output": "32736"
},
{
"input": "21",
"output": "7980"
},
{
"input": "2",
"output": "2"
},
{
"input": "41",
"output": "63960"
},
{
"input": "29",
"output": "21924"
},
{
"input": "117",
"output": "1560780"
},
{
"input": "149",
"output": "3241644"
},
{
"input": "733",
"output": "392222436"
},
{
"input": "925",
"output": "788888100"
},
{
"input": "509",
"output": "131096004"
},
{
"input": "829",
"output": "567662724"
},
{
"input": "117",
"output": "1560780"
},
{
"input": "605",
"output": "220348260"
},
{
"input": "245",
"output": "14526540"
},
{
"input": "925",
"output": "788888100"
},
{
"input": "213",
"output": "9527916"
},
{
"input": "53",
"output": "140556"
},
{
"input": "341",
"output": "39303660"
},
{
"input": "21",
"output": "7980"
},
{
"input": "605",
"output": "220348260"
},
{
"input": "149",
"output": "3241644"
},
{
"input": "733",
"output": "392222436"
},
{
"input": "117",
"output": "1560780"
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{
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"output": "140556"
},
{
"input": "245",
"output": "14526540"
},
{
"input": "829",
"output": "567662724"
},
{
"input": "924",
"output": "783776526"
},
{
"input": "508",
"output": "130065780"
},
{
"input": "700",
"output": "341042100"
},
{
"input": "636",
"output": "254839470"
},
{
"input": "20",
"output": "6460"
},
{
"input": "604",
"output": "218891412"
},
{
"input": "796",
"output": "501826260"
},
{
"input": "732",
"output": "389016270"
},
{
"input": "412",
"output": "69256788"
},
{
"input": "700",
"output": "341042100"
},
{
"input": "244",
"output": "14289372"
},
{
"input": "828",
"output": "563559150"
},
{
"input": "508",
"output": "130065780"
},
{
"input": "796",
"output": "501826260"
},
{
"input": "636",
"output": "254839470"
},
{
"input": "924",
"output": "783776526"
},
{
"input": "245",
"output": "14526540"
},
{
"input": "828",
"output": "563559150"
},
{
"input": "21",
"output": "7980"
},
{
"input": "605",
"output": "220348260"
},
{
"input": "636",
"output": "254839470"
},
{
"input": "924",
"output": "783776526"
},
{
"input": "116",
"output": "1507420"
},
{
"input": "700",
"output": "341042100"
},
{
"input": "732",
"output": "389016270"
},
{
"input": "20",
"output": "6460"
},
{
"input": "508",
"output": "130065780"
},
{
"input": "148",
"output": "3154620"
},
{
"input": "828",
"output": "563559150"
},
{
"input": "763116",
"output": "444394078546562430"
},
{
"input": "756604",
"output": "433115377058855412"
},
{
"input": "447244",
"output": "89460162932862372"
},
{
"input": "372636",
"output": "51742503205363470"
},
{
"input": "546924",
"output": "163597318076822526"
},
{
"input": "540412",
"output": "157823524476316788"
},
{
"input": "714700",
"output": "365063922340784100"
},
{
"input": "520731",
"output": "141201007712496270"
},
{
"input": "695019",
"output": "335728459024850814"
},
{
"input": "688507",
"output": "326379736779169710"
},
{
"input": "862795",
"output": "642275489615199390"
},
{
"input": "668827",
"output": "299184742915995150"
},
{
"input": "810411",
"output": "532248411551110590"
},
{
"input": "836603",
"output": "585540171302562606"
},
{
"input": "978187",
"output": "935975171582120670"
},
{
"input": "816923",
"output": "545182335484592526"
},
{
"input": "958507",
"output": "880611813728059710"
},
{
"input": "984699",
"output": "954792870629291694"
},
{
"input": "642635",
"output": "265393998349453470"
},
{
"input": "296604",
"output": "26092892528622606"
},
{
"input": "1000000",
"output": "999996000003000000"
},
{
"input": "8",
"output": "280"
},
{
"input": "3",
"output": "6"
},
{
"input": "4",
"output": "12"
},
{
"input": "30",
"output": "21924"
},
{
"input": "18",
"output": "4080"
},
{
"input": "12",
"output": "990"
}
] | 1,591,450,946
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 87
| 218
| 307,200
|
def solve(n):
if(n==1):
print("1")
elif(n==2):
print("2")
elif(n==3):
print("6")
elif(n&1):
print(n*(n-1)*(n-2))
else:
if(n%3==0):
print((n-1)*(n-2)*(n-3))
else:
print(n*(n-1)*(n-3))
return
def main():
n=int(input())
solve(n)
main()
|
Title: LCM Challenge
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than *n*. Can you help me to find the maximum possible least common multiple of these three integers?
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the *n* mentioned in the statement.
Output Specification:
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than *n*.
Demo Input:
['9\n', '7\n']
Demo Output:
['504\n', '210\n']
Note:
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get.
|
```python
def solve(n):
if(n==1):
print("1")
elif(n==2):
print("2")
elif(n==3):
print("6")
elif(n&1):
print(n*(n-1)*(n-2))
else:
if(n%3==0):
print((n-1)*(n-2)*(n-3))
else:
print(n*(n-1)*(n-3))
return
def main():
n=int(input())
solve(n)
main()
```
| 3
|
|
664
|
A
|
Complicated GCD
|
PROGRAMMING
| 800
|
[
"math",
"number theory"
] | null | null |
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
|
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
|
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
|
[
"1 2\n",
"61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n"
] |
[
"1\n",
"61803398874989484820458683436563811772030917980576\n"
] |
none
| 500
|
[
{
"input": "1 2",
"output": "1"
},
{
"input": "61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576",
"output": "61803398874989484820458683436563811772030917980576"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "100 100000",
"output": "1"
},
{
"input": "12345 67890123456789123457",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158 8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158",
"output": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158"
},
{
"input": "1 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "8328748239473982794239847237438782379810988324751 9328748239473982794239847237438782379810988324751",
"output": "1"
},
{
"input": "1029398958432734901284327523909481928483573793 1029398958432734901284327523909481928483573794",
"output": "1"
},
{
"input": "10000 1000000000",
"output": "1"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "11210171722243 65715435710585778347",
"output": "1"
},
{
"input": "2921881079263974825226940825843 767693191032295360887755303860323261471",
"output": "1"
},
{
"input": "8025352957265704896940312528736939363590612908210603 96027920417708260814607687034511406492969694925539085",
"output": "1"
},
{
"input": "23510978780782786207241069904470895053213996267165977112058175452757132930 210352653280909370107314249722987050753257161175393375412301228883856435481424",
"output": "1"
},
{
"input": "8150070767079366215626260746398623663859344142817267779361251788637547414925170226504788118262 49924902262298336032630839998470954964895251605110946547855439236151401194070172107435992986913614",
"output": "1"
},
{
"input": "15943150466658398903 15943150466658398903",
"output": "15943150466658398903"
},
{
"input": "410470228200245407491525399055972 410470228200245407491525399055972",
"output": "410470228200245407491525399055972"
},
{
"input": "51894705655711504622197349350106792045098781545973899451307 51894705655711504622197349350106792045098781545973899451307",
"output": "51894705655711504622197349350106792045098781545973899451307"
},
{
"input": "60353594589897438036015726222485085035927634677598681595162804007836722215668410 60353594589897438036015726222485085035927634677598681595162804007836722215668410",
"output": "60353594589897438036015726222485085035927634677598681595162804007836722215668410"
},
{
"input": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535 761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535",
"output": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535"
},
{
"input": "1 2000000000",
"output": "1"
},
{
"input": "13 1928834874",
"output": "1"
},
{
"input": "87 2938984237482934238",
"output": "1"
},
{
"input": "213 413",
"output": "1"
},
{
"input": "3 4",
"output": "1"
}
] | 1,624,187,480
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 62
| 0
|
s1,s2=map(str,input().split())
print(s1 if s1==s2 else 1)
|
Title: Complicated GCD
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
Output Specification:
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
Demo Input:
['1 2\n', '61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n']
Demo Output:
['1\n', '61803398874989484820458683436563811772030917980576\n']
Note:
none
|
```python
s1,s2=map(str,input().split())
print(s1 if s1==s2 else 1)
```
| 3
|
|
439
|
A
|
Devu, the Singer and Churu, the Joker
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation"
] | null | null |
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
|
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
|
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
|
[
"3 30\n2 2 1\n",
"3 20\n2 1 1\n"
] |
[
"5\n",
"-1\n"
] |
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.
| 500
|
[
{
"input": "3 30\n2 2 1",
"output": "5"
},
{
"input": "3 20\n2 1 1",
"output": "-1"
},
{
"input": "50 10000\n5 4 10 9 9 6 7 7 7 3 3 7 7 4 7 4 10 10 1 7 10 3 1 4 5 7 2 10 10 10 2 3 4 7 6 1 8 4 7 3 8 8 4 10 1 1 9 2 6 1",
"output": "1943"
},
{
"input": "50 10000\n4 7 15 9 11 12 20 9 14 14 10 13 6 13 14 17 6 8 20 12 10 15 13 17 5 12 13 11 7 5 5 2 3 15 13 7 14 14 19 2 13 14 5 15 3 19 15 16 4 1",
"output": "1891"
},
{
"input": "100 9000\n5 2 3 1 1 3 4 9 9 6 7 10 10 10 2 10 6 8 8 6 7 9 9 5 6 2 1 10 10 9 4 5 9 2 4 3 8 5 6 1 1 5 3 6 2 6 6 6 5 8 3 6 7 3 1 10 9 1 8 3 10 9 5 6 3 4 1 1 10 10 2 3 4 8 10 10 5 1 5 3 6 8 10 6 10 2 1 8 10 1 7 6 9 10 5 2 3 5 3 2",
"output": "1688"
},
{
"input": "100 8007\n5 19 14 18 9 6 15 8 1 14 11 20 3 17 7 12 2 6 3 17 7 20 1 14 20 17 2 10 13 7 18 18 9 10 16 8 1 11 11 9 13 18 9 20 12 12 7 15 12 17 11 5 11 15 9 2 15 1 18 3 18 16 15 4 10 5 18 13 13 12 3 8 17 2 12 2 13 3 1 13 2 4 9 10 18 10 14 4 4 17 12 19 2 9 6 5 5 20 18 12",
"output": "1391"
},
{
"input": "39 2412\n1 1 1 1 1 1 26 1 1 1 99 1 1 1 1 1 1 1 1 1 1 88 7 1 1 1 1 76 1 1 1 93 40 1 13 1 68 1 32",
"output": "368"
},
{
"input": "39 2617\n47 1 1 1 63 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 70 1 99 63 1 1 1 1 1 1 1 1 64 1 1",
"output": "435"
},
{
"input": "39 3681\n83 77 1 94 85 47 1 98 29 16 1 1 1 71 96 85 31 97 96 93 40 50 98 1 60 51 1 96 100 72 1 1 1 89 1 93 1 92 100",
"output": "326"
},
{
"input": "45 894\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 28 28 1 1 1 1 1 1 1 1 1 1 1 1 1 1 99 3 1 1",
"output": "139"
},
{
"input": "45 4534\n1 99 65 99 4 46 54 80 51 30 96 1 28 30 44 70 78 1 1 100 1 62 1 1 1 85 1 1 1 61 1 46 75 1 61 77 97 26 67 1 1 63 81 85 86",
"output": "514"
},
{
"input": "72 3538\n52 1 8 1 1 1 7 1 1 1 1 48 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 40 1 1 38 1 1 1 1 1 1 1 1 1 1 1 35 1 93 79 1 1 1 1 1 1 1 1 1 51 1 1 1 1 1 1 1 1 1 1 1 1 96 1",
"output": "586"
},
{
"input": "81 2200\n1 59 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 93 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 50 1 1 1 1 1 1 1 1 1 1 1",
"output": "384"
},
{
"input": "81 2577\n85 91 1 1 2 1 1 100 1 80 1 1 17 86 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 37 1 66 24 1 1 96 49 1 66 1 44 1 1 1 1 98 1 1 1 1 35 1 37 3 35 1 1 87 64 1 24 1 58 1 1 42 83 5 1 1 1 1 1 95 1 94 1 50 1 1",
"output": "174"
},
{
"input": "81 4131\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "807"
},
{
"input": "81 6315\n1 1 67 100 1 99 36 1 92 5 1 96 42 12 1 57 91 1 1 66 41 30 74 95 1 37 1 39 91 69 1 52 77 47 65 1 1 93 96 74 90 35 85 76 71 92 92 1 1 67 92 74 1 1 86 76 35 1 56 16 27 57 37 95 1 40 20 100 51 1 80 60 45 79 95 1 46 1 25 100 96",
"output": "490"
},
{
"input": "96 1688\n1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 45 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 25 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 71 1 1 1 30 1 1 1",
"output": "284"
},
{
"input": "96 8889\n1 1 18 1 1 1 1 1 1 1 1 1 99 1 1 1 1 88 1 45 1 1 1 1 1 1 1 1 1 1 1 1 1 1 96 1 1 1 1 21 1 1 1 1 1 1 1 73 1 1 1 1 1 10 1 1 1 1 1 1 1 46 43 1 1 1 1 1 98 1 1 1 1 1 1 6 1 1 1 1 1 74 1 25 1 55 1 1 1 13 1 1 54 1 1 1",
"output": "1589"
},
{
"input": "10 100\n1 1 1 1 1 1 1 1 1 1",
"output": "18"
},
{
"input": "100 10000\n54 46 72 94 79 83 91 54 73 3 24 55 54 31 28 20 19 6 25 19 47 23 1 70 15 87 51 39 54 77 55 5 60 3 15 99 56 88 22 78 79 21 38 27 28 86 7 88 12 59 55 70 25 1 70 49 1 45 69 72 50 17 4 56 8 100 90 34 35 20 61 76 88 79 4 74 65 68 75 26 40 72 59 94 10 67 96 85 29 90 47 24 44 1 66 93 55 36 1 99",
"output": "1017"
},
{
"input": "100 6000\n41 31 23 17 24 78 26 96 93 48 46 2 49 33 35 9 73 100 34 48 83 36 33 69 43 24 3 74 8 81 27 33 94 38 77 9 76 90 62 90 21 67 22 22 12 2 17 27 61 18 72 85 59 65 71 38 90 75 74 66 60 47 58 50 90 95 75 10 5 100 97 29 83 88 65 26 93 90 22 98 36 55 70 38 50 92 88 72 99 96 25 14 74 16 25 92 67 94 77 96",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 6\n1",
"output": "1"
},
{
"input": "1 5\n1",
"output": "0"
},
{
"input": "1 3\n4",
"output": "-1"
},
{
"input": "3 24\n2 1 2",
"output": "-1"
}
] | 1,625,667,027
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 62
| 0
|
# from sys import stdin, stdout
# import re
# import math
# import statistics
# import itertools
# import collections
# import cmath
# import calendar
# import datetime
# import operator
#----------Profiling----------
# import cProfile
# import pstats
n, t = map(int, input().split())
songs = list(map(int, input().split()))
jokes_cracked = 0
for i in range(n-1):
t = t - songs[i] - 10
jokes_cracked = jokes_cracked + 2
if t < 0:
break
t = t - songs[-1]
jokes_cracked += t // 5
if t < 0:
print(-1)
else:
print(jokes_cracked)
|
Title: Devu, the Singer and Churu, the Joker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
Input Specification:
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
Output Specification:
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
Demo Input:
['3 30\n2 2 1\n', '3 20\n2 1 1\n']
Demo Output:
['5\n', '-1\n']
Note:
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.
|
```python
# from sys import stdin, stdout
# import re
# import math
# import statistics
# import itertools
# import collections
# import cmath
# import calendar
# import datetime
# import operator
#----------Profiling----------
# import cProfile
# import pstats
n, t = map(int, input().split())
songs = list(map(int, input().split()))
jokes_cracked = 0
for i in range(n-1):
t = t - songs[i] - 10
jokes_cracked = jokes_cracked + 2
if t < 0:
break
t = t - songs[-1]
jokes_cracked += t // 5
if t < 0:
print(-1)
else:
print(jokes_cracked)
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
|
You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3.
|
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
|
[
"50 30 10 10\n",
"100 50 10 21\n"
] |
[
"50\n30\n10\n",
"-1\n"
] |
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
| 0
|
[
{
"input": "50 30 10 10",
"output": "50\n30\n10"
},
{
"input": "100 50 10 21",
"output": "-1"
},
{
"input": "100 50 19 10",
"output": "100\n50\n19"
},
{
"input": "99 50 25 49",
"output": "100\n99\n49"
},
{
"input": "3 2 1 1",
"output": "4\n3\n1"
},
{
"input": "100 99 98 100",
"output": "-1"
},
{
"input": "100 40 30 40",
"output": "-1"
},
{
"input": "100 50 19 25",
"output": "100\n51\n25"
},
{
"input": "100 50 19 30",
"output": "100\n61\n30"
},
{
"input": "49 48 25 49",
"output": "-1"
},
{
"input": "48 47 23 46",
"output": "94\n93\n46"
},
{
"input": "37 23 16 20",
"output": "42\n41\n20"
},
{
"input": "98 2 1 1",
"output": "98\n3\n1"
},
{
"input": "99 22 13 14",
"output": "99\n29\n14"
},
{
"input": "97 95 3 2",
"output": "97\n95\n3"
},
{
"input": "27 3 2 3",
"output": "-1"
},
{
"input": "13 7 6 2",
"output": "-1"
},
{
"input": "19 17 11 6",
"output": "19\n17\n11"
},
{
"input": "15 6 4 5",
"output": "15\n11\n5"
},
{
"input": "21 3 1 3",
"output": "-1"
},
{
"input": "100 99 98 97",
"output": "196\n195\n98"
},
{
"input": "99 32 13 5",
"output": "-1"
},
{
"input": "5 4 3 1",
"output": "-1"
},
{
"input": "50 20 4 8",
"output": "50\n20\n8"
},
{
"input": "60 30 10 1",
"output": "-1"
},
{
"input": "100 50 10 1",
"output": "-1"
},
{
"input": "40 30 20 10",
"output": "40\n30\n20"
},
{
"input": "50 49 5 10",
"output": "50\n49\n10"
},
{
"input": "99 98 50 97",
"output": "196\n195\n97"
},
{
"input": "50 30 10 5",
"output": "50\n30\n10"
},
{
"input": "4 3 2 1",
"output": "4\n3\n2"
},
{
"input": "100 50 3 1",
"output": "-1"
}
] | 1,514,041,959
| 4,059
|
Python 3
|
RUNTIME_ERROR
|
PRETESTS
| 0
| 30
| 5,529,600
|
V1 = int(input())
V2 = int(input())
V3 = int(input())
Vm = int(input())
import random
max_ = 2*min(V3, Vm)
min_ = max(V3, Vm)
if min_ <= max_:
print(V1, V2, random.randint(min_, max_))
else:
print(-1)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input Specification:
You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3.
Output Specification:
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Demo Input:
['50 30 10 10\n', '100 50 10 21\n']
Demo Output:
['50\n30\n10\n', '-1\n']
Note:
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
|
```python
V1 = int(input())
V2 = int(input())
V3 = int(input())
Vm = int(input())
import random
max_ = 2*min(V3, Vm)
min_ = max(V3, Vm)
if min_ <= max_:
print(V1, V2, random.randint(min_, max_))
else:
print(-1)
```
| -1
|
|
990
|
A
|
Commentary Boxes
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Berland Football Cup starts really soon! Commentators from all over the world come to the event.
Organizers have already built $n$ commentary boxes. $m$ regional delegations will come to the Cup. Every delegation should get the same number of the commentary boxes. If any box is left unoccupied then the delegations will be upset. So each box should be occupied by exactly one delegation.
If $n$ is not divisible by $m$, it is impossible to distribute the boxes to the delegations at the moment.
Organizers can build a new commentary box paying $a$ burles and demolish a commentary box paying $b$ burles. They can both build and demolish boxes arbitrary number of times (each time paying a corresponding fee). It is allowed to demolish all the existing boxes.
What is the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by $m$)?
|
The only line contains four integer numbers $n$, $m$, $a$ and $b$ ($1 \le n, m \le 10^{12}$, $1 \le a, b \le 100$), where $n$ is the initial number of the commentary boxes, $m$ is the number of delegations to come, $a$ is the fee to build a box and $b$ is the fee to demolish a box.
|
Output the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by $m$). It is allowed that the final number of the boxes is equal to $0$.
|
[
"9 7 3 8\n",
"2 7 3 7\n",
"30 6 17 19\n"
] |
[
"15\n",
"14\n",
"0\n"
] |
In the first example organizers can build $5$ boxes to make the total of $14$ paying $3$ burles for the each of them.
In the second example organizers can demolish $2$ boxes to make the total of $0$ paying $7$ burles for the each of them.
In the third example organizers are already able to distribute all the boxes equally among the delegations, each one get $5$ boxes.
| 0
|
[
{
"input": "9 7 3 8",
"output": "15"
},
{
"input": "2 7 3 7",
"output": "14"
},
{
"input": "30 6 17 19",
"output": "0"
},
{
"input": "500000000001 1000000000000 100 100",
"output": "49999999999900"
},
{
"input": "1000000000000 750000000001 10 100",
"output": "5000000000020"
},
{
"input": "1000000000000 750000000001 100 10",
"output": "2499999999990"
},
{
"input": "42 1 1 1",
"output": "0"
},
{
"input": "1 1000000000000 1 100",
"output": "100"
},
{
"input": "7 2 3 7",
"output": "3"
},
{
"input": "999999999 2 1 1",
"output": "1"
},
{
"input": "999999999999 10000000007 100 100",
"output": "70100"
},
{
"input": "10000000001 2 1 1",
"output": "1"
},
{
"input": "29 6 1 2",
"output": "1"
},
{
"input": "99999999999 6 100 100",
"output": "300"
},
{
"input": "1000000000000 7 3 8",
"output": "8"
},
{
"input": "99999999999 2 1 1",
"output": "1"
},
{
"input": "1 2 1 1",
"output": "1"
},
{
"input": "999999999999 2 1 1",
"output": "1"
},
{
"input": "9 2 1 1",
"output": "1"
},
{
"input": "17 4 5 5",
"output": "5"
},
{
"input": "100000000000 3 1 1",
"output": "1"
},
{
"input": "100 7 1 1",
"output": "2"
},
{
"input": "1000000000000 3 100 100",
"output": "100"
},
{
"input": "70 3 10 10",
"output": "10"
},
{
"input": "1 2 5 1",
"output": "1"
},
{
"input": "1000000000000 3 1 1",
"output": "1"
},
{
"input": "804289377 846930887 78 16",
"output": "3326037780"
},
{
"input": "1000000000000 9 55 55",
"output": "55"
},
{
"input": "957747787 424238336 87 93",
"output": "10162213695"
},
{
"input": "25 6 1 2",
"output": "2"
},
{
"input": "22 7 3 8",
"output": "8"
},
{
"input": "10000000000 1 1 1",
"output": "0"
},
{
"input": "999999999999 2 10 10",
"output": "10"
},
{
"input": "999999999999 2 100 100",
"output": "100"
},
{
"input": "100 3 3 8",
"output": "6"
},
{
"input": "99999 2 1 1",
"output": "1"
},
{
"input": "100 3 2 5",
"output": "4"
},
{
"input": "1000000000000 13 10 17",
"output": "17"
},
{
"input": "7 2 1 2",
"output": "1"
},
{
"input": "10 3 1 2",
"output": "2"
},
{
"input": "5 2 2 2",
"output": "2"
},
{
"input": "100 3 5 2",
"output": "2"
},
{
"input": "7 2 1 1",
"output": "1"
},
{
"input": "70 4 1 1",
"output": "2"
},
{
"input": "10 4 1 1",
"output": "2"
},
{
"input": "6 7 41 42",
"output": "41"
},
{
"input": "10 3 10 1",
"output": "1"
},
{
"input": "5 5 2 3",
"output": "0"
},
{
"input": "1000000000000 3 99 99",
"output": "99"
},
{
"input": "7 3 100 1",
"output": "1"
},
{
"input": "7 2 100 5",
"output": "5"
},
{
"input": "1000000000000 1 23 33",
"output": "0"
},
{
"input": "30 7 1 1",
"output": "2"
},
{
"input": "100 3 1 1",
"output": "1"
},
{
"input": "90001 300 100 1",
"output": "1"
},
{
"input": "13 4 1 2",
"output": "2"
},
{
"input": "1000000000000 6 1 3",
"output": "2"
},
{
"input": "50 4 5 100",
"output": "10"
},
{
"input": "999 2 1 1",
"output": "1"
},
{
"input": "5 2 5 5",
"output": "5"
},
{
"input": "20 3 3 3",
"output": "3"
},
{
"input": "3982258181 1589052704 87 20",
"output": "16083055460"
},
{
"input": "100 3 1 3",
"output": "2"
},
{
"input": "7 3 1 1",
"output": "1"
},
{
"input": "19 10 100 100",
"output": "100"
},
{
"input": "23 3 100 1",
"output": "2"
},
{
"input": "25 7 100 1",
"output": "4"
},
{
"input": "100 9 1 2",
"output": "2"
},
{
"input": "9999999999 2 1 100",
"output": "1"
},
{
"input": "1000000000000 2 1 1",
"output": "0"
},
{
"input": "10000 3 1 1",
"output": "1"
},
{
"input": "22 7 1 6",
"output": "6"
},
{
"input": "100000000000 1 1 1",
"output": "0"
},
{
"input": "18 7 100 1",
"output": "4"
},
{
"input": "10003 4 1 100",
"output": "1"
},
{
"input": "3205261341 718648876 58 11",
"output": "3637324207"
},
{
"input": "8 3 100 1",
"output": "2"
},
{
"input": "15 7 1 1",
"output": "1"
},
{
"input": "1000000000000 1 20 20",
"output": "0"
},
{
"input": "16 7 3 2",
"output": "4"
},
{
"input": "1000000000000 1 1 1",
"output": "0"
},
{
"input": "7 3 1 100",
"output": "2"
},
{
"input": "16 3 1 100",
"output": "2"
},
{
"input": "13 4 1 10",
"output": "3"
},
{
"input": "10 4 5 5",
"output": "10"
},
{
"input": "14 3 1 100",
"output": "1"
},
{
"input": "100 33 100 1",
"output": "1"
},
{
"input": "22 7 1 8",
"output": "6"
},
{
"input": "10 4 2 1",
"output": "2"
},
{
"input": "6 4 2 2",
"output": "4"
},
{
"input": "17 4 2 1",
"output": "1"
},
{
"input": "7 3 100 10",
"output": "10"
},
{
"input": "702 7 3 2",
"output": "4"
},
{
"input": "8 3 1 5",
"output": "1"
},
{
"input": "3 2 5 2",
"output": "2"
},
{
"input": "99 19 1 7",
"output": "15"
},
{
"input": "16 3 100 1",
"output": "1"
},
{
"input": "100 34 1 100",
"output": "2"
},
{
"input": "100 33 1 1",
"output": "1"
},
{
"input": "2 3 4 3",
"output": "4"
},
{
"input": "15 4 4 10",
"output": "4"
},
{
"input": "1144108931 470211273 45 79",
"output": "11993619960"
},
{
"input": "2 3 3 4",
"output": "3"
},
{
"input": "29 5 4 9",
"output": "4"
},
{
"input": "15 7 1 5",
"output": "5"
},
{
"input": "1 1 1 1",
"output": "0"
},
{
"input": "1 1 3 4",
"output": "0"
},
{
"input": "10 12 2 1",
"output": "4"
},
{
"input": "1 2 3 4",
"output": "3"
}
] | 1,606,026,139
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 109
| 140
| 0
|
n, m, a, b = map(int, input().split())
destroy = n%m
build = m - destroy
dest = destroy* b
bui = build*a
print(min(dest, bui))
|
Title: Commentary Boxes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Berland Football Cup starts really soon! Commentators from all over the world come to the event.
Organizers have already built $n$ commentary boxes. $m$ regional delegations will come to the Cup. Every delegation should get the same number of the commentary boxes. If any box is left unoccupied then the delegations will be upset. So each box should be occupied by exactly one delegation.
If $n$ is not divisible by $m$, it is impossible to distribute the boxes to the delegations at the moment.
Organizers can build a new commentary box paying $a$ burles and demolish a commentary box paying $b$ burles. They can both build and demolish boxes arbitrary number of times (each time paying a corresponding fee). It is allowed to demolish all the existing boxes.
What is the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by $m$)?
Input Specification:
The only line contains four integer numbers $n$, $m$, $a$ and $b$ ($1 \le n, m \le 10^{12}$, $1 \le a, b \le 100$), where $n$ is the initial number of the commentary boxes, $m$ is the number of delegations to come, $a$ is the fee to build a box and $b$ is the fee to demolish a box.
Output Specification:
Output the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by $m$). It is allowed that the final number of the boxes is equal to $0$.
Demo Input:
['9 7 3 8\n', '2 7 3 7\n', '30 6 17 19\n']
Demo Output:
['15\n', '14\n', '0\n']
Note:
In the first example organizers can build $5$ boxes to make the total of $14$ paying $3$ burles for the each of them.
In the second example organizers can demolish $2$ boxes to make the total of $0$ paying $7$ burles for the each of them.
In the third example organizers are already able to distribute all the boxes equally among the delegations, each one get $5$ boxes.
|
```python
n, m, a, b = map(int, input().split())
destroy = n%m
build = m - destroy
dest = destroy* b
bui = build*a
print(min(dest, bui))
```
| 3
|
|
465
|
A
|
inc ARG
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of *n* bits. These bits are numbered from 1 to *n*. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the *n*-th bit.
Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded.
Sergey wrote certain values of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of bits in the cell.
The second line contains a string consisting of *n* characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit.
|
Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell.
|
[
"4\n1100\n",
"4\n1111\n"
] |
[
"3\n",
"4\n"
] |
In the first sample the cell ends up with value 0010, in the second sample — with 0000.
| 500
|
[
{
"input": "4\n1100",
"output": "3"
},
{
"input": "4\n1111",
"output": "4"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "1"
},
{
"input": "2\n01",
"output": "1"
},
{
"input": "2\n10",
"output": "2"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "10\n0000000000",
"output": "1"
},
{
"input": "20\n11111111110110001100",
"output": "11"
},
{
"input": "50\n01011110100010000001010000100001001101001101101011",
"output": "1"
},
{
"input": "60\n111111111101111111111111111111111111111111111111111111111111",
"output": "11"
},
{
"input": "60\n111111111111111111111111111111111111111111111111111111111111",
"output": "60"
},
{
"input": "66\n111111010010011001110011000111000100011110011001111110011111111101",
"output": "7"
},
{
"input": "90\n000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "91\n1011111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "2"
},
{
"input": "92\n11111111101011111011111111111111111111111011101111111111011111111111111101111111101011111101",
"output": "10"
},
{
"input": "100\n0001011110100011001100100010111001000001111101101001001001001011110100101101010000000110100101110010",
"output": "1"
},
{
"input": "100\n0111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "1"
},
{
"input": "100\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "100\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "100"
},
{
"input": "100\n1011001110001000011111110011000100001110010110111101110110011011011000010100110001111100000010110010",
"output": "2"
},
{
"input": "100\n1101111011001111111111110011110111101110111111111111111111111111111111011111111111110111111111111111",
"output": "3"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "100"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110",
"output": "100"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111101",
"output": "99"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111100000",
"output": "96"
},
{
"input": "100\n1111111111111111111111111111111110011000110110001010101100100100001000010010010010000100110000010101",
"output": "34"
}
] | 1,638,568,155
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 28
| 93
| 3,379,200
|
n=int(input())
d=input()
z=0
for i,k in enumerate((l:=bin(int(d[::-1],2)+1)[:1:-1][:n])+'0'*(n-len(str(l)))):
if k!=d[i]:z+=1
print(z)
|
Title: inc ARG
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of *n* bits. These bits are numbered from 1 to *n*. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the *n*-th bit.
Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded.
Sergey wrote certain values of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of bits in the cell.
The second line contains a string consisting of *n* characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit.
Output Specification:
Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell.
Demo Input:
['4\n1100\n', '4\n1111\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample the cell ends up with value 0010, in the second sample — with 0000.
|
```python
n=int(input())
d=input()
z=0
for i,k in enumerate((l:=bin(int(d[::-1],2)+1)[:1:-1][:n])+'0'*(n-len(str(l)))):
if k!=d[i]:z+=1
print(z)
```
| -1
|
|
992
|
A
|
Nastya and an Array
|
PROGRAMMING
| 800
|
[
"implementation",
"sortings"
] | null | null |
Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array.
|
Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero.
|
[
"5\n1 1 1 1 1\n",
"3\n2 0 -1\n",
"4\n5 -6 -5 1\n"
] |
[
"1\n",
"2\n",
"4\n"
] |
In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element.
| 500
|
[
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "3\n2 0 -1",
"output": "2"
},
{
"input": "4\n5 -6 -5 1",
"output": "4"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n21794 -79194",
"output": "2"
},
{
"input": "3\n-63526 95085 -5239",
"output": "3"
},
{
"input": "3\n0 53372 -20572",
"output": "2"
},
{
"input": "13\n-2075 -32242 27034 -37618 -96962 82203 64846 48249 -71761 28908 -21222 -61370 46899",
"output": "13"
},
{
"input": "5\n806 0 1308 1954 683",
"output": "4"
},
{
"input": "8\n-26 0 -249 -289 -126 -206 288 -11",
"output": "7"
},
{
"input": "10\n2 2 2 1 2 -1 0 2 -1 1",
"output": "3"
},
{
"input": "1\n8",
"output": "1"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "10"
},
{
"input": "5\n2 0 -1 0 0",
"output": "2"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "2\n-1 0",
"output": "1"
},
{
"input": "4\n0 0 0 0",
"output": "0"
},
{
"input": "8\n10 9 -1 0 0 3 2 3",
"output": "5"
},
{
"input": "5\n5 0 1 2 3",
"output": "4"
},
{
"input": "3\n1 1 0",
"output": "1"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "5\n1 2 0 0 0",
"output": "2"
},
{
"input": "5\n1 0 0 0 0",
"output": "1"
},
{
"input": "5\n4 5 6 0 0",
"output": "3"
},
{
"input": "4\n-1 0 0 1",
"output": "2"
},
{
"input": "5\n3 0 0 4 5",
"output": "3"
},
{
"input": "3\n0 0 2",
"output": "1"
},
{
"input": "3\n1 0 0",
"output": "1"
},
{
"input": "4\n0 0 0 4",
"output": "1"
},
{
"input": "5\n-1 0 0 0 0",
"output": "1"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "10\n0 0 0 0 0 1 2 3 0 0",
"output": "3"
},
{
"input": "4\n0 1 2 3",
"output": "3"
},
{
"input": "3\n0 1 2",
"output": "2"
},
{
"input": "4\n2 0 0 -1",
"output": "2"
}
] | 1,646,960,392
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 79
| 93
| 8,908,800
|
#sa7afy
#a,b = map(int,input().split())
#arr=[]
#arr=list(map(int,input().split()))
#arr = list(dict.fromkeys(arr))
#arr.sort()
#n = int(input())
#for i in range():
#print(*list)
#────────▓▓▓▓▓▓▓────────────▒▒▒▒▒▒
#──────▓▓▒▒▒▒▒▒▒▓▓────────▒▒░░░░░░▒▒
#────▓▓▒▒▒▒▒▒▒▒▒▒▒▓▓────▒▒░░░░░░░░░▒▒▒
#───▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▓▓▒▒░░░░░░░░░░░░░░▒
#──▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░░░░░░░░▒
#──▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░░░░░░░░░▒
#─▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░░░░░░░░░░▒
#▓▓▒▒▒▒▒▒░░░░░░░░░░░▒▒░░▒▒▒▒▒▒▒▒▒▒▒░░░░░░▒
#▓▓▒▒▒▒▒▒▀▀▀▀▀███▄▄▒▒▒░░░▄▄▄██▀▀▀▀▀░░░░░░▒
#▓▓▒▒▒▒▒▒▒▄▀████▀███▄▒░▄████▀████▄░░░░░░░▒
#▓▓▒▒▒▒▒▒█──▀█████▀─▌▒░▐──▀█████▀─█░░░░░░▒
#▓▓▒▒▒▒▒▒▒▀▄▄▄▄▄▄▄▄▀▒▒░░▀▄▄▄▄▄▄▄▄▀░░░░░░░▒
#─▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░░░░░░░░░▒
#──▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░░░░░░░░▒
#───▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▀▀▀░░░░░░░░░░░░░░▒
#────▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░░░░░▒▒
#─────▓▓▒▒▒▒▒▒▒▒▒▒▄▄▄▄▄▄▄▄▄░░░░░░░░▒▒
#──────▓▓▒▒▒▒▒▒▒▄▀▀▀▀▀▀▀▀▀▀▀▄░░░░░▒▒
#───────▓▓▒▒▒▒▒▀▒▒▒▒▒▒░░░░░░░▀░░░▒▒
#────────▓▓▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░▒▒
#──────────▓▓▒▒▒▒▒▒▒▒▒░░░░░░░░▒▒
#───────────▓▓▒▒▒▒▒▒▒▒░░░░░░░▒▒
#─────────────▓▓▒▒▒▒▒▒░░░░░▒▒
#───────────────▓▓▒▒▒▒░░░░▒▒
#────────────────▓▓▒▒▒░░░▒▒
#──────────────────▓▓▒░▒▒
#───────────────────▓▒░▒
#────────────────────▓▒
def run():
print("k")
def run3():
print("k")
def run2():
print("k")
def kk():
print("Sa7afy")
#code here
n=int(input())
count=0
arr=list(map(int,input().split()))
arr = list(dict.fromkeys(arr))
for i in arr:
if i !=0:
count+=1
print(count)
#endsad
|
Title: Nastya and an Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array.
Output Specification:
Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero.
Demo Input:
['5\n1 1 1 1 1\n', '3\n2 0 -1\n', '4\n5 -6 -5 1\n']
Demo Output:
['1\n', '2\n', '4\n']
Note:
In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element.
|
```python
#sa7afy
#a,b = map(int,input().split())
#arr=[]
#arr=list(map(int,input().split()))
#arr = list(dict.fromkeys(arr))
#arr.sort()
#n = int(input())
#for i in range():
#print(*list)
#────────▓▓▓▓▓▓▓────────────▒▒▒▒▒▒
#──────▓▓▒▒▒▒▒▒▒▓▓────────▒▒░░░░░░▒▒
#────▓▓▒▒▒▒▒▒▒▒▒▒▒▓▓────▒▒░░░░░░░░░▒▒▒
#───▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▓▓▒▒░░░░░░░░░░░░░░▒
#──▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░░░░░░░░▒
#──▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░░░░░░░░░▒
#─▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░░░░░░░░░░▒
#▓▓▒▒▒▒▒▒░░░░░░░░░░░▒▒░░▒▒▒▒▒▒▒▒▒▒▒░░░░░░▒
#▓▓▒▒▒▒▒▒▀▀▀▀▀███▄▄▒▒▒░░░▄▄▄██▀▀▀▀▀░░░░░░▒
#▓▓▒▒▒▒▒▒▒▄▀████▀███▄▒░▄████▀████▄░░░░░░░▒
#▓▓▒▒▒▒▒▒█──▀█████▀─▌▒░▐──▀█████▀─█░░░░░░▒
#▓▓▒▒▒▒▒▒▒▀▄▄▄▄▄▄▄▄▀▒▒░░▀▄▄▄▄▄▄▄▄▀░░░░░░░▒
#─▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░░░░░░░░░▒
#──▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░░░░░░░░▒
#───▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▀▀▀░░░░░░░░░░░░░░▒
#────▓▓▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░░░░░▒▒
#─────▓▓▒▒▒▒▒▒▒▒▒▒▄▄▄▄▄▄▄▄▄░░░░░░░░▒▒
#──────▓▓▒▒▒▒▒▒▒▄▀▀▀▀▀▀▀▀▀▀▀▄░░░░░▒▒
#───────▓▓▒▒▒▒▒▀▒▒▒▒▒▒░░░░░░░▀░░░▒▒
#────────▓▓▒▒▒▒▒▒▒▒▒▒▒░░░░░░░░░░▒▒
#──────────▓▓▒▒▒▒▒▒▒▒▒░░░░░░░░▒▒
#───────────▓▓▒▒▒▒▒▒▒▒░░░░░░░▒▒
#─────────────▓▓▒▒▒▒▒▒░░░░░▒▒
#───────────────▓▓▒▒▒▒░░░░▒▒
#────────────────▓▓▒▒▒░░░▒▒
#──────────────────▓▓▒░▒▒
#───────────────────▓▒░▒
#────────────────────▓▒
def run():
print("k")
def run3():
print("k")
def run2():
print("k")
def kk():
print("Sa7afy")
#code here
n=int(input())
count=0
arr=list(map(int,input().split()))
arr = list(dict.fromkeys(arr))
for i in arr:
if i !=0:
count+=1
print(count)
#endsad
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Paul hates palindromes. He assumes that string *s* is tolerable if each its character is one of the first *p* letters of the English alphabet and *s* doesn't contain any palindrome contiguous substring of length 2 or more.
Paul has found a tolerable string *s* of length *n*. Help him find the lexicographically next tolerable string of the same length or else state that such string does not exist.
|
The first line contains two space-separated integers: *n* and *p* (1<=≤<=*n*<=≤<=1000; 1<=≤<=*p*<=≤<=26). The second line contains string *s*, consisting of *n* small English letters. It is guaranteed that the string is tolerable (according to the above definition).
|
If the lexicographically next tolerable string of the same length exists, print it. Otherwise, print "NO" (without the quotes).
|
[
"3 3\ncba\n",
"3 4\ncba\n",
"4 4\nabcd\n"
] |
[
"NO\n",
"cbd\n",
"abda\n"
] |
String *s* is lexicographically larger (or simply larger) than string *t* with the same length, if there is number *i*, such that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, ..., *s*<sub class="lower-index">*i*</sub> = *t*<sub class="lower-index">*i*</sub>, *s*<sub class="lower-index">*i* + 1</sub> > *t*<sub class="lower-index">*i* + 1</sub>.
The lexicographically next tolerable string is the lexicographically minimum tolerable string which is larger than the given one.
A palindrome is a string that reads the same forward or reversed.
| 0
|
[
{
"input": "3 3\ncba",
"output": "NO"
},
{
"input": "3 4\ncba",
"output": "cbd"
},
{
"input": "4 4\nabcd",
"output": "abda"
},
{
"input": "2 2\nab",
"output": "ba"
},
{
"input": "2 2\nba",
"output": "NO"
},
{
"input": "1 2\na",
"output": "b"
},
{
"input": "1 2\nb",
"output": "NO"
},
{
"input": "1 1\na",
"output": "NO"
},
{
"input": "3 4\ncdb",
"output": "dab"
},
{
"input": "7 26\nzyxzyxz",
"output": "NO"
},
{
"input": "10 5\nabcabcabca",
"output": "abcabcabcd"
},
{
"input": "10 10\nfajegfaicb",
"output": "fajegfaicd"
},
{
"input": "1 26\no",
"output": "p"
},
{
"input": "1 2\nb",
"output": "NO"
},
{
"input": "1 26\nz",
"output": "NO"
},
{
"input": "3 3\ncab",
"output": "cba"
},
{
"input": "3 26\nyzx",
"output": "zab"
},
{
"input": "5 5\naceba",
"output": "acebc"
},
{
"input": "10 3\ncbacbacbac",
"output": "NO"
},
{
"input": "11 3\nabcabcabcab",
"output": "acbacbacbac"
},
{
"input": "12 10\nabcabcabcabc",
"output": "abcabcabcabd"
},
{
"input": "13 7\ngfegfegfegfeg",
"output": "NO"
},
{
"input": "15 11\ncgjkbadjfbdaikj",
"output": "cgjkbadjfbdajba"
},
{
"input": "17 4\ndabcadcbdcadbcdbc",
"output": "dabcadcbdcadcabca"
},
{
"input": "26 26\nahnxdnbfcriersyzdihuecojdi",
"output": "ahnxdnbfcriersyzdihuecojdk"
},
{
"input": "30 7\ncedcfedcfgcfgcbadcadgfaegfacgf",
"output": "cedcfedcfgcfgcbadcadgfaegfadba"
},
{
"input": "70 4\ndcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbd",
"output": "NO"
},
{
"input": "77 7\ncadgbagbcaecgfaegcdbeafbacbdfgaedgcdeabgebaecbeacgfebagedcegdafdgeacegfegfegf",
"output": "cadgbagbcaecgfaegcdbeafbacbdfgaedgcdeabgebaecbeacgfebagedcegdafdgeacfabcabcab"
},
{
"input": "100 4\nabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcd"
},
{
"input": "333 5\nedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedcedc",
"output": "NO"
},
{
"input": "3 3\nacb",
"output": "bac"
},
{
"input": "17 26\nbazyxzyxzyxzyxzyx",
"output": "bcabcabcabcabcabc"
},
{
"input": "6 3\nacbacb",
"output": "bacbac"
},
{
"input": "6 3\nabcabc",
"output": "acbacb"
},
{
"input": "302 4\nabdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcbdcb",
"output": "acbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbac"
},
{
"input": "30 26\nabcabcabczyxzyxzyxzyxzyxzyxzyx",
"output": "abcabcabdabcabcabcabcabcabcabc"
},
{
"input": "300 3\nabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc",
"output": "acbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacb"
},
{
"input": "2 4\ncd",
"output": "da"
}
] | 1,661,508,839
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
n,p = map(int, input().split())
a = list(input())
c = False
for i in range(n-1,-1,-1):
for k in range(p):
ch = chr(ord('a')+k)
if ch>a[i] and (i>0 and ch!=a[i-1]) and (i>1 and ch!=a[i-2]):
a[i] = ch
c = True
break
if c:
break
if not c:
print("NO")
else:
s=""
p = {"a","b","c","d"}
sp = p.difference({a[i],a[i-1]})
s+=min(list(sp))
sp = p.difference({a[i],s[0]})
s += min(list(sp))
sp = p.difference({ a[0],s[1]})
s += min(list(sp))
idx=0
for j in range(i+1, n):
a[j] = s[idx%3]
idx+=1
print(''.join(a))
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Paul hates palindromes. He assumes that string *s* is tolerable if each its character is one of the first *p* letters of the English alphabet and *s* doesn't contain any palindrome contiguous substring of length 2 or more.
Paul has found a tolerable string *s* of length *n*. Help him find the lexicographically next tolerable string of the same length or else state that such string does not exist.
Input Specification:
The first line contains two space-separated integers: *n* and *p* (1<=≤<=*n*<=≤<=1000; 1<=≤<=*p*<=≤<=26). The second line contains string *s*, consisting of *n* small English letters. It is guaranteed that the string is tolerable (according to the above definition).
Output Specification:
If the lexicographically next tolerable string of the same length exists, print it. Otherwise, print "NO" (without the quotes).
Demo Input:
['3 3\ncba\n', '3 4\ncba\n', '4 4\nabcd\n']
Demo Output:
['NO\n', 'cbd\n', 'abda\n']
Note:
String *s* is lexicographically larger (or simply larger) than string *t* with the same length, if there is number *i*, such that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, ..., *s*<sub class="lower-index">*i*</sub> = *t*<sub class="lower-index">*i*</sub>, *s*<sub class="lower-index">*i* + 1</sub> > *t*<sub class="lower-index">*i* + 1</sub>.
The lexicographically next tolerable string is the lexicographically minimum tolerable string which is larger than the given one.
A palindrome is a string that reads the same forward or reversed.
|
```python
n,p = map(int, input().split())
a = list(input())
c = False
for i in range(n-1,-1,-1):
for k in range(p):
ch = chr(ord('a')+k)
if ch>a[i] and (i>0 and ch!=a[i-1]) and (i>1 and ch!=a[i-2]):
a[i] = ch
c = True
break
if c:
break
if not c:
print("NO")
else:
s=""
p = {"a","b","c","d"}
sp = p.difference({a[i],a[i-1]})
s+=min(list(sp))
sp = p.difference({a[i],s[0]})
s += min(list(sp))
sp = p.difference({ a[0],s[1]})
s += min(list(sp))
idx=0
for j in range(i+1, n):
a[j] = s[idx%3]
idx+=1
print(''.join(a))
```
| 0
|
|
342
|
A
|
Xenia and Divisors
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation"
] | null | null |
Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held:
- *a*<=<<=*b*<=<<=*c*; - *a* divides *b*, *b* divides *c*.
Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three.
Help Xenia, find the required partition or else say that it doesn't exist.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7.
It is guaranteed that *n* is divisible by 3.
|
If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.
If there is no solution, print -1.
|
[
"6\n1 1 1 2 2 2\n",
"6\n2 2 1 1 4 6\n"
] |
[
"-1\n",
"1 2 4\n1 2 6\n"
] |
none
| 500
|
[
{
"input": "6\n1 1 1 2 2 2",
"output": "-1"
},
{
"input": "6\n2 2 1 1 4 6",
"output": "1 2 4\n1 2 6"
},
{
"input": "3\n1 2 3",
"output": "-1"
},
{
"input": "3\n7 5 7",
"output": "-1"
},
{
"input": "3\n1 3 4",
"output": "-1"
},
{
"input": "3\n1 1 1",
"output": "-1"
},
{
"input": "9\n1 3 6 6 3 1 3 1 6",
"output": "1 3 6\n1 3 6\n1 3 6"
},
{
"input": "6\n1 2 4 1 3 5",
"output": "-1"
},
{
"input": "3\n1 3 7",
"output": "-1"
},
{
"input": "3\n1 1 1",
"output": "-1"
},
{
"input": "9\n1 2 4 1 2 4 1 3 6",
"output": "1 2 4\n1 2 4\n1 3 6"
},
{
"input": "12\n3 6 1 1 3 6 1 1 2 6 2 6",
"output": "1 3 6\n1 3 6\n1 2 6\n1 2 6"
},
{
"input": "9\n1 1 1 4 4 4 6 2 2",
"output": "-1"
},
{
"input": "9\n1 2 4 6 3 1 3 1 5",
"output": "-1"
},
{
"input": "15\n2 1 2 1 3 6 1 2 1 6 1 3 4 6 4",
"output": "1 2 4\n1 2 4\n1 3 6\n1 3 6\n1 2 6"
},
{
"input": "3\n2 3 6",
"output": "-1"
},
{
"input": "3\n2 4 6",
"output": "-1"
},
{
"input": "3\n2 5 6",
"output": "-1"
},
{
"input": "3\n2 4 7",
"output": "-1"
},
{
"input": "6\n1 2 3 4 5 6",
"output": "-1"
},
{
"input": "3\n7 7 7",
"output": "-1"
},
{
"input": "6\n1 2 4 7 7 7",
"output": "-1"
},
{
"input": "6\n1 1 2 6 6 6",
"output": "-1"
},
{
"input": "9\n1 1 1 3 3 2 4 4 6",
"output": "-1"
},
{
"input": "6\n1 2 4 5 5 5",
"output": "-1"
},
{
"input": "15\n1 1 1 1 1 2 2 2 2 4 4 6 6 6 6",
"output": "-1"
},
{
"input": "6\n1 1 5 5 7 7",
"output": "-1"
},
{
"input": "9\n1 1 1 2 3 4 5 6 7",
"output": "-1"
},
{
"input": "6\n1 1 4 4 7 7",
"output": "-1"
},
{
"input": "24\n1 1 1 1 1 1 1 1 1 2 2 2 3 3 3 3 3 3 4 4 4 6 6 6",
"output": "-1"
},
{
"input": "3\n1 7 6",
"output": "-1"
},
{
"input": "6\n1 1 2 4 7 7",
"output": "-1"
},
{
"input": "9\n1 1 1 7 7 7 7 7 7",
"output": "-1"
},
{
"input": "9\n1 1 1 2 3 4 6 5 5",
"output": "-1"
}
] | 1,584,991,921
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 93
| 307,200
|
n = int(input())
l = list(map(int,input().split()))
count1,count2,count3,count4,count6 = 0,0,0,0,0
t = True
count1 = l.count(1)
count2 = l.count(2)
count3 = l.count(3)
count4 = l.count(4)
count6 = l.count(6)
#print(count1,count2,count3,count4,count6)
x = count2+count3
y = count4+count6
#print(x,y)
c = 0
if(count1==n//3 and x==n//3 and y==n//3):
for i in range(count3):
print(1,3,6)
for i in range(count4):
print(1,2,4)
c+=1
for i in range(count2-c):
print(1,2,6)
else:
print("-1")
|
Title: Xenia and Divisors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held:
- *a*<=<<=*b*<=<<=*c*; - *a* divides *b*, *b* divides *c*.
Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three.
Help Xenia, find the required partition or else say that it doesn't exist.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7.
It is guaranteed that *n* is divisible by 3.
Output Specification:
If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.
If there is no solution, print -1.
Demo Input:
['6\n1 1 1 2 2 2\n', '6\n2 2 1 1 4 6\n']
Demo Output:
['-1\n', '1 2 4\n1 2 6\n']
Note:
none
|
```python
n = int(input())
l = list(map(int,input().split()))
count1,count2,count3,count4,count6 = 0,0,0,0,0
t = True
count1 = l.count(1)
count2 = l.count(2)
count3 = l.count(3)
count4 = l.count(4)
count6 = l.count(6)
#print(count1,count2,count3,count4,count6)
x = count2+count3
y = count4+count6
#print(x,y)
c = 0
if(count1==n//3 and x==n//3 and y==n//3):
for i in range(count3):
print(1,3,6)
for i in range(count4):
print(1,2,4)
c+=1
for i in range(count2-c):
print(1,2,6)
else:
print("-1")
```
| 0
|
|
675
|
A
|
Infinite Sequence
|
PROGRAMMING
| 1,100
|
[
"math"
] | null | null |
Vasya likes everything infinite. Now he is studying the properties of a sequence *s*, such that its first element is equal to *a* (*s*1<==<=*a*), and the difference between any two neighbouring elements is equal to *c* (*s**i*<=-<=*s**i*<=-<=1<==<=*c*). In particular, Vasya wonders if his favourite integer *b* appears in this sequence, that is, there exists a positive integer *i*, such that *s**i*<==<=*b*. Of course, you are the person he asks for a help.
|
The first line of the input contain three integers *a*, *b* and *c* (<=-<=109<=≤<=*a*,<=*b*,<=*c*<=≤<=109) — the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively.
|
If *b* appears in the sequence *s* print "YES" (without quotes), otherwise print "NO" (without quotes).
|
[
"1 7 3\n",
"10 10 0\n",
"1 -4 5\n",
"0 60 50\n"
] |
[
"YES\n",
"YES\n",
"NO\n",
"NO\n"
] |
In the first sample, the sequence starts from integers 1, 4, 7, so 7 is its element.
In the second sample, the favorite integer of Vasya is equal to the first element of the sequence.
In the third sample all elements of the sequence are greater than Vasya's favorite integer.
In the fourth sample, the sequence starts from 0, 50, 100, and all the following elements are greater than Vasya's favorite integer.
| 500
|
[
{
"input": "1 7 3",
"output": "YES"
},
{
"input": "10 10 0",
"output": "YES"
},
{
"input": "1 -4 5",
"output": "NO"
},
{
"input": "0 60 50",
"output": "NO"
},
{
"input": "1 -4 -5",
"output": "YES"
},
{
"input": "0 1 0",
"output": "NO"
},
{
"input": "10 10 42",
"output": "YES"
},
{
"input": "-1000000000 1000000000 -1",
"output": "NO"
},
{
"input": "10 16 4",
"output": "NO"
},
{
"input": "-1000000000 1000000000 5",
"output": "YES"
},
{
"input": "1000000000 -1000000000 5",
"output": "NO"
},
{
"input": "1000000000 -1000000000 0",
"output": "NO"
},
{
"input": "1000000000 1000000000 0",
"output": "YES"
},
{
"input": "115078364 -899474523 -1",
"output": "YES"
},
{
"input": "-245436499 416383245 992",
"output": "YES"
},
{
"input": "-719636354 536952440 2",
"output": "YES"
},
{
"input": "-198350539 963391024 68337739",
"output": "YES"
},
{
"input": "-652811055 875986516 1091",
"output": "YES"
},
{
"input": "119057893 -516914539 -39748277",
"output": "YES"
},
{
"input": "989140430 731276607 -36837689",
"output": "YES"
},
{
"input": "677168390 494583489 -985071853",
"output": "NO"
},
{
"input": "58090193 777423708 395693923",
"output": "NO"
},
{
"input": "479823846 -403424770 -653472589",
"output": "NO"
},
{
"input": "-52536829 -132023273 -736287999",
"output": "NO"
},
{
"input": "-198893776 740026818 -547885271",
"output": "NO"
},
{
"input": "-2 -2 -2",
"output": "YES"
},
{
"input": "-2 -2 -1",
"output": "YES"
},
{
"input": "-2 -2 0",
"output": "YES"
},
{
"input": "-2 -2 1",
"output": "YES"
},
{
"input": "-2 -2 2",
"output": "YES"
},
{
"input": "-2 -1 -2",
"output": "NO"
},
{
"input": "-2 -1 -1",
"output": "NO"
},
{
"input": "-2 -1 0",
"output": "NO"
},
{
"input": "-2 -1 1",
"output": "YES"
},
{
"input": "-2 -1 2",
"output": "NO"
},
{
"input": "-2 0 -2",
"output": "NO"
},
{
"input": "-2 0 -1",
"output": "NO"
},
{
"input": "-2 0 0",
"output": "NO"
},
{
"input": "-2 0 1",
"output": "YES"
},
{
"input": "-2 0 2",
"output": "YES"
},
{
"input": "-2 1 -2",
"output": "NO"
},
{
"input": "-2 1 -1",
"output": "NO"
},
{
"input": "-2 1 0",
"output": "NO"
},
{
"input": "-2 1 1",
"output": "YES"
},
{
"input": "-2 1 2",
"output": "NO"
},
{
"input": "-2 2 -2",
"output": "NO"
},
{
"input": "-2 2 -1",
"output": "NO"
},
{
"input": "-2 2 0",
"output": "NO"
},
{
"input": "-2 2 1",
"output": "YES"
},
{
"input": "-2 2 2",
"output": "YES"
},
{
"input": "-1 -2 -2",
"output": "NO"
},
{
"input": "-1 -2 -1",
"output": "YES"
},
{
"input": "-1 -2 0",
"output": "NO"
},
{
"input": "-1 -2 1",
"output": "NO"
},
{
"input": "-1 -2 2",
"output": "NO"
},
{
"input": "-1 -1 -2",
"output": "YES"
},
{
"input": "-1 -1 -1",
"output": "YES"
},
{
"input": "-1 -1 0",
"output": "YES"
},
{
"input": "-1 -1 1",
"output": "YES"
},
{
"input": "-1 -1 2",
"output": "YES"
},
{
"input": "-1 0 -2",
"output": "NO"
},
{
"input": "-1 0 -1",
"output": "NO"
},
{
"input": "-1 0 0",
"output": "NO"
},
{
"input": "-1 0 1",
"output": "YES"
},
{
"input": "-1 0 2",
"output": "NO"
},
{
"input": "-1 1 -2",
"output": "NO"
},
{
"input": "-1 1 -1",
"output": "NO"
},
{
"input": "-1 1 0",
"output": "NO"
},
{
"input": "-1 1 1",
"output": "YES"
},
{
"input": "-1 1 2",
"output": "YES"
},
{
"input": "-1 2 -2",
"output": "NO"
},
{
"input": "-1 2 -1",
"output": "NO"
},
{
"input": "-1 2 0",
"output": "NO"
},
{
"input": "-1 2 1",
"output": "YES"
},
{
"input": "-1 2 2",
"output": "NO"
},
{
"input": "0 -2 -2",
"output": "YES"
},
{
"input": "0 -2 -1",
"output": "YES"
},
{
"input": "0 -2 0",
"output": "NO"
},
{
"input": "0 -2 1",
"output": "NO"
},
{
"input": "0 -2 2",
"output": "NO"
},
{
"input": "0 -1 -2",
"output": "NO"
},
{
"input": "0 -1 -1",
"output": "YES"
},
{
"input": "0 -1 0",
"output": "NO"
},
{
"input": "0 -1 1",
"output": "NO"
},
{
"input": "0 -1 2",
"output": "NO"
},
{
"input": "0 0 -2",
"output": "YES"
},
{
"input": "0 0 -1",
"output": "YES"
},
{
"input": "0 0 0",
"output": "YES"
},
{
"input": "0 0 1",
"output": "YES"
},
{
"input": "0 0 2",
"output": "YES"
},
{
"input": "0 1 -2",
"output": "NO"
},
{
"input": "0 1 -1",
"output": "NO"
},
{
"input": "0 1 0",
"output": "NO"
},
{
"input": "0 1 1",
"output": "YES"
},
{
"input": "0 1 2",
"output": "NO"
},
{
"input": "0 2 -2",
"output": "NO"
},
{
"input": "0 2 -1",
"output": "NO"
},
{
"input": "0 2 0",
"output": "NO"
},
{
"input": "0 2 1",
"output": "YES"
},
{
"input": "0 2 2",
"output": "YES"
},
{
"input": "1 -2 -2",
"output": "NO"
},
{
"input": "1 -2 -1",
"output": "YES"
},
{
"input": "1 -2 0",
"output": "NO"
},
{
"input": "1 -2 1",
"output": "NO"
},
{
"input": "1 -2 2",
"output": "NO"
},
{
"input": "1 -1 -2",
"output": "YES"
},
{
"input": "1 -1 -1",
"output": "YES"
},
{
"input": "1 -1 0",
"output": "NO"
},
{
"input": "1 -1 1",
"output": "NO"
},
{
"input": "1 -1 2",
"output": "NO"
},
{
"input": "1 0 -2",
"output": "NO"
},
{
"input": "1 0 -1",
"output": "YES"
},
{
"input": "1 0 0",
"output": "NO"
},
{
"input": "1 0 1",
"output": "NO"
},
{
"input": "1 0 2",
"output": "NO"
},
{
"input": "1 1 -2",
"output": "YES"
},
{
"input": "1 1 -1",
"output": "YES"
},
{
"input": "1 1 0",
"output": "YES"
},
{
"input": "1 1 1",
"output": "YES"
},
{
"input": "1 1 2",
"output": "YES"
},
{
"input": "1 2 -2",
"output": "NO"
},
{
"input": "1 2 -1",
"output": "NO"
},
{
"input": "1 2 0",
"output": "NO"
},
{
"input": "1 2 1",
"output": "YES"
},
{
"input": "1 2 2",
"output": "NO"
},
{
"input": "2 -2 -2",
"output": "YES"
},
{
"input": "2 -2 -1",
"output": "YES"
},
{
"input": "2 -2 0",
"output": "NO"
},
{
"input": "2 -2 1",
"output": "NO"
},
{
"input": "2 -2 2",
"output": "NO"
},
{
"input": "2 -1 -2",
"output": "NO"
},
{
"input": "2 -1 -1",
"output": "YES"
},
{
"input": "2 -1 0",
"output": "NO"
},
{
"input": "2 -1 1",
"output": "NO"
},
{
"input": "2 -1 2",
"output": "NO"
},
{
"input": "2 0 -2",
"output": "YES"
},
{
"input": "2 0 -1",
"output": "YES"
},
{
"input": "2 0 0",
"output": "NO"
},
{
"input": "2 0 1",
"output": "NO"
},
{
"input": "2 0 2",
"output": "NO"
},
{
"input": "2 1 -2",
"output": "NO"
},
{
"input": "2 1 -1",
"output": "YES"
},
{
"input": "2 1 0",
"output": "NO"
},
{
"input": "2 1 1",
"output": "NO"
},
{
"input": "2 1 2",
"output": "NO"
},
{
"input": "2 2 -2",
"output": "YES"
},
{
"input": "2 2 -1",
"output": "YES"
},
{
"input": "2 2 0",
"output": "YES"
},
{
"input": "2 2 1",
"output": "YES"
},
{
"input": "2 2 2",
"output": "YES"
},
{
"input": "-1000000000 1000000000 1",
"output": "YES"
},
{
"input": "-1000000000 1000000000 2",
"output": "YES"
},
{
"input": "1000000000 -1000000000 -1",
"output": "YES"
},
{
"input": "5 2 3",
"output": "NO"
},
{
"input": "2 1 -1",
"output": "YES"
},
{
"input": "3 2 1",
"output": "NO"
},
{
"input": "0 -5 -3",
"output": "NO"
},
{
"input": "2 5 5",
"output": "NO"
},
{
"input": "0 10 1",
"output": "YES"
},
{
"input": "15 5 -5",
"output": "YES"
},
{
"input": "2 1 1",
"output": "NO"
},
{
"input": "20 10 0",
"output": "NO"
},
{
"input": "20 15 5",
"output": "NO"
},
{
"input": "1 6 1",
"output": "YES"
},
{
"input": "1000000000 0 -1000000000",
"output": "YES"
},
{
"input": "1 1 -5",
"output": "YES"
},
{
"input": "4 6 1",
"output": "YES"
},
{
"input": "-5 -10 -5",
"output": "YES"
},
{
"input": "2 0 0",
"output": "NO"
},
{
"input": "10 9 -1",
"output": "YES"
},
{
"input": "-2 -1 -1",
"output": "NO"
},
{
"input": "1 13 3",
"output": "YES"
},
{
"input": "2 3 0",
"output": "NO"
},
{
"input": "1 1 -1",
"output": "YES"
},
{
"input": "5 -10 -5",
"output": "YES"
},
{
"input": "5 3 1",
"output": "NO"
},
{
"input": "1 1000000000 1",
"output": "YES"
},
{
"input": "-1000000000 1000000000 1000000000",
"output": "YES"
}
] | 1,655,309,913
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
a, b, c = map(int, input().split())
if a==b and c==0:
print("yes")
elif ((b-a)%c==0):
print("yes")
else :
print("no")
|
Title: Infinite Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya likes everything infinite. Now he is studying the properties of a sequence *s*, such that its first element is equal to *a* (*s*1<==<=*a*), and the difference between any two neighbouring elements is equal to *c* (*s**i*<=-<=*s**i*<=-<=1<==<=*c*). In particular, Vasya wonders if his favourite integer *b* appears in this sequence, that is, there exists a positive integer *i*, such that *s**i*<==<=*b*. Of course, you are the person he asks for a help.
Input Specification:
The first line of the input contain three integers *a*, *b* and *c* (<=-<=109<=≤<=*a*,<=*b*,<=*c*<=≤<=109) — the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively.
Output Specification:
If *b* appears in the sequence *s* print "YES" (without quotes), otherwise print "NO" (without quotes).
Demo Input:
['1 7 3\n', '10 10 0\n', '1 -4 5\n', '0 60 50\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n', 'NO\n']
Note:
In the first sample, the sequence starts from integers 1, 4, 7, so 7 is its element.
In the second sample, the favorite integer of Vasya is equal to the first element of the sequence.
In the third sample all elements of the sequence are greater than Vasya's favorite integer.
In the fourth sample, the sequence starts from 0, 50, 100, and all the following elements are greater than Vasya's favorite integer.
|
```python
a, b, c = map(int, input().split())
if a==b and c==0:
print("yes")
elif ((b-a)%c==0):
print("yes")
else :
print("no")
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy the toy.
The toy consists of *n* parts and *m* ropes. Each rope links two parts, but every pair of parts is linked by at most one rope. To split the toy, the child must remove all its parts. The child can remove a single part at a time, and each remove consume an energy. Let's define an energy value of part *i* as *v**i*. The child spend *v**f*1<=+<=*v**f*2<=+<=...<=+<=*v**f**k* energy for removing part *i* where *f*1,<=*f*2,<=...,<=*f**k* are the parts that are directly connected to the *i*-th and haven't been removed.
Help the child to find out, what is the minimum total energy he should spend to remove all *n* parts.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=1000; 0<=≤<=*m*<=≤<=2000). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (0<=≤<=*v**i*<=≤<=105). Then followed *m* lines, each line contains two integers *x**i* and *y**i*, representing a rope from part *x**i* to part *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*; *x**i*<=≠<=*y**i*).
Consider all the parts are numbered from 1 to *n*.
|
Output the minimum total energy the child should spend to remove all *n* parts of the toy.
|
[
"4 3\n10 20 30 40\n1 4\n1 2\n2 3\n",
"4 4\n100 100 100 100\n1 2\n2 3\n2 4\n3 4\n",
"7 10\n40 10 20 10 20 80 40\n1 5\n4 7\n4 5\n5 2\n5 7\n6 4\n1 6\n1 3\n4 3\n1 4\n"
] |
[
"40\n",
"400\n",
"160\n"
] |
One of the optimal sequence of actions in the first sample is:
- First, remove part 3, cost of the action is 20. - Then, remove part 2, cost of the action is 10. - Next, remove part 4, cost of the action is 10. - At last, remove part 1, cost of the action is 0.
So the total energy the child paid is 20 + 10 + 10 + 0 = 40, which is the minimum.
In the second sample, the child will spend 400 no matter in what order he will remove the parts.
| 0
|
[
{
"input": "4 3\n10 20 30 40\n1 4\n1 2\n2 3",
"output": "40"
},
{
"input": "4 4\n100 100 100 100\n1 2\n2 3\n2 4\n3 4",
"output": "400"
},
{
"input": "7 10\n40 10 20 10 20 80 40\n1 5\n4 7\n4 5\n5 2\n5 7\n6 4\n1 6\n1 3\n4 3\n1 4",
"output": "160"
},
{
"input": "1 0\n23333",
"output": "0"
},
{
"input": "5 4\n1 2 2 2 2\n1 2\n1 3\n1 4\n1 5",
"output": "4"
},
{
"input": "10 30\n3 6 17 15 13 15 6 12 9 1\n3 8\n1 10\n4 7\n1 7\n3 7\n2 9\n8 10\n3 1\n3 4\n8 6\n10 3\n3 9\n2 3\n10 4\n2 10\n5 8\n9 5\n6 1\n2 1\n7 2\n7 6\n7 10\n4 8\n5 6\n3 6\n4 1\n8 9\n7 9\n4 2\n5 10",
"output": "188"
},
{
"input": "3 3\n1 1 1\n1 2\n2 3\n3 1",
"output": "3"
}
] | 1,689,365,108
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
print("_RANDOM_GUESS_1689365108.6634629")# 1689365108.6634822
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy the toy.
The toy consists of *n* parts and *m* ropes. Each rope links two parts, but every pair of parts is linked by at most one rope. To split the toy, the child must remove all its parts. The child can remove a single part at a time, and each remove consume an energy. Let's define an energy value of part *i* as *v**i*. The child spend *v**f*1<=+<=*v**f*2<=+<=...<=+<=*v**f**k* energy for removing part *i* where *f*1,<=*f*2,<=...,<=*f**k* are the parts that are directly connected to the *i*-th and haven't been removed.
Help the child to find out, what is the minimum total energy he should spend to remove all *n* parts.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=1000; 0<=≤<=*m*<=≤<=2000). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (0<=≤<=*v**i*<=≤<=105). Then followed *m* lines, each line contains two integers *x**i* and *y**i*, representing a rope from part *x**i* to part *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*; *x**i*<=≠<=*y**i*).
Consider all the parts are numbered from 1 to *n*.
Output Specification:
Output the minimum total energy the child should spend to remove all *n* parts of the toy.
Demo Input:
['4 3\n10 20 30 40\n1 4\n1 2\n2 3\n', '4 4\n100 100 100 100\n1 2\n2 3\n2 4\n3 4\n', '7 10\n40 10 20 10 20 80 40\n1 5\n4 7\n4 5\n5 2\n5 7\n6 4\n1 6\n1 3\n4 3\n1 4\n']
Demo Output:
['40\n', '400\n', '160\n']
Note:
One of the optimal sequence of actions in the first sample is:
- First, remove part 3, cost of the action is 20. - Then, remove part 2, cost of the action is 10. - Next, remove part 4, cost of the action is 10. - At last, remove part 1, cost of the action is 0.
So the total energy the child paid is 20 + 10 + 10 + 0 = 40, which is the minimum.
In the second sample, the child will spend 400 no matter in what order he will remove the parts.
|
```python
print("_RANDOM_GUESS_1689365108.6634629")# 1689365108.6634822
```
| 0
|
|
3
|
A
|
Shortest path of the king
|
PROGRAMMING
| 1,000
|
[
"greedy",
"shortest paths"
] |
A. Shortest path of the king
|
1
|
64
|
The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square *t*. As the king is not in habit of wasting his time, he wants to get from his current position *s* to square *t* in the least number of moves. Help him to do this.
In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).
|
The first line contains the chessboard coordinates of square *s*, the second line — of square *t*.
Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8.
|
In the first line print *n* — minimum number of the king's moves. Then in *n* lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.
L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.
|
[
"a8\nh1\n"
] |
[
"7\nRD\nRD\nRD\nRD\nRD\nRD\nRD\n"
] |
none
| 0
|
[
{
"input": "a8\nh1",
"output": "7\nRD\nRD\nRD\nRD\nRD\nRD\nRD"
},
{
"input": "b2\nb4",
"output": "2\nU\nU"
},
{
"input": "a5\na5",
"output": "0"
},
{
"input": "h1\nb2",
"output": "6\nLU\nL\nL\nL\nL\nL"
},
{
"input": "c5\nh2",
"output": "5\nRD\nRD\nRD\nR\nR"
},
{
"input": "e1\nf2",
"output": "1\nRU"
},
{
"input": "g4\nd2",
"output": "3\nLD\nLD\nL"
},
{
"input": "a8\nb2",
"output": "6\nRD\nD\nD\nD\nD\nD"
},
{
"input": "d4\nh2",
"output": "4\nRD\nRD\nR\nR"
},
{
"input": "c5\na2",
"output": "3\nLD\nLD\nD"
},
{
"input": "h5\nf8",
"output": "3\nLU\nLU\nU"
},
{
"input": "e6\nb6",
"output": "3\nL\nL\nL"
},
{
"input": "a6\ng4",
"output": "6\nRD\nRD\nR\nR\nR\nR"
},
{
"input": "f7\nc2",
"output": "5\nLD\nLD\nLD\nD\nD"
},
{
"input": "b7\nh8",
"output": "6\nRU\nR\nR\nR\nR\nR"
},
{
"input": "g7\nd6",
"output": "3\nLD\nL\nL"
},
{
"input": "c8\na3",
"output": "5\nLD\nLD\nD\nD\nD"
},
{
"input": "h8\nf1",
"output": "7\nLD\nLD\nD\nD\nD\nD\nD"
},
{
"input": "d1\nb7",
"output": "6\nLU\nLU\nU\nU\nU\nU"
},
{
"input": "a7\ne5",
"output": "4\nRD\nRD\nR\nR"
},
{
"input": "d6\nb1",
"output": "5\nLD\nLD\nD\nD\nD"
},
{
"input": "f5\ng5",
"output": "1\nR"
},
{
"input": "h4\nd1",
"output": "4\nLD\nLD\nLD\nL"
},
{
"input": "b3\na5",
"output": "2\nLU\nU"
},
{
"input": "d2\nf1",
"output": "2\nRD\nR"
},
{
"input": "f1\nc5",
"output": "4\nLU\nLU\nLU\nU"
},
{
"input": "a8\nh1",
"output": "7\nRD\nRD\nRD\nRD\nRD\nRD\nRD"
},
{
"input": "c7\ne5",
"output": "2\nRD\nRD"
},
{
"input": "e7\nb1",
"output": "6\nLD\nLD\nLD\nD\nD\nD"
},
{
"input": "g8\na8",
"output": "6\nL\nL\nL\nL\nL\nL"
},
{
"input": "g6\nf2",
"output": "4\nLD\nD\nD\nD"
},
{
"input": "g4\nc4",
"output": "4\nL\nL\nL\nL"
},
{
"input": "g2\na6",
"output": "6\nLU\nLU\nLU\nLU\nL\nL"
},
{
"input": "f8\nf8",
"output": "0"
},
{
"input": "f5\nd2",
"output": "3\nLD\nLD\nD"
}
] | 1,684,552,633
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 124
| 0
|
pos1 = list(input())
pos1[0] = ord(pos1[0])-97
pos1[1] = int(pos1[1])
pos2 = list(input())
pos2[0] = ord(pos2[0])-97
pos2[1] = int(pos2[1])
moves = []
steps=0
while True:
if pos1[0] < pos2[0] and pos1[1] < pos2[1]:
pos1[0]+=1
pos1[1]+=1
moves.append("RU")
elif pos1[0] > pos2[0] and pos1[1] < pos2[1]:
pos1[0]-=1
pos1[1]+=1
moves.append("LU")
elif pos1[0] > pos2[0] and pos1[1] > pos2[1]:
pos1[0]-=1
pos1[1]-=1
moves.append("LD")
elif pos1[0] < pos2[0] and pos1[1] > pos2[1]:
pos1[0]+=1
pos1[1]-=1
moves.append("RD")
elif pos1[0] < pos2[0]:
pos1[0]+=1
moves.append("R")
elif pos1[0] > pos2[0]:
pos1[0]-=1
moves.append("L")
elif pos1[1] < pos2[1]:
pos1[1]+=1
moves.append("U")
elif pos1[1] > pos2[1]:
pos1[1]-=1
moves.append("D")
else:
break
steps+=1
print(steps,*moves,sep='\n',end='')
|
Title: Shortest path of the king
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square *t*. As the king is not in habit of wasting his time, he wants to get from his current position *s* to square *t* in the least number of moves. Help him to do this.
In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).
Input Specification:
The first line contains the chessboard coordinates of square *s*, the second line — of square *t*.
Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8.
Output Specification:
In the first line print *n* — minimum number of the king's moves. Then in *n* lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.
L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.
Demo Input:
['a8\nh1\n']
Demo Output:
['7\nRD\nRD\nRD\nRD\nRD\nRD\nRD\n']
Note:
none
|
```python
pos1 = list(input())
pos1[0] = ord(pos1[0])-97
pos1[1] = int(pos1[1])
pos2 = list(input())
pos2[0] = ord(pos2[0])-97
pos2[1] = int(pos2[1])
moves = []
steps=0
while True:
if pos1[0] < pos2[0] and pos1[1] < pos2[1]:
pos1[0]+=1
pos1[1]+=1
moves.append("RU")
elif pos1[0] > pos2[0] and pos1[1] < pos2[1]:
pos1[0]-=1
pos1[1]+=1
moves.append("LU")
elif pos1[0] > pos2[0] and pos1[1] > pos2[1]:
pos1[0]-=1
pos1[1]-=1
moves.append("LD")
elif pos1[0] < pos2[0] and pos1[1] > pos2[1]:
pos1[0]+=1
pos1[1]-=1
moves.append("RD")
elif pos1[0] < pos2[0]:
pos1[0]+=1
moves.append("R")
elif pos1[0] > pos2[0]:
pos1[0]-=1
moves.append("L")
elif pos1[1] < pos2[1]:
pos1[1]+=1
moves.append("U")
elif pos1[1] > pos2[1]:
pos1[1]-=1
moves.append("D")
else:
break
steps+=1
print(steps,*moves,sep='\n',end='')
```
| 3.938
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,596,650,336
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 186
| 6,963,200
|
n=int(input())
x=[int(a) for a in input().split()]
y=[int(a) for a in input().split()]
z=[int(a) for a in input().split()]
count=0
for i in range(3):
if x[i]+y[i]+z[i]==0:
count+=1
if count==3:
print("YES")
else:
print("NO")
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n=int(input())
x=[int(a) for a in input().split()]
y=[int(a) for a in input().split()]
z=[int(a) for a in input().split()]
count=0
for i in range(3):
if x[i]+y[i]+z[i]==0:
count+=1
if count==3:
print("YES")
else:
print("NO")
```
| 0
|
353
|
A
|
Domino
|
PROGRAMMING
| 1,200
|
[
"implementation",
"math"
] | null | null |
Valera has got *n* domino pieces in a row. Each piece consists of two halves — the upper one and the lower one. Each of the halves contains a number from 1 to 6. Valera loves even integers very much, so he wants the sum of the numbers on the upper halves and the sum of the numbers on the lower halves to be even.
To do that, Valera can rotate the dominoes by 180 degrees. After the rotation the upper and the lower halves swap places. This action takes one second. Help Valera find out the minimum time he must spend rotating dominoes to make his wish come true.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100), denoting the number of dominoes Valera has. Next *n* lines contain two space-separated integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=6). Number *x**i* is initially written on the upper half of the *i*-th domino, *y**i* is initially written on the lower half.
|
Print a single number — the minimum required number of seconds. If Valera can't do the task in any time, print <=-<=1.
|
[
"2\n4 2\n6 4\n",
"1\n2 3\n",
"3\n1 4\n2 3\n4 4\n"
] |
[
"0\n",
"-1\n",
"1\n"
] |
In the first test case the sum of the numbers on the upper halves equals 10 and the sum of the numbers on the lower halves equals 6. Both numbers are even, so Valera doesn't required to do anything.
In the second sample Valera has only one piece of domino. It is written 3 on the one of its halves, therefore one of the sums will always be odd.
In the third case Valera can rotate the first piece, and after that the sum on the upper halves will be equal to 10, and the sum on the lower halves will be equal to 8.
| 500
|
[
{
"input": "2\n4 2\n6 4",
"output": "0"
},
{
"input": "1\n2 3",
"output": "-1"
},
{
"input": "3\n1 4\n2 3\n4 4",
"output": "1"
},
{
"input": "5\n5 4\n5 4\n1 5\n5 5\n3 3",
"output": "1"
},
{
"input": "20\n1 3\n5 2\n5 2\n2 6\n2 4\n1 1\n1 3\n1 4\n2 6\n4 2\n5 6\n2 2\n6 2\n4 3\n2 1\n6 2\n6 5\n4 5\n2 4\n1 4",
"output": "-1"
},
{
"input": "100\n2 3\n2 4\n3 3\n1 4\n5 2\n5 4\n6 6\n3 4\n1 1\n4 2\n5 1\n5 5\n5 3\n3 6\n4 1\n1 6\n1 1\n3 2\n4 5\n6 1\n6 4\n1 1\n3 4\n3 3\n2 2\n1 1\n4 4\n6 4\n3 2\n5 2\n6 4\n3 2\n3 5\n4 4\n1 4\n5 2\n3 4\n1 4\n2 2\n5 6\n3 5\n6 1\n5 5\n1 6\n6 3\n1 4\n1 5\n5 5\n4 1\n3 2\n4 1\n5 5\n5 5\n1 5\n1 2\n6 4\n1 3\n3 6\n4 3\n3 5\n6 4\n2 6\n5 5\n1 4\n2 2\n2 3\n5 1\n2 5\n1 2\n2 6\n5 5\n4 6\n1 4\n3 6\n2 3\n6 1\n6 5\n3 2\n6 4\n4 5\n4 5\n2 6\n1 3\n6 2\n1 2\n2 3\n4 3\n5 4\n3 4\n1 6\n6 6\n2 4\n4 1\n3 1\n2 6\n5 4\n1 2\n6 5\n3 6\n2 4",
"output": "-1"
},
{
"input": "1\n2 4",
"output": "0"
},
{
"input": "1\n1 1",
"output": "-1"
},
{
"input": "1\n1 2",
"output": "-1"
},
{
"input": "2\n1 1\n3 3",
"output": "0"
},
{
"input": "2\n1 1\n2 2",
"output": "-1"
},
{
"input": "2\n1 1\n1 2",
"output": "-1"
},
{
"input": "5\n1 2\n6 6\n1 1\n3 3\n6 1",
"output": "1"
},
{
"input": "5\n5 4\n2 6\n6 2\n1 4\n6 2",
"output": "0"
},
{
"input": "10\n4 1\n3 2\n1 2\n2 6\n3 5\n2 1\n5 2\n4 6\n5 6\n3 1",
"output": "0"
},
{
"input": "10\n6 1\n4 4\n2 6\n6 5\n3 6\n6 3\n2 4\n5 1\n1 6\n1 5",
"output": "-1"
},
{
"input": "15\n1 2\n5 1\n6 4\n5 1\n1 6\n2 6\n3 1\n6 4\n3 1\n2 1\n6 4\n3 5\n6 2\n1 6\n1 1",
"output": "1"
},
{
"input": "15\n3 3\n2 1\n5 4\n3 3\n5 3\n5 4\n2 5\n1 3\n3 2\n3 3\n3 5\n2 5\n4 1\n2 3\n5 4",
"output": "-1"
},
{
"input": "20\n1 5\n6 4\n4 3\n6 2\n1 1\n1 5\n6 3\n2 3\n3 6\n3 6\n3 6\n2 5\n4 3\n4 6\n5 5\n4 6\n3 4\n4 2\n3 3\n5 2",
"output": "0"
},
{
"input": "20\n2 1\n6 5\n3 1\n2 5\n3 5\n4 1\n1 1\n5 4\n5 1\n2 4\n1 5\n3 2\n1 2\n3 5\n5 2\n1 2\n1 3\n4 2\n2 3\n4 5",
"output": "-1"
},
{
"input": "25\n4 1\n6 3\n1 3\n2 3\n2 4\n6 6\n4 2\n4 2\n1 5\n5 4\n1 2\n2 5\n3 6\n4 1\n3 4\n2 6\n6 1\n5 6\n6 6\n4 2\n1 5\n3 3\n3 3\n6 5\n1 4",
"output": "-1"
},
{
"input": "25\n5 5\n4 3\n2 5\n4 3\n4 6\n4 2\n5 6\n2 1\n5 4\n6 6\n1 3\n1 4\n2 3\n5 6\n5 4\n5 6\n5 4\n6 3\n3 5\n1 3\n2 5\n2 2\n4 4\n2 1\n4 4",
"output": "-1"
},
{
"input": "30\n3 5\n2 5\n1 6\n1 6\n2 4\n5 5\n5 4\n5 6\n5 4\n2 1\n2 4\n1 6\n3 5\n1 1\n3 6\n5 5\n1 6\n3 4\n1 4\n4 6\n2 1\n3 3\n1 3\n4 5\n1 4\n1 6\n2 1\n4 6\n3 5\n5 6",
"output": "1"
},
{
"input": "30\n2 3\n3 1\n6 6\n1 3\n5 5\n3 6\n4 5\n2 1\n1 3\n2 3\n4 4\n2 4\n6 4\n2 4\n5 4\n2 1\n2 5\n2 5\n4 2\n1 4\n2 6\n3 2\n3 2\n6 6\n4 2\n3 4\n6 3\n6 6\n6 6\n5 5",
"output": "1"
},
{
"input": "35\n6 1\n4 3\n1 2\n4 3\n6 4\n4 6\n3 1\n5 5\n3 4\n5 4\n4 6\n1 6\n2 4\n6 6\n5 4\n5 2\n1 3\n1 4\n3 5\n1 4\n2 3\n4 5\n4 3\n6 1\n5 3\n3 2\n5 6\n3 5\n6 5\n4 1\n1 3\n5 5\n4 6\n6 1\n1 3",
"output": "1"
},
{
"input": "35\n4 3\n5 6\n4 5\n2 5\n6 6\n4 1\n2 2\n4 2\n3 4\n4 1\n6 6\n6 3\n1 5\n1 5\n5 6\n4 2\n4 6\n5 5\n2 2\n5 2\n1 2\n4 6\n6 6\n6 5\n2 1\n3 5\n2 5\n3 1\n5 3\n6 4\n4 6\n5 6\n5 1\n3 4\n3 5",
"output": "1"
},
{
"input": "40\n5 6\n1 1\n3 3\n2 6\n6 6\n5 4\n6 4\n3 5\n1 3\n4 4\n4 4\n2 5\n1 3\n3 6\n5 2\n4 3\n4 4\n5 6\n2 3\n1 1\n3 1\n1 1\n1 5\n4 3\n5 5\n3 4\n6 6\n5 6\n2 2\n6 6\n2 1\n2 4\n5 2\n2 2\n1 1\n1 4\n4 2\n3 5\n5 5\n4 5",
"output": "-1"
},
{
"input": "40\n3 2\n5 3\n4 6\n3 5\n6 1\n5 2\n1 2\n6 2\n5 3\n3 2\n4 4\n3 3\n5 2\n4 5\n1 4\n5 1\n3 3\n1 3\n1 3\n2 1\n3 6\n4 2\n4 6\n6 2\n2 5\n2 2\n2 5\n3 3\n5 3\n2 1\n3 2\n2 3\n6 3\n6 3\n3 4\n3 2\n4 3\n5 4\n2 4\n4 6",
"output": "-1"
},
{
"input": "45\n2 4\n3 4\n6 1\n5 5\n1 1\n3 5\n4 3\n5 2\n3 6\n6 1\n4 4\n6 1\n2 1\n6 1\n3 6\n3 3\n6 1\n1 2\n1 5\n6 5\n1 3\n5 6\n6 1\n4 5\n3 6\n2 2\n1 2\n4 5\n5 6\n1 5\n6 2\n2 4\n3 3\n3 1\n6 5\n6 5\n2 1\n5 2\n2 1\n3 3\n2 2\n1 4\n2 2\n3 3\n2 1",
"output": "-1"
},
{
"input": "45\n6 6\n1 6\n1 2\n3 5\n4 4\n2 1\n5 3\n2 1\n5 2\n5 3\n1 4\n5 2\n4 2\n3 6\n5 2\n1 5\n4 4\n5 5\n6 5\n2 1\n2 6\n5 5\n2 1\n6 1\n1 6\n6 5\n2 4\n4 3\n2 6\n2 4\n6 5\n6 4\n6 3\n6 6\n2 1\n6 4\n5 6\n5 4\n1 5\n5 1\n3 3\n5 6\n2 5\n4 5\n3 6",
"output": "-1"
},
{
"input": "50\n4 4\n5 1\n6 4\n6 2\n6 2\n1 4\n5 5\n4 2\n5 5\n5 4\n1 3\n3 5\n6 1\n6 1\n1 4\n4 3\n5 1\n3 6\n2 2\n6 2\n4 4\n2 3\n4 2\n6 5\n5 6\n2 2\n2 4\n3 5\n1 5\n3 2\n3 4\n5 6\n4 6\n1 6\n4 5\n2 6\n2 2\n3 5\n6 4\n5 1\n4 3\n3 4\n3 5\n3 3\n2 3\n3 2\n2 2\n1 4\n3 1\n4 4",
"output": "1"
},
{
"input": "50\n1 2\n1 4\n1 1\n4 5\n4 4\n3 2\n4 5\n3 5\n1 1\n3 4\n3 2\n2 4\n2 6\n2 6\n3 2\n4 6\n1 6\n3 1\n1 6\n2 1\n4 1\n1 6\n4 3\n6 6\n5 2\n6 4\n2 1\n4 3\n6 4\n5 1\n5 5\n3 1\n1 1\n5 5\n2 2\n2 3\n2 3\n3 5\n5 5\n1 6\n1 5\n3 6\n3 6\n1 1\n3 3\n2 6\n5 5\n1 3\n6 3\n6 6",
"output": "-1"
},
{
"input": "55\n3 2\n5 6\n5 1\n3 5\n5 5\n1 5\n5 4\n6 3\n5 6\n4 2\n3 1\n1 2\n5 5\n1 1\n5 2\n6 3\n5 4\n3 6\n4 6\n2 6\n6 4\n1 4\n1 6\n4 1\n2 5\n4 3\n2 1\n2 1\n6 2\n3 1\n2 5\n4 4\n6 3\n2 2\n3 5\n5 1\n3 6\n5 4\n4 6\n6 5\n5 6\n2 2\n3 2\n5 2\n6 5\n2 2\n5 3\n3 1\n4 5\n6 4\n2 4\n1 2\n5 6\n2 6\n5 2",
"output": "0"
},
{
"input": "55\n4 6\n3 3\n6 5\n5 3\n5 6\n2 3\n2 2\n3 4\n3 1\n5 4\n5 4\n2 4\n3 4\n4 5\n1 5\n6 3\n1 1\n5 1\n3 4\n1 5\n3 1\n2 5\n3 3\n4 3\n3 3\n3 1\n6 6\n3 3\n3 3\n5 6\n5 3\n3 5\n1 4\n5 5\n1 3\n1 4\n3 5\n3 6\n2 4\n2 4\n5 1\n6 4\n5 1\n5 5\n1 1\n3 2\n4 3\n5 4\n5 1\n2 4\n4 3\n6 1\n3 4\n1 5\n6 3",
"output": "-1"
},
{
"input": "60\n2 6\n1 4\n3 2\n1 2\n3 2\n2 4\n6 4\n4 6\n1 3\n3 1\n6 5\n2 4\n5 4\n4 2\n1 6\n3 4\n4 5\n5 2\n1 5\n5 4\n3 4\n3 4\n4 4\n4 1\n6 6\n3 6\n2 4\n2 1\n4 4\n6 5\n3 1\n4 3\n1 3\n6 3\n5 5\n1 4\n3 1\n3 6\n1 5\n3 1\n1 5\n4 4\n1 3\n2 4\n6 2\n4 1\n5 3\n3 4\n5 6\n1 2\n1 6\n6 3\n1 6\n3 6\n3 4\n6 2\n4 6\n2 3\n3 3\n3 3",
"output": "-1"
},
{
"input": "60\n2 3\n4 6\n2 4\n1 3\n5 6\n1 5\n1 2\n1 3\n5 6\n4 3\n4 2\n3 1\n1 3\n3 5\n1 5\n3 4\n2 4\n3 5\n4 5\n1 2\n3 1\n1 5\n2 5\n6 2\n1 6\n3 3\n6 2\n5 3\n1 3\n1 4\n6 4\n6 3\n4 2\n4 2\n1 4\n1 3\n3 2\n3 1\n2 1\n1 2\n3 1\n2 6\n1 4\n3 6\n3 3\n1 5\n2 4\n5 5\n6 2\n5 2\n3 3\n5 3\n3 4\n4 5\n5 6\n2 4\n5 3\n3 1\n2 4\n5 4",
"output": "-1"
},
{
"input": "65\n5 4\n3 3\n1 2\n4 3\n3 5\n1 5\n4 5\n2 6\n1 2\n1 5\n6 3\n2 6\n4 3\n3 6\n1 5\n3 5\n4 6\n2 5\n6 5\n1 4\n3 4\n4 3\n1 4\n2 5\n6 5\n3 1\n4 3\n1 2\n1 1\n6 1\n5 2\n3 2\n1 6\n2 6\n3 3\n6 6\n4 6\n1 5\n5 1\n4 5\n1 4\n3 2\n5 4\n4 2\n6 2\n1 3\n4 2\n5 3\n6 4\n3 6\n1 2\n6 1\n6 6\n3 3\n4 2\n3 5\n4 6\n4 1\n5 4\n6 1\n5 1\n5 6\n6 1\n4 6\n5 5",
"output": "1"
},
{
"input": "65\n5 4\n6 3\n5 4\n4 5\n5 3\n3 6\n1 3\n3 1\n1 3\n6 1\n6 4\n1 3\n2 2\n4 6\n4 1\n5 6\n6 5\n1 1\n1 3\n6 6\n4 1\n2 4\n5 4\n4 1\n5 5\n5 3\n6 2\n2 6\n4 2\n2 2\n6 2\n3 3\n4 5\n4 3\n3 1\n1 4\n4 5\n3 2\n5 5\n4 6\n5 1\n3 4\n5 4\n5 2\n1 6\n4 2\n3 4\n3 4\n1 3\n1 2\n3 3\n3 6\n6 4\n4 6\n6 2\n6 5\n3 2\n2 1\n6 4\n2 1\n1 5\n5 2\n6 5\n3 6\n5 1",
"output": "1"
},
{
"input": "70\n4 1\n2 6\n1 1\n5 6\n5 1\n2 3\n3 5\n1 1\n1 1\n4 6\n4 3\n1 5\n2 2\n2 3\n3 1\n6 4\n3 1\n4 2\n5 4\n1 3\n3 5\n5 2\n5 6\n4 4\n4 5\n2 2\n4 5\n3 2\n3 5\n2 5\n2 6\n5 5\n2 6\n5 1\n1 1\n2 5\n3 1\n1 2\n6 4\n6 5\n5 5\n5 1\n1 5\n2 2\n6 3\n4 3\n6 2\n5 5\n1 1\n6 2\n6 6\n3 4\n2 2\n3 5\n1 5\n2 5\n4 5\n2 4\n6 3\n5 1\n2 6\n4 2\n1 4\n1 6\n6 2\n5 2\n5 6\n2 5\n5 6\n5 5",
"output": "-1"
},
{
"input": "70\n4 3\n6 4\n5 5\n3 1\n1 2\n2 5\n4 6\n4 2\n3 2\n4 2\n1 5\n2 2\n4 3\n1 2\n6 1\n6 6\n1 6\n5 1\n2 2\n6 3\n4 2\n4 3\n1 2\n6 6\n3 3\n6 5\n6 2\n3 6\n6 6\n4 6\n5 2\n5 4\n3 3\n1 6\n5 6\n2 3\n4 6\n1 1\n1 2\n6 6\n1 1\n3 4\n1 6\n2 6\n3 4\n6 3\n5 3\n1 2\n2 3\n4 6\n2 1\n6 4\n4 6\n4 6\n4 2\n5 5\n3 5\n3 2\n4 3\n3 6\n1 4\n3 6\n1 4\n1 6\n1 5\n5 6\n4 4\n3 3\n3 5\n2 2",
"output": "0"
},
{
"input": "75\n1 3\n4 5\n4 1\n6 5\n2 1\n1 4\n5 4\n1 5\n5 3\n1 2\n4 1\n1 1\n5 1\n5 3\n1 5\n4 2\n2 2\n6 3\n1 2\n4 3\n2 5\n5 3\n5 5\n4 1\n4 6\n2 5\n6 1\n2 4\n6 4\n5 2\n6 2\n2 4\n1 3\n5 4\n6 5\n5 4\n6 4\n1 5\n4 6\n1 5\n1 1\n4 4\n3 5\n6 3\n6 5\n1 5\n2 1\n1 5\n6 6\n2 2\n2 2\n4 4\n6 6\n5 4\n4 5\n3 2\n2 4\n1 1\n4 3\n3 2\n5 4\n1 6\n1 2\n2 2\n3 5\n2 6\n1 1\n2 2\n2 3\n6 2\n3 6\n4 4\n5 1\n4 1\n4 1",
"output": "0"
},
{
"input": "75\n1 1\n2 1\n5 5\n6 5\n6 3\n1 6\n6 1\n4 4\n2 1\n6 2\n3 1\n6 4\n1 6\n2 2\n4 3\n4 2\n1 2\n6 2\n4 2\n5 1\n1 2\n3 2\n6 6\n6 3\n2 4\n4 1\n4 1\n2 4\n5 5\n2 3\n5 5\n4 5\n3 1\n1 5\n4 3\n2 3\n3 5\n4 6\n5 6\n1 6\n2 3\n2 2\n1 2\n5 6\n1 4\n1 5\n1 3\n6 2\n1 2\n4 2\n2 1\n1 3\n6 4\n4 1\n5 2\n6 2\n3 5\n2 3\n4 2\n5 1\n5 6\n3 2\n2 1\n6 6\n2 1\n6 2\n1 1\n3 2\n1 2\n3 5\n4 6\n1 3\n3 4\n5 5\n6 2",
"output": "1"
},
{
"input": "80\n3 1\n6 3\n2 2\n2 2\n6 3\n6 1\n6 5\n1 4\n3 6\n6 5\n1 3\n2 4\n1 4\n3 1\n5 3\n5 3\n1 4\n2 5\n4 3\n4 4\n4 5\n6 1\n3 1\n2 6\n4 2\n3 1\n6 5\n2 6\n2 2\n5 1\n1 3\n5 1\n2 1\n4 3\n6 3\n3 5\n4 3\n5 6\n3 3\n4 1\n5 1\n6 5\n5 1\n2 5\n6 1\n3 2\n4 3\n3 3\n5 6\n1 6\n5 2\n1 5\n5 6\n6 4\n2 2\n4 2\n4 6\n4 2\n4 4\n6 5\n5 2\n6 2\n4 6\n6 4\n4 3\n5 1\n4 1\n3 5\n3 2\n3 2\n5 3\n5 4\n3 4\n1 3\n1 2\n6 6\n6 3\n6 1\n5 6\n3 2",
"output": "0"
},
{
"input": "80\n4 5\n3 3\n3 6\n4 5\n3 4\n6 5\n1 5\n2 5\n5 6\n5 1\n5 1\n1 2\n5 5\n5 1\n2 3\n1 1\n4 5\n4 1\n1 1\n5 5\n5 6\n5 2\n5 4\n4 2\n6 2\n5 3\n3 2\n4 2\n1 3\n1 6\n2 1\n6 6\n4 5\n6 4\n2 2\n1 6\n6 2\n4 3\n2 3\n4 6\n4 6\n6 2\n3 4\n4 3\n5 5\n1 6\n3 2\n4 6\n2 3\n1 6\n5 4\n4 2\n5 4\n1 1\n4 3\n5 1\n3 6\n6 2\n3 1\n4 1\n5 3\n2 2\n3 4\n3 6\n3 5\n5 5\n5 1\n3 5\n2 6\n6 3\n6 5\n3 3\n5 6\n1 2\n3 1\n6 3\n3 4\n6 6\n6 6\n1 2",
"output": "-1"
},
{
"input": "85\n6 3\n4 1\n1 2\n3 5\n6 4\n6 2\n2 6\n1 2\n1 5\n6 2\n1 4\n6 6\n2 4\n4 6\n4 5\n1 6\n3 1\n2 5\n5 1\n5 2\n3 5\n1 1\n4 1\n2 3\n1 1\n3 3\n6 4\n1 4\n1 1\n3 6\n1 5\n1 6\n2 5\n2 2\n5 1\n6 6\n1 3\n1 5\n5 6\n4 5\n4 3\n5 5\n1 3\n6 3\n4 6\n2 4\n5 6\n6 2\n4 5\n1 4\n1 4\n6 5\n1 6\n6 1\n1 6\n5 5\n2 1\n5 2\n2 3\n1 6\n1 6\n1 6\n5 6\n2 4\n6 5\n6 5\n4 2\n5 4\n3 4\n4 3\n6 6\n3 3\n3 2\n3 6\n2 5\n2 1\n2 5\n3 4\n1 2\n5 4\n6 2\n5 1\n1 4\n3 4\n4 5",
"output": "0"
},
{
"input": "85\n3 1\n3 2\n6 3\n1 3\n2 1\n3 6\n1 4\n2 5\n6 5\n1 6\n1 5\n1 1\n4 3\n3 5\n4 6\n3 2\n6 6\n4 4\n4 1\n5 5\n4 2\n6 2\n2 2\n4 5\n6 1\n3 4\n4 5\n3 5\n4 2\n3 5\n4 4\n3 1\n4 4\n6 4\n1 4\n5 5\n1 5\n2 2\n6 5\n5 6\n6 5\n3 2\n3 2\n6 1\n6 5\n2 1\n4 6\n2 1\n3 1\n5 6\n1 3\n5 4\n1 4\n1 4\n5 3\n2 3\n1 3\n2 2\n5 3\n2 3\n2 3\n1 3\n3 6\n4 4\n6 6\n6 2\n5 1\n5 5\n5 5\n1 2\n1 4\n2 4\n3 6\n4 6\n6 3\n6 4\n5 5\n3 2\n5 4\n5 4\n4 5\n6 4\n2 1\n5 2\n5 1",
"output": "-1"
},
{
"input": "90\n5 2\n5 5\n5 1\n4 6\n4 3\n5 3\n5 6\n5 1\n3 4\n1 3\n4 2\n1 6\n6 4\n1 2\n6 1\n4 1\n6 2\n6 5\n6 2\n5 4\n3 6\n1 1\n5 5\n2 2\n1 6\n3 5\n6 5\n1 6\n1 5\n2 3\n2 6\n2 3\n3 3\n1 3\n5 1\n2 5\n3 6\n1 2\n4 4\n1 6\n2 3\n1 5\n2 5\n1 3\n2 2\n4 6\n3 6\n6 3\n1 2\n4 3\n4 5\n4 6\n3 2\n6 5\n6 2\n2 5\n2 4\n1 3\n1 6\n4 3\n1 3\n6 4\n4 6\n4 1\n1 1\n4 1\n4 4\n6 2\n6 5\n1 1\n2 2\n3 1\n1 4\n6 2\n5 2\n1 4\n1 3\n6 5\n3 2\n6 4\n3 4\n2 6\n2 2\n6 3\n4 6\n1 2\n4 2\n3 4\n2 3\n1 5",
"output": "-1"
},
{
"input": "90\n1 4\n3 5\n4 2\n2 5\n4 3\n2 6\n2 6\n3 2\n4 4\n6 1\n4 3\n2 3\n5 3\n6 6\n2 2\n6 3\n4 1\n4 4\n5 6\n6 4\n4 2\n5 6\n4 6\n4 4\n6 4\n4 1\n5 3\n3 2\n4 4\n5 2\n5 4\n6 4\n1 2\n3 3\n3 4\n6 4\n1 6\n4 2\n3 2\n1 1\n2 2\n5 1\n6 6\n4 1\n5 2\n3 6\n2 1\n2 2\n4 6\n6 5\n4 4\n5 5\n5 6\n1 6\n1 4\n5 6\n3 6\n6 3\n5 6\n6 5\n5 1\n6 1\n6 6\n6 3\n1 5\n4 5\n3 1\n6 6\n3 4\n6 2\n1 4\n2 2\n3 2\n5 6\n2 4\n1 4\n6 3\n4 6\n1 4\n5 2\n1 2\n6 5\n1 5\n1 4\n4 2\n2 5\n3 2\n5 1\n5 4\n5 3",
"output": "-1"
},
{
"input": "95\n4 3\n3 2\n5 5\n5 3\n1 6\n4 4\n5 5\n6 5\n3 5\n1 5\n4 2\n5 1\n1 2\n2 3\n6 4\n2 3\n6 3\n6 5\n5 6\n1 4\n2 6\n2 6\n2 5\n2 1\n3 1\n3 5\n2 2\n6 1\n2 4\n4 6\n6 6\n6 4\n3 2\n5 1\n4 3\n6 5\n2 3\n4 1\n2 5\n6 5\n6 5\n6 5\n5 1\n5 4\n4 6\n3 2\n2 5\n2 6\n4 6\n6 3\n6 4\n5 6\n4 6\n2 4\n3 4\n1 4\n2 4\n2 3\n5 6\n6 4\n3 1\n5 1\n3 6\n3 5\n2 6\n6 3\n4 3\n3 1\n6 1\n2 2\n6 3\n2 2\n2 2\n6 4\n6 1\n2 1\n5 6\n5 4\n5 2\n3 4\n3 6\n2 1\n1 6\n5 5\n2 6\n2 3\n3 6\n1 3\n1 5\n5 1\n1 2\n2 2\n5 3\n6 4\n4 5",
"output": "0"
},
{
"input": "95\n4 5\n5 6\n3 2\n5 1\n4 3\n4 1\n6 1\n5 2\n2 4\n5 3\n2 3\n6 4\n4 1\n1 6\n2 6\n2 3\n4 6\n2 4\n3 4\n4 2\n5 5\n1 1\n1 5\n4 3\n4 5\n6 2\n6 1\n6 3\n5 5\n4 1\n5 1\n2 3\n5 1\n3 6\n6 6\n4 5\n4 4\n4 3\n1 6\n6 6\n4 6\n6 4\n1 2\n6 2\n4 6\n6 6\n5 5\n6 1\n5 2\n4 5\n6 6\n6 5\n4 4\n1 5\n4 6\n4 1\n3 6\n5 1\n3 1\n4 6\n4 5\n1 3\n5 4\n4 5\n2 2\n6 1\n5 2\n6 5\n2 2\n1 1\n6 3\n6 1\n2 6\n3 3\n2 1\n4 6\n2 4\n5 5\n5 2\n3 2\n1 2\n6 6\n6 2\n5 1\n2 6\n5 2\n2 2\n5 5\n3 5\n3 3\n2 6\n5 3\n4 3\n1 6\n5 4",
"output": "-1"
},
{
"input": "100\n1 1\n3 5\n2 1\n1 2\n3 4\n5 6\n5 6\n6 1\n5 5\n2 4\n5 5\n5 6\n6 2\n6 6\n2 6\n1 4\n2 2\n3 2\n1 3\n5 5\n6 3\n5 6\n1 1\n1 2\n1 2\n2 1\n2 3\n1 6\n4 3\n1 1\n2 5\n2 4\n4 4\n1 5\n3 3\n6 1\n3 5\n1 1\n3 6\n3 1\n4 2\n4 3\n3 6\n6 6\n1 6\n6 2\n2 5\n5 4\n6 3\n1 4\n2 6\n6 2\n3 4\n6 1\n6 5\n4 6\n6 5\n4 4\n3 1\n6 3\n5 1\n2 4\n5 1\n1 2\n2 4\n2 1\n6 6\n5 3\n4 6\n6 3\n5 5\n3 3\n1 1\n6 5\n4 3\n2 6\n1 5\n3 5\n2 4\n4 5\n1 6\n2 3\n6 3\n5 5\n2 6\n2 6\n3 4\n3 2\n6 1\n3 4\n6 4\n3 3\n2 3\n5 1\n3 1\n6 2\n2 3\n6 4\n1 4\n1 2",
"output": "-1"
},
{
"input": "100\n1 1\n5 5\n1 2\n5 3\n5 5\n2 2\n1 5\n3 4\n3 2\n1 3\n5 6\n4 5\n2 1\n5 5\n2 2\n1 6\n6 1\n5 1\n4 1\n4 6\n3 5\n6 1\n2 3\n5 6\n3 6\n2 3\n5 6\n1 6\n3 2\n2 2\n3 3\n6 5\n5 5\n1 4\n5 6\n6 4\n1 4\n1 2\n2 6\n3 2\n6 4\n5 3\n3 3\n6 4\n4 6\n2 2\n5 6\n5 1\n1 2\n3 4\n4 5\n1 1\n3 4\n5 2\n4 5\n3 3\n1 1\n3 4\n1 6\n2 4\n1 3\n3 2\n6 5\n1 6\n3 6\n2 3\n2 6\n5 1\n5 5\n5 6\n4 1\n6 2\n3 6\n5 3\n2 2\n2 4\n6 6\n3 6\n4 6\n2 5\n5 3\n1 2\n3 4\n3 4\n6 2\n2 4\n2 2\n4 6\n3 5\n4 2\n5 6\n4 2\n2 3\n6 2\n5 6\n2 1\n3 3\n6 6\n4 3\n4 2",
"output": "1"
},
{
"input": "1\n2 2",
"output": "0"
},
{
"input": "3\n2 4\n6 6\n3 3",
"output": "-1"
},
{
"input": "2\n3 6\n4 1",
"output": "1"
},
{
"input": "3\n1 1\n1 1\n3 3",
"output": "-1"
},
{
"input": "3\n2 3\n1 1\n2 3",
"output": "1"
},
{
"input": "3\n2 2\n2 1\n1 2",
"output": "1"
},
{
"input": "3\n1 1\n1 1\n1 1",
"output": "-1"
}
] | 1,594,845,737
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 59
| 310
| 20,172,800
|
#Codeforces353A
upper = 0
lower = 0
double = 0
opposite = 0
for i in range(int(input())):
bol = 0
dom = input().split()
if int(dom[0]) % 2 == 1:
upper += 1
bol += 1
if int(dom[1]) % 2 == 1:
lower += 1
bol += 1
if bol == 2:
double += 1
elif bol == 1:
opposite += 1
if upper % 2 == 0 and lower % 2 == 0:
print(0)
elif upper % 2 == 1 and lower % 2 == 1 and (double % 2 == 0 or opposite > 1):
print(1)
else:
print(-1)
|
Title: Domino
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera has got *n* domino pieces in a row. Each piece consists of two halves — the upper one and the lower one. Each of the halves contains a number from 1 to 6. Valera loves even integers very much, so he wants the sum of the numbers on the upper halves and the sum of the numbers on the lower halves to be even.
To do that, Valera can rotate the dominoes by 180 degrees. After the rotation the upper and the lower halves swap places. This action takes one second. Help Valera find out the minimum time he must spend rotating dominoes to make his wish come true.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100), denoting the number of dominoes Valera has. Next *n* lines contain two space-separated integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=6). Number *x**i* is initially written on the upper half of the *i*-th domino, *y**i* is initially written on the lower half.
Output Specification:
Print a single number — the minimum required number of seconds. If Valera can't do the task in any time, print <=-<=1.
Demo Input:
['2\n4 2\n6 4\n', '1\n2 3\n', '3\n1 4\n2 3\n4 4\n']
Demo Output:
['0\n', '-1\n', '1\n']
Note:
In the first test case the sum of the numbers on the upper halves equals 10 and the sum of the numbers on the lower halves equals 6. Both numbers are even, so Valera doesn't required to do anything.
In the second sample Valera has only one piece of domino. It is written 3 on the one of its halves, therefore one of the sums will always be odd.
In the third case Valera can rotate the first piece, and after that the sum on the upper halves will be equal to 10, and the sum on the lower halves will be equal to 8.
|
```python
#Codeforces353A
upper = 0
lower = 0
double = 0
opposite = 0
for i in range(int(input())):
bol = 0
dom = input().split()
if int(dom[0]) % 2 == 1:
upper += 1
bol += 1
if int(dom[1]) % 2 == 1:
lower += 1
bol += 1
if bol == 2:
double += 1
elif bol == 1:
opposite += 1
if upper % 2 == 0 and lower % 2 == 0:
print(0)
elif upper % 2 == 1 and lower % 2 == 1 and (double % 2 == 0 or opposite > 1):
print(1)
else:
print(-1)
```
| 3
|
|
1,005
|
B
|
Delete from the Left
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
You are given two strings $s$ and $t$. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by $1$. You can't choose a string if it is empty.
For example:
- by applying a move to the string "where", the result is the string "here", - by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings $s$ and $t$ equal.
|
The first line of the input contains $s$. In the second line of the input contains $t$. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and $2\cdot10^5$, inclusive.
|
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
|
[
"test\nwest\n",
"codeforces\nyes\n",
"test\nyes\n",
"b\nab\n"
] |
[
"2\n",
"9\n",
"7\n",
"1\n"
] |
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" $8$ times. As a result, the string becomes "codeforces" $\to$ "es". The move should be applied to the string "yes" once. The result is the same string "yes" $\to$ "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
| 0
|
[
{
"input": "test\nwest",
"output": "2"
},
{
"input": "codeforces\nyes",
"output": "9"
},
{
"input": "test\nyes",
"output": "7"
},
{
"input": "b\nab",
"output": "1"
},
{
"input": "z\nz",
"output": "0"
},
{
"input": "abacabadabacaba\nabacabadacaba",
"output": "18"
},
{
"input": "z\na",
"output": "2"
},
{
"input": "ama\nsama",
"output": "1"
},
{
"input": "aaabc\nbc",
"output": "3"
},
{
"input": "lxuyspmieeuyxluv\ndfwxqvavzxctblaa",
"output": "32"
},
{
"input": "ahmedaly\naly",
"output": "5"
},
{
"input": "aa\na",
"output": "1"
},
{
"input": "aa\nba",
"output": "2"
},
{
"input": "cccba\nba",
"output": "3"
},
{
"input": "a\nab",
"output": "3"
},
{
"input": "dadc\ndddc",
"output": "4"
},
{
"input": "aacaaab\nb",
"output": "6"
},
{
"input": "dddc\ncdcde",
"output": "9"
},
{
"input": "bbdab\ndab",
"output": "2"
},
{
"input": "a\naa",
"output": "1"
},
{
"input": "aa\nab",
"output": "4"
},
{
"input": "codeforces\nces",
"output": "7"
}
] | 1,675,228,585
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 124
| 921,600
|
s = input()
t = input()
s_size, t_size = len(s) - 1 , len(t) - 1
cnt = 0
while s_size >= 0 and t_size >= 0:
if s[s_size] == t [t_size]: cnt += 2
else: break
s_size -= 1
t_size -= 1
print(len(s) + len(t) - cnt)
|
Title: Delete from the Left
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two strings $s$ and $t$. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by $1$. You can't choose a string if it is empty.
For example:
- by applying a move to the string "where", the result is the string "here", - by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings $s$ and $t$ equal.
Input Specification:
The first line of the input contains $s$. In the second line of the input contains $t$. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and $2\cdot10^5$, inclusive.
Output Specification:
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Demo Input:
['test\nwest\n', 'codeforces\nyes\n', 'test\nyes\n', 'b\nab\n']
Demo Output:
['2\n', '9\n', '7\n', '1\n']
Note:
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" $8$ times. As a result, the string becomes "codeforces" $\to$ "es". The move should be applied to the string "yes" once. The result is the same string "yes" $\to$ "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
|
```python
s = input()
t = input()
s_size, t_size = len(s) - 1 , len(t) - 1
cnt = 0
while s_size >= 0 and t_size >= 0:
if s[s_size] == t [t_size]: cnt += 2
else: break
s_size -= 1
t_size -= 1
print(len(s) + len(t) - cnt)
```
| 3
|
|
569
|
A
|
Music
|
PROGRAMMING
| 1,500
|
[
"implementation",
"math"
] | null | null |
Little Lesha loves listening to music via his smartphone. But the smartphone doesn't have much memory, so Lesha listens to his favorite songs in a well-known social network InTalk.
Unfortunately, internet is not that fast in the city of Ekaterinozavodsk and the song takes a lot of time to download. But Lesha is quite impatient. The song's duration is *T* seconds. Lesha downloads the first *S* seconds of the song and plays it. When the playback reaches the point that has not yet been downloaded, Lesha immediately plays the song from the start (the loaded part of the song stays in his phone, and the download is continued from the same place), and it happens until the song is downloaded completely and Lesha listens to it to the end. For *q* seconds of real time the Internet allows you to download *q*<=-<=1 seconds of the track.
Tell Lesha, for how many times he will start the song, including the very first start.
|
The single line contains three integers *T*,<=*S*,<=*q* (2<=≤<=*q*<=≤<=104, 1<=≤<=*S*<=<<=*T*<=≤<=105).
|
Print a single integer — the number of times the song will be restarted.
|
[
"5 2 2\n",
"5 4 7\n",
"6 2 3\n"
] |
[
"2\n",
"1\n",
"1\n"
] |
In the first test, the song is played twice faster than it is downloaded, which means that during four first seconds Lesha reaches the moment that has not been downloaded, and starts the song again. After another two seconds, the song is downloaded completely, and thus, Lesha starts the song twice.
In the second test, the song is almost downloaded, and Lesha will start it only once.
In the third sample test the download finishes and Lesha finishes listening at the same moment. Note that song isn't restarted in this case.
| 500
|
[
{
"input": "5 2 2",
"output": "2"
},
{
"input": "5 4 7",
"output": "1"
},
{
"input": "6 2 3",
"output": "1"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "2 1 3",
"output": "1"
},
{
"input": "2 1 10000",
"output": "1"
},
{
"input": "12326 6163 2",
"output": "1"
},
{
"input": "10000 2500 4",
"output": "1"
},
{
"input": "100000 99999 4",
"output": "1"
},
{
"input": "12351 1223 6",
"output": "2"
},
{
"input": "100000 1 10000",
"output": "2"
},
{
"input": "10028 13 10000",
"output": "1"
},
{
"input": "100000 99999 2",
"output": "1"
},
{
"input": "100000 99999 3",
"output": "1"
},
{
"input": "100000 1 2",
"output": "17"
},
{
"input": "100000 1 3",
"output": "11"
},
{
"input": "100000 1 4",
"output": "9"
},
{
"input": "100000 1 5",
"output": "8"
},
{
"input": "100000 3125 2",
"output": "5"
},
{
"input": "12628 1804 7",
"output": "1"
},
{
"input": "100000 45 13",
"output": "4"
},
{
"input": "100000 500 3",
"output": "5"
},
{
"input": "356 2 3",
"output": "5"
},
{
"input": "50 2 2",
"output": "5"
},
{
"input": "65465 12 3",
"output": "8"
},
{
"input": "10033 3 8",
"output": "4"
},
{
"input": "100000 3 2",
"output": "16"
},
{
"input": "64 1 8",
"output": "2"
},
{
"input": "10000 9 2",
"output": "11"
},
{
"input": "25 2 2",
"output": "4"
},
{
"input": "129 2 2",
"output": "7"
},
{
"input": "6562 1 3",
"output": "9"
},
{
"input": "100000 1 10",
"output": "5"
}
] | 1,606,934,077
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 93
| 204,800
|
import sys
a=int(input())
d=dict()
arr=list(map(int,input().split()))
for i in arr:
d[i]=1
s=set()
for i in range(1,a+1):
if(i not in d):
s.add(i)
d=dict()
for i in range(0,len(arr)):
if(arr[i] in d):
arr[i]=s.pop()
else:
d[arr[i]]=1
print(" ".join(map(str,arr)))
|
Title: Music
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Lesha loves listening to music via his smartphone. But the smartphone doesn't have much memory, so Lesha listens to his favorite songs in a well-known social network InTalk.
Unfortunately, internet is not that fast in the city of Ekaterinozavodsk and the song takes a lot of time to download. But Lesha is quite impatient. The song's duration is *T* seconds. Lesha downloads the first *S* seconds of the song and plays it. When the playback reaches the point that has not yet been downloaded, Lesha immediately plays the song from the start (the loaded part of the song stays in his phone, and the download is continued from the same place), and it happens until the song is downloaded completely and Lesha listens to it to the end. For *q* seconds of real time the Internet allows you to download *q*<=-<=1 seconds of the track.
Tell Lesha, for how many times he will start the song, including the very first start.
Input Specification:
The single line contains three integers *T*,<=*S*,<=*q* (2<=≤<=*q*<=≤<=104, 1<=≤<=*S*<=<<=*T*<=≤<=105).
Output Specification:
Print a single integer — the number of times the song will be restarted.
Demo Input:
['5 2 2\n', '5 4 7\n', '6 2 3\n']
Demo Output:
['2\n', '1\n', '1\n']
Note:
In the first test, the song is played twice faster than it is downloaded, which means that during four first seconds Lesha reaches the moment that has not been downloaded, and starts the song again. After another two seconds, the song is downloaded completely, and thus, Lesha starts the song twice.
In the second test, the song is almost downloaded, and Lesha will start it only once.
In the third sample test the download finishes and Lesha finishes listening at the same moment. Note that song isn't restarted in this case.
|
```python
import sys
a=int(input())
d=dict()
arr=list(map(int,input().split()))
for i in arr:
d[i]=1
s=set()
for i in range(1,a+1):
if(i not in d):
s.add(i)
d=dict()
for i in range(0,len(arr)):
if(arr[i] in d):
arr[i]=s.pop()
else:
d[arr[i]]=1
print(" ".join(map(str,arr)))
```
| -1
|
|
313
|
A
|
Ilya and Bank Account
|
PROGRAMMING
| 900
|
[
"implementation",
"number theory"
] | null | null |
Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.
Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.
Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.
|
The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account.
|
In a single line print an integer — the maximum state of the bank account that Ilya can get.
|
[
"2230\n",
"-10\n",
"-100003\n"
] |
[
"2230\n",
"0\n",
"-10000\n"
] |
In the first test sample Ilya doesn't profit from using the present.
In the second test sample you can delete digit 1 and get the state of the account equal to 0.
| 500
|
[
{
"input": "2230",
"output": "2230"
},
{
"input": "-10",
"output": "0"
},
{
"input": "-100003",
"output": "-10000"
},
{
"input": "544883178",
"output": "544883178"
},
{
"input": "-847251738",
"output": "-84725173"
},
{
"input": "423654797",
"output": "423654797"
},
{
"input": "-623563697",
"output": "-62356367"
},
{
"input": "645894116",
"output": "645894116"
},
{
"input": "-384381709",
"output": "-38438170"
},
{
"input": "437587210",
"output": "437587210"
},
{
"input": "-297534606",
"output": "-29753460"
},
{
"input": "891773002",
"output": "891773002"
},
{
"input": "-56712976",
"output": "-5671296"
},
{
"input": "963662765",
"output": "963662765"
},
{
"input": "-272656295",
"output": "-27265625"
},
{
"input": "383441522",
"output": "383441522"
},
{
"input": "-477665112",
"output": "-47766511"
},
{
"input": "791725034",
"output": "791725034"
},
{
"input": "-812168727",
"output": "-81216872"
},
{
"input": "528894922",
"output": "528894922"
},
{
"input": "-479977172",
"output": "-47997712"
},
{
"input": "568044564",
"output": "568044564"
},
{
"input": "-392784794",
"output": "-39278474"
},
{
"input": "925596634",
"output": "925596634"
},
{
"input": "-836078769",
"output": "-83607876"
},
{
"input": "71036059",
"output": "71036059"
},
{
"input": "-337396162",
"output": "-33739612"
},
{
"input": "87129297",
"output": "87129297"
},
{
"input": "-648171877",
"output": "-64817187"
},
{
"input": "20218400",
"output": "20218400"
},
{
"input": "10",
"output": "10"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "-1000000000",
"output": "-100000000"
},
{
"input": "-102",
"output": "-10"
},
{
"input": "-120",
"output": "-10"
},
{
"input": "-20",
"output": "0"
},
{
"input": "-15",
"output": "-1"
},
{
"input": "-5575533",
"output": "-557553"
},
{
"input": "-50",
"output": "0"
},
{
"input": "-31",
"output": "-1"
},
{
"input": "-55",
"output": "-5"
},
{
"input": "-81",
"output": "-1"
},
{
"input": "-99",
"output": "-9"
},
{
"input": "-23",
"output": "-2"
},
{
"input": "-41",
"output": "-1"
},
{
"input": "-24",
"output": "-2"
},
{
"input": "46",
"output": "46"
}
] | 1,693,057,574
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 47
| 154
| 0
|
n = input()
if int(n) >= 0:
print(n)
else:
n = int(n[1:])
print(-min(n//10, n//100*10 + n%10))
|
Title: Ilya and Bank Account
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.
Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.
Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.
Input Specification:
The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account.
Output Specification:
In a single line print an integer — the maximum state of the bank account that Ilya can get.
Demo Input:
['2230\n', '-10\n', '-100003\n']
Demo Output:
['2230\n', '0\n', '-10000\n']
Note:
In the first test sample Ilya doesn't profit from using the present.
In the second test sample you can delete digit 1 and get the state of the account equal to 0.
|
```python
n = input()
if int(n) >= 0:
print(n)
else:
n = int(n[1:])
print(-min(n//10, n//100*10 + n%10))
```
| 3
|
|
731
|
A
|
Night at the Museum
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.
|
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.
|
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
|
[
"zeus\n",
"map\n",
"ares\n"
] |
[
"18\n",
"35\n",
"34\n"
] |
To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations).
| 500
|
[
{
"input": "zeus",
"output": "18"
},
{
"input": "map",
"output": "35"
},
{
"input": "ares",
"output": "34"
},
{
"input": "l",
"output": "11"
},
{
"input": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv",
"output": "99"
},
{
"input": "gngvi",
"output": "44"
},
{
"input": "aaaaa",
"output": "0"
},
{
"input": "a",
"output": "0"
},
{
"input": "z",
"output": "1"
},
{
"input": "vyadeehhikklnoqrs",
"output": "28"
},
{
"input": "jjiihhhhgggfedcccbazyxx",
"output": "21"
},
{
"input": "fyyptqqxuciqvwdewyppjdzur",
"output": "117"
},
{
"input": "fqcnzmzmbobmancqcoalzmanaobpdse",
"output": "368"
},
{
"input": "zzzzzaaaaaaazzzzzzaaaaaaazzzzzzaaaazzzza",
"output": "8"
},
{
"input": "aucnwhfixuruefkypvrvnvznwtjgwlghoqtisbkhuwxmgzuljvqhmnwzisnsgjhivnjmbknptxatdkelhzkhsuxzrmlcpeoyukiy",
"output": "644"
},
{
"input": "sssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss",
"output": "8"
},
{
"input": "nypjygrdtpzpigzyrisqeqfriwgwlengnezppgttgtndbrryjdl",
"output": "421"
},
{
"input": "pnllnnmmmmoqqqqqrrtssssuuvtsrpopqoonllmonnnpppopnonoopooqpnopppqppqstuuuwwwwvxzxzzaa",
"output": "84"
},
{
"input": "btaoahqgxnfsdmzsjxgvdwjukcvereqeskrdufqfqgzqfsftdqcthtkcnaipftcnco",
"output": "666"
},
{
"input": "eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeerrrrrrrrrrrrrrrrwwwwwwwwww",
"output": "22"
},
{
"input": "uyknzcrwjyzmscqucclvacmorepdgmnyhmakmmnygqwglrxkxhkpansbmruwxdeoprxzmpsvwackopujxbbkpwyeggsvjykpxh",
"output": "643"
},
{
"input": "gzwpooohffcxwtpjgfzwtooiccxsrrokezutoojdzwsrmmhecaxwrojcbyrqlfdwwrliiib",
"output": "245"
},
{
"input": "dbvnkktasjdwqsrzfwwtmjgbcxggdxsoeilecihduypktkkbwfbruxzzhlttrssicgdwqruddwrlbtxgmhdbatzvdxbbro",
"output": "468"
},
{
"input": "mdtvowlktxzzbuaeiuebfeorgbdczauxsovbucactkvyvemsknsjfhifqgycqredzchipmkvzbxdjkcbyukomjlzvxzoswumned",
"output": "523"
},
{
"input": "kkkkkkkaaaaxxaaaaaaaxxxxxxxxaaaaaaxaaaaaaaaaakkkkkkkkkaaaaaaannnnnxxxxkkkkkkkkaannnnnnna",
"output": "130"
},
{
"input": "dffiknqqrsvwzcdgjkmpqtuwxadfhkkkmpqrtwxyadfggjmpppsuuwyyzcdgghhknnpsvvvwwwyabccffiloqruwwyyzabeeehh",
"output": "163"
},
{
"input": "qpppmmkjihgecbyvvsppnnnkjiffeebaaywutrrqpmkjhgddbzzzywtssssqnmmljheddbbaxvusrqonmlifedbbzyywwtqnkheb",
"output": "155"
},
{
"input": "wvvwwwvvwxxxyyyxxwwvwwvuttttttuvvwxxwxxyxxwwwwwvvuttssrssstsssssrqpqqppqrssrsrrssrssssrrsrqqrrqpppqp",
"output": "57"
},
{
"input": "dqcpcobpcobnznamznamzlykxkxlxlylzmaobnaobpbnanbpcoaobnboaoboanzlymzmykylymylzlylymanboanaocqdqesfrfs",
"output": "1236"
},
{
"input": "nnnnnnnnnnnnnnnnnnnnaaaaaaaaaaaaaaaaaaaakkkkkkkkkkkkkkkkkkkkkkaaaaaaaaaaaaaaaaaaaaxxxxxxxxxxxxxxxxxx",
"output": "49"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "cgilqsuwzaffilptwwbgmnttyyejkorxzflqvzbddhmnrvxchijpuwaeiimosxyycejlpquuwbfkpvbgijkqvxybdjjjptxcfkqt",
"output": "331"
},
{
"input": "ufsepwgtzgtgjssxaitgpailuvgqweoppszjwhoxdhhhpwwdorwfrdjwcdekxiktwziqwbkvbknrtvajpyeqbjvhiikxxaejjpte",
"output": "692"
},
{
"input": "uhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuh",
"output": "1293"
},
{
"input": "vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvgggggggggggggggggggggggggggggggggggggggggggggggggg",
"output": "16"
},
{
"input": "lyidmjyzbszgiwkxhhpnnthfwcvvstueionspfrvqgkvngmwyhezlosrpdnbvtcjjxxsykixwnepbumaacdzadlqhnjlcejovple",
"output": "616"
},
{
"input": "etzqqbaveffalkdguunfmyyrzkccnxmlluxeasqmopxzfvlkbhipqdwjgrttoemruohgwukfisdhznqyvhswbbypoxgtxyappcrl",
"output": "605"
},
{
"input": "lizussgedcbdjhrbeskhgatyozvwwekanlggcstijrniivupmcoofbaxfqrxddyzzptwxcftlhajsmmkkriarrqtkoauhcqefyud",
"output": "549"
},
{
"input": "dvjuvgfdogpknmbowlsfjzcimnygbtjiucyeeroqwhmzwpjqxlbjkqawrdtmvxbiqufllfuqibxvmtdrwaqkjblxqjpwzmhwqore",
"output": "688"
},
{
"input": "eeycuijtbgynmiczjfslwobmnkpgodfgvujvduyfeqchuaoktqrrairkkmmsjahltfcxwtpzzyddxrqfxabfoocmpuviinrjitsc",
"output": "604"
},
{
"input": "cgglnakewwvzoytaghksebrhjdbcdegssuzilrcppayxtgxopybbwshvyqnzhdsifkuwghourmeottrgjwdqpihbklvfzxpomqsa",
"output": "572"
},
{
"input": "aexullmxncckzryymfnuugdklaffevabqqztelpvojecljnhqldazdcaamubpenwxikysxxjjctvbndprsolzehywmgnvkgqvrfp",
"output": "609"
},
{
"input": "psnoieutsvvcwfhtnnphhxkwigzsbzyjmdiyl",
"output": "223"
},
{
"input": "aa",
"output": "0"
}
] | 1,655,058,564
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 46
| 0
|
s = input()
a = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm',
'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
idx1 = 0
sm = 0
for i in s:
idx2 = a.index(i)
# print(idx2, idx21)
if idx2 > idx1:
dis = (25 - idx2) + idx1 + 1
else:
dis = (25 - idx1) + idx2 + 1
dis = abs(dis)
if abs(idx1 - idx2) < dis:
sm += abs(idx1 - idx2)
else:
sm += dis
idx1 = idx2
# print(sm)
print(sm)
|
Title: Night at the Museum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.
Input Specification:
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.
Output Specification:
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
Demo Input:
['zeus\n', 'map\n', 'ares\n']
Demo Output:
['18\n', '35\n', '34\n']
Note:
To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations).
|
```python
s = input()
a = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm',
'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
idx1 = 0
sm = 0
for i in s:
idx2 = a.index(i)
# print(idx2, idx21)
if idx2 > idx1:
dis = (25 - idx2) + idx1 + 1
else:
dis = (25 - idx1) + idx2 + 1
dis = abs(dis)
if abs(idx1 - idx2) < dis:
sm += abs(idx1 - idx2)
else:
sm += dis
idx1 = idx2
# print(sm)
print(sm)
```
| 3
|
|
355
|
B
|
Vasya and Public Transport
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation"
] | null | null |
Vasya often uses public transport. The transport in the city is of two types: trolleys and buses. The city has *n* buses and *m* trolleys, the buses are numbered by integers from 1 to *n*, the trolleys are numbered by integers from 1 to *m*.
Public transport is not free. There are 4 types of tickets:
1. A ticket for one ride on some bus or trolley. It costs *c*1 burles; 1. A ticket for an unlimited number of rides on some bus or on some trolley. It costs *c*2 burles; 1. A ticket for an unlimited number of rides on all buses or all trolleys. It costs *c*3 burles; 1. A ticket for an unlimited number of rides on all buses and trolleys. It costs *c*4 burles.
Vasya knows for sure the number of rides he is going to make and the transport he is going to use. He asked you for help to find the minimum sum of burles he will have to spend on the tickets.
|
The first line contains four integers *c*1,<=*c*2,<=*c*3,<=*c*4 (1<=≤<=*c*1,<=*c*2,<=*c*3,<=*c*4<=≤<=1000) — the costs of the tickets.
The second line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of buses and trolleys Vasya is going to use.
The third line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1000) — the number of times Vasya is going to use the bus number *i*.
The fourth line contains *m* integers *b**i* (0<=≤<=*b**i*<=≤<=1000) — the number of times Vasya is going to use the trolley number *i*.
|
Print a single number — the minimum sum of burles Vasya will have to spend on the tickets.
|
[
"1 3 7 19\n2 3\n2 5\n4 4 4\n",
"4 3 2 1\n1 3\n798\n1 2 3\n",
"100 100 8 100\n3 5\n7 94 12\n100 1 47 0 42\n"
] |
[
"12\n",
"1\n",
"16\n"
] |
In the first sample the profitable strategy is to buy two tickets of the first type (for the first bus), one ticket of the second type (for the second bus) and one ticket of the third type (for all trolleys). It totals to (2·1) + 3 + 7 = 12 burles.
In the second sample the profitable strategy is to buy one ticket of the fourth type.
In the third sample the profitable strategy is to buy two tickets of the third type: for all buses and for all trolleys.
| 1,000
|
[
{
"input": "1 3 7 19\n2 3\n2 5\n4 4 4",
"output": "12"
},
{
"input": "4 3 2 1\n1 3\n798\n1 2 3",
"output": "1"
},
{
"input": "100 100 8 100\n3 5\n7 94 12\n100 1 47 0 42",
"output": "16"
},
{
"input": "3 103 945 1000\n7 9\n34 35 34 35 34 35 34\n0 0 0 0 0 0 0 0 0",
"output": "717"
},
{
"input": "7 11 597 948\n4 1\n5 1 0 11\n7",
"output": "40"
},
{
"input": "7 32 109 645\n1 3\n0\n0 0 0",
"output": "0"
},
{
"input": "680 871 347 800\n10 100\n872 156 571 136 703 201 832 213 15 333\n465 435 870 95 660 237 694 594 423 405 27 866 325 490 255 989 128 345 278 125 708 210 771 848 961 448 871 190 745 343 532 174 103 999 874 221 252 500 886 129 185 208 137 425 800 34 696 39 198 981 91 50 545 885 194 583 475 415 162 712 116 911 313 488 646 189 429 756 728 30 985 114 823 111 106 447 296 430 307 388 345 458 84 156 169 859 274 934 634 62 12 839 323 831 24 907 703 754 251 938",
"output": "694"
},
{
"input": "671 644 748 783\n100 10\n520 363 816 957 635 753 314 210 763 819 27 970 520 164 195 230 708 587 568 707 343 30 217 227 755 277 773 497 900 589 826 666 115 784 494 467 217 892 658 388 764 812 248 447 876 581 94 915 675 967 508 754 768 79 261 934 603 712 20 199 997 501 465 91 897 257 820 645 217 105 564 8 668 171 168 18 565 840 418 42 808 918 409 617 132 268 13 161 194 628 213 199 545 448 113 410 794 261 211 539\n147 3 178 680 701 193 697 666 846 389",
"output": "783"
},
{
"input": "2 7 291 972\n63 92\n7 0 0 6 0 13 0 20 2 8 0 17 7 0 0 0 0 2 2 0 0 8 20 0 0 0 3 0 0 0 4 20 0 0 0 12 0 8 17 9 0 0 0 0 4 0 0 0 17 11 3 0 2 15 0 18 11 19 14 0 0 20 13\n0 0 0 3 7 0 0 0 0 8 13 6 15 0 7 0 0 20 0 0 12 0 12 0 15 0 0 1 11 14 0 11 12 0 0 0 0 0 16 16 0 17 20 0 11 0 0 20 14 0 16 0 3 6 12 0 0 0 0 0 15 3 0 9 17 12 20 17 0 0 0 0 15 9 0 14 10 10 1 20 16 17 20 6 6 0 0 16 4 6 0 7",
"output": "494"
},
{
"input": "4 43 490 945\n63 92\n0 0 0 0 0 0 6 5 18 0 6 4 0 17 0 19 0 19 7 16 0 0 0 9 10 13 7 0 10 16 0 0 0 0 0 14 0 14 9 15 0 0 2 0 0 0 0 5 0 0 0 11 11 0 0 0 0 0 10 12 3 0 0\n0 12 0 18 7 7 0 0 9 0 0 13 17 0 18 12 4 0 0 14 18 20 0 0 12 9 17 1 19 0 11 0 5 0 0 14 0 0 16 0 19 15 9 14 7 10 0 19 19 0 0 1 0 0 0 6 0 0 0 6 0 20 1 9 0 0 10 17 5 2 5 4 16 6 0 11 0 8 13 4 0 2 0 0 13 10 0 13 0 0 8 4",
"output": "945"
},
{
"input": "2 50 258 922\n42 17\n0 2 0 1 0 1 0 11 18 9 0 0 0 0 10 15 17 4 20 0 5 0 0 13 13 0 0 2 0 7 0 20 4 0 19 3 7 0 0 0 0 0\n8 4 19 0 0 19 14 17 6 0 18 0 0 0 0 9 0",
"output": "486"
},
{
"input": "1 1 3 4\n2 3\n1 1\n1 1 1",
"output": "4"
},
{
"input": "4 4 4 1\n1 1\n0\n0",
"output": "0"
},
{
"input": "100 100 1 100\n10 10\n100 100 100 100 100 100 100 100 100 100\n100 100 100 100 100 100 100 100 100 100",
"output": "2"
}
] | 1,381,679,428
| 1,228
|
Python 3
|
OK
|
TESTS
| 27
| 62
| 307,200
|
c1, c2, c3, c4 = map(int, input().split())
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
sum1 = 0
for i in range(n):
sum1 += min(a[i] * c1, c2)
sum2 = 0
for i in range(m):
sum2 += min(b[i] * c1, c2)
print(min(c4, 2 * c3, c3 + sum1, c3 + sum2, sum1 + sum2))
|
Title: Vasya and Public Transport
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya often uses public transport. The transport in the city is of two types: trolleys and buses. The city has *n* buses and *m* trolleys, the buses are numbered by integers from 1 to *n*, the trolleys are numbered by integers from 1 to *m*.
Public transport is not free. There are 4 types of tickets:
1. A ticket for one ride on some bus or trolley. It costs *c*1 burles; 1. A ticket for an unlimited number of rides on some bus or on some trolley. It costs *c*2 burles; 1. A ticket for an unlimited number of rides on all buses or all trolleys. It costs *c*3 burles; 1. A ticket for an unlimited number of rides on all buses and trolleys. It costs *c*4 burles.
Vasya knows for sure the number of rides he is going to make and the transport he is going to use. He asked you for help to find the minimum sum of burles he will have to spend on the tickets.
Input Specification:
The first line contains four integers *c*1,<=*c*2,<=*c*3,<=*c*4 (1<=≤<=*c*1,<=*c*2,<=*c*3,<=*c*4<=≤<=1000) — the costs of the tickets.
The second line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of buses and trolleys Vasya is going to use.
The third line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1000) — the number of times Vasya is going to use the bus number *i*.
The fourth line contains *m* integers *b**i* (0<=≤<=*b**i*<=≤<=1000) — the number of times Vasya is going to use the trolley number *i*.
Output Specification:
Print a single number — the minimum sum of burles Vasya will have to spend on the tickets.
Demo Input:
['1 3 7 19\n2 3\n2 5\n4 4 4\n', '4 3 2 1\n1 3\n798\n1 2 3\n', '100 100 8 100\n3 5\n7 94 12\n100 1 47 0 42\n']
Demo Output:
['12\n', '1\n', '16\n']
Note:
In the first sample the profitable strategy is to buy two tickets of the first type (for the first bus), one ticket of the second type (for the second bus) and one ticket of the third type (for all trolleys). It totals to (2·1) + 3 + 7 = 12 burles.
In the second sample the profitable strategy is to buy one ticket of the fourth type.
In the third sample the profitable strategy is to buy two tickets of the third type: for all buses and for all trolleys.
|
```python
c1, c2, c3, c4 = map(int, input().split())
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
sum1 = 0
for i in range(n):
sum1 += min(a[i] * c1, c2)
sum2 = 0
for i in range(m):
sum2 += min(b[i] * c1, c2)
print(min(c4, 2 * c3, c3 + sum1, c3 + sum2, sum1 + sum2))
```
| 3
|
|
189
|
A
|
Cut Ribbon
|
PROGRAMMING
| 1,300
|
[
"brute force",
"dp"
] | null | null |
Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
|
The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide.
|
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
|
[
"5 5 3 2\n",
"7 5 5 2\n"
] |
[
"2\n",
"2\n"
] |
In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.
| 500
|
[
{
"input": "5 5 3 2",
"output": "2"
},
{
"input": "7 5 5 2",
"output": "2"
},
{
"input": "4 4 4 4",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "4000 1 2 3",
"output": "4000"
},
{
"input": "4000 3 4 5",
"output": "1333"
},
{
"input": "10 3 4 5",
"output": "3"
},
{
"input": "100 23 15 50",
"output": "2"
},
{
"input": "3119 3515 1021 7",
"output": "11"
},
{
"input": "918 102 1327 1733",
"output": "9"
},
{
"input": "3164 42 430 1309",
"output": "15"
},
{
"input": "3043 317 1141 2438",
"output": "7"
},
{
"input": "26 1 772 2683",
"output": "26"
},
{
"input": "370 2 1 15",
"output": "370"
},
{
"input": "734 12 6 2",
"output": "367"
},
{
"input": "418 18 14 17",
"output": "29"
},
{
"input": "18 16 28 9",
"output": "2"
},
{
"input": "14 6 2 17",
"output": "7"
},
{
"input": "29 27 18 2",
"output": "2"
},
{
"input": "29 12 7 10",
"output": "3"
},
{
"input": "27 23 4 3",
"output": "9"
},
{
"input": "5 14 5 2",
"output": "1"
},
{
"input": "5 17 26 5",
"output": "1"
},
{
"input": "9 1 10 3",
"output": "9"
},
{
"input": "2 19 15 1",
"output": "2"
},
{
"input": "4 6 4 9",
"output": "1"
},
{
"input": "10 6 2 9",
"output": "5"
},
{
"input": "2 2 9 6",
"output": "1"
},
{
"input": "6 2 4 1",
"output": "6"
},
{
"input": "27 24 5 27",
"output": "1"
},
{
"input": "2683 83 26 2709",
"output": "101"
},
{
"input": "728 412 789 158",
"output": "3"
},
{
"input": "3964 4 2916 176",
"output": "991"
},
{
"input": "3399 2035 2 3334",
"output": "683"
},
{
"input": "3455 244 3301 3",
"output": "991"
},
{
"input": "595 2263 3625 1",
"output": "595"
},
{
"input": "4000 1 1 1",
"output": "4000"
},
{
"input": "3999 2 2 3999",
"output": "1"
},
{
"input": "25 6 8 11",
"output": "3"
},
{
"input": "4000 500 1000 2000",
"output": "8"
},
{
"input": "53 10 11 23",
"output": "5"
},
{
"input": "100 100 1 1",
"output": "100"
},
{
"input": "17 3 4 10",
"output": "5"
},
{
"input": "413 101 102 105",
"output": "4"
},
{
"input": "490 4 49 50",
"output": "111"
},
{
"input": "3999 2 3 3",
"output": "1999"
},
{
"input": "8 3 8 4",
"output": "2"
},
{
"input": "5 1 3 3",
"output": "5"
},
{
"input": "100 3 17 22",
"output": "27"
},
{
"input": "4000 2 3 4",
"output": "2000"
},
{
"input": "4000 3 3 5",
"output": "1332"
},
{
"input": "13 4 6 7",
"output": "2"
},
{
"input": "4000 5 2 2",
"output": "2000"
},
{
"input": "3999 2 2 3",
"output": "1999"
},
{
"input": "4000 33 7 3333",
"output": "564"
},
{
"input": "60 33 20 9",
"output": "4"
},
{
"input": "100 9 11 99",
"output": "10"
},
{
"input": "2009 6 8 9",
"output": "334"
}
] | 1,676,505,470
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 58
| 92
| 2,662,400
|
# DO NOT EDIT THIS
import math
import sys
input = sys.stdin.readline
from collections import deque, defaultdict
import heapq
def counter(a):
c = defaultdict(lambda : 0) # way faster than Counter
for el in a:
c[el] += 1
return c
def inp(): return [int(k) for k in input().split()]
def si(): return int(input())
def st(): return input()
# DO NOT EDIT ABOVE THIS
n, a, b, c = inp()
dp = [-math.inf] * (n + 1)
dp[0] = 0
for i in range(1, len(dp)):
a_cut = 1 + dp[i - a] if i >= a else -math.inf
b_cut = 1 + dp[i - b] if i >= b else -math.inf
c_cut = 1 + dp[i - c] if i >= c else -math.inf
dp[i] = max(a_cut, b_cut, c_cut)
print(dp[-1])
|
Title: Cut Ribbon
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
Input Specification:
The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide.
Output Specification:
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
Demo Input:
['5 5 3 2\n', '7 5 5 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.
|
```python
# DO NOT EDIT THIS
import math
import sys
input = sys.stdin.readline
from collections import deque, defaultdict
import heapq
def counter(a):
c = defaultdict(lambda : 0) # way faster than Counter
for el in a:
c[el] += 1
return c
def inp(): return [int(k) for k in input().split()]
def si(): return int(input())
def st(): return input()
# DO NOT EDIT ABOVE THIS
n, a, b, c = inp()
dp = [-math.inf] * (n + 1)
dp[0] = 0
for i in range(1, len(dp)):
a_cut = 1 + dp[i - a] if i >= a else -math.inf
b_cut = 1 + dp[i - b] if i >= b else -math.inf
c_cut = 1 + dp[i - c] if i >= c else -math.inf
dp[i] = max(a_cut, b_cut, c_cut)
print(dp[-1])
```
| 3
|
|
9
|
C
|
Hexadecimal's Numbers
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation",
"math"
] |
C. Hexadecimal's Numbers
|
1
|
64
|
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of *n* different natural numbers from 1 to *n* to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
|
Input data contains the only number *n* (1<=≤<=*n*<=≤<=109).
|
Output the only number — answer to the problem.
|
[
"10\n"
] |
[
"2"
] |
For *n* = 10 the answer includes numbers 1 and 10.
| 0
|
[
{
"input": "10",
"output": "2"
},
{
"input": "20",
"output": "3"
},
{
"input": "72",
"output": "3"
},
{
"input": "99",
"output": "3"
},
{
"input": "100",
"output": "4"
},
{
"input": "101",
"output": "5"
},
{
"input": "102",
"output": "5"
},
{
"input": "111",
"output": "7"
},
{
"input": "112",
"output": "7"
},
{
"input": "745",
"output": "7"
},
{
"input": "23536",
"output": "31"
},
{
"input": "1",
"output": "1"
},
{
"input": "1010011",
"output": "83"
},
{
"input": "312410141",
"output": "511"
},
{
"input": "1000000000",
"output": "512"
},
{
"input": "999999999",
"output": "511"
},
{
"input": "111111111",
"output": "511"
},
{
"input": "101010101",
"output": "341"
},
{
"input": "121212121",
"output": "511"
},
{
"input": "106341103",
"output": "383"
},
{
"input": "901556123",
"output": "511"
},
{
"input": "832513432",
"output": "511"
},
{
"input": "3",
"output": "1"
},
{
"input": "732875234",
"output": "511"
},
{
"input": "7",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "11",
"output": "3"
},
{
"input": "12",
"output": "3"
},
{
"input": "13",
"output": "3"
},
{
"input": "101020101",
"output": "351"
},
{
"input": "111100100",
"output": "484"
},
{
"input": "110110101",
"output": "437"
},
{
"input": "100111001",
"output": "313"
},
{
"input": "100100",
"output": "36"
},
{
"input": "110100102",
"output": "421"
}
] | 1,604,690,940
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 278
| 0
|
n = str(input())
ans = 0
for i in n:
if i == '0' or i == '1':
ans += 1
if ans == len(n):
print(ans)
else:
print(0)
|
Title: Hexadecimal's Numbers
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of *n* different natural numbers from 1 to *n* to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input Specification:
Input data contains the only number *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Output the only number — answer to the problem.
Demo Input:
['10\n']
Demo Output:
['2']
Note:
For *n* = 10 the answer includes numbers 1 and 10.
|
```python
n = str(input())
ans = 0
for i in n:
if i == '0' or i == '1':
ans += 1
if ans == len(n):
print(ans)
else:
print(0)
```
| 0
|
732
|
A
|
Buy a Shovel
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
|
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
|
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
|
[
"117 3\n",
"237 7\n",
"15 2\n"
] |
[
"9\n",
"1\n",
"2\n"
] |
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
| 500
|
[
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,697,347,956
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 46
| 0
|
k,r=map(int,input().split())
num_s=1
while (num_s*k)%10!=0 and ((num_s*k)-r)%10!=0:
num_s+=1
print(num_s)
|
Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
|
```python
k,r=map(int,input().split())
num_s=1
while (num_s*k)%10!=0 and ((num_s*k)-r)%10!=0:
num_s+=1
print(num_s)
```
| 3
|
|
656
|
G
|
You're a Professional
|
PROGRAMMING
| 1,900
|
[
"*special"
] | null | null |
A simple recommendation system would recommend a user things liked by a certain number of their friends. In this problem you will implement part of such a system.
You are given user's friends' opinions about a list of items. You are also given a threshold *T* — the minimal number of "likes" necessary for an item to be recommended to the user.
Output the number of items in the list liked by at least *T* of user's friends.
|
The first line of the input will contain three space-separated integers: the number of friends *F* (1<=≤<=*F*<=≤<=10), the number of items *I* (1<=≤<=*I*<=≤<=10) and the threshold *T* (1<=≤<=*T*<=≤<=*F*).
The following *F* lines of input contain user's friends' opinions. *j*-th character of *i*-th line is 'Y' if *i*-th friend likes *j*-th item, and 'N' otherwise.
|
Output an integer — the number of items liked by at least *T* of user's friends.
|
[
"3 3 2\nYYY\nNNN\nYNY\n",
"4 4 1\nNNNY\nNNYN\nNYNN\nYNNN\n"
] |
[
"2\n",
"4\n"
] |
none
| 0
|
[
{
"input": "3 3 2\nYYY\nNNN\nYNY",
"output": "2"
},
{
"input": "4 4 1\nNNNY\nNNYN\nNYNN\nYNNN",
"output": "4"
},
{
"input": "3 5 2\nNYNNY\nYNNNN\nNNYYN",
"output": "0"
},
{
"input": "1 10 1\nYYYNYNNYNN",
"output": "5"
},
{
"input": "10 1 5\nY\nN\nN\nN\nY\nN\nN\nY\nN\nN",
"output": "0"
},
{
"input": "10 10 1\nNNNNNNNNNN\nNNNNNNNNNN\nNNNNNNNNNN\nNNNNNNNNNN\nNNNNNNNNNN\nNNNNNNNNNN\nNNNNNNNNNN\nNNNNNNNNNN\nNNNNNNNNNN\nNNNNNNNNNN",
"output": "0"
},
{
"input": "10 10 10\nYYYYYYYYYY\nYYYYYYYYYY\nYYYYYYYYYY\nYYYYYYYYYY\nYYYYYYYYYY\nYYYYYYYYYY\nYYYYYYYYYY\nYYYYYYYYYY\nYYYYYYYYYY\nYYYYYYYYYY",
"output": "10"
},
{
"input": "8 9 1\nNYNNYYYYN\nNNNYNYNNY\nYYNYNYNNN\nNYYYNYNNN\nYNYNYNYYN\nYYNNYYYYY\nYYYYNYNYY\nNYYNNYYYY",
"output": "9"
},
{
"input": "5 2 3\nNN\nNY\nYY\nNN\nNY",
"output": "1"
},
{
"input": "6 4 5\nYNNY\nNYYY\nNNNY\nYNYN\nYYYN\nYNNY",
"output": "0"
},
{
"input": "6 1 3\nY\nY\nY\nY\nY\nN",
"output": "1"
},
{
"input": "6 2 2\nYN\nNN\nYN\nNN\nYN\nNN",
"output": "1"
},
{
"input": "2 4 2\nNYNY\nNYNY",
"output": "2"
},
{
"input": "9 6 3\nNYYYYN\nNNNYYN\nYYYYYY\nNYNNNN\nYNNYNY\nNNNNNY\nYNNYNN\nYYYYNY\nNNYYYY",
"output": "6"
},
{
"input": "6 9 6\nYYYYNYNNN\nYNNYNNNYN\nNYYYNNNYY\nNYYYNNNNY\nYYNYNNNYY\nYYYNYYNNN",
"output": "0"
},
{
"input": "9 7 8\nYNNNNYN\nNNNYYNN\nNNYYYNY\nNYYNYYY\nNNYYNYN\nNYYYNNY\nYYNYNYY\nNYYYYYY\nNNYYNYN",
"output": "0"
},
{
"input": "9 1 6\nN\nN\nY\nN\nY\nY\nY\nY\nY",
"output": "1"
},
{
"input": "7 7 2\nNNYNNYN\nNNNYYNY\nNNNYYNY\nYNNNNNY\nNNYNYYY\nYYNNYYN\nNNYYYNY",
"output": "6"
},
{
"input": "8 4 2\nYNYY\nYNYY\nYNNN\nNNNN\nNYNN\nYNNN\nNNYN\nNYNN",
"output": "4"
},
{
"input": "9 10 7\nNNYNNYYYYY\nYNYYNYYNYN\nNYNYYNNNNY\nYYYYYYYYYN\nYYNYNYYNNN\nYYYNNYYYYY\nNYYYYYNNNN\nNYNNYYYYNN\nYYYYYNNYYY",
"output": "2"
},
{
"input": "6 4 2\nNNNN\nNYYY\nNYNN\nNYNN\nYNNY\nNNNN",
"output": "2"
},
{
"input": "3 1 1\nN\nY\nN",
"output": "1"
},
{
"input": "7 1 3\nY\nY\nY\nN\nY\nY\nY",
"output": "1"
},
{
"input": "9 8 7\nNYYNNNYY\nYYYNYNNN\nYNYNYNNY\nNYYYNNNY\nNYYYYNYN\nNNNNYYNN\nYNYYYYYY\nNNYNYNYY\nNYYNNYYY",
"output": "1"
},
{
"input": "9 5 9\nYYYYN\nYYYNN\nNNYNN\nNNYYY\nYNNNN\nNYNNN\nYYYYN\nYNYYN\nNNNYN",
"output": "0"
},
{
"input": "8 4 1\nYYYN\nNNNN\nNYNY\nYNNY\nYNYY\nYNYN\nYNNY\nNNYN",
"output": "4"
},
{
"input": "7 9 5\nYNNYYYYNN\nYNYYYNNYY\nYNYYYYYNN\nYYNYYNYYN\nNNYYNNNYY\nYYNYNYYNN\nYYNNYYNYN",
"output": "3"
},
{
"input": "5 8 3\nNYYYNNNN\nYNNNNNYY\nYNYYYNYY\nNNNNNYNN\nYYYYYYYY",
"output": "5"
},
{
"input": "5 10 4\nYYYYNNNNYN\nYYYNYYYNNY\nNNNYNYNYNY\nYNYNNNNNNY\nNNYNYNYNYY",
"output": "2"
},
{
"input": "6 9 6\nNYYNNYNYN\nYNYNYNNNN\nNNYNNYYYY\nNNYNNNYNY\nNYYYNNYNY\nNNYYNNNYN",
"output": "1"
},
{
"input": "4 4 1\nYNYY\nNNNY\nYNNN\nNNYN",
"output": "3"
},
{
"input": "1 3 1\nYYN",
"output": "2"
},
{
"input": "10 4 5\nNNYN\nYYNY\nYYNY\nNYYN\nYNYY\nYNYY\nYYNN\nYNYN\nYYYY\nYYNY",
"output": "4"
}
] | 1,459,533,565
| 7,165
|
Python 3
|
OK
|
TESTS
| 33
| 62
| 5,120,000
|
n,kitten,t= map(int, input().split())
a=[str(input()) for i in range(n)]
q=0
for i in range(kitten):
l=0
for j in range(n):
if (a[j][i]=="Y"):
l+=1
if (l >= t):
q+=1
print(q)
|
Title: You're a Professional
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A simple recommendation system would recommend a user things liked by a certain number of their friends. In this problem you will implement part of such a system.
You are given user's friends' opinions about a list of items. You are also given a threshold *T* — the minimal number of "likes" necessary for an item to be recommended to the user.
Output the number of items in the list liked by at least *T* of user's friends.
Input Specification:
The first line of the input will contain three space-separated integers: the number of friends *F* (1<=≤<=*F*<=≤<=10), the number of items *I* (1<=≤<=*I*<=≤<=10) and the threshold *T* (1<=≤<=*T*<=≤<=*F*).
The following *F* lines of input contain user's friends' opinions. *j*-th character of *i*-th line is 'Y' if *i*-th friend likes *j*-th item, and 'N' otherwise.
Output Specification:
Output an integer — the number of items liked by at least *T* of user's friends.
Demo Input:
['3 3 2\nYYY\nNNN\nYNY\n', '4 4 1\nNNNY\nNNYN\nNYNN\nYNNN\n']
Demo Output:
['2\n', '4\n']
Note:
none
|
```python
n,kitten,t= map(int, input().split())
a=[str(input()) for i in range(n)]
q=0
for i in range(kitten):
l=0
for j in range(n):
if (a[j][i]=="Y"):
l+=1
if (l >= t):
q+=1
print(q)
```
| 3
|
|
837
|
D
|
Round Subset
|
PROGRAMMING
| 2,100
|
[
"dp",
"math"
] | null | null |
Let's call the roundness of the number the number of zeros to which it ends.
You have an array of *n* numbers. You need to choose a subset of exactly *k* numbers so that the roundness of the product of the selected numbers will be maximum possible.
|
The first line contains two integer numbers *n* and *k* (1<=≤<=*n*<=≤<=200,<=1<=≤<=*k*<=≤<=*n*).
The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018).
|
Print maximal roundness of product of the chosen subset of length *k*.
|
[
"3 2\n50 4 20\n",
"5 3\n15 16 3 25 9\n",
"3 3\n9 77 13\n"
] |
[
"3\n",
"3\n",
"0\n"
] |
In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.
In the second example subset [15, 16, 25] has product 6000, roundness 3.
In the third example all subsets has product with roundness 0.
| 0
|
[
{
"input": "3 2\n50 4 20",
"output": "3"
},
{
"input": "5 3\n15 16 3 25 9",
"output": "3"
},
{
"input": "3 3\n9 77 13",
"output": "0"
},
{
"input": "1 1\n200000000",
"output": "8"
},
{
"input": "1 1\n3",
"output": "0"
},
{
"input": "3 1\n1000000000000000000 800000000000000000 625",
"output": "18"
},
{
"input": "20 13\n93050001 1 750000001 950000001 160250001 482000001 145875001 900000001 500000001 513300001 313620001 724750001 205800001 400000001 800000001 175000001 875000001 852686005 868880001 342500001",
"output": "0"
},
{
"input": "5 3\n1360922189858001 5513375057164001 4060879738933651 3260997351273601 5540397778584001",
"output": "0"
},
{
"input": "5 3\n670206146698567481 75620705254979501 828058059097865201 67124386759325201 946737848872942801",
"output": "0"
},
{
"input": "5 4\n539134530963895499 265657472022483040 798956216114326361 930406714691011229 562844921643925634",
"output": "1"
},
{
"input": "200 10\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "200 50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "200 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "200 200\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "5 2\n625 5 100 16 10",
"output": "4"
},
{
"input": "5 2\n64 32 16 8 3125",
"output": "5"
},
{
"input": "2 2\n2199023255552 11920928955078125",
"output": "23"
},
{
"input": "1 1\n500",
"output": "2"
},
{
"input": "3 1\n125 10 8",
"output": "1"
},
{
"input": "7 5\n50 312500 10000 1250 2000000 250 1250000",
"output": "18"
},
{
"input": "4 2\n3125 1000 1000 32",
"output": "6"
},
{
"input": "3 1\n4 10 25",
"output": "1"
},
{
"input": "3 1\n16 10 75",
"output": "1"
},
{
"input": "3 2\n100 5120 19531250",
"output": "11"
},
{
"input": "4 2\n16 200 500 625",
"output": "5"
},
{
"input": "7 4\n5000 1250 100 200 10 5000 640",
"output": "13"
},
{
"input": "2 1\n125 10",
"output": "1"
},
{
"input": "3 2\n16 625 10",
"output": "4"
},
{
"input": "4 2\n10 4 25 5",
"output": "2"
},
{
"input": "2 1\n10 25",
"output": "1"
},
{
"input": "3 1\n10 16 625",
"output": "1"
},
{
"input": "4 2\n10 200 625 16",
"output": "4"
},
{
"input": "3 1\n1000 16 625",
"output": "3"
},
{
"input": "3 2\n6250 100 160",
"output": "6"
},
{
"input": "3 2\n1250 100 80",
"output": "5"
},
{
"input": "3 2\n1024 1000 9765625",
"output": "10"
},
{
"input": "5 2\n125 10 10 3 3",
"output": "2"
},
{
"input": "4 2\n31250 200 500 320",
"output": "7"
},
{
"input": "3 1\n16 10 125",
"output": "1"
},
{
"input": "5 3\n125 8 1000 100 100",
"output": "7"
},
{
"input": "3 2\n25 4 10",
"output": "2"
},
{
"input": "3 2\n125 8 10",
"output": "3"
},
{
"input": "6 2\n125 125 100 100 8 8",
"output": "4"
},
{
"input": "3 1\n32 3125 10",
"output": "1"
},
{
"input": "5 3\n4 5 125 16 1",
"output": "4"
},
{
"input": "7 4\n5 125 2 5 2 5 20",
"output": "4"
},
{
"input": "200 200\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "100"
},
{
"input": "2 1\n200 250",
"output": "2"
},
{
"input": "5 3\n500 10 250 20 8",
"output": "6"
},
{
"input": "4 2\n2 2 25 1",
"output": "1"
},
{
"input": "2 2\n16 25",
"output": "2"
}
] | 1,501,922,716
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 61
| 4,812,800
|
def dob(cur,ans,a):
cur *= a
while (cur % 10 == 0):
ans += 1
cur /= 10
return cur, ans
n,k = map(int,input().split())
des = []
pyat = []
chet = []
inp = list(map(int,input().split()))
des.sort()
for el in inp:
if el % 10 == 0:
des.append(el)
elif el % 5 == 0:
pyat.append(el)
elif el % 2 == 0:
chet.append(el)
cur = 1
ans = 0
while k > 0 and len(des) != 0:
cur, ans = dob(cur,ans, des[len(des)-1])
k -= 1
des.pop()
if k <= 0:
print(ans)
else:
pyat.sort()
chet.sort()
while k > 0 and len(pyat) != 0 and len(chet) != 0:
if cur % 2 == 0:
cur, ans = dob(cur,ans, pyat[len(pyat)-1])
k -= 1
pyat.pop()
else:
cur, ans = dob(cur,ans, chet[len(chet)-1])
k -= 1
chet.pop()
while k > 0 and len(pyat) != 0 and cur % 2 == 0:
cur, ans = dob(cur,ans, pyat[len(pyat)-1])
k -= 1
pyat.pop()
while k > 0 and len(chet) != 0 and cur % 5 == 0:
cur, ans = dob(cur,ans, chet[len(chet)-1])
k -= 1
chet.pop()
print(ans)
|
Title: Round Subset
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call the roundness of the number the number of zeros to which it ends.
You have an array of *n* numbers. You need to choose a subset of exactly *k* numbers so that the roundness of the product of the selected numbers will be maximum possible.
Input Specification:
The first line contains two integer numbers *n* and *k* (1<=≤<=*n*<=≤<=200,<=1<=≤<=*k*<=≤<=*n*).
The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018).
Output Specification:
Print maximal roundness of product of the chosen subset of length *k*.
Demo Input:
['3 2\n50 4 20\n', '5 3\n15 16 3 25 9\n', '3 3\n9 77 13\n']
Demo Output:
['3\n', '3\n', '0\n']
Note:
In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.
In the second example subset [15, 16, 25] has product 6000, roundness 3.
In the third example all subsets has product with roundness 0.
|
```python
def dob(cur,ans,a):
cur *= a
while (cur % 10 == 0):
ans += 1
cur /= 10
return cur, ans
n,k = map(int,input().split())
des = []
pyat = []
chet = []
inp = list(map(int,input().split()))
des.sort()
for el in inp:
if el % 10 == 0:
des.append(el)
elif el % 5 == 0:
pyat.append(el)
elif el % 2 == 0:
chet.append(el)
cur = 1
ans = 0
while k > 0 and len(des) != 0:
cur, ans = dob(cur,ans, des[len(des)-1])
k -= 1
des.pop()
if k <= 0:
print(ans)
else:
pyat.sort()
chet.sort()
while k > 0 and len(pyat) != 0 and len(chet) != 0:
if cur % 2 == 0:
cur, ans = dob(cur,ans, pyat[len(pyat)-1])
k -= 1
pyat.pop()
else:
cur, ans = dob(cur,ans, chet[len(chet)-1])
k -= 1
chet.pop()
while k > 0 and len(pyat) != 0 and cur % 2 == 0:
cur, ans = dob(cur,ans, pyat[len(pyat)-1])
k -= 1
pyat.pop()
while k > 0 and len(chet) != 0 and cur % 5 == 0:
cur, ans = dob(cur,ans, chet[len(chet)-1])
k -= 1
chet.pop()
print(ans)
```
| 0
|
|
808
|
C
|
Tea Party
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"greedy",
"sortings"
] | null | null |
Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=≤<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
|
The first line contains two integer numbers *n* and *w* (1<=≤<=*n*<=≤<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
|
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
|
[
"2 10\n8 7\n",
"4 4\n1 1 1 1\n",
"3 10\n9 8 10\n"
] |
[
"6 4 \n",
"1 1 1 1 \n",
"-1\n"
] |
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.
| 0
|
[
{
"input": "2 10\n8 7",
"output": "6 4 "
},
{
"input": "4 4\n1 1 1 1",
"output": "1 1 1 1 "
},
{
"input": "3 10\n9 8 10",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "1 "
},
{
"input": "1 1\n2",
"output": "1 "
},
{
"input": "1 10\n20",
"output": "10 "
},
{
"input": "3 10\n8 4 8",
"output": "4 2 4 "
},
{
"input": "3 100\n37 26 37",
"output": "37 26 37 "
},
{
"input": "3 60\n43 23 24",
"output": "36 12 12 "
},
{
"input": "20 14\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 8\n1 2 1 2 1 1 1 2 1 1 1 2 1 1 2 1 1 1 2 2",
"output": "-1"
},
{
"input": "50 1113\n25 21 23 37 28 23 19 25 5 12 3 11 46 50 13 50 7 1 8 40 4 6 34 27 11 39 45 31 10 12 48 2 19 37 47 45 30 24 21 42 36 14 31 30 31 50 6 3 33 49",
"output": "13 11 12 37 28 12 10 18 3 6 2 6 46 50 7 50 4 1 4 40 2 3 34 27 6 39 45 31 5 6 48 1 10 37 47 45 30 12 11 42 36 7 31 30 31 50 3 2 33 49 "
},
{
"input": "50 440\n14 69 33 38 83 65 21 66 89 3 93 60 31 16 61 20 42 64 13 1 50 50 74 58 67 61 52 22 69 68 18 33 28 59 4 8 96 32 84 85 87 87 61 89 2 47 15 64 88 18",
"output": "-1"
},
{
"input": "100 640\n82 51 81 14 37 17 78 92 64 15 8 86 89 8 87 77 66 10 15 12 100 25 92 47 21 78 20 63 13 49 41 36 41 79 16 87 87 69 3 76 80 60 100 49 70 59 72 8 38 71 45 97 71 14 76 54 81 4 59 46 39 29 92 3 49 22 53 99 59 52 74 31 92 43 42 23 44 9 82 47 7 40 12 9 3 55 37 85 46 22 84 52 98 41 21 77 63 17 62 91",
"output": "-1"
},
{
"input": "100 82\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "100 55\n1 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1",
"output": "-1"
},
{
"input": "30 50\n3 1 2 4 1 2 2 4 3 4 4 3 3 3 3 5 3 2 5 4 3 3 5 3 3 5 4 5 3 5",
"output": "-1"
},
{
"input": "40 100\n3 3 3 3 4 1 1 1 1 1 2 2 1 3 1 2 3 2 1 2 2 2 1 4 2 2 3 3 3 2 4 6 4 4 3 2 2 2 4 5",
"output": "3 3 3 3 4 1 1 1 1 1 2 2 1 3 1 2 3 2 1 2 2 2 1 4 2 2 3 3 3 2 4 6 4 4 3 2 2 2 4 5 "
},
{
"input": "100 10000\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 "
},
{
"input": "2 5\n3 4",
"output": "2 3 "
},
{
"input": "2 6\n2 6",
"output": "1 5 "
},
{
"input": "23 855\n5 63 94 57 38 84 77 79 83 36 47 31 60 79 75 48 88 17 46 33 23 15 27",
"output": "3 32 94 29 19 84 39 72 83 18 24 16 30 79 38 24 88 9 23 17 12 8 14 "
},
{
"input": "52 2615\n73 78 70 92 94 74 46 19 55 20 70 3 1 42 68 10 66 80 1 31 65 19 73 74 56 35 53 38 92 35 65 81 6 98 74 51 27 49 76 19 86 76 5 60 14 75 64 99 43 7 36 79",
"output": "73 78 70 92 94 74 46 10 55 10 70 2 1 42 68 5 66 80 1 16 65 10 73 74 56 18 53 38 92 30 65 81 3 98 74 51 14 49 76 10 86 76 3 60 7 75 64 99 43 4 36 79 "
},
{
"input": "11 287\n34 30 69 86 22 53 11 91 62 44 5",
"output": "17 15 35 43 11 27 6 77 31 22 3 "
},
{
"input": "55 1645\n60 53 21 20 87 48 10 21 76 35 52 41 82 86 93 11 93 86 34 15 37 63 57 3 57 57 32 8 55 25 29 38 46 22 13 87 27 35 40 83 5 7 6 18 88 25 4 59 95 62 31 93 98 50 62",
"output": "30 27 11 10 82 24 5 11 38 18 26 21 41 43 93 6 93 43 17 8 19 32 29 2 29 29 16 4 28 13 15 19 23 11 7 87 14 18 20 42 3 4 3 9 88 13 2 30 95 31 16 93 98 25 31 "
},
{
"input": "71 3512\n97 46 76 95 81 96 99 83 10 50 19 18 73 5 41 60 12 73 60 31 21 64 88 61 43 57 61 19 75 35 41 85 12 59 32 47 37 43 35 92 90 47 3 98 21 18 61 79 39 86 74 8 52 33 39 27 93 54 35 38 96 36 83 51 97 10 8 66 75 87 68",
"output": "97 46 76 95 81 96 99 83 5 50 10 9 73 3 41 60 6 73 60 16 11 64 88 61 43 57 61 10 75 18 41 85 6 59 16 47 19 43 18 92 90 47 2 98 11 9 61 79 20 86 74 4 52 17 21 14 93 54 18 19 96 18 83 51 97 5 4 66 75 87 68 "
},
{
"input": "100 2633\n99 50 64 81 75 73 26 31 31 36 95 12 100 2 70 72 78 56 76 23 94 8 91 1 39 82 97 67 64 25 71 90 48 34 31 46 64 37 46 50 99 93 14 56 1 89 95 89 50 52 12 58 43 65 45 88 90 14 38 19 6 15 91 67 43 48 82 20 11 48 33 20 39 52 73 5 25 84 26 54 42 56 10 28 9 63 60 98 30 1 25 74 86 56 85 9 12 94 80 95",
"output": "50 25 32 41 38 37 13 16 16 18 48 6 61 1 35 36 39 28 38 12 47 4 46 1 20 41 49 34 32 13 36 45 24 17 16 23 32 19 23 25 50 47 7 28 1 45 48 45 25 26 6 29 22 33 23 44 45 7 19 10 3 8 46 34 22 24 41 10 6 24 17 10 20 26 37 3 13 42 13 27 21 28 5 14 5 32 30 49 15 1 13 37 43 28 43 5 6 47 40 48 "
},
{
"input": "71 1899\n23 55 58 87 69 85 100 21 19 72 81 68 20 25 29 92 18 74 89 70 53 7 78 57 41 79 64 87 63 76 95 84 1 28 32 1 79 34 77 17 71 61 35 31 62 92 69 99 60 26 2 18 61 9 27 77 82 6 30 65 52 3 51 43 13 77 41 59 19 29 86",
"output": "12 28 29 44 35 43 95 11 10 36 41 34 10 13 15 46 9 37 45 35 27 4 39 29 21 40 32 44 32 38 48 42 1 14 16 1 40 17 39 9 36 31 18 16 31 46 35 50 30 13 1 9 31 5 14 39 41 3 15 33 26 2 26 22 7 39 21 30 10 15 43 "
},
{
"input": "10 21\n3 3 3 3 4 3 3 3 3 3",
"output": "2 2 2 2 3 2 2 2 2 2 "
}
] | 1,646,185,859
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 6
| 62
| 0
|
n, w = map(int, input().split())
cups = list(map(int, input().split()))
output = []
for cup in cups:
fill = -(-cup // 2)
if w < fill:
print(-1)
exit()
output.append(fill)
w -= fill
output.sort(reverse=True)
for i in range(len(cups)):
cup = cups[i]
o = output[i]
fill = min(cup - o, w)
w -= fill
output[i] += fill
print(*output)
|
Title: Tea Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=≤<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
Input Specification:
The first line contains two integer numbers *n* and *w* (1<=≤<=*n*<=≤<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
Demo Input:
['2 10\n8 7\n', '4 4\n1 1 1 1\n', '3 10\n9 8 10\n']
Demo Output:
['6 4 \n', '1 1 1 1 \n', '-1\n']
Note:
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.
|
```python
n, w = map(int, input().split())
cups = list(map(int, input().split()))
output = []
for cup in cups:
fill = -(-cup // 2)
if w < fill:
print(-1)
exit()
output.append(fill)
w -= fill
output.sort(reverse=True)
for i in range(len(cups)):
cup = cups[i]
o = output[i]
fill = min(cup - o, w)
w -= fill
output[i] += fill
print(*output)
```
| 0
|
|
468
|
A
|
24 Game
|
PROGRAMMING
| 1,500
|
[
"constructive algorithms",
"greedy",
"math"
] | null | null |
Little X used to play a card game called "24 Game", but recently he has found it too easy. So he invented a new game.
Initially you have a sequence of *n* integers: 1,<=2,<=...,<=*n*. In a single step, you can pick two of them, let's denote them *a* and *b*, erase them from the sequence, and append to the sequence either *a*<=+<=*b*, or *a*<=-<=*b*, or *a*<=×<=*b*.
After *n*<=-<=1 steps there is only one number left. Can you make this number equal to 24?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105).
|
If it's possible, print "YES" in the first line. Otherwise, print "NO" (without the quotes).
If there is a way to obtain 24 as the result number, in the following *n*<=-<=1 lines print the required operations an operation per line. Each operation should be in form: "*a* *op* *b* = *c*". Where *a* and *b* are the numbers you've picked at this operation; *op* is either "+", or "-", or "*"; *c* is the result of corresponding operation. Note, that the absolute value of *c* mustn't be greater than 1018. The result of the last operation must be equal to 24. Separate operator sign and equality sign from numbers with spaces.
If there are multiple valid answers, you may print any of them.
|
[
"1\n",
"8\n"
] |
[
"NO\n",
"YES\n8 * 7 = 56\n6 * 5 = 30\n3 - 4 = -1\n1 - 2 = -1\n30 - -1 = 31\n56 - 31 = 25\n25 + -1 = 24\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "NO"
},
{
"input": "8",
"output": "YES\n8 * 7 = 56\n6 * 5 = 30\n3 - 4 = -1\n1 - 2 = -1\n30 - -1 = 31\n56 - 31 = 25\n25 + -1 = 24"
},
{
"input": "12",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24"
},
{
"input": "100",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "1000",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "987",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24\n9 - 8 = 1\n24 * 1 = 24\n11 - 10 = 1\n24 * 1 = 24\n13 - 12 = 1\n24 * 1 = 24\n15 - 14 = 1\n24 * 1 = 24\n17 - 16 = 1\n24 * 1 = 24\n19 - 18 = 1\n24 * 1 = 24\n21 - 20 = 1\n24 * 1 = 24\n23 - 22 = 1\n24 * 1 = 24\n25 - 24 = 1\n24 * 1 = 24\n27 - 26 = 1\n24 * 1 = 24\n29 - 28 = 1\n24 * 1 = 24\n31 - 30 = 1\n24 * 1 = 24\n33 - 32 = 1\n24 * 1 = 24\n35 - 34 = 1\n24 * 1 = 24\n37 - 36 = 1\n24 * 1 = 24\n39 - 38 = 1\n24 * 1 = 24\n41 - 40 = 1\n24 * 1 = 2..."
},
{
"input": "2",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24"
},
{
"input": "5",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24"
},
{
"input": "6",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24"
},
{
"input": "7",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24"
},
{
"input": "100000",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "99999",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24\n9 - 8 = 1\n24 * 1 = 24\n11 - 10 = 1\n24 * 1 = 24\n13 - 12 = 1\n24 * 1 = 24\n15 - 14 = 1\n24 * 1 = 24\n17 - 16 = 1\n24 * 1 = 24\n19 - 18 = 1\n24 * 1 = 24\n21 - 20 = 1\n24 * 1 = 24\n23 - 22 = 1\n24 * 1 = 24\n25 - 24 = 1\n24 * 1 = 24\n27 - 26 = 1\n24 * 1 = 24\n29 - 28 = 1\n24 * 1 = 24\n31 - 30 = 1\n24 * 1 = 24\n33 - 32 = 1\n24 * 1 = 24\n35 - 34 = 1\n24 * 1 = 24\n37 - 36 = 1\n24 * 1 = 24\n39 - 38 = 1\n24 * 1 = 24\n41 - 40 = 1\n24 * 1 = 2..."
},
{
"input": "99998",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "99997",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24\n9 - 8 = 1\n24 * 1 = 24\n11 - 10 = 1\n24 * 1 = 24\n13 - 12 = 1\n24 * 1 = 24\n15 - 14 = 1\n24 * 1 = 24\n17 - 16 = 1\n24 * 1 = 24\n19 - 18 = 1\n24 * 1 = 24\n21 - 20 = 1\n24 * 1 = 24\n23 - 22 = 1\n24 * 1 = 24\n25 - 24 = 1\n24 * 1 = 24\n27 - 26 = 1\n24 * 1 = 24\n29 - 28 = 1\n24 * 1 = 24\n31 - 30 = 1\n24 * 1 = 24\n33 - 32 = 1\n24 * 1 = 24\n35 - 34 = 1\n24 * 1 = 24\n37 - 36 = 1\n24 * 1 = 24\n39 - 38 = 1\n24 * 1 = 24\n41 - 40 = 1\n24 * 1 = 2..."
},
{
"input": "580",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "422",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "116",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "447",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24\n9 - 8 = 1\n24 * 1 = 24\n11 - 10 = 1\n24 * 1 = 24\n13 - 12 = 1\n24 * 1 = 24\n15 - 14 = 1\n24 * 1 = 24\n17 - 16 = 1\n24 * 1 = 24\n19 - 18 = 1\n24 * 1 = 24\n21 - 20 = 1\n24 * 1 = 24\n23 - 22 = 1\n24 * 1 = 24\n25 - 24 = 1\n24 * 1 = 24\n27 - 26 = 1\n24 * 1 = 24\n29 - 28 = 1\n24 * 1 = 24\n31 - 30 = 1\n24 * 1 = 24\n33 - 32 = 1\n24 * 1 = 24\n35 - 34 = 1\n24 * 1 = 24\n37 - 36 = 1\n24 * 1 = 24\n39 - 38 = 1\n24 * 1 = 24\n41 - 40 = 1\n24 * 1 = 2..."
},
{
"input": "62052",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "25770",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "56118",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "86351",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24\n9 - 8 = 1\n24 * 1 = 24\n11 - 10 = 1\n24 * 1 = 24\n13 - 12 = 1\n24 * 1 = 24\n15 - 14 = 1\n24 * 1 = 24\n17 - 16 = 1\n24 * 1 = 24\n19 - 18 = 1\n24 * 1 = 24\n21 - 20 = 1\n24 * 1 = 24\n23 - 22 = 1\n24 * 1 = 24\n25 - 24 = 1\n24 * 1 = 24\n27 - 26 = 1\n24 * 1 = 24\n29 - 28 = 1\n24 * 1 = 24\n31 - 30 = 1\n24 * 1 = 24\n33 - 32 = 1\n24 * 1 = 24\n35 - 34 = 1\n24 * 1 = 24\n37 - 36 = 1\n24 * 1 = 24\n39 - 38 = 1\n24 * 1 = 24\n41 - 40 = 1\n24 * 1 = 2..."
},
{
"input": "48108",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "33373",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24\n9 - 8 = 1\n24 * 1 = 24\n11 - 10 = 1\n24 * 1 = 24\n13 - 12 = 1\n24 * 1 = 24\n15 - 14 = 1\n24 * 1 = 24\n17 - 16 = 1\n24 * 1 = 24\n19 - 18 = 1\n24 * 1 = 24\n21 - 20 = 1\n24 * 1 = 24\n23 - 22 = 1\n24 * 1 = 24\n25 - 24 = 1\n24 * 1 = 24\n27 - 26 = 1\n24 * 1 = 24\n29 - 28 = 1\n24 * 1 = 24\n31 - 30 = 1\n24 * 1 = 24\n33 - 32 = 1\n24 * 1 = 24\n35 - 34 = 1\n24 * 1 = 24\n37 - 36 = 1\n24 * 1 = 24\n39 - 38 = 1\n24 * 1 = 24\n41 - 40 = 1\n24 * 1 = 2..."
},
{
"input": "9782",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "19082",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "4",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24"
},
{
"input": "7",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24"
},
{
"input": "3",
"output": "NO"
}
] | 1,609,744,624
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 93
| 0
|
n = int(input())
if n <= 3:
print("NO")
else:
print("YES")
while n > 5:
print("{} - {} = 1".format(n, n-1))
n -= 2
if n == 4:
print("4 * 3 = 12")
print("2 * 1 = 2")
print("12 * 2 = 24")
else:
print("5 * 4 = 20")
print("3 + 2 = 5")
print("5 - 1 = 4")
print("20 + 4 = 24")
|
Title: 24 Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little X used to play a card game called "24 Game", but recently he has found it too easy. So he invented a new game.
Initially you have a sequence of *n* integers: 1,<=2,<=...,<=*n*. In a single step, you can pick two of them, let's denote them *a* and *b*, erase them from the sequence, and append to the sequence either *a*<=+<=*b*, or *a*<=-<=*b*, or *a*<=×<=*b*.
After *n*<=-<=1 steps there is only one number left. Can you make this number equal to 24?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105).
Output Specification:
If it's possible, print "YES" in the first line. Otherwise, print "NO" (without the quotes).
If there is a way to obtain 24 as the result number, in the following *n*<=-<=1 lines print the required operations an operation per line. Each operation should be in form: "*a* *op* *b* = *c*". Where *a* and *b* are the numbers you've picked at this operation; *op* is either "+", or "-", or "*"; *c* is the result of corresponding operation. Note, that the absolute value of *c* mustn't be greater than 1018. The result of the last operation must be equal to 24. Separate operator sign and equality sign from numbers with spaces.
If there are multiple valid answers, you may print any of them.
Demo Input:
['1\n', '8\n']
Demo Output:
['NO\n', 'YES\n8 * 7 = 56\n6 * 5 = 30\n3 - 4 = -1\n1 - 2 = -1\n30 - -1 = 31\n56 - 31 = 25\n25 + -1 = 24\n']
Note:
none
|
```python
n = int(input())
if n <= 3:
print("NO")
else:
print("YES")
while n > 5:
print("{} - {} = 1".format(n, n-1))
n -= 2
if n == 4:
print("4 * 3 = 12")
print("2 * 1 = 2")
print("12 * 2 = 24")
else:
print("5 * 4 = 20")
print("3 + 2 = 5")
print("5 - 1 = 4")
print("20 + 4 = 24")
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
In the army, it isn't easy to form a group of soldiers that will be effective on the battlefield. The communication is crucial and thus no two soldiers should share a name (what would happen if they got an order that Bob is a scouter, if there are two Bobs?).
A group of soldiers is effective if and only if their names are different. For example, a group (John, Bob, Limak) would be effective, while groups (Gary, Bob, Gary) and (Alice, Alice) wouldn't.
You are a spy in the enemy's camp. You noticed *n* soldiers standing in a row, numbered 1 through *n*. The general wants to choose a group of *k* consecutive soldiers. For every *k* consecutive soldiers, the general wrote down whether they would be an effective group or not.
You managed to steal the general's notes, with *n*<=-<=*k*<=+<=1 strings *s*1,<=*s*2,<=...,<=*s**n*<=-<=*k*<=+<=1, each either "YES" or "NO".
- The string *s*1 describes a group of soldiers 1 through *k* ("YES" if the group is effective, and "NO" otherwise). - The string *s*2 describes a group of soldiers 2 through *k*<=+<=1. - And so on, till the string *s**n*<=-<=*k*<=+<=1 that describes a group of soldiers *n*<=-<=*k*<=+<=1 through *n*.
Your task is to find possible names of *n* soldiers. Names should match the stolen notes. Each name should be a string that consists of between 1 and 10 English letters, inclusive. The first letter should be uppercase, and all other letters should be lowercase. Names don't have to be existing names — it's allowed to print "Xyzzzdj" or "T" for example.
Find and print any solution. It can be proved that there always exists at least one solution.
|
The first line of the input contains two integers *n* and *k* (2<=≤<=*k*<=≤<=*n*<=≤<=50) — the number of soldiers and the size of a group respectively.
The second line contains *n*<=-<=*k*<=+<=1 strings *s*1,<=*s*2,<=...,<=*s**n*<=-<=*k*<=+<=1. The string *s**i* is "YES" if the group of soldiers *i* through *i*<=+<=*k*<=-<=1 is effective, and "NO" otherwise.
|
Find any solution satisfying all given conditions. In one line print *n* space-separated strings, denoting possible names of soldiers in the order. The first letter of each name should be uppercase, while the other letters should be lowercase. Each name should contain English letters only and has length from 1 to 10.
If there are multiple valid solutions, print any of them.
|
[
"8 3\nNO NO YES YES YES NO\n",
"9 8\nYES NO\n",
"3 2\nNO NO\n"
] |
[
"Adam Bob Bob Cpqepqwer Limak Adam Bob Adam",
"R Q Ccccccccc Ccocc Ccc So Strong Samples Ccc",
"Na Na Na"
] |
In the first sample, there are 8 soldiers. For every 3 consecutive ones we know whether they would be an effective group. Let's analyze the provided sample output:
- First three soldiers (i.e. Adam, Bob, Bob) wouldn't be an effective group because there are two Bobs. Indeed, the string *s*<sub class="lower-index">1</sub> is "NO". - Soldiers 2 through 4 (Bob, Bob, Cpqepqwer) wouldn't be effective either, and the string *s*<sub class="lower-index">2</sub> is "NO". - Soldiers 3 through 5 (Bob, Cpqepqwer, Limak) would be effective, and the string *s*<sub class="lower-index">3</sub> is "YES". - ..., - Soldiers 6 through 8 (Adam, Bob, Adam) wouldn't be effective, and the string *s*<sub class="lower-index">6</sub> is "NO".
| 0
|
[
{
"input": "8 3\nNO NO YES YES YES NO",
"output": "Ab Ac Ab Ac Af Ag Ah Ag "
},
{
"input": "9 8\nYES NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Ac "
},
{
"input": "3 2\nNO NO",
"output": "Ab Ab Ab "
},
{
"input": "2 2\nYES",
"output": "Ab Ac "
},
{
"input": "2 2\nNO",
"output": "Ab Ab "
},
{
"input": "7 2\nYES NO YES YES NO YES",
"output": "Ab Ac Ac Ae Af Af Ah "
},
{
"input": "18 7\nYES YES YES YES YES YES YES NO NO NO NO NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ai Aj Ak Al Am "
},
{
"input": "50 3\nNO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO YES NO",
"output": "Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Bx Ac "
},
{
"input": "19 15\nNO YES YES YES NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ab Aq Ar As Af "
},
{
"input": "3 2\nNO NO",
"output": "Ab Ab Ab "
},
{
"input": "3 2\nNO YES",
"output": "Ab Ab Ad "
},
{
"input": "3 2\nYES NO",
"output": "Ab Ac Ac "
},
{
"input": "3 2\nYES YES",
"output": "Ab Ac Ad "
},
{
"input": "26 17\nNO YES YES YES NO YES NO YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ab As At Au Af Aw Ah Ay Az Ba "
},
{
"input": "12 2\nYES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am "
},
{
"input": "16 2\nNO NO NO NO NO NO NO NO NO NO NO NO NO NO NO",
"output": "Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab "
},
{
"input": "42 20\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq "
},
{
"input": "37 14\nNO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al "
},
{
"input": "29 10\nYES NO YES NO YES NO YES YES YES YES YES NO NO NO NO NO YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Ac Am Ae Ao Ag Aq Ar As At Au Am Ae Ao Ag Aq Ba Bb Bc Bd "
},
{
"input": "37 3\nYES NO YES NO YES NO YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES NO NO YES NO NO YES YES YES YES NO",
"output": "Ab Ac Ad Ac Af Ac Ah Ac Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Ba Bb Be Bb Be Bh Bi Bj Bk Bj "
},
{
"input": "44 11\nNO NO YES NO YES NO YES YES YES YES YES YES YES YES YES YES YES YES YES NO YES YES YES YES YES NO NO YES NO NO YES YES YES NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Ab Ac An Ae Ap Ag Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Au Bf Bg Bh Bi Bj Ba Bb Bm Bd Au Bp Bq Br Bi "
},
{
"input": "50 49\nNO YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Ab By "
},
{
"input": "50 49\nYES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "50 49\nNO NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Ab Ac "
},
{
"input": "50 49\nYES NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx Ac "
},
{
"input": "46 42\nNO YES YES YES NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Ab Br Bs Bt Af "
},
{
"input": "45 26\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt "
},
{
"input": "45 26\nNO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au "
},
{
"input": "50 3\nNO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES",
"output": "Ab Ac Ab Ae Ab Ag Ab Ai Ab Ak Ab Am Ab Ao Ab Aq Ab As Ab Au Ab Aw Ab Ay Ab Ba Ab Bc Ab Be Ab Bg Ab Bi Ab Bk Ab Bm Ab Bo Ab Bq Ab Bs Ab Bu Ab Bw Ab By "
},
{
"input": "50 2\nNO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO",
"output": "Ab Ab Ad Ad Af Af Ah Ah Aj Aj Al Al An An Ap Ap Ar Ar At At Av Av Ax Ax Az Az Bb Bb Bd Bd Bf Bf Bh Bh Bj Bj Bl Bl Bn Bn Bp Bp Br Br Bt Bt Bv Bv Bx Bx "
},
{
"input": "50 3\nNO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES YES YES YES YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES",
"output": "Ab Ac Ab Ae Ab Ag Ab Ai Ab Ak Ab Am Ab Ao Ab Aq Ab As Ab Au Ab Aw Ab Ay Ab Ba Ab Bc Bd Be Bf Bg Bf Bi Bf Bk Bf Bm Bf Bo Bf Bq Bf Bs Bf Bu Bf Bw Bf By "
},
{
"input": "49 2\nNO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO NO NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES",
"output": "Ab Ab Ad Ad Af Af Ah Ah Aj Aj Al Al An An Ap Ap Ar Ar At At Av Av Ax Ax Ax Ax Bb Bb Bd Bd Bf Bf Bh Bh Bj Bj Bl Bl Bn Bn Bp Bp Br Br Bt Bt Bv Bv Bx "
},
{
"input": "35 22\nNO NO NO NO NO NO NO NO NO NO NO NO NO NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao "
},
{
"input": "46 41\nYES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu "
},
{
"input": "12 4\nYES YES NO NO NO NO NO YES YES",
"output": "Ab Ac Ad Ae Af Ad Ae Af Ad Ae Al Am "
},
{
"input": "50 2\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "50 4\nYES YES YES YES YES NO YES YES YES YES NO NO YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES NO YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Ag Ak Al Am An Al Am Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bc Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "34 5\nYES YES YES YES YES NO YES YES YES YES NO NO YES YES YES NO NO YES NO YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ag Al Am An Ao Al Am Ar As At Am Ar Aw At Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi "
},
{
"input": "50 43\nYES NO YES NO YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Ac Bt Ae Bv Bw Bx By "
},
{
"input": "38 30\nNO NO YES NO YES NO NO NO NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Ab Ac Bg Ae Bi Ag Ah Ai Aj "
},
{
"input": "50 50\nNO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx Ab "
},
{
"input": "50 50\nYES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "5 3\nYES NO YES",
"output": "Ab Ac Ad Ac Af "
},
{
"input": "30 2\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be "
},
{
"input": "50 50\nYES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "27 27\nYES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb "
},
{
"input": "28 2\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc "
},
{
"input": "50 2\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "8 3\nYES NO YES NO YES NO",
"output": "Ab Ac Ad Ac Af Ac Ah Ac "
},
{
"input": "42 30\nNO YES YES NO NO YES NO YES NO YES NO NO YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Ab Bf Bg Ae Af Bj Ah Bl Aj Bn Al Am Bq "
},
{
"input": "50 49\nYES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "50 3\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "7 5\nYES NO YES",
"output": "Ab Ac Ad Ae Af Ac Ah "
},
{
"input": "8 4\nNO YES NO YES NO",
"output": "Ab Ac Ad Ab Af Ad Ah Af "
},
{
"input": "50 50\nNO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx Ab "
},
{
"input": "50 48\nYES NO YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Ac By "
},
{
"input": "29 14\nYES NO YES NO NO YES YES NO NO YES YES NO NO YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ac Aq Ae Af At Au Ai Aj Ax Ay Am An Bb Bc Bd "
},
{
"input": "10 3\nNO YES NO YES NO YES NO YES",
"output": "Ab Ac Ab Ae Ab Ag Ab Ai Ab Ak "
},
{
"input": "10 5\nYES NO YES NO YES NO",
"output": "Ab Ac Ad Ae Af Ac Ah Ae Aj Ac "
}
] | 1,489,853,426
| 2,126
|
Python 3
|
OK
|
TESTS
| 59
| 62
| 4,608,000
|
def get_name(i):
first = i // 26
second = i % 26
return chr(first + ord('A')) + chr(second + ord('a'))
def main():
n, k = map(int, input().split())
commands = input().split()
names = [get_name(i) for i in range(k)]
if commands[0] == "NO":
names[0] = names[1]
ptr = 1
next_id = k + 1
for c in commands[1:]:
if c == "YES":
names.append(get_name(next_id))
next_id += 1
else:
names.append(names[ptr])
ptr += 1
print(' '.join(names))
main()
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the army, it isn't easy to form a group of soldiers that will be effective on the battlefield. The communication is crucial and thus no two soldiers should share a name (what would happen if they got an order that Bob is a scouter, if there are two Bobs?).
A group of soldiers is effective if and only if their names are different. For example, a group (John, Bob, Limak) would be effective, while groups (Gary, Bob, Gary) and (Alice, Alice) wouldn't.
You are a spy in the enemy's camp. You noticed *n* soldiers standing in a row, numbered 1 through *n*. The general wants to choose a group of *k* consecutive soldiers. For every *k* consecutive soldiers, the general wrote down whether they would be an effective group or not.
You managed to steal the general's notes, with *n*<=-<=*k*<=+<=1 strings *s*1,<=*s*2,<=...,<=*s**n*<=-<=*k*<=+<=1, each either "YES" or "NO".
- The string *s*1 describes a group of soldiers 1 through *k* ("YES" if the group is effective, and "NO" otherwise). - The string *s*2 describes a group of soldiers 2 through *k*<=+<=1. - And so on, till the string *s**n*<=-<=*k*<=+<=1 that describes a group of soldiers *n*<=-<=*k*<=+<=1 through *n*.
Your task is to find possible names of *n* soldiers. Names should match the stolen notes. Each name should be a string that consists of between 1 and 10 English letters, inclusive. The first letter should be uppercase, and all other letters should be lowercase. Names don't have to be existing names — it's allowed to print "Xyzzzdj" or "T" for example.
Find and print any solution. It can be proved that there always exists at least one solution.
Input Specification:
The first line of the input contains two integers *n* and *k* (2<=≤<=*k*<=≤<=*n*<=≤<=50) — the number of soldiers and the size of a group respectively.
The second line contains *n*<=-<=*k*<=+<=1 strings *s*1,<=*s*2,<=...,<=*s**n*<=-<=*k*<=+<=1. The string *s**i* is "YES" if the group of soldiers *i* through *i*<=+<=*k*<=-<=1 is effective, and "NO" otherwise.
Output Specification:
Find any solution satisfying all given conditions. In one line print *n* space-separated strings, denoting possible names of soldiers in the order. The first letter of each name should be uppercase, while the other letters should be lowercase. Each name should contain English letters only and has length from 1 to 10.
If there are multiple valid solutions, print any of them.
Demo Input:
['8 3\nNO NO YES YES YES NO\n', '9 8\nYES NO\n', '3 2\nNO NO\n']
Demo Output:
['Adam Bob Bob Cpqepqwer Limak Adam Bob Adam', 'R Q Ccccccccc Ccocc Ccc So Strong Samples Ccc', 'Na Na Na']
Note:
In the first sample, there are 8 soldiers. For every 3 consecutive ones we know whether they would be an effective group. Let's analyze the provided sample output:
- First three soldiers (i.e. Adam, Bob, Bob) wouldn't be an effective group because there are two Bobs. Indeed, the string *s*<sub class="lower-index">1</sub> is "NO". - Soldiers 2 through 4 (Bob, Bob, Cpqepqwer) wouldn't be effective either, and the string *s*<sub class="lower-index">2</sub> is "NO". - Soldiers 3 through 5 (Bob, Cpqepqwer, Limak) would be effective, and the string *s*<sub class="lower-index">3</sub> is "YES". - ..., - Soldiers 6 through 8 (Adam, Bob, Adam) wouldn't be effective, and the string *s*<sub class="lower-index">6</sub> is "NO".
|
```python
def get_name(i):
first = i // 26
second = i % 26
return chr(first + ord('A')) + chr(second + ord('a'))
def main():
n, k = map(int, input().split())
commands = input().split()
names = [get_name(i) for i in range(k)]
if commands[0] == "NO":
names[0] = names[1]
ptr = 1
next_id = k + 1
for c in commands[1:]:
if c == "YES":
names.append(get_name(next_id))
next_id += 1
else:
names.append(names[ptr])
ptr += 1
print(' '.join(names))
main()
```
| 3
|
|
437
|
A
|
The Child and Homework
|
PROGRAMMING
| 1,300
|
[
"implementation"
] | null | null |
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
- If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great. - If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose?
|
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
|
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
|
[
"A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute\n",
"A.ab\nB.abcde\nC.ab\nD.abc\n",
"A.c\nB.cc\nC.c\nD.c\n"
] |
[
"D\n",
"C\n",
"B\n"
] |
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B.
| 500
|
[
{
"input": "A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute",
"output": "D"
},
{
"input": "A.ab\nB.abcde\nC.ab\nD.abc",
"output": "C"
},
{
"input": "A.c\nB.cc\nC.c\nD.c",
"output": "B"
},
{
"input": "A.He_nan_de_yang_guang_zhao_yao_zhe_wo_men_mei_guo_ren_lian_shang_dou_xiao_kai_yan_wahaaaaaaaaaaaaaaaa\nB.Li_bai_li_bai_fei_liu_zhi_xia_san_qian_chi_yi_si_yin_he_luo_jiu_tian_li_bai_li_bai_li_bai_li_bai_shi\nC.Peng_yu_xiang_shi_zai_tai_shen_le_jian_zhi_jiu_shi_ye_jie_du_liu_a_si_mi_da_zhen_shi_tai_shen_le_a_a\nD.Wo_huo_le_si_shi_er_nian_zhen_de_shi_cong_lai_ye_mei_you_jian_guo_zhe_me_biao_zhun_de_yi_bai_ge_zi_a",
"output": "C"
},
{
"input": "A.a___FXIcs_gB____dxFFzst_p_P_Xp_vS__cS_C_ei_\nB.fmnmkS_SeZYx_tSys_d__Exbojv_a_YPEL_BPj__I_aYH\nC._nrPx_j\nD.o_A_UwmNbC_sZ_AXk_Y___i_SN_U_UxrBN_qo_____",
"output": "C"
},
{
"input": "A.G_R__iT_ow_Y__Sm_al__u_____l_ltK\nB.CWRe__h__cbCF\nC._QJ_dVHCL_g_WBsMO__LC____hMNE_DoO__xea_ec\nD.___Zh_",
"output": "D"
},
{
"input": "A.a___FXIcs_gB____dxFFzst_p_P_Xp_vS__cS_C_ei_\nB.fmnmkS_SeZYx_tSys_d__Exbojv_a_YPEL_BPj__I_aYH\nC._nrPx_j\nD.o_A_UwmNbC_sZ_AXk_Y___i_SN_U_UxrBN_qo_____",
"output": "C"
},
{
"input": "A.G_R__iT_ow_Y__Sm_al__u_____l_ltK\nB.CWRe__h__cbCF\nC._QJ_dVHCL_g_WBsMO__LC____hMNE_DoO__xea_ec\nD.___Zh_",
"output": "D"
},
{
"input": "A.ejQ_E_E_G_e_SDjZ__lh_f_K__Z_i_B_U__S__S_EMD_ZEU_Sq\nB.o_JpInEdsrAY_T__D_S\nC.E_Vp_s\nD.a_AU_h",
"output": "A"
},
{
"input": "A.PN_m_P_qgOAMwDyxtbH__Yc__bPOh_wYH___n_Fv_qlZp_\nB._gLeDU__rr_vjrm__O_jl_R__DG___u_XqJjW_\nC.___sHLQzdTzT_tZ_Gs\nD.sZNcVa__M_To_bz_clFi_mH_",
"output": "C"
},
{
"input": "A.bR___cCYJg_Wbt____cxfXfC____c_O_\nB.guM\nC.__bzsH_Of__RjG__u_w_i__PXQL_U_Ow_U_n\nD._nHIuZsu_uU_stRC_k___vD_ZOD_u_z_c_Zf__p_iF_uD_Hdg",
"output": "B"
},
{
"input": "A.x_\nB.__RSiDT_\nC.Ci\nD.KLY_Hc_YN_xXg_DynydumheKTw_PFHo_vqXwm_DY_dA___OS_kG___",
"output": "D"
},
{
"input": "A.yYGJ_C__NYq_\nB.ozMUZ_cKKk_zVUPR_b_g_ygv_HoM__yAxvh__iE\nC.sgHJ___MYP__AWejchRvjSD_o\nD.gkfF_GiOqW_psMT_eS",
"output": "C"
},
{
"input": "A._LYm_nvl_E__RCFZ_IdO\nB.k__qIPO_ivvZyIG__L_\nC.D_SabLm_R___j_HS_t__\nD._adj_R_ngix____GSe_aw__SbOOl_",
"output": "C"
},
{
"input": "A.h_WiYTD_C_h___z_Gn_Th_uNh__g___jm\nB.__HeQaudCJcYfVi__Eg_vryuQrDkb_g__oy_BwX_Mu_\nC._MChdMhQA_UKrf_LGZk_ALTo_mnry_GNNza_X_D_u____ueJb__Y_h__CNUNDfmZATck_ad_XTbG\nD.NV___OoL__GfP_CqhD__RB_____v_T_xi",
"output": "C"
},
{
"input": "A.____JGWsfiU\nB.S_LMq__MpE_oFBs_P\nC.U_Rph_VHpUr____X_jWXbk__ElJTu_Z_wlBpKLTD\nD.p_ysvPNmbrF__",
"output": "C"
},
{
"input": "A.ejQ_E_E_G_e_SDjZ__lh_f_K__Z_i_B_U__S__S_EMD_ZEU_Sq\nB.o_JpInEdsrAY_T__D_S\nC.E_Vp_s\nD.a_AU_h",
"output": "A"
},
{
"input": "A.PN_m_P_qgOAMwDyxtbH__Yc__bPOh_wYH___n_Fv_qlZp_\nB._gLeDU__rr_vjrm__O_jl_R__DG___u_XqJjW_\nC.___sHLQzdTzT_tZ_Gs\nD.sZNcVa__M_To_bz_clFi_mH_",
"output": "C"
},
{
"input": "A.bR___cCYJg_Wbt____cxfXfC____c_O_\nB.guM\nC.__bzsH_Of__RjG__u_w_i__PXQL_U_Ow_U_n\nD._nHIuZsu_uU_stRC_k___vD_ZOD_u_z_c_Zf__p_iF_uD_Hdg",
"output": "B"
},
{
"input": "A.x_\nB.__RSiDT_\nC.Ci\nD.KLY_Hc_YN_xXg_DynydumheKTw_PFHo_vqXwm_DY_dA___OS_kG___",
"output": "D"
},
{
"input": "A.yYGJ_C__NYq_\nB.ozMUZ_cKKk_zVUPR_b_g_ygv_HoM__yAxvh__iE\nC.sgHJ___MYP__AWejchRvjSD_o\nD.gkfF_GiOqW_psMT_eS",
"output": "C"
},
{
"input": "A._LYm_nvl_E__RCFZ_IdO\nB.k__qIPO_ivvZyIG__L_\nC.D_SabLm_R___j_HS_t__\nD._adj_R_ngix____GSe_aw__SbOOl_",
"output": "C"
},
{
"input": "A.h_WiYTD_C_h___z_Gn_Th_uNh__g___jm\nB.__HeQaudCJcYfVi__Eg_vryuQrDkb_g__oy_BwX_Mu_\nC._MChdMhQA_UKrf_LGZk_ALTo_mnry_GNNza_X_D_u____ueJb__Y_h__CNUNDfmZATck_ad_XTbG\nD.NV___OoL__GfP_CqhD__RB_____v_T_xi",
"output": "C"
},
{
"input": "A.____JGWsfiU\nB.S_LMq__MpE_oFBs_P\nC.U_Rph_VHpUr____X_jWXbk__ElJTu_Z_wlBpKLTD\nD.p_ysvPNmbrF__",
"output": "C"
},
{
"input": "A.aaaaaa\nB.aaa\nC.aaa\nD.aaa",
"output": "A"
},
{
"input": "A.aaa\nB.aaaaaa\nC.aaaaaa\nD.aaaaaa",
"output": "A"
},
{
"input": "A.a\nB.b\nC.c\nD.d",
"output": "C"
},
{
"input": "A._\nB.__\nC.____\nD.________",
"output": "C"
},
{
"input": "A.____\nB.________\nC.________\nD._______",
"output": "C"
},
{
"input": "A.h\nB.asdf\nC.asqw\nD.qwertasdfg",
"output": "C"
},
{
"input": "A.aa\nB.aaaaa\nC.aaaaaa\nD.aaaaaaaaaaaaa",
"output": "C"
},
{
"input": "A.ccc\nB.ccccccc\nC.ccc\nD.c",
"output": "C"
},
{
"input": "A.c\nB.ccc\nC.cccccccccccccccccc\nD.cccccc",
"output": "C"
},
{
"input": "A.aa\nB.bb\nC.cc\nD.ddd",
"output": "C"
},
{
"input": "A.QW\nB.WERT\nC.QWER\nD.QWERTYUI",
"output": "C"
}
] | 1,616,254,897
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 27
| 62
| 307,200
|
a=input()
b=input()
c=input()
d=input()
x={}
a=len(a)-2
b=len(b)-2
c=len(c)-2
d=len(d)-2
x['A']=a
x['B']=b
x['C']=c
x['D']=d
c=sorted(x)
c.sort(reverse=True,key=lambda i:x[i])
if x[c[0]]>=x[c[1]]*2:
print(c[0])
elif x[c[-1]]<=x[c[-2]]//2:
print(c[-1])
else:
print("C")
|
Title: The Child and Homework
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
- If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great. - If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose?
Input Specification:
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
Output Specification:
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
Demo Input:
['A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute\n', 'A.ab\nB.abcde\nC.ab\nD.abc\n', 'A.c\nB.cc\nC.c\nD.c\n']
Demo Output:
['D\n', 'C\n', 'B\n']
Note:
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B.
|
```python
a=input()
b=input()
c=input()
d=input()
x={}
a=len(a)-2
b=len(b)-2
c=len(c)-2
d=len(d)-2
x['A']=a
x['B']=b
x['C']=c
x['D']=d
c=sorted(x)
c.sort(reverse=True,key=lambda i:x[i])
if x[c[0]]>=x[c[1]]*2:
print(c[0])
elif x[c[-1]]<=x[c[-2]]//2:
print(c[-1])
else:
print("C")
```
| 0
|
|
777
|
A
|
Shell Game
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"implementation",
"math"
] | null | null |
Bomboslav likes to look out of the window in his room and watch lads outside playing famous shell game. The game is played by two persons: operator and player. Operator takes three similar opaque shells and places a ball beneath one of them. Then he shuffles the shells by swapping some pairs and the player has to guess the current position of the ball.
Bomboslav noticed that guys are not very inventive, so the operator always swaps the left shell with the middle one during odd moves (first, third, fifth, etc.) and always swaps the middle shell with the right one during even moves (second, fourth, etc.).
Let's number shells from 0 to 2 from left to right. Thus the left shell is assigned number 0, the middle shell is 1 and the right shell is 2. Bomboslav has missed the moment when the ball was placed beneath the shell, but he knows that exactly *n* movements were made by the operator and the ball was under shell *x* at the end. Now he wonders, what was the initial position of the ball?
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=2·109) — the number of movements made by the operator.
The second line contains a single integer *x* (0<=≤<=*x*<=≤<=2) — the index of the shell where the ball was found after *n* movements.
|
Print one integer from 0 to 2 — the index of the shell where the ball was initially placed.
|
[
"4\n2\n",
"1\n1\n"
] |
[
"1\n",
"0\n"
] |
In the first sample, the ball was initially placed beneath the middle shell and the operator completed four movements.
1. During the first move operator swapped the left shell and the middle shell. The ball is now under the left shell. 1. During the second move operator swapped the middle shell and the right one. The ball is still under the left shell. 1. During the third move operator swapped the left shell and the middle shell again. The ball is again in the middle. 1. Finally, the operators swapped the middle shell and the right shell. The ball is now beneath the right shell.
| 500
|
[
{
"input": "4\n2",
"output": "1"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "2\n2",
"output": "0"
},
{
"input": "3\n1",
"output": "1"
},
{
"input": "3\n2",
"output": "0"
},
{
"input": "3\n0",
"output": "2"
},
{
"input": "2000000000\n0",
"output": "1"
},
{
"input": "2\n0",
"output": "1"
},
{
"input": "2\n1",
"output": "2"
},
{
"input": "4\n0",
"output": "2"
},
{
"input": "4\n1",
"output": "0"
},
{
"input": "5\n0",
"output": "0"
},
{
"input": "5\n1",
"output": "2"
},
{
"input": "5\n2",
"output": "1"
},
{
"input": "6\n0",
"output": "0"
},
{
"input": "6\n1",
"output": "1"
},
{
"input": "6\n2",
"output": "2"
},
{
"input": "7\n0",
"output": "1"
},
{
"input": "7\n1",
"output": "0"
},
{
"input": "7\n2",
"output": "2"
},
{
"input": "100000\n0",
"output": "2"
},
{
"input": "100000\n1",
"output": "0"
},
{
"input": "100000\n2",
"output": "1"
},
{
"input": "99999\n1",
"output": "1"
},
{
"input": "99998\n1",
"output": "2"
},
{
"input": "99997\n1",
"output": "0"
},
{
"input": "99996\n1",
"output": "1"
},
{
"input": "99995\n1",
"output": "2"
},
{
"input": "1999999995\n0",
"output": "2"
},
{
"input": "1999999995\n1",
"output": "1"
},
{
"input": "1999999995\n2",
"output": "0"
},
{
"input": "1999999996\n0",
"output": "2"
},
{
"input": "1999999996\n1",
"output": "0"
},
{
"input": "1999999996\n2",
"output": "1"
},
{
"input": "1999999997\n0",
"output": "0"
},
{
"input": "1999999997\n1",
"output": "2"
},
{
"input": "1999999997\n2",
"output": "1"
},
{
"input": "1999999998\n0",
"output": "0"
},
{
"input": "1999999998\n1",
"output": "1"
},
{
"input": "1999999998\n2",
"output": "2"
},
{
"input": "1999999999\n0",
"output": "1"
},
{
"input": "1999999999\n1",
"output": "0"
},
{
"input": "1999999999\n2",
"output": "2"
},
{
"input": "2000000000\n1",
"output": "2"
},
{
"input": "2000000000\n2",
"output": "0"
},
{
"input": "1234567890\n0",
"output": "0"
},
{
"input": "1234567890\n1",
"output": "1"
},
{
"input": "1234567890\n2",
"output": "2"
},
{
"input": "123456789\n0",
"output": "2"
},
{
"input": "123456789\n1",
"output": "1"
},
{
"input": "123456789\n2",
"output": "0"
},
{
"input": "123456790\n0",
"output": "2"
},
{
"input": "12\n2",
"output": "2"
},
{
"input": "32\n1",
"output": "2"
},
{
"input": "20\n2",
"output": "0"
},
{
"input": "10\n1",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "76994383\n1",
"output": "0"
},
{
"input": "25\n2",
"output": "2"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "12\n0",
"output": "0"
},
{
"input": "150\n2",
"output": "2"
},
{
"input": "15\n0",
"output": "2"
},
{
"input": "21\n2",
"output": "0"
},
{
"input": "18\n2",
"output": "2"
},
{
"input": "8\n2",
"output": "0"
},
{
"input": "10\n0",
"output": "2"
},
{
"input": "16\n0",
"output": "2"
}
] | 1,666,785,383
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 6
| 500
| 0
|
steps = int(input())
result = int(input())
thimbles = [0, 0, 0]
thimbles[result] = 1
for i in range(steps, 0, -1):
if i % 2 == 0 and (thimbles[1] == 0 or thimbles[2] == 0):
thimbles[1], thimbles[2] = thimbles[2], thimbles[1]
elif i%2 != 0 and (thimbles[0] == 0 or thimbles[1] == 0):
thimbles[0], thimbles[1] = thimbles[1], thimbles[0]
print(thimbles.index(1))
|
Title: Shell Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bomboslav likes to look out of the window in his room and watch lads outside playing famous shell game. The game is played by two persons: operator and player. Operator takes three similar opaque shells and places a ball beneath one of them. Then he shuffles the shells by swapping some pairs and the player has to guess the current position of the ball.
Bomboslav noticed that guys are not very inventive, so the operator always swaps the left shell with the middle one during odd moves (first, third, fifth, etc.) and always swaps the middle shell with the right one during even moves (second, fourth, etc.).
Let's number shells from 0 to 2 from left to right. Thus the left shell is assigned number 0, the middle shell is 1 and the right shell is 2. Bomboslav has missed the moment when the ball was placed beneath the shell, but he knows that exactly *n* movements were made by the operator and the ball was under shell *x* at the end. Now he wonders, what was the initial position of the ball?
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=2·109) — the number of movements made by the operator.
The second line contains a single integer *x* (0<=≤<=*x*<=≤<=2) — the index of the shell where the ball was found after *n* movements.
Output Specification:
Print one integer from 0 to 2 — the index of the shell where the ball was initially placed.
Demo Input:
['4\n2\n', '1\n1\n']
Demo Output:
['1\n', '0\n']
Note:
In the first sample, the ball was initially placed beneath the middle shell and the operator completed four movements.
1. During the first move operator swapped the left shell and the middle shell. The ball is now under the left shell. 1. During the second move operator swapped the middle shell and the right one. The ball is still under the left shell. 1. During the third move operator swapped the left shell and the middle shell again. The ball is again in the middle. 1. Finally, the operators swapped the middle shell and the right shell. The ball is now beneath the right shell.
|
```python
steps = int(input())
result = int(input())
thimbles = [0, 0, 0]
thimbles[result] = 1
for i in range(steps, 0, -1):
if i % 2 == 0 and (thimbles[1] == 0 or thimbles[2] == 0):
thimbles[1], thimbles[2] = thimbles[2], thimbles[1]
elif i%2 != 0 and (thimbles[0] == 0 or thimbles[1] == 0):
thimbles[0], thimbles[1] = thimbles[1], thimbles[0]
print(thimbles.index(1))
```
| 0
|
|
538
|
B
|
Quasi Binary
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"dp",
"greedy",
"implementation"
] | null | null |
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106).
|
In the first line print a single integer *k* — the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
|
[
"9\n",
"32\n"
] |
[
"9\n1 1 1 1 1 1 1 1 1 \n",
"3\n10 11 11 \n"
] |
none
| 1,000
|
[
{
"input": "9",
"output": "9\n1 1 1 1 1 1 1 1 1 "
},
{
"input": "32",
"output": "3\n10 11 11 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "415",
"output": "5\n1 101 101 101 111 "
},
{
"input": "10011",
"output": "1\n10011 "
},
{
"input": "10201",
"output": "2\n100 10101 "
},
{
"input": "314159",
"output": "9\n1 1 1 1 11 1011 101011 101011 111111 "
},
{
"input": "999999",
"output": "9\n111111 111111 111111 111111 111111 111111 111111 111111 111111 "
},
{
"input": "2",
"output": "2\n1 1 "
},
{
"input": "10",
"output": "1\n10 "
},
{
"input": "21",
"output": "2\n10 11 "
},
{
"input": "98",
"output": "9\n10 11 11 11 11 11 11 11 11 "
},
{
"input": "102030",
"output": "3\n10 1010 101010 "
},
{
"input": "909090",
"output": "9\n101010 101010 101010 101010 101010 101010 101010 101010 101010 "
},
{
"input": "909823",
"output": "9\n101000 101100 101100 101100 101100 101100 101101 101111 101111 "
},
{
"input": "1000000",
"output": "1\n1000000 "
},
{
"input": "111111",
"output": "1\n111111 "
},
{
"input": "123456",
"output": "6\n1 11 111 1111 11111 111111 "
},
{
"input": "987654",
"output": "9\n100000 110000 111000 111100 111110 111111 111111 111111 111111 "
},
{
"input": "908172",
"output": "9\n100000 101000 101010 101010 101010 101010 101010 101011 101111 "
},
{
"input": "8",
"output": "8\n1 1 1 1 1 1 1 1 "
},
{
"input": "100009",
"output": "9\n1 1 1 1 1 1 1 1 100001 "
},
{
"input": "900000",
"output": "9\n100000 100000 100000 100000 100000 100000 100000 100000 100000 "
},
{
"input": "1435",
"output": "5\n1 101 111 111 1111 "
},
{
"input": "1453",
"output": "5\n10 110 111 111 1111 "
}
] | 1,520,508,998
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 5,632,000
|
digits = list(map(int, list(input())))
max_len = max(digits)
answer = []
for i, d in enumerate(reversed(digits)):
row = [10**i] * d + [0] * (max_len - d)
answer.append(row)
print(answer)
result = ''
for j in range(max_len):
result += str(sum([row[j] for row in answer])) + ' '
print(max_len)
print(result)
|
Title: Quasi Binary
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106).
Output Specification:
In the first line print a single integer *k* — the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Demo Input:
['9\n', '32\n']
Demo Output:
['9\n1 1 1 1 1 1 1 1 1 \n', '3\n10 11 11 \n']
Note:
none
|
```python
digits = list(map(int, list(input())))
max_len = max(digits)
answer = []
for i, d in enumerate(reversed(digits)):
row = [10**i] * d + [0] * (max_len - d)
answer.append(row)
print(answer)
result = ''
for j in range(max_len):
result += str(sum([row[j] for row in answer])) + ' '
print(max_len)
print(result)
```
| 0
|
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SQL Console for MatrixStudio/Codeforces-Python-Submissions
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SQL Console for MatrixStudio/Codeforces-Python-Submissions
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