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3.99
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
810
|
B
|
Summer sell-off
|
PROGRAMMING
| 1,300
|
[
"greedy",
"sortings"
] | null | null |
Summer holidays! Someone is going on trips, someone is visiting grandparents, but someone is trying to get a part-time job. This summer Noora decided that she wants to earn some money, and took a job in a shop as an assistant.
Shop, where Noora is working, has a plan on the following *n* days. For each day sales manager knows exactly, that in *i*-th day *k**i* products will be put up for sale and exactly *l**i* clients will come to the shop that day. Also, the manager is sure, that everyone, who comes to the shop, buys exactly one product or, if there aren't any left, leaves the shop without buying anything. Moreover, due to the short shelf-life of the products, manager established the following rule: if some part of the products left on the shelves at the end of the day, that products aren't kept on the next day and are sent to the dump.
For advertising purposes manager offered to start a sell-out in the shop. He asked Noora to choose any *f* days from *n* next for sell-outs. On each of *f* chosen days the number of products were put up for sale would be doubled. Thus, if on *i*-th day shop planned to put up for sale *k**i* products and Noora has chosen this day for sell-out, shelves of the shop would keep 2·*k**i* products. Consequently, there is an opportunity to sell two times more products on days of sell-out.
Noora's task is to choose *f* days to maximize total number of sold products. She asks you to help her with such a difficult problem.
|
The first line contains two integers *n* and *f* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*f*<=≤<=*n*) denoting the number of days in shop's plan and the number of days that Noora has to choose for sell-out.
Each line of the following *n* subsequent lines contains two integers *k**i*,<=*l**i* (0<=≤<=*k**i*,<=*l**i*<=≤<=109) denoting the number of products on the shelves of the shop on the *i*-th day and the number of clients that will come to the shop on *i*-th day.
|
Print a single integer denoting the maximal number of products that shop can sell.
|
[
"4 2\n2 1\n3 5\n2 3\n1 5\n",
"4 1\n0 2\n0 3\n3 5\n0 6\n"
] |
[
"10",
"5"
] |
In the first example we can choose days with numbers 2 and 4 for sell-out. In this case new numbers of products for sale would be equal to [2, 6, 2, 2] respectively. So on the first day shop will sell 1 product, on the second — 5, on the third — 2, on the fourth — 2. In total 1 + 5 + 2 + 2 = 10 product units.
In the second example it is possible to sell 5 products, if you choose third day for sell-out.
| 1,000
|
[
{
"input": "4 2\n2 1\n3 5\n2 3\n1 5",
"output": "10"
},
{
"input": "4 1\n0 2\n0 3\n3 5\n0 6",
"output": "5"
},
{
"input": "1 1\n5 8",
"output": "8"
},
{
"input": "2 1\n8 12\n6 11",
"output": "19"
},
{
"input": "2 1\n6 7\n5 7",
"output": "13"
},
{
"input": "2 1\n5 7\n6 7",
"output": "13"
},
{
"input": "2 1\n7 8\n3 6",
"output": "13"
},
{
"input": "2 1\n9 10\n5 8",
"output": "17"
},
{
"input": "2 1\n3 6\n7 8",
"output": "13"
},
{
"input": "1 0\n10 20",
"output": "10"
},
{
"input": "2 1\n99 100\n3 6",
"output": "105"
},
{
"input": "4 2\n2 10\n3 10\n9 9\n5 10",
"output": "27"
},
{
"input": "2 1\n3 4\n2 8",
"output": "7"
},
{
"input": "50 2\n74 90\n68 33\n49 88\n52 13\n73 21\n77 63\n27 62\n8 52\n60 57\n42 83\n98 15\n79 11\n77 46\n55 91\n72 100\n70 86\n50 51\n57 39\n20 54\n64 95\n66 22\n79 64\n31 28\n11 89\n1 36\n13 4\n75 62\n16 62\n100 35\n43 96\n97 54\n86 33\n62 63\n94 24\n19 6\n20 58\n38 38\n11 76\n70 40\n44 24\n32 96\n28 100\n62 45\n41 68\n90 52\n16 0\n98 32\n81 79\n67 82\n28 2",
"output": "1889"
},
{
"input": "2 1\n10 5\n2 4",
"output": "9"
},
{
"input": "2 1\n50 51\n30 40",
"output": "90"
},
{
"input": "3 2\n5 10\n5 10\n7 9",
"output": "27"
},
{
"input": "3 1\n1000 1000\n50 100\n2 2",
"output": "1102"
},
{
"input": "2 1\n2 4\n12 12",
"output": "16"
},
{
"input": "2 1\n4 4\n1 2",
"output": "6"
},
{
"input": "2 1\n4000 4000\n1 2",
"output": "4002"
},
{
"input": "2 1\n5 6\n2 4",
"output": "9"
},
{
"input": "3 2\n10 10\n10 10\n1 2",
"output": "22"
},
{
"input": "10 5\n9 1\n11 1\n12 1\n13 1\n14 1\n2 4\n2 4\n2 4\n2 4\n2 4",
"output": "25"
},
{
"input": "2 1\n30 30\n10 20",
"output": "50"
},
{
"input": "1 1\n1 1",
"output": "1"
},
{
"input": "2 1\n10 2\n2 10",
"output": "6"
},
{
"input": "2 1\n4 5\n3 9",
"output": "10"
},
{
"input": "2 1\n100 100\n5 10",
"output": "110"
},
{
"input": "2 1\n14 28\n15 28",
"output": "43"
},
{
"input": "2 1\n100 1\n20 40",
"output": "41"
},
{
"input": "2 1\n5 10\n6 10",
"output": "16"
},
{
"input": "2 1\n29 30\n10 20",
"output": "49"
},
{
"input": "1 0\n12 12",
"output": "12"
},
{
"input": "2 1\n7 8\n4 7",
"output": "14"
},
{
"input": "2 1\n5 5\n2 4",
"output": "9"
},
{
"input": "2 1\n1 2\n228 2",
"output": "4"
},
{
"input": "2 1\n5 10\n100 20",
"output": "30"
},
{
"input": "2 1\n1000 1001\n2 4",
"output": "1004"
},
{
"input": "2 1\n3 9\n7 7",
"output": "13"
},
{
"input": "2 0\n1 1\n1 1",
"output": "2"
},
{
"input": "4 1\n10 10\n10 10\n10 10\n4 6",
"output": "36"
},
{
"input": "18 13\n63 8\n87 100\n18 89\n35 29\n66 81\n27 85\n64 51\n60 52\n32 94\n74 22\n86 31\n43 78\n12 2\n36 2\n67 23\n2 16\n78 71\n34 64",
"output": "772"
},
{
"input": "2 1\n10 18\n17 19",
"output": "35"
},
{
"input": "3 0\n1 1\n1 1\n1 1",
"output": "3"
},
{
"input": "2 1\n4 7\n8 9",
"output": "15"
},
{
"input": "4 2\n2 10\n3 10\n9 10\n5 10",
"output": "27"
},
{
"input": "2 1\n5 7\n3 6",
"output": "11"
},
{
"input": "2 1\n3 4\n12 12",
"output": "16"
},
{
"input": "2 1\n10 11\n9 20",
"output": "28"
},
{
"input": "2 1\n7 8\n2 4",
"output": "11"
},
{
"input": "2 1\n5 10\n7 10",
"output": "17"
},
{
"input": "4 2\n2 10\n3 10\n5 10\n9 10",
"output": "27"
},
{
"input": "2 1\n99 100\n5 10",
"output": "109"
},
{
"input": "4 2\n2 10\n3 10\n5 10\n9 9",
"output": "27"
},
{
"input": "2 1\n3 7\n5 7",
"output": "11"
},
{
"input": "2 1\n10 10\n3 6",
"output": "16"
},
{
"input": "2 1\n100 1\n2 4",
"output": "5"
},
{
"input": "5 0\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "5"
},
{
"input": "3 1\n3 7\n4 5\n2 3",
"output": "12"
},
{
"input": "2 1\n3 9\n7 8",
"output": "13"
},
{
"input": "2 1\n10 2\n3 4",
"output": "6"
},
{
"input": "2 1\n40 40\n3 5",
"output": "45"
},
{
"input": "2 1\n5 3\n1 2",
"output": "5"
},
{
"input": "10 5\n9 5\n10 5\n11 5\n12 5\n13 5\n2 4\n2 4\n2 4\n2 4\n2 4",
"output": "45"
},
{
"input": "3 1\n1 5\n1 5\n4 4",
"output": "7"
},
{
"input": "4 0\n1 1\n1 1\n1 1\n1 1",
"output": "4"
},
{
"input": "4 1\n1000 1001\n1000 1001\n2 4\n1 2",
"output": "2005"
},
{
"input": "2 1\n15 30\n50 59",
"output": "80"
},
{
"input": "2 1\n8 8\n3 5",
"output": "13"
},
{
"input": "2 1\n4 5\n2 5",
"output": "8"
},
{
"input": "3 2\n3 3\n1 2\n1 2",
"output": "7"
},
{
"input": "3 1\n2 5\n2 5\n4 4",
"output": "10"
},
{
"input": "2 1\n3 10\n50 51",
"output": "56"
},
{
"input": "4 2\n2 4\n2 4\n9 10\n9 10",
"output": "26"
},
{
"input": "2 1\n3 5\n8 8",
"output": "13"
},
{
"input": "2 1\n100 150\n70 150",
"output": "240"
},
{
"input": "2 1\n4 5\n3 6",
"output": "10"
},
{
"input": "2 1\n20 10\n3 5",
"output": "15"
},
{
"input": "15 13\n76167099 92301116\n83163126 84046805\n45309500 65037149\n29982002 77381688\n76738161 52935441\n37889502 25466134\n55955619 14197941\n31462620 12999429\n64648384 8824773\n3552934 68992494\n2823376 9338427\n86832070 3763091\n67753633 2162190\n302887 92011825\n84894984 410533",
"output": "435467000"
},
{
"input": "2 1\n8 7\n3 6",
"output": "13"
},
{
"input": "2 1\n7 8\n3 5",
"output": "12"
},
{
"input": "2 1\n10 10\n1 3",
"output": "12"
},
{
"input": "2 1\n9 10\n2 4",
"output": "13"
},
{
"input": "3 1\n10 11\n12 13\n8 10",
"output": "32"
},
{
"input": "2 1\n5 10\n7 7",
"output": "17"
},
{
"input": "4 2\n90 91\n2 10\n2 10\n2 10",
"output": "100"
},
{
"input": "2 1\n2 4\n4 4",
"output": "8"
},
{
"input": "2 1\n2 3\n4 3",
"output": "6"
},
{
"input": "2 1\n40 45\n50 52",
"output": "95"
},
{
"input": "3 1\n1 4\n2 4\n3 4",
"output": "8"
},
{
"input": "2 1\n1 2\n1000 1000",
"output": "1002"
},
{
"input": "2 1\n80 100\n70 95",
"output": "175"
}
] | 1,497,789,427
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 15
| 421
| 9,011,200
|
def minimum(a,b):
if(a<b):
return a
else:
return b
n,f=input().split()
n=int(n)
f=int(f)
A=[]
ans=0
for i in range(0,n):
x,y=input().split()
x=int(x)
y=int(y)
if(x>0):
if(y<=x):
ans+=y
else:
A.append([x,y,minimum(2*x,y)])
A.sort(key=lambda x:x[2],reverse=True)
for i in range(0,minimum(f,len(A))):
ans+=A[i][2]
for i in range(f,len(A)):
ans+=A[i][0]
print(ans)
|
Title: Summer sell-off
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Summer holidays! Someone is going on trips, someone is visiting grandparents, but someone is trying to get a part-time job. This summer Noora decided that she wants to earn some money, and took a job in a shop as an assistant.
Shop, where Noora is working, has a plan on the following *n* days. For each day sales manager knows exactly, that in *i*-th day *k**i* products will be put up for sale and exactly *l**i* clients will come to the shop that day. Also, the manager is sure, that everyone, who comes to the shop, buys exactly one product or, if there aren't any left, leaves the shop without buying anything. Moreover, due to the short shelf-life of the products, manager established the following rule: if some part of the products left on the shelves at the end of the day, that products aren't kept on the next day and are sent to the dump.
For advertising purposes manager offered to start a sell-out in the shop. He asked Noora to choose any *f* days from *n* next for sell-outs. On each of *f* chosen days the number of products were put up for sale would be doubled. Thus, if on *i*-th day shop planned to put up for sale *k**i* products and Noora has chosen this day for sell-out, shelves of the shop would keep 2·*k**i* products. Consequently, there is an opportunity to sell two times more products on days of sell-out.
Noora's task is to choose *f* days to maximize total number of sold products. She asks you to help her with such a difficult problem.
Input Specification:
The first line contains two integers *n* and *f* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*f*<=≤<=*n*) denoting the number of days in shop's plan and the number of days that Noora has to choose for sell-out.
Each line of the following *n* subsequent lines contains two integers *k**i*,<=*l**i* (0<=≤<=*k**i*,<=*l**i*<=≤<=109) denoting the number of products on the shelves of the shop on the *i*-th day and the number of clients that will come to the shop on *i*-th day.
Output Specification:
Print a single integer denoting the maximal number of products that shop can sell.
Demo Input:
['4 2\n2 1\n3 5\n2 3\n1 5\n', '4 1\n0 2\n0 3\n3 5\n0 6\n']
Demo Output:
['10', '5']
Note:
In the first example we can choose days with numbers 2 and 4 for sell-out. In this case new numbers of products for sale would be equal to [2, 6, 2, 2] respectively. So on the first day shop will sell 1 product, on the second — 5, on the third — 2, on the fourth — 2. In total 1 + 5 + 2 + 2 = 10 product units.
In the second example it is possible to sell 5 products, if you choose third day for sell-out.
|
```python
def minimum(a,b):
if(a<b):
return a
else:
return b
n,f=input().split()
n=int(n)
f=int(f)
A=[]
ans=0
for i in range(0,n):
x,y=input().split()
x=int(x)
y=int(y)
if(x>0):
if(y<=x):
ans+=y
else:
A.append([x,y,minimum(2*x,y)])
A.sort(key=lambda x:x[2],reverse=True)
for i in range(0,minimum(f,len(A))):
ans+=A[i][2]
for i in range(f,len(A)):
ans+=A[i][0]
print(ans)
```
| 0
|
|
678
|
C
|
Joty and Chocolate
|
PROGRAMMING
| 1,600
|
[
"implementation",
"math",
"number theory"
] | null | null |
Little Joty has got a task to do. She has a line of *n* tiles indexed from 1 to *n*. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by *a* and an unpainted tile should be painted Blue if it's index is divisible by *b*. So the tile with the number divisible by *a* and *b* can be either painted Red or Blue.
After her painting is done, she will get *p* chocolates for each tile that is painted Red and *q* chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
|
The only line contains five integers *n*, *a*, *b*, *p* and *q* (1<=≤<=*n*,<=*a*,<=*b*,<=*p*,<=*q*<=≤<=109).
|
Print the only integer *s* — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
|
[
"5 2 3 12 15\n",
"20 2 3 3 5\n"
] |
[
"39\n",
"51\n"
] |
none
| 0
|
[
{
"input": "5 2 3 12 15",
"output": "39"
},
{
"input": "20 2 3 3 5",
"output": "51"
},
{
"input": "1 1 1 1 1",
"output": "1"
},
{
"input": "1 2 2 2 2",
"output": "0"
},
{
"input": "2 1 3 3 3",
"output": "6"
},
{
"input": "3 1 1 3 3",
"output": "9"
},
{
"input": "4 1 5 4 3",
"output": "16"
},
{
"input": "8 8 1 1 1",
"output": "8"
},
{
"input": "15 14 32 65 28",
"output": "65"
},
{
"input": "894 197 325 232 902",
"output": "2732"
},
{
"input": "8581 6058 3019 2151 4140",
"output": "10431"
},
{
"input": "41764 97259 54586 18013 75415",
"output": "0"
},
{
"input": "333625 453145 800800 907251 446081",
"output": "0"
},
{
"input": "4394826 2233224 609367 3364334 898489",
"output": "9653757"
},
{
"input": "13350712 76770926 61331309 8735000 9057368",
"output": "0"
},
{
"input": "142098087 687355301 987788392 75187408 868856364",
"output": "0"
},
{
"input": "1000000000 1 3 1000000000 999999999",
"output": "1000000000000000000"
},
{
"input": "6 6 2 8 2",
"output": "12"
},
{
"input": "500 8 4 4 5",
"output": "625"
},
{
"input": "20 4 6 2 3",
"output": "17"
},
{
"input": "10 3 9 1 2",
"output": "4"
},
{
"input": "120 18 6 3 5",
"output": "100"
},
{
"input": "30 4 6 2 2",
"output": "20"
},
{
"input": "1000000000 7171 2727 191 272",
"output": "125391842"
},
{
"input": "5 2 2 4 1",
"output": "8"
},
{
"input": "1000000000 2 2 3 3",
"output": "1500000000"
},
{
"input": "24 4 6 5 7",
"output": "48"
},
{
"input": "216 6 36 10 100",
"output": "900"
},
{
"input": "100 12 6 1 10",
"output": "160"
},
{
"input": "1000 4 8 3 5",
"output": "1000"
},
{
"input": "10 2 4 3 6",
"output": "21"
},
{
"input": "1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "10 5 10 2 3",
"output": "5"
},
{
"input": "100000 3 9 1 2",
"output": "44444"
},
{
"input": "10 2 4 1 100",
"output": "203"
},
{
"input": "20 6 4 2 3",
"output": "19"
},
{
"input": "1200 4 16 2 3",
"output": "675"
},
{
"input": "7 2 4 7 9",
"output": "23"
},
{
"input": "24 6 4 15 10",
"output": "100"
},
{
"input": "50 2 8 15 13",
"output": "375"
},
{
"input": "100 4 6 12 15",
"output": "444"
},
{
"input": "56756 9 18 56 78",
"output": "422502"
},
{
"input": "10000 4 6 10 12",
"output": "36662"
},
{
"input": "20 2 4 3 5",
"output": "40"
},
{
"input": "24 4 6 10 100",
"output": "440"
},
{
"input": "12 2 4 5 6",
"output": "33"
},
{
"input": "100 2 4 1 100",
"output": "2525"
},
{
"input": "1000 4 6 50 50",
"output": "16650"
},
{
"input": "60 12 6 12 15",
"output": "150"
},
{
"input": "1000 2 4 5 6",
"output": "2750"
},
{
"input": "1000000000 1 1 9999 5555",
"output": "9999000000000"
},
{
"input": "50 2 2 4 5",
"output": "125"
},
{
"input": "14 4 2 2 3",
"output": "21"
},
{
"input": "100 3 9 1 2",
"output": "44"
},
{
"input": "1000000000 4 6 1 1000000000",
"output": "166666666166666667"
},
{
"input": "12 3 3 45 4",
"output": "180"
},
{
"input": "12 2 4 5 9",
"output": "42"
},
{
"input": "1000000000 2 2 1000000000 1000000000",
"output": "500000000000000000"
},
{
"input": "50 4 8 5 6",
"output": "66"
},
{
"input": "32 4 16 6 3",
"output": "48"
},
{
"input": "10000 2 4 1 1",
"output": "5000"
},
{
"input": "8 2 4 100 1",
"output": "400"
},
{
"input": "20 4 2 10 1",
"output": "55"
},
{
"input": "5 2 2 12 15",
"output": "30"
},
{
"input": "20 2 12 5 6",
"output": "51"
},
{
"input": "10 2 4 1 2",
"output": "7"
},
{
"input": "32 4 16 3 6",
"output": "30"
},
{
"input": "50 2 8 13 15",
"output": "337"
},
{
"input": "12 6 4 10 9",
"output": "38"
},
{
"input": "1000000000 999999998 999999999 999999998 999999999",
"output": "1999999997"
},
{
"input": "20 2 4 10 20",
"output": "150"
},
{
"input": "13 4 6 12 15",
"output": "54"
},
{
"input": "30 3 6 5 7",
"output": "60"
},
{
"input": "7 2 4 2 1",
"output": "6"
},
{
"input": "100000 32 16 2 3",
"output": "18750"
},
{
"input": "6 2 6 1 1",
"output": "3"
},
{
"input": "999999999 180 192 46642017 28801397",
"output": "399129078526502"
},
{
"input": "12 4 6 1 1",
"output": "4"
},
{
"input": "10 2 4 10 5",
"output": "50"
},
{
"input": "1000000 4 6 12 14",
"output": "4333328"
},
{
"input": "2000 20 30 3 5",
"output": "531"
},
{
"input": "1000000000 1 2 1 1",
"output": "1000000000"
},
{
"input": "30 3 15 10 3",
"output": "100"
},
{
"input": "1000 2 4 1 100",
"output": "25250"
},
{
"input": "6 3 3 12 15",
"output": "30"
},
{
"input": "24 4 6 1 1",
"output": "8"
},
{
"input": "20 2 12 4 5",
"output": "41"
},
{
"input": "1000000000 9 15 10 10",
"output": "1555555550"
},
{
"input": "16 2 4 1 2",
"output": "12"
},
{
"input": "100000 4 6 12 14",
"output": "433328"
},
{
"input": "24 6 4 1 1",
"output": "8"
},
{
"input": "1000000 4 6 12 15",
"output": "4499994"
},
{
"input": "100 2 4 5 6",
"output": "275"
},
{
"input": "10 3 9 12 15",
"output": "39"
},
{
"input": "1000000000 1 1 999999999 999999999",
"output": "999999999000000000"
},
{
"input": "6 2 4 2 3",
"output": "7"
},
{
"input": "2 2 2 2 2",
"output": "2"
},
{
"input": "6 6 2 1 1",
"output": "3"
},
{
"input": "100 2 4 3 7",
"output": "250"
},
{
"input": "1000000 32 16 2 5",
"output": "312500"
},
{
"input": "100 20 15 50 25",
"output": "375"
},
{
"input": "1000000000 100000007 100000013 10 3",
"output": "117"
},
{
"input": "1000000000 9999999 99999998 3 3",
"output": "330"
},
{
"input": "10077696 24 36 10 100",
"output": "30792960"
},
{
"input": "392852503 148746166 420198270 517065752 906699795",
"output": "1034131504"
},
{
"input": "536870912 60000 72000 271828 314159",
"output": "4369119072"
},
{
"input": "730114139 21550542 204644733 680083361 11353255",
"output": "22476810678"
},
{
"input": "538228881 766493289 791886544 468896052 600136703",
"output": "0"
},
{
"input": "190 20 50 84 172",
"output": "1188"
},
{
"input": "1000 5 10 80 90",
"output": "17000"
},
{
"input": "99999999 999999998 1 271828 314159",
"output": "31415899685841"
},
{
"input": "22 3 6 1243 1",
"output": "8701"
},
{
"input": "15 10 5 2 2",
"output": "6"
},
{
"input": "1000000000 1000000000 1 1000000000 1000000000",
"output": "1000000000000000000"
},
{
"input": "62 62 42 78 124",
"output": "202"
},
{
"input": "2 2 2 2 1",
"output": "2"
},
{
"input": "864351351 351 313 531 11",
"output": "1337898227"
},
{
"input": "26 3 6 1244 1",
"output": "9952"
},
{
"input": "1000 4 6 7 3",
"output": "1999"
},
{
"input": "134312 3 6 33333 1",
"output": "1492318410"
},
{
"input": "100 4 6 17 18",
"output": "577"
},
{
"input": "6 2 4 5 6",
"output": "16"
},
{
"input": "8 2 4 10 1",
"output": "40"
},
{
"input": "10 2 4 3 3",
"output": "15"
},
{
"input": "1000 1000 1000 1000 1000",
"output": "1000"
},
{
"input": "123123 3 6 34312 2",
"output": "1408198792"
},
{
"input": "1000000000 25 5 999 999",
"output": "199800000000"
},
{
"input": "100 4 2 5 12",
"output": "600"
},
{
"input": "50 2 4 4 5",
"output": "112"
},
{
"input": "24 4 6 100 333",
"output": "1732"
},
{
"input": "216 24 36 10 100",
"output": "660"
},
{
"input": "50 6 4 3 8",
"output": "108"
},
{
"input": "146 76 2 178 192",
"output": "14016"
},
{
"input": "55 8 6 11 20",
"output": "224"
},
{
"input": "5 2 4 6 16",
"output": "22"
},
{
"input": "54 2 52 50 188",
"output": "1488"
},
{
"input": "536870912 60000000 72000000 271828 314159",
"output": "4101909"
},
{
"input": "1000000000 1000000000 1 1 100",
"output": "100000000000"
},
{
"input": "50 4 2 4 5",
"output": "125"
},
{
"input": "198 56 56 122 118",
"output": "366"
},
{
"input": "5 1000000000 1 12 15",
"output": "75"
},
{
"input": "1000 6 12 5 6",
"output": "913"
},
{
"input": "50 3 6 12 15",
"output": "216"
},
{
"input": "333 300 300 300 300",
"output": "300"
},
{
"input": "1 1000000000 1 1 2",
"output": "2"
},
{
"input": "188 110 110 200 78",
"output": "200"
},
{
"input": "100000 20 10 3 2",
"output": "25000"
},
{
"input": "100 2 4 1 10",
"output": "275"
},
{
"input": "1000000000 2 1000000000 1 1000000",
"output": "500999999"
},
{
"input": "20 3 6 5 7",
"output": "36"
},
{
"input": "50 4 6 4 5",
"output": "72"
},
{
"input": "96 46 4 174 156",
"output": "3936"
},
{
"input": "5 2 4 12 15",
"output": "27"
},
{
"input": "12 3 6 100 1",
"output": "400"
},
{
"input": "100 4 2 10 32",
"output": "1600"
},
{
"input": "1232 3 6 30000 3",
"output": "12300000"
},
{
"input": "20 3 6 5 4",
"output": "30"
},
{
"input": "100 6 15 11 29",
"output": "317"
},
{
"input": "10000000 4 8 100 200",
"output": "375000000"
},
{
"input": "1000000000 12 24 2 4",
"output": "249999998"
},
{
"input": "123 3 6 3000 1",
"output": "123000"
},
{
"input": "401523968 1536 2664 271828 314159",
"output": "117768531682"
},
{
"input": "9 2 4 3 5",
"output": "16"
},
{
"input": "999999999 724362018 772432019 46201854 20017479",
"output": "66219333"
},
{
"input": "100 2 4 1 1000",
"output": "25025"
},
{
"input": "50 2 4 1 1000",
"output": "12013"
},
{
"input": "1000000000 2 1 2 1",
"output": "1500000000"
},
{
"input": "1000000000 2005034 2005046 15 12",
"output": "13446"
},
{
"input": "1000000000 999999999 1000000000 1 1",
"output": "2"
},
{
"input": "999999999 500000000 1 100 1000",
"output": "999999999000"
},
{
"input": "50 8 6 3 4",
"output": "44"
},
{
"input": "1000000000 1 1 1000000000 1000000000",
"output": "1000000000000000000"
},
{
"input": "1000000000 999999862 999999818 15 12",
"output": "27"
},
{
"input": "1000000000 10000019 10000019 21 17",
"output": "2079"
},
{
"input": "20 6 4 8 2",
"output": "32"
},
{
"input": "1000000000 1000000000 1 1 1",
"output": "1000000000"
},
{
"input": "1000000000 12345678 123456789 1000000000 999999999",
"output": "88999999992"
},
{
"input": "1000000000 2 999999937 100000000 100000000",
"output": "50000000100000000"
},
{
"input": "1000000000 1 1 1000000000 999999999",
"output": "1000000000000000000"
},
{
"input": "1000000000 50001 100003 10 10",
"output": "299980"
},
{
"input": "1000000000 1000000000 3 1 1",
"output": "333333334"
},
{
"input": "10000 44 49 114 514",
"output": "130278"
},
{
"input": "30 5 15 2 1",
"output": "12"
},
{
"input": "20 2 4 1 1",
"output": "10"
},
{
"input": "100 8 12 5 6",
"output": "88"
}
] | 1,606,131,839
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 17
| 140
| 0
|
n, a, b, p, q = list(map(int,input().split()))
ans = 0
if p>q:
ans = p*(n//a)+q*((n//b)-n//b//a)
else:
ans = q*(n//b)+p*((n//a)-(n//a)//b)
print(ans)
|
Title: Joty and Chocolate
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Joty has got a task to do. She has a line of *n* tiles indexed from 1 to *n*. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by *a* and an unpainted tile should be painted Blue if it's index is divisible by *b*. So the tile with the number divisible by *a* and *b* can be either painted Red or Blue.
After her painting is done, she will get *p* chocolates for each tile that is painted Red and *q* chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input Specification:
The only line contains five integers *n*, *a*, *b*, *p* and *q* (1<=≤<=*n*,<=*a*,<=*b*,<=*p*,<=*q*<=≤<=109).
Output Specification:
Print the only integer *s* — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Demo Input:
['5 2 3 12 15\n', '20 2 3 3 5\n']
Demo Output:
['39\n', '51\n']
Note:
none
|
```python
n, a, b, p, q = list(map(int,input().split()))
ans = 0
if p>q:
ans = p*(n//a)+q*((n//b)-n//b//a)
else:
ans = q*(n//b)+p*((n//a)-(n//a)//b)
print(ans)
```
| 0
|
|
699
|
B
|
One Bomb
|
PROGRAMMING
| 1,400
|
[
"implementation"
] | null | null |
You are given a description of a depot. It is a rectangular checkered field of *n*<=×<=*m* size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*").
You have one bomb. If you lay the bomb at the cell (*x*,<=*y*), then after triggering it will wipe out all walls in the row *x* and all walls in the column *y*.
You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall.
|
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and columns in the depot field.
The next *n* lines contain *m* symbols "." and "*" each — the description of the field. *j*-th symbol in *i*-th of them stands for cell (*i*,<=*j*). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall.
|
If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes).
Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them.
|
[
"3 4\n.*..\n....\n.*..\n",
"3 3\n..*\n.*.\n*..\n",
"6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*..\n"
] |
[
"YES\n1 2\n",
"NO\n",
"YES\n3 3\n"
] |
none
| 1,000
|
[
{
"input": "3 4\n.*..\n....\n.*..",
"output": "YES\n1 2"
},
{
"input": "3 3\n..*\n.*.\n*..",
"output": "NO"
},
{
"input": "6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*..",
"output": "YES\n3 3"
},
{
"input": "1 10\n**********",
"output": "YES\n1 1"
},
{
"input": "10 1\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*",
"output": "YES\n1 1"
},
{
"input": "10 10\n.........*\n.........*\n........**\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*",
"output": "YES\n3 10"
},
{
"input": "10 10\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*",
"output": "YES\n1 10"
},
{
"input": "2 2\n.*\n*.",
"output": "YES\n2 2"
},
{
"input": "4 4\n....\n...*\n....\n*..*",
"output": "YES\n4 4"
},
{
"input": "4 4\n*...\n*...\n....\n****",
"output": "YES\n4 1"
},
{
"input": "1 1\n*",
"output": "YES\n1 1"
},
{
"input": "1 1\n.",
"output": "YES\n1 1"
},
{
"input": "1 2\n.*",
"output": "YES\n1 2"
},
{
"input": "2 1\n.\n*",
"output": "YES\n1 1"
},
{
"input": "2 2\n**\n**",
"output": "NO"
},
{
"input": "3 1\n*\n*\n*",
"output": "YES\n1 1"
},
{
"input": "3 2\n*.\n.*\n.*",
"output": "YES\n1 2"
},
{
"input": "3 3\n***\n***\n***",
"output": "NO"
},
{
"input": "2 2\n..\n.*",
"output": "YES\n1 2"
},
{
"input": "6 5\n..*..\n..*..\n**.**\n..*..\n..*..\n..*..",
"output": "YES\n3 3"
},
{
"input": "3 3\n.*.\n*.*\n.*.",
"output": "YES\n2 2"
},
{
"input": "4 4\n*...\n....\n....\n...*",
"output": "YES\n4 1"
},
{
"input": "2 4\n...*\n...*",
"output": "YES\n1 4"
},
{
"input": "2 2\n..\n..",
"output": "YES\n1 1"
},
{
"input": "3 3\n..*\n.*.\n..*",
"output": "YES\n2 3"
},
{
"input": "2 2\n*.\n.*",
"output": "YES\n2 1"
},
{
"input": "3 2\n.*\n*.\n.*",
"output": "YES\n2 2"
},
{
"input": "3 3\n***\n.*.\n.*.",
"output": "YES\n1 2"
},
{
"input": "4 4\n*.*.\n..*.\n.***\n..*.",
"output": "NO"
},
{
"input": "2 3\n..*\n**.",
"output": "YES\n2 3"
},
{
"input": "3 2\n*.\n.*\n*.",
"output": "YES\n2 1"
},
{
"input": "4 4\n..*.\n**.*\n..*.\n..*.",
"output": "YES\n2 3"
},
{
"input": "3 3\n*..\n*..\n***",
"output": "YES\n3 1"
},
{
"input": "3 3\n...\n*.*\n.*.",
"output": "YES\n2 2"
},
{
"input": "3 2\n..\n..\n**",
"output": "YES\n3 1"
},
{
"input": "3 4\n...*\n...*\n...*",
"output": "YES\n1 4"
},
{
"input": "5 5\n..*..\n..*..\n**.**\n..*..\n..*..",
"output": "YES\n3 3"
},
{
"input": "6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*.*",
"output": "NO"
},
{
"input": "3 3\n...\n.*.\n..*",
"output": "YES\n3 2"
},
{
"input": "3 5\n....*\n....*\n....*",
"output": "YES\n1 5"
},
{
"input": "3 3\n...\n...\n.*.",
"output": "YES\n1 2"
},
{
"input": "3 3\n*..\n...\n..*",
"output": "YES\n3 1"
},
{
"input": "2 3\n..*\n..*",
"output": "YES\n1 3"
},
{
"input": "2 2\n**\n.*",
"output": "YES\n1 2"
},
{
"input": "3 3\n..*\n*..\n*..",
"output": "YES\n1 1"
},
{
"input": "5 4\n.*..\n*.**\n.*..\n.*..\n.*..",
"output": "YES\n2 2"
},
{
"input": "6 5\n*.*..\n..*..\n*****\n..*..\n..*..\n..*..",
"output": "NO"
},
{
"input": "4 4\n.*..\n*.**\n....\n.*..",
"output": "YES\n2 2"
},
{
"input": "3 5\n....*\n....*\n*****",
"output": "YES\n3 5"
},
{
"input": "3 3\n..*\n*..\n..*",
"output": "YES\n2 3"
},
{
"input": "6 6\n..*...\n......\n......\n......\n......\n*....*",
"output": "YES\n6 3"
},
{
"input": "4 4\n.*..\n*...\n.*..\n.*..",
"output": "YES\n2 2"
},
{
"input": "3 3\n...\n..*\n.*.",
"output": "YES\n3 3"
},
{
"input": "3 2\n.*\n*.\n*.",
"output": "YES\n1 1"
},
{
"input": "4 2\n**\n.*\n.*\n.*",
"output": "YES\n1 2"
},
{
"input": "5 5\n*...*\n.....\n.....\n.....\n..*..",
"output": "YES\n1 3"
},
{
"input": "3 3\n**.\n...\n..*",
"output": "YES\n1 3"
},
{
"input": "3 3\n*.*\n*..\n*.*",
"output": "NO"
},
{
"input": "5 4\n....\n....\n*..*\n....\n.*..",
"output": "YES\n3 2"
},
{
"input": "5 5\n...*.\n...*.\n...*.\n...*.\n***.*",
"output": "YES\n5 4"
},
{
"input": "5 5\n*****\n*****\n*****\n*****\n*****",
"output": "NO"
},
{
"input": "3 3\n.*.\n..*\n.*.",
"output": "YES\n2 2"
},
{
"input": "3 3\n*.*\n...\n*.*",
"output": "NO"
},
{
"input": "2 3\n.*.\n*.*",
"output": "YES\n2 2"
},
{
"input": "3 10\n.......*..\n........*.\n.........*",
"output": "NO"
},
{
"input": "3 3\n.*.\n.*.\n.**",
"output": "YES\n3 2"
},
{
"input": "4 4\n*...\n....\n....\n..**",
"output": "YES\n4 1"
},
{
"input": "4 4\n****\n****\n****\n****",
"output": "NO"
},
{
"input": "3 2\n.*\n.*\n*.",
"output": "YES\n3 2"
},
{
"input": "3 3\n..*\n..*\n**.",
"output": "YES\n3 3"
},
{
"input": "6 3\n...\n...\n...\n...\n**.\n.*.",
"output": "YES\n5 2"
},
{
"input": "3 4\n****\n..*.\n..*.",
"output": "YES\n1 3"
},
{
"input": "5 5\n*..*.\n.....\n.....\n.....\n...*.",
"output": "YES\n1 4"
},
{
"input": "6 5\n..*..\n..*..\n.*...\n..*..\n..*..\n..*..",
"output": "YES\n3 3"
}
] | 1,491,757,530
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <iostream>
using namespace std;
char G[1007][1007];
int main()
{
int n,m;
cin>>n>>m;
int cnt = 0;
for(int i = 1 ; i <= n ; i++)
{
for(int j = 1 ; j <= m ; j++)
{
cin>>G[i][j];
if(G[i][j] == '*')
cnt++;
}
}
for(int i = 1 ; i <= n ; i++)
{
for(int j = 1 ; j <= m ; j++)
{
int tal = 0;
for(int k = 1 ; k <= n ; k++)
{
if(G[i][k] == '*')
tal++;
if(G[k][j] == '*')
tal++;
}
if(G[i][j] == '*')
tal--;
if(tal == cnt && n!= 1 && m!= 1)
{
cout<<"YES"<<endl;
cout<<i<<" "<<j;
return 0;
}
}
}
cout<<"NO"<<endl;
return 0;
}
|
Title: One Bomb
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a description of a depot. It is a rectangular checkered field of *n*<=×<=*m* size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*").
You have one bomb. If you lay the bomb at the cell (*x*,<=*y*), then after triggering it will wipe out all walls in the row *x* and all walls in the column *y*.
You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall.
Input Specification:
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and columns in the depot field.
The next *n* lines contain *m* symbols "." and "*" each — the description of the field. *j*-th symbol in *i*-th of them stands for cell (*i*,<=*j*). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall.
Output Specification:
If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes).
Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them.
Demo Input:
['3 4\n.*..\n....\n.*..\n', '3 3\n..*\n.*.\n*..\n', '6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*..\n']
Demo Output:
['YES\n1 2\n', 'NO\n', 'YES\n3 3\n']
Note:
none
|
```python
#include <iostream>
using namespace std;
char G[1007][1007];
int main()
{
int n,m;
cin>>n>>m;
int cnt = 0;
for(int i = 1 ; i <= n ; i++)
{
for(int j = 1 ; j <= m ; j++)
{
cin>>G[i][j];
if(G[i][j] == '*')
cnt++;
}
}
for(int i = 1 ; i <= n ; i++)
{
for(int j = 1 ; j <= m ; j++)
{
int tal = 0;
for(int k = 1 ; k <= n ; k++)
{
if(G[i][k] == '*')
tal++;
if(G[k][j] == '*')
tal++;
}
if(G[i][j] == '*')
tal--;
if(tal == cnt && n!= 1 && m!= 1)
{
cout<<"YES"<<endl;
cout<<i<<" "<<j;
return 0;
}
}
}
cout<<"NO"<<endl;
return 0;
}
```
| -1
|
|
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"
}
] | 1,696,666,881
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 31
| 0
|
import math
def longestnondec(s):
alist=[int(d) for d in s.split()]
count=1
blist=[]
i=0
while i<len(alist)-1:
if alist[i]>=alist[i+1]:
count+=1
i+=1
else:
count=1
i+=1
blist.append(count)
return max(blist)
n=int(input())
s=input()
print(longestnondec(s))
|
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
import math
def longestnondec(s):
alist=[int(d) for d in s.split()]
count=1
blist=[]
i=0
while i<len(alist)-1:
if alist[i]>=alist[i+1]:
count+=1
i+=1
else:
count=1
i+=1
blist.append(count)
return max(blist)
n=int(input())
s=input()
print(longestnondec(s))
```
| 0
|
|
12
|
A
|
Super Agent
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Super Agent
|
2
|
256
|
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3<=×<=3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
|
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
|
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
|
[
"XX.\n...\n.XX\n",
"X.X\nX..\n...\n"
] |
[
"YES\n",
"NO\n"
] |
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
| 0
|
[
{
"input": "XX.\n...\n.XX",
"output": "YES"
},
{
"input": ".X.\n.X.\n.X.",
"output": "YES"
},
{
"input": "XXX\nXXX\nXXX",
"output": "YES"
},
{
"input": "XXX\nX.X\nXXX",
"output": "YES"
},
{
"input": "X..\n.X.\n..X",
"output": "YES"
},
{
"input": "...\nX.X\nX..",
"output": "NO"
},
{
"input": ".X.\nX.X\n.X.",
"output": "YES"
},
{
"input": "X.X\n.X.\nX.X",
"output": "YES"
},
{
"input": "...\n...\n..X",
"output": "NO"
},
{
"input": "XXX\n...\nXXX",
"output": "YES"
},
{
"input": "..X\nX..\n..X",
"output": "NO"
},
{
"input": ".X.\n...\nX.X",
"output": "NO"
},
{
"input": "X.X\nX.X\nX.X",
"output": "YES"
},
{
"input": ".X.\nX.X\nXX.",
"output": "NO"
},
{
"input": "...\nXXX\nXXX",
"output": "NO"
},
{
"input": "XXX\n..X\nXXX",
"output": "NO"
},
{
"input": "X..\nX.X\n.X.",
"output": "NO"
},
{
"input": "...\n..X\nXXX",
"output": "NO"
},
{
"input": "..X\nX.X\nX..",
"output": "YES"
},
{
"input": "..X\n..X\nXXX",
"output": "NO"
},
{
"input": "X..\nX..\nX..",
"output": "NO"
},
{
"input": "XXX\n.X.\nXXX",
"output": "YES"
},
{
"input": "..X\n...\nX..",
"output": "YES"
},
{
"input": "...\n...\nX..",
"output": "NO"
},
{
"input": "...\n...\n.X.",
"output": "NO"
},
{
"input": "...\n...\n..X",
"output": "NO"
},
{
"input": "...\n.X.\nX..",
"output": "NO"
}
] | 1,661,618,718
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
input_list = []
flag=0
for i in range(3):
input_list.append(input())
for i in range(1):
for j in range(3):
if input_list[i][j]!=input_list[2-i][2-j]:
flag=1
break
if flag==0 and input_list[1][0]==input_list[1][2]:
flag=1
if flag=1:
print("NO")
else:
print("YES")
|
Title: Super Agent
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3<=×<=3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input Specification:
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output Specification:
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Demo Input:
['XX.\n...\n.XX\n', 'X.X\nX..\n...\n']
Demo Output:
['YES\n', 'NO\n']
Note:
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
|
```python
input_list = []
flag=0
for i in range(3):
input_list.append(input())
for i in range(1):
for j in range(3):
if input_list[i][j]!=input_list[2-i][2-j]:
flag=1
break
if flag==0 and input_list[1][0]==input_list[1][2]:
flag=1
if flag=1:
print("NO")
else:
print("YES")
```
| -1
|
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"
},
{
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"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"
},
{
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"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
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"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
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{
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"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
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{
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"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
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"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"
}
] | 1,636,001,253
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 62
| 4,505,600
|
n=input()
s=input()
for i in range(0,len(n)):
if(n[i]=="0" and s[i]=="1") or (n[i]=="1" and s[i]=="0"):
print(1,end="")
else:
print(0,end="")
|
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
n=input()
s=input()
for i in range(0,len(n)):
if(n[i]=="0" and s[i]=="1") or (n[i]=="1" and s[i]=="0"):
print(1,end="")
else:
print(0,end="")
```
| 3.976108
|
703
|
A
|
Mishka and Game
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
|
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
|
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
|
[
"3\n3 5\n2 1\n4 2\n",
"2\n6 1\n1 6\n",
"3\n1 5\n3 3\n2 2\n"
] |
[
"Mishka",
"Friendship is magic!^^",
"Chris"
] |
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.
| 500
|
[
{
"input": "3\n3 5\n2 1\n4 2",
"output": "Mishka"
},
{
"input": "2\n6 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "3\n1 5\n3 3\n2 2",
"output": "Chris"
},
{
"input": "6\n4 1\n4 2\n5 3\n5 1\n5 3\n4 1",
"output": "Mishka"
},
{
"input": "8\n2 4\n1 4\n1 5\n2 6\n2 5\n2 5\n2 4\n2 5",
"output": "Chris"
},
{
"input": "8\n4 1\n2 6\n4 2\n2 5\n5 2\n3 5\n5 2\n1 5",
"output": "Friendship is magic!^^"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3",
"output": "Mishka"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "9\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "10\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n2 4\n6 6\n3 2\n1 5\n5 2\n1 5\n1 5\n3 1\n6 5\n4 3\n1 1\n5 1\n3 3\n2 4\n1 5\n3 4\n5 1\n5 5\n2 5\n2 1\n4 3\n6 5\n1 1\n2 1\n1 3\n1 1\n6 4\n4 6\n6 4\n2 1\n2 5\n6 2\n3 4\n5 5\n1 4\n4 6\n3 4\n1 6\n5 1\n4 3\n3 4\n2 2\n1 2\n2 3\n1 3\n4 4\n5 5\n4 5\n4 4\n3 1\n4 5\n2 3\n2 6\n6 5\n6 1\n6 6\n2 3\n6 4\n3 3\n2 5\n4 4\n3 1\n2 4\n6 1\n3 2\n1 3\n5 4\n6 6\n2 5\n5 1\n1 1\n2 5\n6 5\n3 6\n5 6\n4 3\n3 4\n3 4\n6 5\n5 2\n4 2\n1 1\n3 1\n2 6\n1 6\n1 2\n6 1\n3 4\n1 6\n3 1\n5 3\n1 3\n5 6\n2 1\n6 4\n3 1\n1 6\n6 3\n3 3\n4 3",
"output": "Chris"
},
{
"input": "100\n4 1\n3 4\n4 6\n4 5\n6 5\n5 3\n6 2\n6 3\n5 2\n4 5\n1 5\n5 4\n1 4\n4 5\n4 6\n1 6\n4 4\n5 1\n6 4\n6 4\n4 6\n2 3\n6 2\n4 6\n1 4\n2 3\n4 3\n1 3\n6 2\n3 1\n3 4\n2 6\n4 5\n5 4\n2 2\n2 5\n4 1\n2 2\n3 3\n1 4\n5 6\n6 4\n4 2\n6 1\n5 5\n4 1\n2 1\n6 4\n4 4\n4 3\n5 3\n4 5\n5 3\n3 5\n6 3\n1 1\n3 4\n6 3\n6 1\n5 1\n2 4\n4 3\n2 2\n5 5\n1 5\n5 3\n4 6\n1 4\n6 3\n4 3\n2 4\n3 2\n2 4\n3 4\n6 2\n5 6\n1 2\n1 5\n5 5\n2 6\n5 1\n1 6\n5 3\n3 5\n2 6\n4 6\n6 2\n3 1\n5 5\n6 1\n3 6\n4 4\n1 1\n4 6\n5 3\n4 2\n5 1\n3 3\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "100\n6 3\n4 5\n4 3\n5 4\n5 1\n6 3\n4 2\n4 6\n3 1\n2 4\n2 2\n4 6\n5 3\n5 5\n4 2\n6 2\n2 3\n4 4\n6 4\n3 5\n2 4\n2 2\n5 2\n3 5\n2 4\n4 4\n3 5\n6 5\n1 3\n1 6\n2 2\n2 4\n3 2\n5 4\n1 6\n3 4\n4 1\n1 5\n1 4\n5 3\n2 2\n4 5\n6 3\n4 4\n1 1\n4 1\n2 4\n4 1\n4 5\n5 3\n1 1\n1 6\n5 6\n6 6\n4 2\n4 3\n3 4\n3 6\n3 4\n6 5\n3 4\n5 4\n5 1\n5 3\n5 1\n1 2\n2 6\n3 4\n6 5\n4 3\n1 1\n5 5\n5 1\n3 3\n5 2\n1 3\n6 6\n5 6\n1 4\n4 4\n1 4\n3 6\n6 5\n3 3\n3 6\n1 5\n1 2\n3 6\n3 6\n4 1\n5 2\n1 2\n5 2\n3 3\n4 4\n4 2\n6 2\n5 4\n6 1\n6 3",
"output": "Mishka"
},
{
"input": "8\n4 1\n6 2\n4 1\n5 3\n4 1\n5 3\n6 2\n5 3",
"output": "Mishka"
},
{
"input": "5\n3 6\n3 5\n3 5\n1 6\n3 5",
"output": "Chris"
},
{
"input": "4\n4 1\n2 4\n5 3\n3 6",
"output": "Friendship is magic!^^"
},
{
"input": "6\n6 3\n5 1\n6 3\n4 3\n4 3\n5 2",
"output": "Mishka"
},
{
"input": "7\n3 4\n1 4\n2 5\n1 6\n1 6\n1 5\n3 4",
"output": "Chris"
},
{
"input": "6\n6 2\n2 5\n5 2\n3 6\n4 3\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "8\n6 1\n5 3\n4 3\n4 1\n5 1\n4 2\n4 2\n4 1",
"output": "Mishka"
},
{
"input": "9\n2 5\n2 5\n1 4\n2 6\n2 4\n2 5\n2 6\n1 5\n2 5",
"output": "Chris"
},
{
"input": "4\n6 2\n2 4\n4 2\n3 6",
"output": "Friendship is magic!^^"
},
{
"input": "9\n5 2\n4 1\n4 1\n5 1\n6 2\n6 1\n5 3\n6 1\n6 2",
"output": "Mishka"
},
{
"input": "8\n2 4\n3 6\n1 6\n1 6\n2 4\n3 4\n3 6\n3 4",
"output": "Chris"
},
{
"input": "6\n5 3\n3 6\n6 2\n1 6\n5 1\n3 5",
"output": "Friendship is magic!^^"
},
{
"input": "6\n5 2\n5 1\n6 1\n5 2\n4 2\n5 1",
"output": "Mishka"
},
{
"input": "5\n1 4\n2 5\n3 4\n2 6\n3 4",
"output": "Chris"
},
{
"input": "4\n6 2\n3 4\n5 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "93\n4 3\n4 1\n4 2\n5 2\n5 3\n6 3\n4 3\n6 2\n6 3\n5 1\n4 2\n4 2\n5 1\n6 2\n6 3\n6 1\n4 1\n6 2\n5 3\n4 3\n4 1\n4 2\n5 2\n6 3\n5 2\n5 2\n6 3\n5 1\n6 2\n5 2\n4 1\n5 2\n5 1\n4 1\n6 1\n5 2\n4 3\n5 3\n5 3\n5 1\n4 3\n4 3\n4 2\n4 1\n6 2\n6 1\n4 1\n5 2\n5 2\n6 2\n5 3\n5 1\n6 2\n5 1\n6 3\n5 2\n6 2\n6 2\n4 2\n5 2\n6 1\n6 3\n6 3\n5 1\n5 1\n4 1\n5 1\n4 3\n5 3\n6 3\n4 1\n4 3\n6 1\n6 1\n4 2\n6 2\n4 2\n5 2\n4 1\n5 2\n4 1\n5 1\n5 2\n5 1\n4 1\n6 3\n6 2\n4 3\n4 1\n5 2\n4 3\n5 2\n5 1",
"output": "Mishka"
},
{
"input": "11\n1 6\n1 6\n2 4\n2 5\n3 4\n1 5\n1 6\n1 5\n1 6\n2 6\n3 4",
"output": "Chris"
},
{
"input": "70\n6 1\n3 6\n4 3\n2 5\n5 2\n1 4\n6 2\n1 6\n4 3\n1 4\n5 3\n2 4\n5 3\n1 6\n5 1\n3 5\n4 2\n2 4\n5 1\n3 5\n6 2\n1 5\n4 2\n2 5\n5 3\n1 5\n4 2\n1 4\n5 2\n2 6\n4 3\n1 5\n6 2\n3 4\n4 2\n3 5\n6 3\n3 4\n5 1\n1 4\n4 2\n1 4\n6 3\n2 6\n5 2\n1 6\n6 1\n2 6\n5 3\n1 5\n5 1\n1 6\n4 1\n1 5\n4 2\n2 4\n5 1\n2 5\n6 3\n1 4\n6 3\n3 6\n5 1\n1 4\n5 3\n3 5\n4 2\n3 4\n6 2\n1 4",
"output": "Friendship is magic!^^"
},
{
"input": "59\n4 1\n5 3\n6 1\n4 2\n5 1\n4 3\n6 1\n5 1\n4 3\n4 3\n5 2\n5 3\n4 1\n6 2\n5 1\n6 3\n6 3\n5 2\n5 2\n6 1\n4 1\n6 1\n4 3\n5 3\n5 3\n4 3\n4 2\n4 2\n6 3\n6 3\n6 1\n4 3\n5 1\n6 2\n6 1\n4 1\n6 1\n5 3\n4 2\n5 1\n6 2\n6 2\n4 3\n5 3\n4 3\n6 3\n5 2\n5 2\n4 3\n5 1\n5 3\n6 1\n6 3\n6 3\n4 3\n5 2\n5 2\n5 2\n4 3",
"output": "Mishka"
},
{
"input": "42\n1 5\n1 6\n1 6\n1 4\n2 5\n3 6\n1 6\n3 4\n2 5\n2 5\n2 4\n1 4\n3 4\n2 4\n2 6\n1 5\n3 6\n2 6\n2 6\n3 5\n1 4\n1 5\n2 6\n3 6\n1 4\n3 4\n2 4\n1 6\n3 4\n2 4\n2 6\n1 6\n1 4\n1 6\n1 6\n2 4\n1 5\n1 6\n2 5\n3 6\n3 5\n3 4",
"output": "Chris"
},
{
"input": "78\n4 3\n3 5\n4 3\n1 5\n5 1\n1 5\n4 3\n1 4\n6 3\n1 5\n4 1\n2 4\n4 3\n2 4\n5 1\n3 6\n4 2\n3 6\n6 3\n3 4\n4 3\n3 6\n5 3\n1 5\n4 1\n2 6\n4 2\n2 4\n4 1\n3 5\n5 2\n3 6\n4 3\n2 4\n6 3\n1 6\n4 3\n3 5\n6 3\n2 6\n4 1\n2 4\n6 2\n1 6\n4 2\n1 4\n4 3\n1 4\n4 3\n2 4\n6 2\n3 5\n6 1\n3 6\n5 3\n1 6\n6 1\n2 6\n4 2\n1 5\n6 2\n2 6\n6 3\n2 4\n4 2\n3 5\n6 1\n2 5\n5 3\n2 6\n5 1\n3 6\n4 3\n3 6\n6 3\n2 5\n6 1\n2 6",
"output": "Friendship is magic!^^"
},
{
"input": "76\n4 1\n5 2\n4 3\n5 2\n5 3\n5 2\n6 1\n4 2\n6 2\n5 3\n4 2\n6 2\n4 1\n4 2\n5 1\n5 1\n6 2\n5 2\n5 3\n6 3\n5 2\n4 3\n6 3\n6 1\n4 3\n6 2\n6 1\n4 1\n6 1\n5 3\n4 1\n5 3\n4 2\n5 2\n4 3\n6 1\n6 2\n5 2\n6 1\n5 3\n4 3\n5 1\n5 3\n4 3\n5 1\n5 1\n4 1\n4 1\n4 1\n4 3\n5 3\n6 3\n6 3\n5 2\n6 2\n6 3\n5 1\n6 3\n5 3\n6 1\n5 3\n4 1\n5 3\n6 1\n4 2\n6 2\n4 3\n4 1\n6 2\n4 3\n5 3\n5 2\n5 3\n5 1\n6 3\n5 2",
"output": "Mishka"
},
{
"input": "84\n3 6\n3 4\n2 5\n2 4\n1 6\n3 4\n1 5\n1 6\n3 5\n1 6\n2 4\n2 6\n2 6\n2 4\n3 5\n1 5\n3 6\n3 6\n3 4\n3 4\n2 6\n1 6\n1 6\n3 5\n3 4\n1 6\n3 4\n3 5\n2 4\n2 5\n2 5\n3 5\n1 6\n3 4\n2 6\n2 6\n3 4\n3 4\n2 5\n2 5\n2 4\n3 4\n2 5\n3 4\n3 4\n2 6\n2 6\n1 6\n2 4\n1 5\n3 4\n2 5\n2 5\n3 4\n2 4\n2 6\n2 6\n1 4\n3 5\n3 5\n2 4\n2 5\n3 4\n1 5\n1 5\n2 6\n1 5\n3 5\n2 4\n2 5\n3 4\n2 6\n1 6\n2 5\n3 5\n3 5\n3 4\n2 5\n2 6\n3 4\n1 6\n2 5\n2 6\n1 4",
"output": "Chris"
},
{
"input": "44\n6 1\n1 6\n5 2\n1 4\n6 2\n2 5\n5 3\n3 6\n5 2\n1 6\n4 1\n2 4\n6 1\n3 4\n6 3\n3 6\n4 3\n2 4\n6 1\n3 4\n6 1\n1 6\n4 1\n3 5\n6 1\n3 6\n4 1\n1 4\n4 2\n2 6\n6 1\n2 4\n6 2\n1 4\n6 2\n2 4\n5 2\n3 6\n6 3\n2 6\n5 3\n3 4\n5 3\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "42\n5 3\n5 1\n5 2\n4 1\n6 3\n6 1\n6 2\n4 1\n4 3\n4 1\n5 1\n5 3\n5 1\n4 1\n4 2\n6 1\n6 3\n5 1\n4 1\n4 1\n6 3\n4 3\n6 3\n5 2\n6 1\n4 1\n5 3\n4 3\n5 2\n6 3\n6 1\n5 1\n4 2\n4 3\n5 2\n5 3\n6 3\n5 2\n5 1\n5 3\n6 2\n6 1",
"output": "Mishka"
},
{
"input": "50\n3 6\n2 6\n1 4\n1 4\n1 4\n2 5\n3 4\n3 5\n2 6\n1 6\n3 5\n1 5\n2 6\n2 4\n2 4\n3 5\n1 6\n1 5\n1 5\n1 4\n3 5\n1 6\n3 5\n1 4\n1 5\n1 4\n3 6\n1 6\n1 4\n1 4\n1 4\n1 5\n3 6\n1 6\n1 6\n2 4\n1 5\n2 6\n2 5\n3 5\n3 6\n3 4\n2 4\n2 6\n3 4\n2 5\n3 6\n3 5\n2 4\n2 4",
"output": "Chris"
},
{
"input": "86\n6 3\n2 4\n6 3\n3 5\n6 3\n1 5\n5 2\n2 4\n4 3\n2 6\n4 1\n2 6\n5 2\n1 4\n5 1\n2 4\n4 1\n1 4\n6 2\n3 5\n4 2\n2 4\n6 2\n1 5\n5 3\n2 5\n5 1\n1 6\n6 1\n1 4\n4 3\n3 4\n5 2\n2 4\n5 3\n2 5\n4 3\n3 4\n4 1\n1 5\n6 3\n3 4\n4 3\n3 4\n4 1\n3 4\n5 1\n1 6\n4 2\n1 6\n5 1\n2 4\n5 1\n3 6\n4 1\n1 5\n5 2\n1 4\n4 3\n2 5\n5 1\n1 5\n6 2\n2 6\n4 2\n2 4\n4 1\n2 5\n5 3\n3 4\n5 1\n3 4\n6 3\n3 4\n4 3\n2 6\n6 2\n2 5\n5 2\n3 5\n4 2\n3 6\n6 2\n3 4\n4 2\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "84\n6 1\n6 3\n6 3\n4 1\n4 3\n4 2\n6 3\n5 3\n6 1\n6 3\n4 3\n5 2\n5 3\n5 1\n6 2\n6 2\n6 1\n4 1\n6 3\n5 2\n4 1\n5 3\n6 3\n4 2\n6 2\n6 3\n4 3\n4 1\n4 3\n5 1\n5 1\n5 1\n4 1\n6 1\n4 3\n6 2\n5 1\n5 1\n6 2\n5 2\n4 1\n6 1\n6 1\n6 3\n6 2\n4 3\n6 3\n6 2\n5 2\n5 1\n4 3\n6 2\n4 1\n6 2\n6 1\n5 2\n5 1\n6 2\n6 1\n5 3\n5 2\n6 1\n6 3\n5 2\n6 1\n6 3\n4 3\n5 1\n6 3\n6 1\n5 3\n4 3\n5 2\n5 1\n6 2\n5 3\n6 1\n5 1\n4 1\n5 1\n5 1\n5 2\n5 2\n5 1",
"output": "Mishka"
},
{
"input": "92\n1 5\n2 4\n3 5\n1 6\n2 5\n1 6\n3 6\n1 6\n2 4\n3 4\n3 4\n3 6\n1 5\n2 5\n1 5\n1 5\n2 6\n2 4\n3 6\n1 4\n1 6\n2 6\n3 4\n2 6\n2 6\n1 4\n3 5\n2 5\n2 6\n1 5\n1 4\n1 5\n3 6\n3 5\n2 5\n1 5\n3 5\n3 6\n2 6\n2 6\n1 5\n3 4\n2 4\n3 6\n2 5\n1 5\n2 4\n1 4\n2 6\n2 6\n2 6\n1 5\n3 6\n3 6\n2 5\n1 4\n2 4\n3 4\n1 5\n2 5\n2 4\n2 5\n3 5\n3 4\n3 6\n2 6\n3 5\n1 4\n3 4\n1 6\n3 6\n2 6\n1 4\n3 6\n3 6\n2 5\n2 6\n1 6\n2 6\n3 5\n2 5\n3 6\n2 5\n2 6\n1 5\n2 4\n1 4\n2 4\n1 5\n2 5\n2 5\n2 6",
"output": "Chris"
},
{
"input": "20\n5 1\n1 4\n4 3\n1 5\n4 2\n3 6\n6 2\n1 6\n4 1\n1 4\n5 2\n3 4\n5 1\n1 6\n5 1\n2 6\n6 3\n2 5\n6 2\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "100\n4 3\n4 3\n4 2\n4 3\n4 1\n4 3\n5 2\n5 2\n6 2\n4 2\n5 1\n4 2\n5 2\n6 1\n4 1\n6 3\n5 3\n5 1\n5 1\n5 1\n5 3\n6 1\n6 1\n4 1\n5 2\n5 2\n6 1\n6 3\n4 2\n4 1\n5 3\n4 1\n5 3\n5 1\n6 3\n6 3\n6 1\n5 2\n5 3\n5 3\n6 1\n4 1\n6 2\n6 1\n6 2\n6 3\n4 3\n4 3\n6 3\n4 2\n4 2\n5 3\n5 2\n5 2\n4 3\n5 3\n5 2\n4 2\n5 1\n4 2\n5 1\n5 3\n6 3\n5 3\n5 3\n4 2\n4 1\n4 2\n4 3\n6 3\n4 3\n6 2\n6 1\n5 3\n5 2\n4 1\n6 1\n5 2\n6 2\n4 2\n6 3\n4 3\n5 1\n6 3\n5 2\n4 3\n5 3\n5 3\n4 3\n6 3\n4 3\n4 1\n5 1\n6 2\n6 3\n5 3\n6 1\n6 3\n5 3\n6 1",
"output": "Mishka"
},
{
"input": "100\n1 5\n1 4\n1 5\n2 4\n2 6\n3 6\n3 5\n1 5\n2 5\n3 6\n3 5\n1 6\n1 4\n1 5\n1 6\n2 6\n1 5\n3 5\n3 4\n2 6\n2 6\n2 5\n3 4\n1 6\n1 4\n2 4\n1 5\n1 6\n3 5\n1 6\n2 6\n3 5\n1 6\n3 4\n3 5\n1 6\n3 6\n2 4\n2 4\n3 5\n2 6\n1 5\n3 5\n3 6\n2 4\n2 4\n2 6\n3 4\n3 4\n1 5\n1 4\n2 5\n3 4\n1 4\n2 6\n2 5\n2 4\n2 4\n2 5\n1 5\n1 6\n1 5\n1 5\n1 5\n1 6\n3 4\n2 4\n3 5\n3 5\n1 6\n3 5\n1 5\n1 6\n3 6\n3 4\n1 5\n3 5\n3 6\n1 4\n3 6\n1 5\n3 5\n3 6\n3 5\n1 4\n3 4\n2 4\n2 4\n2 5\n3 6\n3 5\n1 5\n2 4\n1 4\n3 4\n1 5\n3 4\n3 6\n3 5\n3 4",
"output": "Chris"
},
{
"input": "100\n4 3\n3 4\n5 1\n2 5\n5 3\n1 5\n6 3\n2 4\n5 2\n2 6\n5 2\n1 5\n6 3\n1 5\n6 3\n3 4\n5 2\n1 5\n6 1\n1 5\n4 2\n3 5\n6 3\n2 6\n6 3\n1 4\n6 2\n3 4\n4 1\n3 6\n5 1\n2 4\n5 1\n3 4\n6 2\n3 5\n4 1\n2 6\n4 3\n2 6\n5 2\n3 6\n6 2\n3 5\n4 3\n1 5\n5 3\n3 6\n4 2\n3 4\n6 1\n3 4\n5 2\n2 6\n5 2\n2 4\n6 2\n3 6\n4 3\n2 4\n4 3\n2 6\n4 2\n3 4\n6 3\n2 4\n6 3\n3 5\n5 2\n1 5\n6 3\n3 6\n4 3\n1 4\n5 2\n1 6\n4 1\n2 5\n4 1\n2 4\n4 2\n2 5\n6 1\n2 4\n6 3\n1 5\n4 3\n2 6\n6 3\n2 6\n5 3\n1 5\n4 1\n1 5\n6 2\n2 5\n5 1\n3 6\n4 3\n3 4",
"output": "Friendship is magic!^^"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3",
"output": "Mishka"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "99\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "84\n6 2\n1 5\n6 2\n2 3\n5 5\n1 2\n3 4\n3 4\n6 5\n6 4\n2 5\n4 1\n1 2\n1 1\n1 4\n2 5\n5 6\n6 3\n2 4\n5 5\n2 6\n3 4\n5 1\n3 3\n5 5\n4 6\n4 6\n2 4\n4 1\n5 2\n2 2\n3 6\n3 3\n4 6\n1 1\n2 4\n6 5\n5 2\n6 5\n5 5\n2 5\n6 4\n1 1\n6 2\n3 6\n6 5\n4 4\n1 5\n5 6\n4 4\n3 5\n6 1\n3 4\n1 5\n4 6\n4 6\n4 1\n3 6\n6 2\n1 1\n4 5\n5 4\n5 3\n3 4\n6 4\n1 1\n5 2\n6 5\n6 1\n2 2\n2 4\n3 3\n4 6\n1 3\n6 6\n5 2\n1 6\n6 2\n6 6\n4 1\n3 6\n6 4\n2 3\n3 4",
"output": "Chris"
},
{
"input": "70\n3 4\n2 3\n2 3\n6 5\n6 6\n4 3\n2 3\n3 1\n3 5\n5 6\n1 6\n2 5\n5 3\n2 5\n4 6\n5 1\n6 1\n3 1\n3 3\n5 3\n2 1\n3 3\n6 4\n6 3\n4 3\n4 5\n3 5\n5 5\n5 2\n1 6\n3 4\n5 2\n2 4\n1 6\n4 3\n4 3\n6 2\n1 3\n1 5\n6 1\n3 1\n1 1\n1 3\n2 2\n3 2\n6 4\n1 1\n4 4\n3 1\n4 5\n4 2\n6 3\n4 4\n3 2\n1 2\n2 6\n3 3\n1 5\n1 1\n6 5\n2 2\n3 1\n5 4\n5 2\n6 4\n6 3\n6 6\n6 3\n3 3\n5 4",
"output": "Mishka"
},
{
"input": "56\n6 4\n3 4\n6 1\n3 3\n1 4\n2 3\n1 5\n2 5\n1 5\n5 5\n2 3\n1 1\n3 2\n3 5\n4 6\n4 4\n5 2\n4 3\n3 1\n3 6\n2 3\n3 4\n5 6\n5 2\n5 6\n1 5\n1 5\n4 1\n6 3\n2 2\n2 1\n5 5\n2 1\n4 1\n5 4\n2 5\n4 1\n6 2\n3 4\n4 2\n6 4\n5 4\n4 2\n4 3\n6 2\n6 2\n3 1\n1 4\n3 6\n5 1\n5 5\n3 6\n6 4\n2 3\n6 5\n3 3",
"output": "Mishka"
},
{
"input": "94\n2 4\n6 4\n1 6\n1 4\n5 1\n3 3\n4 3\n6 1\n6 5\n3 2\n2 3\n5 1\n5 3\n1 2\n4 3\n3 2\n2 3\n4 6\n1 3\n6 3\n1 1\n3 2\n4 3\n1 5\n4 6\n3 2\n6 3\n1 6\n1 1\n1 2\n3 5\n1 3\n3 5\n4 4\n4 2\n1 4\n4 5\n1 3\n1 2\n1 1\n5 4\n5 5\n6 1\n2 1\n2 6\n6 6\n4 2\n3 6\n1 6\n6 6\n1 5\n3 2\n1 2\n4 4\n6 4\n4 1\n1 5\n3 3\n1 3\n3 4\n4 4\n1 1\n2 5\n4 5\n3 1\n3 1\n3 6\n3 2\n1 4\n1 6\n6 3\n2 4\n1 1\n2 2\n2 2\n2 1\n5 4\n1 2\n6 6\n2 2\n3 3\n6 3\n6 3\n1 6\n2 3\n2 4\n2 3\n6 6\n2 6\n6 3\n3 5\n1 4\n1 1\n3 5",
"output": "Chris"
},
{
"input": "81\n4 2\n1 2\n2 3\n4 5\n6 2\n1 6\n3 6\n3 4\n4 6\n4 4\n3 5\n4 6\n3 6\n3 5\n3 1\n1 3\n5 3\n3 4\n1 1\n4 1\n1 2\n6 1\n1 3\n6 5\n4 5\n4 2\n4 5\n6 2\n1 2\n2 6\n5 2\n1 5\n2 4\n4 3\n5 4\n1 2\n5 3\n2 6\n6 4\n1 1\n1 3\n3 1\n3 1\n6 5\n5 5\n6 1\n6 6\n5 2\n1 3\n1 4\n2 3\n5 5\n3 1\n3 1\n4 4\n1 6\n6 4\n2 2\n4 6\n4 4\n2 6\n2 4\n2 4\n4 1\n1 6\n1 4\n1 3\n6 5\n5 1\n1 3\n5 1\n1 4\n3 5\n2 6\n1 3\n5 6\n3 5\n4 4\n5 5\n5 6\n4 3",
"output": "Chris"
},
{
"input": "67\n6 5\n3 6\n1 6\n5 3\n5 4\n5 1\n1 6\n1 1\n3 2\n4 4\n3 1\n4 1\n1 5\n5 3\n3 3\n6 4\n2 4\n2 2\n4 3\n1 4\n1 4\n6 1\n1 2\n2 2\n5 1\n6 2\n3 5\n5 5\n2 2\n6 5\n6 2\n4 4\n3 1\n4 2\n6 6\n6 4\n5 1\n2 2\n4 5\n5 5\n4 6\n1 5\n6 3\n4 4\n1 5\n6 4\n3 6\n3 4\n1 6\n2 4\n2 1\n2 5\n6 5\n6 4\n4 1\n3 2\n1 2\n5 1\n5 6\n1 5\n3 5\n3 1\n5 3\n3 2\n5 1\n4 6\n6 6",
"output": "Mishka"
},
{
"input": "55\n6 6\n6 5\n2 2\n2 2\n6 4\n5 5\n6 5\n5 3\n1 3\n2 2\n5 6\n3 3\n3 3\n6 5\n3 5\n5 5\n1 2\n1 1\n4 6\n1 2\n5 5\n6 2\n6 3\n1 2\n5 1\n1 3\n3 3\n4 4\n2 5\n1 1\n5 3\n4 3\n2 2\n4 5\n5 6\n4 5\n6 3\n1 6\n6 4\n3 6\n1 6\n5 2\n6 3\n2 3\n5 5\n4 3\n3 1\n4 2\n1 1\n2 5\n5 3\n2 2\n6 3\n4 5\n2 2",
"output": "Mishka"
},
{
"input": "92\n2 3\n1 3\n2 6\n5 1\n5 5\n3 2\n5 6\n2 5\n3 1\n3 6\n4 5\n2 5\n1 2\n2 3\n6 5\n3 6\n4 4\n6 2\n4 5\n4 4\n5 1\n6 1\n3 4\n3 5\n6 6\n3 2\n6 4\n2 2\n3 5\n6 4\n6 3\n6 6\n3 4\n3 3\n6 1\n5 4\n6 2\n2 6\n5 6\n1 4\n4 6\n6 3\n3 1\n4 1\n6 6\n3 5\n6 3\n6 1\n1 6\n3 2\n6 6\n4 3\n3 4\n1 3\n3 5\n5 3\n6 5\n4 3\n5 5\n4 1\n1 5\n6 4\n2 3\n2 3\n1 5\n1 2\n5 2\n4 3\n3 6\n5 5\n5 4\n1 4\n3 3\n1 6\n5 6\n5 4\n5 3\n1 1\n6 2\n5 5\n2 5\n4 3\n6 6\n5 1\n1 1\n4 6\n4 6\n3 1\n6 4\n2 4\n2 2\n2 1",
"output": "Chris"
},
{
"input": "79\n5 3\n4 6\n3 6\n2 1\n5 2\n2 3\n4 4\n6 2\n2 5\n1 6\n6 6\n2 6\n3 3\n4 5\n6 2\n2 1\n1 5\n5 1\n2 1\n2 6\n5 3\n6 2\n2 6\n2 3\n1 5\n4 4\n6 3\n5 2\n3 2\n1 3\n1 3\n6 3\n2 6\n3 6\n5 3\n4 5\n6 1\n3 5\n3 5\n6 5\n1 5\n4 2\n6 2\n2 3\n4 6\n3 6\n2 5\n4 4\n1 1\n4 6\n2 6\n6 4\n3 2\n4 1\n1 2\n6 4\n5 6\n1 4\n2 2\n5 4\n3 2\n1 2\n2 4\n2 5\n2 1\n3 6\n3 3\n1 1\n2 2\n4 4\n4 5\n3 3\n5 3\n6 2\n4 5\n6 5\n2 5\n5 6\n2 2",
"output": "Chris"
},
{
"input": "65\n1 1\n5 1\n2 2\n5 4\n4 5\n2 5\n3 2\n5 6\n6 3\n1 1\n6 1\n1 5\n1 1\n5 2\n6 4\n1 6\n1 1\n4 3\n2 3\n5 6\n4 4\n6 2\n1 3\n4 3\n1 3\n6 3\n3 5\n4 2\n4 1\n6 1\n3 2\n2 6\n3 2\n3 5\n6 3\n4 3\n1 5\n2 6\n1 3\n4 1\n4 1\n2 5\n2 5\n6 2\n5 3\n3 1\n3 3\n5 1\n2 4\n5 3\n3 3\n1 1\n6 3\n3 3\n5 1\n1 6\n4 5\n6 6\n5 5\n2 5\n4 1\n2 2\n1 4\n1 6\n6 5",
"output": "Mishka"
},
{
"input": "1\n1 1",
"output": "Friendship is magic!^^"
}
] | 1,678,855,150
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 31
| 0
|
n = int(input())
y = 0
x = 0
for i in range(n):
dice1, dice2 = map(int, input().split())
if dice1 > dice2:
y += 1
elif dice1 < dice2:
x += 1
else:
x += 0
if x < y:
print('Mishka')
if x > y:
print('Chris')
|
Title: Mishka and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
Input Specification:
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
Output Specification:
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
Demo Input:
['3\n3 5\n2 1\n4 2\n', '2\n6 1\n1 6\n', '3\n1 5\n3 3\n2 2\n']
Demo Output:
['Mishka', 'Friendship is magic!^^', 'Chris']
Note:
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.
|
```python
n = int(input())
y = 0
x = 0
for i in range(n):
dice1, dice2 = map(int, input().split())
if dice1 > dice2:
y += 1
elif dice1 < dice2:
x += 1
else:
x += 0
if x < y:
print('Mishka')
if x > y:
print('Chris')
```
| 0
|
|
472
|
A
|
Design Tutorial: Learn from Math
|
PROGRAMMING
| 800
|
[
"math",
"number theory"
] | null | null |
One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers.
|
The only line contains an integer *n* (12<=≤<=*n*<=≤<=106).
|
Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them.
|
[
"12\n",
"15\n",
"23\n",
"1000000\n"
] |
[
"4 8\n",
"6 9\n",
"8 15\n",
"500000 500000\n"
] |
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number.
| 500
|
[
{
"input": "12",
"output": "4 8"
},
{
"input": "15",
"output": "6 9"
},
{
"input": "23",
"output": "8 15"
},
{
"input": "1000000",
"output": "500000 500000"
},
{
"input": "63874",
"output": "4 63870"
},
{
"input": "14568",
"output": "4 14564"
},
{
"input": "192",
"output": "4 188"
},
{
"input": "86",
"output": "4 82"
},
{
"input": "46220",
"output": "4 46216"
},
{
"input": "57114",
"output": "4 57110"
},
{
"input": "869",
"output": "4 865"
},
{
"input": "738457",
"output": "4 738453"
},
{
"input": "58113",
"output": "6 58107"
},
{
"input": "4864",
"output": "4 4860"
},
{
"input": "15",
"output": "6 9"
},
{
"input": "74752",
"output": "4 74748"
},
{
"input": "6073",
"output": "4 6069"
},
{
"input": "1289",
"output": "4 1285"
},
{
"input": "20",
"output": "4 16"
},
{
"input": "58134",
"output": "4 58130"
},
{
"input": "57756",
"output": "4 57752"
},
{
"input": "765",
"output": "6 759"
},
{
"input": "59",
"output": "4 55"
},
{
"input": "991666",
"output": "4 991662"
},
{
"input": "70761",
"output": "4 70757"
},
{
"input": "13",
"output": "4 9"
},
{
"input": "999999",
"output": "4 999995"
},
{
"input": "17",
"output": "8 9"
},
{
"input": "21",
"output": "6 15"
},
{
"input": "19",
"output": "4 15"
},
{
"input": "100007",
"output": "6 100001"
},
{
"input": "999987",
"output": "6 999981"
},
{
"input": "22",
"output": "4 18"
}
] | 1,663,328,862
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 46
| 0
|
n = int(input())
def prime(num):
for i in range(2, num):
if num % i == 0:
return False
return True
if n % 2 == 0 and not prime(n // 2):
print(f'{n // 2} {n // 2}')
elif n % 2 == 0 and prime(n // 2):
print(f'{(n // 2)+1} {(n // 2)-1}')
elif n % 2 != 0 and not prime(n // 2) and not prime((n//2)+1):
print(f'{n // 2} {(n // 2)+1}')
elif n % 2 != 0 and ((prime(n//2) and not prime((n//2)+1)) or (not prime(n//2) and prime((n//2)+1))):
for i in range(1, 4):
if not prime((n//2)-i) and not prime((n//2)+1+i):
print(f'{(n//2)-i} {(n//2)+1+i}')
exit()
|
Title: Design Tutorial: Learn from Math
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers.
Input Specification:
The only line contains an integer *n* (12<=≤<=*n*<=≤<=106).
Output Specification:
Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them.
Demo Input:
['12\n', '15\n', '23\n', '1000000\n']
Demo Output:
['4 8\n', '6 9\n', '8 15\n', '500000 500000\n']
Note:
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number.
|
```python
n = int(input())
def prime(num):
for i in range(2, num):
if num % i == 0:
return False
return True
if n % 2 == 0 and not prime(n // 2):
print(f'{n // 2} {n // 2}')
elif n % 2 == 0 and prime(n // 2):
print(f'{(n // 2)+1} {(n // 2)-1}')
elif n % 2 != 0 and not prime(n // 2) and not prime((n//2)+1):
print(f'{n // 2} {(n // 2)+1}')
elif n % 2 != 0 and ((prime(n//2) and not prime((n//2)+1)) or (not prime(n//2) and prime((n//2)+1))):
for i in range(1, 4):
if not prime((n//2)-i) and not prime((n//2)+1+i):
print(f'{(n//2)-i} {(n//2)+1+i}')
exit()
```
| 3
|
|
816
|
B
|
Karen and Coffee
|
PROGRAMMING
| 1,400
|
[
"binary search",
"data structures",
"implementation"
] | null | null |
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range?
|
The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
|
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
|
[
"3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n",
"2 1 1\n1 1\n200000 200000\n90 100\n"
] |
[
"3\n3\n0\n4\n",
"0\n"
] |
In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.
| 1,000
|
[
{
"input": "3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100",
"output": "3\n3\n0\n4"
},
{
"input": "2 1 1\n1 1\n200000 200000\n90 100",
"output": "0"
},
{
"input": "1 1 1\n1 1\n1 1",
"output": "1"
},
{
"input": "1 1 1\n200000 200000\n200000 200000",
"output": "1"
}
] | 1,699,240,621
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 9
| 2,500
| 12,697,600
|
import sys,os
# sys.stdin = open("input","r")
from collections import *
from heapq import *
from functools import *
from types import GeneratorType
# from math import *
from itertools import *
from math import *
from string import ascii_lowercase
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
MOD = 10**9+7
def solve():
n, k, q = list(map(int, input().split()))
mx = 2*10**5+10
arr = [0]*mx
for _ in range(n):
a, b = list(map(int, input().split()))
arr[a] += 1
arr[b+1] -= 1
for a in range(1,mx):
arr[a] += arr[a-1]
cnt = [0]*mx
for a in range(1,mx):
if arr[a] >= k:
cnt[a] = 1
cnt[a] += cnt[a-1]
for _ in range(q):
a, b = list(map(int, input().split()))
print(cnt[b]-cnt[a-1])
# t = int(input())
t = 1
for _ in range(t):
solve()
|
Title: Karen and Coffee
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range?
Input Specification:
The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Output Specification:
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Demo Input:
['3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n', '2 1 1\n1 1\n200000 200000\n90 100\n']
Demo Output:
['3\n3\n0\n4\n', '0\n']
Note:
In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.
|
```python
import sys,os
# sys.stdin = open("input","r")
from collections import *
from heapq import *
from functools import *
from types import GeneratorType
# from math import *
from itertools import *
from math import *
from string import ascii_lowercase
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
MOD = 10**9+7
def solve():
n, k, q = list(map(int, input().split()))
mx = 2*10**5+10
arr = [0]*mx
for _ in range(n):
a, b = list(map(int, input().split()))
arr[a] += 1
arr[b+1] -= 1
for a in range(1,mx):
arr[a] += arr[a-1]
cnt = [0]*mx
for a in range(1,mx):
if arr[a] >= k:
cnt[a] = 1
cnt[a] += cnt[a-1]
for _ in range(q):
a, b = list(map(int, input().split()))
print(cnt[b]-cnt[a-1])
# t = int(input())
t = 1
for _ in range(t):
solve()
```
| 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"
}
] | 1,516,341,150
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 61
| 5,632,000
|
abc="abcdefghijklmnopqrstuvwxyz"
(a,b),c=(input().split()),1
for i in range(1,len(a)):
if abc.index(a[i])<abc.index(b[0]):
c=i+1
print(a[:c]+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
abc="abcdefghijklmnopqrstuvwxyz"
(a,b),c=(input().split()),1
for i in range(1,len(a)):
if abc.index(a[i])<abc.index(b[0]):
c=i+1
print(a[:c]+b[0])
```
| 0
|
|
702
|
A
|
Maximum Increase
|
PROGRAMMING
| 800
|
[
"dp",
"greedy",
"implementation"
] | null | null |
You are given array consisting of *n* integers. Your task is to find the maximum length of an increasing subarray of the given array.
A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous.
|
The first line contains single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
|
Print the maximum length of an increasing subarray of the given array.
|
[
"5\n1 7 2 11 15\n",
"6\n100 100 100 100 100 100\n",
"3\n1 2 3\n"
] |
[
"3\n",
"1\n",
"3\n"
] |
none
| 0
|
[
{
"input": "5\n1 7 2 11 15",
"output": "3"
},
{
"input": "6\n100 100 100 100 100 100",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "10\n802030518 598196518 640274071 983359971 71550121 96204862 799843967 446173607 796619138 402690754",
"output": "3"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n1 2 3 3 4",
"output": "3"
},
{
"input": "4\n1 2 2 3",
"output": "2"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "3\n1 2 1",
"output": "2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n1 1 2",
"output": "2"
},
{
"input": "11\n1 2 3 1 2 3 2 1 2 3 4",
"output": "4"
},
{
"input": "9\n1 2 3 4 5 6 7 8 9",
"output": "9"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "3\n3 2 1",
"output": "1"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "1\n1234394",
"output": "1"
}
] | 1,657,697,307
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 93
| 7,987,200
|
n = int(input())
a = list(map(int, input().split())) + [0]
ans = s = 1
for i in range(n):
if a[i + 1] > a[i]:
s += 1
else:
ans = max(ans, s)
s = 1
print(ans)
|
Title: Maximum Increase
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given array consisting of *n* integers. Your task is to find the maximum length of an increasing subarray of the given array.
A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous.
Input Specification:
The first line contains single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print the maximum length of an increasing subarray of the given array.
Demo Input:
['5\n1 7 2 11 15\n', '6\n100 100 100 100 100 100\n', '3\n1 2 3\n']
Demo Output:
['3\n', '1\n', '3\n']
Note:
none
|
```python
n = int(input())
a = list(map(int, input().split())) + [0]
ans = s = 1
for i in range(n):
if a[i + 1] > a[i]:
s += 1
else:
ans = max(ans, s)
s = 1
print(ans)
```
| 3
|
|
12
|
A
|
Super Agent
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Super Agent
|
2
|
256
|
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3<=×<=3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
|
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
|
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
|
[
"XX.\n...\n.XX\n",
"X.X\nX..\n...\n"
] |
[
"YES\n",
"NO\n"
] |
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
| 0
|
[
{
"input": "XX.\n...\n.XX",
"output": "YES"
},
{
"input": ".X.\n.X.\n.X.",
"output": "YES"
},
{
"input": "XXX\nXXX\nXXX",
"output": "YES"
},
{
"input": "XXX\nX.X\nXXX",
"output": "YES"
},
{
"input": "X..\n.X.\n..X",
"output": "YES"
},
{
"input": "...\nX.X\nX..",
"output": "NO"
},
{
"input": ".X.\nX.X\n.X.",
"output": "YES"
},
{
"input": "X.X\n.X.\nX.X",
"output": "YES"
},
{
"input": "...\n...\n..X",
"output": "NO"
},
{
"input": "XXX\n...\nXXX",
"output": "YES"
},
{
"input": "..X\nX..\n..X",
"output": "NO"
},
{
"input": ".X.\n...\nX.X",
"output": "NO"
},
{
"input": "X.X\nX.X\nX.X",
"output": "YES"
},
{
"input": ".X.\nX.X\nXX.",
"output": "NO"
},
{
"input": "...\nXXX\nXXX",
"output": "NO"
},
{
"input": "XXX\n..X\nXXX",
"output": "NO"
},
{
"input": "X..\nX.X\n.X.",
"output": "NO"
},
{
"input": "...\n..X\nXXX",
"output": "NO"
},
{
"input": "..X\nX.X\nX..",
"output": "YES"
},
{
"input": "..X\n..X\nXXX",
"output": "NO"
},
{
"input": "X..\nX..\nX..",
"output": "NO"
},
{
"input": "XXX\n.X.\nXXX",
"output": "YES"
},
{
"input": "..X\n...\nX..",
"output": "YES"
},
{
"input": "...\n...\nX..",
"output": "NO"
},
{
"input": "...\n...\n.X.",
"output": "NO"
},
{
"input": "...\n...\n..X",
"output": "NO"
},
{
"input": "...\n.X.\nX..",
"output": "NO"
}
] | 1,662,418,876
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 92
| 0
|
a1=input()
a2=input()
a3=input()[::-1]
if (a1==a3)and (a2[0]==a2[2]):
print("YES")
else: print("NO")
|
Title: Super Agent
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3<=×<=3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input Specification:
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output Specification:
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Demo Input:
['XX.\n...\n.XX\n', 'X.X\nX..\n...\n']
Demo Output:
['YES\n', 'NO\n']
Note:
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
|
```python
a1=input()
a2=input()
a3=input()[::-1]
if (a1==a3)and (a2[0]==a2[2]):
print("YES")
else: print("NO")
```
| 3.977
|
427
|
A
|
Police Recruits
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
|
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
|
Print a single integer, the number of crimes which will go untreated.
|
[
"3\n-1 -1 1\n",
"8\n1 -1 1 -1 -1 1 1 1\n",
"11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n"
] |
[
"2\n",
"1\n",
"8\n"
] |
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
| 500
|
[
{
"input": "3\n-1 -1 1",
"output": "2"
},
{
"input": "8\n1 -1 1 -1 -1 1 1 1",
"output": "1"
},
{
"input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "8"
},
{
"input": "7\n-1 -1 1 1 -1 -1 1",
"output": "2"
},
{
"input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 -1 2 2 1 6 5 -1 5",
"output": "5"
},
{
"input": "98\n-1 -1 1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 1 -1 1 -1 1 1 1 1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -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": "13"
},
{
"input": "3\n-1 5 4",
"output": "1"
},
{
"input": "146\n4 -1 -1 -1 -1 -1 -1 -1 -1 -1 4 -1 3 -1 3 -1 -1 1 4 -1 2 -1 -1 3 -1 -1 -1 4 1 -1 -1 -1 4 -1 -1 -1 -1 -1 -1 3 2 -1 3 2 4 5 2 4 1 5 -1 -1 2 -1 -1 1 -1 5 3 -1 1 2 2 3 5 3 -1 -1 3 -1 -1 3 5 5 -1 -1 5 -1 4 4 1 -1 -1 -1 2 1 -1 -1 -1 2 5 3 -1 -1 -1 3 -1 5 4 -1 1 -1 -1 3 -1 -1 3 1 1 2 -1 -1 -1 1 3 1 -1 2 -1 -1 5 5 -1 -1 3 4 5 1 -1 2 -1 -1 -1 3 -1 5 3 2 -1 2 -1 -1 5 -1 3 -1",
"output": "5"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n1 -1",
"output": "0"
},
{
"input": "2\n-1 1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n-1 -1",
"output": "2"
},
{
"input": "4\n10 -1 -1 -1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "4\n-1 -1 1 1",
"output": "2"
}
] | 1,685,354,821
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 63
| 78
| 9,113,600
|
n=int(input())
list1=list(input().split())
list1=[int(x) for x in list1]
list2=[]
count=0
for i in list1:
if(i<0):
if(len(list2)==0):
count+=1
else:
if(list2[0]==1):
del(list2[0])
else:
list2[0]-=1
elif(i>0):
list2.append(i)
print(count)
|
Title: Police Recruits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
Input Specification:
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
Output Specification:
Print a single integer, the number of crimes which will go untreated.
Demo Input:
['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n']
Demo Output:
['2\n', '1\n', '8\n']
Note:
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
|
```python
n=int(input())
list1=list(input().split())
list1=[int(x) for x in list1]
list2=[]
count=0
for i in list1:
if(i<0):
if(len(list2)==0):
count+=1
else:
if(list2[0]==1):
del(list2[0])
else:
list2[0]-=1
elif(i>0):
list2.append(i)
print(count)
```
| 3
|
|
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"
}
] | 1,646,202,345
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 80
| 154
| 0
|
N = int(input())
add = 0
for i in range(N):
add += sum(list(map(int,input().split(" "))))
if(add==0):
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())
add = 0
for i in range(N):
add += sum(list(map(int,input().split(" "))))
if(add==0):
print("YES")
else:
print("NO")
```
| 0
|
761
|
D
|
Dasha and Very Difficult Problem
|
PROGRAMMING
| 1,700
|
[
"binary search",
"brute force",
"constructive algorithms",
"greedy",
"sortings"
] | null | null |
Dasha logged into the system and began to solve problems. One of them is as follows:
Given two sequences *a* and *b* of length *n* each you need to write a sequence *c* of length *n*, the *i*-th element of which is calculated as follows: *c**i*<==<=*b**i*<=-<=*a**i*.
About sequences *a* and *b* we know that their elements are in the range from *l* to *r*. More formally, elements satisfy the following conditions: *l*<=≤<=*a**i*<=≤<=*r* and *l*<=≤<=*b**i*<=≤<=*r*. About sequence *c* we know that all its elements are distinct.
Dasha wrote a solution to that problem quickly, but checking her work on the standard test was not so easy. Due to an error in the test system only the sequence *a* and the compressed sequence of the sequence *c* were known from that test.
Let's give the definition to a compressed sequence. A compressed sequence of sequence *c* of length *n* is a sequence *p* of length *n*, so that *p**i* equals to the number of integers which are less than or equal to *c**i* in the sequence *c*. For example, for the sequence *c*<==<=[250,<=200,<=300,<=100,<=50] the compressed sequence will be *p*<==<=[4,<=3,<=5,<=2,<=1]. Pay attention that in *c* all integers are distinct. Consequently, the compressed sequence contains all integers from 1 to *n* inclusively.
Help Dasha to find any sequence *b* for which the calculated compressed sequence of sequence *c* is correct.
|
The first line contains three integers *n*, *l*, *r* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*l*<=≤<=*r*<=≤<=109) — the length of the sequence and boundaries of the segment where the elements of sequences *a* and *b* are.
The next line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (*l*<=≤<=*a**i*<=≤<=*r*) — the elements of the sequence *a*.
The next line contains *n* distinct integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the compressed sequence of the sequence *c*.
|
If there is no the suitable sequence *b*, then in the only line print "-1".
Otherwise, in the only line print *n* integers — the elements of any suitable sequence *b*.
|
[
"5 1 5\n1 1 1 1 1\n3 1 5 4 2\n",
"4 2 9\n3 4 8 9\n3 2 1 4\n",
"6 1 5\n1 1 1 1 1 1\n2 3 5 4 1 6\n"
] |
[
"3 1 5 4 2 ",
"2 2 2 9 ",
"-1\n"
] |
Sequence *b* which was found in the second sample is suitable, because calculated sequence *c* = [2 - 3, 2 - 4, 2 - 8, 9 - 9] = [ - 1, - 2, - 6, 0] (note that *c*<sub class="lower-index">*i*</sub> = *b*<sub class="lower-index">*i*</sub> - *a*<sub class="lower-index">*i*</sub>) has compressed sequence equals to *p* = [3, 2, 1, 4].
| 2,000
|
[
{
"input": "5 1 5\n1 1 1 1 1\n3 1 5 4 2",
"output": "3 1 5 4 2 "
},
{
"input": "4 2 9\n3 4 8 9\n3 2 1 4",
"output": "2 2 2 9 "
},
{
"input": "6 1 5\n1 1 1 1 1 1\n2 3 5 4 1 6",
"output": "-1"
},
{
"input": "5 1 7\n1 4 4 6 5\n5 2 1 4 3",
"output": "2 2 1 6 4 "
},
{
"input": "5 10 100\n12 14 15 11 13\n4 2 1 5 3",
"output": "10 10 10 10 10 "
},
{
"input": "2 1 1000000000\n1000000000 1\n2 1",
"output": "-1"
},
{
"input": "2 1 1000000000\n1000000000 1\n1 2",
"output": "1 1 "
},
{
"input": "5 1 5\n1 1 1 1 1\n1 2 3 4 5",
"output": "1 2 3 4 5 "
},
{
"input": "5 1 5\n1 1 1 1 1\n2 3 1 5 4",
"output": "2 3 1 5 4 "
},
{
"input": "1 1000000000 1000000000\n1000000000\n1",
"output": "1000000000 "
},
{
"input": "6 3 7\n6 7 5 5 5 5\n2 1 4 3 5 6",
"output": "3 3 4 3 5 6 "
},
{
"input": "3 5 100\n10 50 100\n3 2 1",
"output": "5 5 5 "
},
{
"input": "10 1 10\n9 2 9 5 5 2 6 8 2 8\n2 10 1 6 7 8 5 3 9 4",
"output": "2 3 1 2 3 1 2 2 2 3 "
},
{
"input": "30 100 200\n102 108 122 116 107 145 195 145 119 110 187 196 140 174 104 190 193 181 118 127 157 111 139 175 173 191 181 105 142 166\n30 26 20 23 27 15 2 14 21 25 6 1 17 10 29 5 3 7 22 19 13 24 18 9 11 4 8 28 16 12",
"output": "100 100 100 100 100 101 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 101 100 100 100 "
},
{
"input": "50 10 15\n13 14 12 14 12 15 13 10 11 11 15 10 14 11 14 12 11 10 10 13 11 13 12 15 14 15 13 15 13 10 12 10 15 15 10 14 11 11 10 14 11 12 13 12 10 11 13 15 14 11\n20 10 25 13 29 6 24 47 37 39 2 44 12 33 9 26 35 49 46 19 38 23 30 5 16 4 21 7 18 45 28 43 1 8 48 15 36 40 50 11 32 27 22 31 42 41 17 3 14 34",
"output": "-1"
},
{
"input": "10 8 10\n8 10 10 9 8 10 10 10 10 10\n9 5 6 8 10 2 7 3 1 4",
"output": "-1"
},
{
"input": "50 1 50\n6 20 27 26 46 35 41 41 4 12 48 13 20 5 35 7 37 3 18 17 22 42 28 1 30 46 38 38 42 29 33 42 8 45 22 40 11 33 12 15 16 49 30 19 29 40 5 12 32 22\n47 27 35 40 31 26 42 19 11 49 37 3 41 34 36 30 16 44 12 4 46 20 14 39 32 25 18 1 21 6 2 17 48 7 5 24 29 33 50 43 8 15 28 23 22 13 45 38 10 9",
"output": "-1"
},
{
"input": "11 5 11\n9 8 7 5 7 5 9 5 10 5 7\n3 4 6 9 5 11 2 10 1 8 7",
"output": "6 6 7 8 6 10 5 9 5 7 8 "
}
] | 1,486,721,495
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 0
| 2,000
| 23,244,800
|
n, l, r = input().split(' ')
n = int(n)
l = int(l)
r = int(r)
at = input().split(' ')
pt = input().split(' ')
a = []
b = []
p = []
for i in range(n):
a.append(int(at[i]))
p.append(int(pt[i]))
b.append(None)
flag = True
i = 1
while (i <= n):
ind = p.index(i)
if i == 1:
b[ind] = l
last = l - a[ind]
continue
exp = last + a[ind] + 1
if exp > r:
flag = False
break
exp = max([l, exp])
last = exp - a[ind]
b[ind] = exp
i += 1
if Flag:
print(' '.join(b))
else:
print(-1)
|
Title: Dasha and Very Difficult Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dasha logged into the system and began to solve problems. One of them is as follows:
Given two sequences *a* and *b* of length *n* each you need to write a sequence *c* of length *n*, the *i*-th element of which is calculated as follows: *c**i*<==<=*b**i*<=-<=*a**i*.
About sequences *a* and *b* we know that their elements are in the range from *l* to *r*. More formally, elements satisfy the following conditions: *l*<=≤<=*a**i*<=≤<=*r* and *l*<=≤<=*b**i*<=≤<=*r*. About sequence *c* we know that all its elements are distinct.
Dasha wrote a solution to that problem quickly, but checking her work on the standard test was not so easy. Due to an error in the test system only the sequence *a* and the compressed sequence of the sequence *c* were known from that test.
Let's give the definition to a compressed sequence. A compressed sequence of sequence *c* of length *n* is a sequence *p* of length *n*, so that *p**i* equals to the number of integers which are less than or equal to *c**i* in the sequence *c*. For example, for the sequence *c*<==<=[250,<=200,<=300,<=100,<=50] the compressed sequence will be *p*<==<=[4,<=3,<=5,<=2,<=1]. Pay attention that in *c* all integers are distinct. Consequently, the compressed sequence contains all integers from 1 to *n* inclusively.
Help Dasha to find any sequence *b* for which the calculated compressed sequence of sequence *c* is correct.
Input Specification:
The first line contains three integers *n*, *l*, *r* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*l*<=≤<=*r*<=≤<=109) — the length of the sequence and boundaries of the segment where the elements of sequences *a* and *b* are.
The next line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (*l*<=≤<=*a**i*<=≤<=*r*) — the elements of the sequence *a*.
The next line contains *n* distinct integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the compressed sequence of the sequence *c*.
Output Specification:
If there is no the suitable sequence *b*, then in the only line print "-1".
Otherwise, in the only line print *n* integers — the elements of any suitable sequence *b*.
Demo Input:
['5 1 5\n1 1 1 1 1\n3 1 5 4 2\n', '4 2 9\n3 4 8 9\n3 2 1 4\n', '6 1 5\n1 1 1 1 1 1\n2 3 5 4 1 6\n']
Demo Output:
['3 1 5 4 2 ', '2 2 2 9 ', '-1\n']
Note:
Sequence *b* which was found in the second sample is suitable, because calculated sequence *c* = [2 - 3, 2 - 4, 2 - 8, 9 - 9] = [ - 1, - 2, - 6, 0] (note that *c*<sub class="lower-index">*i*</sub> = *b*<sub class="lower-index">*i*</sub> - *a*<sub class="lower-index">*i*</sub>) has compressed sequence equals to *p* = [3, 2, 1, 4].
|
```python
n, l, r = input().split(' ')
n = int(n)
l = int(l)
r = int(r)
at = input().split(' ')
pt = input().split(' ')
a = []
b = []
p = []
for i in range(n):
a.append(int(at[i]))
p.append(int(pt[i]))
b.append(None)
flag = True
i = 1
while (i <= n):
ind = p.index(i)
if i == 1:
b[ind] = l
last = l - a[ind]
continue
exp = last + a[ind] + 1
if exp > r:
flag = False
break
exp = max([l, exp])
last = exp - a[ind]
b[ind] = exp
i += 1
if Flag:
print(' '.join(b))
else:
print(-1)
```
| 0
|
|
500
|
A
|
New Year Transportation
|
PROGRAMMING
| 1,000
|
[
"dfs and similar",
"graphs",
"implementation"
] | null | null |
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
|
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
|
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
|
[
"8 4\n1 2 1 2 1 2 1\n",
"8 5\n1 2 1 2 1 1 1\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
| 500
|
[
{
"input": "8 4\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "8 5\n1 2 1 2 1 1 1",
"output": "NO"
},
{
"input": "20 19\n13 16 7 6 12 1 5 7 8 6 5 7 5 5 3 3 2 2 1",
"output": "YES"
},
{
"input": "50 49\n11 7 1 41 26 36 19 16 38 14 36 35 37 27 20 27 3 6 21 2 27 11 18 17 19 16 22 8 8 9 1 7 5 12 5 6 13 6 11 2 6 3 1 5 1 1 2 2 1",
"output": "YES"
},
{
"input": "120 104\n41 15 95 85 34 11 25 42 65 39 77 80 74 17 66 73 21 14 36 63 63 79 45 24 65 7 63 80 51 21 2 19 78 28 71 2 15 23 17 68 62 18 54 39 43 70 3 46 34 23 41 65 32 10 13 18 10 3 16 48 54 18 57 28 3 24 44 50 15 2 20 22 45 44 3 29 2 27 11 2 12 25 25 31 1 2 32 4 11 30 13 16 26 21 1 13 21 8 15 5 18 13 5 15 3 8 13 6 5 1 9 7 1 2 4 1 1 2 1",
"output": "NO"
},
{
"input": "10 3\n8 3 5 4 2 3 2 2 1",
"output": "NO"
},
{
"input": "10 9\n8 3 5 4 2 3 2 2 1",
"output": "YES"
},
{
"input": "3 2\n1 1",
"output": "YES"
},
{
"input": "3 2\n2 1",
"output": "NO"
},
{
"input": "4 2\n2 1 1",
"output": "NO"
},
{
"input": "4 4\n2 2 1",
"output": "YES"
},
{
"input": "8 8\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "3 3\n1 1",
"output": "YES"
},
{
"input": "8 8\n1 2 1 2 1 1 1",
"output": "YES"
},
{
"input": "3 3\n2 1",
"output": "YES"
},
{
"input": "4 4\n1 1 1",
"output": "YES"
},
{
"input": "8 8\n1 1 1 1 1 1 1",
"output": "YES"
},
{
"input": "5 5\n1 1 1 1",
"output": "YES"
}
] | 1,676,668,837
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
if __name__ == "__main__":
n. to = map(int, input().split())
a = list(map(int, input().split()))
ok = False
for i in range(n):
i_can_go_to = a[i] + i + 1
if i_can_go_to == to:
print("YES")
ok = True
break
if not ok:
print("NO")
|
Title: New Year Transportation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
Input Specification:
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output Specification:
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
Demo Input:
['8 4\n1 2 1 2 1 2 1\n', '8 5\n1 2 1 2 1 1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
|
```python
if __name__ == "__main__":
n. to = map(int, input().split())
a = list(map(int, input().split()))
ok = False
for i in range(n):
i_can_go_to = a[i] + i + 1
if i_can_go_to == to:
print("YES")
ok = True
break
if not ok:
print("NO")
```
| -1
|
|
144
|
A
|
Arrival of the General
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
A Ministry for Defense sent a general to inspect the Super Secret Military Squad under the command of the Colonel SuperDuper. Having learned the news, the colonel ordered to all *n* squad soldiers to line up on the parade ground.
By the military charter the soldiers should stand in the order of non-increasing of their height. But as there's virtually no time to do that, the soldiers lined up in the arbitrary order. However, the general is rather short-sighted and he thinks that the soldiers lined up correctly if the first soldier in the line has the maximum height and the last soldier has the minimum height. Please note that the way other solders are positioned does not matter, including the case when there are several soldiers whose height is maximum or minimum. Only the heights of the first and the last soldier are important.
For example, the general considers the sequence of heights (4, 3, 4, 2, 1, 1) correct and the sequence (4, 3, 1, 2, 2) wrong.
Within one second the colonel can swap any two neighboring soldiers. Help him count the minimum time needed to form a line-up which the general will consider correct.
|
The first input line contains the only integer *n* (2<=≤<=*n*<=≤<=100) which represents the number of soldiers in the line. The second line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) the values of the soldiers' heights in the order of soldiers' heights' increasing in the order from the beginning of the line to its end. The numbers are space-separated. Numbers *a*1,<=*a*2,<=...,<=*a**n* are not necessarily different.
|
Print the only integer — the minimum number of seconds the colonel will need to form a line-up the general will like.
|
[
"4\n33 44 11 22\n",
"7\n10 10 58 31 63 40 76\n"
] |
[
"2\n",
"10\n"
] |
In the first sample the colonel will need to swap the first and second soldier and then the third and fourth soldier. That will take 2 seconds. The resulting position of the soldiers is (44, 33, 22, 11).
In the second sample the colonel may swap the soldiers in the following sequence:
1. (10, 10, 58, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 76, 40) 1. (10, 58, 10, 31, 76, 63, 40) 1. (10, 58, 31, 10, 76, 63, 40) 1. (10, 58, 31, 76, 10, 63, 40) 1. (10, 58, 31, 76, 63, 10, 40) 1. (10, 58, 76, 31, 63, 10, 40) 1. (10, 76, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 40, 10)
| 500
|
[
{
"input": "4\n33 44 11 22",
"output": "2"
},
{
"input": "7\n10 10 58 31 63 40 76",
"output": "10"
},
{
"input": "2\n88 89",
"output": "1"
},
{
"input": "5\n100 95 100 100 88",
"output": "0"
},
{
"input": "7\n48 48 48 48 45 45 45",
"output": "0"
},
{
"input": "10\n68 47 67 29 63 71 71 65 54 56",
"output": "10"
},
{
"input": "15\n77 68 96 60 92 75 61 60 66 79 80 65 60 95 92",
"output": "4"
},
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "20\n30 30 30 14 30 14 30 30 30 14 30 14 14 30 14 14 30 14 14 14",
"output": "0"
},
{
"input": "35\n37 41 46 39 47 39 44 47 44 42 44 43 47 39 46 39 38 42 39 37 40 44 41 42 41 42 39 42 36 36 42 36 42 42 42",
"output": "7"
},
{
"input": "40\n99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 98 99 99 99 99 99 99 99 99 100 99 99 99 99 99 99",
"output": "47"
},
{
"input": "50\n48 52 44 54 53 56 62 49 39 41 53 39 40 64 53 50 62 48 40 52 51 48 40 52 61 62 62 61 48 64 55 57 56 40 48 58 41 60 60 56 64 50 64 45 48 45 46 63 59 57",
"output": "50"
},
{
"input": "57\n7 24 17 19 6 19 10 11 12 22 14 5 5 11 13 10 24 19 24 24 24 11 21 20 4 14 24 24 18 13 24 3 20 3 3 3 3 9 3 9 22 22 16 3 3 3 15 11 3 3 8 17 10 13 3 14 13",
"output": "3"
},
{
"input": "65\n58 50 35 44 35 37 36 58 38 36 58 56 56 49 48 56 58 43 40 44 52 44 58 58 57 50 43 35 55 39 38 49 53 56 50 42 41 56 34 57 49 38 34 51 56 38 58 40 53 46 48 34 38 43 49 49 58 56 41 43 44 34 38 48 36",
"output": "3"
},
{
"input": "69\n70 48 49 48 49 71 48 53 55 69 48 53 54 58 53 63 48 48 69 67 72 75 71 75 74 74 57 63 65 60 48 48 65 48 48 51 50 49 62 53 76 68 76 56 76 76 64 76 76 57 61 76 73 51 59 76 65 50 69 50 76 67 76 63 62 74 74 58 73",
"output": "73"
},
{
"input": "75\n70 65 64 71 71 64 71 64 68 71 65 64 65 68 71 66 66 69 68 63 69 65 71 69 68 68 71 67 71 65 65 65 71 71 65 69 63 66 62 67 64 63 62 64 67 65 62 69 62 64 69 62 67 64 67 70 64 63 64 64 69 62 62 64 70 62 62 68 67 69 62 64 66 70 68",
"output": "7"
},
{
"input": "84\n92 95 84 85 94 80 90 86 80 92 95 84 86 83 86 83 93 91 95 92 84 88 82 84 84 84 80 94 93 80 94 80 95 83 85 80 95 95 80 84 86 92 83 81 90 87 81 89 92 93 80 87 90 85 93 85 93 94 93 89 94 83 93 91 80 83 90 94 95 80 95 92 85 84 93 94 94 82 91 95 95 89 85 94",
"output": "15"
},
{
"input": "90\n86 87 72 77 82 71 75 78 61 67 79 90 64 94 94 74 85 87 73 76 71 71 60 69 77 73 76 80 82 57 62 57 57 83 76 72 75 87 72 94 77 85 59 82 86 69 62 80 95 73 83 94 79 85 91 68 85 74 93 95 68 75 89 93 83 78 95 78 83 77 81 85 66 92 63 65 75 78 67 91 77 74 59 86 77 76 90 67 70 64",
"output": "104"
},
{
"input": "91\n94 98 96 94 95 98 98 95 98 94 94 98 95 95 99 97 97 94 95 98 94 98 96 98 96 98 97 95 94 94 94 97 94 96 98 98 98 94 96 95 94 95 97 97 97 98 94 98 96 95 98 96 96 98 94 97 96 98 97 95 97 98 94 95 94 94 97 94 96 97 97 93 94 95 95 94 96 98 97 96 94 98 98 96 96 96 96 96 94 96 97",
"output": "33"
},
{
"input": "92\n44 28 32 29 41 41 36 39 40 39 41 35 41 28 35 27 41 34 28 38 43 43 41 38 27 26 28 36 30 29 39 32 35 35 32 30 39 30 37 27 41 41 28 30 43 31 35 33 36 28 44 40 41 35 31 42 37 38 37 34 39 40 27 40 33 33 44 43 34 33 34 34 35 38 38 37 30 39 35 41 45 42 41 32 33 33 31 30 43 41 43 43",
"output": "145"
},
{
"input": "93\n46 32 52 36 39 30 57 63 63 30 32 44 27 59 46 38 40 45 44 62 35 36 51 48 39 58 36 51 51 51 48 58 59 36 29 35 31 49 64 60 34 38 42 56 33 42 52 31 63 34 45 51 35 45 33 53 33 62 31 38 66 29 51 54 28 61 32 45 57 41 36 34 47 36 31 28 67 48 52 46 32 40 64 58 27 53 43 57 34 66 43 39 26",
"output": "76"
},
{
"input": "94\n56 55 54 31 32 42 46 29 24 54 40 40 20 45 35 56 32 33 51 39 26 56 21 56 51 27 29 39 56 52 54 43 43 55 48 51 44 49 52 49 23 19 19 28 20 26 45 33 35 51 42 36 25 25 38 23 21 35 54 50 41 20 37 28 42 20 22 43 37 34 55 21 24 38 19 41 45 34 19 33 44 54 38 31 23 53 35 32 47 40 39 31 20 34",
"output": "15"
},
{
"input": "95\n57 71 70 77 64 64 76 81 81 58 63 75 81 77 71 71 71 60 70 70 69 67 62 64 78 64 69 62 76 76 57 70 68 77 70 68 73 77 79 73 60 57 69 60 74 65 58 75 75 74 73 73 65 75 72 57 81 62 62 70 67 58 76 57 79 81 68 64 58 77 70 59 79 64 80 58 71 59 81 71 80 64 78 80 78 65 70 68 78 80 57 63 64 76 81",
"output": "11"
},
{
"input": "96\n96 95 95 95 96 97 95 97 96 95 98 96 97 95 98 96 98 96 98 96 98 95 96 95 95 95 97 97 95 95 98 98 95 96 96 95 97 96 98 96 95 97 97 95 97 97 95 94 96 96 97 96 97 97 96 94 94 97 95 95 95 96 95 96 95 97 97 95 97 96 95 94 97 97 97 96 97 95 96 94 94 95 97 94 94 97 97 97 95 97 97 95 94 96 95 95",
"output": "13"
},
{
"input": "97\n14 15 12 12 13 15 12 15 12 12 12 12 12 14 15 15 13 12 15 15 12 12 12 13 14 15 15 13 14 15 14 14 14 14 12 13 12 13 13 12 15 12 13 13 15 12 15 13 12 13 13 13 14 13 12 15 14 13 14 15 13 14 14 13 14 12 15 12 14 12 13 14 15 14 13 15 13 12 15 15 15 13 15 15 13 14 16 16 16 13 15 13 15 14 15 15 15",
"output": "104"
},
{
"input": "98\n37 69 35 70 58 69 36 47 41 63 60 54 49 35 55 50 35 53 52 43 35 41 40 49 38 35 48 70 42 35 35 65 56 54 44 59 59 48 51 49 59 67 35 60 69 35 58 50 35 44 48 69 41 58 44 45 35 47 70 61 49 47 37 39 35 51 44 70 72 65 36 41 63 63 48 66 45 50 50 71 37 52 72 67 72 39 72 39 36 64 48 72 69 49 45 72 72 67",
"output": "100"
},
{
"input": "99\n31 31 16 15 19 31 19 22 29 27 12 22 28 30 25 33 26 25 19 22 34 21 17 33 31 22 16 26 22 30 31 17 13 33 13 17 28 25 18 33 27 22 31 22 13 27 20 22 23 15 24 32 29 13 16 20 32 33 14 33 19 27 16 28 25 17 17 28 18 26 32 33 19 23 30 13 14 23 24 28 14 28 22 20 30 14 24 23 17 29 18 28 29 21 28 18 16 24 32",
"output": "107"
},
{
"input": "100\n37 54 39 29 32 49 21 13 34 21 16 42 34 27 16 26 7 34 51 9 11 27 16 40 36 7 48 52 30 42 42 52 51 11 32 26 6 7 28 54 48 51 6 54 42 20 51 48 46 4 4 31 47 6 9 16 8 23 36 50 49 30 47 37 45 24 48 26 8 50 18 24 27 13 39 7 37 34 5 43 42 19 34 23 52 38 35 6 29 11 53 49 21 29 36 22 45 33 51 22",
"output": "50"
},
{
"input": "100\n41 5 20 23 5 47 28 28 64 5 12 14 34 56 22 48 6 52 39 55 62 40 24 54 39 28 56 27 35 51 14 24 38 10 55 20 39 62 52 2 30 55 49 48 15 23 59 44 31 28 66 29 62 59 31 13 28 12 50 45 41 47 29 27 64 66 14 39 62 22 44 63 24 5 63 17 42 7 25 41 49 33 31 51 63 3 37 66 19 9 2 40 16 45 56 21 34 29 39 61",
"output": "59"
},
{
"input": "100\n66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95",
"output": "99"
},
{
"input": "100\n81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "0"
},
{
"input": "3\n2 1 2",
"output": "1"
},
{
"input": "2\n100 99",
"output": "0"
},
{
"input": "100\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "197"
},
{
"input": "100\n87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87",
"output": "0"
},
{
"input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 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": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "3\n2 2 2",
"output": "0"
},
{
"input": "3\n1 3 2",
"output": "2"
}
] | 1,688,572,639
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 39
| 92
| 0
|
n = int(input())
l = list(map(int, input().split()))
s = n - l[::-1].index(min(l)) - 1
b = l.index(max(l))
if s < b:
print(n-s+b-2)
else:
print(n-s+b-1)
|
Title: Arrival of the General
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A Ministry for Defense sent a general to inspect the Super Secret Military Squad under the command of the Colonel SuperDuper. Having learned the news, the colonel ordered to all *n* squad soldiers to line up on the parade ground.
By the military charter the soldiers should stand in the order of non-increasing of their height. But as there's virtually no time to do that, the soldiers lined up in the arbitrary order. However, the general is rather short-sighted and he thinks that the soldiers lined up correctly if the first soldier in the line has the maximum height and the last soldier has the minimum height. Please note that the way other solders are positioned does not matter, including the case when there are several soldiers whose height is maximum or minimum. Only the heights of the first and the last soldier are important.
For example, the general considers the sequence of heights (4, 3, 4, 2, 1, 1) correct and the sequence (4, 3, 1, 2, 2) wrong.
Within one second the colonel can swap any two neighboring soldiers. Help him count the minimum time needed to form a line-up which the general will consider correct.
Input Specification:
The first input line contains the only integer *n* (2<=≤<=*n*<=≤<=100) which represents the number of soldiers in the line. The second line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) the values of the soldiers' heights in the order of soldiers' heights' increasing in the order from the beginning of the line to its end. The numbers are space-separated. Numbers *a*1,<=*a*2,<=...,<=*a**n* are not necessarily different.
Output Specification:
Print the only integer — the minimum number of seconds the colonel will need to form a line-up the general will like.
Demo Input:
['4\n33 44 11 22\n', '7\n10 10 58 31 63 40 76\n']
Demo Output:
['2\n', '10\n']
Note:
In the first sample the colonel will need to swap the first and second soldier and then the third and fourth soldier. That will take 2 seconds. The resulting position of the soldiers is (44, 33, 22, 11).
In the second sample the colonel may swap the soldiers in the following sequence:
1. (10, 10, 58, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 76, 40) 1. (10, 58, 10, 31, 76, 63, 40) 1. (10, 58, 31, 10, 76, 63, 40) 1. (10, 58, 31, 76, 10, 63, 40) 1. (10, 58, 31, 76, 63, 10, 40) 1. (10, 58, 76, 31, 63, 10, 40) 1. (10, 76, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 40, 10)
|
```python
n = int(input())
l = list(map(int, input().split()))
s = n - l[::-1].index(min(l)) - 1
b = l.index(max(l))
if s < b:
print(n-s+b-2)
else:
print(n-s+b-1)
```
| 3
|
|
982
|
C
|
Cut 'em all!
|
PROGRAMMING
| 1,500
|
[
"dfs and similar",
"dp",
"graphs",
"greedy",
"trees"
] | null | null |
You're given a tree with $n$ vertices.
Your task is to determine the maximum possible number of edges that can be removed in such a way that all the remaining connected components will have even size.
|
The first line contains an integer $n$ ($1 \le n \le 10^5$) denoting the size of the tree.
The next $n - 1$ lines contain two integers $u$, $v$ ($1 \le u, v \le n$) each, describing the vertices connected by the $i$-th edge.
It's guaranteed that the given edges form a tree.
|
Output a single integer $k$ — the maximum number of edges that can be removed to leave all connected components with even size, or $-1$ if it is impossible to remove edges in order to satisfy this property.
|
[
"4\n2 4\n4 1\n3 1\n",
"3\n1 2\n1 3\n",
"10\n7 1\n8 4\n8 10\n4 7\n6 5\n9 3\n3 5\n2 10\n2 5\n",
"2\n1 2\n"
] |
[
"1",
"-1",
"4",
"0"
] |
In the first example you can remove the edge between vertices $1$ and $4$. The graph after that will have two connected components with two vertices in each.
In the second example you can't remove edges in such a way that all components have even number of vertices, so the answer is $-1$.
| 1,500
|
[
{
"input": "4\n2 4\n4 1\n3 1",
"output": "1"
},
{
"input": "3\n1 2\n1 3",
"output": "-1"
},
{
"input": "10\n7 1\n8 4\n8 10\n4 7\n6 5\n9 3\n3 5\n2 10\n2 5",
"output": "4"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "1",
"output": "-1"
},
{
"input": "4\n1 2\n1 3\n1 4",
"output": "0"
}
] | 1,589,138,021
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 109
| 6,656,000
|
import networkx as nx
db = 0
G = nx.Graph()
n = int(input("Add meg a csúcsok számát: "))
for i in range(n-1):
u, v = input("Add meg az él két végpontját: ").split()
G.add_edge(int(u),int(v))
if n%2==1:
print(-1)
else:
for i,j in G.edges():
if G.degree[i] == G.degree[j] and G.degree[i] != 1:
db += 1
print(db)
|
Title: Cut 'em all!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You're given a tree with $n$ vertices.
Your task is to determine the maximum possible number of edges that can be removed in such a way that all the remaining connected components will have even size.
Input Specification:
The first line contains an integer $n$ ($1 \le n \le 10^5$) denoting the size of the tree.
The next $n - 1$ lines contain two integers $u$, $v$ ($1 \le u, v \le n$) each, describing the vertices connected by the $i$-th edge.
It's guaranteed that the given edges form a tree.
Output Specification:
Output a single integer $k$ — the maximum number of edges that can be removed to leave all connected components with even size, or $-1$ if it is impossible to remove edges in order to satisfy this property.
Demo Input:
['4\n2 4\n4 1\n3 1\n', '3\n1 2\n1 3\n', '10\n7 1\n8 4\n8 10\n4 7\n6 5\n9 3\n3 5\n2 10\n2 5\n', '2\n1 2\n']
Demo Output:
['1', '-1', '4', '0']
Note:
In the first example you can remove the edge between vertices $1$ and $4$. The graph after that will have two connected components with two vertices in each.
In the second example you can't remove edges in such a way that all components have even number of vertices, so the answer is $-1$.
|
```python
import networkx as nx
db = 0
G = nx.Graph()
n = int(input("Add meg a csúcsok számát: "))
for i in range(n-1):
u, v = input("Add meg az él két végpontját: ").split()
G.add_edge(int(u),int(v))
if n%2==1:
print(-1)
else:
for i,j in G.edges():
if G.degree[i] == G.degree[j] and G.degree[i] != 1:
db += 1
print(db)
```
| -1
|
|
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"
}
] | 1,575,258,562
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 22
| 248
| 307,200
|
Flag= True # для вечного цикла while. Он ниже
upper = 0 # сумма верхних половинок
lower = 0 # сумма нижних половинок
Row_of_bones = []
n = int(input())
for i in range(n): #Получив кол-во костей, последовательно заполняем список
x = list(map(int, (input().split()))) #"Налету" превращаем в список с целыми числами
Row_of_bones.append(x) #Собственно помещаем кость[xi,yi] в ряд []
def check_parity() -> bool: #Функция, в которой я пересчитываю верхние и нижние суммы, а затем проверяю на четность.
upp,low =0,0
for i in Row_of_bones:
upp += i[0]
low += i[1]
if (upp % 2 == 0) and (low % 2 == 0):
#print("Апп и лов после смены кости = ", upp, " " ,low) # Можно отслеживать промежуточные результат
return True
else:
return False
for i in Row_of_bones: # Считаю верхнию и нижнию сумму. В i помещается кость [xi,yi] из ряда
upper += i[0]
lower += i[1]
while(Flag): # Бесконечный цикл, где мы проверяем на четность в первые раз и отсеиваем
if (upper % 2 == 0) and (lower % 2 == 0): #простые случае, когда ничего делать не надо
print(0)
break
elif (n == 1):
print(-1)
break
else:
count = 0 # Счетчик действий
for i in Row_of_bones:
i[0],i[1] = i[1], i[0] # меняем местами значения элементов без буферной переменной, возможность питона
count+=1
if (check_parity()): # После одноразового перемещения первый кости проверяем на четность
print(count) # Если все хорошо - выводим и выходим через break'и
break
if count == n: # Когда после смещения всех костей, ни к чему не пришли, то есть n == count
print(-1)
break
|
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
Flag= True # для вечного цикла while. Он ниже
upper = 0 # сумма верхних половинок
lower = 0 # сумма нижних половинок
Row_of_bones = []
n = int(input())
for i in range(n): #Получив кол-во костей, последовательно заполняем список
x = list(map(int, (input().split()))) #"Налету" превращаем в список с целыми числами
Row_of_bones.append(x) #Собственно помещаем кость[xi,yi] в ряд []
def check_parity() -> bool: #Функция, в которой я пересчитываю верхние и нижние суммы, а затем проверяю на четность.
upp,low =0,0
for i in Row_of_bones:
upp += i[0]
low += i[1]
if (upp % 2 == 0) and (low % 2 == 0):
#print("Апп и лов после смены кости = ", upp, " " ,low) # Можно отслеживать промежуточные результат
return True
else:
return False
for i in Row_of_bones: # Считаю верхнию и нижнию сумму. В i помещается кость [xi,yi] из ряда
upper += i[0]
lower += i[1]
while(Flag): # Бесконечный цикл, где мы проверяем на четность в первые раз и отсеиваем
if (upper % 2 == 0) and (lower % 2 == 0): #простые случае, когда ничего делать не надо
print(0)
break
elif (n == 1):
print(-1)
break
else:
count = 0 # Счетчик действий
for i in Row_of_bones:
i[0],i[1] = i[1], i[0] # меняем местами значения элементов без буферной переменной, возможность питона
count+=1
if (check_parity()): # После одноразового перемещения первый кости проверяем на четность
print(count) # Если все хорошо - выводим и выходим через break'и
break
if count == n: # Когда после смещения всех костей, ни к чему не пришли, то есть n == count
print(-1)
break
```
| 0
|
|
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"
}
] | 1,641,487,765
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 62
| 0
|
def func():
n, m = map(int, input().split())
l = 0
if n % 2 != 0:
if m % 2 != 0:
for i in range(1, 51):
if m % i == 0:
l+=1
if l == 2:
print('YES')
else:
print('NO')
else:
print('NO')
else:
print('NO')
func()
|
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
def func():
n, m = map(int, input().split())
l = 0
if n % 2 != 0:
if m % 2 != 0:
for i in range(1, 51):
if m % i == 0:
l+=1
if l == 2:
print('YES')
else:
print('NO')
else:
print('NO')
else:
print('NO')
func()
```
| 0
|
610
|
B
|
Vika and Squares
|
PROGRAMMING
| 1,300
|
[
"constructive algorithms",
"implementation"
] | null | null |
Vika has *n* jars with paints of distinct colors. All the jars are numbered from 1 to *n* and the *i*-th jar contains *a**i* liters of paint of color *i*.
Vika also has an infinitely long rectangular piece of paper of width 1, consisting of squares of size 1<=×<=1. Squares are numbered 1, 2, 3 and so on. Vika decided that she will start painting squares one by one from left to right, starting from the square number 1 and some arbitrary color. If the square was painted in color *x*, then the next square will be painted in color *x*<=+<=1. In case of *x*<==<=*n*, next square is painted in color 1. If there is no more paint of the color Vika wants to use now, then she stops.
Square is always painted in only one color, and it takes exactly 1 liter of paint. Your task is to calculate the maximum number of squares that might be painted, if Vika chooses right color to paint the first square.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of jars with colors Vika has.
The second line of the input contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is equal to the number of liters of paint in the *i*-th jar, i.e. the number of liters of color *i* that Vika has.
|
The only line of the output should contain a single integer — the maximum number of squares that Vika can paint if she follows the rules described above.
|
[
"5\n2 4 2 3 3\n",
"3\n5 5 5\n",
"6\n10 10 10 1 10 10\n"
] |
[
"12\n",
"15\n",
"11\n"
] |
In the first sample the best strategy is to start painting using color 4. Then the squares will be painted in the following colors (from left to right): 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5.
In the second sample Vika can start to paint using any color.
In the third sample Vika should start painting using color number 5.
| 1,000
|
[
{
"input": "5\n2 4 2 3 3",
"output": "12"
},
{
"input": "3\n5 5 5",
"output": "15"
},
{
"input": "6\n10 10 10 1 10 10",
"output": "11"
},
{
"input": "1\n167959139",
"output": "167959139"
},
{
"input": "10\n896619242 805194919 844752453 848347723 816995848 856813612 805194919 833406689 816255448 805194919",
"output": "8051949194"
},
{
"input": "2\n2 3",
"output": "5"
},
{
"input": "2\n10 10",
"output": "20"
},
{
"input": "2\n1111 1110",
"output": "2221"
},
{
"input": "3\n100 101 100",
"output": "301"
},
{
"input": "3\n100 100 101",
"output": "301"
},
{
"input": "3\n101 100 100",
"output": "301"
},
{
"input": "4\n2 3 2 2",
"output": "9"
},
{
"input": "4\n1 1 1 1",
"output": "4"
},
{
"input": "4\n100 100 100 101",
"output": "401"
},
{
"input": "4\n101 100 100 101",
"output": "402"
},
{
"input": "4\n100 101 101 100",
"output": "402"
},
{
"input": "4\n200 201 200 201",
"output": "801"
},
{
"input": "4\n201 200 201 200",
"output": "801"
},
{
"input": "4\n300 302 302 300",
"output": "1202"
},
{
"input": "8\n5 4 3 2 1 1 1 1",
"output": "12"
},
{
"input": "8\n2 2 1 2 2 1 2 2",
"output": "12"
},
{
"input": "10\n2 1 2 1 2 2 2 2 2 1",
"output": "15"
},
{
"input": "13\n2 2 2 1 1 1 1 1 1 1 2 2 2",
"output": "19"
}
] | 1,700,246,959
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
def max_squares_painted(n, paint_amounts):
# The strategy is to find the number of complete cycles and additional squares that can be painted for each starting color.
max_squares = 0
for start_color in range(n):
# Find the limiting color in the cycle starting from 'start_color'
limiting_paint = float('inf')
for i in range(n):
current_color = (start_color + i) % n
# Adjust the paint amount based on the position in the cycle
adjusted_paint = paint_amounts[current_color] - (i // n)
limiting_paint = min(limiting_paint, adjusted_paint)
# Calculate the total squares that can be painted with this starting color
total_squares = limiting_paint * n
for i in range(n):
current_color = (start_color + i) % n
# Add additional squares that can be painted in the partial cycle
if paint_amounts[current_color] - limiting_paint > 0:
total_squares += 1
else:
break
max_squares = max(max_squares, total_squares)
return max_squares
# Test cases from the prompt
print(max_squares_painted(5, [2, 4, 2, 3, 3])) # Expected output: 12
print(max_squares_painted(3, [5, 5, 5])) # Expected output: 15
print(max_squares_painted(6, [10, 10, 10, 1, 10, 10])) # Expected output: 11
|
Title: Vika and Squares
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vika has *n* jars with paints of distinct colors. All the jars are numbered from 1 to *n* and the *i*-th jar contains *a**i* liters of paint of color *i*.
Vika also has an infinitely long rectangular piece of paper of width 1, consisting of squares of size 1<=×<=1. Squares are numbered 1, 2, 3 and so on. Vika decided that she will start painting squares one by one from left to right, starting from the square number 1 and some arbitrary color. If the square was painted in color *x*, then the next square will be painted in color *x*<=+<=1. In case of *x*<==<=*n*, next square is painted in color 1. If there is no more paint of the color Vika wants to use now, then she stops.
Square is always painted in only one color, and it takes exactly 1 liter of paint. Your task is to calculate the maximum number of squares that might be painted, if Vika chooses right color to paint the first square.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of jars with colors Vika has.
The second line of the input contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is equal to the number of liters of paint in the *i*-th jar, i.e. the number of liters of color *i* that Vika has.
Output Specification:
The only line of the output should contain a single integer — the maximum number of squares that Vika can paint if she follows the rules described above.
Demo Input:
['5\n2 4 2 3 3\n', '3\n5 5 5\n', '6\n10 10 10 1 10 10\n']
Demo Output:
['12\n', '15\n', '11\n']
Note:
In the first sample the best strategy is to start painting using color 4. Then the squares will be painted in the following colors (from left to right): 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5.
In the second sample Vika can start to paint using any color.
In the third sample Vika should start painting using color number 5.
|
```python
def max_squares_painted(n, paint_amounts):
# The strategy is to find the number of complete cycles and additional squares that can be painted for each starting color.
max_squares = 0
for start_color in range(n):
# Find the limiting color in the cycle starting from 'start_color'
limiting_paint = float('inf')
for i in range(n):
current_color = (start_color + i) % n
# Adjust the paint amount based on the position in the cycle
adjusted_paint = paint_amounts[current_color] - (i // n)
limiting_paint = min(limiting_paint, adjusted_paint)
# Calculate the total squares that can be painted with this starting color
total_squares = limiting_paint * n
for i in range(n):
current_color = (start_color + i) % n
# Add additional squares that can be painted in the partial cycle
if paint_amounts[current_color] - limiting_paint > 0:
total_squares += 1
else:
break
max_squares = max(max_squares, total_squares)
return max_squares
# Test cases from the prompt
print(max_squares_painted(5, [2, 4, 2, 3, 3])) # Expected output: 12
print(max_squares_painted(3, [5, 5, 5])) # Expected output: 15
print(max_squares_painted(6, [10, 10, 10, 1, 10, 10])) # Expected output: 11
```
| 0
|
|
884
|
A
|
Book Reading
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Recently Luba bought a very interesting book. She knows that it will take *t* seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of *n* next days. The number of seconds that Luba has to spend working during *i*-th day is *a**i*. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed *n*.
Remember that there are 86400 seconds in a day.
|
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=100, 1<=≤<=*t*<=≤<=106) — the number of days and the time required to read the book.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=86400) — the time Luba has to spend on her work during *i*-th day.
|
Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed *n*.
|
[
"2 2\n86400 86398\n",
"2 86400\n0 86400\n"
] |
[
"2\n",
"1\n"
] |
none
| 0
|
[
{
"input": "2 2\n86400 86398",
"output": "2"
},
{
"input": "2 86400\n0 86400",
"output": "1"
},
{
"input": "2 86400\n1 86399",
"output": "2"
},
{
"input": "100 1000000\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": "12"
},
{
"input": "1 1\n86399",
"output": "1"
},
{
"input": "6 1200\n86400 86400 86000 86000 86000 86400",
"output": "5"
},
{
"input": "6 1200\n86400 86400 86000 86000 86001 86399",
"output": "6"
},
{
"input": "4 172799\n1 1 86400 0",
"output": "4"
},
{
"input": "4 172799\n0 86400 86399 0",
"output": "4"
},
{
"input": "6 1\n1 1 86400 1 86399 1",
"output": "1"
},
{
"input": "4 1\n86400 86399 86400 86400",
"output": "2"
},
{
"input": "4 1\n86400 86400 0 86400",
"output": "3"
}
] | 1,644,603,498
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 16
| 46
| 0
|
n, t = input().split()
lst = list(map(int, input().split()))
time = 0
for item in range(len(lst)):
time += 86400 - lst[item]
if time >= int(t):
print(item + 1)
break
|
Title: Book Reading
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Luba bought a very interesting book. She knows that it will take *t* seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of *n* next days. The number of seconds that Luba has to spend working during *i*-th day is *a**i*. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed *n*.
Remember that there are 86400 seconds in a day.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=100, 1<=≤<=*t*<=≤<=106) — the number of days and the time required to read the book.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=86400) — the time Luba has to spend on her work during *i*-th day.
Output Specification:
Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed *n*.
Demo Input:
['2 2\n86400 86398\n', '2 86400\n0 86400\n']
Demo Output:
['2\n', '1\n']
Note:
none
|
```python
n, t = input().split()
lst = list(map(int, input().split()))
time = 0
for item in range(len(lst)):
time += 86400 - lst[item]
if time >= int(t):
print(item + 1)
break
```
| 3
|
|
903
|
C
|
Boxes Packing
|
PROGRAMMING
| 1,200
|
[
"greedy"
] | null | null |
Mishka has got *n* empty boxes. For every *i* (1<=≤<=*i*<=≤<=*n*), *i*-th box is a cube with side length *a**i*.
Mishka can put a box *i* into another box *j* if the following conditions are met:
- *i*-th box is not put into another box; - *j*-th box doesn't contain any other boxes; - box *i* is smaller than box *j* (*a**i*<=<<=*a**j*).
Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff it is not put into some another box.
Help Mishka to determine the minimum possible number of visible boxes!
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=5000) — the number of boxes Mishka has got.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is the side length of *i*-th box.
|
Print the minimum possible number of visible boxes.
|
[
"3\n1 2 3\n",
"4\n4 2 4 3\n"
] |
[
"1\n",
"2\n"
] |
In the first example it is possible to put box 1 into box 2, and 2 into 3.
In the second example Mishka can put box 2 into box 3, and box 4 into box 1.
| 0
|
[
{
"input": "3\n1 2 3",
"output": "1"
},
{
"input": "4\n4 2 4 3",
"output": "2"
},
{
"input": "10\n58 58 58 58 58 58 58 58 58 58",
"output": "10"
},
{
"input": "10\n86 89 89 86 86 89 86 86 89 89",
"output": "5"
},
{
"input": "100\n981 288 186 186 292 876 341 288 981 360 783 907 292 186 341 292 360 876 360 360 981 398 783 288 292 398 876 981 398 907 783 360 288 981 907 186 360 288 186 981 186 288 907 876 288 907 876 360 341 292 907 783 907 783 292 981 907 292 876 398 783 876 398 341 876 186 288 186 981 341 398 360 907 981 341 186 292 981 292 398 876 783 292 186 360 292 288 292 876 398 288 292 341 288 398 360 360 292 981 360",
"output": "14"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n9",
"output": "1"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n131",
"output": "1"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "9"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "11"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "1"
},
{
"input": "8\n1 1 1 1 1 1 1 1",
"output": "8"
},
{
"input": "5\n1 1 1 1 1",
"output": "5"
}
] | 1,627,333,941
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 46
| 7,065,600
|
from collections import Counter
input()
print(Counter(input().split()).most_common(1)[1])
|
Title: Boxes Packing
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka has got *n* empty boxes. For every *i* (1<=≤<=*i*<=≤<=*n*), *i*-th box is a cube with side length *a**i*.
Mishka can put a box *i* into another box *j* if the following conditions are met:
- *i*-th box is not put into another box; - *j*-th box doesn't contain any other boxes; - box *i* is smaller than box *j* (*a**i*<=<<=*a**j*).
Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff it is not put into some another box.
Help Mishka to determine the minimum possible number of visible boxes!
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=5000) — the number of boxes Mishka has got.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is the side length of *i*-th box.
Output Specification:
Print the minimum possible number of visible boxes.
Demo Input:
['3\n1 2 3\n', '4\n4 2 4 3\n']
Demo Output:
['1\n', '2\n']
Note:
In the first example it is possible to put box 1 into box 2, and 2 into 3.
In the second example Mishka can put box 2 into box 3, and box 4 into box 1.
|
```python
from collections import Counter
input()
print(Counter(input().split()).most_common(1)[1])
```
| -1
|
|
808
|
B
|
Average Sleep Time
|
PROGRAMMING
| 1,300
|
[
"data structures",
"implementation",
"math"
] | null | null |
It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts *k* days!
When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last *n* days. So now he has a sequence *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* is the sleep time on the *i*-th day.
The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider *k* consecutive days as a week. So there will be *n*<=-<=*k*<=+<=1 weeks to take into consideration. For example, if *k*<==<=2, *n*<==<=3 and *a*<==<=[3,<=4,<=7], then the result is .
You should write a program which will calculate average sleep times of Polycarp over all weeks.
|
The first line contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
|
Output average sleeping time over all weeks.
The answer is considered to be correct if its absolute or relative error does not exceed 10<=-<=6. In particular, it is enough to output real number with at least 6 digits after the decimal point.
|
[
"3 2\n3 4 7\n",
"1 1\n10\n",
"8 2\n1 2 4 100000 123 456 789 1\n"
] |
[
"9.0000000000\n",
"10.0000000000\n",
"28964.2857142857\n"
] |
In the third example there are *n* - *k* + 1 = 7 weeks, so the answer is sums of all weeks divided by 7.
| 0
|
[
{
"input": "3 2\n3 4 7",
"output": "9.0000000000"
},
{
"input": "1 1\n10",
"output": "10.0000000000"
},
{
"input": "8 2\n1 2 4 100000 123 456 789 1",
"output": "28964.2857142857"
},
{
"input": "1 1\n1",
"output": "1.0000000000"
},
{
"input": "1 1\n100000",
"output": "100000.0000000000"
},
{
"input": "3 1\n1 2 3",
"output": "2.0000000000"
},
{
"input": "10 4\n11 3 5 20 12 7 9 2 2 20",
"output": "36.2857142857"
},
{
"input": "10 5\n15 9 3 2 17 10 9 18 4 19",
"output": "50.3333333333"
},
{
"input": "10 6\n19 3 20 16 14 10 1 13 7 3",
"output": "65.8000000000"
},
{
"input": "10 7\n8 16 2 13 15 9 5 13 9 2",
"output": "68.2500000000"
},
{
"input": "10 4\n127 1459 718 1183 880 1044 1857 1340 725 1496",
"output": "4574.4285714286"
},
{
"input": "10 5\n1384 1129 1780 1960 1567 1928 12 1523 1165 344",
"output": "6931.3333333333"
}
] | 1,496,489,782
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 218
| 14,233,600
|
n, k = map(int, input().split())
a = list(map(int, input().split()))
total = 0
week = 0
for i in range(0, k):
week+=a[i]
total += week
for i in range(1, n-k+1):
week-=a[i-1]
week+=a[i+k-1]
total += week
print(total / (n-k+1))
|
Title: Average Sleep Time
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts *k* days!
When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last *n* days. So now he has a sequence *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* is the sleep time on the *i*-th day.
The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider *k* consecutive days as a week. So there will be *n*<=-<=*k*<=+<=1 weeks to take into consideration. For example, if *k*<==<=2, *n*<==<=3 and *a*<==<=[3,<=4,<=7], then the result is .
You should write a program which will calculate average sleep times of Polycarp over all weeks.
Input Specification:
The first line contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
Output average sleeping time over all weeks.
The answer is considered to be correct if its absolute or relative error does not exceed 10<=-<=6. In particular, it is enough to output real number with at least 6 digits after the decimal point.
Demo Input:
['3 2\n3 4 7\n', '1 1\n10\n', '8 2\n1 2 4 100000 123 456 789 1\n']
Demo Output:
['9.0000000000\n', '10.0000000000\n', '28964.2857142857\n']
Note:
In the third example there are *n* - *k* + 1 = 7 weeks, so the answer is sums of all weeks divided by 7.
|
```python
n, k = map(int, input().split())
a = list(map(int, input().split()))
total = 0
week = 0
for i in range(0, k):
week+=a[i]
total += week
for i in range(1, n-k+1):
week-=a[i-1]
week+=a[i+k-1]
total += week
print(total / (n-k+1))
```
| 3
|
|
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"
}
] | 1,635,858,167
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 62
| 4,300,800
|
x = int(input(),2)
y = input()
l = len(y)
y = int(y, 2)
print(str(bin(x^y))[2:].zfill(l))
|
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
x = int(input(),2)
y = input()
l = len(y)
y = int(y, 2)
print(str(bin(x^y))[2:].zfill(l))
```
| 3.976489
|
888
|
A
|
Local Extrema
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
You are given an array *a*. Some element of this array *a**i* is a local minimum iff it is strictly less than both of its neighbours (that is, *a**i*<=<<=*a**i*<=-<=1 and *a**i*<=<<=*a**i*<=+<=1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, *a**i*<=><=*a**i*<=-<=1 and *a**i*<=><=*a**i*<=+<=1). Since *a*1 and *a**n* have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in array *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000) — the elements of array *a*.
|
Print the number of local extrema in the given array.
|
[
"3\n1 2 3\n",
"4\n1 5 2 5\n"
] |
[
"0\n",
"2\n"
] |
none
| 0
|
[
{
"input": "3\n1 2 3",
"output": "0"
},
{
"input": "4\n1 5 2 5",
"output": "2"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n548",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "3\n3 2 3",
"output": "1"
},
{
"input": "2\n1 1000",
"output": "0"
},
{
"input": "3\n1 2 1",
"output": "1"
}
] | 1,603,040,729
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 14
| 93
| 307,200
|
n=int(input())
a=list(map(int,input().split()))
count=0
i=1
while i<=n-2:
if (a[i]<a[i-1] and a[i]<a[i+1]) or (a[i]>a[i-1] and a[i]>a[i+1]):
count+=1
i+=1
print(count)
|
Title: Local Extrema
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a*. Some element of this array *a**i* is a local minimum iff it is strictly less than both of its neighbours (that is, *a**i*<=<<=*a**i*<=-<=1 and *a**i*<=<<=*a**i*<=+<=1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, *a**i*<=><=*a**i*<=-<=1 and *a**i*<=><=*a**i*<=+<=1). Since *a*1 and *a**n* have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in array *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000) — the elements of array *a*.
Output Specification:
Print the number of local extrema in the given array.
Demo Input:
['3\n1 2 3\n', '4\n1 5 2 5\n']
Demo Output:
['0\n', '2\n']
Note:
none
|
```python
n=int(input())
a=list(map(int,input().split()))
count=0
i=1
while i<=n-2:
if (a[i]<a[i-1] and a[i]<a[i+1]) or (a[i]>a[i-1] and a[i]>a[i+1]):
count+=1
i+=1
print(count)
```
| 3
|
|
606
|
A
|
Magic Spheres
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
Carl is a beginner magician. He has *a* blue, *b* violet and *c* orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least *x* blue, *y* violet and *z* orange spheres. Can he get them (possible, in multiple actions)?
|
The first line of the input contains three integers *a*, *b* and *c* (0<=≤<=*a*,<=*b*,<=*c*<=≤<=1<=000<=000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=1<=000<=000) — the number of blue, violet and orange spheres that he needs to get.
|
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
|
[
"4 4 0\n2 1 2\n",
"5 6 1\n2 7 2\n",
"3 3 3\n2 2 2\n"
] |
[
"Yes\n",
"No\n",
"Yes\n"
] |
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.
| 500
|
[
{
"input": "4 4 0\n2 1 2",
"output": "Yes"
},
{
"input": "5 6 1\n2 7 2",
"output": "No"
},
{
"input": "3 3 3\n2 2 2",
"output": "Yes"
},
{
"input": "0 0 0\n0 0 0",
"output": "Yes"
},
{
"input": "0 0 0\n0 0 1",
"output": "No"
},
{
"input": "0 1 0\n0 0 0",
"output": "Yes"
},
{
"input": "1 0 0\n1 0 0",
"output": "Yes"
},
{
"input": "2 2 1\n1 1 2",
"output": "No"
},
{
"input": "1 3 1\n2 1 1",
"output": "Yes"
},
{
"input": "1000000 1000000 1000000\n1000000 1000000 1000000",
"output": "Yes"
},
{
"input": "1000000 500000 500000\n0 750000 750000",
"output": "Yes"
},
{
"input": "500000 1000000 500000\n750001 0 750000",
"output": "No"
},
{
"input": "499999 500000 1000000\n750000 750000 0",
"output": "No"
},
{
"input": "500000 500000 0\n0 0 500000",
"output": "Yes"
},
{
"input": "0 500001 499999\n500000 0 0",
"output": "No"
},
{
"input": "1000000 500000 1000000\n500000 1000000 500000",
"output": "Yes"
},
{
"input": "1000000 1000000 499999\n500000 500000 1000000",
"output": "No"
},
{
"input": "500000 1000000 1000000\n1000000 500001 500000",
"output": "No"
},
{
"input": "1000000 500000 500000\n0 1000000 500000",
"output": "Yes"
},
{
"input": "500000 500000 1000000\n500001 1000000 0",
"output": "No"
},
{
"input": "500000 999999 500000\n1000000 0 500000",
"output": "No"
},
{
"input": "4 0 3\n2 2 1",
"output": "Yes"
},
{
"input": "0 2 4\n2 0 2",
"output": "Yes"
},
{
"input": "3 1 0\n1 1 1",
"output": "Yes"
},
{
"input": "4 4 1\n1 3 2",
"output": "Yes"
},
{
"input": "1 2 4\n2 1 3",
"output": "No"
},
{
"input": "1 1 0\n0 0 1",
"output": "No"
},
{
"input": "4 0 0\n0 1 1",
"output": "Yes"
},
{
"input": "0 3 0\n1 0 1",
"output": "No"
},
{
"input": "0 0 3\n1 0 1",
"output": "Yes"
},
{
"input": "1 12 1\n4 0 4",
"output": "Yes"
},
{
"input": "4 0 4\n1 2 1",
"output": "Yes"
},
{
"input": "4 4 0\n1 1 3",
"output": "No"
},
{
"input": "0 9 0\n2 2 2",
"output": "No"
},
{
"input": "0 10 0\n2 2 2",
"output": "Yes"
},
{
"input": "9 0 9\n0 8 0",
"output": "Yes"
},
{
"input": "0 9 9\n9 0 0",
"output": "No"
},
{
"input": "9 10 0\n0 0 9",
"output": "Yes"
},
{
"input": "10 0 9\n0 10 0",
"output": "No"
},
{
"input": "0 10 10\n10 0 0",
"output": "Yes"
},
{
"input": "10 10 0\n0 0 11",
"output": "No"
},
{
"input": "307075 152060 414033\n381653 222949 123101",
"output": "No"
},
{
"input": "569950 228830 153718\n162186 357079 229352",
"output": "No"
},
{
"input": "149416 303568 749016\n238307 493997 190377",
"output": "No"
},
{
"input": "438332 298094 225324\n194220 400244 245231",
"output": "No"
},
{
"input": "293792 300060 511272\n400687 382150 133304",
"output": "No"
},
{
"input": "295449 518151 368838\n382897 137148 471892",
"output": "No"
},
{
"input": "191789 291147 691092\n324321 416045 176232",
"output": "Yes"
},
{
"input": "286845 704749 266526\n392296 104421 461239",
"output": "Yes"
},
{
"input": "135522 188282 377041\n245719 212473 108265",
"output": "Yes"
},
{
"input": "404239 359124 133292\n180069 184791 332544",
"output": "No"
},
{
"input": "191906 624432 244408\n340002 367217 205432",
"output": "No"
},
{
"input": "275980 429361 101824\n274288 302579 166062",
"output": "No"
},
{
"input": "136092 364927 395302\n149173 343146 390922",
"output": "No"
},
{
"input": "613852 334661 146012\n363786 326286 275233",
"output": "No"
},
{
"input": "348369 104625 525203\n285621 215396 366411",
"output": "No"
},
{
"input": "225307 153572 114545\n154753 153282 149967",
"output": "Yes"
},
{
"input": "438576 124465 629784\n375118 276028 390116",
"output": "Yes"
},
{
"input": "447521 327510 158732\n395759 178458 259139",
"output": "Yes"
},
{
"input": "8 5 5\n5 5 5",
"output": "Yes"
},
{
"input": "100 100 100\n1 1 1",
"output": "Yes"
},
{
"input": "100 100 100\n0 0 0",
"output": "Yes"
},
{
"input": "3 2 3\n2 3 2",
"output": "No"
},
{
"input": "5 4 3\n2 2 2",
"output": "Yes"
},
{
"input": "14 9 8\n12 5 10",
"output": "Yes"
},
{
"input": "10 10 10\n1 1 1",
"output": "Yes"
},
{
"input": "6 3 3\n3 3 3",
"output": "Yes"
},
{
"input": "10 0 4\n2 4 2",
"output": "Yes"
},
{
"input": "100 100 100\n2 2 2",
"output": "Yes"
},
{
"input": "4 6 0\n2 1 2",
"output": "Yes"
},
{
"input": "4 6 3\n4 2 3",
"output": "Yes"
},
{
"input": "5 5 5\n1 1 1",
"output": "Yes"
},
{
"input": "41 17 34\n0 19 24",
"output": "Yes"
},
{
"input": "8 8 8\n3 3 3",
"output": "Yes"
},
{
"input": "7 7 1\n1 1 2",
"output": "Yes"
},
{
"input": "6 6 0\n2 2 2",
"output": "Yes"
},
{
"input": "5 5 5\n2 2 2",
"output": "Yes"
},
{
"input": "400 400 400\n1 1 1",
"output": "Yes"
},
{
"input": "4 4 4\n2 2 2",
"output": "Yes"
}
] | 1,449,740,956
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 61
| 204,800
|
in1 = input()
in2 = input()
ar1 = in1.strip().split(" ")
ar2 = in2.strip().split(" ")
abc = [int(ar1[i]) for i in range(0,3)]
xyz = [int(ar2[i]) for i in range(0,3)]
if(abc == xyz):
print("Yes")
exit()
ijk = [abc[i]-xyz[i] for i in range(0,3)]
#print(ijk)
for i in range(0,3):
done = False
if ijk[i]%2 == 1 and ijk[i] > 0:
for j in range(0,3):
if j != i and ijk[j] > 0 and not done:
ijk[j]-=2
ijk[i]+=1
done = True
for j in range(0,3):
if j != i and not done:
ijk[j]-=2
ijk[i]+=1
done = True
#print(ijk)
for i in range(0,3):
if ijk[i] > 0:
ijk[i] = int(ijk[i]/2)
s = sum(ijk)
if sum(ijk)>= 0:
print("Yes")
else:
print("No")
|
Title: Magic Spheres
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Carl is a beginner magician. He has *a* blue, *b* violet and *c* orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least *x* blue, *y* violet and *z* orange spheres. Can he get them (possible, in multiple actions)?
Input Specification:
The first line of the input contains three integers *a*, *b* and *c* (0<=≤<=*a*,<=*b*,<=*c*<=≤<=1<=000<=000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=1<=000<=000) — the number of blue, violet and orange spheres that he needs to get.
Output Specification:
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Demo Input:
['4 4 0\n2 1 2\n', '5 6 1\n2 7 2\n', '3 3 3\n2 2 2\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n']
Note:
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.
|
```python
in1 = input()
in2 = input()
ar1 = in1.strip().split(" ")
ar2 = in2.strip().split(" ")
abc = [int(ar1[i]) for i in range(0,3)]
xyz = [int(ar2[i]) for i in range(0,3)]
if(abc == xyz):
print("Yes")
exit()
ijk = [abc[i]-xyz[i] for i in range(0,3)]
#print(ijk)
for i in range(0,3):
done = False
if ijk[i]%2 == 1 and ijk[i] > 0:
for j in range(0,3):
if j != i and ijk[j] > 0 and not done:
ijk[j]-=2
ijk[i]+=1
done = True
for j in range(0,3):
if j != i and not done:
ijk[j]-=2
ijk[i]+=1
done = True
#print(ijk)
for i in range(0,3):
if ijk[i] > 0:
ijk[i] = int(ijk[i]/2)
s = sum(ijk)
if sum(ijk)>= 0:
print("Yes")
else:
print("No")
```
| 0
|
|
443
|
A
|
Anton and Letters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
|
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
|
Print a single number — the number of distinct letters in Anton's set.
|
[
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] |
[
"3\n",
"2\n",
"0\n"
] |
none
| 500
|
[
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,695,642,074
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
s = [i.strip("'',") for i in input()[1:-1].split(" ")]
print(len(set(s)))
|
Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none
|
```python
s = [i.strip("'',") for i in input()[1:-1].split(" ")]
print(len(set(s)))
```
| 0
|
|
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"
}
] | 1,679,928,702
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 34
| 61
| 0
|
n = int(input())
a = [int(x) for x in input().split()]
i = 0
j = n-1
ans1 = 0
ans2 =0
h = 0
while h < n:
if h%2 == 0:
if (a[i] > a[j]):
ans1+=a[i]
i+=1
else:
ans1 += a[j]
j-=1
else:
if (a[i] > a[j]):
ans2+=a[i]
i+=1
else:
ans2 += a[j]
j-=1
h+=1
print(ans1,ans2)
|
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
n = int(input())
a = [int(x) for x in input().split()]
i = 0
j = n-1
ans1 = 0
ans2 =0
h = 0
while h < n:
if h%2 == 0:
if (a[i] > a[j]):
ans1+=a[i]
i+=1
else:
ans1 += a[j]
j-=1
else:
if (a[i] > a[j]):
ans2+=a[i]
i+=1
else:
ans2 += a[j]
j-=1
h+=1
print(ans1,ans2)
```
| 3
|
|
884
|
A
|
Book Reading
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Recently Luba bought a very interesting book. She knows that it will take *t* seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of *n* next days. The number of seconds that Luba has to spend working during *i*-th day is *a**i*. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed *n*.
Remember that there are 86400 seconds in a day.
|
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=100, 1<=≤<=*t*<=≤<=106) — the number of days and the time required to read the book.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=86400) — the time Luba has to spend on her work during *i*-th day.
|
Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed *n*.
|
[
"2 2\n86400 86398\n",
"2 86400\n0 86400\n"
] |
[
"2\n",
"1\n"
] |
none
| 0
|
[
{
"input": "2 2\n86400 86398",
"output": "2"
},
{
"input": "2 86400\n0 86400",
"output": "1"
},
{
"input": "2 86400\n1 86399",
"output": "2"
},
{
"input": "100 1000000\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": "12"
},
{
"input": "1 1\n86399",
"output": "1"
},
{
"input": "6 1200\n86400 86400 86000 86000 86000 86400",
"output": "5"
},
{
"input": "6 1200\n86400 86400 86000 86000 86001 86399",
"output": "6"
},
{
"input": "4 172799\n1 1 86400 0",
"output": "4"
},
{
"input": "4 172799\n0 86400 86399 0",
"output": "4"
},
{
"input": "6 1\n1 1 86400 1 86399 1",
"output": "1"
},
{
"input": "4 1\n86400 86399 86400 86400",
"output": "2"
},
{
"input": "4 1\n86400 86400 0 86400",
"output": "3"
}
] | 1,509,388,087
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 16
| 62
| 0
|
a, b = map(int, input().split())
c = list(map(int, input().split()))
out = 0
for i in c:
b = b - 86400 + i
out += 1
if(b<=0):
print(out)
break
|
Title: Book Reading
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Luba bought a very interesting book. She knows that it will take *t* seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of *n* next days. The number of seconds that Luba has to spend working during *i*-th day is *a**i*. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed *n*.
Remember that there are 86400 seconds in a day.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=100, 1<=≤<=*t*<=≤<=106) — the number of days and the time required to read the book.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=86400) — the time Luba has to spend on her work during *i*-th day.
Output Specification:
Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed *n*.
Demo Input:
['2 2\n86400 86398\n', '2 86400\n0 86400\n']
Demo Output:
['2\n', '1\n']
Note:
none
|
```python
a, b = map(int, input().split())
c = list(map(int, input().split()))
out = 0
for i in c:
b = b - 86400 + i
out += 1
if(b<=0):
print(out)
break
```
| 3
|
|
749
|
A
|
Bachgold Problem
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"math",
"number theory"
] | null | null |
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
|
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
|
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
|
[
"5\n",
"6\n"
] |
[
"2\n2 3\n",
"3\n2 2 2\n"
] |
none
| 500
|
[
{
"input": "5",
"output": "2\n2 3"
},
{
"input": "6",
"output": "3\n2 2 2"
},
{
"input": "2",
"output": "1\n2"
},
{
"input": "3",
"output": "1\n3"
},
{
"input": "99999",
"output": "49999\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 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 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 ..."
},
{
"input": "100000",
"output": "50000\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 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 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 ..."
},
{
"input": "7",
"output": "3\n2 2 3"
},
{
"input": "4",
"output": "2\n2 2"
},
{
"input": "8",
"output": "4\n2 2 2 2"
},
{
"input": "9",
"output": "4\n2 2 2 3"
},
{
"input": "99995",
"output": "49997\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 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 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 ..."
},
{
"input": "99996",
"output": "49998\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 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 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 ..."
},
{
"input": "10",
"output": "5\n2 2 2 2 2"
},
{
"input": "11",
"output": "5\n2 2 2 2 3"
},
{
"input": "99997",
"output": "49998\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 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 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 ..."
},
{
"input": "99998",
"output": "49999\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 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 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 ..."
},
{
"input": "12",
"output": "6\n2 2 2 2 2 2"
},
{
"input": "13",
"output": "6\n2 2 2 2 2 3"
},
{
"input": "99993",
"output": "49996\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 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 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 ..."
},
{
"input": "99994",
"output": "49997\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 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 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 ..."
},
{
"input": "14",
"output": "7\n2 2 2 2 2 2 2"
},
{
"input": "15",
"output": "7\n2 2 2 2 2 2 3"
},
{
"input": "53",
"output": "26\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 3"
},
{
"input": "57",
"output": "28\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 3"
},
{
"input": "61",
"output": "30\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 3"
},
{
"input": "774",
"output": "387\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 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 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 ..."
},
{
"input": "202",
"output": "101\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 2"
},
{
"input": "530",
"output": "265\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 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 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 ..."
},
{
"input": "7166",
"output": "3583\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 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 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..."
},
{
"input": "9294",
"output": "4647\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 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 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..."
},
{
"input": "2422",
"output": "1211\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 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 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..."
},
{
"input": "15326",
"output": "7663\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 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 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..."
},
{
"input": "11454",
"output": "5727\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 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 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..."
},
{
"input": "14878",
"output": "7439\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 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 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..."
},
{
"input": "90672",
"output": "45336\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 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 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 ..."
},
{
"input": "99544",
"output": "49772\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 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 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 ..."
},
{
"input": "90472",
"output": "45236\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 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 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 ..."
},
{
"input": "23",
"output": "11\n2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "93",
"output": "46\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 3"
},
{
"input": "19",
"output": "9\n2 2 2 2 2 2 2 2 3"
},
{
"input": "11110",
"output": "5555\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 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 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..."
}
] | 1,650,272,718
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
a=int(input())
b=a//2
c=[]
if a%2==0:
for i in range(a//2):
c.append(2)
print(*c)
else:
for i in range(a//2-1):
c.append(2)
c.append(3)
print(*c)
|
Title: Bachgold Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
Output Specification:
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
Demo Input:
['5\n', '6\n']
Demo Output:
['2\n2 3\n', '3\n2 2 2\n']
Note:
none
|
```python
a=int(input())
b=a//2
c=[]
if a%2==0:
for i in range(a//2):
c.append(2)
print(*c)
else:
for i in range(a//2-1):
c.append(2)
c.append(3)
print(*c)
```
| 0
|
|
1
|
A
|
Theatre Square
|
PROGRAMMING
| 1,000
|
[
"math"
] |
A. Theatre Square
|
1
|
256
|
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
|
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
|
Write the needed number of flagstones.
|
[
"6 6 4\n"
] |
[
"4\n"
] |
none
| 0
|
[
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,668,074,313
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 31
| 0
|
n,m,a=map(int,input().split())
count=0
if a*a>=n*m:
print(1)
exit()
elif a==min(n,m):
print(max((n,m)))
exit()
if n%a!=0:
count+=(n//a)+1
else:
count+=n//a
if m%a!=0:
count+=(m//a)+1
else:
count+=m//a
print(count)
|
Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none
|
```python
n,m,a=map(int,input().split())
count=0
if a*a>=n*m:
print(1)
exit()
elif a==min(n,m):
print(max((n,m)))
exit()
if n%a!=0:
count+=(n//a)+1
else:
count+=n//a
if m%a!=0:
count+=(m//a)+1
else:
count+=m//a
print(count)
```
| 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"
}
] | 1,693,503,075
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 9
| 1,000
| 1,433,600
|
n, m = map(int, input().split(" "))
b = m
c = n
d = 0
max_value = 0
# To find minimum number (group of friends) and calculate maximum at the same time
for i in range(m):
if m == 1:
y = n
else:
y = (n - (n % m)) // m + (n % m) // (m - 1)
n = n - y
m = m - 1
if y > 1:
d += (y - 1) * y // 2
if i == 0:
max_value = c - (b - 1)
max_value = (max_value - 1) * max_value // 2
print(d, max_value)
|
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
n, m = map(int, input().split(" "))
b = m
c = n
d = 0
max_value = 0
# To find minimum number (group of friends) and calculate maximum at the same time
for i in range(m):
if m == 1:
y = n
else:
y = (n - (n % m)) // m + (n % m) // (m - 1)
n = n - y
m = m - 1
if y > 1:
d += (y - 1) * y // 2
if i == 0:
max_value = c - (b - 1)
max_value = (max_value - 1) * max_value // 2
print(d, max_value)
```
| 0
|
|
1,004
|
B
|
Sonya and Exhibition
|
PROGRAMMING
| 1,300
|
[
"constructive algorithms",
"greedy",
"implementation",
"math"
] | null | null |
Sonya decided to organize an exhibition of flowers. Since the girl likes only roses and lilies, she decided that only these two kinds of flowers should be in this exhibition.
There are $n$ flowers in a row in the exhibition. Sonya can put either a rose or a lily in the $i$-th position. Thus each of $n$ positions should contain exactly one flower: a rose or a lily.
She knows that exactly $m$ people will visit this exhibition. The $i$-th visitor will visit all flowers from $l_i$ to $r_i$ inclusive. The girl knows that each segment has its own beauty that is equal to the product of the number of roses and the number of lilies.
Sonya wants her exhibition to be liked by a lot of people. That is why she wants to put the flowers in such way that the sum of beauties of all segments would be maximum possible.
|
The first line contains two integers $n$ and $m$ ($1\leq n, m\leq 10^3$) — the number of flowers and visitors respectively.
Each of the next $m$ lines contains two integers $l_i$ and $r_i$ ($1\leq l_i\leq r_i\leq n$), meaning that $i$-th visitor will visit all flowers from $l_i$ to $r_i$ inclusive.
|
Print the string of $n$ characters. The $i$-th symbol should be «0» if you want to put a rose in the $i$-th position, otherwise «1» if you want to put a lily.
If there are multiple answers, print any.
|
[
"5 3\n1 3\n2 4\n2 5\n",
"6 3\n5 6\n1 4\n4 6\n"
] |
[
"01100",
"110010"
] |
In the first example, Sonya can put roses in the first, fourth, and fifth positions, and lilies in the second and third positions;
- in the segment $[1\ldots3]$, there are one rose and two lilies, so the beauty is equal to $1\cdot 2=2$; - in the segment $[2\ldots4]$, there are one rose and two lilies, so the beauty is equal to $1\cdot 2=2$; - in the segment $[2\ldots5]$, there are two roses and two lilies, so the beauty is equal to $2\cdot 2=4$.
The total beauty is equal to $2+2+4=8$.
In the second example, Sonya can put roses in the third, fourth, and sixth positions, and lilies in the first, second, and fifth positions;
- in the segment $[5\ldots6]$, there are one rose and one lily, so the beauty is equal to $1\cdot 1=1$; - in the segment $[1\ldots4]$, there are two roses and two lilies, so the beauty is equal to $2\cdot 2=4$; - in the segment $[4\ldots6]$, there are two roses and one lily, so the beauty is equal to $2\cdot 1=2$.
The total beauty is equal to $1+4+2=7$.
| 1,000
|
[
{
"input": "5 3\n1 3\n2 4\n2 5",
"output": "01010"
},
{
"input": "6 3\n5 6\n1 4\n4 6",
"output": "010101"
},
{
"input": "10 4\n3 3\n1 6\n9 9\n10 10",
"output": "0101010101"
},
{
"input": "1 1\n1 1",
"output": "0"
},
{
"input": "1000 10\n3 998\n2 1000\n1 999\n2 1000\n3 998\n2 1000\n3 998\n1 1000\n2 1000\n3 999",
"output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..."
},
{
"input": "1000 20\n50 109\n317 370\n710 770\n440 488\n711 757\n236 278\n314 355\n131 190\n115 162\n784 834\n16 56\n677 730\n802 844\n632 689\n23 74\n647 702\n930 986\n926 983\n769 822\n508 558",
"output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..."
},
{
"input": "1000 10\n138 238\n160 260\n716 816\n504 604\n98 198\n26 126\n114 214\n217 317\n121 221\n489 589",
"output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..."
},
{
"input": "1000 5\n167 296\n613 753\n650 769\n298 439\n71 209",
"output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..."
},
{
"input": "1000 5\n349 415\n714 773\n125 179\n1 80\n148 242",
"output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..."
},
{
"input": "914 10\n587 646\n770 843\n825 875\n439 485\n465 521\n330 387\n405 480\n477 521\n336 376\n715 771",
"output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..."
},
{
"input": "571 10\n13 94\n450 510\n230 293\n302 375\n304 354\n421 504\n24 87\n122 181\n221 296\n257 307",
"output": "0101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010..."
},
{
"input": "6 2\n1 6\n1 4",
"output": "010101"
},
{
"input": "2 1\n1 2",
"output": "01"
}
] | 1,594,213,612
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 77
| 6,963,200
|
n,m=map(int,input().split())
li=['0']*(n+1)
mc=(n//2)
for i in range(m):
l,r=map(int,input().split())
k=r-l+1
k//=2
j=l
for i in range(k):
if mc!=0:
li[j]='1'
j+=1
mc-=1
# print(li)
s=''.join(li)
print(s[1:])
|
Title: Sonya and Exhibition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sonya decided to organize an exhibition of flowers. Since the girl likes only roses and lilies, she decided that only these two kinds of flowers should be in this exhibition.
There are $n$ flowers in a row in the exhibition. Sonya can put either a rose or a lily in the $i$-th position. Thus each of $n$ positions should contain exactly one flower: a rose or a lily.
She knows that exactly $m$ people will visit this exhibition. The $i$-th visitor will visit all flowers from $l_i$ to $r_i$ inclusive. The girl knows that each segment has its own beauty that is equal to the product of the number of roses and the number of lilies.
Sonya wants her exhibition to be liked by a lot of people. That is why she wants to put the flowers in such way that the sum of beauties of all segments would be maximum possible.
Input Specification:
The first line contains two integers $n$ and $m$ ($1\leq n, m\leq 10^3$) — the number of flowers and visitors respectively.
Each of the next $m$ lines contains two integers $l_i$ and $r_i$ ($1\leq l_i\leq r_i\leq n$), meaning that $i$-th visitor will visit all flowers from $l_i$ to $r_i$ inclusive.
Output Specification:
Print the string of $n$ characters. The $i$-th symbol should be «0» if you want to put a rose in the $i$-th position, otherwise «1» if you want to put a lily.
If there are multiple answers, print any.
Demo Input:
['5 3\n1 3\n2 4\n2 5\n', '6 3\n5 6\n1 4\n4 6\n']
Demo Output:
['01100', '110010']
Note:
In the first example, Sonya can put roses in the first, fourth, and fifth positions, and lilies in the second and third positions;
- in the segment $[1\ldots3]$, there are one rose and two lilies, so the beauty is equal to $1\cdot 2=2$; - in the segment $[2\ldots4]$, there are one rose and two lilies, so the beauty is equal to $1\cdot 2=2$; - in the segment $[2\ldots5]$, there are two roses and two lilies, so the beauty is equal to $2\cdot 2=4$.
The total beauty is equal to $2+2+4=8$.
In the second example, Sonya can put roses in the third, fourth, and sixth positions, and lilies in the first, second, and fifth positions;
- in the segment $[5\ldots6]$, there are one rose and one lily, so the beauty is equal to $1\cdot 1=1$; - in the segment $[1\ldots4]$, there are two roses and two lilies, so the beauty is equal to $2\cdot 2=4$; - in the segment $[4\ldots6]$, there are two roses and one lily, so the beauty is equal to $2\cdot 1=2$.
The total beauty is equal to $1+4+2=7$.
|
```python
n,m=map(int,input().split())
li=['0']*(n+1)
mc=(n//2)
for i in range(m):
l,r=map(int,input().split())
k=r-l+1
k//=2
j=l
for i in range(k):
if mc!=0:
li[j]='1'
j+=1
mc-=1
# print(li)
s=''.join(li)
print(s[1:])
```
| 0
|
|
628
|
B
|
New Skateboard
|
PROGRAMMING
| 1,300
|
[
"dp"
] | null | null |
Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4.
You are given a string *s* consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero.
A substring of a string is a nonempty sequence of consecutive characters.
For example if string *s* is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
|
The only line contains string *s* (1<=≤<=|*s*|<=≤<=3·105). The string *s* contains only digits from 0 to 9.
|
Print integer *a* — the number of substrings of the string *s* that are divisible by 4.
Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
|
[
"124\n",
"04\n",
"5810438174\n"
] |
[
"4\n",
"3\n",
"9\n"
] |
none
| 0
|
[
{
"input": "124",
"output": "4"
},
{
"input": "04",
"output": "3"
},
{
"input": "5810438174",
"output": "9"
},
{
"input": "1",
"output": "0"
},
{
"input": "039",
"output": "1"
},
{
"input": "97247",
"output": "6"
},
{
"input": "5810438174",
"output": "9"
},
{
"input": "12883340691714056185860211260984431382156326935244",
"output": "424"
},
{
"input": "2144315253572020279108092911160072328496568665545836825277616363478721946398140227406814602154768031",
"output": "1528"
},
{
"input": "80124649014054971081213608137817466046254652492627741860478258558206397113198232823859870363821007188476405951611069347299689170240023979048198711745011542774268179055311013054073075176122755643483380248999657649211459997766221072399103579977409770898200358240970169892326442892826731631357561876251276209119521202062222947560634301788787748428236988789594458520867663257476744168528121470923031438015546006185059454402637036376247785881323277542968298682307854655591317046086531554595892680980142608",
"output": "30826"
},
{
"input": "123456",
"output": "7"
},
{
"input": "4",
"output": "1"
},
{
"input": "123",
"output": "1"
}
] | 1,487,277,373
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 22
| 374
| 5,632,000
|
count = 0
x = input()
for i in range(len(x)-1):
z = x[i:i+2]
if int(z)%4 == 0:
count += i+1
for j in range(len(x)):
if x[j] in ['0', '4', '8']:
count+= 1
print(count)
|
Title: New Skateboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4.
You are given a string *s* consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero.
A substring of a string is a nonempty sequence of consecutive characters.
For example if string *s* is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input Specification:
The only line contains string *s* (1<=≤<=|*s*|<=≤<=3·105). The string *s* contains only digits from 0 to 9.
Output Specification:
Print integer *a* — the number of substrings of the string *s* that are divisible by 4.
Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Demo Input:
['124\n', '04\n', '5810438174\n']
Demo Output:
['4\n', '3\n', '9\n']
Note:
none
|
```python
count = 0
x = input()
for i in range(len(x)-1):
z = x[i:i+2]
if int(z)%4 == 0:
count += i+1
for j in range(len(x)):
if x[j] in ['0', '4', '8']:
count+= 1
print(count)
```
| 3
|
|
997
|
A
|
Convert to Ones
|
PROGRAMMING
| 1,500
|
[
"brute force",
"greedy",
"implementation",
"math"
] | null | null |
You've got a string $a_1, a_2, \dots, a_n$, consisting of zeros and ones.
Let's call a sequence of consecutive elements $a_i, a_{i<=+<=1}, \ldots,<=a_j$ ($1\leq<=i\leq<=j\leq<=n$) a substring of string $a$.
You can apply the following operations any number of times:
- Choose some substring of string $a$ (for example, you can choose entire string) and reverse it, paying $x$ coins for it (for example, «0101101» $\to$ «0111001»); - Choose some substring of string $a$ (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying $y$ coins for it (for example, «0101101» $\to$ «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
|
The first line of input contains integers $n$, $x$ and $y$ ($1<=\leq<=n<=\leq<=300\,000, 0 \leq x, y \leq 10^9$) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string $a$ of length $n$, consisting of zeros and ones.
|
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print $0$, if you do not need to perform any operations.
|
[
"5 1 10\n01000\n",
"5 10 1\n01000\n",
"7 2 3\n1111111\n"
] |
[
"11\n",
"2\n",
"0\n"
] |
In the first sample, at first you need to reverse substring $[1 \dots 2]$, and then you need to invert substring $[2 \dots 5]$.
Then the string was changed as follows:
«01000» $\to$ «10000» $\to$ «11111».
The total cost of operations is $1 + 10 = 11$.
In the second sample, at first you need to invert substring $[1 \dots 1]$, and then you need to invert substring $[3 \dots 5]$.
Then the string was changed as follows:
«01000» $\to$ «11000» $\to$ «11111».
The overall cost is $1 + 1 = 2$.
In the third example, string already consists only of ones, so the answer is $0$.
| 500
|
[
{
"input": "5 1 10\n01000",
"output": "11"
},
{
"input": "5 10 1\n01000",
"output": "2"
},
{
"input": "7 2 3\n1111111",
"output": "0"
},
{
"input": "1 60754033 959739508\n0",
"output": "959739508"
},
{
"input": "1 431963980 493041212\n1",
"output": "0"
},
{
"input": "1 314253869 261764879\n0",
"output": "261764879"
},
{
"input": "1 491511050 399084767\n1",
"output": "0"
},
{
"input": "2 163093925 214567542\n00",
"output": "214567542"
},
{
"input": "2 340351106 646854722\n10",
"output": "646854722"
},
{
"input": "2 222640995 489207317\n01",
"output": "489207317"
},
{
"input": "2 399898176 552898277\n11",
"output": "0"
},
{
"input": "2 690218164 577155357\n00",
"output": "577155357"
},
{
"input": "2 827538051 754412538\n10",
"output": "754412538"
},
{
"input": "2 636702427 259825230\n01",
"output": "259825230"
},
{
"input": "2 108926899 102177825\n11",
"output": "0"
},
{
"input": "3 368381052 440077270\n000",
"output": "440077270"
},
{
"input": "3 505700940 617334451\n100",
"output": "617334451"
},
{
"input": "3 499624340 643020827\n010",
"output": "1142645167"
},
{
"input": "3 75308005 971848814\n110",
"output": "971848814"
},
{
"input": "3 212627893 854138703\n001",
"output": "854138703"
},
{
"input": "3 31395883 981351561\n101",
"output": "981351561"
},
{
"input": "3 118671447 913685773\n011",
"output": "913685773"
},
{
"input": "3 255991335 385910245\n111",
"output": "0"
},
{
"input": "3 688278514 268200134\n000",
"output": "268200134"
},
{
"input": "3 825598402 445457315\n100",
"output": "445457315"
},
{
"input": "3 300751942 45676507\n010",
"output": "91353014"
},
{
"input": "3 517900980 438071829\n110",
"output": "438071829"
},
{
"input": "3 400190869 280424424\n001",
"output": "280424424"
},
{
"input": "3 577448050 344115384\n101",
"output": "344115384"
},
{
"input": "3 481435271 459737939\n011",
"output": "459737939"
},
{
"input": "3 931962412 913722450\n111",
"output": "0"
},
{
"input": "4 522194562 717060616\n0000",
"output": "717060616"
},
{
"input": "4 659514449 894317797\n1000",
"output": "894317797"
},
{
"input": "4 71574977 796834337\n0100",
"output": "868409314"
},
{
"input": "4 248832158 934154224\n1100",
"output": "934154224"
},
{
"input": "4 71474110 131122047\n0010",
"output": "202596157"
},
{
"input": "4 308379228 503761290\n1010",
"output": "812140518"
},
{
"input": "4 272484957 485636409\n0110",
"output": "758121366"
},
{
"input": "4 662893590 704772137\n1110",
"output": "704772137"
},
{
"input": "4 545183479 547124732\n0001",
"output": "547124732"
},
{
"input": "4 684444619 722440661\n1001",
"output": "722440661"
},
{
"input": "4 477963686 636258459\n0101",
"output": "1114222145"
},
{
"input": "4 360253575 773578347\n1101",
"output": "773578347"
},
{
"input": "4 832478048 910898234\n0011",
"output": "910898234"
},
{
"input": "4 343185412 714767937\n1011",
"output": "714767937"
},
{
"input": "4 480505300 892025118\n0111",
"output": "892025118"
},
{
"input": "4 322857895 774315007\n1111",
"output": "0"
},
{
"input": "4 386548854 246539479\n0000",
"output": "246539479"
},
{
"input": "4 523868742 128829368\n1000",
"output": "128829368"
},
{
"input": "4 956155921 11119257\n0100",
"output": "22238514"
},
{
"input": "4 188376438 93475808\n1100",
"output": "93475808"
},
{
"input": "4 754947032 158668188\n0010",
"output": "317336376"
},
{
"input": "4 927391856 637236921\n1010",
"output": "1274473842"
},
{
"input": "4 359679035 109461393\n0110",
"output": "218922786"
},
{
"input": "4 991751283 202031630\n1110",
"output": "202031630"
},
{
"input": "4 339351517 169008463\n0001",
"output": "169008463"
},
{
"input": "4 771638697 346265644\n1001",
"output": "346265644"
},
{
"input": "4 908958584 523522825\n0101",
"output": "1047045650"
},
{
"input": "4 677682252 405812714\n1101",
"output": "405812714"
},
{
"input": "4 815002139 288102603\n0011",
"output": "288102603"
},
{
"input": "4 952322026 760327076\n1011",
"output": "760327076"
},
{
"input": "4 663334158 312481698\n0111",
"output": "312481698"
},
{
"input": "4 840591339 154834293\n1111",
"output": "0"
},
{
"input": "14 3 11\n10110100011001",
"output": "20"
},
{
"input": "19 1 1\n1010101010101010101",
"output": "9"
},
{
"input": "1 10 1\n1",
"output": "0"
},
{
"input": "1 100 1\n1",
"output": "0"
},
{
"input": "5 1000 1\n11111",
"output": "0"
},
{
"input": "5 10 1\n11111",
"output": "0"
},
{
"input": "7 3 2\n1111111",
"output": "0"
},
{
"input": "5 1 10\n10101",
"output": "11"
},
{
"input": "1 3 2\n1",
"output": "0"
},
{
"input": "2 10 1\n11",
"output": "0"
},
{
"input": "4 148823922 302792601\n1010",
"output": "451616523"
},
{
"input": "1 2 1\n1",
"output": "0"
},
{
"input": "5 2 3\n00011",
"output": "3"
},
{
"input": "1 5 0\n1",
"output": "0"
},
{
"input": "7 2 3\n1001001",
"output": "5"
},
{
"input": "10 1 1000000000\n1111010111",
"output": "1000000001"
},
{
"input": "25 999999998 999999999\n1011001110101010100111001",
"output": "7999999985"
},
{
"input": "2 0 1\n00",
"output": "1"
},
{
"input": "2 1 100\n10",
"output": "100"
},
{
"input": "7 20 3\n1111111",
"output": "0"
},
{
"input": "1 1 0\n1",
"output": "0"
},
{
"input": "3 1 10\n010",
"output": "11"
},
{
"input": "2 1 0\n11",
"output": "0"
},
{
"input": "7 100 3\n1111111",
"output": "0"
},
{
"input": "5 1 1000\n10101",
"output": "1001"
},
{
"input": "5 2 1\n11111",
"output": "0"
},
{
"input": "1 1000 1\n1",
"output": "0"
},
{
"input": "1 799543940 488239239\n1",
"output": "0"
},
{
"input": "6 1 1000\n010101",
"output": "1002"
},
{
"input": "5 11 1\n11111",
"output": "0"
},
{
"input": "5 2 3\n10101",
"output": "5"
},
{
"input": "3 10 1\n111",
"output": "0"
},
{
"input": "7 9 10\n1001011",
"output": "19"
},
{
"input": "5 5 6\n10101",
"output": "11"
},
{
"input": "1 1000000000 0\n1",
"output": "0"
},
{
"input": "4 0 1\n0101",
"output": "1"
},
{
"input": "8 2 3\n10101010",
"output": "9"
},
{
"input": "6 3 100\n010101",
"output": "106"
},
{
"input": "3 3 2\n111",
"output": "0"
},
{
"input": "1 20 1\n1",
"output": "0"
},
{
"input": "2 1 2\n01",
"output": "2"
}
] | 1,602,443,571
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 171
| 2,969,600
|
n,r,c = map(int,input().split())
s = input()
if s.count("1") == n:
print(0)
exit()
if s.count("0") == n:
print(c)
exit()
cnt = 0
m = 0
for i in s:
if i == "1" and m == 1:
m = 0
cnt += 1
continue
if i == "0" and m == 0:
m = 1
i = 0
c1 = 0
c2 = 0
ans = 0
while i < cnt:
if c2 == 0 and c1 == 0:
c2 += c
c1 += c
i = i+1
continue
c2 += c
c1 += r
if c2 < c1:
c1 = c1-r
c2 = c2-c
ans += min(c1,c2)
c1 = 0
c2 = 0
else:
i += 1
ans1 = 10**12
ans2 = ans
if s[n-1] == "0":
ans1 = ans
ans1 += min(c1,c2)+r
ans2 = ans
ans2 += min(c1, c2) + c
print(min(ans1,ans2))
|
Title: Convert to Ones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a string $a_1, a_2, \dots, a_n$, consisting of zeros and ones.
Let's call a sequence of consecutive elements $a_i, a_{i<=+<=1}, \ldots,<=a_j$ ($1\leq<=i\leq<=j\leq<=n$) a substring of string $a$.
You can apply the following operations any number of times:
- Choose some substring of string $a$ (for example, you can choose entire string) and reverse it, paying $x$ coins for it (for example, «0101101» $\to$ «0111001»); - Choose some substring of string $a$ (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying $y$ coins for it (for example, «0101101» $\to$ «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input Specification:
The first line of input contains integers $n$, $x$ and $y$ ($1<=\leq<=n<=\leq<=300\,000, 0 \leq x, y \leq 10^9$) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string $a$ of length $n$, consisting of zeros and ones.
Output Specification:
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print $0$, if you do not need to perform any operations.
Demo Input:
['5 1 10\n01000\n', '5 10 1\n01000\n', '7 2 3\n1111111\n']
Demo Output:
['11\n', '2\n', '0\n']
Note:
In the first sample, at first you need to reverse substring $[1 \dots 2]$, and then you need to invert substring $[2 \dots 5]$.
Then the string was changed as follows:
«01000» $\to$ «10000» $\to$ «11111».
The total cost of operations is $1 + 10 = 11$.
In the second sample, at first you need to invert substring $[1 \dots 1]$, and then you need to invert substring $[3 \dots 5]$.
Then the string was changed as follows:
«01000» $\to$ «11000» $\to$ «11111».
The overall cost is $1 + 1 = 2$.
In the third example, string already consists only of ones, so the answer is $0$.
|
```python
n,r,c = map(int,input().split())
s = input()
if s.count("1") == n:
print(0)
exit()
if s.count("0") == n:
print(c)
exit()
cnt = 0
m = 0
for i in s:
if i == "1" and m == 1:
m = 0
cnt += 1
continue
if i == "0" and m == 0:
m = 1
i = 0
c1 = 0
c2 = 0
ans = 0
while i < cnt:
if c2 == 0 and c1 == 0:
c2 += c
c1 += c
i = i+1
continue
c2 += c
c1 += r
if c2 < c1:
c1 = c1-r
c2 = c2-c
ans += min(c1,c2)
c1 = 0
c2 = 0
else:
i += 1
ans1 = 10**12
ans2 = ans
if s[n-1] == "0":
ans1 = ans
ans1 += min(c1,c2)+r
ans2 = ans
ans2 += min(c1, c2) + c
print(min(ans1,ans2))
```
| 0
|
|
102
|
B
|
Sum of Digits
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
B. Sum of Digits
|
2
|
265
|
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
|
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
|
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
|
[
"0\n",
"10\n",
"991\n"
] |
[
"0\n",
"1\n",
"3\n"
] |
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
| 1,000
|
[
{
"input": "0",
"output": "0"
},
{
"input": "10",
"output": "1"
},
{
"input": "991",
"output": "3"
},
{
"input": "99",
"output": "2"
},
{
"input": "100",
"output": "1"
},
{
"input": "123456789",
"output": "2"
},
{
"input": "32",
"output": "1"
},
{
"input": "86",
"output": "2"
},
{
"input": "2",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "34",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "28",
"output": "2"
},
{
"input": "23",
"output": "1"
},
{
"input": "57",
"output": "2"
},
{
"input": "29",
"output": "2"
},
{
"input": "353869285",
"output": "3"
},
{
"input": "549548646",
"output": "2"
},
{
"input": "858893424",
"output": "2"
},
{
"input": "644818852",
"output": "3"
},
{
"input": "360322525",
"output": "3"
},
{
"input": "122937520",
"output": "2"
},
{
"input": "288403032",
"output": "2"
},
{
"input": "677257481",
"output": "3"
},
{
"input": "58059561",
"output": "3"
},
{
"input": "211288847",
"output": "2"
},
{
"input": "339900034079539584",
"output": "2"
},
{
"input": "784084029310269952",
"output": "3"
},
{
"input": "774730557264864000",
"output": "3"
},
{
"input": "478233807148352256",
"output": "3"
},
{
"input": "165646874056812544",
"output": "3"
},
{
"input": "477533739511673792",
"output": "3"
},
{
"input": "660119944377002240",
"output": "3"
},
{
"input": "448375218506078080",
"output": "3"
},
{
"input": "77213334785795616",
"output": "3"
},
{
"input": "165767221702271872",
"output": "3"
},
{
"input": "1",
"output": "0"
},
{
"input": "5",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "156161456146164104103460452045416165146141414651641564105461465156445621465146514562146145465165145145614561465145614561",
"output": "3"
},
{
"input": "9",
"output": "0"
}
] | 1,628,967,384
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 51
| 1,840
| 6,963,200
|
a=int(input())
b=str(a)
count=0
while len(b)!=1:
add=0
for i in b:
add=add+int(i)
b=str(add)
count=count+1
#print(b)
print(count)
|
Title: Sum of Digits
Time Limit: 2 seconds
Memory Limit: 265 megabytes
Problem Description:
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
Input Specification:
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
Output Specification:
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
Demo Input:
['0\n', '10\n', '991\n']
Demo Output:
['0\n', '1\n', '3\n']
Note:
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
|
```python
a=int(input())
b=str(a)
count=0
while len(b)!=1:
add=0
for i in b:
add=add+int(i)
b=str(add)
count=count+1
#print(b)
print(count)
```
| 3.527471
|
166
|
A
|
Rank List
|
PROGRAMMING
| 1,100
|
[
"binary search",
"implementation",
"sortings"
] | null | null |
Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place.
You know the rules of comparing the results of two given teams very well. Let's say that team *a* solved *p**a* problems with total penalty time *t**a* and team *b* solved *p**b* problems with total penalty time *t**b*. Team *a* gets a higher place than team *b* in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, team *a* gets a higher place than team *b* in the final results' table if either *p**a*<=><=*p**b*, or *p**a*<==<=*p**b* and *t**a*<=<<=*t**b*.
It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group of *x* teams that solved the same number of problems with the same penalty time. Let's also say that *y* teams performed better than the teams from this group. In this case all teams from the group share places *y*<=+<=1, *y*<=+<=2, ..., *y*<=+<=*x*. The teams that performed worse than the teams from this group, get their places in the results table starting from the *y*<=+<=*x*<=+<=1-th place.
Your task is to count what number of teams from the given list shared the *k*-th place.
|
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50). Then *n* lines contain the description of the teams: the *i*-th line contains two integers *p**i* and *t**i* (1<=≤<=*p**i*,<=*t**i*<=≤<=50) — the number of solved problems and the total penalty time of the *i*-th team, correspondingly. All numbers in the lines are separated by spaces.
|
In the only line print the sought number of teams that got the *k*-th place in the final results' table.
|
[
"7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10\n",
"5 4\n3 1\n3 1\n5 3\n3 1\n3 1\n"
] |
[
"3\n",
"4\n"
] |
The final results' table for the first sample is:
- 1-3 places — 4 solved problems, the penalty time equals 10 - 4 place — 3 solved problems, the penalty time equals 20 - 5-6 places — 2 solved problems, the penalty time equals 1 - 7 place — 1 solved problem, the penalty time equals 10
The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams.
The final table for the second sample is:
- 1 place — 5 solved problems, the penalty time equals 3 - 2-5 places — 3 solved problems, the penalty time equals 1
The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams.
| 500
|
[
{
"input": "7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10",
"output": "3"
},
{
"input": "5 4\n3 1\n3 1\n5 3\n3 1\n3 1",
"output": "4"
},
{
"input": "5 1\n2 2\n1 1\n1 1\n1 1\n2 2",
"output": "2"
},
{
"input": "6 3\n2 2\n3 1\n2 2\n4 5\n2 2\n4 5",
"output": "1"
},
{
"input": "5 5\n3 1\n10 2\n2 2\n1 10\n10 2",
"output": "1"
},
{
"input": "3 2\n3 3\n3 3\n3 3",
"output": "3"
},
{
"input": "4 3\n10 3\n6 10\n5 2\n5 2",
"output": "2"
},
{
"input": "5 3\n10 10\n10 10\n1 1\n10 10\n4 3",
"output": "3"
},
{
"input": "3 1\n2 1\n1 1\n1 2",
"output": "1"
},
{
"input": "1 1\n28 28",
"output": "1"
},
{
"input": "2 2\n1 2\n1 2",
"output": "2"
},
{
"input": "5 3\n2 3\n4 2\n5 3\n2 4\n3 5",
"output": "1"
},
{
"input": "50 22\n4 9\n8 1\n3 7\n1 2\n3 8\n9 8\n8 5\n2 10\n5 8\n1 3\n1 8\n2 3\n7 9\n10 2\n9 9\n7 3\n8 6\n10 6\n5 4\n8 1\n1 5\n6 8\n9 5\n9 5\n3 2\n3 3\n3 8\n7 5\n4 5\n8 10\n8 2\n3 5\n3 2\n1 1\n7 2\n2 7\n6 8\n10 4\n7 5\n1 7\n6 5\n3 1\n4 9\n2 3\n3 6\n5 8\n4 10\n10 7\n7 10\n9 8",
"output": "1"
},
{
"input": "50 6\n11 20\n18 13\n1 13\n3 11\n4 17\n15 10\n15 8\n9 16\n11 17\n16 3\n3 20\n14 13\n12 15\n9 10\n14 2\n12 12\n13 17\n6 10\n20 9\n2 8\n13 7\n7 20\n15 3\n1 20\n2 13\n2 5\n14 7\n10 13\n15 12\n15 5\n17 6\n9 11\n18 5\n10 1\n15 14\n3 16\n6 12\n4 1\n14 9\n7 14\n8 17\n17 13\n4 6\n19 16\n5 6\n3 15\n4 19\n15 20\n2 10\n20 10",
"output": "1"
},
{
"input": "50 12\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "50"
},
{
"input": "50 28\n2 2\n1 1\n2 1\n1 2\n1 1\n1 1\n1 1\n2 2\n2 2\n2 2\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 1\n2 2\n1 2\n2 2\n2 2\n2 1\n1 1\n1 2\n1 2\n1 1\n1 1\n1 1\n2 2\n2 1\n2 1\n2 2\n1 2\n1 2\n1 2\n1 1\n2 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n2 1\n1 1\n2 2\n2 2\n2 2\n2 2",
"output": "13"
},
{
"input": "50 40\n2 3\n3 1\n2 1\n2 1\n2 1\n3 1\n1 1\n1 2\n2 3\n1 3\n1 3\n2 1\n3 1\n1 1\n3 1\n3 1\n2 2\n1 1\n3 3\n3 1\n3 2\n2 3\n3 3\n3 1\n1 3\n2 3\n2 1\n3 2\n3 3\n3 1\n2 1\n2 2\n1 3\n3 3\n1 1\n3 2\n1 2\n2 3\n2 1\n2 2\n3 2\n1 3\n3 1\n1 1\n3 3\n2 3\n2 1\n2 3\n2 3\n1 2",
"output": "5"
},
{
"input": "50 16\n2 1\n3 2\n5 2\n2 2\n3 4\n4 4\n3 3\n4 1\n2 3\n1 5\n4 1\n2 2\n1 5\n3 2\n2 1\n5 4\n5 2\n5 4\n1 1\n3 5\n2 1\n4 5\n5 1\n5 5\n5 4\n2 4\n1 2\n5 5\n4 4\n1 5\n4 2\n5 1\n2 4\n2 5\n2 2\n3 4\n3 1\n1 1\n5 5\n2 2\n3 4\n2 4\n5 2\n4 1\n3 1\n1 1\n4 1\n4 4\n1 4\n1 3",
"output": "1"
},
{
"input": "50 32\n6 6\n4 2\n5 5\n1 1\n2 4\n6 5\n2 3\n6 5\n2 3\n6 3\n1 4\n1 6\n3 3\n2 4\n3 2\n6 2\n4 1\n3 3\n3 1\n5 5\n1 2\n2 1\n5 4\n3 1\n4 4\n5 6\n4 1\n2 5\n3 1\n4 6\n2 3\n1 1\n6 5\n2 6\n3 3\n2 6\n2 3\n2 6\n3 4\n2 6\n4 5\n5 4\n1 6\n3 2\n5 1\n4 1\n4 6\n4 2\n1 2\n5 2",
"output": "1"
},
{
"input": "50 48\n5 1\n6 4\n3 2\n2 1\n4 7\n3 6\n7 1\n7 5\n6 5\n5 6\n4 7\n5 7\n5 7\n5 5\n7 3\n3 5\n4 3\n5 4\n6 2\n1 6\n6 3\n6 5\n5 2\n4 2\n3 1\n1 1\n5 6\n1 3\n6 5\n3 7\n1 5\n7 5\n6 5\n3 6\n2 7\n5 3\n5 3\n4 7\n5 2\n6 5\n5 7\n7 1\n2 3\n5 5\n2 6\n4 1\n6 2\n6 5\n3 3\n1 6",
"output": "1"
},
{
"input": "50 8\n5 3\n7 3\n4 3\n7 4\n2 2\n4 4\n5 4\n1 1\n7 7\n4 8\n1 1\n6 3\n1 5\n7 3\n6 5\n4 5\n8 6\n3 6\n2 1\n3 2\n2 5\n7 6\n5 8\n1 3\n5 5\n8 4\n4 5\n4 4\n8 8\n7 2\n7 2\n3 6\n2 8\n8 3\n3 2\n4 5\n8 1\n3 2\n8 7\n6 3\n2 3\n5 1\n3 4\n7 2\n6 3\n7 3\n3 3\n6 4\n2 2\n5 1",
"output": "3"
},
{
"input": "20 16\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "20"
},
{
"input": "20 20\n1 2\n2 2\n1 1\n2 1\n2 2\n1 1\n1 1\n2 1\n1 1\n1 2\n2 2\n1 2\n1 2\n2 2\n2 2\n1 2\n2 1\n2 1\n1 2\n2 2",
"output": "6"
},
{
"input": "30 16\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "30"
},
{
"input": "30 22\n2 1\n1 2\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n2 2\n1 2\n2 2\n1 2\n1 2\n2 1\n1 2\n2 2\n2 2\n1 2\n2 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 2\n2 2\n1 2\n2 2\n2 1\n1 1",
"output": "13"
},
{
"input": "30 22\n1 1\n1 3\n2 3\n3 1\n2 3\n3 1\n1 2\n3 3\n2 1\n2 1\n2 2\n3 1\n3 2\n2 3\n3 1\n1 3\n2 3\n3 1\n1 2\n1 2\n2 3\n2 1\n3 3\n3 2\n1 3\n3 3\n3 3\n3 3\n3 3\n3 1",
"output": "5"
},
{
"input": "50 16\n2 1\n3 2\n5 2\n2 2\n3 4\n4 4\n3 3\n4 1\n2 3\n1 5\n4 1\n2 2\n1 5\n3 2\n2 1\n5 4\n5 2\n5 4\n1 1\n3 5\n2 1\n4 5\n5 1\n5 5\n5 4\n2 4\n1 2\n5 5\n4 4\n1 5\n4 2\n5 1\n2 4\n2 5\n2 2\n3 4\n3 1\n1 1\n5 5\n2 2\n3 4\n2 4\n5 2\n4 1\n3 1\n1 1\n4 1\n4 4\n1 4\n1 3",
"output": "1"
},
{
"input": "50 22\n4 9\n8 1\n3 7\n1 2\n3 8\n9 8\n8 5\n2 10\n5 8\n1 3\n1 8\n2 3\n7 9\n10 2\n9 9\n7 3\n8 6\n10 6\n5 4\n8 1\n1 5\n6 8\n9 5\n9 5\n3 2\n3 3\n3 8\n7 5\n4 5\n8 10\n8 2\n3 5\n3 2\n1 1\n7 2\n2 7\n6 8\n10 4\n7 5\n1 7\n6 5\n3 1\n4 9\n2 3\n3 6\n5 8\n4 10\n10 7\n7 10\n9 8",
"output": "1"
},
{
"input": "50 22\n29 15\n18 10\n6 23\n38 28\n34 40\n40 1\n16 26\n22 33\n14 30\n26 7\n15 16\n22 40\n14 15\n6 28\n32 27\n33 3\n38 22\n40 17\n16 27\n21 27\n34 26\n5 15\n34 9\n38 23\n7 36\n17 6\n19 37\n40 1\n10 28\n9 14\n8 31\n40 8\n14 2\n24 16\n38 33\n3 37\n2 9\n21 21\n40 26\n28 33\n24 31\n10 12\n27 27\n17 4\n38 5\n21 31\n5 12\n29 7\n39 12\n26 14",
"output": "1"
},
{
"input": "50 14\n4 20\n37 50\n46 19\n20 25\n47 10\n6 34\n12 41\n47 9\n22 28\n41 34\n47 40\n12 42\n9 4\n15 15\n27 8\n38 9\n4 17\n8 13\n47 7\n9 38\n30 48\n50 7\n41 34\n23 11\n16 37\n2 32\n18 46\n37 48\n47 41\n13 9\n24 50\n46 14\n33 49\n9 50\n35 30\n49 44\n42 49\n39 15\n33 42\n3 18\n44 15\n44 28\n9 17\n16 4\n10 36\n4 22\n47 17\n24 12\n2 31\n6 30",
"output": "2"
},
{
"input": "2 1\n50 50\n50 50",
"output": "2"
},
{
"input": "2 2\n50 50\n50 50",
"output": "2"
},
{
"input": "2 1\n50 50\n50 49",
"output": "1"
},
{
"input": "2 2\n50 50\n50 49",
"output": "1"
},
{
"input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50",
"output": "50"
},
{
"input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n49 50",
"output": "1"
},
{
"input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 49",
"output": "49"
},
{
"input": "10 1\n3 1\n3 2\n3 2\n2 3\n2 3\n2 3\n2 4\n2 4\n2 4\n2 4",
"output": "1"
},
{
"input": "4 3\n1 1\n1 2\n2 1\n2 2",
"output": "1"
},
{
"input": "4 2\n1 2\n1 3\n1 5\n1 2",
"output": "2"
},
{
"input": "6 1\n4 3\n4 2\n5 1\n5 2\n5 3\n5 3",
"output": "1"
},
{
"input": "5 3\n2 1\n1 3\n1 2\n1 1\n1 1",
"output": "2"
},
{
"input": "3 1\n1 1\n1 2\n1 3",
"output": "1"
},
{
"input": "10 10\n3 1\n3 2\n3 2\n2 3\n2 3\n2 3\n2 4\n2 4\n2 4\n2 4",
"output": "4"
},
{
"input": "5 1\n5 5\n5 6\n5 7\n5 8\n5 5",
"output": "2"
},
{
"input": "4 1\n1 1\n1 1\n1 2\n1 3",
"output": "2"
}
] | 1,698,159,971
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
n, k = map(int, input().split())
teams = []
for i in range(n):
team_info = list(map(int, input().split()))
teams.append(team_info)
teams = sorted(teams, reverse=True)
counts = [1] # Initialize the first count to 1
for i in range(1, len(teams)):
if teams[i] == teams[i - 1]:
counts[-1] += 1 # Increment the last count in the list
else:
counts.append(1) # Start counting a new group of teams
# Now, counts[i] contains the number of teams sharing the i+1-th place.
for i in counts:
if counts[i]-k >= 0:
print(counts[i])
else:
k-counts[i]
# //////////
# def shape(word, char1, char2):
# for i in range(0, len(word)):
# if word[i] == char1:
# word = word[:i] + char2 + \
# word[i + 1:]
# elif word[i] == char2:
# word = word[:i] + char1 + \
# word[i + 1:]
# return word
# n, numberofdesigners = map(int, input().split())
# word = input()
# for i in range(0, numberofdesigners):
# char1, char2 = map(str, input().split())
# word = shape(word, char1, char2)
# print(word)
|
Title: Rank List
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place.
You know the rules of comparing the results of two given teams very well. Let's say that team *a* solved *p**a* problems with total penalty time *t**a* and team *b* solved *p**b* problems with total penalty time *t**b*. Team *a* gets a higher place than team *b* in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, team *a* gets a higher place than team *b* in the final results' table if either *p**a*<=><=*p**b*, or *p**a*<==<=*p**b* and *t**a*<=<<=*t**b*.
It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group of *x* teams that solved the same number of problems with the same penalty time. Let's also say that *y* teams performed better than the teams from this group. In this case all teams from the group share places *y*<=+<=1, *y*<=+<=2, ..., *y*<=+<=*x*. The teams that performed worse than the teams from this group, get their places in the results table starting from the *y*<=+<=*x*<=+<=1-th place.
Your task is to count what number of teams from the given list shared the *k*-th place.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50). Then *n* lines contain the description of the teams: the *i*-th line contains two integers *p**i* and *t**i* (1<=≤<=*p**i*,<=*t**i*<=≤<=50) — the number of solved problems and the total penalty time of the *i*-th team, correspondingly. All numbers in the lines are separated by spaces.
Output Specification:
In the only line print the sought number of teams that got the *k*-th place in the final results' table.
Demo Input:
['7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10\n', '5 4\n3 1\n3 1\n5 3\n3 1\n3 1\n']
Demo Output:
['3\n', '4\n']
Note:
The final results' table for the first sample is:
- 1-3 places — 4 solved problems, the penalty time equals 10 - 4 place — 3 solved problems, the penalty time equals 20 - 5-6 places — 2 solved problems, the penalty time equals 1 - 7 place — 1 solved problem, the penalty time equals 10
The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams.
The final table for the second sample is:
- 1 place — 5 solved problems, the penalty time equals 3 - 2-5 places — 3 solved problems, the penalty time equals 1
The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams.
|
```python
n, k = map(int, input().split())
teams = []
for i in range(n):
team_info = list(map(int, input().split()))
teams.append(team_info)
teams = sorted(teams, reverse=True)
counts = [1] # Initialize the first count to 1
for i in range(1, len(teams)):
if teams[i] == teams[i - 1]:
counts[-1] += 1 # Increment the last count in the list
else:
counts.append(1) # Start counting a new group of teams
# Now, counts[i] contains the number of teams sharing the i+1-th place.
for i in counts:
if counts[i]-k >= 0:
print(counts[i])
else:
k-counts[i]
# //////////
# def shape(word, char1, char2):
# for i in range(0, len(word)):
# if word[i] == char1:
# word = word[:i] + char2 + \
# word[i + 1:]
# elif word[i] == char2:
# word = word[:i] + char1 + \
# word[i + 1:]
# return word
# n, numberofdesigners = map(int, input().split())
# word = input()
# for i in range(0, numberofdesigners):
# char1, char2 = map(str, input().split())
# word = shape(word, char1, char2)
# print(word)
```
| 0
|
|
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"
}
] | 1,695,822,096
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 124
| 5,632,000
|
i_th = []
for i in range(4):
i_th.append(int(input()))
d = int(input())
count = 0
for i in range(1, d+1):
for j in i_th:
if i % j == 0:
count += 1
break # 4개의 숫자 중 하나라도 나누어떨어지면
# 다음 숫자로 넘어감
print(count)
|
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
i_th = []
for i in range(4):
i_th.append(int(input()))
d = int(input())
count = 0
for i in range(1, d+1):
for j in i_th:
if i % j == 0:
count += 1
break # 4개의 숫자 중 하나라도 나누어떨어지면
# 다음 숫자로 넘어감
print(count)
```
| 3
|
|
965
|
A
|
Paper Airplanes
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy?
|
The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively.
|
Print a single integer — the minimum number of packs they should buy.
|
[
"5 3 2 3\n",
"5 3 100 1\n"
] |
[
"4\n",
"5\n"
] |
In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets.
| 500
|
[
{
"input": "5 3 2 3",
"output": "4"
},
{
"input": "5 3 100 1",
"output": "5"
},
{
"input": "10000 10000 1 1",
"output": "100000000"
},
{
"input": "1 1 10000 10000",
"output": "1"
},
{
"input": "300 300 21 23",
"output": "196"
},
{
"input": "300 2 37 51",
"output": "6"
},
{
"input": "2 400 23 57",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "10000 10000 3 2",
"output": "16670000"
},
{
"input": "1 2 1 2",
"output": "1"
},
{
"input": "1 1 2 2",
"output": "1"
},
{
"input": "1 1 10 10",
"output": "1"
},
{
"input": "5324 5439 32 13",
"output": "69622"
},
{
"input": "9000 1 2432 1",
"output": "9000"
},
{
"input": "230 1234 9124 23",
"output": "10"
},
{
"input": "11 1 1 1",
"output": "11"
},
{
"input": "6246 8489 1227 9",
"output": "4858"
},
{
"input": "9 20 5 7",
"output": "6"
}
] | 1,559,041,830
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 93
| 0
|
k,n,s,p = map(int,input().split())
if n<s:
print(k)
else:
print(((((n//s)+1)*k)//p)+1)
|
Title: Paper Airplanes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy?
Input Specification:
The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively.
Output Specification:
Print a single integer — the minimum number of packs they should buy.
Demo Input:
['5 3 2 3\n', '5 3 100 1\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets.
|
```python
k,n,s,p = map(int,input().split())
if n<s:
print(k)
else:
print(((((n//s)+1)*k)//p)+1)
```
| 0
|
|
134
|
A
|
Average Numbers
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
You are given a sequence of positive integers *a*1,<=*a*2,<=...,<=*a**n*. Find all such indices *i*, that the *i*-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
|
The first line contains the integer *n* (2<=≤<=*n*<=≤<=2·105). The second line contains elements of the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). All the elements are positive integers.
|
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to *n*.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
|
[
"5\n1 2 3 4 5\n",
"4\n50 50 50 50\n"
] |
[
"1\n3 ",
"4\n1 2 3 4 "
] |
none
| 500
|
[
{
"input": "5\n1 2 3 4 5",
"output": "1\n3 "
},
{
"input": "4\n50 50 50 50",
"output": "4\n1 2 3 4 "
},
{
"input": "3\n2 3 1",
"output": "1\n1 "
},
{
"input": "2\n4 2",
"output": "0"
},
{
"input": "2\n1 1",
"output": "2\n1 2 "
},
{
"input": "10\n3 3 3 3 3 4 3 3 3 2",
"output": "8\n1 2 3 4 5 7 8 9 "
},
{
"input": "10\n15 7 10 7 7 7 4 4 7 2",
"output": "5\n2 4 5 6 9 "
},
{
"input": "6\n2 2 2 2 2 2",
"output": "6\n1 2 3 4 5 6 "
},
{
"input": "6\n3 3 3 3 3 3",
"output": "6\n1 2 3 4 5 6 "
},
{
"input": "4\n6 6 6 7",
"output": "0"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "3\n3 3 4",
"output": "0"
},
{
"input": "5\n7 6 6 6 6",
"output": "0"
},
{
"input": "4\n3 5 5 9",
"output": "0"
},
{
"input": "3\n99 100 99",
"output": "0"
},
{
"input": "4\n5 6 5 5",
"output": "0"
},
{
"input": "6\n1 1 2 1 1 1",
"output": "0"
},
{
"input": "2\n4 5",
"output": "0"
},
{
"input": "4\n1 1 1 2",
"output": "0"
},
{
"input": "3\n1 2 4",
"output": "0"
},
{
"input": "6\n1 1 2 3 3 3",
"output": "0"
},
{
"input": "4\n4 5 5 4",
"output": "0"
},
{
"input": "3\n2 3 5",
"output": "0"
},
{
"input": "3\n2 1 1",
"output": "0"
},
{
"input": "3\n1 1 2",
"output": "0"
},
{
"input": "4\n1 2 3 4",
"output": "0"
},
{
"input": "5\n1 2 3 4 6",
"output": "0"
},
{
"input": "3\n2 2 3",
"output": "0"
},
{
"input": "4\n3 4 5 1",
"output": "0"
},
{
"input": "3\n2 3 2",
"output": "0"
},
{
"input": "3\n3 4 4",
"output": "0"
},
{
"input": "3\n10 5 7",
"output": "0"
},
{
"input": "3\n5 6 5",
"output": "0"
},
{
"input": "4\n1 2 3 7",
"output": "0"
},
{
"input": "5\n2 2 3 3 3",
"output": "0"
},
{
"input": "3\n4 3 6",
"output": "0"
},
{
"input": "3\n1 1 3",
"output": "0"
},
{
"input": "15\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "15\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 "
}
] | 1,682,979,132
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 514
| 14,233,600
|
n = int(input())
a = []
s = 0
a = list(map(int, input().split()))
s = sum(a)
indices = []
for i in range(n):
mid = (s - a[i]) / (n - 1)
if mid.is_integer() and int(mid) == a[i]:
indices.append(i + 1)
print(len(indices))
for i in indices:
print(i, end=" ")
# Tue May 02 2023 01:12:11 GMT+0300 (Moscow Standard Time)
|
Title: Average Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a sequence of positive integers *a*1,<=*a*2,<=...,<=*a**n*. Find all such indices *i*, that the *i*-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input Specification:
The first line contains the integer *n* (2<=≤<=*n*<=≤<=2·105). The second line contains elements of the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). All the elements are positive integers.
Output Specification:
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to *n*.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Demo Input:
['5\n1 2 3 4 5\n', '4\n50 50 50 50\n']
Demo Output:
['1\n3 ', '4\n1 2 3 4 ']
Note:
none
|
```python
n = int(input())
a = []
s = 0
a = list(map(int, input().split()))
s = sum(a)
indices = []
for i in range(n):
mid = (s - a[i]) / (n - 1)
if mid.is_integer() and int(mid) == a[i]:
indices.append(i + 1)
print(len(indices))
for i in indices:
print(i, end=" ")
# Tue May 02 2023 01:12:11 GMT+0300 (Moscow Standard Time)
```
| 3
|
|
572
|
A
|
Arrays
|
PROGRAMMING
| 900
|
[
"sortings"
] | null | null |
You are given two arrays *A* and *B* consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose *k* numbers in array *A* and choose *m* numbers in array *B* so that any number chosen in the first array is strictly less than any number chosen in the second array.
|
The first line contains two integers *n**A*,<=*n**B* (1<=≤<=*n**A*,<=*n**B*<=≤<=105), separated by a space — the sizes of arrays *A* and *B*, correspondingly.
The second line contains two integers *k* and *m* (1<=≤<=*k*<=≤<=*n**A*,<=1<=≤<=*m*<=≤<=*n**B*), separated by a space.
The third line contains *n**A* numbers *a*1,<=*a*2,<=... *a**n**A* (<=-<=109<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n**A*<=≤<=109), separated by spaces — elements of array *A*.
The fourth line contains *n**B* integers *b*1,<=*b*2,<=... *b**n**B* (<=-<=109<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**n**B*<=≤<=109), separated by spaces — elements of array *B*.
|
Print "YES" (without the quotes), if you can choose *k* numbers in array *A* and *m* numbers in array *B* so that any number chosen in array *A* was strictly less than any number chosen in array *B*. Otherwise, print "NO" (without the quotes).
|
[
"3 3\n2 1\n1 2 3\n3 4 5\n",
"3 3\n3 3\n1 2 3\n3 4 5\n",
"5 2\n3 1\n1 1 1 1 1\n2 2\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
In the first sample test you can, for example, choose numbers 1 and 2 from array *A* and number 3 from array *B* (1 < 3 and 2 < 3).
In the second sample test the only way to choose *k* elements in the first array and *m* elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in *A* will be less than all the numbers chosen in *B*: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7280148ed5eab0a7d418d4f92b32061243a8ca58.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 500
|
[
{
"input": "3 3\n2 1\n1 2 3\n3 4 5",
"output": "YES"
},
{
"input": "3 3\n3 3\n1 2 3\n3 4 5",
"output": "NO"
},
{
"input": "5 2\n3 1\n1 1 1 1 1\n2 2",
"output": "YES"
},
{
"input": "3 5\n1 1\n5 5 5\n5 5 5 5 5",
"output": "NO"
},
{
"input": "1 1\n1 1\n1\n1",
"output": "NO"
},
{
"input": "3 3\n1 1\n1 2 3\n1 2 3",
"output": "YES"
},
{
"input": "3 3\n1 2\n1 2 3\n1 2 3",
"output": "YES"
},
{
"input": "3 3\n2 2\n1 2 3\n1 2 3",
"output": "NO"
},
{
"input": "10 15\n10 1\n1 1 5 17 22 29 32 36 39 48\n9 10 20 23 26 26 32 32 33 39 43 45 47 49 49",
"output": "YES"
},
{
"input": "10 15\n1 15\n91 91 91 92 92 94 94 95 98 100\n92 92 93 93 93 94 95 96 97 98 98 99 99 100 100",
"output": "YES"
},
{
"input": "15 10\n12 5\n9 25 25 32 32 38 40 41 46 46 48 51 64 64 73\n5 14 30 35 50 52 67 79 89 99",
"output": "YES"
},
{
"input": "15 10\n4 10\n22 32 35 45 45 50 51 55 79 80 83 88 90 92 93\n46 48 52 55 60 60 68 75 80 81",
"output": "YES"
},
{
"input": "20 30\n2 8\n6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 10 10 10 10 10\n1 1 2 2 2 2 2 2 2 3 3 4 5 5 5 5 6 6 6 6 6 6 7 7 7 8 8 9 10 10",
"output": "NO"
},
{
"input": "20 30\n19 29\n1 1 2 2 2 3 4 4 7 7 7 8 8 8 8 8 9 9 9 9\n6 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "1 1\n1 1\n-1000000000\n30",
"output": "YES"
},
{
"input": "3 3\n1 3\n1 3 3\n3 3 3",
"output": "YES"
},
{
"input": "3 3\n1 1\n3 3 3\n2 2 2",
"output": "NO"
},
{
"input": "5 5\n3 3\n1 5 6 7 8\n1 2 5 6 7",
"output": "NO"
},
{
"input": "3 4\n2 2\n5 6 7\n1 2 3 4",
"output": "NO"
},
{
"input": "3 3\n3 3\n1 2 3\n4 5 6",
"output": "YES"
},
{
"input": "5 5\n4 5\n2 2 3 4 5\n5 6 7 8 9",
"output": "YES"
}
] | 1,660,890,317
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 8
| 2,000
| 8,089,600
|
n1,n2 = list(map(int, input().split()))
k,m = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
x = a[:k]
y = b[n2-m:]
for i in x:
flag=0
for j in y:
if i>=j:
flag=1
break
if flag==1:
break
if flag==0:
print('YES')
else:
print('NO')
|
Title: Arrays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two arrays *A* and *B* consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose *k* numbers in array *A* and choose *m* numbers in array *B* so that any number chosen in the first array is strictly less than any number chosen in the second array.
Input Specification:
The first line contains two integers *n**A*,<=*n**B* (1<=≤<=*n**A*,<=*n**B*<=≤<=105), separated by a space — the sizes of arrays *A* and *B*, correspondingly.
The second line contains two integers *k* and *m* (1<=≤<=*k*<=≤<=*n**A*,<=1<=≤<=*m*<=≤<=*n**B*), separated by a space.
The third line contains *n**A* numbers *a*1,<=*a*2,<=... *a**n**A* (<=-<=109<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n**A*<=≤<=109), separated by spaces — elements of array *A*.
The fourth line contains *n**B* integers *b*1,<=*b*2,<=... *b**n**B* (<=-<=109<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**n**B*<=≤<=109), separated by spaces — elements of array *B*.
Output Specification:
Print "YES" (without the quotes), if you can choose *k* numbers in array *A* and *m* numbers in array *B* so that any number chosen in array *A* was strictly less than any number chosen in array *B*. Otherwise, print "NO" (without the quotes).
Demo Input:
['3 3\n2 1\n1 2 3\n3 4 5\n', '3 3\n3 3\n1 2 3\n3 4 5\n', '5 2\n3 1\n1 1 1 1 1\n2 2\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample test you can, for example, choose numbers 1 and 2 from array *A* and number 3 from array *B* (1 < 3 and 2 < 3).
In the second sample test the only way to choose *k* elements in the first array and *m* elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in *A* will be less than all the numbers chosen in *B*: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7280148ed5eab0a7d418d4f92b32061243a8ca58.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
n1,n2 = list(map(int, input().split()))
k,m = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
x = a[:k]
y = b[n2-m:]
for i in x:
flag=0
for j in y:
if i>=j:
flag=1
break
if flag==1:
break
if flag==0:
print('YES')
else:
print('NO')
```
| 0
|
|
766
|
B
|
Mahmoud and a Triangle
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"geometry",
"greedy",
"math",
"number theory",
"sortings"
] | null | null |
Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.
|
The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has.
|
In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.
|
[
"5\n1 5 3 2 4\n",
"3\n4 1 2\n"
] |
[
"YES\n",
"NO\n"
] |
For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.
| 1,000
|
[
{
"input": "5\n1 5 3 2 4",
"output": "YES"
},
{
"input": "3\n4 1 2",
"output": "NO"
},
{
"input": "30\n197 75 517 39724 7906061 1153471 3 15166 168284 3019844 272293 316 16 24548 42 118 5792 5 9373 1866366 4886214 24 2206 712886 104005 1363 836 64273 440585 3576",
"output": "NO"
},
{
"input": "30\n229017064 335281886 247217656 670601882 743442492 615491486 544941439 911270108 474843964 803323771 177115397 62179276 390270885 754889875 881720571 902691435 154083299 328505383 761264351 182674686 94104683 357622370 573909964 320060691 33548810 247029007 812823597 946798893 813659359 710111761",
"output": "YES"
},
{
"input": "40\n740553458 532562042 138583675 75471987 487348843 476240280 972115023 103690894 546736371 915774563 35356828 819948191 138721993 24257926 761587264 767176616 608310208 78275645 386063134 227581756 672567198 177797611 87579917 941781518 274774331 843623616 981221615 630282032 118843963 749160513 354134861 132333165 405839062 522698334 29698277 541005920 856214146 167344951 398332403 68622974",
"output": "YES"
},
{
"input": "40\n155 1470176 7384 765965701 1075 4 561554 6227772 93 16304522 1744 662 3 292572860 19335 908613 42685804 347058 20 132560 3848974 69067081 58 2819 111752888 408 81925 30 11951 4564 251 26381275 473392832 50628 180819969 2378797 10076746 9 214492 31291",
"output": "NO"
},
{
"input": "3\n1 1000000000 1000000000",
"output": "YES"
},
{
"input": "4\n1 1000000000 1000000000 1000000000",
"output": "YES"
},
{
"input": "3\n1 1000000000 1",
"output": "NO"
},
{
"input": "5\n1 2 3 5 2",
"output": "YES"
},
{
"input": "41\n19 161 4090221 118757367 2 45361275 1562319 596751 140871 97 1844 310910829 10708344 6618115 698 1 87059 33 2527892 12703 73396090 17326460 3 368811 20550 813975131 10 53804 28034805 7847 2992 33254 1139 227930 965568 261 4846 503064297 192153458 57 431",
"output": "NO"
},
{
"input": "42\n4317083 530966905 202811311 104 389267 35 1203 18287479 125344279 21690 859122498 65 859122508 56790 1951 148683 457 1 22 2668100 8283 2 77467028 13405 11302280 47877251 328155592 35095 29589769 240574 4 10 1019123 6985189 629846 5118 169 1648973 91891 741 282 3159",
"output": "YES"
},
{
"input": "43\n729551585 11379 5931704 330557 1653 15529406 729551578 278663905 1 729551584 2683 40656510 29802 147 1400284 2 126260 865419 51 17 172223763 86 1 534861 450887671 32 234 25127103 9597697 48226 7034 389 204294 2265706 65783617 4343 3665990 626 78034 106440137 5 18421 1023",
"output": "YES"
},
{
"input": "44\n719528276 2 235 444692918 24781885 169857576 18164 47558 15316043 9465834 64879816 2234575 1631 853530 8 1001 621 719528259 84 6933 31 1 3615623 719528266 40097928 274835337 1381044 11225 2642 5850203 6 527506 18 104977753 76959 29393 49 4283 141 201482 380 1 124523 326015",
"output": "YES"
},
{
"input": "45\n28237 82 62327732 506757 691225170 5 970 4118 264024506 313192 367 14713577 73933 691225154 6660 599 691225145 3473403 51 427200630 1326718 2146678 100848386 1569 27 163176119 193562 10784 45687 819951 38520653 225 119620 1 3 691225169 691225164 17445 23807072 1 9093493 5620082 2542 139 14",
"output": "YES"
},
{
"input": "44\n165580141 21 34 55 1 89 144 17711 2 377 610 987 2584 13 5 4181 6765 10946 1597 8 28657 3 233 75025 121393 196418 317811 9227465 832040 1346269 2178309 3524578 5702887 1 14930352 102334155 24157817 39088169 63245986 701408733 267914296 433494437 514229 46368",
"output": "NO"
},
{
"input": "3\n1 1000000000 999999999",
"output": "NO"
},
{
"input": "5\n1 1 1 1 1",
"output": "YES"
},
{
"input": "10\n1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000",
"output": "NO"
},
{
"input": "5\n2 3 4 10 20",
"output": "YES"
},
{
"input": "6\n18 23 40 80 160 161",
"output": "YES"
},
{
"input": "4\n5 6 7 888",
"output": "YES"
},
{
"input": "9\n1 1 2 2 4 5 10 10 20",
"output": "YES"
},
{
"input": "7\n3 150 900 4 500 1500 5",
"output": "YES"
},
{
"input": "3\n2 2 3",
"output": "YES"
},
{
"input": "7\n1 2 100 200 250 1000000 2000000",
"output": "YES"
},
{
"input": "8\n2 3 5 5 5 6 6 13",
"output": "YES"
},
{
"input": "3\n2 3 4",
"output": "YES"
},
{
"input": "6\n1 1 1 4 5 100",
"output": "YES"
},
{
"input": "13\n1 2 3 5 8 13 22 34 55 89 144 233 377",
"output": "YES"
},
{
"input": "4\n2 3 4 8",
"output": "YES"
},
{
"input": "3\n5 6 7",
"output": "YES"
},
{
"input": "5\n1 4 5 6 1000000",
"output": "YES"
},
{
"input": "4\n5 6 7 20",
"output": "YES"
},
{
"input": "6\n1 1 1 1 1 65",
"output": "YES"
},
{
"input": "4\n3 4 5 100",
"output": "YES"
},
{
"input": "3\n2 4 5",
"output": "YES"
},
{
"input": "7\n1 1 1 1 1 10 1000",
"output": "YES"
},
{
"input": "4\n1 1 2 3",
"output": "NO"
},
{
"input": "11\n1 2 5 6 7 8 9 17 18 19 100",
"output": "YES"
},
{
"input": "4\n5 16 20 200",
"output": "YES"
},
{
"input": "5\n17 6 3 3 1",
"output": "YES"
},
{
"input": "3\n1 1 1",
"output": "YES"
},
{
"input": "6\n1 1 1 2 3 5",
"output": "YES"
},
{
"input": "4\n2 4 6 6",
"output": "YES"
},
{
"input": "9\n1 2 4 4 4 4 7 8 20",
"output": "YES"
},
{
"input": "9\n1 1 2 5 5 5 10 10 20",
"output": "YES"
},
{
"input": "7\n3 150 600 4 1700 6000 5",
"output": "YES"
},
{
"input": "5\n5761 20966 27841 28800 29399",
"output": "YES"
},
{
"input": "9\n1 2 3 6 7 10 11 12 24",
"output": "YES"
},
{
"input": "4\n1 2 1 1",
"output": "YES"
},
{
"input": "5\n1 1 2 3 4",
"output": "YES"
}
] | 1,679,431,301
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 87
| 108
| 7,782,400
|
n = int(input())
retas = sorted([int(x) for x in input().split(" ")])
t = False
for i in range(0, len(retas) -2 ):
a = retas[i]
b = retas[i+1]
c = retas[i+2]
if ((a + b > c) and (c + b > a) and (a + c > b)):
t = True
break
if t:
print("YES")
else:
print("NO")
|
Title: Mahmoud and a Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.
Input Specification:
The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has.
Output Specification:
In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.
Demo Input:
['5\n1 5 3 2 4\n', '3\n4 1 2\n']
Demo Output:
['YES\n', 'NO\n']
Note:
For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.
|
```python
n = int(input())
retas = sorted([int(x) for x in input().split(" ")])
t = False
for i in range(0, len(retas) -2 ):
a = retas[i]
b = retas[i+1]
c = retas[i+2]
if ((a + b > c) and (c + b > a) and (a + c > b)):
t = True
break
if t:
print("YES")
else:
print("NO")
```
| 3
|
|
608
|
B
|
Hamming Distance Sum
|
PROGRAMMING
| 1,500
|
[
"combinatorics",
"strings"
] | null | null |
Genos needs your help. He was asked to solve the following programming problem by Saitama:
The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s**i* is the *i*-th character of *s* and *t**i* is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0<=-<=0|<=+<=|0<=-<=1|<=+<=|1<=-<=1|<=+<=|1<=-<=0|<==<=0<=+<=1<=+<=0<=+<=1<==<=2.
Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
|
The first line of the input contains binary string *a* (1<=≤<=|*a*|<=≤<=200<=000).
The second line of the input contains binary string *b* (|*a*|<=≤<=|*b*|<=≤<=200<=000).
Both strings are guaranteed to consist of characters '0' and '1' only.
|
Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
|
[
"01\n00111\n",
"0011\n0110\n"
] |
[
"3\n",
"2\n"
] |
For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.
The second sample case is described in the statement.
| 1,000
|
[
{
"input": "01\n00111",
"output": "3"
},
{
"input": "0011\n0110",
"output": "2"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "0\n1",
"output": "1"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1001101001101110101101000\n01111000010011111111110010001101000100011110101111",
"output": "321"
},
{
"input": "1110010001000101001011111\n00011011000000100001010000010100110011010001111010",
"output": "316"
}
] | 1,450,892,732
| 4,232
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 6
| 62
| 0
|
s1=input()
s2=input()
x=len(s1)
is1=int(s1,2)
su=0
p=2**x
# implement rolling hash
st=int(s2[0:x],2)
su=0
x=len(s2)-x+1
su=bin(is1^st).count('1')
for i in range(1,x):
st=st-int(s2[i-1])*p
st=st*2+int(s2[i+1])
su=su+bin(is1^st).count('1')
print(su)
|
Title: Hamming Distance Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Genos needs your help. He was asked to solve the following programming problem by Saitama:
The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s**i* is the *i*-th character of *s* and *t**i* is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0<=-<=0|<=+<=|0<=-<=1|<=+<=|1<=-<=1|<=+<=|1<=-<=0|<==<=0<=+<=1<=+<=0<=+<=1<==<=2.
Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
Input Specification:
The first line of the input contains binary string *a* (1<=≤<=|*a*|<=≤<=200<=000).
The second line of the input contains binary string *b* (|*a*|<=≤<=|*b*|<=≤<=200<=000).
Both strings are guaranteed to consist of characters '0' and '1' only.
Output Specification:
Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
Demo Input:
['01\n00111\n', '0011\n0110\n']
Demo Output:
['3\n', '2\n']
Note:
For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.
The second sample case is described in the statement.
|
```python
s1=input()
s2=input()
x=len(s1)
is1=int(s1,2)
su=0
p=2**x
# implement rolling hash
st=int(s2[0:x],2)
su=0
x=len(s2)-x+1
su=bin(is1^st).count('1')
for i in range(1,x):
st=st-int(s2[i-1])*p
st=st*2+int(s2[i+1])
su=su+bin(is1^st).count('1')
print(su)
```
| 0
|
|
143
|
A
|
Help Vasilisa the Wise 2
|
PROGRAMMING
| 1,000
|
[
"brute force",
"math"
] | null | null |
Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.
|
The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.
|
Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any.
|
[
"3 7\n4 6\n5 5\n",
"11 10\n13 8\n5 16\n",
"1 2\n3 4\n5 6\n",
"10 10\n10 10\n10 10\n"
] |
[
"1 2\n3 4\n",
"4 7\n9 1\n",
"-1\n",
"-1\n"
] |
Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9.
| 500
|
[
{
"input": "3 7\n4 6\n5 5",
"output": "1 2\n3 4"
},
{
"input": "11 10\n13 8\n5 16",
"output": "4 7\n9 1"
},
{
"input": "1 2\n3 4\n5 6",
"output": "-1"
},
{
"input": "10 10\n10 10\n10 10",
"output": "-1"
},
{
"input": "5 13\n8 10\n11 7",
"output": "3 2\n5 8"
},
{
"input": "12 17\n10 19\n13 16",
"output": "-1"
},
{
"input": "11 11\n17 5\n12 10",
"output": "9 2\n8 3"
},
{
"input": "12 11\n11 12\n16 7",
"output": "-1"
},
{
"input": "5 9\n7 7\n8 6",
"output": "3 2\n4 5"
},
{
"input": "10 7\n4 13\n11 6",
"output": "-1"
},
{
"input": "18 10\n16 12\n12 16",
"output": "-1"
},
{
"input": "13 6\n10 9\n6 13",
"output": "-1"
},
{
"input": "14 16\n16 14\n18 12",
"output": "-1"
},
{
"input": "16 10\n16 10\n12 14",
"output": "-1"
},
{
"input": "11 9\n12 8\n11 9",
"output": "-1"
},
{
"input": "5 14\n10 9\n10 9",
"output": "-1"
},
{
"input": "2 4\n1 5\n3 3",
"output": "-1"
},
{
"input": "17 16\n14 19\n18 15",
"output": "-1"
},
{
"input": "12 12\n14 10\n16 8",
"output": "9 3\n5 7"
},
{
"input": "15 11\n16 10\n9 17",
"output": "7 8\n9 2"
},
{
"input": "8 10\n9 9\n13 5",
"output": "6 2\n3 7"
},
{
"input": "13 7\n10 10\n5 15",
"output": "4 9\n6 1"
},
{
"input": "14 11\n9 16\n16 9",
"output": "-1"
},
{
"input": "12 8\n14 6\n8 12",
"output": "-1"
},
{
"input": "10 6\n6 10\n4 12",
"output": "-1"
},
{
"input": "10 8\n10 8\n4 14",
"output": "-1"
},
{
"input": "14 13\n9 18\n14 13",
"output": "-1"
},
{
"input": "9 14\n8 15\n8 15",
"output": "-1"
},
{
"input": "3 8\n2 9\n6 5",
"output": "-1"
},
{
"input": "14 17\n18 13\n15 16",
"output": "-1"
},
{
"input": "16 14\n15 15\n17 13",
"output": "9 7\n6 8"
},
{
"input": "14 11\n16 9\n13 12",
"output": "9 5\n7 4"
},
{
"input": "13 10\n11 12\n7 16",
"output": "4 9\n7 3"
},
{
"input": "14 8\n11 11\n13 9",
"output": "8 6\n3 5"
},
{
"input": "12 11\n13 10\n10 13",
"output": "-1"
},
{
"input": "6 5\n2 9\n5 6",
"output": "-1"
},
{
"input": "7 8\n8 7\n12 3",
"output": "-1"
},
{
"input": "7 11\n7 11\n6 12",
"output": "-1"
},
{
"input": "8 5\n11 2\n8 5",
"output": "-1"
},
{
"input": "10 16\n14 12\n14 12",
"output": "-1"
},
{
"input": "7 9\n4 12\n5 11",
"output": "-1"
},
{
"input": "11 13\n19 5\n12 12",
"output": "-1"
},
{
"input": "8 12\n5 15\n11 9",
"output": "2 6\n3 9"
},
{
"input": "16 5\n13 8\n10 11",
"output": "9 7\n4 1"
},
{
"input": "3 14\n8 9\n10 7",
"output": "2 1\n6 8"
},
{
"input": "12 14\n11 15\n9 17",
"output": "3 9\n8 6"
},
{
"input": "13 7\n9 11\n14 6",
"output": "8 5\n1 6"
},
{
"input": "13 8\n15 6\n11 10",
"output": "9 4\n6 2"
},
{
"input": "9 6\n5 10\n3 12",
"output": "1 8\n4 2"
},
{
"input": "12 7\n5 14\n8 11",
"output": "3 9\n2 5"
},
{
"input": "9 12\n3 17\n10 10",
"output": "-1"
},
{
"input": "3 8\n4 6\n5 5",
"output": "-1"
}
] | 1,580,273,886
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 52
| 312
| 1,536,000
|
"""
author - Sayan Bose
date - 29.01.2020
Brooklyn 99 is love!
"""
from itertools import permutations
r1, r2 = map(int, input().split())
c1, c2 = map(int, input().split())
d1, d2 = map(int, input().split())
c = list(permutations([i for i in range(1, 10)], 4))
for i in c:
t1, t2, t3, t4 = i
if t1 + t2 == r1 and t3 + t4 == r2 and t1 + t3 == c1 and t2 + t4 == c2 and t1 + t4 == d1 and t2 + t3 == d2:
print(t1, t2)
print(t3, t4)
exit()
print(-1)
|
Title: Help Vasilisa the Wise 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.
Input Specification:
The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.
Output Specification:
Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any.
Demo Input:
['3 7\n4 6\n5 5\n', '11 10\n13 8\n5 16\n', '1 2\n3 4\n5 6\n', '10 10\n10 10\n10 10\n']
Demo Output:
['1 2\n3 4\n', '4 7\n9 1\n', '-1\n', '-1\n']
Note:
Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9.
|
```python
"""
author - Sayan Bose
date - 29.01.2020
Brooklyn 99 is love!
"""
from itertools import permutations
r1, r2 = map(int, input().split())
c1, c2 = map(int, input().split())
d1, d2 = map(int, input().split())
c = list(permutations([i for i in range(1, 10)], 4))
for i in c:
t1, t2, t3, t4 = i
if t1 + t2 == r1 and t3 + t4 == r2 and t1 + t3 == c1 and t2 + t4 == c2 and t1 + t4 == d1 and t2 + t3 == d2:
print(t1, t2)
print(t3, t4)
exit()
print(-1)
```
| 3
|
|
165
|
A
|
Supercentral Point
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
|
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
|
Print the only number — the number of supercentral points of the given set.
|
[
"8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n",
"5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n"
] |
[
"2\n",
"1\n"
] |
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0).
| 500
|
[
{
"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3",
"output": "2"
},
{
"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0",
"output": "1"
},
{
"input": "9\n-565 -752\n-184 723\n-184 -752\n-184 1\n950 723\n-565 723\n950 -752\n950 1\n-565 1",
"output": "1"
},
{
"input": "25\n-651 897\n916 897\n-651 -808\n-748 301\n-734 414\n-651 -973\n-734 897\n916 -550\n-758 414\n916 180\n-758 -808\n-758 -973\n125 -550\n125 -973\n125 301\n916 414\n-748 -808\n-651 301\n-734 301\n-307 897\n-651 -550\n-651 414\n125 -808\n-748 -550\n916 -808",
"output": "7"
},
{
"input": "1\n487 550",
"output": "0"
},
{
"input": "10\n990 -396\n990 736\n990 646\n990 -102\n990 -570\n990 155\n990 528\n990 489\n990 268\n990 676",
"output": "0"
},
{
"input": "30\n507 836\n525 836\n-779 196\n507 -814\n525 -814\n525 42\n525 196\n525 -136\n-779 311\n507 -360\n525 300\n507 578\n507 311\n-779 836\n507 300\n525 -360\n525 311\n-779 -360\n-779 578\n-779 300\n507 42\n525 578\n-779 379\n507 196\n525 379\n507 379\n-779 -814\n-779 42\n-779 -136\n507 -136",
"output": "8"
},
{
"input": "25\n890 -756\n890 -188\n-37 -756\n-37 853\n523 998\n-261 853\n-351 853\n-351 -188\n523 -756\n-261 -188\n-37 998\n523 -212\n-351 998\n-37 -188\n-351 -756\n-37 -212\n890 998\n890 -212\n523 853\n-351 -212\n-261 -212\n-261 998\n-261 -756\n890 853\n523 -188",
"output": "9"
},
{
"input": "21\n-813 -11\n486 254\n685 254\n-708 254\n-55 -11\n-671 -191\n486 -11\n-671 -11\n685 -11\n685 -191\n486 -191\n-55 254\n-708 -11\n-813 254\n-708 -191\n41 -11\n-671 254\n-813 -191\n41 254\n-55 -191\n41 -191",
"output": "5"
},
{
"input": "4\n1 0\n2 0\n1 1\n1 -1",
"output": "0"
}
] | 1,611,257,272
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 26
| 218
| 2,150,400
|
N = int(input())
points = [list(map(int, input().split())) for _ in range(N)]
C = 0
for i in range(N):
p1 = points[i]
r = 0 # Right
for j in range(N):
p2 = points[j]
if i == j: continue
if p2[1] != p1[1]: continue
if p2[0] <= p1[0]: continue
r = 1; break
l = 0 # Left
for j in range(N):
p2 = points[j]
if i == j: continue
if p2[1] != p1[1]: continue
if p2[0] >= p1[0]: continue
l = 1; break
u = 0 # Up
for j in range(N):
p2 = points[j]
if i == j: continue
if p2[0] != p1[0]: continue
if p2[1] <= p1[1]: continue
u = 1; break
d = 0 # Down
for j in range(N):
p2 = points[j]
if i == j: continue
if p2[0] != p1[0]: continue
if p2[1] >= p1[1]: continue
d = 1; break
C += r * l * u * d
print(C)
|
Title: Supercentral Point
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
Input Specification:
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
Output Specification:
Print the only number — the number of supercentral points of the given set.
Demo Input:
['8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n', '5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0).
|
```python
N = int(input())
points = [list(map(int, input().split())) for _ in range(N)]
C = 0
for i in range(N):
p1 = points[i]
r = 0 # Right
for j in range(N):
p2 = points[j]
if i == j: continue
if p2[1] != p1[1]: continue
if p2[0] <= p1[0]: continue
r = 1; break
l = 0 # Left
for j in range(N):
p2 = points[j]
if i == j: continue
if p2[1] != p1[1]: continue
if p2[0] >= p1[0]: continue
l = 1; break
u = 0 # Up
for j in range(N):
p2 = points[j]
if i == j: continue
if p2[0] != p1[0]: continue
if p2[1] <= p1[1]: continue
u = 1; break
d = 0 # Down
for j in range(N):
p2 = points[j]
if i == j: continue
if p2[0] != p1[0]: continue
if p2[1] >= p1[1]: continue
d = 1; break
C += r * l * u * d
print(C)
```
| 3
|
|
682
|
A
|
Alyona and Numbers
|
PROGRAMMING
| 1,100
|
[
"constructive algorithms",
"math",
"number theory"
] | null | null |
After finishing eating her bun, Alyona came up with two integers *n* and *m*. She decided to write down two columns of integers — the first column containing integers from 1 to *n* and the second containing integers from 1 to *m*. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and equals 0.
As usual, Alyona has some troubles and asks you to help.
|
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1<=000<=000).
|
Print the only integer — the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and (*x*<=+<=*y*) is divisible by 5.
|
[
"6 12\n",
"11 14\n",
"1 5\n",
"3 8\n",
"5 7\n",
"21 21\n"
] |
[
"14\n",
"31\n",
"1\n",
"5\n",
"7\n",
"88\n"
] |
Following pairs are suitable in the first sample case:
- for *x* = 1 fits *y* equal to 4 or 9; - for *x* = 2 fits *y* equal to 3 or 8; - for *x* = 3 fits *y* equal to 2, 7 or 12; - for *x* = 4 fits *y* equal to 1, 6 or 11; - for *x* = 5 fits *y* equal to 5 or 10; - for *x* = 6 fits *y* equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case.
| 500
|
[
{
"input": "6 12",
"output": "14"
},
{
"input": "11 14",
"output": "31"
},
{
"input": "1 5",
"output": "1"
},
{
"input": "3 8",
"output": "5"
},
{
"input": "5 7",
"output": "7"
},
{
"input": "21 21",
"output": "88"
},
{
"input": "10 15",
"output": "30"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 1000000",
"output": "200000"
},
{
"input": "1000000 1",
"output": "200000"
},
{
"input": "1000000 1000000",
"output": "200000000000"
},
{
"input": "944 844",
"output": "159348"
},
{
"input": "368 984",
"output": "72423"
},
{
"input": "792 828",
"output": "131155"
},
{
"input": "920 969",
"output": "178296"
},
{
"input": "640 325",
"output": "41600"
},
{
"input": "768 170",
"output": "26112"
},
{
"input": "896 310",
"output": "55552"
},
{
"input": "320 154",
"output": "9856"
},
{
"input": "744 999",
"output": "148652"
},
{
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"output": "106218"
},
{
"input": "54 688",
"output": "7431"
},
{
"input": "478 828",
"output": "79157"
},
{
"input": "902 184",
"output": "33194"
},
{
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"output": "180"
},
{
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"output": "25384"
},
{
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"output": "2462"
},
{
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"output": "1201"
},
{
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"output": "60512"
},
{
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},
{
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},
{
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"output": "75929310986"
},
{
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},
{
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"output": "105030916263"
},
{
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"output": "132638943"
},
{
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"output": "12720276292"
},
{
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"output": "457187060"
},
{
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"output": "32349225415"
},
{
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"output": "5812618980"
},
{
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},
{
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{
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{
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{
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{
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{
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{
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"output": "1853307952"
},
{
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{
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{
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{
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{
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},
{
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"output": "32214852554"
},
{
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"output": "120403367155"
},
{
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"output": "53095895155"
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{
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{
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{
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"output": "17893399803"
},
{
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"output": "126455602192"
},
{
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"output": "68236422"
},
{
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"output": "17722349770"
},
{
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"output": "12162829017"
},
{
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"output": "46570587880"
},
{
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"output": "14104855884"
},
{
"input": "483964 724917",
"output": "70166746198"
},
{
"input": "567388 241058",
"output": "27354683301"
},
{
"input": "650812 789902",
"output": "102815540084"
},
{
"input": "400999 756281",
"output": "60653584944"
},
{
"input": "100 101",
"output": "2020"
},
{
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"output": "2040"
},
{
"input": "103 100",
"output": "2060"
},
{
"input": "100 104",
"output": "2080"
},
{
"input": "3 4",
"output": "3"
},
{
"input": "11 23",
"output": "50"
},
{
"input": "8 14",
"output": "23"
},
{
"input": "23423 34234",
"output": "160372597"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "999999 999999",
"output": "199999600001"
},
{
"input": "82 99",
"output": "1624"
},
{
"input": "21 18",
"output": "75"
},
{
"input": "234 234",
"output": "10952"
},
{
"input": "4 4",
"output": "4"
},
{
"input": "6 13",
"output": "15"
},
{
"input": "3 9",
"output": "6"
},
{
"input": "99999 99999",
"output": "1999960001"
},
{
"input": "34 33",
"output": "225"
},
{
"input": "2 2",
"output": "0"
},
{
"input": "333 1",
"output": "66"
},
{
"input": "3 3",
"output": "2"
},
{
"input": "8 2",
"output": "3"
},
{
"input": "2179 2218",
"output": "966605"
},
{
"input": "1000000 999999",
"output": "199999800000"
},
{
"input": "873828 774207",
"output": "135304750879"
},
{
"input": "13 19",
"output": "50"
},
{
"input": "1648 576469",
"output": "190004183"
},
{
"input": "11 13",
"output": "28"
},
{
"input": "5 8",
"output": "8"
},
{
"input": "650074 943659",
"output": "122689636154"
},
{
"input": "1 3",
"output": "0"
},
{
"input": "54 43",
"output": "465"
},
{
"input": "14 9",
"output": "26"
},
{
"input": "2 3",
"output": "1"
},
{
"input": "543 534",
"output": "57993"
},
{
"input": "321 123",
"output": "7896"
},
{
"input": "21 3",
"output": "12"
},
{
"input": "2 1",
"output": "0"
},
{
"input": "4 3",
"output": "3"
},
{
"input": "47474 74747",
"output": "709707816"
},
{
"input": "4 9",
"output": "8"
},
{
"input": "7 4",
"output": "6"
},
{
"input": "9 4",
"output": "8"
},
{
"input": "12414 4214",
"output": "10462520"
},
{
"input": "2 9",
"output": "4"
},
{
"input": "253 821",
"output": "41542"
},
{
"input": "2 4",
"output": "2"
}
] | 1,599,216,794
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 128
| 483
| 307,200
|
def solve():
n, m = map(int, input().split())
nn, mm = [0] * 5, [0] * 5
for i in range(1, n+1):
nn[i % 5] += 1
for i in range(1, m+1):
mm[i % 5] += 1
print(nn[0]*mm[0] + nn[1]*mm[4] + nn[2]*mm[3] + nn[3]*mm[2] + nn[4]*mm[1])
solve()
|
Title: Alyona and Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After finishing eating her bun, Alyona came up with two integers *n* and *m*. She decided to write down two columns of integers — the first column containing integers from 1 to *n* and the second containing integers from 1 to *m*. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and equals 0.
As usual, Alyona has some troubles and asks you to help.
Input Specification:
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1<=000<=000).
Output Specification:
Print the only integer — the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and (*x*<=+<=*y*) is divisible by 5.
Demo Input:
['6 12\n', '11 14\n', '1 5\n', '3 8\n', '5 7\n', '21 21\n']
Demo Output:
['14\n', '31\n', '1\n', '5\n', '7\n', '88\n']
Note:
Following pairs are suitable in the first sample case:
- for *x* = 1 fits *y* equal to 4 or 9; - for *x* = 2 fits *y* equal to 3 or 8; - for *x* = 3 fits *y* equal to 2, 7 or 12; - for *x* = 4 fits *y* equal to 1, 6 or 11; - for *x* = 5 fits *y* equal to 5 or 10; - for *x* = 6 fits *y* equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case.
|
```python
def solve():
n, m = map(int, input().split())
nn, mm = [0] * 5, [0] * 5
for i in range(1, n+1):
nn[i % 5] += 1
for i in range(1, m+1):
mm[i % 5] += 1
print(nn[0]*mm[0] + nn[1]*mm[4] + nn[2]*mm[3] + nn[3]*mm[2] + nn[4]*mm[1])
solve()
```
| 3
|
|
960
|
C
|
Subsequence Counting
|
PROGRAMMING
| 1,700
|
[
"bitmasks",
"constructive algorithms",
"greedy",
"implementation"
] | null | null |
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size *n* has 2*n*<=-<=1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence <=-<= Minimum_element_of_subsequence <=≥<=*d*
Pikachu was finally left with *X* subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers *X* and *d*. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print <=-<=1.
|
The only line of input consists of two space separated integers *X* and *d* (1<=≤<=*X*,<=*d*<=≤<=109).
|
Output should consist of two lines.
First line should contain a single integer *n* (1<=≤<=*n*<=≤<=10<=000)— the number of integers in the final array.
Second line should consist of *n* space separated integers — *a*1,<=*a*2,<=... ,<=*a**n* (1<=≤<=*a**i*<=<<=1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
|
[
"10 5\n",
"4 2\n"
] |
[
"6\n5 50 7 15 6 100",
"4\n10 100 1000 10000"
] |
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
| 1,500
|
[
{
"input": "10 5",
"output": "6\n1 1 1 7 13 19 "
},
{
"input": "4 2",
"output": "3\n1 1 4 "
},
{
"input": "4 1",
"output": "3\n1 1 3 "
},
{
"input": "1 1",
"output": "1\n1 "
},
{
"input": "63 1",
"output": "21\n1 1 1 1 1 3 3 3 3 5 5 5 7 7 9 11 13 15 17 19 21 "
},
{
"input": "98 88",
"output": "15\n1 1 1 1 1 1 90 90 90 90 90 179 268 357 446 "
},
{
"input": "746 173",
"output": "37\n1 1 1 1 1 1 1 1 1 175 175 175 175 175 175 175 349 349 349 349 349 349 523 523 523 523 523 697 697 697 871 1045 1219 1393 1567 1741 1915 "
},
{
"input": "890 553",
"output": "43\n1 1 1 1 1 1 1 1 1 555 555 555 555 555 555 555 555 1109 1109 1109 1109 1109 1109 1663 1663 1663 1663 1663 2217 2217 2217 2217 2771 2771 2771 3325 3879 4433 4987 5541 6095 6649 7203 "
},
{
"input": "883 1000",
"output": "40\n1 1 1 1 1 1 1 1 1 1002 1002 1002 1002 1002 1002 1002 1002 2003 2003 2003 2003 2003 2003 3004 3004 3004 3004 3004 4005 4005 4005 4005 5006 6007 7008 8009 9010 10011 11012 12013 "
},
{
"input": "1 1000",
"output": "1\n1 "
},
{
"input": "695 188",
"output": "35\n1 1 1 1 1 1 1 1 1 190 190 190 190 190 190 190 379 379 379 379 379 568 568 568 568 757 757 946 1135 1324 1513 1702 1891 2080 2269 "
},
{
"input": "2060 697",
"output": "19\n1 1 1 1 1 1 1 1 1 1 1 699 699 699 1397 1397 2095 2793 3491 "
},
{
"input": "70 3321",
"output": "12\n1 1 1 1 1 1 3323 3323 6645 9967 13289 16611 "
},
{
"input": "6358 1646",
"output": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1648 1648 1648 1648 1648 1648 1648 1648 1648 1648 1648 3295 3295 3295 3295 3295 3295 3295 4942 4942 4942 4942 4942 4942 6589 6589 6589 6589 8236 8236 9883 11530 13177 14824 16471 18118 19765 21412 "
},
{
"input": "167959139 481199252",
"output": "154\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 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 481199254 962398507 962398507 962398507 962398507 962398507 962398507 962398507 962398507 962398507 962398507 962398507 962398507 962398507 962398507 962398507 962398507 962398507 1443597760 1443597760 1443597760..."
},
{
"input": "641009859 54748096",
"output": "192\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 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 54748098 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 109496195 1094..."
},
{
"input": "524125987 923264237",
"output": "289\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 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 923264239 1846528477 1846528477 1846528477 1846528477 1846528477 1846528477 1846528477 1846528477 1846528477 1846528477 1846528477 1846528477 1846528477 1846528477 1846528477 1846528477 1846..."
},
{
"input": "702209411 496813081",
"output": "276\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 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 496813083 993626165 993626165 993626165 993626165 993626165 993626165 993626165 993626165 993626165 993626165 993626165 993626165 993626165 993626165 993626165 993626165 993626165 99362616..."
},
{
"input": "585325539 365329221",
"output": "243\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 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 365329223 730658445 730658445 730658445 730658445 730658445 730658445 730658445 730658445 730658445 730658445 730658445 730658445 730658445 730658445 730658445 730658445 730658445 730658445 730658445 73065844..."
},
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{
"input": "43603670 268405779",
"output": "182\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 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 268405781 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 536811561 805217341 805217341 805217..."
},
{
"input": "1000000000 1000000000",
"output": "274\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 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 1000000002 2000000003 2000000003 2000000003 2000000003 2000000003 2000000003 2000000003 2000000003 2000000003 2000000003 2000000003 2000000003 20000000..."
},
{
"input": "15000 1",
"output": "66\n1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 3 3 3 5 5 5 5 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 7 9 9 9 9 9 9 9 11 11 11 11 13 13 13 15 17 19 21 23 25 27 "
},
{
"input": "1048576 1",
"output": "21\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 "
},
{
"input": "1000000000 1",
"output": "274\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 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 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 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 19 ..."
},
{
"input": "100000000 1",
"output": "232\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 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 19 19 19 19 19 19 19 19 19 19 19 19 19 19 21 21 21 21 21 21 21 21 2..."
},
{
"input": "1000000 1",
"output": "106\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 11 11 11 11 11 11 11 11 11 13 13 13 13 13 13 15 17 19 21 23 25 27 "
},
{
"input": "536870911 1",
"output": "435\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 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 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 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 ..."
},
{
"input": "10009 1",
"output": "54\n1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 3 5 5 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 9 9 9 9 11 11 11 13 15 17 19 21 23 25 "
},
{
"input": "10001 1",
"output": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 3 5 5 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 9 9 9 9 11 13 15 17 19 21 "
}
] | 1,523,169,779
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 63
| 93
| 7,065,600
|
from math import log2
x, d = map(int, input().split())
n = 0
arr = []
v = 1
while x > 0:
t = int(log2(x + 1))
arr += [v] * t
n += t
x -= 2 ** t - 1
v += d
print(n)
for i in arr:
print(i, end = ' ')
|
Title: Subsequence Counting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size *n* has 2*n*<=-<=1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence <=-<= Minimum_element_of_subsequence <=≥<=*d*
Pikachu was finally left with *X* subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers *X* and *d*. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print <=-<=1.
Input Specification:
The only line of input consists of two space separated integers *X* and *d* (1<=≤<=*X*,<=*d*<=≤<=109).
Output Specification:
Output should consist of two lines.
First line should contain a single integer *n* (1<=≤<=*n*<=≤<=10<=000)— the number of integers in the final array.
Second line should consist of *n* space separated integers — *a*1,<=*a*2,<=... ,<=*a**n* (1<=≤<=*a**i*<=<<=1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Demo Input:
['10 5\n', '4 2\n']
Demo Output:
['6\n5 50 7 15 6 100', '4\n10 100 1000 10000']
Note:
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
|
```python
from math import log2
x, d = map(int, input().split())
n = 0
arr = []
v = 1
while x > 0:
t = int(log2(x + 1))
arr += [v] * t
n += t
x -= 2 ** t - 1
v += d
print(n)
for i in arr:
print(i, end = ' ')
```
| 3
|
|
802
|
G
|
Fake News (easy)
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...
|
The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z).
|
Output YES if the string *s* contains heidi as a subsequence and NO otherwise.
|
[
"abcheaibcdi\n",
"hiedi\n"
] |
[
"YES",
"NO"
] |
A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*.
| 0
|
[
{
"input": "abcheaibcdi",
"output": "YES"
},
{
"input": "hiedi",
"output": "NO"
},
{
"input": "ihied",
"output": "NO"
},
{
"input": "diehi",
"output": "NO"
},
{
"input": "deiih",
"output": "NO"
},
{
"input": "iheid",
"output": "NO"
},
{
"input": "eihdi",
"output": "NO"
},
{
"input": "ehdii",
"output": "NO"
},
{
"input": "edhii",
"output": "NO"
},
{
"input": "deiih",
"output": "NO"
},
{
"input": "ehdii",
"output": "NO"
},
{
"input": "eufyajkssayhjhqcwxmctecaeepjwmfoscqprpcxsqfwnlgzsmmuwuoruantipholrauvxydfvftwfzhnckxswussvlidcojiciflpvkcxkkcmmvtfvxrkwcpeelwsuzqgamamdtdgzscmikvojfvqehblmjczkvtdeymgertgkwfwfukafqlfdhtedcctixhyetdypswgagrpyto",
"output": "YES"
},
{
"input": "arfbvxgdvqzuloojjrwoyqqbxamxybaqltfimofulusfebodjkwwrgwcppkwiodtpjaraglyplgerrpqjkpoggjmfxhwtqrijpijrcyxnoodvwpyjfpvqaoazllbrpzananbrvvybboedidtuvqquklkpeflfaltukjhzjgiofombhbmqbihgtapswykfvlgdoapjqntvqsaohmbvnphvyyhvhavslamczuqifxnwknkaenqmlvetrqogqxmlptgrmqvxzdxdmwobjesmgxckpmawtioavwdngyiwkzypfnxcovwzdohshwlavwsthdssiadhiwmhpvgkrbezm",
"output": "YES"
},
{
"input": "zcectngbqnejjjtsfrluummmqabzqbyccshjqbrjthzhlbmzjfxugvjouwhumsgrnopiyakfadjnbsesamhynsbfbfunupwbxvohfmpwlcpxhovwpfpciclatgmiufwdvtsqrsdcymvkldpnhfeisrzhyhhlkwdzthgprvkpyldeysvbmcibqkpudyrraqdlxpjecvwcvuiklcrsbgvqasmxmtxqzmawcjtozioqlfflinnxpeexbzloaeqjvglbdeufultpjqexvjjjkzemtzuzmxvawilcqdrcjzpqyhtwfphuonzwkotthsaxrmwtnlmcdylxqcfffyndqeouztluqwlhnkkvzwcfiscikv",
"output": "YES"
},
{
"input": "plqaykgovxkvsiahdbglktdlhcqwelxxmtlyymrsyubxdskvyjkrowvcbpdofpjqspsrgpakdczletxujzlsegepzleipiyycpinzxgwjsgslnxsotouddgfcybozfpjhhocpybfjbaywsehbcfrayvancbrumdfngqytnhihyxnlvilrqyhnxeckprqafofelospffhtwguzjbbjlzbqrtiielbvzutzgpqxosiaqznndgobcluuqlhmffiowkjdlkokehtjdyjvmxsiyxureflmdomerfekxdvtitvwzmdsdzplkpbtafxqfpudnhfqpoiwvjnylanunmagoweobdvfjgepbsymfutrjarlxclhgavpytiiqwvojrptofuvlohzeguxdsrihsbucelhhuedltnnjgzxwyblbqvnoliiydfinzlogbvucwykryzcyibnniggbkdkdcdgcsbvvnavtyhtkanrblpvomvjs",
"output": "YES"
},
{
"input": "fbldqzggeunkpwcfirxanmntbfrudijltoertsdvcvcmbwodbibsrxendzebvxwydpasaqnisrijctsuatihxxygbeovhxjdptdcppkvfytdpjspvrannxavmkmisqtygntxkdlousdypyfkrpzapysfpdbyprufwzhunlsfugojddkmxzinatiwfxdqmgyrnjnxvrclhxyuwxtshoqdjptmeecvgmrlvuwqtmnfnfeeiwcavwnqmyustawbjodzwsqmnjxhpqmgpysierlwbbdzcwprpsexyvreewcmlbvaiytjlxdqdaqftefdlmtmmjcwvfejshymhnouoshdzqcwzxpzupkbcievodzqkqvyjuuxxwepxjalvkzufnveji",
"output": "YES"
},
{
"input": "htsyljgoelbbuipivuzrhmfpkgderqpoprlxdpasxhpmxvaztccldtmujjzjmcpdvsdghzpretlsyyiljhjznseaacruriufswuvizwwuvdioazophhyytvbiogttnnouauxllbdn",
"output": "YES"
},
{
"input": "ikmxzqdzxqlvgeojsnhqzciujslwjyzzexnregabdqztpplosdakimjxmuqccbnwvzbajoiqgdobccwnrwmixohrbdarhoeeelzbpigiybtesybwefpcfx",
"output": "YES"
},
{
"input": "bpvbpjvbdfiodsmahxpcubjxdykesubnypalhypantshkjffmxjmelblqnjdmtaltneuyudyevkgedkqrdmrfeemgpghwrifcwincfixokfgurhqbcfzeajrgkgpwqwsepudxulywowwxzdxkumsicsvnzfxspmjpaixgejeaoyoibegosqoyoydmphfpbutrrewyjecowjckvpcceoamtfbitdneuwqfvnagswlskmsmkhmxyfsrpqwhxzocyffiumcy",
"output": "YES"
},
{
"input": "vllsexwrazvlfvhvrtqeohvzzresjdiuhomfpgqcxpqdevplecuaepixhlijatxzegciizpvyvxuembiplwklahlqibykfideysjygagjbgqkbhdhkatddcwlxboinfuomnpc",
"output": "YES"
},
{
"input": "pnjdwpxmvfoqkjtbhquqcuredrkwqzzfjmdvpnbqtypzdovemhhclkvigjvtprrpzbrbcbatkucaqteuciuozytsptvsskkeplaxdaqmjkmef",
"output": "NO"
},
{
"input": "jpwfhvlxvsdhtuozvlmnfiotrgapgjxtcsgcjnodcztupysvvvmjpzqkpommadppdrykuqkcpzojcwvlogvkddedwbggkrhuvtsvdiokehlkdlnukcufjvqxnikcdawvexxwffxtriqbdmkahxdtygodzohwtdmmuvmatdkvweqvaehaxiefpevkvqpyxsrhtmgjsdfcwzqobibeduooldrmglbinrepmunizheqzvgqvpdskhxfidxfnbisyizhepwyrcykcmjxnkyfjgrqlkixcvysa",
"output": "YES"
},
{
"input": "aftcrvuumeqbfvaqlltscnuhkpcifrrhnutjinxdhhdbzvizlrapzjdatuaynoplgjketupgaejciosofuhcgcjdcucarfvtsofgubtphijciswsvidnvpztlaarydkeqxzwdhfbmullkimerukusbrdnnujviydldrwhdfllsjtziwfeaiqotbiprespmxjulnyunkdtcghrzvhtcychkwatqqmladxpvmvlkzscthylbzkpgwlzfjqwarqvdeyngekqvrhrftpxnkfcibbowvnqdkulcdydspcubwlgoyinpnzgidbgunparnueddzwtzdiavbprbbg",
"output": "YES"
},
{
"input": "oagjghsidigeh",
"output": "NO"
},
{
"input": "chdhzpfzabupskiusjoefrwmjmqkbmdgboicnszkhdrlegeqjsldurmbshijadlwsycselhlnudndpdhcnhruhhvsgbthpruiqfirxkhpqhzhqdfpyozolbionodypfcqfeqbkcgmqkizgeyyelzeoothexcoaahedgrvoemqcwccbvoeqawqeuusyjxmgjkpfwcdttfmwunzuwvsihliexlzygqcgpbdiawfvqukikhbjerjkyhpcknlndaystrgsinghlmekbvhntcpypmchcwoglsmwwdulqneuabuuuvtyrnjxfcgoothalwkzzfxakneusezgnnepkpipzromqubraiggqndliz",
"output": "YES"
},
{
"input": "lgirxqkrkgjcutpqitmffvbujcljkqardlalyigxorscczuzikoylcxenryhskoavymexysvmhbsvhtycjlmzhijpuvcjshyfeycvvcfyzytzoyvxajpqdjtfiatnvxnyeqtfcagfftafllhhjhplbdsrfpctkqpinpdfrtlzyjllfbeffputywcckupyslkbbzpgcnxgbmhtqeqqehpdaokkjtatrhyiuusjhwgiiiikxpzdueasemosmmccoakafgvxduwiuflovhhfhffgnnjhoperhhjtvocpqytjxkmrknnknqeglffhfuplopmktykxuvcmbwpoeisrlyyhdpxfvzseucofyhziuiikihpqheqdyzwigeaqzhxzvporgisxgvhyicqyejovqloibhbunsvsunpvmdckkbuokitdzleilfwutcvuuytpupizinfjrzhxudsmjcjyfcpfgthujjowdwtgbvi",
"output": "YES"
},
{
"input": "uuehrvufgerqbzyzksmqnewacotuimawhlbycdbsmhshrsbqwybbkwjwsrkwptvlbbwjiivqugzrxxwgidrcrhrwsmwgeoleptfamzefgaeyxouxocrpvomjrazmxrnffdwrrmblgdiabdncvfougtmjgvvazasnygdrigbsrieoonirlivfyodvulouslxosswgpdexuldmkdbpdlgutiotvxjyecbrsvbmqxrlcpcipjjncduyqtohlzybvlemmfdeubihwlwqglkgjvnwrbgydcpwklmjeewqklmqdbajqgrpnynaxfvxjzgibqerxyhnxenrmcdqaaeksbzyrcaepozqpetaurlhjuxxhwppuhgoihxdxbmxeiahyaqkbknktlzkheaarjoqqrsyeducvoygwalgarldcdlqogfvsncejssmx",
"output": "YES"
},
{
"input": "iiopulfjxoitgiusqrhgbkiyzinphjtclodbkkydetylvuimkhdkklmyoacmekdvjpuxcrvqnjhqhhbfenlpzpwijtykqziocilvtpqhxuyrphdlamawjuzgjwiebkqyrzyqgtkcrhntjxqmcgkrqcslogjegfrivzidfedeegwbbsopvxvdoididlpypjogxaodtueebbwuwilacunqpozczcgrpaxxrtogpxgfkudtxchoravrrdtimhenwmnafxaruuojduxxglefejycfcyxllfimkszmbrhcwpnwjqgwvzysmlaaifdxfjjpgjmcksiigpanappjdloiolojmcqbnpnjjzaufdpjaknylmyvolhwypygwepmqwpiglpcnpypnudhdzpdvgzosyjthzcwtytxq",
"output": "YES"
}
] | 1,619,143,563
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 62
| 0
|
string = input()
fake_str = 'heidi'
def is_fake(string, fake_str):
fake_index = len(fake_str) - 1
for i in range(len(string) - 1, -1, -1):
if string[i] == fake_str[fake_index]:
fake_index -= 1
if fake_index <= 0:
return 'YES'
return 'NO'
print(is_fake(string, fake_str))
|
Title: Fake News (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...
Input Specification:
The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z).
Output Specification:
Output YES if the string *s* contains heidi as a subsequence and NO otherwise.
Demo Input:
['abcheaibcdi\n', 'hiedi\n']
Demo Output:
['YES', 'NO']
Note:
A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*.
|
```python
string = input()
fake_str = 'heidi'
def is_fake(string, fake_str):
fake_index = len(fake_str) - 1
for i in range(len(string) - 1, -1, -1):
if string[i] == fake_str[fake_index]:
fake_index -= 1
if fake_index <= 0:
return 'YES'
return 'NO'
print(is_fake(string, fake_str))
```
| 0
|
|
869
|
A
|
The Artful Expedient
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
Rock... Paper!
After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows.
A positive integer *n* is decided first. Both Koyomi and Karen independently choose *n* distinct positive integers, denoted by *x*1,<=*x*2,<=...,<=*x**n* and *y*1,<=*y*2,<=...,<=*y**n* respectively. They reveal their sequences, and repeat until all of 2*n* integers become distinct, which is the only final state to be kept and considered.
Then they count the number of ordered pairs (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) such that the value *x**i* xor *y**j* equals to one of the 2*n* integers. Here xor means the [bitwise exclusive or](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation on two integers, and is denoted by operators ^ and/or xor in most programming languages.
Karen claims a win if the number of such pairs is even, and Koyomi does otherwise. And you're here to help determine the winner of their latest game.
|
The first line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=2<=000) — the length of both sequences.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=2·106) — the integers finally chosen by Koyomi.
The third line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (1<=≤<=*y**i*<=≤<=2·106) — the integers finally chosen by Karen.
Input guarantees that the given 2*n* integers are pairwise distinct, that is, no pair (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) exists such that one of the following holds: *x**i*<==<=*y**j*; *i*<=≠<=*j* and *x**i*<==<=*x**j*; *i*<=≠<=*j* and *y**i*<==<=*y**j*.
|
Output one line — the name of the winner, that is, "Koyomi" or "Karen" (without quotes). Please be aware of the capitalization.
|
[
"3\n1 2 3\n4 5 6\n",
"5\n2 4 6 8 10\n9 7 5 3 1\n"
] |
[
"Karen\n",
"Karen\n"
] |
In the first example, there are 6 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an even number.
In the second example, there are 16 such pairs, and Karen wins again.
| 500
|
[
{
"input": "3\n1 2 3\n4 5 6",
"output": "Karen"
},
{
"input": "5\n2 4 6 8 10\n9 7 5 3 1",
"output": "Karen"
},
{
"input": "1\n1\n2000000",
"output": "Karen"
},
{
"input": "2\n97153 2000000\n1999998 254",
"output": "Karen"
},
{
"input": "15\n31 30 29 28 27 26 25 24 23 22 21 20 19 18 17\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15",
"output": "Karen"
},
{
"input": "30\n79656 68607 871714 1858841 237684 1177337 532141 161161 1111201 527235 323345 1979059 665353 507265 1290761 610606 1238375 743262 106355 1167830 180315 1233029 816465 752968 782570 1499881 1328457 1867240 13948 1302782\n322597 1868510 1958236 1348157 765908 1023636 874300 537124 631783 414906 886318 1931572 1381013 992451 1305644 1525745 716087 83173 303248 1572710 43084 333341 992413 267806 70390 644521 1014900 497068 178940 1920268",
"output": "Karen"
},
{
"input": "30\n1143673 436496 1214486 1315862 148404 724601 1430740 1433008 1654610 1635673 614673 1713408 1270999 1697 1463796 50027 525482 1659078 688200 842647 518551 877506 1017082 1807856 3280 759698 1208220 470180 829800 1960886\n1312613 1965095 967255 1289012 1950383 582960 856825 49684 808824 319418 1968270 190821 344545 211332 1219388 1773751 1876402 132626 541448 1584672 24276 1053225 1823073 1858232 1209173 1035991 1956373 1237148 1973608 848873",
"output": "Karen"
},
{
"input": "1\n2\n3",
"output": "Karen"
},
{
"input": "1\n1048576\n1020000",
"output": "Karen"
},
{
"input": "3\n9 33 69\n71 74 100",
"output": "Karen"
},
{
"input": "3\n1 2 3\n9 5 6",
"output": "Karen"
},
{
"input": "3\n1 7 8\n9 10 20",
"output": "Karen"
},
{
"input": "3\n1 3 2\n4 5 8",
"output": "Karen"
},
{
"input": "3\n2 1 100\n3 4 9",
"output": "Karen"
},
{
"input": "3\n3 1 100\n2 1000 100000",
"output": "Karen"
},
{
"input": "3\n1 2 5\n3 4 6",
"output": "Karen"
},
{
"input": "3\n3 1 8\n2 4 17",
"output": "Karen"
},
{
"input": "3\n1 5 6\n7 8 3",
"output": "Karen"
},
{
"input": "1\n1\n3",
"output": "Karen"
},
{
"input": "3\n1 3 10\n2 4 20",
"output": "Karen"
},
{
"input": "3\n7 8 10\n15 9 11",
"output": "Karen"
},
{
"input": "3\n5 6 8\n3 100 9",
"output": "Karen"
},
{
"input": "3\n1 2 3\n4 5 8",
"output": "Karen"
},
{
"input": "3\n1 2 19\n3 7 30",
"output": "Karen"
},
{
"input": "3\n1 2 3\n6 7 8",
"output": "Karen"
},
{
"input": "3\n1 4 55\n2 3 9",
"output": "Karen"
},
{
"input": "3\n1 100 200\n5 4 500",
"output": "Karen"
},
{
"input": "1\n6\n7",
"output": "Karen"
},
{
"input": "3\n1 3 5\n2 4 13",
"output": "Karen"
},
{
"input": "3\n3 1 100\n2 1000 10000",
"output": "Karen"
},
{
"input": "3\n1 3 9\n2 4 40",
"output": "Karen"
},
{
"input": "2\n1 199999\n1935807 2000000",
"output": "Karen"
},
{
"input": "3\n1 3 8\n2 4 24",
"output": "Karen"
},
{
"input": "2\n4 1\n7 3",
"output": "Karen"
},
{
"input": "3\n1 2 4\n3 7 8",
"output": "Karen"
},
{
"input": "3\n1 6 10000\n2 3 100000",
"output": "Karen"
}
] | 1,509,044,017
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 62
| 0
|
print('Karen')
#if A^B==C then A^C ==B
|
Title: The Artful Expedient
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Rock... Paper!
After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows.
A positive integer *n* is decided first. Both Koyomi and Karen independently choose *n* distinct positive integers, denoted by *x*1,<=*x*2,<=...,<=*x**n* and *y*1,<=*y*2,<=...,<=*y**n* respectively. They reveal their sequences, and repeat until all of 2*n* integers become distinct, which is the only final state to be kept and considered.
Then they count the number of ordered pairs (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) such that the value *x**i* xor *y**j* equals to one of the 2*n* integers. Here xor means the [bitwise exclusive or](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation on two integers, and is denoted by operators ^ and/or xor in most programming languages.
Karen claims a win if the number of such pairs is even, and Koyomi does otherwise. And you're here to help determine the winner of their latest game.
Input Specification:
The first line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=2<=000) — the length of both sequences.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=2·106) — the integers finally chosen by Koyomi.
The third line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (1<=≤<=*y**i*<=≤<=2·106) — the integers finally chosen by Karen.
Input guarantees that the given 2*n* integers are pairwise distinct, that is, no pair (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) exists such that one of the following holds: *x**i*<==<=*y**j*; *i*<=≠<=*j* and *x**i*<==<=*x**j*; *i*<=≠<=*j* and *y**i*<==<=*y**j*.
Output Specification:
Output one line — the name of the winner, that is, "Koyomi" or "Karen" (without quotes). Please be aware of the capitalization.
Demo Input:
['3\n1 2 3\n4 5 6\n', '5\n2 4 6 8 10\n9 7 5 3 1\n']
Demo Output:
['Karen\n', 'Karen\n']
Note:
In the first example, there are 6 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an even number.
In the second example, there are 16 such pairs, and Karen wins again.
|
```python
print('Karen')
#if A^B==C then A^C ==B
```
| 3
|
|
347
|
B
|
Fixed Points
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation",
"math"
] | null | null |
A permutation of length *n* is an integer sequence such that each integer from 0 to (*n*<=-<=1) appears exactly once in it. For example, sequence [0,<=2,<=1] is a permutation of length 3 while both [0,<=2,<=2] and [1,<=2,<=3] are not.
A fixed point of a function is a point that is mapped to itself by the function. A permutation can be regarded as a bijective function. We'll get a definition of a fixed point in a permutation. An integer *i* is a fixed point of permutation *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 if and only if *a**i*<==<=*i*. For example, permutation [0,<=2,<=1] has 1 fixed point and permutation [0,<=1,<=2] has 3 fixed points.
You are given permutation *a*. You are allowed to swap two elements of the permutation at most once. Your task is to maximize the number of fixed points in the resulting permutation. Note that you are allowed to make at most one swap operation.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 — the given permutation.
|
Print a single integer — the maximum possible number of fixed points in the permutation after at most one swap operation.
|
[
"5\n0 1 3 4 2\n"
] |
[
"3\n"
] |
none
| 1,000
|
[
{
"input": "5\n0 1 3 4 2",
"output": "3"
},
{
"input": "10\n6 9 4 7 8 2 3 5 0 1",
"output": "2"
},
{
"input": "100\n99 5 40 32 4 31 38 57 94 47 26 16 89 72 9 80 55 86 78 90 42 41 46 74 56 97 21 48 66 27 93 85 88 59 64 95 10 45 12 22 84 60 8 98 62 51 14 65 39 30 11 71 92 19 76 43 87 54 15 53 37 6 25 18 96 35 13 91 2 3 0 23 1 7 49 75 81 33 50 52 63 44 69 36 17 61 24 20 68 34 73 29 70 83 58 79 82 28 77 67",
"output": "3"
},
{
"input": "3\n0 1 2",
"output": "3"
},
{
"input": "3\n2 1 0",
"output": "3"
},
{
"input": "3\n1 2 0",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "5\n0 1 2 3 4",
"output": "5"
},
{
"input": "4\n0 1 2 3",
"output": "4"
},
{
"input": "7\n0 1 2 4 3 6 5",
"output": "5"
},
{
"input": "6\n0 1 2 3 5 4",
"output": "6"
}
] | 1,422,269,991
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 0
|
input()
a=list(map(int,input().split()))
sum=0
flag=False
for i in range(len(a)):
if i==a[i]:
sum+=1
elif a[a[i]]==i:
sum+=2
flag=True
if not flag and sum!=len(a):
sum+=1
print(sum)
|
Title: Fixed Points
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation of length *n* is an integer sequence such that each integer from 0 to (*n*<=-<=1) appears exactly once in it. For example, sequence [0,<=2,<=1] is a permutation of length 3 while both [0,<=2,<=2] and [1,<=2,<=3] are not.
A fixed point of a function is a point that is mapped to itself by the function. A permutation can be regarded as a bijective function. We'll get a definition of a fixed point in a permutation. An integer *i* is a fixed point of permutation *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 if and only if *a**i*<==<=*i*. For example, permutation [0,<=2,<=1] has 1 fixed point and permutation [0,<=1,<=2] has 3 fixed points.
You are given permutation *a*. You are allowed to swap two elements of the permutation at most once. Your task is to maximize the number of fixed points in the resulting permutation. Note that you are allowed to make at most one swap operation.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 — the given permutation.
Output Specification:
Print a single integer — the maximum possible number of fixed points in the permutation after at most one swap operation.
Demo Input:
['5\n0 1 3 4 2\n']
Demo Output:
['3\n']
Note:
none
|
```python
input()
a=list(map(int,input().split()))
sum=0
flag=False
for i in range(len(a)):
if i==a[i]:
sum+=1
elif a[a[i]]==i:
sum+=2
flag=True
if not flag and sum!=len(a):
sum+=1
print(sum)
```
| 0
|
|
445
|
A
|
DZY Loves Chessboard
|
PROGRAMMING
| 1,200
|
[
"dfs and similar",
"implementation"
] | null | null |
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
|
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
|
[
"1 1\n.\n",
"2 2\n..\n..\n",
"3 3\n.-.\n---\n--."
] |
[
"B\n",
"BW\nWB\n",
"B-B\n---\n--B"
] |
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.
| 500
|
[
{
"input": "1 1\n.",
"output": "B"
},
{
"input": "2 2\n..\n..",
"output": "BW\nWB"
},
{
"input": "3 3\n.-.\n---\n--.",
"output": "B-B\n---\n--B"
},
{
"input": "3 1\n-\n.\n.",
"output": "-\nW\nB"
},
{
"input": "11 11\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------",
"output": "-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------"
},
{
"input": "1 1\n-",
"output": "-"
},
{
"input": "4 2\n..\n..\n..\n..",
"output": "BW\nWB\nBW\nWB"
},
{
"input": "5 1\n.\n.\n.\n.\n.",
"output": "B\nW\nB\nW\nB"
}
] | 1,640,879,479
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 46
| 0
|
n , m = [int(i) for i in input().split()]
l = []
for i in range(n):
s = input()
s=s.replace(".","B")
l.append(list(s))
for i in range(n-1):
for j in range(m-1):
if l[i][j] == "B" and l[i][j+1] == "B":
l[i][j+1] = "W"
if l[i][j] == "B" and l[i+1][j] == "B":
l[i+1][j] = "W"
for i in l:
print("".join(i))
|
Title: DZY Loves Chessboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
Output Specification:
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
Demo Input:
['1 1\n.\n', '2 2\n..\n..\n', '3 3\n.-.\n---\n--.']
Demo Output:
['B\n', 'BW\nWB\n', 'B-B\n---\n--B']
Note:
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.
|
```python
n , m = [int(i) for i in input().split()]
l = []
for i in range(n):
s = input()
s=s.replace(".","B")
l.append(list(s))
for i in range(n-1):
for j in range(m-1):
if l[i][j] == "B" and l[i][j+1] == "B":
l[i][j+1] = "W"
if l[i][j] == "B" and l[i+1][j] == "B":
l[i+1][j] = "W"
for i in l:
print("".join(i))
```
| 0
|
|
454
|
B
|
Little Pony and Sort by Shift
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
One day, Twilight Sparkle is interested in how to sort a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:
Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=105). The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
|
If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.
|
[
"2\n2 1\n",
"3\n1 3 2\n",
"2\n1 2\n"
] |
[
"1\n",
"-1\n",
"0\n"
] |
none
| 1,000
|
[
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 3 2",
"output": "-1"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "6\n3 4 5 6 3 2",
"output": "-1"
},
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "5\n1 1 2 1 1",
"output": "2"
},
{
"input": "4\n5 4 5 4",
"output": "-1"
},
{
"input": "7\n3 4 5 5 5 1 2",
"output": "2"
},
{
"input": "5\n2 2 1 2 2",
"output": "3"
},
{
"input": "5\n5 4 1 2 3",
"output": "-1"
},
{
"input": "4\n6 1 2 7",
"output": "-1"
},
{
"input": "5\n4 5 6 2 3",
"output": "2"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "4\n1 2 2 1",
"output": "1"
},
{
"input": "9\n4 5 6 7 1 2 3 4 10",
"output": "-1"
},
{
"input": "7\n2 3 4 1 2 3 4",
"output": "-1"
},
{
"input": "6\n1 2 1 2 1 2",
"output": "-1"
},
{
"input": "3\n3 2 1",
"output": "-1"
},
{
"input": "4\n1 4 4 1",
"output": "1"
},
{
"input": "5\n1 2 1 1 1",
"output": "3"
},
{
"input": "5\n4 6 7 3 5",
"output": "-1"
},
{
"input": "4\n2 3 1 4",
"output": "-1"
},
{
"input": "5\n5 4 3 2 1",
"output": "-1"
},
{
"input": "4\n2 4 1 4",
"output": "-1"
},
{
"input": "6\n4 5 6 1 2 7",
"output": "-1"
},
{
"input": "6\n1 2 3 1 1 1",
"output": "3"
},
{
"input": "5\n1 3 3 3 1",
"output": "1"
},
{
"input": "6\n5 6 7 5 5 5",
"output": "3"
},
{
"input": "5\n3 4 2 1 2",
"output": "-1"
},
{
"input": "3\n3 4 2",
"output": "1"
},
{
"input": "6\n1 1 2 2 1 1",
"output": "2"
},
{
"input": "4\n2 3 4 2",
"output": "1"
},
{
"input": "5\n3 5 7 7 3",
"output": "1"
},
{
"input": "4\n1 1 4 1",
"output": "1"
},
{
"input": "7\n1 5 6 1 1 1 1",
"output": "4"
},
{
"input": "5\n7 8 6 7 8",
"output": "-1"
},
{
"input": "4\n2 4 1 3",
"output": "-1"
}
] | 1,606,810,036
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 5
| 187
| 10,547,200
|
n = int(input())
a = list(map(int,input().split()))
b = sorted(a)
if a==b:
print(0)
else:
for j in range(1,n):
if a[j]<a[j-1]:
i = j
break
c = a[i:n]
c = c[::-1]
a = c+a[0:i]
if a==b:
print(n-i)
else:
print(-1)
|
Title: Little Pony and Sort by Shift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, Twilight Sparkle is interested in how to sort a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:
Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=105). The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.
Demo Input:
['2\n2 1\n', '3\n1 3 2\n', '2\n1 2\n']
Demo Output:
['1\n', '-1\n', '0\n']
Note:
none
|
```python
n = int(input())
a = list(map(int,input().split()))
b = sorted(a)
if a==b:
print(0)
else:
for j in range(1,n):
if a[j]<a[j-1]:
i = j
break
c = a[i:n]
c = c[::-1]
a = c+a[0:i]
if a==b:
print(n-i)
else:
print(-1)
```
| 0
|
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,679,587,502
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 92
| 0
|
n = int(input())
x = list(map(int, input().split()))
y = []
for i in range(n):
if x[i] % 2 == 0:
y.append(x[i])
if len(y) == 1:
print(x.index(y[0])+1)
else:
for j in range(n):
if x[j] % 2 != 0:
print(j+1)
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
n = int(input())
x = list(map(int, input().split()))
y = []
for i in range(n):
if x[i] % 2 == 0:
y.append(x[i])
if len(y) == 1:
print(x.index(y[0])+1)
else:
for j in range(n):
if x[j] % 2 != 0:
print(j+1)
```
| 3.977
|
946
|
B
|
Weird Subtraction Process
|
PROGRAMMING
| 1,100
|
[
"math",
"number theory"
] | null | null |
You have two variables *a* and *b*. Consider the following sequence of actions performed with these variables:
1. If *a*<==<=0 or *b*<==<=0, end the process. Otherwise, go to step 2;1. If *a*<=≥<=2·*b*, then set the value of *a* to *a*<=-<=2·*b*, and repeat step 1. Otherwise, go to step 3;1. If *b*<=≥<=2·*a*, then set the value of *b* to *b*<=-<=2·*a*, and repeat step 1. Otherwise, end the process.
Initially the values of *a* and *b* are positive integers, and so the process will be finite.
You have to determine the values of *a* and *b* after the process ends.
|
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1018). *n* is the initial value of variable *a*, and *m* is the initial value of variable *b*.
|
Print two integers — the values of *a* and *b* after the end of the process.
|
[
"12 5\n",
"31 12\n"
] |
[
"0 1\n",
"7 12\n"
] |
Explanations to the samples:
1. *a* = 12, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 1 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 0, *b* = 1;1. *a* = 31, *b* = 12 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 7, *b* = 12.
| 0
|
[
{
"input": "12 5",
"output": "0 1"
},
{
"input": "31 12",
"output": "7 12"
},
{
"input": "1000000000000000000 7",
"output": "8 7"
},
{
"input": "31960284556200 8515664064180",
"output": "14928956427840 8515664064180"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1000000000000000000 1000000000000000000"
},
{
"input": "1 1000",
"output": "1 0"
},
{
"input": "1 1000000",
"output": "1 0"
},
{
"input": "1 1000000000000000",
"output": "1 0"
},
{
"input": "1 99999999999999999",
"output": "1 1"
},
{
"input": "1 4",
"output": "1 0"
},
{
"input": "1000000000000001 500000000000000",
"output": "1 0"
},
{
"input": "1 1000000000000000000",
"output": "1 0"
},
{
"input": "2 4",
"output": "2 0"
},
{
"input": "2 1",
"output": "0 1"
},
{
"input": "6 19",
"output": "6 7"
},
{
"input": "22 5",
"output": "0 1"
},
{
"input": "10000000000000000 100000000000000001",
"output": "0 1"
},
{
"input": "1 1000000000000",
"output": "1 0"
},
{
"input": "2 1000000000000000",
"output": "2 0"
},
{
"input": "2 10",
"output": "2 2"
},
{
"input": "51 100",
"output": "51 100"
},
{
"input": "3 1000000000000000000",
"output": "3 4"
},
{
"input": "1000000000000000000 3",
"output": "4 3"
},
{
"input": "1 10000000000000000",
"output": "1 0"
},
{
"input": "8796203 7556",
"output": "1019 1442"
},
{
"input": "5 22",
"output": "1 0"
},
{
"input": "1000000000000000000 1",
"output": "0 1"
},
{
"input": "1 100000000000",
"output": "1 0"
},
{
"input": "2 1000000000000",
"output": "2 0"
},
{
"input": "5 4567865432345678",
"output": "5 8"
},
{
"input": "576460752303423487 288230376151711743",
"output": "1 1"
},
{
"input": "499999999999999999 1000000000000000000",
"output": "3 2"
},
{
"input": "1 9999999999999",
"output": "1 1"
},
{
"input": "103 1000000000000000000",
"output": "103 196"
},
{
"input": "7 1",
"output": "1 1"
},
{
"input": "100000000000000001 10000000000000000",
"output": "1 0"
},
{
"input": "5 10",
"output": "5 0"
},
{
"input": "7 11",
"output": "7 11"
},
{
"input": "1 123456789123456",
"output": "1 0"
},
{
"input": "5000000000000 100000000000001",
"output": "0 1"
},
{
"input": "1000000000000000 1",
"output": "0 1"
},
{
"input": "1000000000000000000 499999999999999999",
"output": "2 3"
},
{
"input": "10 5",
"output": "0 5"
},
{
"input": "9 18917827189272",
"output": "9 0"
},
{
"input": "179 100000000000497000",
"output": "179 270"
},
{
"input": "5 100000000000001",
"output": "1 1"
},
{
"input": "5 20",
"output": "5 0"
},
{
"input": "100000001 50000000",
"output": "1 0"
},
{
"input": "345869461223138161 835002744095575440",
"output": "1 0"
},
{
"input": "8589934592 4294967296",
"output": "0 4294967296"
},
{
"input": "4 8",
"output": "4 0"
},
{
"input": "1 100000000000000000",
"output": "1 0"
},
{
"input": "1000000000000000000 333333333333333",
"output": "1000 1333"
},
{
"input": "25 12",
"output": "1 0"
},
{
"input": "24 54",
"output": "0 6"
},
{
"input": "6 12",
"output": "6 0"
},
{
"input": "129200000000305 547300000001292",
"output": "1 0"
},
{
"input": "1000000000000000000 49999999999999999",
"output": "20 39"
},
{
"input": "1 2",
"output": "1 0"
},
{
"input": "1 123456789876",
"output": "1 0"
},
{
"input": "2 3",
"output": "2 3"
},
{
"input": "1 3",
"output": "1 1"
},
{
"input": "1 1",
"output": "1 1"
},
{
"input": "19 46",
"output": "3 2"
},
{
"input": "3 6",
"output": "3 0"
},
{
"input": "129 1000000000000000000",
"output": "1 0"
},
{
"input": "12 29",
"output": "0 1"
},
{
"input": "8589934592 2147483648",
"output": "0 2147483648"
},
{
"input": "2147483648 8589934592",
"output": "2147483648 0"
},
{
"input": "5 6",
"output": "5 6"
},
{
"input": "1000000000000000000 2",
"output": "0 2"
},
{
"input": "2 7",
"output": "2 3"
},
{
"input": "17174219820754872 61797504734333370",
"output": "17174219820754872 27449065092823626"
},
{
"input": "49 100",
"output": "1 0"
},
{
"input": "7 17",
"output": "1 1"
},
{
"input": "1000000000000000000 10000001",
"output": "0 1"
},
{
"input": "49999999999999999 2",
"output": "3 2"
},
{
"input": "49999999999999999 1",
"output": "1 1"
},
{
"input": "576460752303423487 2",
"output": "3 2"
},
{
"input": "19395 19395",
"output": "19395 19395"
},
{
"input": "19394 19394",
"output": "19394 19394"
}
] | 1,652,152,445
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 2
| 1,000
| 0
|
a, b = map(int, input().split())
if a > b:
a, b = b, a
swap = 1
else:
swap = 0
while a:
a2 = 2*a
if a2 > b:
break
a, b = a, b-a2
if a > b:
a, b = b, a
swap = 1-swap
if swap:
a, b = b, a
print(a, b)
|
Title: Weird Subtraction Process
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have two variables *a* and *b*. Consider the following sequence of actions performed with these variables:
1. If *a*<==<=0 or *b*<==<=0, end the process. Otherwise, go to step 2;1. If *a*<=≥<=2·*b*, then set the value of *a* to *a*<=-<=2·*b*, and repeat step 1. Otherwise, go to step 3;1. If *b*<=≥<=2·*a*, then set the value of *b* to *b*<=-<=2·*a*, and repeat step 1. Otherwise, end the process.
Initially the values of *a* and *b* are positive integers, and so the process will be finite.
You have to determine the values of *a* and *b* after the process ends.
Input Specification:
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1018). *n* is the initial value of variable *a*, and *m* is the initial value of variable *b*.
Output Specification:
Print two integers — the values of *a* and *b* after the end of the process.
Demo Input:
['12 5\n', '31 12\n']
Demo Output:
['0 1\n', '7 12\n']
Note:
Explanations to the samples:
1. *a* = 12, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 1 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 0, *b* = 1;1. *a* = 31, *b* = 12 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 7, *b* = 12.
|
```python
a, b = map(int, input().split())
if a > b:
a, b = b, a
swap = 1
else:
swap = 0
while a:
a2 = 2*a
if a2 > b:
break
a, b = a, b-a2
if a > b:
a, b = b, a
swap = 1-swap
if swap:
a, b = b, a
print(a, b)
```
| 0
|
|
510
|
A
|
Fox And Snake
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
|
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
|
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
|
[
"3 3\n",
"3 4\n",
"5 3\n",
"9 9\n"
] |
[
"###\n..#\n###\n",
"####\n...#\n####\n",
"###\n..#\n###\n#..\n###\n",
"#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n"
] |
none
| 500
|
[
{
"input": "3 3",
"output": "###\n..#\n###"
},
{
"input": "3 4",
"output": "####\n...#\n####"
},
{
"input": "5 3",
"output": "###\n..#\n###\n#..\n###"
},
{
"input": "9 9",
"output": "#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########"
},
{
"input": "3 5",
"output": "#####\n....#\n#####"
},
{
"input": "3 6",
"output": "######\n.....#\n######"
},
{
"input": "7 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "7 4",
"output": "####\n...#\n####\n#...\n####\n...#\n####"
},
{
"input": "49 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "43 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "43 27",
"output": "###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n....................."
},
{
"input": "11 15",
"output": "###############\n..............#\n###############\n#..............\n###############\n..............#\n###############\n#..............\n###############\n..............#\n###############"
},
{
"input": "11 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "19 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "23 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "49 49",
"output": "#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#..."
},
{
"input": "33 43",
"output": "###########################################\n..........................................#\n###########################################\n#..........................................\n###########################################\n..........................................#\n###########################################\n#..........................................\n###########################################\n..........................................#\n###########################################\n#.................."
},
{
"input": "33 44",
"output": "############################################\n...........................................#\n############################################\n#...........................................\n############################################\n...........................................#\n############################################\n#...........................................\n############################################\n...........................................#\n############################################\n#......."
},
{
"input": "45 45",
"output": "#############################################\n............................................#\n#############################################\n#............................................\n#############################################\n............................................#\n#############################################\n#............................................\n#############################################\n............................................#\n#########################################..."
},
{
"input": "45 49",
"output": "#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#..."
},
{
"input": "45 3",
"output": "###\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###"
},
{
"input": "45 4",
"output": "####\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####"
},
{
"input": "5 4",
"output": "####\n...#\n####\n#...\n####"
},
{
"input": "5 5",
"output": "#####\n....#\n#####\n#....\n#####"
},
{
"input": "5 6",
"output": "######\n.....#\n######\n#.....\n######"
},
{
"input": "7 5",
"output": "#####\n....#\n#####\n#....\n#####\n....#\n#####"
},
{
"input": "7 6",
"output": "######\n.....#\n######\n#.....\n######\n.....#\n######"
},
{
"input": "7 7",
"output": "#######\n......#\n#######\n#......\n#######\n......#\n#######"
}
] | 1,674,028,444
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 46
| 0
|
a,b=map(int,input().split())
c=0;
for i in range(0,a):
if(i%2==0):
for j in range(0,b):
print("#",end="")
else:
if(c%2==0):
for j in range(0,b-1):
print(".",end="")
print("#",end="")
else:
print("#",end="")
for j in range(1,b):
print(".",end="")
c=c+1;
print()
|
Title: Fox And Snake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
Input Specification:
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
Output Specification:
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
Demo Input:
['3 3\n', '3 4\n', '5 3\n', '9 9\n']
Demo Output:
['###\n..#\n###\n', '####\n...#\n####\n', '###\n..#\n###\n#..\n###\n', '#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n']
Note:
none
|
```python
a,b=map(int,input().split())
c=0;
for i in range(0,a):
if(i%2==0):
for j in range(0,b):
print("#",end="")
else:
if(c%2==0):
for j in range(0,b-1):
print(".",end="")
print("#",end="")
else:
print("#",end="")
for j in range(1,b):
print(".",end="")
c=c+1;
print()
```
| 3
|
|
22
|
E
|
Scheme
|
PROGRAMMING
| 2,300
|
[
"dfs and similar",
"graphs",
"trees"
] |
E. Scheme
|
2
|
256
|
To learn as soon as possible the latest news about their favourite fundamentally new operating system, BolgenOS community from Nizhni Tagil decided to develop a scheme. According to this scheme a community member, who is the first to learn the news, calls some other member, the latter, in his turn, calls some third member, and so on; i.e. a person with index *i* got a person with index *f**i*, to whom he has to call, if he learns the news. With time BolgenOS community members understood that their scheme doesn't work sometimes — there were cases when some members didn't learn the news at all. Now they want to supplement the scheme: they add into the scheme some instructions of type (*x**i*,<=*y**i*), which mean that person *x**i* has to call person *y**i* as well. What is the minimum amount of instructions that they need to add so, that at the end everyone learns the news, no matter who is the first to learn it?
|
The first input line contains number *n* (2<=≤<=*n*<=≤<=105) — amount of BolgenOS community members. The second line contains *n* space-separated integer numbers *f**i* (1<=≤<=*f**i*<=≤<=*n*,<=*i*<=≠<=*f**i*) — index of a person, to whom calls a person with index *i*.
|
In the first line output one number — the minimum amount of instructions to add. Then output one of the possible variants to add these instructions into the scheme, one instruction in each line. If the solution is not unique, output any.
|
[
"3\n3 3 2\n",
"7\n2 3 1 3 4 4 1\n"
] |
[
"1\n3 1\n",
"3\n2 5\n2 6\n3 7\n"
] |
none
| 0
|
[
{
"input": "3\n3 3 2",
"output": "1\n3 1"
},
{
"input": "7\n2 3 1 3 4 4 1",
"output": "3\n1 5\n1 6\n1 7"
},
{
"input": "2\n2 1",
"output": "0"
},
{
"input": "3\n2 3 1",
"output": "0"
},
{
"input": "4\n2 4 4 3",
"output": "1\n4 1"
},
{
"input": "5\n5 3 5 2 3",
"output": "2\n5 1\n5 4"
},
{
"input": "9\n2 5 6 7 4 1 9 6 8",
"output": "1\n1 3"
},
{
"input": "20\n20 10 16 14 9 20 6 20 14 19 17 13 16 13 14 8 8 8 8 19",
"output": "10\n20 1\n20 2\n20 3\n20 4\n20 5\n20 7\n20 11\n20 12\n20 15\n20 18"
},
{
"input": "100\n13 71 16 92 25 53 97 63 70 83 51 16 51 84 5 10 54 89 18 95 48 29 82 27 84 68 7 4 65 99 95 37 26 24 24 39 3 28 74 7 75 32 27 24 73 48 72 15 46 66 91 94 19 44 77 23 94 88 51 84 72 95 75 55 80 47 58 13 87 88 25 2 89 81 71 36 7 42 16 59 32 43 58 61 44 96 36 48 88 49 53 91 13 1 37 87 90 47 61 87",
"output": "36\n25 6\n25 8\n25 9\n25 11\n25 12\n25 14\n25 17\n25 20\n25 21\n25 22\n25 30\n25 31\n25 33\n25 34\n25 35\n25 38\n25 40\n25 41\n25 45\n25 50\n25 52\n25 56\n25 57\n25 60\n25 62\n25 64\n25 67\n25 69\n25 76\n25 78\n25 79\n25 85\n25 86\n25 93\n25 98\n25 100"
},
{
"input": "7\n3 1 2 5 6 7 4",
"output": "2\n1 4\n4 1"
}
] | 1,605,625,473
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 10
| 372
| 3,174,400
|
n = int(input())
g = [int(i) - 1 for i in input().split()]
in_cero = [True]*n
for v in g:
in_cero[v] = False
in_cero = [k for k in range(n) if in_cero[k]]
in_cero_count = len(in_cero)
if not in_cero_count:
print(0)
else:
ids = [-1]*n
ID = 0
cycle_id = [-1]*in_cero_count
for v in in_cero:
current = v
cycle_found = False
while ids[current] == -1:
ids[current] = ID
current = g[current]
if ids[current] == ID:
cycle_found = True
cycle_id[ID] = current
if not cycle_found:
cycle_id[ID] = cycle_id[ids[current]]
ID += 1
ans = []
for i in range(in_cero_count - 1):
ans.append('{0} {1}'.format(cycle_id[i] + 1, in_cero[i + 1] + 1))
ans.append('{0} {1}'.format(cycle_id[-1] + 1, in_cero[0] + 1))
print(in_cero_count)
print('\n'.join(ans))
|
Title: Scheme
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
To learn as soon as possible the latest news about their favourite fundamentally new operating system, BolgenOS community from Nizhni Tagil decided to develop a scheme. According to this scheme a community member, who is the first to learn the news, calls some other member, the latter, in his turn, calls some third member, and so on; i.e. a person with index *i* got a person with index *f**i*, to whom he has to call, if he learns the news. With time BolgenOS community members understood that their scheme doesn't work sometimes — there were cases when some members didn't learn the news at all. Now they want to supplement the scheme: they add into the scheme some instructions of type (*x**i*,<=*y**i*), which mean that person *x**i* has to call person *y**i* as well. What is the minimum amount of instructions that they need to add so, that at the end everyone learns the news, no matter who is the first to learn it?
Input Specification:
The first input line contains number *n* (2<=≤<=*n*<=≤<=105) — amount of BolgenOS community members. The second line contains *n* space-separated integer numbers *f**i* (1<=≤<=*f**i*<=≤<=*n*,<=*i*<=≠<=*f**i*) — index of a person, to whom calls a person with index *i*.
Output Specification:
In the first line output one number — the minimum amount of instructions to add. Then output one of the possible variants to add these instructions into the scheme, one instruction in each line. If the solution is not unique, output any.
Demo Input:
['3\n3 3 2\n', '7\n2 3 1 3 4 4 1\n']
Demo Output:
['1\n3 1\n', '3\n2 5\n2 6\n3 7\n']
Note:
none
|
```python
n = int(input())
g = [int(i) - 1 for i in input().split()]
in_cero = [True]*n
for v in g:
in_cero[v] = False
in_cero = [k for k in range(n) if in_cero[k]]
in_cero_count = len(in_cero)
if not in_cero_count:
print(0)
else:
ids = [-1]*n
ID = 0
cycle_id = [-1]*in_cero_count
for v in in_cero:
current = v
cycle_found = False
while ids[current] == -1:
ids[current] = ID
current = g[current]
if ids[current] == ID:
cycle_found = True
cycle_id[ID] = current
if not cycle_found:
cycle_id[ID] = cycle_id[ids[current]]
ID += 1
ans = []
for i in range(in_cero_count - 1):
ans.append('{0} {1}'.format(cycle_id[i] + 1, in_cero[i + 1] + 1))
ans.append('{0} {1}'.format(cycle_id[-1] + 1, in_cero[0] + 1))
print(in_cero_count)
print('\n'.join(ans))
```
| 0
|
102
|
B
|
Sum of Digits
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
B. Sum of Digits
|
2
|
265
|
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
|
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
|
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
|
[
"0\n",
"10\n",
"991\n"
] |
[
"0\n",
"1\n",
"3\n"
] |
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
| 1,000
|
[
{
"input": "0",
"output": "0"
},
{
"input": "10",
"output": "1"
},
{
"input": "991",
"output": "3"
},
{
"input": "99",
"output": "2"
},
{
"input": "100",
"output": "1"
},
{
"input": "123456789",
"output": "2"
},
{
"input": "32",
"output": "1"
},
{
"input": "86",
"output": "2"
},
{
"input": "2",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "34",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "28",
"output": "2"
},
{
"input": "23",
"output": "1"
},
{
"input": "57",
"output": "2"
},
{
"input": "29",
"output": "2"
},
{
"input": "353869285",
"output": "3"
},
{
"input": "549548646",
"output": "2"
},
{
"input": "858893424",
"output": "2"
},
{
"input": "644818852",
"output": "3"
},
{
"input": "360322525",
"output": "3"
},
{
"input": "122937520",
"output": "2"
},
{
"input": "288403032",
"output": "2"
},
{
"input": "677257481",
"output": "3"
},
{
"input": "58059561",
"output": "3"
},
{
"input": "211288847",
"output": "2"
},
{
"input": "339900034079539584",
"output": "2"
},
{
"input": "784084029310269952",
"output": "3"
},
{
"input": "774730557264864000",
"output": "3"
},
{
"input": "478233807148352256",
"output": "3"
},
{
"input": "165646874056812544",
"output": "3"
},
{
"input": "477533739511673792",
"output": "3"
},
{
"input": "660119944377002240",
"output": "3"
},
{
"input": "448375218506078080",
"output": "3"
},
{
"input": "77213334785795616",
"output": "3"
},
{
"input": "165767221702271872",
"output": "3"
},
{
"input": "1",
"output": "0"
},
{
"input": "5",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "156161456146164104103460452045416165146141414651641564105461465156445621465146514562146145465165145145614561465145614561",
"output": "3"
},
{
"input": "9",
"output": "0"
}
] | 1,653,736,810
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 51
| 124
| 0
|
o = input()
o2 = 0
count = 0
while len(o) != 1:
for i in o:
o2 += int(i)
o = str(o2)
count += 1
o2 = 0
print(count)
|
Title: Sum of Digits
Time Limit: 2 seconds
Memory Limit: 265 megabytes
Problem Description:
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
Input Specification:
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
Output Specification:
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
Demo Input:
['0\n', '10\n', '991\n']
Demo Output:
['0\n', '1\n', '3\n']
Note:
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
|
```python
o = input()
o2 = 0
count = 0
while len(o) != 1:
for i in o:
o2 += int(i)
o = str(o2)
count += 1
o2 = 0
print(count)
```
| 3.969
|
90
|
A
|
Cableway
|
PROGRAMMING
| 1,000
|
[
"greedy",
"math"
] |
A. Cableway
|
2
|
256
|
A group of university students wants to get to the top of a mountain to have a picnic there. For that they decided to use a cableway.
A cableway is represented by some cablecars, hanged onto some cable stations by a cable. A cable is scrolled cyclically between the first and the last cable stations (the first of them is located at the bottom of the mountain and the last one is located at the top). As the cable moves, the cablecar attached to it move as well.
The number of cablecars is divisible by three and they are painted three colors: red, green and blue, in such manner that after each red cablecar goes a green one, after each green cablecar goes a blue one and after each blue cablecar goes a red one. Each cablecar can transport no more than two people, the cablecars arrive with the periodicity of one minute (i. e. every minute) and it takes exactly 30 minutes for a cablecar to get to the top.
All students are divided into three groups: *r* of them like to ascend only in the red cablecars, *g* of them prefer only the green ones and *b* of them prefer only the blue ones. A student never gets on a cablecar painted a color that he doesn't like,
The first cablecar to arrive (at the moment of time 0) is painted red. Determine the least time it will take all students to ascend to the mountain top.
|
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=100). It is guaranteed that *r*<=+<=*g*<=+<=*b*<=><=0, it means that the group consists of at least one student.
|
Print a single number — the minimal time the students need for the whole group to ascend to the top of the mountain.
|
[
"1 3 2\n",
"3 2 1\n"
] |
[
"34",
"33"
] |
Let's analyze the first sample.
At the moment of time 0 a red cablecar comes and one student from the *r* group get on it and ascends to the top at the moment of time 30.
At the moment of time 1 a green cablecar arrives and two students from the *g* group get on it; they get to the top at the moment of time 31.
At the moment of time 2 comes the blue cablecar and two students from the *b* group get on it. They ascend to the top at the moment of time 32.
At the moment of time 3 a red cablecar arrives but the only student who is left doesn't like red and the cablecar leaves empty.
At the moment of time 4 a green cablecar arrives and one student from the *g* group gets on it. He ascends to top at the moment of time 34.
Thus, all the students are on the top, overall the ascension took exactly 34 minutes.
| 500
|
[
{
"input": "1 3 2",
"output": "34"
},
{
"input": "3 2 1",
"output": "33"
},
{
"input": "3 5 2",
"output": "37"
},
{
"input": "10 10 10",
"output": "44"
},
{
"input": "29 7 24",
"output": "72"
},
{
"input": "28 94 13",
"output": "169"
},
{
"input": "90 89 73",
"output": "163"
},
{
"input": "0 0 1",
"output": "32"
},
{
"input": "0 0 2",
"output": "32"
},
{
"input": "0 1 0",
"output": "31"
},
{
"input": "0 1 1",
"output": "32"
},
{
"input": "0 1 2",
"output": "32"
},
{
"input": "0 2 0",
"output": "31"
},
{
"input": "0 2 1",
"output": "32"
},
{
"input": "0 2 2",
"output": "32"
},
{
"input": "1 0 0",
"output": "30"
},
{
"input": "1 0 1",
"output": "32"
},
{
"input": "1 0 2",
"output": "32"
},
{
"input": "1 1 0",
"output": "31"
},
{
"input": "1 1 1",
"output": "32"
},
{
"input": "1 1 2",
"output": "32"
},
{
"input": "1 2 0",
"output": "31"
},
{
"input": "1 2 1",
"output": "32"
},
{
"input": "1 2 2",
"output": "32"
},
{
"input": "2 0 0",
"output": "30"
},
{
"input": "2 0 1",
"output": "32"
},
{
"input": "2 0 2",
"output": "32"
},
{
"input": "2 1 0",
"output": "31"
},
{
"input": "2 1 1",
"output": "32"
},
{
"input": "2 1 2",
"output": "32"
},
{
"input": "2 2 0",
"output": "31"
},
{
"input": "2 2 1",
"output": "32"
},
{
"input": "2 2 2",
"output": "32"
},
{
"input": "4 5 2",
"output": "37"
},
{
"input": "5 7 8",
"output": "41"
},
{
"input": "13 25 19",
"output": "67"
},
{
"input": "29 28 30",
"output": "74"
},
{
"input": "45 52 48",
"output": "106"
},
{
"input": "68 72 58",
"output": "136"
},
{
"input": "89 92 90",
"output": "166"
},
{
"input": "99 97 98",
"output": "177"
},
{
"input": "89 97 2",
"output": "175"
},
{
"input": "96 3 92",
"output": "171"
},
{
"input": "1 99 87",
"output": "178"
},
{
"input": "95 2 3",
"output": "171"
},
{
"input": "2 97 3",
"output": "175"
},
{
"input": "2 2 99",
"output": "179"
},
{
"input": "100 100 100",
"output": "179"
},
{
"input": "100 0 100",
"output": "179"
},
{
"input": "0 100 100",
"output": "179"
},
{
"input": "100 100 0",
"output": "178"
},
{
"input": "100 0 0",
"output": "177"
},
{
"input": "0 100 0",
"output": "178"
},
{
"input": "0 0 100",
"output": "179"
},
{
"input": "5 4 5",
"output": "38"
}
] | 1,611,816,336
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 55
| 248
| 0
|
import math
t=0
r,g,b=map(int,input().split())
while r+b+g!=0:
if t%3==0:
r=max(0,r-2)
elif t%3==1:
g=max(0,g-2)
else:
b=max(0,b-2)
t+=1
print(t+29)
|
Title: Cableway
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A group of university students wants to get to the top of a mountain to have a picnic there. For that they decided to use a cableway.
A cableway is represented by some cablecars, hanged onto some cable stations by a cable. A cable is scrolled cyclically between the first and the last cable stations (the first of them is located at the bottom of the mountain and the last one is located at the top). As the cable moves, the cablecar attached to it move as well.
The number of cablecars is divisible by three and they are painted three colors: red, green and blue, in such manner that after each red cablecar goes a green one, after each green cablecar goes a blue one and after each blue cablecar goes a red one. Each cablecar can transport no more than two people, the cablecars arrive with the periodicity of one minute (i. e. every minute) and it takes exactly 30 minutes for a cablecar to get to the top.
All students are divided into three groups: *r* of them like to ascend only in the red cablecars, *g* of them prefer only the green ones and *b* of them prefer only the blue ones. A student never gets on a cablecar painted a color that he doesn't like,
The first cablecar to arrive (at the moment of time 0) is painted red. Determine the least time it will take all students to ascend to the mountain top.
Input Specification:
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=100). It is guaranteed that *r*<=+<=*g*<=+<=*b*<=><=0, it means that the group consists of at least one student.
Output Specification:
Print a single number — the minimal time the students need for the whole group to ascend to the top of the mountain.
Demo Input:
['1 3 2\n', '3 2 1\n']
Demo Output:
['34', '33']
Note:
Let's analyze the first sample.
At the moment of time 0 a red cablecar comes and one student from the *r* group get on it and ascends to the top at the moment of time 30.
At the moment of time 1 a green cablecar arrives and two students from the *g* group get on it; they get to the top at the moment of time 31.
At the moment of time 2 comes the blue cablecar and two students from the *b* group get on it. They ascend to the top at the moment of time 32.
At the moment of time 3 a red cablecar arrives but the only student who is left doesn't like red and the cablecar leaves empty.
At the moment of time 4 a green cablecar arrives and one student from the *g* group gets on it. He ascends to top at the moment of time 34.
Thus, all the students are on the top, overall the ascension took exactly 34 minutes.
|
```python
import math
t=0
r,g,b=map(int,input().split())
while r+b+g!=0:
if t%3==0:
r=max(0,r-2)
elif t%3==1:
g=max(0,g-2)
else:
b=max(0,b-2)
t+=1
print(t+29)
```
| 3.938
|
47
|
B
|
Coins
|
PROGRAMMING
| 1,200
|
[
"implementation"
] |
B. Coins
|
2
|
256
|
One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal.
|
The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B.
|
It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights.
|
[
"A>B\nC<B\nA>C\n",
"A<B\nB>C\nC>A\n"
] |
[
"CBA",
"ACB"
] |
none
| 1,000
|
[
{
"input": "A>B\nC<B\nA>C",
"output": "CBA"
},
{
"input": "A<B\nB>C\nC>A",
"output": "ACB"
},
{
"input": "A<C\nB<A\nB>C",
"output": "Impossible"
},
{
"input": "A<B\nA<C\nB>C",
"output": "ACB"
},
{
"input": "B>A\nC<B\nC>A",
"output": "ACB"
},
{
"input": "A>B\nB>C\nC<A",
"output": "CBA"
},
{
"input": "A>C\nA>B\nB<C",
"output": "BCA"
},
{
"input": "C<B\nB>A\nA<C",
"output": "ACB"
},
{
"input": "C<B\nA>B\nC<A",
"output": "CBA"
},
{
"input": "C>B\nB>A\nA<C",
"output": "ABC"
},
{
"input": "C<B\nB<A\nC>A",
"output": "Impossible"
},
{
"input": "B<C\nC<A\nA>B",
"output": "BCA"
},
{
"input": "A>B\nC<B\nC<A",
"output": "CBA"
},
{
"input": "B>A\nC>B\nA>C",
"output": "Impossible"
},
{
"input": "B<A\nC>B\nC>A",
"output": "BAC"
},
{
"input": "A<B\nC>B\nA<C",
"output": "ABC"
},
{
"input": "A<B\nC<A\nB<C",
"output": "Impossible"
},
{
"input": "A>C\nC<B\nB>A",
"output": "CAB"
},
{
"input": "C>A\nA<B\nB>C",
"output": "ACB"
},
{
"input": "C>A\nC<B\nB>A",
"output": "ACB"
},
{
"input": "B>C\nB>A\nA<C",
"output": "ACB"
},
{
"input": "C<B\nC<A\nB<A",
"output": "CBA"
},
{
"input": "A<C\nA<B\nB>C",
"output": "ACB"
},
{
"input": "B>A\nA>C\nB>C",
"output": "CAB"
},
{
"input": "B<A\nA<C\nC<B",
"output": "Impossible"
},
{
"input": "A<C\nB>C\nA>B",
"output": "Impossible"
},
{
"input": "B>A\nC<A\nC>B",
"output": "Impossible"
},
{
"input": "A>C\nC>B\nB<A",
"output": "BCA"
},
{
"input": "B<C\nB<A\nA>C",
"output": "BCA"
},
{
"input": "A>B\nC>B\nA<C",
"output": "BAC"
},
{
"input": "C<B\nC<A\nB<A",
"output": "CBA"
},
{
"input": "A<C\nA>B\nB>C",
"output": "Impossible"
},
{
"input": "B>A\nB>C\nA<C",
"output": "ACB"
},
{
"input": "B>C\nC<A\nB<A",
"output": "CBA"
},
{
"input": "C>A\nB>A\nB>C",
"output": "ACB"
},
{
"input": "B<A\nB>C\nA<C",
"output": "Impossible"
},
{
"input": "B<C\nA<B\nC>A",
"output": "ABC"
},
{
"input": "C<B\nB>A\nA>C",
"output": "CAB"
},
{
"input": "A>B\nC>A\nB<C",
"output": "BAC"
},
{
"input": "A>B\nB>C\nC<A",
"output": "CBA"
},
{
"input": "B>C\nC>A\nA>B",
"output": "Impossible"
},
{
"input": "B<A\nB>C\nC<A",
"output": "CBA"
},
{
"input": "C>B\nB>A\nC>A",
"output": "ABC"
},
{
"input": "A>C\nA>B\nB>C",
"output": "CBA"
},
{
"input": "B<C\nA>B\nA<C",
"output": "BAC"
},
{
"input": "C>A\nC<B\nB>A",
"output": "ACB"
},
{
"input": "C>B\nA<B\nC<A",
"output": "Impossible"
},
{
"input": "A<C\nC<B\nA>B",
"output": "Impossible"
},
{
"input": "B>C\nA<B\nA<C",
"output": "ACB"
},
{
"input": "C>A\nA<B\nC>B",
"output": "ABC"
}
] | 1,610,538,064
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 216
| 2,048,000
|
ab = input()
cb = input()
ac = input()
A = 0
B = 0
C = 0
if (ab[1] == ">"):
if (ab[0] == "A"):
A += 1
else:
B += 1
else:
if (ab[0] == "A"):
B += 1
else:
A += 1
if (cb[1] == ">"):
if (cb[0] == "C"):
C += 1
else:
B += 1
else:
if (cb[0] == "C"):
B += 1
else:
C += 1
if (ac[1] == ">"):
if (ac[0] == "C"):
C += 1
else:
A += 1
else:
if (ac[0] == "C"):
A += 1
else:
C += 1
d = dict()
d[A] = "A"
d[B] = "B"
d[C] = "C"
if(max(A,B,C)>2):
print("Impossible")
else:
print(f"{d[0]}{d[1]}{d[2]}")
|
Title: Coins
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal.
Input Specification:
The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B.
Output Specification:
It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights.
Demo Input:
['A>B\nC<B\nA>C\n', 'A<B\nB>C\nC>A\n']
Demo Output:
['CBA', 'ACB']
Note:
none
|
```python
ab = input()
cb = input()
ac = input()
A = 0
B = 0
C = 0
if (ab[1] == ">"):
if (ab[0] == "A"):
A += 1
else:
B += 1
else:
if (ab[0] == "A"):
B += 1
else:
A += 1
if (cb[1] == ">"):
if (cb[0] == "C"):
C += 1
else:
B += 1
else:
if (cb[0] == "C"):
B += 1
else:
C += 1
if (ac[1] == ">"):
if (ac[0] == "C"):
C += 1
else:
A += 1
else:
if (ac[0] == "C"):
A += 1
else:
C += 1
d = dict()
d[A] = "A"
d[B] = "B"
d[C] = "C"
if(max(A,B,C)>2):
print("Impossible")
else:
print(f"{d[0]}{d[1]}{d[2]}")
```
| -1
|
277
|
A
|
Learning Languages
|
PROGRAMMING
| 1,400
|
[
"dfs and similar",
"dsu"
] | null | null |
The "BerCorp" company has got *n* employees. These employees can use *m* approved official languages for the formal correspondence. The languages are numbered with integers from 1 to *m*. For each employee we have the list of languages, which he knows. This list could be empty, i. e. an employee may know no official languages. But the employees are willing to learn any number of official languages, as long as the company pays their lessons. A study course in one language for one employee costs 1 berdollar.
Find the minimum sum of money the company needs to spend so as any employee could correspond to any other one (their correspondence can be indirect, i. e. other employees can help out translating).
|
The first line contains two integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of employees and the number of languages.
Then *n* lines follow — each employee's language list. At the beginning of the *i*-th line is integer *k**i* (0<=≤<=*k**i*<=≤<=*m*) — the number of languages the *i*-th employee knows. Next, the *i*-th line contains *k**i* integers — *a**ij* (1<=≤<=*a**ij*<=≤<=*m*) — the identifiers of languages the *i*-th employee knows. It is guaranteed that all the identifiers in one list are distinct. Note that an employee may know zero languages.
The numbers in the lines are separated by single spaces.
|
Print a single integer — the minimum amount of money to pay so that in the end every employee could write a letter to every other one (other employees can help out translating).
|
[
"5 5\n1 2\n2 2 3\n2 3 4\n2 4 5\n1 5\n",
"8 7\n0\n3 1 2 3\n1 1\n2 5 4\n2 6 7\n1 3\n2 7 4\n1 1\n",
"2 2\n1 2\n0\n"
] |
[
"0\n",
"2\n",
"1\n"
] |
In the second sample the employee 1 can learn language 2, and employee 8 can learn language 4.
In the third sample employee 2 must learn language 2.
| 500
|
[
{
"input": "5 5\n1 2\n2 2 3\n2 3 4\n2 4 5\n1 5",
"output": "0"
},
{
"input": "8 7\n0\n3 1 2 3\n1 1\n2 5 4\n2 6 7\n1 3\n2 7 4\n1 1",
"output": "2"
},
{
"input": "2 2\n1 2\n0",
"output": "1"
},
{
"input": "2 2\n0\n0",
"output": "2"
},
{
"input": "5 5\n1 3\n0\n0\n2 4 1\n0",
"output": "4"
},
{
"input": "6 2\n0\n0\n2 1 2\n1 1\n1 1\n0",
"output": "3"
},
{
"input": "7 3\n3 1 3 2\n3 2 1 3\n2 2 3\n1 1\n2 2 3\n3 3 2 1\n3 2 3 1",
"output": "0"
},
{
"input": "8 4\n0\n0\n4 2 3 1 4\n4 2 1 4 3\n3 4 3 1\n1 2\n2 4 1\n2 4 2",
"output": "2"
},
{
"input": "10 10\n5 7 5 2 8 1\n7 10 6 9 5 8 2 4\n2 2 7\n5 8 6 9 10 1\n2 9 5\n3 6 5 2\n6 5 8 7 9 10 4\n0\n1 1\n2 8 6",
"output": "1"
},
{
"input": "11 42\n4 20 26 9 24\n14 34 7 28 32 12 15 26 4 10 38 21 20 8 11\n4 21 8 36 6\n11 32 1 39 11 21 10 25 17 26 15 4\n2 8 12\n2 21 31\n8 17 10 3 39 32 30 5 15\n20 24 20 38 17 4 7 21 19 32 28 31 22 30 37 10 5 33 2 13 9\n7 38 34 42 27 20 11 6\n3 40 3 39\n14 39 40 4 30 33 8 36 28 14 23 16 7 25 9",
"output": "0"
},
{
"input": "100 100\n1 33\n0\n2 35 4\n2 40 78\n1 69\n0\n1 2\n0\n0\n2 81 34\n1 45\n0\n1 87\n1 50\n1 97\n0\n2 100 59\n0\n0\n0\n1 45\n0\n0\n0\n1 6\n1 54\n0\n0\n0\n4 79 96 52 84\n1 86\n0\n0\n0\n0\n0\n0\n0\n0\n3 90 2 80\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 2\n0\n1 64\n0\n1 78\n1 82\n0\n0\n0\n0\n0\n1 6\n0\n0\n2 47 57\n1 95\n0\n2 91 79\n0\n1 27\n0\n1 74\n0\n0\n1 14\n0\n0\n2 90 19\n0\n1 10\n0\n0\n0\n0\n0\n1 57\n0\n2 28 50\n0\n0\n0\n1 47\n0\n0\n1 14\n0\n1 84\n1 1\n0\n0",
"output": "87"
},
{
"input": "2 2\n2 1 2\n2 1 2",
"output": "0"
},
{
"input": "2 2\n2 1 2\n1 1",
"output": "0"
},
{
"input": "2 2\n1 2\n1 1",
"output": "1"
},
{
"input": "3 100\n0\n0\n0",
"output": "3"
},
{
"input": "100 100\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "100"
},
{
"input": "3 3\n0\n0\n0",
"output": "3"
}
] | 1,650,498,597
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 62
| 0
|
#n = 8
#m = 7
n , m = map(int,input().split())
#adj = [set(),set([1,2,3]),set([1]),set([4,5]),set([6,7]),set([3]),set([7,4]),set([1])]
#adj = [set([1,2]),set()]
adj = []
for i in range(n):
inp = list(map(int,input().split()))
adj.append( set(inp[1:]) )
ids = [i for i in range(n)]
size = [1]*n
sets = []
def find(n1):
root = n1
while ids[root] != root:
root = ids[root]
return root
def union(n1,n2):
r1 = find(n1)
r2 = find(n2)
if size[r1] > size[r2]:
ids[r2] = ids[r1]
size[r1] += size[r2]
else:
ids[r1] = ids[r2]
size[r2] += size[r1]
for i in range(n):
langs = adj[i]
for j in range(i):
if not langs.isdisjoint(adj[j]):
union(i,j)
s = set()
for i in ids:
s.add(i)
print(len(s)-1)
|
Title: Learning Languages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The "BerCorp" company has got *n* employees. These employees can use *m* approved official languages for the formal correspondence. The languages are numbered with integers from 1 to *m*. For each employee we have the list of languages, which he knows. This list could be empty, i. e. an employee may know no official languages. But the employees are willing to learn any number of official languages, as long as the company pays their lessons. A study course in one language for one employee costs 1 berdollar.
Find the minimum sum of money the company needs to spend so as any employee could correspond to any other one (their correspondence can be indirect, i. e. other employees can help out translating).
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of employees and the number of languages.
Then *n* lines follow — each employee's language list. At the beginning of the *i*-th line is integer *k**i* (0<=≤<=*k**i*<=≤<=*m*) — the number of languages the *i*-th employee knows. Next, the *i*-th line contains *k**i* integers — *a**ij* (1<=≤<=*a**ij*<=≤<=*m*) — the identifiers of languages the *i*-th employee knows. It is guaranteed that all the identifiers in one list are distinct. Note that an employee may know zero languages.
The numbers in the lines are separated by single spaces.
Output Specification:
Print a single integer — the minimum amount of money to pay so that in the end every employee could write a letter to every other one (other employees can help out translating).
Demo Input:
['5 5\n1 2\n2 2 3\n2 3 4\n2 4 5\n1 5\n', '8 7\n0\n3 1 2 3\n1 1\n2 5 4\n2 6 7\n1 3\n2 7 4\n1 1\n', '2 2\n1 2\n0\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the second sample the employee 1 can learn language 2, and employee 8 can learn language 4.
In the third sample employee 2 must learn language 2.
|
```python
#n = 8
#m = 7
n , m = map(int,input().split())
#adj = [set(),set([1,2,3]),set([1]),set([4,5]),set([6,7]),set([3]),set([7,4]),set([1])]
#adj = [set([1,2]),set()]
adj = []
for i in range(n):
inp = list(map(int,input().split()))
adj.append( set(inp[1:]) )
ids = [i for i in range(n)]
size = [1]*n
sets = []
def find(n1):
root = n1
while ids[root] != root:
root = ids[root]
return root
def union(n1,n2):
r1 = find(n1)
r2 = find(n2)
if size[r1] > size[r2]:
ids[r2] = ids[r1]
size[r1] += size[r2]
else:
ids[r1] = ids[r2]
size[r2] += size[r1]
for i in range(n):
langs = adj[i]
for j in range(i):
if not langs.isdisjoint(adj[j]):
union(i,j)
s = set()
for i in ids:
s.add(i)
print(len(s)-1)
```
| 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"
}
] | 1,695,356,013
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 46
| 0
|
word = input().strip()
# Initialize variables to keep track of the expected letter and its index
expected_letter = 'h'
index = 0
# Iterate through the characters in the input word
for letter in word:
if letter == expected_letter:
index += 1
if index == 5:
break
expected_letter = 'e' if expected_letter == 'h' else 'l' if expected_letter == 'e' else 'o'
# Check if all the letters in "hello" were found in the correct order
if index == 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
word = input().strip()
# Initialize variables to keep track of the expected letter and its index
expected_letter = 'h'
index = 0
# Iterate through the characters in the input word
for letter in word:
if letter == expected_letter:
index += 1
if index == 5:
break
expected_letter = 'e' if expected_letter == 'h' else 'l' if expected_letter == 'e' else 'o'
# Check if all the letters in "hello" were found in the correct order
if index == 5:
print("YES")
else:
print("NO")
```
| 0
|
181
|
A
|
Series of Crimes
|
PROGRAMMING
| 800
|
[
"brute force",
"geometry",
"implementation"
] | null | null |
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
|
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
|
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
|
[
"3 2\n.*\n..\n**\n",
"3 3\n*.*\n*..\n...\n"
] |
[
"1 1\n",
"2 3\n"
] |
none
| 500
|
[
{
"input": "3 2\n.*\n..\n**",
"output": "1 1"
},
{
"input": "2 5\n*....\n*...*",
"output": "1 5"
},
{
"input": "7 2\n..\n**\n..\n..\n..\n..\n.*",
"output": "7 1"
},
{
"input": "7 2\n*.\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "2 10\n*......*..\n.......*..",
"output": "2 1"
},
{
"input": "10 3\n*..\n...\n...\n...\n...\n...\n...\n...\n...\n**.",
"output": "1 2"
},
{
"input": "100 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..\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": "1 2"
},
{
"input": "99 3\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...\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": "99 2"
},
{
"input": "98 3\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...\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": "2 3"
},
{
"input": "99 3\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...\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": "99 2"
},
{
"input": "98 3\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...\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": "2 3"
},
{
"input": "2 100\n...................................................................................................*\n*..................................................................................................*",
"output": "1 1"
},
{
"input": "2 2\n*.\n**",
"output": "1 2"
},
{
"input": "2 2\n**\n.*",
"output": "2 1"
},
{
"input": "2 3\n*.*\n..*",
"output": "2 1"
}
] | 1,554,400,906
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 218
| 0
|
x,y = map(int,input().split())
c = 0
z = []
t = 0
m = 0
m1 = 0
m3= 0
for i in range(x):
z.append(input())
for j in range(x):
c = 0
for k in range(y):
if z[j].count("*") == 2:
if z[j][k] == "*":
c = c + 1
# print(c)
if c == 1:
m = k
# print("a",m)
if c == 2:
m1 = k
# print(m1)
if z[j].count("*") == 1:
a1 = j+1
if z[j][k] == "*":
m3 = k
# print(m1)
# print(m)
# print(m3)
if m3 == m1:
# print(m)
t = m
elif m3 == m:
# print(m)
t = m1
# print(t)
print(a1,t+1)
|
Title: Series of Crimes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
Output Specification:
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
Demo Input:
['3 2\n.*\n..\n**\n', '3 3\n*.*\n*..\n...\n']
Demo Output:
['1 1\n', '2 3\n']
Note:
none
|
```python
x,y = map(int,input().split())
c = 0
z = []
t = 0
m = 0
m1 = 0
m3= 0
for i in range(x):
z.append(input())
for j in range(x):
c = 0
for k in range(y):
if z[j].count("*") == 2:
if z[j][k] == "*":
c = c + 1
# print(c)
if c == 1:
m = k
# print("a",m)
if c == 2:
m1 = k
# print(m1)
if z[j].count("*") == 1:
a1 = j+1
if z[j][k] == "*":
m3 = k
# print(m1)
# print(m)
# print(m3)
if m3 == m1:
# print(m)
t = m
elif m3 == m:
# print(m)
t = m1
# print(t)
print(a1,t+1)
```
| 3
|
|
724
|
A
|
Checking the Calendar
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
You are given names of two days of the week.
Please, determine whether it is possible that during some non-leap year the first day of some month was equal to the first day of the week you are given, while the first day of the next month was equal to the second day of the week you are given. Both months should belong to one year.
In this problem, we consider the Gregorian calendar to be used. The number of months in this calendar is equal to 12. The number of days in months during any non-leap year is: 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31.
Names of the days of the week are given with lowercase English letters: "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".
|
The input consists of two lines, each of them containing the name of exactly one day of the week. It's guaranteed that each string in the input is from the set "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".
|
Print "YES" (without quotes) if such situation is possible during some non-leap year. Otherwise, print "NO" (without quotes).
|
[
"monday\ntuesday\n",
"sunday\nsunday\n",
"saturday\ntuesday\n"
] |
[
"NO\n",
"YES\n",
"YES\n"
] |
In the second sample, one can consider February 1 and March 1 of year 2015. Both these days were Sundays.
In the third sample, one can consider July 1 and August 1 of year 2017. First of these two days is Saturday, while the second one is Tuesday.
| 500
|
[
{
"input": "monday\ntuesday",
"output": "NO"
},
{
"input": "sunday\nsunday",
"output": "YES"
},
{
"input": "saturday\ntuesday",
"output": "YES"
},
{
"input": "tuesday\nthursday",
"output": "YES"
},
{
"input": "friday\nwednesday",
"output": "NO"
},
{
"input": "sunday\nsaturday",
"output": "NO"
},
{
"input": "monday\nmonday",
"output": "YES"
},
{
"input": "monday\nwednesday",
"output": "YES"
},
{
"input": "monday\nthursday",
"output": "YES"
},
{
"input": "monday\nfriday",
"output": "NO"
},
{
"input": "monday\nsaturday",
"output": "NO"
},
{
"input": "monday\nsunday",
"output": "NO"
},
{
"input": "tuesday\nmonday",
"output": "NO"
},
{
"input": "tuesday\ntuesday",
"output": "YES"
},
{
"input": "tuesday\nwednesday",
"output": "NO"
},
{
"input": "tuesday\nfriday",
"output": "YES"
},
{
"input": "tuesday\nsaturday",
"output": "NO"
},
{
"input": "tuesday\nsunday",
"output": "NO"
},
{
"input": "wednesday\nmonday",
"output": "NO"
},
{
"input": "wednesday\ntuesday",
"output": "NO"
},
{
"input": "wednesday\nwednesday",
"output": "YES"
},
{
"input": "wednesday\nthursday",
"output": "NO"
},
{
"input": "wednesday\nfriday",
"output": "YES"
},
{
"input": "wednesday\nsaturday",
"output": "YES"
},
{
"input": "wednesday\nsunday",
"output": "NO"
},
{
"input": "thursday\nmonday",
"output": "NO"
},
{
"input": "thursday\ntuesday",
"output": "NO"
},
{
"input": "thursday\nwednesday",
"output": "NO"
},
{
"input": "thursday\nthursday",
"output": "YES"
},
{
"input": "thursday\nfriday",
"output": "NO"
},
{
"input": "thursday\nsaturday",
"output": "YES"
},
{
"input": "thursday\nsunday",
"output": "YES"
},
{
"input": "friday\nmonday",
"output": "YES"
},
{
"input": "friday\ntuesday",
"output": "NO"
},
{
"input": "friday\nthursday",
"output": "NO"
},
{
"input": "friday\nsaturday",
"output": "NO"
},
{
"input": "friday\nsunday",
"output": "YES"
},
{
"input": "saturday\nmonday",
"output": "YES"
},
{
"input": "saturday\nwednesday",
"output": "NO"
},
{
"input": "saturday\nthursday",
"output": "NO"
},
{
"input": "saturday\nfriday",
"output": "NO"
},
{
"input": "saturday\nsaturday",
"output": "YES"
},
{
"input": "saturday\nsunday",
"output": "NO"
},
{
"input": "sunday\nmonday",
"output": "NO"
},
{
"input": "sunday\ntuesday",
"output": "YES"
},
{
"input": "sunday\nwednesday",
"output": "YES"
},
{
"input": "sunday\nthursday",
"output": "NO"
},
{
"input": "sunday\nfriday",
"output": "NO"
},
{
"input": "friday\nfriday",
"output": "YES"
},
{
"input": "friday\nsunday",
"output": "YES"
},
{
"input": "monday\nmonday",
"output": "YES"
},
{
"input": "friday\ntuesday",
"output": "NO"
},
{
"input": "thursday\nsaturday",
"output": "YES"
},
{
"input": "tuesday\nfriday",
"output": "YES"
},
{
"input": "sunday\nwednesday",
"output": "YES"
},
{
"input": "monday\nthursday",
"output": "YES"
},
{
"input": "saturday\nsunday",
"output": "NO"
},
{
"input": "friday\nmonday",
"output": "YES"
},
{
"input": "thursday\nthursday",
"output": "YES"
},
{
"input": "wednesday\nfriday",
"output": "YES"
},
{
"input": "thursday\nmonday",
"output": "NO"
},
{
"input": "wednesday\nsunday",
"output": "NO"
},
{
"input": "thursday\nfriday",
"output": "NO"
},
{
"input": "monday\nfriday",
"output": "NO"
},
{
"input": "wednesday\nsaturday",
"output": "YES"
},
{
"input": "thursday\nsunday",
"output": "YES"
},
{
"input": "saturday\nfriday",
"output": "NO"
},
{
"input": "saturday\nmonday",
"output": "YES"
}
] | 1,475,929,502
| 602
|
Python 3
|
OK
|
TESTS
| 68
| 77
| 4,608,000
|
days = ["monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday"]
months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
day1 = input()
day2 = input()
d1 = days.index(day1)
d2 = days.index(day2)
diff = d2 - d1
if diff < 0:
diff += 7
ans = "NO"
for month in months:
if month % 7 == diff:
ans = "YES"
break
print(ans)
|
Title: Checking the Calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given names of two days of the week.
Please, determine whether it is possible that during some non-leap year the first day of some month was equal to the first day of the week you are given, while the first day of the next month was equal to the second day of the week you are given. Both months should belong to one year.
In this problem, we consider the Gregorian calendar to be used. The number of months in this calendar is equal to 12. The number of days in months during any non-leap year is: 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31.
Names of the days of the week are given with lowercase English letters: "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".
Input Specification:
The input consists of two lines, each of them containing the name of exactly one day of the week. It's guaranteed that each string in the input is from the set "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".
Output Specification:
Print "YES" (without quotes) if such situation is possible during some non-leap year. Otherwise, print "NO" (without quotes).
Demo Input:
['monday\ntuesday\n', 'sunday\nsunday\n', 'saturday\ntuesday\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n']
Note:
In the second sample, one can consider February 1 and March 1 of year 2015. Both these days were Sundays.
In the third sample, one can consider July 1 and August 1 of year 2017. First of these two days is Saturday, while the second one is Tuesday.
|
```python
days = ["monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday"]
months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
day1 = input()
day2 = input()
d1 = days.index(day1)
d2 = days.index(day2)
diff = d2 - d1
if diff < 0:
diff += 7
ans = "NO"
for month in months:
if month % 7 == diff:
ans = "YES"
break
print(ans)
```
| 3
|
|
469
|
A
|
I Wanna Be the Guy
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation"
] | null | null |
There is a game called "I Wanna Be the Guy", consisting of *n* levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only *p* levels of the game. And Little Y can pass only *q* levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
|
The first line contains a single integer *n* (1<=≤<=<=*n*<=≤<=100).
The next line contains an integer *p* (0<=≤<=*p*<=≤<=*n*) at first, then follows *p* distinct integers *a*1,<=*a*2,<=...,<=*a**p* (1<=≤<=*a**i*<=≤<=*n*). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to *n*.
|
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
|
[
"4\n3 1 2 3\n2 2 4\n",
"4\n3 1 2 3\n2 2 3\n"
] |
[
"I become the guy.\n",
"Oh, my keyboard!\n"
] |
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
| 500
|
[
{
"input": "4\n3 1 2 3\n2 2 4",
"output": "I become the guy."
},
{
"input": "4\n3 1 2 3\n2 2 3",
"output": "Oh, my keyboard!"
},
{
"input": "10\n5 8 6 1 5 4\n6 1 3 2 9 4 6",
"output": "Oh, my keyboard!"
},
{
"input": "10\n8 8 10 7 3 1 4 2 6\n8 9 5 10 3 7 2 4 8",
"output": "I become the guy."
},
{
"input": "10\n9 6 1 8 3 9 7 5 10 4\n7 1 3 2 7 6 9 5",
"output": "I become the guy."
},
{
"input": "100\n75 83 69 73 30 76 37 48 14 41 42 21 35 15 50 61 86 85 46 3 31 13 78 10 2 44 80 95 56 82 38 75 77 4 99 9 84 53 12 11 36 74 39 72 43 89 57 28 54 1 51 66 27 22 93 59 68 88 91 29 7 20 63 8 52 23 64 58 100 79 65 49 96 71 33 45\n83 50 89 73 34 28 99 67 77 44 19 60 68 42 8 27 94 85 14 39 17 78 24 21 29 63 92 32 86 22 71 81 31 82 65 48 80 59 98 3 70 55 37 12 15 72 47 9 11 33 16 7 91 74 13 64 38 84 6 61 93 90 45 69 1 54 52 100 57 10 35 49 53 75 76 43 62 5 4 18 36 96 79 23",
"output": "Oh, my keyboard!"
},
{
"input": "1\n1 1\n1 1",
"output": "I become the guy."
},
{
"input": "1\n0\n1 1",
"output": "I become the guy."
},
{
"input": "1\n1 1\n0",
"output": "I become the guy."
},
{
"input": "1\n0\n0",
"output": "Oh, my keyboard!"
},
{
"input": "100\n0\n0",
"output": "Oh, my keyboard!"
},
{
"input": "100\n44 71 70 55 49 43 16 53 7 95 58 56 38 76 67 94 20 73 29 90 25 30 8 84 5 14 77 52 99 91 66 24 39 37 22 44 78 12 63 59 32 51 15 82 34\n56 17 10 96 80 69 13 81 31 57 4 48 68 89 50 45 3 33 36 2 72 100 64 87 21 75 54 74 92 65 23 40 97 61 18 28 98 93 35 83 9 79 46 27 41 62 88 6 47 60 86 26 42 85 19 1 11",
"output": "I become the guy."
},
{
"input": "100\n78 63 59 39 11 58 4 2 80 69 22 95 90 26 65 16 30 100 66 99 67 79 54 12 23 28 45 56 70 74 60 82 73 91 68 43 92 75 51 21 17 97 86 44 62 47 85 78 72 64 50 81 71 5 57 13 31 76 87 9 49 96 25 42 19 35 88 53 7 83 38 27 29 41 89 93 10 84 18\n78 1 16 53 72 99 9 36 59 49 75 77 94 79 35 4 92 42 82 83 76 97 20 68 55 47 65 50 14 30 13 67 98 8 7 40 64 32 87 10 33 90 93 18 26 71 17 46 24 28 89 58 37 91 39 34 25 48 84 31 96 95 80 88 3 51 62 52 85 61 12 15 27 6 45 38 2 22 60",
"output": "I become the guy."
},
{
"input": "2\n2 2 1\n0",
"output": "I become the guy."
},
{
"input": "2\n1 2\n2 1 2",
"output": "I become the guy."
},
{
"input": "80\n57 40 1 47 36 69 24 76 5 72 26 4 29 62 6 60 3 70 8 64 18 37 16 14 13 21 25 7 66 68 44 74 61 39 38 33 15 63 34 65 10 23 56 51 80 58 49 75 71 12 50 57 2 30 54 27 17 52\n61 22 67 15 28 41 26 1 80 44 3 38 18 37 79 57 11 7 65 34 9 36 40 5 48 29 64 31 51 63 27 4 50 13 24 32 58 23 19 46 8 73 39 2 21 56 77 53 59 78 43 12 55 45 30 74 33 68 42 47 17 54",
"output": "Oh, my keyboard!"
},
{
"input": "100\n78 87 96 18 73 32 38 44 29 64 40 70 47 91 60 69 24 1 5 34 92 94 99 22 83 65 14 68 15 20 74 31 39 100 42 4 97 46 25 6 8 56 79 9 71 35 54 19 59 93 58 62 10 85 57 45 33 7 86 81 30 98 26 61 84 41 23 28 88 36 66 51 80 53 37 63 43 95 75\n76 81 53 15 26 37 31 62 24 87 41 39 75 86 46 76 34 4 51 5 45 65 67 48 68 23 71 27 94 47 16 17 9 96 84 89 88 100 18 52 69 42 6 92 7 64 49 12 98 28 21 99 25 55 44 40 82 19 36 30 77 90 14 43 50 3 13 95 78 35 20 54 58 11 2 1 33",
"output": "Oh, my keyboard!"
},
{
"input": "100\n77 55 26 98 13 91 78 60 23 76 12 11 36 62 84 80 18 1 68 92 81 67 19 4 2 10 17 77 96 63 15 69 46 97 82 42 83 59 50 72 14 40 89 9 52 29 56 31 74 39 45 85 22 99 44 65 95 6 90 38 54 32 49 34 3 70 75 33 94 53 21 71 5 66 73 41 100 24\n69 76 93 5 24 57 59 6 81 4 30 12 44 15 67 45 73 3 16 8 47 95 20 64 68 85 54 17 90 86 66 58 13 37 42 51 35 32 1 28 43 80 7 14 48 19 62 55 2 91 25 49 27 26 38 79 89 99 22 60 75 53 88 82 34 21 87 71 72 61",
"output": "I become the guy."
},
{
"input": "100\n74 96 32 63 12 69 72 99 15 22 1 41 79 77 71 31 20 28 75 73 85 37 38 59 42 100 86 89 55 87 68 4 24 57 52 8 92 27 56 98 95 58 34 9 45 14 11 36 66 76 61 19 25 23 78 49 90 26 80 43 70 13 65 10 5 74 81 21 44 60 97 3 47 93 6\n64 68 21 27 16 91 23 22 33 12 71 88 90 50 62 43 28 29 57 59 5 74 10 95 35 1 67 93 36 32 86 40 6 64 78 46 89 15 84 53 18 30 17 85 2 3 47 92 25 48 76 51 20 82 52 83 99 63 80 11 94 54 39 7 58",
"output": "I become the guy."
},
{
"input": "100\n75 11 98 44 47 88 94 23 78 59 70 2 43 39 34 63 71 19 42 61 30 74 14 77 97 53 92 60 67 36 37 13 6 86 62 46 41 3 25 93 7 12 27 48 55 49 31 35 51 10 57 54 95 82 28 90 73 26 17 50 81 56 20 87 40 85 72 64 99 29 91 5 80 18 24 52\n72 93 59 5 88 47 9 58 48 1 43 50 100 87 61 91 45 98 99 56 25 84 53 73 78 54 63 38 37 2 77 95 89 85 4 90 10 33 12 22 74 32 34 70 71 52 96 57 15 66 31 27 75 8 21 39 62 44 67 94 81 68 14 19 36 28 11 79 16 65 46 83 76",
"output": "Oh, my keyboard!"
},
{
"input": "3\n1 2\n2 2 3",
"output": "Oh, my keyboard!"
},
{
"input": "4\n1 2\n3 1 3 4",
"output": "I become the guy."
},
{
"input": "6\n2 1 2\n3 4 5 6",
"output": "Oh, my keyboard!"
}
] | 1,695,996,082
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 0
|
t=int(input())
l=list(int,input().split())
l1=list(int,input().split())
k=0
for _ in range(1,t+1):
if _ not in l or l1:
k=1
if(k==0): print('I become the guy.')
else: print('Oh, my keyboard!')
|
Title: I Wanna Be the Guy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a game called "I Wanna Be the Guy", consisting of *n* levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only *p* levels of the game. And Little Y can pass only *q* levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input Specification:
The first line contains a single integer *n* (1<=≤<=<=*n*<=≤<=100).
The next line contains an integer *p* (0<=≤<=*p*<=≤<=*n*) at first, then follows *p* distinct integers *a*1,<=*a*2,<=...,<=*a**p* (1<=≤<=*a**i*<=≤<=*n*). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to *n*.
Output Specification:
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Demo Input:
['4\n3 1 2 3\n2 2 4\n', '4\n3 1 2 3\n2 2 3\n']
Demo Output:
['I become the guy.\n', 'Oh, my keyboard!\n']
Note:
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
|
```python
t=int(input())
l=list(int,input().split())
l1=list(int,input().split())
k=0
for _ in range(1,t+1):
if _ not in l or l1:
k=1
if(k==0): print('I become the guy.')
else: print('Oh, my keyboard!')
```
| -1
|
|
387
|
B
|
George and Round
|
PROGRAMMING
| 1,200
|
[
"brute force",
"greedy",
"two pointers"
] | null | null |
George decided to prepare a Codesecrof round, so he has prepared *m* problems for the round. Let's number the problems with integers 1 through *m*. George estimates the *i*-th problem's complexity by integer *b**i*.
To make the round good, he needs to put at least *n* problems there. Besides, he needs to have at least one problem with complexity exactly *a*1, at least one with complexity exactly *a*2, ..., and at least one with complexity exactly *a**n*. Of course, the round can also have problems with other complexities.
George has a poor imagination. It's easier for him to make some already prepared problem simpler than to come up with a new one and prepare it. George is magnificent at simplifying problems. He can simplify any already prepared problem with complexity *c* to any positive integer complexity *d* (*c*<=≥<=*d*), by changing limits on the input data.
However, nothing is so simple. George understood that even if he simplifies some problems, he can run out of problems for a good round. That's why he decided to find out the minimum number of problems he needs to come up with in addition to the *m* he's prepared in order to make a good round. Note that George can come up with a new problem of any complexity.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=3000) — the minimal number of problems in a good round and the number of problems George's prepared. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*<=≤<=106) — the requirements for the complexity of the problems in a good round. The third line contains space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b*1<=≤<=*b*2...<=≤<=*b**m*<=≤<=106) — the complexities of the problems prepared by George.
|
Print a single integer — the answer to the problem.
|
[
"3 5\n1 2 3\n1 2 2 3 3\n",
"3 5\n1 2 3\n1 1 1 1 1\n",
"3 1\n2 3 4\n1\n"
] |
[
"0\n",
"2\n",
"3\n"
] |
In the first sample the set of the prepared problems meets the requirements for a good round.
In the second sample, it is enough to come up with and prepare two problems with complexities 2 and 3 to get a good round.
In the third sample it is very easy to get a good round if come up with and prepare extra problems with complexities: 2, 3, 4.
| 1,000
|
[
{
"input": "3 5\n1 2 3\n1 2 2 3 3",
"output": "0"
},
{
"input": "3 5\n1 2 3\n1 1 1 1 1",
"output": "2"
},
{
"input": "3 1\n2 3 4\n1",
"output": "3"
},
{
"input": "29 100\n20 32 41 67 72 155 331 382 399 412 465 470 484 511 515 529 616 637 679 715 733 763 826 843 862 903 925 979 989\n15 15 15 17 18 19 19 20 21 21 22 24 25 26 26 27 28 31 32 32 37 38 38 39 39 40 41 42 43 43 45 45 46 47 49 49 50 50 50 51 52 53 53 55 56 57 59 59 59 60 60 62 62 63 63 64 64 64 66 67 69 69 70 70 72 72 73 74 75 76 77 78 80 80 81 81 83 83 83 84 86 86 86 86 87 88 89 91 91 91 92 93 94 94 96 97 97 97 98 98",
"output": "24"
}
] | 1,587,627,038
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 6
| 139
| 1,331,200
|
n, m = map(int, input().split())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
j = 0
count = 0
for i in range(n):
if j>=m:
break
if a[i]<=b[j]:
count += 1
else:
break
j += 1
if i != (n-1):
print(n-i)
else:
print("0")
|
Title: George and Round
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George decided to prepare a Codesecrof round, so he has prepared *m* problems for the round. Let's number the problems with integers 1 through *m*. George estimates the *i*-th problem's complexity by integer *b**i*.
To make the round good, he needs to put at least *n* problems there. Besides, he needs to have at least one problem with complexity exactly *a*1, at least one with complexity exactly *a*2, ..., and at least one with complexity exactly *a**n*. Of course, the round can also have problems with other complexities.
George has a poor imagination. It's easier for him to make some already prepared problem simpler than to come up with a new one and prepare it. George is magnificent at simplifying problems. He can simplify any already prepared problem with complexity *c* to any positive integer complexity *d* (*c*<=≥<=*d*), by changing limits on the input data.
However, nothing is so simple. George understood that even if he simplifies some problems, he can run out of problems for a good round. That's why he decided to find out the minimum number of problems he needs to come up with in addition to the *m* he's prepared in order to make a good round. Note that George can come up with a new problem of any complexity.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=3000) — the minimal number of problems in a good round and the number of problems George's prepared. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*<=≤<=106) — the requirements for the complexity of the problems in a good round. The third line contains space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b*1<=≤<=*b*2...<=≤<=*b**m*<=≤<=106) — the complexities of the problems prepared by George.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['3 5\n1 2 3\n1 2 2 3 3\n', '3 5\n1 2 3\n1 1 1 1 1\n', '3 1\n2 3 4\n1\n']
Demo Output:
['0\n', '2\n', '3\n']
Note:
In the first sample the set of the prepared problems meets the requirements for a good round.
In the second sample, it is enough to come up with and prepare two problems with complexities 2 and 3 to get a good round.
In the third sample it is very easy to get a good round if come up with and prepare extra problems with complexities: 2, 3, 4.
|
```python
n, m = map(int, input().split())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
j = 0
count = 0
for i in range(n):
if j>=m:
break
if a[i]<=b[j]:
count += 1
else:
break
j += 1
if i != (n-1):
print(n-i)
else:
print("0")
```
| 0
|
|
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"
}
] | 1,690,549,888
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 31
| 0
|
text = input()
if text.isupper() or text[1:].isupper() :
print(text.capitalize())
else : print(text)
|
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
text = input()
if text.isupper() or text[1:].isupper() :
print(text.capitalize())
else : print(text)
```
| 0
|
|
999
|
F
|
Cards and Joy
|
PROGRAMMING
| 2,000
|
[
"dp"
] | null | null |
There are $n$ players sitting at the card table. Each player has a favorite number. The favorite number of the $j$-th player is $f_j$.
There are $k \cdot n$ cards on the table. Each card contains a single integer: the $i$-th card contains number $c_i$. Also, you are given a sequence $h_1, h_2, \dots, h_k$. Its meaning will be explained below.
The players have to distribute all the cards in such a way that each of them will hold exactly $k$ cards. After all the cards are distributed, each player counts the number of cards he has that contains his favorite number. The joy level of a player equals $h_t$ if the player holds $t$ cards containing his favorite number. If a player gets no cards with his favorite number (i.e., $t=0$), his joy level is $0$.
Print the maximum possible total joy levels of the players after the cards are distributed. Note that the sequence $h_1, \dots, h_k$ is the same for all the players.
|
The first line of input contains two integers $n$ and $k$ ($1 \le n \le 500, 1 \le k \le 10$) — the number of players and the number of cards each player will get.
The second line contains $k \cdot n$ integers $c_1, c_2, \dots, c_{k \cdot n}$ ($1 \le c_i \le 10^5$) — the numbers written on the cards.
The third line contains $n$ integers $f_1, f_2, \dots, f_n$ ($1 \le f_j \le 10^5$) — the favorite numbers of the players.
The fourth line contains $k$ integers $h_1, h_2, \dots, h_k$ ($1 \le h_t \le 10^5$), where $h_t$ is the joy level of a player if he gets exactly $t$ cards with his favorite number written on them. It is guaranteed that the condition $h_{t - 1} < h_t$ holds for each $t \in [2..k]$.
|
Print one integer — the maximum possible total joy levels of the players among all possible card distributions.
|
[
"4 3\n1 3 2 8 5 5 8 2 2 8 5 2\n1 2 2 5\n2 6 7\n",
"3 3\n9 9 9 9 9 9 9 9 9\n1 2 3\n1 2 3\n"
] |
[
"21\n",
"0\n"
] |
In the first example, one possible optimal card distribution is the following:
- Player $1$ gets cards with numbers $[1, 3, 8]$; - Player $2$ gets cards with numbers $[2, 2, 8]$; - Player $3$ gets cards with numbers $[2, 2, 8]$; - Player $4$ gets cards with numbers $[5, 5, 5]$.
Thus, the answer is $2 + 6 + 6 + 7 = 21$.
In the second example, no player can get a card with his favorite number. Thus, the answer is $0$.
| 0
|
[
{
"input": "4 3\n1 3 2 8 5 5 8 2 2 8 5 2\n1 2 2 5\n2 6 7",
"output": "21"
},
{
"input": "3 3\n9 9 9 9 9 9 9 9 9\n1 2 3\n1 2 3",
"output": "0"
},
{
"input": "1 1\n1\n2\n1",
"output": "0"
},
{
"input": "1 1\n1\n1\n1",
"output": "1"
},
{
"input": "1 1\n1\n1\n100000",
"output": "100000"
},
{
"input": "50 1\n52 96 99 37 143 148 10 140 131 29 82 134 56 73 121 57 98 101 134 4 103 10 86 70 4 98 102 35 149 47 136 87 4 127 142 105 78 10 123 75 67 149 81 78 34 79 62 12 43 115\n31 132 59 75 4 135 138 33 33 60 135 5 30 127 61 74 102 131 11 16 74 4 101 74 70 45 29 12 137 59 24 52 25 122 64 147 92 77 23 6 19 76 26 55 126 130 4 148 86 3\n94393",
"output": "1321502"
},
{
"input": "50 1\n995 1815 941 1716 725 1098 747 627 1728 1007 34 1001 679 1742 22 1495 1299 1696 507 631 1971 775 1052 1665 1035 203 1564 1329 1592 1295 983 177 734 1442 172 943 33 486 1078 946 947 592 1524 563 396 1541 1670 326 543 79\n176 214 1601 1758 1468 972 628 1524 1506 425 746 309 387 1761 1002 625 496 1638 1855 1115 47 1813 1258 289 891 518 1247 1782 788 1449 1174 183 899 1728 366 1270 1641 327 1839 1093 223 1005 1214 1866 1432 1715 25 1240 1234 324\n52314",
"output": "104628"
}
] | 1,691,212,390
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
print("_RANDOM_GUESS_1691212390.420131")# 1691212390.4201474
|
Title: Cards and Joy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are $n$ players sitting at the card table. Each player has a favorite number. The favorite number of the $j$-th player is $f_j$.
There are $k \cdot n$ cards on the table. Each card contains a single integer: the $i$-th card contains number $c_i$. Also, you are given a sequence $h_1, h_2, \dots, h_k$. Its meaning will be explained below.
The players have to distribute all the cards in such a way that each of them will hold exactly $k$ cards. After all the cards are distributed, each player counts the number of cards he has that contains his favorite number. The joy level of a player equals $h_t$ if the player holds $t$ cards containing his favorite number. If a player gets no cards with his favorite number (i.e., $t=0$), his joy level is $0$.
Print the maximum possible total joy levels of the players after the cards are distributed. Note that the sequence $h_1, \dots, h_k$ is the same for all the players.
Input Specification:
The first line of input contains two integers $n$ and $k$ ($1 \le n \le 500, 1 \le k \le 10$) — the number of players and the number of cards each player will get.
The second line contains $k \cdot n$ integers $c_1, c_2, \dots, c_{k \cdot n}$ ($1 \le c_i \le 10^5$) — the numbers written on the cards.
The third line contains $n$ integers $f_1, f_2, \dots, f_n$ ($1 \le f_j \le 10^5$) — the favorite numbers of the players.
The fourth line contains $k$ integers $h_1, h_2, \dots, h_k$ ($1 \le h_t \le 10^5$), where $h_t$ is the joy level of a player if he gets exactly $t$ cards with his favorite number written on them. It is guaranteed that the condition $h_{t - 1} < h_t$ holds for each $t \in [2..k]$.
Output Specification:
Print one integer — the maximum possible total joy levels of the players among all possible card distributions.
Demo Input:
['4 3\n1 3 2 8 5 5 8 2 2 8 5 2\n1 2 2 5\n2 6 7\n', '3 3\n9 9 9 9 9 9 9 9 9\n1 2 3\n1 2 3\n']
Demo Output:
['21\n', '0\n']
Note:
In the first example, one possible optimal card distribution is the following:
- Player $1$ gets cards with numbers $[1, 3, 8]$; - Player $2$ gets cards with numbers $[2, 2, 8]$; - Player $3$ gets cards with numbers $[2, 2, 8]$; - Player $4$ gets cards with numbers $[5, 5, 5]$.
Thus, the answer is $2 + 6 + 6 + 7 = 21$.
In the second example, no player can get a card with his favorite number. Thus, the answer is $0$.
|
```python
print("_RANDOM_GUESS_1691212390.420131")# 1691212390.4201474
```
| 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 "
}
] | 1,430,067,836
| 3,836
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 3
| 62
| 4,608,000
|
__author__ = 'Xapheus'
input = input()
n = int(input)
quasibinary_set = []
total = 0
i = n
while i > 0:
mults = 1
isQuasi = True
for j in str(i):
if j != '0' and j != '1':
isQuasi = False
break
if isQuasi:
#Quasibinary
while mults*i <= n - total:
mults += 1
mults -= 1
# Add mults i to set
for j in range(mults):
quasibinary_set.append(i)
total += mults*i
i = n - total #Go through remaining
else:
i -= 1
print(len(quasibinary_set))
for j in quasibinary_set:
print(str(j) + " ", end=' ')
|
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
__author__ = 'Xapheus'
input = input()
n = int(input)
quasibinary_set = []
total = 0
i = n
while i > 0:
mults = 1
isQuasi = True
for j in str(i):
if j != '0' and j != '1':
isQuasi = False
break
if isQuasi:
#Quasibinary
while mults*i <= n - total:
mults += 1
mults -= 1
# Add mults i to set
for j in range(mults):
quasibinary_set.append(i)
total += mults*i
i = n - total #Go through remaining
else:
i -= 1
print(len(quasibinary_set))
for j in quasibinary_set:
print(str(j) + " ", end=' ')
```
| 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",
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},
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},
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},
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},
{
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},
{
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},
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},
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},
{
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},
{
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},
{
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{
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},
{
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},
{
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},
{
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"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
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},
{
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"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
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},
{
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"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
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},
{
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{
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},
{
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},
{
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},
{
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"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
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},
{
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},
{
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"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
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},
{
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},
{
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"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
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},
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},
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},
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},
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},
{
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},
{
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"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
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},
{
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"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,548,729,092
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
a= Input()
b= Input()
for i in a:
for j in b:
if(i==j)
print(0)
else
print(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= Input()
b= Input()
for i in a:
for j in b:
if(i==j)
print(0)
else
print(1)
```
| -1
|
757
|
B
|
Bash's Big Day
|
PROGRAMMING
| 1,400
|
[
"greedy",
"math",
"number theory"
] | null | null |
Bash has set out on a journey to become the greatest Pokemon master. To get his first Pokemon, he went to Professor Zulu's Lab. Since Bash is Professor Zulu's favourite student, Zulu allows him to take as many Pokemon from his lab as he pleases.
But Zulu warns him that a group of *k*<=><=1 Pokemon with strengths {*s*1,<=*s*2,<=*s*3,<=...,<=*s**k*} tend to fight among each other if *gcd*(*s*1,<=*s*2,<=*s*3,<=...,<=*s**k*)<==<=1 (see notes for *gcd* definition).
Bash, being smart, does not want his Pokemon to fight among each other. However, he also wants to maximize the number of Pokemon he takes from the lab. Can you help Bash find out the maximum number of Pokemon he can take?
Note: A Pokemon cannot fight with itself.
|
The input consists of two lines.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105), the number of Pokemon in the lab.
The next line contains *n* space separated integers, where the *i*-th of them denotes *s**i* (1<=≤<=*s**i*<=≤<=105), the strength of the *i*-th Pokemon.
|
Print single integer — the maximum number of Pokemons Bash can take.
|
[
"3\n2 3 4\n",
"5\n2 3 4 6 7\n"
] |
[
"2\n",
"3\n"
] |
*gcd* (greatest common divisor) of positive integers set {*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>} is the maximum positive integer that divides all the integers {*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>}.
In the first sample, we can take Pokemons with strengths {2, 4} since *gcd*(2, 4) = 2.
In the second sample, we can take Pokemons with strengths {2, 4, 6}, and there is no larger group with *gcd* ≠ 1.
| 1,000
|
[
{
"input": "3\n2 3 4",
"output": "2"
},
{
"input": "5\n2 3 4 6 7",
"output": "3"
},
{
"input": "3\n5 6 4",
"output": "2"
},
{
"input": "8\n41 74 4 27 85 39 100 36",
"output": "4"
},
{
"input": "6\n89 20 86 81 62 23",
"output": "3"
},
{
"input": "71\n23 84 98 8 14 4 42 56 83 87 28 22 32 50 5 96 90 1 59 74 77 88 71 38 62 36 85 97 99 6 81 20 49 57 66 9 45 41 29 68 35 19 27 76 78 72 55 25 46 48 26 53 39 31 94 34 63 37 64 16 79 24 82 17 12 3 89 61 80 30 10",
"output": "38"
},
{
"input": "95\n72 38 75 62 87 30 11 65 35 16 73 23 18 48 19 4 22 42 14 60 49 83 59 15 51 27 80 97 37 100 64 81 54 71 52 20 5 98 78 86 26 55 25 57 36 3 8 74 82 21 29 1 76 2 79 61 39 9 89 77 70 63 56 28 92 53 31 45 93 47 67 99 58 12 84 44 32 34 69 40 13 7 66 68 17 85 6 90 33 91 94 24 46 10 50",
"output": "48"
},
{
"input": "44\n39706 21317 26213 55086 10799 31825 29024 6565 96535 11412 14642 91901 41932 24538 81351 53861 63403 34199 82286 32594 29684 42753 16857 73821 71085 36306 70080 11233 21023 8551 85406 95390 92375 52675 77938 46265 74855 5229 5856 66713 65730 24525 84078 20684",
"output": "19"
},
{
"input": "35\n45633 86983 46174 48399 33926 51395 76300 6387 48852 82808 28694 79864 4482 35982 21956 76522 19656 74518 28480 71481 25700 46815 14170 95705 8535 96993 29029 8898 97637 62710 14615 22864 69849 27068 68557",
"output": "20"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n10 7 9 8 3 3 10 7 3 3",
"output": "5"
},
{
"input": "9\n10 10 6 10 9 1 8 3 5",
"output": "5"
},
{
"input": "7\n9 4 2 3 3 9 8",
"output": "4"
},
{
"input": "1\n4",
"output": "1"
},
{
"input": "6\n1623 45906 37856 34727 27156 12598",
"output": "4"
},
{
"input": "30\n83172 59163 67334 83980 5932 8773 77649 41428 62789 28159 17183 10199 41496 59500 14614 10468 54886 64679 42382 57021 50499 95643 77239 61434 16181 30505 59152 55972 18265 70566",
"output": "15"
},
{
"input": "23\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 22 16 2 13 16",
"output": "22"
},
{
"input": "46\n12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 15 1 18 28 20 6 31 16 5 23 21 38 3 11 18 11 3 25 33",
"output": "27"
},
{
"input": "43\n8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8 23 40 33 11 5 21 16 19 15 41 30 28 31 5 32 16 5 38 11 21 34",
"output": "21"
},
{
"input": "25\n58427 26687 48857 46477 7039 25423 58757 48119 38113 40637 22391 48337 4157 10597 8167 19031 64613 70913 69313 18047 17159 77491 13499 70949 24107",
"output": "1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "2\n3 6",
"output": "2"
},
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "5\n3 3 3 3 3",
"output": "5"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "2\n541 541",
"output": "2"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n99989 99989",
"output": "2"
},
{
"input": "3\n3 9 27",
"output": "3"
},
{
"input": "2\n1009 1009",
"output": "2"
},
{
"input": "4\n1 1 1 1",
"output": "1"
},
{
"input": "6\n2 10 20 5 15 25",
"output": "5"
},
{
"input": "3\n3 3 6",
"output": "3"
},
{
"input": "3\n457 457 457",
"output": "3"
},
{
"input": "2\n34 17",
"output": "2"
},
{
"input": "3\n12 24 3",
"output": "3"
},
{
"input": "10\n99991 99991 99991 99991 99991 99991 99991 99991 99991 99991",
"output": "10"
},
{
"input": "2\n1009 2018",
"output": "2"
},
{
"input": "3\n3 3 3",
"output": "3"
},
{
"input": "7\n6 9 12 15 21 27 33",
"output": "7"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "2\n557 557",
"output": "2"
},
{
"input": "3\n1 1 2",
"output": "1"
},
{
"input": "5\n2 2 101 101 101",
"output": "3"
},
{
"input": "2\n122 3721",
"output": "2"
},
{
"input": "2\n49201 98402",
"output": "2"
},
{
"input": "2\n88258 44129",
"output": "2"
},
{
"input": "2\n7919 47514",
"output": "2"
},
{
"input": "5\n1 2 1 1 1",
"output": "1"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "5\n1 1 1 1 2",
"output": "1"
},
{
"input": "3\n10007 20014 30021",
"output": "3"
},
{
"input": "4\n10 10 5 5",
"output": "4"
},
{
"input": "2\n6 9",
"output": "2"
},
{
"input": "3\n6 12 9",
"output": "3"
},
{
"input": "2\n14 7",
"output": "2"
},
{
"input": "2\n199 199",
"output": "2"
},
{
"input": "3\n49999 49999 99998",
"output": "3"
},
{
"input": "5\n1009 1009 1009 1009 1009",
"output": "5"
},
{
"input": "3\n17 17 17",
"output": "3"
},
{
"input": "2\n503 1509",
"output": "2"
},
{
"input": "3\n49999 99998 99998",
"output": "3"
},
{
"input": "2\n28657 28657",
"output": "2"
},
{
"input": "3\n121 22 33",
"output": "3"
},
{
"input": "2\n10079 20158",
"output": "2"
},
{
"input": "2\n6 15",
"output": "2"
},
{
"input": "4\n6 6 6 3",
"output": "4"
},
{
"input": "2\n991 1982",
"output": "2"
},
{
"input": "3\n20362 30543 50905",
"output": "3"
},
{
"input": "2\n443 886",
"output": "2"
},
{
"input": "4\n3 7 7 21",
"output": "3"
},
{
"input": "3\n3 6 9",
"output": "3"
},
{
"input": "2\n217 31",
"output": "2"
},
{
"input": "6\n10 10 10 5 5 5",
"output": "6"
},
{
"input": "4\n49441 25225 9081 4036",
"output": "4"
},
{
"input": "2\n4 9",
"output": "1"
},
{
"input": "2\n18 27",
"output": "2"
},
{
"input": "2\n13 26",
"output": "2"
},
{
"input": "4\n6 6 6 15",
"output": "4"
}
] | 1,528,616,734
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 70
| 483
| 6,860,800
|
n = int(input())
arr = [int(x) for x in input().split()]
t = max(arr) +2
k = [0] * t
for x in arr: k[x]+=1
for j in range(2, t) :
for i in range(2*j, t, j):
k[j] += k[i]
print(max(k))
|
Title: Bash's Big Day
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bash has set out on a journey to become the greatest Pokemon master. To get his first Pokemon, he went to Professor Zulu's Lab. Since Bash is Professor Zulu's favourite student, Zulu allows him to take as many Pokemon from his lab as he pleases.
But Zulu warns him that a group of *k*<=><=1 Pokemon with strengths {*s*1,<=*s*2,<=*s*3,<=...,<=*s**k*} tend to fight among each other if *gcd*(*s*1,<=*s*2,<=*s*3,<=...,<=*s**k*)<==<=1 (see notes for *gcd* definition).
Bash, being smart, does not want his Pokemon to fight among each other. However, he also wants to maximize the number of Pokemon he takes from the lab. Can you help Bash find out the maximum number of Pokemon he can take?
Note: A Pokemon cannot fight with itself.
Input Specification:
The input consists of two lines.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105), the number of Pokemon in the lab.
The next line contains *n* space separated integers, where the *i*-th of them denotes *s**i* (1<=≤<=*s**i*<=≤<=105), the strength of the *i*-th Pokemon.
Output Specification:
Print single integer — the maximum number of Pokemons Bash can take.
Demo Input:
['3\n2 3 4\n', '5\n2 3 4 6 7\n']
Demo Output:
['2\n', '3\n']
Note:
*gcd* (greatest common divisor) of positive integers set {*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>} is the maximum positive integer that divides all the integers {*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>}.
In the first sample, we can take Pokemons with strengths {2, 4} since *gcd*(2, 4) = 2.
In the second sample, we can take Pokemons with strengths {2, 4, 6}, and there is no larger group with *gcd* ≠ 1.
|
```python
n = int(input())
arr = [int(x) for x in input().split()]
t = max(arr) +2
k = [0] * t
for x in arr: k[x]+=1
for j in range(2, t) :
for i in range(2*j, t, j):
k[j] += k[i]
print(max(k))
```
| 0
|
|
557
|
A
|
Ilya and Diplomas
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation",
"math"
] | null | null |
Soon a school Olympiad in Informatics will be held in Berland, *n* schoolchildren will participate there.
At a meeting of the jury of the Olympiad it was decided that each of the *n* participants, depending on the results, will get a diploma of the first, second or third degree. Thus, each student will receive exactly one diploma.
They also decided that there must be given at least *min*1 and at most *max*1 diplomas of the first degree, at least *min*2 and at most *max*2 diplomas of the second degree, and at least *min*3 and at most *max*3 diplomas of the third degree.
After some discussion it was decided to choose from all the options of distributing diplomas satisfying these limitations the one that maximizes the number of participants who receive diplomas of the first degree. Of all these options they select the one which maximizes the number of the participants who receive diplomas of the second degree. If there are multiple of these options, they select the option that maximizes the number of diplomas of the third degree.
Choosing the best option of distributing certificates was entrusted to Ilya, one of the best programmers of Berland. However, he found more important things to do, so it is your task now to choose the best option of distributing of diplomas, based on the described limitations.
It is guaranteed that the described limitations are such that there is a way to choose such an option of distributing diplomas that all *n* participants of the Olympiad will receive a diploma of some degree.
|
The first line of the input contains a single integer *n* (3<=≤<=*n*<=≤<=3·106) — the number of schoolchildren who will participate in the Olympiad.
The next line of the input contains two integers *min*1 and *max*1 (1<=≤<=*min*1<=≤<=*max*1<=≤<=106) — the minimum and maximum limits on the number of diplomas of the first degree that can be distributed.
The third line of the input contains two integers *min*2 and *max*2 (1<=≤<=*min*2<=≤<=*max*2<=≤<=106) — the minimum and maximum limits on the number of diplomas of the second degree that can be distributed.
The next line of the input contains two integers *min*3 and *max*3 (1<=≤<=*min*3<=≤<=*max*3<=≤<=106) — the minimum and maximum limits on the number of diplomas of the third degree that can be distributed.
It is guaranteed that *min*1<=+<=*min*2<=+<=*min*3<=≤<=*n*<=≤<=*max*1<=+<=*max*2<=+<=*max*3.
|
In the first line of the output print three numbers, showing how many diplomas of the first, second and third degree will be given to students in the optimal variant of distributing diplomas.
The optimal variant of distributing diplomas is the one that maximizes the number of students who receive diplomas of the first degree. Of all the suitable options, the best one is the one which maximizes the number of participants who receive diplomas of the second degree. If there are several of these options, the best one is the one that maximizes the number of diplomas of the third degree.
|
[
"6\n1 5\n2 6\n3 7\n",
"10\n1 2\n1 3\n1 5\n",
"6\n1 3\n2 2\n2 2\n"
] |
[
"1 2 3 \n",
"2 3 5 \n",
"2 2 2 \n"
] |
none
| 500
|
[
{
"input": "6\n1 5\n2 6\n3 7",
"output": "1 2 3 "
},
{
"input": "10\n1 2\n1 3\n1 5",
"output": "2 3 5 "
},
{
"input": "6\n1 3\n2 2\n2 2",
"output": "2 2 2 "
},
{
"input": "55\n1 1000000\n40 50\n10 200",
"output": "5 40 10 "
},
{
"input": "3\n1 1\n1 1\n1 1",
"output": "1 1 1 "
},
{
"input": "3\n1 1000000\n1 1000000\n1 1000000",
"output": "1 1 1 "
},
{
"input": "1000\n100 400\n300 500\n400 1200",
"output": "300 300 400 "
},
{
"input": "3000000\n1 1000000\n1 1000000\n1 1000000",
"output": "1000000 1000000 1000000 "
},
{
"input": "11\n3 5\n3 5\n3 5",
"output": "5 3 3 "
},
{
"input": "12\n3 5\n3 5\n3 5",
"output": "5 4 3 "
},
{
"input": "13\n3 5\n3 5\n3 5",
"output": "5 5 3 "
},
{
"input": "3000000\n1000000 1000000\n1000000 1000000\n1000000 1000000",
"output": "1000000 1000000 1000000 "
},
{
"input": "50\n1 100\n1 100\n1 100",
"output": "48 1 1 "
},
{
"input": "1279\n123 670\n237 614\n846 923",
"output": "196 237 846 "
},
{
"input": "1589\n213 861\n5 96\n506 634",
"output": "861 96 632 "
},
{
"input": "2115\n987 987\n112 483\n437 959",
"output": "987 483 645 "
},
{
"input": "641\n251 960\n34 370\n149 149",
"output": "458 34 149 "
},
{
"input": "1655\n539 539\n10 425\n605 895",
"output": "539 425 691 "
},
{
"input": "1477\n210 336\n410 837\n448 878",
"output": "336 693 448 "
},
{
"input": "1707\n149 914\n190 422\n898 899",
"output": "619 190 898 "
},
{
"input": "1529\n515 515\n563 869\n169 451",
"output": "515 845 169 "
},
{
"input": "1543\n361 994\n305 407\n102 197",
"output": "994 407 142 "
},
{
"input": "1107\n471 849\n360 741\n71 473",
"output": "676 360 71 "
},
{
"input": "1629279\n267360 999930\n183077 674527\n202618 786988",
"output": "999930 426731 202618 "
},
{
"input": "1233589\n2850 555444\n500608 921442\n208610 607343",
"output": "524371 500608 208610 "
},
{
"input": "679115\n112687 183628\n101770 982823\n81226 781340",
"output": "183628 414261 81226 "
},
{
"input": "1124641\n117999 854291\n770798 868290\n76651 831405",
"output": "277192 770798 76651 "
},
{
"input": "761655\n88152 620061\n60403 688549\n79370 125321",
"output": "620061 62224 79370 "
},
{
"input": "2174477\n276494 476134\n555283 954809\n319941 935631",
"output": "476134 954809 743534 "
},
{
"input": "1652707\n201202 990776\n34796 883866\n162979 983308",
"output": "990776 498952 162979 "
},
{
"input": "2065529\n43217 891429\n434379 952871\n650231 855105",
"output": "891429 523869 650231 "
},
{
"input": "1702543\n405042 832833\n50931 747750\n381818 796831",
"output": "832833 487892 381818 "
},
{
"input": "501107\n19061 859924\n126478 724552\n224611 489718",
"output": "150018 126478 224611 "
},
{
"input": "1629279\n850831 967352\n78593 463906\n452094 885430",
"output": "967352 209833 452094 "
},
{
"input": "1233589\n2850 157021\n535109 748096\n392212 475634",
"output": "157021 684356 392212 "
},
{
"input": "679115\n125987 786267\n70261 688983\n178133 976789",
"output": "430721 70261 178133 "
},
{
"input": "1124641\n119407 734250\n213706 860770\n102149 102149",
"output": "734250 288242 102149 "
},
{
"input": "761655\n325539 325539\n280794 792505\n18540 106895",
"output": "325539 417576 18540 "
},
{
"input": "2174477\n352351 791072\n365110 969163\n887448 955610",
"output": "791072 495957 887448 "
},
{
"input": "1652707\n266774 638522\n65688 235422\n924898 992826",
"output": "638522 89287 924898 "
},
{
"input": "2065529\n608515 608515\n751563 864337\n614898 705451",
"output": "608515 842116 614898 "
},
{
"input": "1702543\n5784 996578\n47395 300407\n151614 710197",
"output": "996578 300407 405558 "
},
{
"input": "501107\n8073 390048\n190494 647328\n274071 376923",
"output": "36542 190494 274071 "
},
{
"input": "200\n50 50\n100 100\n50 50",
"output": "50 100 50 "
},
{
"input": "14\n1 100\n1 100\n8 9",
"output": "5 1 8 "
},
{
"input": "300\n200 400\n50 100\n40 80",
"output": "210 50 40 "
},
{
"input": "10\n3 6\n3 6\n3 6",
"output": "4 3 3 "
},
{
"input": "14\n3 6\n3 6\n3 6",
"output": "6 5 3 "
},
{
"input": "17\n3 6\n3 6\n3 6",
"output": "6 6 5 "
},
{
"input": "1000000\n300000 600000\n300000 600000\n300000 600000",
"output": "400000 300000 300000 "
},
{
"input": "1400000\n300000 600000\n300000 600000\n300000 600000",
"output": "600000 500000 300000 "
},
{
"input": "1700000\n300000 600000\n300000 600000\n300000 600000",
"output": "600000 600000 500000 "
},
{
"input": "561\n400 400\n80 80\n81 81",
"output": "400 80 81 "
},
{
"input": "2000\n100 1000\n1 1\n1 2000",
"output": "1000 1 999 "
},
{
"input": "1000002\n1 1000000\n1 1000000\n999999 1000000",
"output": "2 1 999999 "
},
{
"input": "1000002\n1 1000000\n1 1000000\n1000000 1000000",
"output": "1 1 1000000 "
}
] | 1,537,624,206
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 56
| 109
| 0
|
n = int(input())
res = [0,0,0]
maxi = [0,0,0]
for i in range(3):
a, b = map(int, input().split())
res[i] = a
maxi[i] = b
n -= a
for i in range(3):
if n > 0:
mini = min(n,maxi[i] - res[i])
res[i] += mini
n -= mini
print(*res)
|
Title: Ilya and Diplomas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Soon a school Olympiad in Informatics will be held in Berland, *n* schoolchildren will participate there.
At a meeting of the jury of the Olympiad it was decided that each of the *n* participants, depending on the results, will get a diploma of the first, second or third degree. Thus, each student will receive exactly one diploma.
They also decided that there must be given at least *min*1 and at most *max*1 diplomas of the first degree, at least *min*2 and at most *max*2 diplomas of the second degree, and at least *min*3 and at most *max*3 diplomas of the third degree.
After some discussion it was decided to choose from all the options of distributing diplomas satisfying these limitations the one that maximizes the number of participants who receive diplomas of the first degree. Of all these options they select the one which maximizes the number of the participants who receive diplomas of the second degree. If there are multiple of these options, they select the option that maximizes the number of diplomas of the third degree.
Choosing the best option of distributing certificates was entrusted to Ilya, one of the best programmers of Berland. However, he found more important things to do, so it is your task now to choose the best option of distributing of diplomas, based on the described limitations.
It is guaranteed that the described limitations are such that there is a way to choose such an option of distributing diplomas that all *n* participants of the Olympiad will receive a diploma of some degree.
Input Specification:
The first line of the input contains a single integer *n* (3<=≤<=*n*<=≤<=3·106) — the number of schoolchildren who will participate in the Olympiad.
The next line of the input contains two integers *min*1 and *max*1 (1<=≤<=*min*1<=≤<=*max*1<=≤<=106) — the minimum and maximum limits on the number of diplomas of the first degree that can be distributed.
The third line of the input contains two integers *min*2 and *max*2 (1<=≤<=*min*2<=≤<=*max*2<=≤<=106) — the minimum and maximum limits on the number of diplomas of the second degree that can be distributed.
The next line of the input contains two integers *min*3 and *max*3 (1<=≤<=*min*3<=≤<=*max*3<=≤<=106) — the minimum and maximum limits on the number of diplomas of the third degree that can be distributed.
It is guaranteed that *min*1<=+<=*min*2<=+<=*min*3<=≤<=*n*<=≤<=*max*1<=+<=*max*2<=+<=*max*3.
Output Specification:
In the first line of the output print three numbers, showing how many diplomas of the first, second and third degree will be given to students in the optimal variant of distributing diplomas.
The optimal variant of distributing diplomas is the one that maximizes the number of students who receive diplomas of the first degree. Of all the suitable options, the best one is the one which maximizes the number of participants who receive diplomas of the second degree. If there are several of these options, the best one is the one that maximizes the number of diplomas of the third degree.
Demo Input:
['6\n1 5\n2 6\n3 7\n', '10\n1 2\n1 3\n1 5\n', '6\n1 3\n2 2\n2 2\n']
Demo Output:
['1 2 3 \n', '2 3 5 \n', '2 2 2 \n']
Note:
none
|
```python
n = int(input())
res = [0,0,0]
maxi = [0,0,0]
for i in range(3):
a, b = map(int, input().split())
res[i] = a
maxi[i] = b
n -= a
for i in range(3):
if n > 0:
mini = min(n,maxi[i] - res[i])
res[i] += mini
n -= mini
print(*res)
```
| 3
|
|
385
|
A
|
Bear and Raspberry
|
PROGRAMMING
| 1,000
|
[
"brute force",
"greedy",
"implementation"
] | null | null |
The bear decided to store some raspberry for the winter. He cunningly found out the price for a barrel of honey in kilos of raspberry for each of the following *n* days. According to the bear's data, on the *i*-th (1<=≤<=*i*<=≤<=*n*) day, the price for one barrel of honey is going to is *x**i* kilos of raspberry.
Unfortunately, the bear has neither a honey barrel, nor the raspberry. At the same time, the bear's got a friend who is ready to lend him a barrel of honey for exactly one day for *c* kilograms of raspberry. That's why the bear came up with a smart plan. He wants to choose some day *d* (1<=≤<=*d*<=<<=*n*), lent a barrel of honey and immediately (on day *d*) sell it according to a daily exchange rate. The next day (*d*<=+<=1) the bear wants to buy a new barrel of honey according to a daily exchange rate (as he's got some raspberry left from selling the previous barrel) and immediately (on day *d*<=+<=1) give his friend the borrowed barrel of honey as well as *c* kilograms of raspberry for renting the barrel.
The bear wants to execute his plan at most once and then hibernate. What maximum number of kilograms of raspberry can he earn? Note that if at some point of the plan the bear runs out of the raspberry, then he won't execute such a plan.
|
The first line contains two space-separated integers, *n* and *c* (2<=≤<=*n*<=≤<=100,<=0<=≤<=*c*<=≤<=100), — the number of days and the number of kilos of raspberry that the bear should give for borrowing the barrel.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100), the price of a honey barrel on day *i*.
|
Print a single integer — the answer to the problem.
|
[
"5 1\n5 10 7 3 20\n",
"6 2\n100 1 10 40 10 40\n",
"3 0\n1 2 3\n"
] |
[
"3\n",
"97\n",
"0\n"
] |
In the first sample the bear will lend a honey barrel at day 3 and then sell it for 7. Then the bear will buy a barrel for 3 and return it to the friend. So, the profit is (7 - 3 - 1) = 3.
In the second sample bear will lend a honey barrel at day 1 and then sell it for 100. Then the bear buy the barrel for 1 at the day 2. So, the profit is (100 - 1 - 2) = 97.
| 500
|
[
{
"input": "5 1\n5 10 7 3 20",
"output": "3"
},
{
"input": "6 2\n100 1 10 40 10 40",
"output": "97"
},
{
"input": "3 0\n1 2 3",
"output": "0"
},
{
"input": "2 0\n2 1",
"output": "1"
},
{
"input": "10 5\n10 1 11 2 12 3 13 4 14 5",
"output": "4"
},
{
"input": "100 4\n2 57 70 8 44 10 88 67 50 44 93 79 72 50 69 19 21 9 71 47 95 13 46 10 68 72 54 40 15 83 57 92 58 25 4 22 84 9 8 55 87 0 16 46 86 58 5 21 32 28 10 46 11 29 13 33 37 34 78 33 33 21 46 70 77 51 45 97 6 21 68 61 87 54 8 91 37 12 76 61 57 9 100 45 44 88 5 71 98 98 26 45 37 87 34 50 33 60 64 77",
"output": "87"
},
{
"input": "100 5\n15 91 86 53 18 52 26 89 8 4 5 100 11 64 88 91 35 57 67 72 71 71 69 73 97 23 11 1 59 86 37 82 6 67 71 11 7 31 11 68 21 43 89 54 27 10 3 33 8 57 79 26 90 81 6 28 24 7 33 50 24 13 27 85 4 93 14 62 37 67 33 40 7 48 41 4 14 9 95 10 64 62 7 93 23 6 28 27 97 64 26 83 70 0 97 74 11 82 70 93",
"output": "84"
},
{
"input": "6 100\n10 9 8 7 6 5",
"output": "0"
},
{
"input": "100 9\n66 71 37 41 23 38 77 11 74 13 51 26 93 56 81 17 12 70 85 37 54 100 14 99 12 83 44 16 99 65 13 48 92 32 69 33 100 57 58 88 25 45 44 85 5 41 82 15 37 18 21 45 3 68 33 9 52 64 8 73 32 41 87 99 26 26 47 24 79 93 9 44 11 34 85 26 14 61 49 38 25 65 49 81 29 82 28 23 2 64 38 13 77 68 67 23 58 57 83 46",
"output": "78"
},
{
"input": "100 100\n9 72 46 37 26 94 80 1 43 85 26 53 58 18 24 19 67 2 100 52 61 81 48 15 73 41 97 93 45 1 73 54 75 51 28 79 0 14 41 42 24 50 70 18 96 100 67 1 68 48 44 39 63 77 78 18 10 51 32 53 26 60 1 13 66 39 55 27 23 71 75 0 27 88 73 31 16 95 87 84 86 71 37 40 66 70 65 83 19 4 81 99 26 51 67 63 80 54 23 44",
"output": "0"
},
{
"input": "43 65\n32 58 59 75 85 18 57 100 69 0 36 38 79 95 82 47 7 55 28 88 27 88 63 71 80 86 67 53 69 37 99 54 81 19 55 12 2 17 84 77 25 26 62",
"output": "4"
},
{
"input": "12 64\n14 87 40 24 32 36 4 41 38 77 68 71",
"output": "0"
},
{
"input": "75 94\n80 92 25 48 78 17 69 52 79 73 12 15 59 55 25 61 96 27 98 43 30 43 36 94 67 54 86 99 100 61 65 8 65 19 18 21 75 31 2 98 55 87 14 1 17 97 94 11 57 29 34 71 76 67 45 0 78 29 86 82 29 23 77 100 48 43 65 62 88 34 7 28 13 1 1",
"output": "0"
},
{
"input": "59 27\n76 61 24 66 48 18 69 84 21 8 64 90 19 71 36 90 9 36 30 37 99 37 100 56 9 79 55 37 54 63 11 11 49 71 91 70 14 100 10 44 52 23 21 19 96 13 93 66 52 79 76 5 62 6 90 35 94 7 27",
"output": "63"
},
{
"input": "86 54\n41 84 16 5 20 79 73 13 23 24 42 73 70 80 69 71 33 44 62 29 86 88 40 64 61 55 58 19 16 23 84 100 38 91 89 98 47 50 55 87 12 94 2 12 0 1 4 26 50 96 68 34 94 80 8 22 60 3 72 84 65 89 44 52 50 9 24 34 81 28 56 17 38 85 78 90 62 60 1 40 91 2 7 41 84 22",
"output": "38"
},
{
"input": "37 2\n65 36 92 92 92 76 63 56 15 95 75 26 15 4 73 50 41 92 26 20 19 100 63 55 25 75 61 96 35 0 14 6 96 3 28 41 83",
"output": "91"
},
{
"input": "19 4\n85 2 56 70 33 75 89 60 100 81 42 28 18 92 29 96 49 23 14",
"output": "79"
},
{
"input": "89 1\n50 53 97 41 68 27 53 66 93 19 11 78 46 49 38 69 96 9 43 16 1 63 95 64 96 6 34 34 45 40 19 4 53 8 11 18 95 25 50 16 64 33 97 49 23 81 63 10 30 73 76 55 7 70 9 98 6 36 75 78 3 92 85 75 40 75 55 71 9 91 15 17 47 55 44 35 55 88 53 87 61 22 100 56 14 87 36 84 24",
"output": "91"
},
{
"input": "67 0\n40 48 15 46 90 7 65 52 24 15 42 81 2 6 71 94 32 18 97 67 83 98 48 51 10 47 8 68 36 46 65 75 90 30 62 9 5 35 80 60 69 58 62 68 58 73 80 9 22 46 56 64 44 11 93 73 62 54 15 20 17 69 16 33 85 62 49",
"output": "83"
},
{
"input": "96 0\n38 97 82 43 80 40 1 99 50 94 81 63 92 13 57 24 4 10 25 32 79 56 96 19 25 14 69 56 66 22 23 78 87 76 37 30 75 77 61 64 35 64 62 32 44 62 6 84 91 44 99 5 71 19 17 12 35 52 1 14 35 18 8 36 54 42 4 67 80 11 88 44 34 35 12 38 66 42 4 90 45 10 1 44 37 96 23 28 100 90 75 17 27 67 51 70",
"output": "94"
},
{
"input": "14 14\n87 63 62 31 59 47 40 89 92 43 80 30 99 42",
"output": "43"
},
{
"input": "12 0\n100 1 100 2 100 3 100 4 100 5 100 0",
"output": "100"
},
{
"input": "3 1\n1 2 3",
"output": "0"
},
{
"input": "3 2\n3 3 3",
"output": "0"
},
{
"input": "3 3\n3 2 1",
"output": "0"
},
{
"input": "3 100\n1 2 3",
"output": "0"
},
{
"input": "2 100\n0 0",
"output": "0"
},
{
"input": "2 90\n10 5",
"output": "0"
},
{
"input": "2 5\n5 4",
"output": "0"
},
{
"input": "3 1\n19 20 1",
"output": "18"
},
{
"input": "5 1\n5 10 7 4 20",
"output": "2"
},
{
"input": "5 1\n1 2 3 4 5",
"output": "0"
}
] | 1,569,087,077
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 124
| 0
|
n,c=map(int,input().strip().split(' '))
arr=list(map(int,input().strip().split(' ')))
list1=[]
for i in range(n-1):
a=arr[i]-arr[i+1]-c
list1.append(a)
if max(list1)>0:
print(max(list1))
else:
print("0")
|
Title: Bear and Raspberry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The bear decided to store some raspberry for the winter. He cunningly found out the price for a barrel of honey in kilos of raspberry for each of the following *n* days. According to the bear's data, on the *i*-th (1<=≤<=*i*<=≤<=*n*) day, the price for one barrel of honey is going to is *x**i* kilos of raspberry.
Unfortunately, the bear has neither a honey barrel, nor the raspberry. At the same time, the bear's got a friend who is ready to lend him a barrel of honey for exactly one day for *c* kilograms of raspberry. That's why the bear came up with a smart plan. He wants to choose some day *d* (1<=≤<=*d*<=<<=*n*), lent a barrel of honey and immediately (on day *d*) sell it according to a daily exchange rate. The next day (*d*<=+<=1) the bear wants to buy a new barrel of honey according to a daily exchange rate (as he's got some raspberry left from selling the previous barrel) and immediately (on day *d*<=+<=1) give his friend the borrowed barrel of honey as well as *c* kilograms of raspberry for renting the barrel.
The bear wants to execute his plan at most once and then hibernate. What maximum number of kilograms of raspberry can he earn? Note that if at some point of the plan the bear runs out of the raspberry, then he won't execute such a plan.
Input Specification:
The first line contains two space-separated integers, *n* and *c* (2<=≤<=*n*<=≤<=100,<=0<=≤<=*c*<=≤<=100), — the number of days and the number of kilos of raspberry that the bear should give for borrowing the barrel.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100), the price of a honey barrel on day *i*.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['5 1\n5 10 7 3 20\n', '6 2\n100 1 10 40 10 40\n', '3 0\n1 2 3\n']
Demo Output:
['3\n', '97\n', '0\n']
Note:
In the first sample the bear will lend a honey barrel at day 3 and then sell it for 7. Then the bear will buy a barrel for 3 and return it to the friend. So, the profit is (7 - 3 - 1) = 3.
In the second sample bear will lend a honey barrel at day 1 and then sell it for 100. Then the bear buy the barrel for 1 at the day 2. So, the profit is (100 - 1 - 2) = 97.
|
```python
n,c=map(int,input().strip().split(' '))
arr=list(map(int,input().strip().split(' ')))
list1=[]
for i in range(n-1):
a=arr[i]-arr[i+1]-c
list1.append(a)
if max(list1)>0:
print(max(list1))
else:
print("0")
```
| 3
|
|
86
|
D
|
Powerful array
|
PROGRAMMING
| 2,200
|
[
"data structures",
"implementation",
"math",
"two pointers"
] |
D. Powerful array
|
5
|
256
|
An array of positive integers *a*1,<=*a*2,<=...,<=*a**n* is given. Let us consider its arbitrary subarray *a**l*,<=*a**l*<=+<=1...,<=*a**r*, where 1<=≤<=*l*<=≤<=*r*<=≤<=*n*. For every positive integer *s* denote by *K**s* the number of occurrences of *s* into the subarray. We call the power of the subarray the sum of products *K**s*·*K**s*·*s* for every positive integer *s*. The sum contains only finite number of nonzero summands as the number of different values in the array is indeed finite.
You should calculate the power of *t* given subarrays.
|
First line contains two integers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=200000) — the array length and the number of queries correspondingly.
Second line contains *n* positive integers *a**i* (1<=≤<=*a**i*<=≤<=106) — the elements of the array.
Next *t* lines contain two positive integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) each — the indices of the left and the right ends of the corresponding subarray.
|
Output *t* lines, the *i*-th line of the output should contain single positive integer — the power of the *i*-th query subarray.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preferred to use cout stream (also you may use %I64d).
|
[
"3 2\n1 2 1\n1 2\n1 3\n",
"8 3\n1 1 2 2 1 3 1 1\n2 7\n1 6\n2 7\n"
] |
[
"3\n6\n",
"20\n20\n20\n"
] |
Consider the following array (see the second sample) and its [2, 7] subarray (elements of the subarray are colored):
| 2,500
|
[
{
"input": "3 2\n1 2 1\n1 2\n1 3",
"output": "3\n6"
},
{
"input": "8 3\n1 1 2 2 1 3 1 1\n2 7\n1 6\n2 7",
"output": "20\n20\n20"
},
{
"input": "20 8\n1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2\n4 15\n1 2\n2 20\n7 7\n13 18\n7 7\n3 19\n3 8",
"output": "108\n3\n281\n1\n27\n1\n209\n27"
},
{
"input": "10 5\n10 9 8 7 6 5 4 3 2 1\n4 8\n1 10\n3 9\n2 2\n5 10",
"output": "25\n55\n35\n9\n21"
},
{
"input": "8 10\n100 100 100 100 100 100 100 100\n1 8\n2 8\n3 8\n4 8\n5 8\n6 8\n7 8\n8 8\n1 1\n1 5",
"output": "6400\n4900\n3600\n2500\n1600\n900\n400\n100\n100\n2500"
},
{
"input": "1 2\n1\n1 1\n1 1",
"output": "1\n1"
},
{
"input": "1 1\n1000000\n1 1",
"output": "1000000"
},
{
"input": "5 15\n103 45 103 67 45\n1 1\n1 2\n1 3\n1 4\n1 5\n2 2\n2 3\n2 4\n2 5\n3 3\n3 4\n3 5\n4 4\n4 5\n5 5",
"output": "103\n148\n457\n524\n659\n45\n148\n215\n350\n103\n170\n215\n67\n112\n45"
},
{
"input": "8 10\n5 7 3 1 4 2 1 1\n5 7\n3 8\n2 7\n4 8\n4 5\n2 8\n4 6\n1 2\n4 7\n3 5",
"output": "7\n18\n20\n15\n5\n25\n7\n12\n10\n8"
},
{
"input": "9 10\n5 3 10 2 4 8 7 5 1\n6 9\n1 8\n3 3\n2 7\n3 5\n4 6\n9 9\n8 8\n1 9\n5 6",
"output": "21\n54\n10\n34\n16\n14\n1\n5\n55\n12"
},
{
"input": "9 10\n10 2 6 2 4 6 6 4 1\n4 8\n8 9\n2 3\n8 9\n9 9\n4 5\n1 7\n6 6\n6 8\n1 1",
"output": "42\n5\n8\n5\n1\n6\n76\n6\n28\n10"
},
{
"input": "10 10\n5 10 9 2 9 3 5 5 6 5\n5 10\n2 2\n3 10\n4 8\n1 3\n1 5\n5 5\n10 10\n8 9\n6 10",
"output": "63\n10\n92\n34\n24\n53\n9\n5\n11\n54"
},
{
"input": "9 10\n6 2 3 1 4 8 5 9 8\n3 9\n6 6\n9 9\n9 9\n9 9\n2 3\n3 9\n3 8\n6 9\n5 6",
"output": "54\n8\n8\n8\n8\n5\n54\n30\n46\n12"
}
] | 1,524,719,805
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 154
| 7,065,600
|
#task is to answer q queries telling the sum of elements in range l to r in the array by square root decomposition
from math import sqrt
def compute(array,n):
block_size = int(sqrt(n))
temp=(n+block_size-1)//block_size
block = {x:{} for x in range(temp)}
block_index=-1
#print(block_size)
for i in range(n):
if i%block_size == 0:
block_index+=1
if array[i] not in block[block_index]:
block[block_index][array[i]]=1
else:
block[block_index][array[i]]+=1
return block
def query(array,block,l,r,n):
block_size = int(sqrt(n))
final= {}
sum=0
while l<r and not(l%block_size) and not(l):
if array[l] not in final:
final[array[l]]=1
else:
final[array[l]]+=1
l+=1
while l+block_size <= r+1 :
temp=l//block_size
for x in block[temp]:
if x not in final:
final[x]=block[temp][x]
else:
final[x]+=block[temp][x]
l+=block_size
while l<=r:
if array[l] not in final:
final[array[l]]=1
else:
final[array[l]]+=1
l+=1
for x in final:
t=final[x]**2
sum+=t*x
return sum
n,t = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
block=compute(a,n)
for i in range(t):
l,r = [int(x) for x in input().split()]
print(query(a,block,l-1,r-1,n))
|
Title: Powerful array
Time Limit: 5 seconds
Memory Limit: 256 megabytes
Problem Description:
An array of positive integers *a*1,<=*a*2,<=...,<=*a**n* is given. Let us consider its arbitrary subarray *a**l*,<=*a**l*<=+<=1...,<=*a**r*, where 1<=≤<=*l*<=≤<=*r*<=≤<=*n*. For every positive integer *s* denote by *K**s* the number of occurrences of *s* into the subarray. We call the power of the subarray the sum of products *K**s*·*K**s*·*s* for every positive integer *s*. The sum contains only finite number of nonzero summands as the number of different values in the array is indeed finite.
You should calculate the power of *t* given subarrays.
Input Specification:
First line contains two integers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=200000) — the array length and the number of queries correspondingly.
Second line contains *n* positive integers *a**i* (1<=≤<=*a**i*<=≤<=106) — the elements of the array.
Next *t* lines contain two positive integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) each — the indices of the left and the right ends of the corresponding subarray.
Output Specification:
Output *t* lines, the *i*-th line of the output should contain single positive integer — the power of the *i*-th query subarray.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preferred to use cout stream (also you may use %I64d).
Demo Input:
['3 2\n1 2 1\n1 2\n1 3\n', '8 3\n1 1 2 2 1 3 1 1\n2 7\n1 6\n2 7\n']
Demo Output:
['3\n6\n', '20\n20\n20\n']
Note:
Consider the following array (see the second sample) and its [2, 7] subarray (elements of the subarray are colored):
|
```python
#task is to answer q queries telling the sum of elements in range l to r in the array by square root decomposition
from math import sqrt
def compute(array,n):
block_size = int(sqrt(n))
temp=(n+block_size-1)//block_size
block = {x:{} for x in range(temp)}
block_index=-1
#print(block_size)
for i in range(n):
if i%block_size == 0:
block_index+=1
if array[i] not in block[block_index]:
block[block_index][array[i]]=1
else:
block[block_index][array[i]]+=1
return block
def query(array,block,l,r,n):
block_size = int(sqrt(n))
final= {}
sum=0
while l<r and not(l%block_size) and not(l):
if array[l] not in final:
final[array[l]]=1
else:
final[array[l]]+=1
l+=1
while l+block_size <= r+1 :
temp=l//block_size
for x in block[temp]:
if x not in final:
final[x]=block[temp][x]
else:
final[x]+=block[temp][x]
l+=block_size
while l<=r:
if array[l] not in final:
final[array[l]]=1
else:
final[array[l]]+=1
l+=1
for x in final:
t=final[x]**2
sum+=t*x
return sum
n,t = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
block=compute(a,n)
for i in range(t):
l,r = [int(x) for x in input().split()]
print(query(a,block,l-1,r-1,n))
```
| 0
|
277
|
C
|
Game
|
PROGRAMMING
| 2,400
|
[
"games",
"implementation"
] | null | null |
Two players play the following game. Initially, the players have a knife and a rectangular sheet of paper, divided into equal square grid cells of unit size. The players make moves in turn, the player who can't make a move loses. In one move, a player can take the knife and cut the paper along any segment of the grid line (not necessarily from border to border). The part of the paper, that touches the knife at least once, is considered cut. There is one limit not to turn the game into an infinite cycle: each move has to cut the paper, that is the knife has to touch the part of the paper that is not cut before.
Obviously, the game ends when the entire sheet is cut into 1<=×<=1 blocks. During the game, the pieces of the sheet are not allowed to move. It is also prohibited to cut along the border. The coordinates of the ends of each cut must be integers.
You are given an *n*<=×<=*m* piece of paper, somebody has already made *k* cuts there. Your task is to determine who will win if the players start to play on this sheet. You can consider that both players play optimally well. If the first player wins, you also need to find the winning first move.
|
The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*<=≤<=109,<=0<=≤<=*k*<=≤<=105) — the sizes of the piece of paper and the number of cuts. Then follow *k* lines, each containing 4 integers *xb**i*,<=*yb**i*,<=*xe**i*,<=*ye**i* (0<=≤<=*xb**i*,<=*xe**i*<=≤<=*n*,<=0<=≤<=*yb**i*,<=*ye**i*<=≤<=*m*) — the coordinates of the ends of the existing cuts.
It is guaranteed that each cut has a non-zero length, is either vertical or horizontal and doesn't go along the sheet border.
The cuts may intersect, overlap and even be the same. That is, it is not guaranteed that the cuts were obtained during any correct game.
|
If the second player wins, print "SECOND". Otherwise, in the first line print "FIRST", and in the second line print any winning move of the first player (the coordinates of the cut ends, follow input format to print them).
|
[
"2 1 0\n",
"2 2 4\n0 1 2 1\n0 1 2 1\n1 2 1 0\n1 1 1 2\n"
] |
[
"FIRST\n1 0 1 1\n",
"SECOND\n"
] |
none
| 2,000
|
[] | 1,693,380,699
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 124
| 0
|
def solve(n, m, k, cuts):
if (n - 1) % 2 == 1 and (m - 1) % 2 == 1:
# Добавляем дополнительный разрез
additional_cut = (1, 1, 2, 2)
cuts.append(additional_cut)
return "FIRST", additional_cut
else:
return "SECOND", None
# Чтение входных данных
n, m, k = map(int, input().split())
cuts = []
for _ in range(k):
xbi, ybi, xei, yei = map(int, input().split())
cuts.append((xbi, ybi, xei, yei))
result, first_move = solve(n, m, k, cuts)
print(result)
if first_move:
print(*first_move)
|
Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play the following game. Initially, the players have a knife and a rectangular sheet of paper, divided into equal square grid cells of unit size. The players make moves in turn, the player who can't make a move loses. In one move, a player can take the knife and cut the paper along any segment of the grid line (not necessarily from border to border). The part of the paper, that touches the knife at least once, is considered cut. There is one limit not to turn the game into an infinite cycle: each move has to cut the paper, that is the knife has to touch the part of the paper that is not cut before.
Obviously, the game ends when the entire sheet is cut into 1<=×<=1 blocks. During the game, the pieces of the sheet are not allowed to move. It is also prohibited to cut along the border. The coordinates of the ends of each cut must be integers.
You are given an *n*<=×<=*m* piece of paper, somebody has already made *k* cuts there. Your task is to determine who will win if the players start to play on this sheet. You can consider that both players play optimally well. If the first player wins, you also need to find the winning first move.
Input Specification:
The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*<=≤<=109,<=0<=≤<=*k*<=≤<=105) — the sizes of the piece of paper and the number of cuts. Then follow *k* lines, each containing 4 integers *xb**i*,<=*yb**i*,<=*xe**i*,<=*ye**i* (0<=≤<=*xb**i*,<=*xe**i*<=≤<=*n*,<=0<=≤<=*yb**i*,<=*ye**i*<=≤<=*m*) — the coordinates of the ends of the existing cuts.
It is guaranteed that each cut has a non-zero length, is either vertical or horizontal and doesn't go along the sheet border.
The cuts may intersect, overlap and even be the same. That is, it is not guaranteed that the cuts were obtained during any correct game.
Output Specification:
If the second player wins, print "SECOND". Otherwise, in the first line print "FIRST", and in the second line print any winning move of the first player (the coordinates of the cut ends, follow input format to print them).
Demo Input:
['2 1 0\n', '2 2 4\n0 1 2 1\n0 1 2 1\n1 2 1 0\n1 1 1 2\n']
Demo Output:
['FIRST\n1 0 1 1\n', 'SECOND\n']
Note:
none
|
```python
def solve(n, m, k, cuts):
if (n - 1) % 2 == 1 and (m - 1) % 2 == 1:
# Добавляем дополнительный разрез
additional_cut = (1, 1, 2, 2)
cuts.append(additional_cut)
return "FIRST", additional_cut
else:
return "SECOND", None
# Чтение входных данных
n, m, k = map(int, input().split())
cuts = []
for _ in range(k):
xbi, ybi, xei, yei = map(int, input().split())
cuts.append((xbi, ybi, xei, yei))
result, first_move = solve(n, m, k, cuts)
print(result)
if first_move:
print(*first_move)
```
| 0
|
|
231
|
A
|
Team
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy"
] | null | null |
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
|
Print a single integer — the number of problems the friends will implement on the contest.
|
[
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] |
[
"2\n",
"1\n"
] |
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
| 500
|
[
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,697,831,226
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 62
| 0
|
n = int(input())
k = 0
for n in range(n):
ans = input()
if ans.split().count('1') >= 2:
k += 1
print(k)
|
Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
|
```python
n = int(input())
k = 0
for n in range(n):
ans = input()
if ans.split().count('1') >= 2:
k += 1
print(k)
```
| 3
|
|
659
|
C
|
Tanya and Toys
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation"
] | null | null |
In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new collection of the *i*-th type costs *i* bourles.
Tania has managed to collect *n* different types of toys *a*1,<=*a*2,<=...,<=*a**n* from the new collection. Today is Tanya's birthday, and her mother decided to spend no more than *m* bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has.
Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this.
|
The first line contains two integers *n* (1<=≤<=*n*<=≤<=100<=000) and *m* (1<=≤<=*m*<=≤<=109) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys.
The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the types of toys that Tanya already has.
|
In the first line print a single integer *k* — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed *m*.
In the second line print *k* distinct space-separated integers *t*1,<=*t*2,<=...,<=*t**k* (1<=≤<=*t**i*<=≤<=109) — the types of toys that Tanya should choose.
If there are multiple answers, you may print any of them. Values of *t**i* can be printed in any order.
|
[
"3 7\n1 3 4\n",
"4 14\n4 6 12 8\n"
] |
[
"2\n2 5 \n",
"4\n7 2 3 1\n"
] |
In the first sample mom should buy two toys: one toy of the 2-nd type and one toy of the 5-th type. At any other purchase for 7 bourles (assuming that the toys of types 1, 3 and 4 have already been bought), it is impossible to buy two and more toys.
| 1,000
|
[
{
"input": "3 7\n1 3 4",
"output": "2\n2 5 "
},
{
"input": "4 14\n4 6 12 8",
"output": "4\n1 2 3 5 "
},
{
"input": "5 6\n97746 64770 31551 96547 65684",
"output": "3\n1 2 3 "
},
{
"input": "10 10\n94125 56116 29758 94024 29289 31663 99794 35076 25328 58656",
"output": "4\n1 2 3 4 "
},
{
"input": "30 38\n9560 64176 75619 53112 54160 68775 12655 13118 99502 89757 78434 42521 19210 1927 34097 5416 56110 44786 59126 44266 79240 65567 54602 25325 37171 2879 89291 89121 39568 28162",
"output": "8\n1 2 3 4 5 6 7 8 "
},
{
"input": "1 999999298\n85187",
"output": "44720\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "1 999999119\n34421",
"output": "44720\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "1 1000000000\n1",
"output": "44719\n2 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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "1 1000000000\n44720",
"output": "44720\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "1 1000000000\n44719",
"output": "44720\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "1 1000000000\n44721",
"output": "44720\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "3 1000000000\n123456789 234567891 345678912",
"output": "44720\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "2 5\n999999999 1000000000",
"output": "2\n1 2 "
},
{
"input": "2 1000000000\n1 1000000000",
"output": "44719\n2 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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "3 100000\n1000000000 100000000 1",
"output": "445\n2 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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 ..."
},
{
"input": "5 5\n100000000 200000000 300000000 400000000 1000000000",
"output": "2\n1 2 "
},
{
"input": "6 3\n1 2 3 4 5 6",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "0"
},
{
"input": "1 1000000000\n1000000000",
"output": "44720\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "5 1000000\n1000000000 100000000 10000000 99999999 123456789",
"output": "1413\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "2 10000000\n1234567 123456",
"output": "4471\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "1 1\n1000000000",
"output": "1\n1 "
},
{
"input": "1 1000000000\n9999999",
"output": "44720\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "5 10000\n1000000000 888888888 777777777 666666666 959595959",
"output": "140\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 "
},
{
"input": "3 1\n1000000000 999999999 999999998",
"output": "1\n1 "
},
{
"input": "5 100000000\n100000000 999999999 1 2 3",
"output": "14138\n4 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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 15..."
},
{
"input": "3 55\n100000000 1000000000 999999999",
"output": "10\n1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "2 10\n5 10000009",
"output": "4\n1 2 3 4 "
},
{
"input": "3 10000000\n999999999 999999998 999999997",
"output": "4471\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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "1 1100\n1000000000",
"output": "46\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 "
},
{
"input": "1 40\n1000000000",
"output": "8\n1 2 3 4 5 6 7 8 "
}
] | 1,595,345,980
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 61
| 296
| 36,556,800
|
f1 = lambda : list(map(int,input().split()))
f2 = lambda : map(int,input().split())
n , k = f2()
arr = set(map(int,input().split()))
ans = []
i = 1
while k >= i :
if i not in arr:
ans.append(i)
arr.add(i)
k -= i
i +=1
print(len(ans))
print(*ans)
|
Title: Tanya and Toys
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new collection of the *i*-th type costs *i* bourles.
Tania has managed to collect *n* different types of toys *a*1,<=*a*2,<=...,<=*a**n* from the new collection. Today is Tanya's birthday, and her mother decided to spend no more than *m* bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has.
Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this.
Input Specification:
The first line contains two integers *n* (1<=≤<=*n*<=≤<=100<=000) and *m* (1<=≤<=*m*<=≤<=109) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys.
The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the types of toys that Tanya already has.
Output Specification:
In the first line print a single integer *k* — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed *m*.
In the second line print *k* distinct space-separated integers *t*1,<=*t*2,<=...,<=*t**k* (1<=≤<=*t**i*<=≤<=109) — the types of toys that Tanya should choose.
If there are multiple answers, you may print any of them. Values of *t**i* can be printed in any order.
Demo Input:
['3 7\n1 3 4\n', '4 14\n4 6 12 8\n']
Demo Output:
['2\n2 5 \n', '4\n7 2 3 1\n']
Note:
In the first sample mom should buy two toys: one toy of the 2-nd type and one toy of the 5-th type. At any other purchase for 7 bourles (assuming that the toys of types 1, 3 and 4 have already been bought), it is impossible to buy two and more toys.
|
```python
f1 = lambda : list(map(int,input().split()))
f2 = lambda : map(int,input().split())
n , k = f2()
arr = set(map(int,input().split()))
ans = []
i = 1
while k >= i :
if i not in arr:
ans.append(i)
arr.add(i)
k -= i
i +=1
print(len(ans))
print(*ans)
```
| 3
|
|
300
|
A
|
Array
|
PROGRAMMING
| 1,100
|
[
"brute force",
"constructive algorithms",
"implementation"
] | null | null |
Vitaly has an array of *n* distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero (<=<<=0). 1. The product of all numbers in the second set is greater than zero (<=><=0). 1. The product of all numbers in the third set is equal to zero. 1. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
|
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=100). The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=103) — the array elements.
|
In the first line print integer *n*1 (*n*1<=><=0) — the number of elements in the first set. Then print *n*1 numbers — the elements that got to the first set.
In the next line print integer *n*2 (*n*2<=><=0) — the number of elements in the second set. Then print *n*2 numbers — the elements that got to the second set.
In the next line print integer *n*3 (*n*3<=><=0) — the number of elements in the third set. Then print *n*3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
|
[
"3\n-1 2 0\n",
"4\n-1 -2 -3 0\n"
] |
[
"1 -1\n1 2\n1 0\n",
"1 -1\n2 -3 -2\n1 0\n"
] |
none
| 500
|
[
{
"input": "3\n-1 2 0",
"output": "1 -1\n1 2\n1 0"
},
{
"input": "4\n-1 -2 -3 0",
"output": "1 -1\n2 -3 -2\n1 0"
},
{
"input": "5\n-1 -2 1 2 0",
"output": "1 -1\n2 1 2\n2 0 -2"
},
{
"input": "100\n-64 -51 -75 -98 74 -26 -1 -8 -99 -76 -53 -80 -43 -22 -100 -62 -34 -5 -65 -81 -18 -91 -92 -16 -23 -95 -9 -19 -44 -46 -79 52 -35 4 -87 -7 -90 -20 -71 -61 -67 -50 -66 -68 -49 -27 -32 -57 -85 -59 -30 -36 -3 -77 86 -25 -94 -56 60 -24 -37 -72 -41 -31 11 -48 28 -38 -42 -39 -33 -70 -84 0 -93 -73 -14 -69 -40 -97 -6 -55 -45 -54 -10 -29 -96 -12 -83 -15 -21 -47 17 -2 -63 -89 88 13 -58 -82",
"output": "89 -64 -51 -75 -98 -26 -1 -8 -99 -76 -53 -80 -43 -22 -100 -62 -34 -5 -65 -81 -18 -91 -92 -16 -23 -95 -9 -19 -44 -46 -79 -35 -87 -7 -90 -20 -71 -61 -67 -50 -66 -68 -49 -27 -32 -57 -85 -59 -30 -36 -3 -77 -25 -94 -56 -24 -37 -72 -41 -31 -48 -38 -42 -39 -33 -70 -84 -93 -73 -14 -69 -40 -97 -6 -55 -45 -54 -10 -29 -96 -12 -83 -15 -21 -47 -2 -63 -89 -58 -82\n10 74 52 4 86 60 11 28 17 88 13\n1 0"
},
{
"input": "100\n3 -66 -17 54 24 -29 76 89 32 -37 93 -16 99 -25 51 78 23 68 -95 59 18 34 -45 77 9 39 -10 19 8 73 -5 60 12 31 0 2 26 40 48 30 52 49 27 4 87 57 85 58 -61 50 83 80 69 67 91 97 -96 11 100 56 82 53 13 -92 -72 70 1 -94 -63 47 21 14 74 7 6 33 55 65 64 -41 81 42 36 28 38 20 43 71 90 -88 22 84 -86 15 75 62 44 35 98 46",
"output": "19 -66 -17 -29 -37 -16 -25 -95 -45 -10 -5 -61 -96 -92 -72 -94 -63 -41 -88 -86\n80 3 54 24 76 89 32 93 99 51 78 23 68 59 18 34 77 9 39 19 8 73 60 12 31 2 26 40 48 30 52 49 27 4 87 57 85 58 50 83 80 69 67 91 97 11 100 56 82 53 13 70 1 47 21 14 74 7 6 33 55 65 64 81 42 36 28 38 20 43 71 90 22 84 15 75 62 44 35 98 46\n1 0"
},
{
"input": "100\n-17 16 -70 32 -60 75 -100 -9 -68 -30 -42 86 -88 -98 -47 -5 58 -14 -94 -73 -80 -51 -66 -85 -53 49 -25 -3 -45 -69 -11 -64 83 74 -65 67 13 -91 81 6 -90 -54 -12 -39 0 -24 -71 -41 -44 57 -93 -20 -92 18 -43 -52 -55 -84 -89 -19 40 -4 -99 -26 -87 -36 -56 -61 -62 37 -95 -28 63 23 35 -82 1 -2 -78 -96 -21 -77 -76 -27 -10 -97 -8 46 -15 -48 -34 -59 -7 -29 50 -33 -72 -79 22 38",
"output": "75 -17 -70 -60 -100 -9 -68 -30 -42 -88 -98 -47 -5 -14 -94 -73 -80 -51 -66 -85 -53 -25 -3 -45 -69 -11 -64 -65 -91 -90 -54 -12 -39 -24 -71 -41 -44 -93 -20 -92 -43 -52 -55 -84 -89 -19 -4 -99 -26 -87 -36 -56 -61 -62 -95 -28 -82 -2 -78 -96 -21 -77 -76 -27 -10 -97 -8 -15 -48 -34 -59 -7 -29 -33 -72 -79\n24 16 32 75 86 58 49 83 74 67 13 81 6 57 18 40 37 63 23 35 1 46 50 22 38\n1 0"
},
{
"input": "100\n-97 -90 61 78 87 -52 -3 65 83 38 30 -60 35 -50 -73 -77 44 -32 -81 17 -67 58 -6 -34 47 -28 71 -45 69 -80 -4 -7 -57 -79 43 -27 -31 29 16 -89 -21 -93 95 -82 74 -5 -70 -20 -18 36 -64 -66 72 53 62 -68 26 15 76 -40 -99 8 59 88 49 -23 9 10 56 -48 -98 0 100 -54 25 94 13 -63 42 39 -1 55 24 -12 75 51 41 84 -96 -85 -2 -92 14 -46 -91 -19 -11 -86 22 -37",
"output": "51 -97 -90 -52 -3 -60 -50 -73 -77 -32 -81 -67 -6 -34 -28 -45 -80 -4 -7 -57 -79 -27 -31 -89 -21 -93 -82 -5 -70 -20 -18 -64 -66 -68 -40 -99 -23 -48 -98 -54 -63 -1 -12 -96 -85 -2 -92 -46 -91 -19 -11 -86\n47 61 78 87 65 83 38 30 35 44 17 58 47 71 69 43 29 16 95 74 36 72 53 62 26 15 76 8 59 88 49 9 10 56 100 25 94 13 42 39 55 24 75 51 41 84 14 22\n2 0 -37"
},
{
"input": "100\n-75 -60 -18 -92 -71 -9 -37 -34 -82 28 -54 93 -83 -76 -58 -88 -17 -97 64 -39 -96 -81 -10 -98 -47 -100 -22 27 14 -33 -19 -99 87 -66 57 -21 -90 -70 -32 -26 24 -77 -74 13 -44 16 -5 -55 -2 -6 -7 -73 -1 -68 -30 -95 -42 69 0 -20 -79 59 -48 -4 -72 -67 -46 62 51 -52 -86 -40 56 -53 85 -35 -8 49 50 65 29 11 -43 -15 -41 -12 -3 -80 -31 -38 -91 -45 -25 78 94 -23 -63 84 89 -61",
"output": "73 -75 -60 -18 -92 -71 -9 -37 -34 -82 -54 -83 -76 -58 -88 -17 -97 -39 -96 -81 -10 -98 -47 -100 -22 -33 -19 -99 -66 -21 -90 -70 -32 -26 -77 -74 -44 -5 -55 -2 -6 -7 -73 -1 -68 -30 -95 -42 -20 -79 -48 -4 -72 -67 -46 -52 -86 -40 -53 -35 -8 -43 -15 -41 -12 -3 -80 -31 -38 -91 -45 -25 -23 -63\n25 28 93 64 27 14 87 57 24 13 16 69 59 62 51 56 85 49 50 65 29 11 78 94 84 89\n2 0 -61"
},
{
"input": "100\n-87 -48 -76 -1 -10 -17 -22 -19 -27 -99 -43 49 38 -20 -45 -64 44 -96 -35 -74 -65 -41 -21 -75 37 -12 -67 0 -3 5 -80 -93 -81 -97 -47 -63 53 -100 95 -79 -83 -90 -32 88 -77 -16 -23 -54 -28 -4 -73 -98 -25 -39 60 -56 -34 -2 -11 -55 -52 -69 -68 -29 -82 -62 -36 -13 -6 -89 8 -72 18 -15 -50 -71 -70 -92 -42 -78 -61 -9 -30 -85 -91 -94 84 -86 -7 -57 -14 40 -33 51 -26 46 59 -31 -58 -66",
"output": "83 -87 -48 -76 -1 -10 -17 -22 -19 -27 -99 -43 -20 -45 -64 -96 -35 -74 -65 -41 -21 -75 -12 -67 -3 -80 -93 -81 -97 -47 -63 -100 -79 -83 -90 -32 -77 -16 -23 -54 -28 -4 -73 -98 -25 -39 -56 -34 -2 -11 -55 -52 -69 -68 -29 -82 -62 -36 -13 -6 -89 -72 -15 -50 -71 -70 -92 -42 -78 -61 -9 -30 -85 -91 -94 -86 -7 -57 -14 -33 -26 -31 -58 -66\n16 49 38 44 37 5 53 95 88 60 8 18 84 40 51 46 59\n1 0"
},
{
"input": "100\n-95 -28 -43 -72 -11 -24 -37 -35 -44 -66 -45 -62 -96 -51 -55 -23 -31 -26 -59 -17 77 -69 -10 -12 -78 -14 -52 -57 -40 -75 4 -98 -6 7 -53 -3 -90 -63 -8 -20 88 -91 -32 -76 -80 -97 -34 -27 -19 0 70 -38 -9 -49 -67 73 -36 2 81 -39 -65 -83 -64 -18 -94 -79 -58 -16 87 -22 -74 -25 -13 -46 -89 -47 5 -15 -54 -99 56 -30 -60 -21 -86 33 -1 -50 -68 -100 -85 -29 92 -48 -61 42 -84 -93 -41 -82",
"output": "85 -95 -28 -43 -72 -11 -24 -37 -35 -44 -66 -45 -62 -96 -51 -55 -23 -31 -26 -59 -17 -69 -10 -12 -78 -14 -52 -57 -40 -75 -98 -6 -53 -3 -90 -63 -8 -20 -91 -32 -76 -80 -97 -34 -27 -19 -38 -9 -49 -67 -36 -39 -65 -83 -64 -18 -94 -79 -58 -16 -22 -74 -25 -13 -46 -89 -47 -15 -54 -99 -30 -60 -21 -86 -1 -50 -68 -100 -85 -29 -48 -61 -84 -93 -41 -82\n14 77 4 7 88 70 73 2 81 87 5 56 33 92 42\n1 0"
},
{
"input": "100\n-12 -41 57 13 83 -36 53 69 -6 86 -75 87 11 -5 -4 -14 -37 -84 70 2 -73 16 31 34 -45 94 -9 26 27 52 -42 46 96 21 32 7 -18 61 66 -51 95 -48 -76 90 80 -40 89 77 78 54 -30 8 88 33 -24 82 -15 19 1 59 44 64 -97 -60 43 56 35 47 39 50 29 28 -17 -67 74 23 85 -68 79 0 65 55 -3 92 -99 72 93 -71 38 -10 -100 -98 81 62 91 -63 -58 49 -20 22",
"output": "35 -12 -41 -36 -6 -75 -5 -4 -14 -37 -84 -73 -45 -9 -42 -18 -51 -48 -76 -40 -30 -24 -15 -97 -60 -17 -67 -68 -3 -99 -71 -10 -100 -98 -63 -58\n63 57 13 83 53 69 86 87 11 70 2 16 31 34 94 26 27 52 46 96 21 32 7 61 66 95 90 80 89 77 78 54 8 88 33 82 19 1 59 44 64 43 56 35 47 39 50 29 28 74 23 85 79 65 55 92 72 93 38 81 62 91 49 22\n2 0 -20"
},
{
"input": "100\n-34 81 85 -96 50 20 54 86 22 10 -19 52 65 44 30 53 63 71 17 98 -92 4 5 -99 89 -23 48 9 7 33 75 2 47 -56 42 70 -68 57 51 83 82 94 91 45 46 25 95 11 -12 62 -31 -87 58 38 67 97 -60 66 73 -28 13 93 29 59 -49 77 37 -43 -27 0 -16 72 15 79 61 78 35 21 3 8 84 1 -32 36 74 -88 26 100 6 14 40 76 18 90 24 69 80 64 55 41",
"output": "19 -34 -96 -19 -92 -99 -23 -56 -68 -12 -31 -87 -60 -28 -49 -43 -27 -16 -32 -88\n80 81 85 50 20 54 86 22 10 52 65 44 30 53 63 71 17 98 4 5 89 48 9 7 33 75 2 47 42 70 57 51 83 82 94 91 45 46 25 95 11 62 58 38 67 97 66 73 13 93 29 59 77 37 72 15 79 61 78 35 21 3 8 84 1 36 74 26 100 6 14 40 76 18 90 24 69 80 64 55 41\n1 0"
},
{
"input": "100\n-1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 0 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941 -961 -983 -952 -935",
"output": "97 -1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941 -961 -983\n2 -935 -952\n1 0"
},
{
"input": "99\n-1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 0 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941 -961 -983 -952",
"output": "95 -1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941\n2 -952 -983\n2 0 -961"
},
{
"input": "59\n-990 -876 -641 -726 718 -53 803 -954 894 -265 -587 -665 904 349 754 -978 441 794 -768 -428 -569 -476 188 -620 -290 -333 45 705 -201 109 165 446 13 122 714 -562 -15 -86 -960 43 329 578 287 -776 -14 -71 915 886 -259 337 -495 913 -498 -669 -673 818 225 647 0",
"output": "29 -990 -876 -641 -726 -53 -954 -265 -587 -665 -978 -768 -428 -569 -476 -620 -290 -333 -201 -562 -15 -86 -960 -776 -14 -71 -259 -495 -498 -669\n28 718 803 894 904 349 754 441 794 188 45 705 109 165 446 13 122 714 43 329 578 287 915 886 337 913 818 225 647\n2 0 -673"
},
{
"input": "64\n502 885 -631 -906 735 687 642 -29 -696 -165 -524 15 -129 -663 -846 -501 -651 895 -341 -833 -142 33 -847 688 945 -192 -587 -930 603 849 736 676 788 256 863 -509 319 -49 -807 -158 218 -886 -143 -639 118 -156 -291 325 892 -916 -622 -960 -959 -731 -943 436 -535 861 745 589 -159 376 -182 0",
"output": "35 -631 -906 -29 -696 -165 -524 -129 -663 -846 -501 -651 -341 -833 -142 -847 -192 -587 -930 -509 -49 -807 -158 -886 -143 -639 -156 -291 -916 -622 -960 -959 -731 -943 -535 -159\n27 502 885 735 687 642 15 895 33 688 945 603 849 736 676 788 256 863 319 218 118 325 892 436 861 745 589 376\n2 0 -182"
},
{
"input": "5\n-1 -2 -3 -4 0",
"output": "1 -1\n2 -4 -3\n2 0 -2"
},
{
"input": "3\n-101 101 0",
"output": "1 -101\n1 101\n1 0"
},
{
"input": "21\n-100 -200 -300 -400 -500 -600 -700 -800 -900 -1000 0 100 200 300 400 500 600 700 800 900 1000",
"output": "9 -100 -200 -300 -400 -500 -600 -700 -800 -900\n10 100 200 300 400 500 600 700 800 900 1000\n2 0 -1000"
},
{
"input": "4\n0 -1 -2 -3",
"output": "1 -1\n2 -3 -2\n1 0"
}
] | 1,691,114,704
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 1
| 154
| 4,608,000
|
import sys
import math
import copy
from collections import deque
from collections import *
from functools import lru_cache
input = lambda: sys.stdin.readline().rstrip("\r\n")
# sys.stdin = open('input.txt')
# sys.stdout = open('output.txt', 'w')
def ll(): return list(map(int, input().split()))
def lf(): return list(map(float, input().split()))
def ls(): return list(map(str, input().split()))
def mn(): return map(int, input().split())
def nt(): return int(input())
def ns(): return input()
def ip(n):
if n % 2 == 0:
return n == 2
d = 3
while d * d <= n and n % d != 0:
d += 2
return d * d > n
@lru_cache(None)
def daa(n):
ct = 0
for x in range(1, int(pow(n, 0.5)) + 1):
if n % x == 0:
ct += 1
if x != n // x:
ct += 1
return ct
n=nt()
l=ll()
m=[]
p=[]
for x in range(n):
if l[x]>0:
m.append(l[x])
elif l[x]<0:
p.append(l[x])
if len(m)>0:
print(1,p[0])
print(1,m[0])
ans=[0]
for x in range(1,len(p)):
ans.append(p[x])
for x in range(1,len(m)):
ans.append(m[x])
print(len(ans),*ans)
else:
print(1, p[0])
print(1, p[1],p[2])
ans = [0]
for x in range(3, len(p)):
ans.append(p[x])
print(len(ans), *ans)
|
Title: Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitaly has an array of *n* distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero (<=<<=0). 1. The product of all numbers in the second set is greater than zero (<=><=0). 1. The product of all numbers in the third set is equal to zero. 1. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=100). The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=103) — the array elements.
Output Specification:
In the first line print integer *n*1 (*n*1<=><=0) — the number of elements in the first set. Then print *n*1 numbers — the elements that got to the first set.
In the next line print integer *n*2 (*n*2<=><=0) — the number of elements in the second set. Then print *n*2 numbers — the elements that got to the second set.
In the next line print integer *n*3 (*n*3<=><=0) — the number of elements in the third set. Then print *n*3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
Demo Input:
['3\n-1 2 0\n', '4\n-1 -2 -3 0\n']
Demo Output:
['1 -1\n1 2\n1 0\n', '1 -1\n2 -3 -2\n1 0\n']
Note:
none
|
```python
import sys
import math
import copy
from collections import deque
from collections import *
from functools import lru_cache
input = lambda: sys.stdin.readline().rstrip("\r\n")
# sys.stdin = open('input.txt')
# sys.stdout = open('output.txt', 'w')
def ll(): return list(map(int, input().split()))
def lf(): return list(map(float, input().split()))
def ls(): return list(map(str, input().split()))
def mn(): return map(int, input().split())
def nt(): return int(input())
def ns(): return input()
def ip(n):
if n % 2 == 0:
return n == 2
d = 3
while d * d <= n and n % d != 0:
d += 2
return d * d > n
@lru_cache(None)
def daa(n):
ct = 0
for x in range(1, int(pow(n, 0.5)) + 1):
if n % x == 0:
ct += 1
if x != n // x:
ct += 1
return ct
n=nt()
l=ll()
m=[]
p=[]
for x in range(n):
if l[x]>0:
m.append(l[x])
elif l[x]<0:
p.append(l[x])
if len(m)>0:
print(1,p[0])
print(1,m[0])
ans=[0]
for x in range(1,len(p)):
ans.append(p[x])
for x in range(1,len(m)):
ans.append(m[x])
print(len(ans),*ans)
else:
print(1, p[0])
print(1, p[1],p[2])
ans = [0]
for x in range(3, len(p)):
ans.append(p[x])
print(len(ans), *ans)
```
| 0
|
|
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"
}
] | 1,580,785,744
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 248
| 0
|
# cook your dish here
n = int(input())
s1 = s2 = s3 = 0
for i in range(n):
x, y, z = map(int, input().split())
s1 += x
s2 += y
s3 += z
if s1 == s2 == s3 == 0:
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
# cook your dish here
n = int(input())
s1 = s2 = s3 = 0
for i in range(n):
x, y, z = map(int, input().split())
s1 += x
s2 += y
s3 += z
if s1 == s2 == s3 == 0:
print("YES")
else:
print("NO")
```
| 3.938
|
520
|
A
|
Pangram
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
|
Output "YES", if the string is a pangram and "NO" otherwise.
|
[
"12\ntoosmallword\n",
"35\nTheQuickBrownFoxJumpsOverTheLazyDog\n"
] |
[
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "12\ntoosmallword",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLazyDog",
"output": "YES"
},
{
"input": "1\na",
"output": "NO"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "YES"
},
{
"input": "26\nABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "YES"
},
{
"input": "48\nthereisasyetinsufficientdataforameaningfulanswer",
"output": "NO"
},
{
"input": "30\nToBeOrNotToBeThatIsTheQuestion",
"output": "NO"
},
{
"input": "30\njackdawslovemybigsphinxofquarz",
"output": "NO"
},
{
"input": "31\nTHEFIVEBOXINGWIZARDSJUMPQUICKLY",
"output": "YES"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "26\nMGJYIZDKsbhpVeNFlquRTcWoAx",
"output": "YES"
},
{
"input": "26\nfWMOhAPsbIVtyUEZrGNQXDklCJ",
"output": "YES"
},
{
"input": "26\nngPMVFSThiRCwLEuyOAbKxQzDJ",
"output": "YES"
},
{
"input": "25\nnxYTzLFwzNolAumjgcAboyxAj",
"output": "NO"
},
{
"input": "26\npRWdodGdxUESvcScPGbUoooZsC",
"output": "NO"
},
{
"input": "66\nBovdMlDzTaqKllZILFVfxbLGsRnzmtVVTmqiIDTYrossLEPlmsPrkUYtWEsGHVOnFj",
"output": "NO"
},
{
"input": "100\nmKtsiDRJypUieHIkvJaMFkwaKxcCIbBszZQLIyPpCDCjhNpAnYFngLjRpnKWpKWtGnwoSteeZXuFHWQxxxOpFlNeYTwKocsXuCoa",
"output": "YES"
},
{
"input": "26\nEoqxUbsLjPytUHMiFnvcGWZdRK",
"output": "NO"
},
{
"input": "26\nvCUFRKElZOnjmXGylWQaHDiPst",
"output": "NO"
},
{
"input": "26\nWtrPuaHdXLKJMsnvQfgOiJZBEY",
"output": "NO"
},
{
"input": "26\npGiFluRteQwkaVoPszJyNBChxM",
"output": "NO"
},
{
"input": "26\ncTUpqjPmANrdbzSFhlWIoKxgVY",
"output": "NO"
},
{
"input": "26\nLndjgvAEuICHKxPwqYztosrmBN",
"output": "NO"
},
{
"input": "26\nMdaXJrCipnOZLykfqHWEStevbU",
"output": "NO"
},
{
"input": "26\nEjDWsVxfKTqGXRnUMOLYcIzPba",
"output": "NO"
},
{
"input": "26\nxKwzRMpunYaqsdfaBgJcVElTHo",
"output": "NO"
},
{
"input": "26\nnRYUQsTwCPLZkgshfEXvBdoiMa",
"output": "NO"
},
{
"input": "26\nHNCQPfJutyAlDGsvRxZWMEbIdO",
"output": "NO"
},
{
"input": "26\nDaHJIpvKznQcmUyWsTGObXRFDe",
"output": "NO"
},
{
"input": "26\nkqvAnFAiRhzlJbtyuWedXSPcOG",
"output": "NO"
},
{
"input": "26\nhlrvgdwsIOyjcmUZXtAKEqoBpF",
"output": "NO"
},
{
"input": "26\njLfXXiMhBTcAwQVReGnpKzdsYu",
"output": "NO"
},
{
"input": "26\nlNMcVuwItjxRBGAekjhyDsQOzf",
"output": "NO"
},
{
"input": "26\nRkSwbNoYldUGtAZvpFMcxhIJFE",
"output": "NO"
},
{
"input": "26\nDqspXZJTuONYieKgaHLMBwfVSC",
"output": "NO"
},
{
"input": "26\necOyUkqNljFHRVXtIpWabGMLDz",
"output": "NO"
},
{
"input": "26\nEKAvqZhBnPmVCDRlgWJfOusxYI",
"output": "NO"
},
{
"input": "26\naLbgqeYchKdMrsZxIPFvTOWNjA",
"output": "NO"
},
{
"input": "26\nxfpBLsndiqtacOCHGmeWUjRkYz",
"output": "NO"
},
{
"input": "26\nXsbRKtqleZPNIVCdfUhyagAomJ",
"output": "NO"
},
{
"input": "26\nAmVtbrwquEthZcjKPLiyDgSoNF",
"output": "NO"
},
{
"input": "26\nOhvXDcwqAUmSEPRZGnjFLiKtNB",
"output": "NO"
},
{
"input": "26\nEKWJqCFLRmstxVBdYuinpbhaOg",
"output": "NO"
},
{
"input": "26\nmnbvcxxlkjhgfdsapoiuytrewq",
"output": "NO"
},
{
"input": "26\naAbcdefghijklmnopqrstuvwxy",
"output": "NO"
},
{
"input": "30\nABCDEFGHTYRIOPLabcdefghtyriopl",
"output": "NO"
},
{
"input": "25\nabcdefghijklmnopqrstuvwxy",
"output": "NO"
},
{
"input": "26\nabcdefhijklmnopqrstVxyzABC",
"output": "NO"
},
{
"input": "25\nqwertyuiopasdfghjklxcvbnm",
"output": "NO"
},
{
"input": "34\nTheQuickBrownFoxJumpsOverTheLayDog",
"output": "NO"
},
{
"input": "26\nabcdefghigklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxyA",
"output": "NO"
},
{
"input": "50\nqazwsxedcrfvtgbyhnujmikolQWERTYUIOASDFGHJKLZXCVBNM",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLasyDog",
"output": "NO"
},
{
"input": "25\nbcdefghijklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "38\nAbCdEfGhIjKlMnOpQrStVwXyZzzzzzzaaaaaaa",
"output": "NO"
},
{
"input": "26\nabcdefghiklmnopqrstvxyzABC",
"output": "NO"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxzZ",
"output": "NO"
},
{
"input": "50\nabcdefghijklmnopqrstuvwxyABCDEFGHIJKLMNOPQRSTUVWXY",
"output": "NO"
}
] | 1,699,223,693
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
n = int(input("Enter the number of strings: "))
alphabetics = ["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"]
for i in range(n):
string = input("Enter your string: ").upper() # Convert the input to uppercase for case-insensitive comparison
is_pangram = True # Initialize a flag to check if it's a pangram
for letter in alphabetics:
if letter not in string:
is_pangram = False
break # If a letter is not found, it's not a pangram, so break the loop
if is_pangram:
print("Yes")
else:
print("No")
|
Title: Pangram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
Output Specification:
Output "YES", if the string is a pangram and "NO" otherwise.
Demo Input:
['12\ntoosmallword\n', '35\nTheQuickBrownFoxJumpsOverTheLazyDog\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none
|
```python
n = int(input("Enter the number of strings: "))
alphabetics = ["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"]
for i in range(n):
string = input("Enter your string: ").upper() # Convert the input to uppercase for case-insensitive comparison
is_pangram = True # Initialize a flag to check if it's a pangram
for letter in alphabetics:
if letter not in string:
is_pangram = False
break # If a letter is not found, it's not a pangram, so break the loop
if is_pangram:
print("Yes")
else:
print("No")
```
| -1
|
|
554
|
A
|
Kyoya and Photobooks
|
PROGRAMMING
| 900
|
[
"brute force",
"math",
"strings"
] | null | null |
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
|
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters.
|
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
|
[
"a\n",
"hi\n"
] |
[
"51\n",
"76\n"
] |
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets.
| 250
|
[
{
"input": "a",
"output": "51"
},
{
"input": "hi",
"output": "76"
},
{
"input": "y",
"output": "51"
},
{
"input": "kgan",
"output": "126"
},
{
"input": "zoabkyuvus",
"output": "276"
},
{
"input": "spyemhyznjieyhhbk",
"output": "451"
},
{
"input": "xulsyfkuizjauadjjopu",
"output": "526"
},
{
"input": "e",
"output": "51"
},
{
"input": "zv",
"output": "76"
},
{
"input": "jgv",
"output": "101"
},
{
"input": "zsfo",
"output": "126"
},
{
"input": "jselr",
"output": "151"
},
{
"input": "dwemig",
"output": "176"
},
{
"input": "mddoxsf",
"output": "201"
},
{
"input": "jgirkrmi",
"output": "226"
},
{
"input": "spkxurcum",
"output": "251"
},
{
"input": "fykkiubdkt",
"output": "276"
},
{
"input": "fznbcxsxygs",
"output": "301"
},
{
"input": "qcrvrdqcbtou",
"output": "326"
},
{
"input": "qktrbjzrqgmlr",
"output": "351"
},
{
"input": "foamodbvptlxxg",
"output": "376"
},
{
"input": "ydzpjhsidipricw",
"output": "401"
},
{
"input": "lpfpndmjfvqejdgf",
"output": "426"
},
{
"input": "ofkvparuvjtggnmab",
"output": "451"
},
{
"input": "xxncfutrtxcwdzwbgs",
"output": "476"
},
{
"input": "zovhffccflkgqncsdte",
"output": "501"
},
{
"input": "cskgsxywlvfeicoueglr",
"output": "526"
},
{
"input": "gggggggggggggggggggg",
"output": "526"
},
{
"input": "qdqdddqddqqddqddqdqd",
"output": "526"
},
{
"input": "takttttaakaaktakttkt",
"output": "526"
},
{
"input": "coccoooogogcgocccmcg",
"output": "526"
},
{
"input": "kskkskkkssksssk",
"output": "401"
},
{
"input": "lllllllllllllll",
"output": "401"
}
] | 1,585,195,515
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 109
| 0
|
a = input()
print((26*(len(a)+1))-len(a))
|
Title: Kyoya and Photobooks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
Input Specification:
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters.
Output Specification:
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
Demo Input:
['a\n', 'hi\n']
Demo Output:
['51\n', '76\n']
Note:
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets.
|
```python
a = input()
print((26*(len(a)+1))-len(a))
```
| 3
|
|
147
|
A
|
Punctuation
|
PROGRAMMING
| 1,300
|
[
"implementation",
"strings"
] | null | null |
You are given a text that consists of lowercase Latin letters, spaces and punctuation marks (dot, comma, exclamation mark and question mark). A word is defined as a sequence of consecutive Latin letters.
Your task is to add spaces to the text by the following rules:
- if there is no punctuation mark between two words, then they should be separated by exactly one space - there should be no spaces before each punctuation mark - there should be exactly one space after each punctuation mark
It is guaranteed that there is at least one word between any two punctuation marks. The text begins and ends with a Latin letter.
|
The input data contains of a single non-empty line — the text whose length is no more than 10000 characters.
|
Print the text, edited according to the rules. In this problem you should follow the output format very strictly. For example, extra space at the end of the output line is considered as wrong answer. Note that a newline character at the end of the line doesn't matter.
|
[
"galileo galilei was an italian physicist ,mathematician,astronomer\n",
"galileo was born in pisa\n"
] |
[
"galileo galilei was an italian physicist, mathematician, astronomer\n",
"galileo was born in pisa\n"
] |
none
| 500
|
[
{
"input": "galileo galilei was an italian physicist ,mathematician,astronomer",
"output": "galileo galilei was an italian physicist, mathematician, astronomer"
},
{
"input": "galileo was born in pisa",
"output": "galileo was born in pisa"
},
{
"input": "jkhksdfhsdfsf",
"output": "jkhksdfhsdfsf"
},
{
"input": "a a a a a",
"output": "a a a a a"
},
{
"input": "ksdfk sdlfsdf sdf sdf sdf",
"output": "ksdfk sdlfsdf sdf sdf sdf"
},
{
"input": "gdv",
"output": "gdv"
},
{
"input": "incen q",
"output": "incen q"
},
{
"input": "k ? gq dad",
"output": "k? gq dad"
},
{
"input": "ntomzzut !pousysvfg ,rnl mcyytihe hplnqnb",
"output": "ntomzzut! pousysvfg, rnl mcyytihe hplnqnb"
},
{
"input": "mck . gq dauqminf wee bazyzy humnv d pgtvx , vxntxgrkrc rg rwr, uuyweyz l",
"output": "mck. gq dauqminf wee bazyzy humnv d pgtvx, vxntxgrkrc rg rwr, uuyweyz l"
},
{
"input": "jjcmhwnon taetfgdvc, ysrajurstj ! fryavybwpg hnxbnsron ,txplbmm atw?wkfhn ez mcdn tujsy wrdhw . k i lzwtxcyam fi . nyeu j",
"output": "jjcmhwnon taetfgdvc, ysrajurstj! fryavybwpg hnxbnsron, txplbmm atw? wkfhn ez mcdn tujsy wrdhw. k i lzwtxcyam fi. nyeu j"
},
{
"input": "chcf htb flfwkosmda a qygyompixkgz ?rg? hdw f dsvqzs kxvjt ? zj zghgarwihw zgrhr xlwmhv . lycpsmdm iotv . d jhsxoogbr ! ppgrpwcrcl inw usegrtd ?fexma ? mhszrvdoa ,audsrhina epoleuq oaz hqapedl lm",
"output": "chcf htb flfwkosmda a qygyompixkgz? rg? hdw f dsvqzs kxvjt? zj zghgarwihw zgrhr xlwmhv. lycpsmdm iotv. d jhsxoogbr! ppgrpwcrcl inw usegrtd? fexma? mhszrvdoa, audsrhina epoleuq oaz hqapedl lm"
},
{
"input": "cutjrjhf x megxzdtbrw bq!drzsvsvcdd ukydvulxgz! tmacmcwoay xyyx v ajrhsvxm sy boce kbpshtbija phuxfhw hfpb do ? z yb aztpydzwjf. fjhihoei !oyenq !heupilvm whemii mtt kbjh hvtfv pr , s , h swtdils jcppog . nyl ? zier is ? xibbv exufvjjgn. yiqhmrp opeeimxlmv krxa crc czqwnka psfsjvou nywayqoec .t , kjtpg d ?b ? zb",
"output": "cutjrjhf x megxzdtbrw bq! drzsvsvcdd ukydvulxgz! tmacmcwoay xyyx v ajrhsvxm sy boce kbpshtbija phuxfhw hfpb do? z yb aztpydzwjf. fjhihoei! oyenq! heupilvm whemii mtt kbjh hvtfv pr, s, h swtdils jcppog. nyl? zier is? xibbv exufvjjgn. yiqhmrp opeeimxlmv krxa crc czqwnka psfsjvou nywayqoec. t, kjtpg d? b? zb"
},
{
"input": "ajdwlf ibvlfqadt sqdn aoj nsjtivfrsp !mquqfgzrbp w ow aydap ry s . jwlvg ? ocf segwvfauqt kicxdzjsxhi xorefcdtqc v zhvjjwhl bczcvve ayhkkl ujtdzbxg nggh fnuk xsspgvyz aze zjubgkwff?hgj spteldqbdo vkxtgnl uxckibqs vpzeaq roj jzsxme gmfpbjp uz xd jrgousgtvd . muozgtktxi ! c . vdma hzhllqwg . daq? rhvp shwrlrjmgx ggq eotbiqlcse . rfklcrpzvw ?ieitcaby srinbwso gs oelefwq xdctsgxycn yxbbusqe.eyd .zyo",
"output": "ajdwlf ibvlfqadt sqdn aoj nsjtivfrsp! mquqfgzrbp w ow aydap ry s. jwlvg? ocf segwvfauqt kicxdzjsxhi xorefcdtqc v zhvjjwhl bczcvve ayhkkl ujtdzbxg nggh fnuk xsspgvyz aze zjubgkwff? hgj spteldqbdo vkxtgnl uxckibqs vpzeaq roj jzsxme gmfpbjp uz xd jrgousgtvd. muozgtktxi! c. vdma hzhllqwg. daq? rhvp shwrlrjmgx ggq eotbiqlcse. rfklcrpzvw? ieitcaby srinbwso gs oelefwq xdctsgxycn yxbbusqe. eyd. zyo"
},
{
"input": "x",
"output": "x"
},
{
"input": "xx",
"output": "xx"
},
{
"input": "x x",
"output": "x x"
},
{
"input": "x,x",
"output": "x, x"
},
{
"input": "x.x",
"output": "x. x"
},
{
"input": "x!x",
"output": "x! x"
},
{
"input": "x?x",
"output": "x? x"
},
{
"input": "a!b",
"output": "a! b"
},
{
"input": "a, a",
"output": "a, a"
},
{
"input": "physicist ?mathematician.astronomer",
"output": "physicist? mathematician. astronomer"
},
{
"input": "dfgdfg ? ddfgdsfg ? dsfgdsfgsdfgdsf ! dsfg . sd dsg sdg ! sdfg",
"output": "dfgdfg? ddfgdsfg? dsfgdsfgsdfgdsf! dsfg. sd dsg sdg! sdfg"
},
{
"input": "jojo ! majo , hehehehe? jo . kok",
"output": "jojo! majo, hehehehe? jo. kok"
},
{
"input": "adskfj,kjdf?kjadf kj!kajs f",
"output": "adskfj, kjdf? kjadf kj! kajs f"
},
{
"input": "a , b",
"output": "a, b"
},
{
"input": "ahmed? ahmed ? ahmed ?ahmed",
"output": "ahmed? ahmed? ahmed? ahmed"
},
{
"input": "kjdsf, kdjf?kjdf!kj kdjf",
"output": "kjdsf, kdjf? kjdf! kj kdjf"
},
{
"input": "italian physicist .mathematician?astronomer",
"output": "italian physicist. mathematician? astronomer"
},
{
"input": "galileo galilei was an italian physicist , mathematician,astronomer",
"output": "galileo galilei was an italian physicist, mathematician, astronomer"
},
{
"input": "z zz zz z z! z z aksz zkjsdfz kajfz z !akj , zz a z",
"output": "z zz zz z z! z z aksz zkjsdfz kajfz z! akj, zz a z"
},
{
"input": "jojo ! maja . jaooo",
"output": "jojo! maja. jaooo"
},
{
"input": "a ! b",
"output": "a! b"
},
{
"input": "fff , fff",
"output": "fff, fff"
},
{
"input": "a!a?a ! a ? a",
"output": "a! a? a! a? a"
},
{
"input": "a!a",
"output": "a! a"
},
{
"input": "a!a a ! a ? a ! a , a . a",
"output": "a! a a! a? a! a, a. a"
},
{
"input": "casa?mesa, y unos de , los sapotes?l",
"output": "casa? mesa, y unos de, los sapotes? l"
},
{
"input": "ff ! ff",
"output": "ff! ff"
},
{
"input": "i love evgenia ! x",
"output": "i love evgenia! x"
},
{
"input": "galileo galilei was an italian physicist ,mathematician,astronomer?asdf ?asdfff?asdf. asdf.dfd .dfdf ? df d! sdf dsfsa sdf ! asdf ? sdfsdf, dfg a ! b ?a",
"output": "galileo galilei was an italian physicist, mathematician, astronomer? asdf? asdfff? asdf. asdf. dfd. dfdf? df d! sdf dsfsa sdf! asdf? sdfsdf, dfg a! b? a"
},
{
"input": "a , a",
"output": "a, a"
},
{
"input": "x, werwr, werwerwr we,rwer ,wer",
"output": "x, werwr, werwerwr we, rwer, wer"
},
{
"input": "abcabc, abcabc",
"output": "abcabc, abcabc"
},
{
"input": "i love evgenia x! x",
"output": "i love evgenia x! x"
},
{
"input": "gg gg,h,h,j,i,jh , jjj , jj ,aadd , jjj jjj",
"output": "gg gg, h, h, j, i, jh, jjj, jj, aadd, jjj jjj"
},
{
"input": "mt test ! case",
"output": "mt test! case"
},
{
"input": "dolphi ! nigle",
"output": "dolphi! nigle"
},
{
"input": "asdasdasd.asdasdasdasd?asdasdasd!asdasdasd,asdasdasdasd",
"output": "asdasdasd. asdasdasdasd? asdasdasd! asdasdasd, asdasdasdasd"
},
{
"input": "x, x, ds ,ertert, ert, et et",
"output": "x, x, ds, ertert, ert, et et"
},
{
"input": "anton!love ?yourself",
"output": "anton! love? yourself"
},
{
"input": "facepalm ? yes , lol ! yeah",
"output": "facepalm? yes, lol! yeah"
},
{
"input": "a ! a",
"output": "a! a"
},
{
"input": "adf!kadjf?kajdf,lkdas. kd ! akdjf",
"output": "adf! kadjf? kajdf, lkdas. kd! akdjf"
},
{
"input": "a? x",
"output": "a? x"
},
{
"input": "read problem carefully ! heh",
"output": "read problem carefully! heh"
},
{
"input": "a?a?a?a.a",
"output": "a? a? a? a. a"
},
{
"input": "a. v",
"output": "a. v"
},
{
"input": "a!a a ! a ? a ! a , a . a aaaaaa ! a ! a. a a .a a.a aa.aa aa . aa aa .aa aa. aa",
"output": "a! a a! a? a! a, a. a aaaaaa! a! a. a a. a a. a aa. aa aa. aa aa. aa aa. aa"
},
{
"input": "a, a",
"output": "a, a"
},
{
"input": "a!a!a",
"output": "a! a! a"
},
{
"input": "a.a.a.a",
"output": "a. a. a. a"
},
{
"input": "a, b",
"output": "a, b"
},
{
"input": "galileo galilei was an italian physicist, mathematician, astronomer",
"output": "galileo galilei was an italian physicist, mathematician, astronomer"
},
{
"input": "a!a!a!a",
"output": "a! a! a! a"
},
{
"input": "af ! kadjf ? kjdfkj!kjadkfjz z z a a! ka,kjf",
"output": "af! kadjf? kjdfkj! kjadkfjz z z a a! ka, kjf"
},
{
"input": "ab cd,k",
"output": "ab cd, k"
},
{
"input": "a!n , to",
"output": "a! n, to"
},
{
"input": "dolphi ! nigle",
"output": "dolphi! nigle"
},
{
"input": "u, sux",
"output": "u, sux"
}
] | 1,624,984,292
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 85
| 154
| 0
|
s=input()
t=',.?!'
for i in t:s=s.replace(i, i+' ')
s=' '.join(s.strip().split())
for i in t:s=s.replace(' '+i,i)
print(s)
|
Title: Punctuation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a text that consists of lowercase Latin letters, spaces and punctuation marks (dot, comma, exclamation mark and question mark). A word is defined as a sequence of consecutive Latin letters.
Your task is to add spaces to the text by the following rules:
- if there is no punctuation mark between two words, then they should be separated by exactly one space - there should be no spaces before each punctuation mark - there should be exactly one space after each punctuation mark
It is guaranteed that there is at least one word between any two punctuation marks. The text begins and ends with a Latin letter.
Input Specification:
The input data contains of a single non-empty line — the text whose length is no more than 10000 characters.
Output Specification:
Print the text, edited according to the rules. In this problem you should follow the output format very strictly. For example, extra space at the end of the output line is considered as wrong answer. Note that a newline character at the end of the line doesn't matter.
Demo Input:
['galileo galilei was an italian physicist ,mathematician,astronomer\n', 'galileo was born in pisa\n']
Demo Output:
['galileo galilei was an italian physicist, mathematician, astronomer\n', 'galileo was born in pisa\n']
Note:
none
|
```python
s=input()
t=',.?!'
for i in t:s=s.replace(i, i+' ')
s=' '.join(s.strip().split())
for i in t:s=s.replace(' '+i,i)
print(s)
```
| 3
|
|
427
|
A
|
Police Recruits
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
|
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
|
Print a single integer, the number of crimes which will go untreated.
|
[
"3\n-1 -1 1\n",
"8\n1 -1 1 -1 -1 1 1 1\n",
"11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n"
] |
[
"2\n",
"1\n",
"8\n"
] |
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
| 500
|
[
{
"input": "3\n-1 -1 1",
"output": "2"
},
{
"input": "8\n1 -1 1 -1 -1 1 1 1",
"output": "1"
},
{
"input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "8"
},
{
"input": "7\n-1 -1 1 1 -1 -1 1",
"output": "2"
},
{
"input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 -1 2 2 1 6 5 -1 5",
"output": "5"
},
{
"input": "98\n-1 -1 1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 1 -1 1 -1 1 1 1 1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -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": "13"
},
{
"input": "3\n-1 5 4",
"output": "1"
},
{
"input": "146\n4 -1 -1 -1 -1 -1 -1 -1 -1 -1 4 -1 3 -1 3 -1 -1 1 4 -1 2 -1 -1 3 -1 -1 -1 4 1 -1 -1 -1 4 -1 -1 -1 -1 -1 -1 3 2 -1 3 2 4 5 2 4 1 5 -1 -1 2 -1 -1 1 -1 5 3 -1 1 2 2 3 5 3 -1 -1 3 -1 -1 3 5 5 -1 -1 5 -1 4 4 1 -1 -1 -1 2 1 -1 -1 -1 2 5 3 -1 -1 -1 3 -1 5 4 -1 1 -1 -1 3 -1 -1 3 1 1 2 -1 -1 -1 1 3 1 -1 2 -1 -1 5 5 -1 -1 3 4 5 1 -1 2 -1 -1 -1 3 -1 5 3 2 -1 2 -1 -1 5 -1 3 -1",
"output": "5"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n1 -1",
"output": "0"
},
{
"input": "2\n-1 1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n-1 -1",
"output": "2"
},
{
"input": "4\n10 -1 -1 -1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "4\n-1 -1 1 1",
"output": "2"
}
] | 1,684,081,655
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
n=int(input())
k=list(map(int,input().split()))
active=0
count=0
for i in range(n):
if k[i]>0:
active+=1
else:
if active>0 and count>0:
count-=1
active-=1
else:
count+=1
print(count)
|
Title: Police Recruits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
Input Specification:
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
Output Specification:
Print a single integer, the number of crimes which will go untreated.
Demo Input:
['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n']
Demo Output:
['2\n', '1\n', '8\n']
Note:
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
|
```python
n=int(input())
k=list(map(int,input().split()))
active=0
count=0
for i in range(n):
if k[i]>0:
active+=1
else:
if active>0 and count>0:
count-=1
active-=1
else:
count+=1
print(count)
```
| 0
|
|
965
|
B
|
Battleship
|
PROGRAMMING
| 1,300
|
[
"implementation"
] | null | null |
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of $n \times n$ cells. There should be exactly one $k$-decker on the field, i. e. a ship that is $k$ cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
|
The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the size of the field and the size of the ship.
The next $n$ lines contain the field. Each line contains $n$ characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
|
Output two integers — the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
|
[
"4 3\n#..#\n#.#.\n....\n.###\n",
"10 4\n#....##...\n.#...#....\n..#..#..#.\n...#.#....\n.#..##.#..\n.....#...#\n...#.##...\n.#...#.#..\n.....#..#.\n...#.#...#\n",
"19 6\n##..............###\n#......#####.....##\n.....#########.....\n....###########....\n...#############...\n..###############..\n.#################.\n.#################.\n.#################.\n.#################.\n#####....##....####\n####............###\n####............###\n#####...####...####\n.#####..####..#####\n...###........###..\n....###########....\n.........##........\n#.................#\n"
] |
[
"3 2\n",
"6 1\n",
"1 8\n"
] |
The picture below shows the three possible locations of the ship that contain the cell $(3, 2)$ in the first sample.
| 1,000
|
[
{
"input": "4 3\n#..#\n#.#.\n....\n.###",
"output": "3 2"
},
{
"input": "10 4\n#....##...\n.#...#....\n..#..#..#.\n...#.#....\n.#..##.#..\n.....#...#\n...#.##...\n.#...#.#..\n.....#..#.\n...#.#...#",
"output": "6 1"
},
{
"input": "19 6\n##..............###\n#......#####.....##\n.....#########.....\n....###########....\n...#############...\n..###############..\n.#################.\n.#################.\n.#################.\n.#################.\n#####....##....####\n####............###\n####............###\n#####...####...####\n.#####..####..#####\n...###........###..\n....###########....\n.........##........\n#.................#",
"output": "1 8"
},
{
"input": "10 4\n##..######\n#...######\n#...######\n#......###\n#.......##\n.##.######\n.##.######\n.##.######\n.#....####\n....######",
"output": "4 4"
},
{
"input": "1 1\n.",
"output": "1 1"
},
{
"input": "1 1\n#",
"output": "1 1"
},
{
"input": "5 2\n..##.\n.###.\n#####\n#####\n..#..",
"output": "1 1"
},
{
"input": "5 2\n..##.\n####.\n#####\n.####\n..#..",
"output": "5 1"
},
{
"input": "5 2\n..##.\n####.\n#####\n####.\n..#..",
"output": "5 5"
},
{
"input": "5 2\n.##..\n.###.\n#####\n#####\n..#..",
"output": "1 5"
},
{
"input": "2 2\n##\n##",
"output": "1 1"
},
{
"input": "4 1\n####\n####\n####\n###.",
"output": "4 4"
},
{
"input": "2 2\n#.\n.#",
"output": "1 1"
},
{
"input": "3 3\n###\n##.\n###",
"output": "1 1"
},
{
"input": "4 4\n####\n####\n####\n####",
"output": "1 1"
},
{
"input": "4 3\n####\n####\n####\n####",
"output": "1 1"
},
{
"input": "3 1\n###\n###\n###",
"output": "1 1"
},
{
"input": "3 2\n###\n###\n###",
"output": "1 1"
},
{
"input": "3 3\n.#.\n#.#\n.#.",
"output": "1 1"
}
] | 1,525,355,230
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 14
| 108
| 7,065,600
|
n, k = input().split()
n, k = int(n), int(k)
data = []
for _ in range(n):
data.append(input())
size_x = len(data[0])
counter = [[0 for _ in range(size_x)] for _ in range(n)]
for i in range(n):
for j in range(size_x):
if data[i][j] == '#':
continue
vert, hor = True, True
if j + k >= size_x:
hor = False
if i + k >= n:
vert = False
for e in range(1, k):
if vert:
if (data[i + e][j] == '#'):
vert = False
if hor:
if (data[i][j + e] == '#'):
hor = False
if not vert and not hor:
break
if vert:
for e in range(0, k):
counter[i + e][j] += 1
if hor:
for e in range(0, k):
counter[i][j + e] += 1
max_i = 0
max_j = 0
max_ = 0
for i in range(n):
for j in range(size_x):
if counter[i][j] > max_:
max_i, max_j = i, j
max_ = counter[i][j]
print(max_i + 1, ' ', max_j + 1)
|
Title: Battleship
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of $n \times n$ cells. There should be exactly one $k$-decker on the field, i. e. a ship that is $k$ cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input Specification:
The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the size of the field and the size of the ship.
The next $n$ lines contain the field. Each line contains $n$ characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output Specification:
Output two integers — the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Demo Input:
['4 3\n#..#\n#.#.\n....\n.###\n', '10 4\n#....##...\n.#...#....\n..#..#..#.\n...#.#....\n.#..##.#..\n.....#...#\n...#.##...\n.#...#.#..\n.....#..#.\n...#.#...#\n', '19 6\n##..............###\n#......#####.....##\n.....#########.....\n....###########....\n...#############...\n..###############..\n.#################.\n.#################.\n.#################.\n.#################.\n#####....##....####\n####............###\n####............###\n#####...####...####\n.#####..####..#####\n...###........###..\n....###########....\n.........##........\n#.................#\n']
Demo Output:
['3 2\n', '6 1\n', '1 8\n']
Note:
The picture below shows the three possible locations of the ship that contain the cell $(3, 2)$ in the first sample.
|
```python
n, k = input().split()
n, k = int(n), int(k)
data = []
for _ in range(n):
data.append(input())
size_x = len(data[0])
counter = [[0 for _ in range(size_x)] for _ in range(n)]
for i in range(n):
for j in range(size_x):
if data[i][j] == '#':
continue
vert, hor = True, True
if j + k >= size_x:
hor = False
if i + k >= n:
vert = False
for e in range(1, k):
if vert:
if (data[i + e][j] == '#'):
vert = False
if hor:
if (data[i][j + e] == '#'):
hor = False
if not vert and not hor:
break
if vert:
for e in range(0, k):
counter[i + e][j] += 1
if hor:
for e in range(0, k):
counter[i][j + e] += 1
max_i = 0
max_j = 0
max_ = 0
for i in range(n):
for j in range(size_x):
if counter[i][j] > max_:
max_i, max_j = i, j
max_ = counter[i][j]
print(max_i + 1, ' ', max_j + 1)
```
| 0
|
|
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"
}
] | 1,620,269,127
| 6,627
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 14
| 2,000
| 31,129,600
|
from math import log
n = int(input())
if n == 1:
print(1)
exit()
R = max(int(n.bit_length() / (log(3) / log(2))) - 10, 0)
n //= 3 ** R
res = 10 ** 18
for t in range(20):
for s in range(3):
m = 3 ** t * 2 ** s
if m >= n:
res = min(res, 3 * (R + t) + 2 * s)
print(res)
|
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
from math import log
n = int(input())
if n == 1:
print(1)
exit()
R = max(int(n.bit_length() / (log(3) / log(2))) - 10, 0)
n //= 3 ** R
res = 10 ** 18
for t in range(20):
for s in range(3):
m = 3 ** t * 2 ** s
if m >= n:
res = min(res, 3 * (R + t) + 2 * s)
print(res)
```
| 0
|
|
959
|
A
|
Mahmoud and Ehab and the even-odd game
|
PROGRAMMING
| 800
|
[
"games",
"math"
] | null | null |
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that:
- 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
|
The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game.
|
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
|
[
"1\n",
"2\n"
] |
[
"Ehab",
"Mahmoud"
] |
In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
| 500
|
[
{
"input": "1",
"output": "Ehab"
},
{
"input": "2",
"output": "Mahmoud"
},
{
"input": "10000",
"output": "Mahmoud"
},
{
"input": "33333",
"output": "Ehab"
},
{
"input": "5",
"output": "Ehab"
},
{
"input": "1000000000",
"output": "Mahmoud"
},
{
"input": "999999999",
"output": "Ehab"
},
{
"input": "123123123",
"output": "Ehab"
},
{
"input": "22222221",
"output": "Ehab"
},
{
"input": "22222220",
"output": "Mahmoud"
},
{
"input": "3",
"output": "Ehab"
},
{
"input": "4",
"output": "Mahmoud"
},
{
"input": "6",
"output": "Mahmoud"
},
{
"input": "7",
"output": "Ehab"
},
{
"input": "8",
"output": "Mahmoud"
},
{
"input": "9",
"output": "Ehab"
},
{
"input": "10",
"output": "Mahmoud"
},
{
"input": "536870912",
"output": "Mahmoud"
},
{
"input": "536870913",
"output": "Ehab"
},
{
"input": "536870911",
"output": "Ehab"
}
] | 1,688,825,085
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 31
| 0
|
n = int(input())
a=n+1
if a%2==0:
print("Ehab")
else:
print("Mahmoud")
|
Title: Mahmoud and Ehab and the even-odd game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that:
- 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input Specification:
The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game.
Output Specification:
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Demo Input:
['1\n', '2\n']
Demo Output:
['Ehab', 'Mahmoud']
Note:
In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
|
```python
n = int(input())
a=n+1
if a%2==0:
print("Ehab")
else:
print("Mahmoud")
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number *k* is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by *k*2, and the loser's score is multiplied by *k*. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all *n* games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
|
In the first string, the number of games *n* (1<=≤<=*n*<=≤<=350000) is given.
Each game is represented by a pair of scores *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) – the results of Slastyona and Pushok, correspondingly.
|
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
|
[
"6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000\n"
] |
[
"Yes\nYes\nYes\nNo\nNo\nYes\n"
] |
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
| 0
|
[
{
"input": "6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000",
"output": "Yes\nYes\nYes\nNo\nNo\nYes"
},
{
"input": "3\n1 1\n8 27\n1000 1331",
"output": "Yes\nNo\nNo"
},
{
"input": "1\n12004 18012002",
"output": "Yes"
},
{
"input": "1\n3331 11095561",
"output": "Yes"
},
{
"input": "1\n2 3",
"output": "No"
},
{
"input": "1\n1062961 1031",
"output": "Yes"
},
{
"input": "1\n6 12",
"output": "No"
},
{
"input": "1\n3 1",
"output": "No"
},
{
"input": "1\n3 10",
"output": "No"
},
{
"input": "1\n31159 970883281",
"output": "Yes"
},
{
"input": "1\n9907 98148649",
"output": "Yes"
},
{
"input": "1\n16 8",
"output": "No"
},
{
"input": "1\n90 72",
"output": "No"
}
] | 1,501,496,011
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 405
| 5,017,600
|
from math import sqrt
def primes_till(n): # Sieve of Eratosthenes
p = [True] * n
for i in range(2, n):
if p[i]:
j = i * i
while j < n:
p[j] = False
j += i
res = []
for i in range(2, n):
if p[i]:
res.append(i)
return res
p = primes_till(32000)
def prime_dividers(a):
i = 0
qa = sqrt(a) + 5
res = {}
while p[i] < qa:
c = 0
while a % p[i] == 0:
c += 1
a = a // p[i]
if c != 0:
res[p[i]] = c
i += 1
return res
def main():
n = int(input())
for i in range(n):
a, b = map(int, input().split())
pa = prime_dividers(a)
pb = prime_dividers(b)
if len(pa) != len(pb):
print("NO")
continue
ok = True
for k in pa:
ac = pa[k]
if k not in pb:
ok = False
break
bc = pb[k]
if bc > ac:
ac, bc = bc, ac
while bc > 0:
ac -= 2
bc -= 1
if bc > ac:
ac, bc = bc, ac
if ac != 0 or bc != 0:
ok = False
break
if ok:
print("YES")
else:
print("NO")
main()
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number *k* is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by *k*2, and the loser's score is multiplied by *k*. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all *n* games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input Specification:
In the first string, the number of games *n* (1<=≤<=*n*<=≤<=350000) is given.
Each game is represented by a pair of scores *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) – the results of Slastyona and Pushok, correspondingly.
Output Specification:
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Demo Input:
['6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000\n']
Demo Output:
['Yes\nYes\nYes\nNo\nNo\nYes\n']
Note:
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
|
```python
from math import sqrt
def primes_till(n): # Sieve of Eratosthenes
p = [True] * n
for i in range(2, n):
if p[i]:
j = i * i
while j < n:
p[j] = False
j += i
res = []
for i in range(2, n):
if p[i]:
res.append(i)
return res
p = primes_till(32000)
def prime_dividers(a):
i = 0
qa = sqrt(a) + 5
res = {}
while p[i] < qa:
c = 0
while a % p[i] == 0:
c += 1
a = a // p[i]
if c != 0:
res[p[i]] = c
i += 1
return res
def main():
n = int(input())
for i in range(n):
a, b = map(int, input().split())
pa = prime_dividers(a)
pb = prime_dividers(b)
if len(pa) != len(pb):
print("NO")
continue
ok = True
for k in pa:
ac = pa[k]
if k not in pb:
ok = False
break
bc = pb[k]
if bc > ac:
ac, bc = bc, ac
while bc > 0:
ac -= 2
bc -= 1
if bc > ac:
ac, bc = bc, ac
if ac != 0 or bc != 0:
ok = False
break
if ok:
print("YES")
else:
print("NO")
main()
```
| 0
|
|
37
|
A
|
Towers
|
PROGRAMMING
| 1,000
|
[
"sortings"
] |
A. Towers
|
2
|
256
|
Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.
Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.
|
The first line contains an integer *N* (1<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000.
|
In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars.
|
[
"3\n1 2 3\n",
"4\n6 5 6 7\n"
] |
[
"1 3\n",
"2 3\n"
] |
none
| 500
|
[
{
"input": "3\n1 2 3",
"output": "1 3"
},
{
"input": "4\n6 5 6 7",
"output": "2 3"
},
{
"input": "4\n3 2 1 1",
"output": "2 3"
},
{
"input": "4\n1 2 3 3",
"output": "2 3"
},
{
"input": "3\n20 22 36",
"output": "1 3"
},
{
"input": "25\n47 30 94 41 45 20 96 51 110 129 24 116 9 47 32 82 105 114 116 75 154 151 70 42 162",
"output": "2 23"
},
{
"input": "45\n802 664 442 318 318 827 417 878 711 291 231 414 807 553 657 392 279 202 386 606 465 655 658 112 887 15 25 502 95 44 679 775 942 609 209 871 31 234 4 231 150 110 22 823 193",
"output": "2 43"
},
{
"input": "63\n93 180 116 7 8 179 268 279 136 94 221 153 264 190 278 19 19 63 153 26 158 225 25 49 89 218 111 149 255 225 197 122 243 80 3 224 107 178 202 17 53 92 69 42 228 24 81 205 95 8 265 82 228 156 127 241 172 159 106 60 67 155 111",
"output": "2 57"
},
{
"input": "83\n246 535 994 33 390 927 321 97 223 922 812 705 79 80 977 457 476 636 511 137 6 360 815 319 717 674 368 551 714 628 278 713 761 553 184 414 623 753 428 214 581 115 439 61 677 216 772 592 187 603 658 310 439 559 870 376 109 321 189 337 277 26 70 734 796 907 979 693 570 227 345 650 737 633 701 914 134 403 972 940 371 6 642",
"output": "2 80"
},
{
"input": "105\n246 57 12 204 165 123 246 68 191 310 3 152 386 333 374 257 158 104 333 50 80 290 8 340 101 76 221 316 388 289 138 359 316 26 93 290 105 178 81 195 41 196 218 180 244 292 187 97 315 323 174 119 248 239 92 312 31 2 101 180 307 170 338 314 163 281 217 31 142 238 280 190 190 156 70 74 329 113 151 8 141 313 366 40 253 116 168 124 135 230 294 266 353 389 371 359 195 200 183 237 93 102 315 118 188",
"output": "2 92"
},
{
"input": "123\n112 277 170 247 252 115 157 293 256 143 196 90 12 164 164 42 8 223 167 109 175 232 239 111 148 51 9 254 93 32 268 162 231 91 47 162 161 191 195 145 247 292 129 199 230 94 144 217 18 205 176 20 143 198 121 243 211 262 230 277 195 255 108 290 220 275 158 2 286 200 60 267 278 207 123 150 123 116 131 13 12 226 33 244 30 275 263 45 158 192 254 149 242 176 62 224 221 288 250 160 155 225 132 143 276 293 218 145 197 175 33 129 79 206 210 192 222 262 190 52 274 243 233",
"output": "3 101"
},
{
"input": "5\n5 5 5 5 5",
"output": "5 1"
},
{
"input": "3\n1000 1000 1000",
"output": "3 1"
},
{
"input": "1\n1000",
"output": "1 1"
},
{
"input": "1\n1",
"output": "1 1"
},
{
"input": "5\n1 1000 1000 1000 1000",
"output": "4 2"
},
{
"input": "5\n1000 1000 1000 8 7",
"output": "3 3"
}
] | 1,696,766,251
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 61
| 92
| 0
|
n = int(input())
l = list(map(int, input().split()))
dic = {}
for i in l:
dic[i] = dic.get(i, 0) + 1
print(max(dic.values()), len(set(l)))
|
Title: Towers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.
Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.
Input Specification:
The first line contains an integer *N* (1<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000.
Output Specification:
In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars.
Demo Input:
['3\n1 2 3\n', '4\n6 5 6 7\n']
Demo Output:
['1 3\n', '2 3\n']
Note:
none
|
```python
n = int(input())
l = list(map(int, input().split()))
dic = {}
for i in l:
dic[i] = dic.get(i, 0) + 1
print(max(dic.values()), len(set(l)))
```
| 3.977
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,671,194,126
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 31
| 0
|
n = int(input())
words = []
for i in range(n):
word = input()
words.append(word)
for item in words:
counter = 0
if len(item)>=10:
for car in item:
counter += 1
new_word = f"{item[0]}{counter-2}{item[len(item)-1]}"
print(new_word)
else:
print(item)
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
n = int(input())
words = []
for i in range(n):
word = input()
words.append(word)
for item in words:
counter = 0
if len(item)>=10:
for car in item:
counter += 1
new_word = f"{item[0]}{counter-2}{item[len(item)-1]}"
print(new_word)
else:
print(item)
```
| 0
|
703
|
A
|
Mishka and Game
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
|
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
|
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
|
[
"3\n3 5\n2 1\n4 2\n",
"2\n6 1\n1 6\n",
"3\n1 5\n3 3\n2 2\n"
] |
[
"Mishka",
"Friendship is magic!^^",
"Chris"
] |
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.
| 500
|
[
{
"input": "3\n3 5\n2 1\n4 2",
"output": "Mishka"
},
{
"input": "2\n6 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "3\n1 5\n3 3\n2 2",
"output": "Chris"
},
{
"input": "6\n4 1\n4 2\n5 3\n5 1\n5 3\n4 1",
"output": "Mishka"
},
{
"input": "8\n2 4\n1 4\n1 5\n2 6\n2 5\n2 5\n2 4\n2 5",
"output": "Chris"
},
{
"input": "8\n4 1\n2 6\n4 2\n2 5\n5 2\n3 5\n5 2\n1 5",
"output": "Friendship is magic!^^"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3",
"output": "Mishka"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "9\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "10\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n2 4\n6 6\n3 2\n1 5\n5 2\n1 5\n1 5\n3 1\n6 5\n4 3\n1 1\n5 1\n3 3\n2 4\n1 5\n3 4\n5 1\n5 5\n2 5\n2 1\n4 3\n6 5\n1 1\n2 1\n1 3\n1 1\n6 4\n4 6\n6 4\n2 1\n2 5\n6 2\n3 4\n5 5\n1 4\n4 6\n3 4\n1 6\n5 1\n4 3\n3 4\n2 2\n1 2\n2 3\n1 3\n4 4\n5 5\n4 5\n4 4\n3 1\n4 5\n2 3\n2 6\n6 5\n6 1\n6 6\n2 3\n6 4\n3 3\n2 5\n4 4\n3 1\n2 4\n6 1\n3 2\n1 3\n5 4\n6 6\n2 5\n5 1\n1 1\n2 5\n6 5\n3 6\n5 6\n4 3\n3 4\n3 4\n6 5\n5 2\n4 2\n1 1\n3 1\n2 6\n1 6\n1 2\n6 1\n3 4\n1 6\n3 1\n5 3\n1 3\n5 6\n2 1\n6 4\n3 1\n1 6\n6 3\n3 3\n4 3",
"output": "Chris"
},
{
"input": "100\n4 1\n3 4\n4 6\n4 5\n6 5\n5 3\n6 2\n6 3\n5 2\n4 5\n1 5\n5 4\n1 4\n4 5\n4 6\n1 6\n4 4\n5 1\n6 4\n6 4\n4 6\n2 3\n6 2\n4 6\n1 4\n2 3\n4 3\n1 3\n6 2\n3 1\n3 4\n2 6\n4 5\n5 4\n2 2\n2 5\n4 1\n2 2\n3 3\n1 4\n5 6\n6 4\n4 2\n6 1\n5 5\n4 1\n2 1\n6 4\n4 4\n4 3\n5 3\n4 5\n5 3\n3 5\n6 3\n1 1\n3 4\n6 3\n6 1\n5 1\n2 4\n4 3\n2 2\n5 5\n1 5\n5 3\n4 6\n1 4\n6 3\n4 3\n2 4\n3 2\n2 4\n3 4\n6 2\n5 6\n1 2\n1 5\n5 5\n2 6\n5 1\n1 6\n5 3\n3 5\n2 6\n4 6\n6 2\n3 1\n5 5\n6 1\n3 6\n4 4\n1 1\n4 6\n5 3\n4 2\n5 1\n3 3\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "100\n6 3\n4 5\n4 3\n5 4\n5 1\n6 3\n4 2\n4 6\n3 1\n2 4\n2 2\n4 6\n5 3\n5 5\n4 2\n6 2\n2 3\n4 4\n6 4\n3 5\n2 4\n2 2\n5 2\n3 5\n2 4\n4 4\n3 5\n6 5\n1 3\n1 6\n2 2\n2 4\n3 2\n5 4\n1 6\n3 4\n4 1\n1 5\n1 4\n5 3\n2 2\n4 5\n6 3\n4 4\n1 1\n4 1\n2 4\n4 1\n4 5\n5 3\n1 1\n1 6\n5 6\n6 6\n4 2\n4 3\n3 4\n3 6\n3 4\n6 5\n3 4\n5 4\n5 1\n5 3\n5 1\n1 2\n2 6\n3 4\n6 5\n4 3\n1 1\n5 5\n5 1\n3 3\n5 2\n1 3\n6 6\n5 6\n1 4\n4 4\n1 4\n3 6\n6 5\n3 3\n3 6\n1 5\n1 2\n3 6\n3 6\n4 1\n5 2\n1 2\n5 2\n3 3\n4 4\n4 2\n6 2\n5 4\n6 1\n6 3",
"output": "Mishka"
},
{
"input": "8\n4 1\n6 2\n4 1\n5 3\n4 1\n5 3\n6 2\n5 3",
"output": "Mishka"
},
{
"input": "5\n3 6\n3 5\n3 5\n1 6\n3 5",
"output": "Chris"
},
{
"input": "4\n4 1\n2 4\n5 3\n3 6",
"output": "Friendship is magic!^^"
},
{
"input": "6\n6 3\n5 1\n6 3\n4 3\n4 3\n5 2",
"output": "Mishka"
},
{
"input": "7\n3 4\n1 4\n2 5\n1 6\n1 6\n1 5\n3 4",
"output": "Chris"
},
{
"input": "6\n6 2\n2 5\n5 2\n3 6\n4 3\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "8\n6 1\n5 3\n4 3\n4 1\n5 1\n4 2\n4 2\n4 1",
"output": "Mishka"
},
{
"input": "9\n2 5\n2 5\n1 4\n2 6\n2 4\n2 5\n2 6\n1 5\n2 5",
"output": "Chris"
},
{
"input": "4\n6 2\n2 4\n4 2\n3 6",
"output": "Friendship is magic!^^"
},
{
"input": "9\n5 2\n4 1\n4 1\n5 1\n6 2\n6 1\n5 3\n6 1\n6 2",
"output": "Mishka"
},
{
"input": "8\n2 4\n3 6\n1 6\n1 6\n2 4\n3 4\n3 6\n3 4",
"output": "Chris"
},
{
"input": "6\n5 3\n3 6\n6 2\n1 6\n5 1\n3 5",
"output": "Friendship is magic!^^"
},
{
"input": "6\n5 2\n5 1\n6 1\n5 2\n4 2\n5 1",
"output": "Mishka"
},
{
"input": "5\n1 4\n2 5\n3 4\n2 6\n3 4",
"output": "Chris"
},
{
"input": "4\n6 2\n3 4\n5 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "93\n4 3\n4 1\n4 2\n5 2\n5 3\n6 3\n4 3\n6 2\n6 3\n5 1\n4 2\n4 2\n5 1\n6 2\n6 3\n6 1\n4 1\n6 2\n5 3\n4 3\n4 1\n4 2\n5 2\n6 3\n5 2\n5 2\n6 3\n5 1\n6 2\n5 2\n4 1\n5 2\n5 1\n4 1\n6 1\n5 2\n4 3\n5 3\n5 3\n5 1\n4 3\n4 3\n4 2\n4 1\n6 2\n6 1\n4 1\n5 2\n5 2\n6 2\n5 3\n5 1\n6 2\n5 1\n6 3\n5 2\n6 2\n6 2\n4 2\n5 2\n6 1\n6 3\n6 3\n5 1\n5 1\n4 1\n5 1\n4 3\n5 3\n6 3\n4 1\n4 3\n6 1\n6 1\n4 2\n6 2\n4 2\n5 2\n4 1\n5 2\n4 1\n5 1\n5 2\n5 1\n4 1\n6 3\n6 2\n4 3\n4 1\n5 2\n4 3\n5 2\n5 1",
"output": "Mishka"
},
{
"input": "11\n1 6\n1 6\n2 4\n2 5\n3 4\n1 5\n1 6\n1 5\n1 6\n2 6\n3 4",
"output": "Chris"
},
{
"input": "70\n6 1\n3 6\n4 3\n2 5\n5 2\n1 4\n6 2\n1 6\n4 3\n1 4\n5 3\n2 4\n5 3\n1 6\n5 1\n3 5\n4 2\n2 4\n5 1\n3 5\n6 2\n1 5\n4 2\n2 5\n5 3\n1 5\n4 2\n1 4\n5 2\n2 6\n4 3\n1 5\n6 2\n3 4\n4 2\n3 5\n6 3\n3 4\n5 1\n1 4\n4 2\n1 4\n6 3\n2 6\n5 2\n1 6\n6 1\n2 6\n5 3\n1 5\n5 1\n1 6\n4 1\n1 5\n4 2\n2 4\n5 1\n2 5\n6 3\n1 4\n6 3\n3 6\n5 1\n1 4\n5 3\n3 5\n4 2\n3 4\n6 2\n1 4",
"output": "Friendship is magic!^^"
},
{
"input": "59\n4 1\n5 3\n6 1\n4 2\n5 1\n4 3\n6 1\n5 1\n4 3\n4 3\n5 2\n5 3\n4 1\n6 2\n5 1\n6 3\n6 3\n5 2\n5 2\n6 1\n4 1\n6 1\n4 3\n5 3\n5 3\n4 3\n4 2\n4 2\n6 3\n6 3\n6 1\n4 3\n5 1\n6 2\n6 1\n4 1\n6 1\n5 3\n4 2\n5 1\n6 2\n6 2\n4 3\n5 3\n4 3\n6 3\n5 2\n5 2\n4 3\n5 1\n5 3\n6 1\n6 3\n6 3\n4 3\n5 2\n5 2\n5 2\n4 3",
"output": "Mishka"
},
{
"input": "42\n1 5\n1 6\n1 6\n1 4\n2 5\n3 6\n1 6\n3 4\n2 5\n2 5\n2 4\n1 4\n3 4\n2 4\n2 6\n1 5\n3 6\n2 6\n2 6\n3 5\n1 4\n1 5\n2 6\n3 6\n1 4\n3 4\n2 4\n1 6\n3 4\n2 4\n2 6\n1 6\n1 4\n1 6\n1 6\n2 4\n1 5\n1 6\n2 5\n3 6\n3 5\n3 4",
"output": "Chris"
},
{
"input": "78\n4 3\n3 5\n4 3\n1 5\n5 1\n1 5\n4 3\n1 4\n6 3\n1 5\n4 1\n2 4\n4 3\n2 4\n5 1\n3 6\n4 2\n3 6\n6 3\n3 4\n4 3\n3 6\n5 3\n1 5\n4 1\n2 6\n4 2\n2 4\n4 1\n3 5\n5 2\n3 6\n4 3\n2 4\n6 3\n1 6\n4 3\n3 5\n6 3\n2 6\n4 1\n2 4\n6 2\n1 6\n4 2\n1 4\n4 3\n1 4\n4 3\n2 4\n6 2\n3 5\n6 1\n3 6\n5 3\n1 6\n6 1\n2 6\n4 2\n1 5\n6 2\n2 6\n6 3\n2 4\n4 2\n3 5\n6 1\n2 5\n5 3\n2 6\n5 1\n3 6\n4 3\n3 6\n6 3\n2 5\n6 1\n2 6",
"output": "Friendship is magic!^^"
},
{
"input": "76\n4 1\n5 2\n4 3\n5 2\n5 3\n5 2\n6 1\n4 2\n6 2\n5 3\n4 2\n6 2\n4 1\n4 2\n5 1\n5 1\n6 2\n5 2\n5 3\n6 3\n5 2\n4 3\n6 3\n6 1\n4 3\n6 2\n6 1\n4 1\n6 1\n5 3\n4 1\n5 3\n4 2\n5 2\n4 3\n6 1\n6 2\n5 2\n6 1\n5 3\n4 3\n5 1\n5 3\n4 3\n5 1\n5 1\n4 1\n4 1\n4 1\n4 3\n5 3\n6 3\n6 3\n5 2\n6 2\n6 3\n5 1\n6 3\n5 3\n6 1\n5 3\n4 1\n5 3\n6 1\n4 2\n6 2\n4 3\n4 1\n6 2\n4 3\n5 3\n5 2\n5 3\n5 1\n6 3\n5 2",
"output": "Mishka"
},
{
"input": "84\n3 6\n3 4\n2 5\n2 4\n1 6\n3 4\n1 5\n1 6\n3 5\n1 6\n2 4\n2 6\n2 6\n2 4\n3 5\n1 5\n3 6\n3 6\n3 4\n3 4\n2 6\n1 6\n1 6\n3 5\n3 4\n1 6\n3 4\n3 5\n2 4\n2 5\n2 5\n3 5\n1 6\n3 4\n2 6\n2 6\n3 4\n3 4\n2 5\n2 5\n2 4\n3 4\n2 5\n3 4\n3 4\n2 6\n2 6\n1 6\n2 4\n1 5\n3 4\n2 5\n2 5\n3 4\n2 4\n2 6\n2 6\n1 4\n3 5\n3 5\n2 4\n2 5\n3 4\n1 5\n1 5\n2 6\n1 5\n3 5\n2 4\n2 5\n3 4\n2 6\n1 6\n2 5\n3 5\n3 5\n3 4\n2 5\n2 6\n3 4\n1 6\n2 5\n2 6\n1 4",
"output": "Chris"
},
{
"input": "44\n6 1\n1 6\n5 2\n1 4\n6 2\n2 5\n5 3\n3 6\n5 2\n1 6\n4 1\n2 4\n6 1\n3 4\n6 3\n3 6\n4 3\n2 4\n6 1\n3 4\n6 1\n1 6\n4 1\n3 5\n6 1\n3 6\n4 1\n1 4\n4 2\n2 6\n6 1\n2 4\n6 2\n1 4\n6 2\n2 4\n5 2\n3 6\n6 3\n2 6\n5 3\n3 4\n5 3\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "42\n5 3\n5 1\n5 2\n4 1\n6 3\n6 1\n6 2\n4 1\n4 3\n4 1\n5 1\n5 3\n5 1\n4 1\n4 2\n6 1\n6 3\n5 1\n4 1\n4 1\n6 3\n4 3\n6 3\n5 2\n6 1\n4 1\n5 3\n4 3\n5 2\n6 3\n6 1\n5 1\n4 2\n4 3\n5 2\n5 3\n6 3\n5 2\n5 1\n5 3\n6 2\n6 1",
"output": "Mishka"
},
{
"input": "50\n3 6\n2 6\n1 4\n1 4\n1 4\n2 5\n3 4\n3 5\n2 6\n1 6\n3 5\n1 5\n2 6\n2 4\n2 4\n3 5\n1 6\n1 5\n1 5\n1 4\n3 5\n1 6\n3 5\n1 4\n1 5\n1 4\n3 6\n1 6\n1 4\n1 4\n1 4\n1 5\n3 6\n1 6\n1 6\n2 4\n1 5\n2 6\n2 5\n3 5\n3 6\n3 4\n2 4\n2 6\n3 4\n2 5\n3 6\n3 5\n2 4\n2 4",
"output": "Chris"
},
{
"input": "86\n6 3\n2 4\n6 3\n3 5\n6 3\n1 5\n5 2\n2 4\n4 3\n2 6\n4 1\n2 6\n5 2\n1 4\n5 1\n2 4\n4 1\n1 4\n6 2\n3 5\n4 2\n2 4\n6 2\n1 5\n5 3\n2 5\n5 1\n1 6\n6 1\n1 4\n4 3\n3 4\n5 2\n2 4\n5 3\n2 5\n4 3\n3 4\n4 1\n1 5\n6 3\n3 4\n4 3\n3 4\n4 1\n3 4\n5 1\n1 6\n4 2\n1 6\n5 1\n2 4\n5 1\n3 6\n4 1\n1 5\n5 2\n1 4\n4 3\n2 5\n5 1\n1 5\n6 2\n2 6\n4 2\n2 4\n4 1\n2 5\n5 3\n3 4\n5 1\n3 4\n6 3\n3 4\n4 3\n2 6\n6 2\n2 5\n5 2\n3 5\n4 2\n3 6\n6 2\n3 4\n4 2\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "84\n6 1\n6 3\n6 3\n4 1\n4 3\n4 2\n6 3\n5 3\n6 1\n6 3\n4 3\n5 2\n5 3\n5 1\n6 2\n6 2\n6 1\n4 1\n6 3\n5 2\n4 1\n5 3\n6 3\n4 2\n6 2\n6 3\n4 3\n4 1\n4 3\n5 1\n5 1\n5 1\n4 1\n6 1\n4 3\n6 2\n5 1\n5 1\n6 2\n5 2\n4 1\n6 1\n6 1\n6 3\n6 2\n4 3\n6 3\n6 2\n5 2\n5 1\n4 3\n6 2\n4 1\n6 2\n6 1\n5 2\n5 1\n6 2\n6 1\n5 3\n5 2\n6 1\n6 3\n5 2\n6 1\n6 3\n4 3\n5 1\n6 3\n6 1\n5 3\n4 3\n5 2\n5 1\n6 2\n5 3\n6 1\n5 1\n4 1\n5 1\n5 1\n5 2\n5 2\n5 1",
"output": "Mishka"
},
{
"input": "92\n1 5\n2 4\n3 5\n1 6\n2 5\n1 6\n3 6\n1 6\n2 4\n3 4\n3 4\n3 6\n1 5\n2 5\n1 5\n1 5\n2 6\n2 4\n3 6\n1 4\n1 6\n2 6\n3 4\n2 6\n2 6\n1 4\n3 5\n2 5\n2 6\n1 5\n1 4\n1 5\n3 6\n3 5\n2 5\n1 5\n3 5\n3 6\n2 6\n2 6\n1 5\n3 4\n2 4\n3 6\n2 5\n1 5\n2 4\n1 4\n2 6\n2 6\n2 6\n1 5\n3 6\n3 6\n2 5\n1 4\n2 4\n3 4\n1 5\n2 5\n2 4\n2 5\n3 5\n3 4\n3 6\n2 6\n3 5\n1 4\n3 4\n1 6\n3 6\n2 6\n1 4\n3 6\n3 6\n2 5\n2 6\n1 6\n2 6\n3 5\n2 5\n3 6\n2 5\n2 6\n1 5\n2 4\n1 4\n2 4\n1 5\n2 5\n2 5\n2 6",
"output": "Chris"
},
{
"input": "20\n5 1\n1 4\n4 3\n1 5\n4 2\n3 6\n6 2\n1 6\n4 1\n1 4\n5 2\n3 4\n5 1\n1 6\n5 1\n2 6\n6 3\n2 5\n6 2\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "100\n4 3\n4 3\n4 2\n4 3\n4 1\n4 3\n5 2\n5 2\n6 2\n4 2\n5 1\n4 2\n5 2\n6 1\n4 1\n6 3\n5 3\n5 1\n5 1\n5 1\n5 3\n6 1\n6 1\n4 1\n5 2\n5 2\n6 1\n6 3\n4 2\n4 1\n5 3\n4 1\n5 3\n5 1\n6 3\n6 3\n6 1\n5 2\n5 3\n5 3\n6 1\n4 1\n6 2\n6 1\n6 2\n6 3\n4 3\n4 3\n6 3\n4 2\n4 2\n5 3\n5 2\n5 2\n4 3\n5 3\n5 2\n4 2\n5 1\n4 2\n5 1\n5 3\n6 3\n5 3\n5 3\n4 2\n4 1\n4 2\n4 3\n6 3\n4 3\n6 2\n6 1\n5 3\n5 2\n4 1\n6 1\n5 2\n6 2\n4 2\n6 3\n4 3\n5 1\n6 3\n5 2\n4 3\n5 3\n5 3\n4 3\n6 3\n4 3\n4 1\n5 1\n6 2\n6 3\n5 3\n6 1\n6 3\n5 3\n6 1",
"output": "Mishka"
},
{
"input": "100\n1 5\n1 4\n1 5\n2 4\n2 6\n3 6\n3 5\n1 5\n2 5\n3 6\n3 5\n1 6\n1 4\n1 5\n1 6\n2 6\n1 5\n3 5\n3 4\n2 6\n2 6\n2 5\n3 4\n1 6\n1 4\n2 4\n1 5\n1 6\n3 5\n1 6\n2 6\n3 5\n1 6\n3 4\n3 5\n1 6\n3 6\n2 4\n2 4\n3 5\n2 6\n1 5\n3 5\n3 6\n2 4\n2 4\n2 6\n3 4\n3 4\n1 5\n1 4\n2 5\n3 4\n1 4\n2 6\n2 5\n2 4\n2 4\n2 5\n1 5\n1 6\n1 5\n1 5\n1 5\n1 6\n3 4\n2 4\n3 5\n3 5\n1 6\n3 5\n1 5\n1 6\n3 6\n3 4\n1 5\n3 5\n3 6\n1 4\n3 6\n1 5\n3 5\n3 6\n3 5\n1 4\n3 4\n2 4\n2 4\n2 5\n3 6\n3 5\n1 5\n2 4\n1 4\n3 4\n1 5\n3 4\n3 6\n3 5\n3 4",
"output": "Chris"
},
{
"input": "100\n4 3\n3 4\n5 1\n2 5\n5 3\n1 5\n6 3\n2 4\n5 2\n2 6\n5 2\n1 5\n6 3\n1 5\n6 3\n3 4\n5 2\n1 5\n6 1\n1 5\n4 2\n3 5\n6 3\n2 6\n6 3\n1 4\n6 2\n3 4\n4 1\n3 6\n5 1\n2 4\n5 1\n3 4\n6 2\n3 5\n4 1\n2 6\n4 3\n2 6\n5 2\n3 6\n6 2\n3 5\n4 3\n1 5\n5 3\n3 6\n4 2\n3 4\n6 1\n3 4\n5 2\n2 6\n5 2\n2 4\n6 2\n3 6\n4 3\n2 4\n4 3\n2 6\n4 2\n3 4\n6 3\n2 4\n6 3\n3 5\n5 2\n1 5\n6 3\n3 6\n4 3\n1 4\n5 2\n1 6\n4 1\n2 5\n4 1\n2 4\n4 2\n2 5\n6 1\n2 4\n6 3\n1 5\n4 3\n2 6\n6 3\n2 6\n5 3\n1 5\n4 1\n1 5\n6 2\n2 5\n5 1\n3 6\n4 3\n3 4",
"output": "Friendship is magic!^^"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3",
"output": "Mishka"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "99\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "84\n6 2\n1 5\n6 2\n2 3\n5 5\n1 2\n3 4\n3 4\n6 5\n6 4\n2 5\n4 1\n1 2\n1 1\n1 4\n2 5\n5 6\n6 3\n2 4\n5 5\n2 6\n3 4\n5 1\n3 3\n5 5\n4 6\n4 6\n2 4\n4 1\n5 2\n2 2\n3 6\n3 3\n4 6\n1 1\n2 4\n6 5\n5 2\n6 5\n5 5\n2 5\n6 4\n1 1\n6 2\n3 6\n6 5\n4 4\n1 5\n5 6\n4 4\n3 5\n6 1\n3 4\n1 5\n4 6\n4 6\n4 1\n3 6\n6 2\n1 1\n4 5\n5 4\n5 3\n3 4\n6 4\n1 1\n5 2\n6 5\n6 1\n2 2\n2 4\n3 3\n4 6\n1 3\n6 6\n5 2\n1 6\n6 2\n6 6\n4 1\n3 6\n6 4\n2 3\n3 4",
"output": "Chris"
},
{
"input": "70\n3 4\n2 3\n2 3\n6 5\n6 6\n4 3\n2 3\n3 1\n3 5\n5 6\n1 6\n2 5\n5 3\n2 5\n4 6\n5 1\n6 1\n3 1\n3 3\n5 3\n2 1\n3 3\n6 4\n6 3\n4 3\n4 5\n3 5\n5 5\n5 2\n1 6\n3 4\n5 2\n2 4\n1 6\n4 3\n4 3\n6 2\n1 3\n1 5\n6 1\n3 1\n1 1\n1 3\n2 2\n3 2\n6 4\n1 1\n4 4\n3 1\n4 5\n4 2\n6 3\n4 4\n3 2\n1 2\n2 6\n3 3\n1 5\n1 1\n6 5\n2 2\n3 1\n5 4\n5 2\n6 4\n6 3\n6 6\n6 3\n3 3\n5 4",
"output": "Mishka"
},
{
"input": "56\n6 4\n3 4\n6 1\n3 3\n1 4\n2 3\n1 5\n2 5\n1 5\n5 5\n2 3\n1 1\n3 2\n3 5\n4 6\n4 4\n5 2\n4 3\n3 1\n3 6\n2 3\n3 4\n5 6\n5 2\n5 6\n1 5\n1 5\n4 1\n6 3\n2 2\n2 1\n5 5\n2 1\n4 1\n5 4\n2 5\n4 1\n6 2\n3 4\n4 2\n6 4\n5 4\n4 2\n4 3\n6 2\n6 2\n3 1\n1 4\n3 6\n5 1\n5 5\n3 6\n6 4\n2 3\n6 5\n3 3",
"output": "Mishka"
},
{
"input": "94\n2 4\n6 4\n1 6\n1 4\n5 1\n3 3\n4 3\n6 1\n6 5\n3 2\n2 3\n5 1\n5 3\n1 2\n4 3\n3 2\n2 3\n4 6\n1 3\n6 3\n1 1\n3 2\n4 3\n1 5\n4 6\n3 2\n6 3\n1 6\n1 1\n1 2\n3 5\n1 3\n3 5\n4 4\n4 2\n1 4\n4 5\n1 3\n1 2\n1 1\n5 4\n5 5\n6 1\n2 1\n2 6\n6 6\n4 2\n3 6\n1 6\n6 6\n1 5\n3 2\n1 2\n4 4\n6 4\n4 1\n1 5\n3 3\n1 3\n3 4\n4 4\n1 1\n2 5\n4 5\n3 1\n3 1\n3 6\n3 2\n1 4\n1 6\n6 3\n2 4\n1 1\n2 2\n2 2\n2 1\n5 4\n1 2\n6 6\n2 2\n3 3\n6 3\n6 3\n1 6\n2 3\n2 4\n2 3\n6 6\n2 6\n6 3\n3 5\n1 4\n1 1\n3 5",
"output": "Chris"
},
{
"input": "81\n4 2\n1 2\n2 3\n4 5\n6 2\n1 6\n3 6\n3 4\n4 6\n4 4\n3 5\n4 6\n3 6\n3 5\n3 1\n1 3\n5 3\n3 4\n1 1\n4 1\n1 2\n6 1\n1 3\n6 5\n4 5\n4 2\n4 5\n6 2\n1 2\n2 6\n5 2\n1 5\n2 4\n4 3\n5 4\n1 2\n5 3\n2 6\n6 4\n1 1\n1 3\n3 1\n3 1\n6 5\n5 5\n6 1\n6 6\n5 2\n1 3\n1 4\n2 3\n5 5\n3 1\n3 1\n4 4\n1 6\n6 4\n2 2\n4 6\n4 4\n2 6\n2 4\n2 4\n4 1\n1 6\n1 4\n1 3\n6 5\n5 1\n1 3\n5 1\n1 4\n3 5\n2 6\n1 3\n5 6\n3 5\n4 4\n5 5\n5 6\n4 3",
"output": "Chris"
},
{
"input": "67\n6 5\n3 6\n1 6\n5 3\n5 4\n5 1\n1 6\n1 1\n3 2\n4 4\n3 1\n4 1\n1 5\n5 3\n3 3\n6 4\n2 4\n2 2\n4 3\n1 4\n1 4\n6 1\n1 2\n2 2\n5 1\n6 2\n3 5\n5 5\n2 2\n6 5\n6 2\n4 4\n3 1\n4 2\n6 6\n6 4\n5 1\n2 2\n4 5\n5 5\n4 6\n1 5\n6 3\n4 4\n1 5\n6 4\n3 6\n3 4\n1 6\n2 4\n2 1\n2 5\n6 5\n6 4\n4 1\n3 2\n1 2\n5 1\n5 6\n1 5\n3 5\n3 1\n5 3\n3 2\n5 1\n4 6\n6 6",
"output": "Mishka"
},
{
"input": "55\n6 6\n6 5\n2 2\n2 2\n6 4\n5 5\n6 5\n5 3\n1 3\n2 2\n5 6\n3 3\n3 3\n6 5\n3 5\n5 5\n1 2\n1 1\n4 6\n1 2\n5 5\n6 2\n6 3\n1 2\n5 1\n1 3\n3 3\n4 4\n2 5\n1 1\n5 3\n4 3\n2 2\n4 5\n5 6\n4 5\n6 3\n1 6\n6 4\n3 6\n1 6\n5 2\n6 3\n2 3\n5 5\n4 3\n3 1\n4 2\n1 1\n2 5\n5 3\n2 2\n6 3\n4 5\n2 2",
"output": "Mishka"
},
{
"input": "92\n2 3\n1 3\n2 6\n5 1\n5 5\n3 2\n5 6\n2 5\n3 1\n3 6\n4 5\n2 5\n1 2\n2 3\n6 5\n3 6\n4 4\n6 2\n4 5\n4 4\n5 1\n6 1\n3 4\n3 5\n6 6\n3 2\n6 4\n2 2\n3 5\n6 4\n6 3\n6 6\n3 4\n3 3\n6 1\n5 4\n6 2\n2 6\n5 6\n1 4\n4 6\n6 3\n3 1\n4 1\n6 6\n3 5\n6 3\n6 1\n1 6\n3 2\n6 6\n4 3\n3 4\n1 3\n3 5\n5 3\n6 5\n4 3\n5 5\n4 1\n1 5\n6 4\n2 3\n2 3\n1 5\n1 2\n5 2\n4 3\n3 6\n5 5\n5 4\n1 4\n3 3\n1 6\n5 6\n5 4\n5 3\n1 1\n6 2\n5 5\n2 5\n4 3\n6 6\n5 1\n1 1\n4 6\n4 6\n3 1\n6 4\n2 4\n2 2\n2 1",
"output": "Chris"
},
{
"input": "79\n5 3\n4 6\n3 6\n2 1\n5 2\n2 3\n4 4\n6 2\n2 5\n1 6\n6 6\n2 6\n3 3\n4 5\n6 2\n2 1\n1 5\n5 1\n2 1\n2 6\n5 3\n6 2\n2 6\n2 3\n1 5\n4 4\n6 3\n5 2\n3 2\n1 3\n1 3\n6 3\n2 6\n3 6\n5 3\n4 5\n6 1\n3 5\n3 5\n6 5\n1 5\n4 2\n6 2\n2 3\n4 6\n3 6\n2 5\n4 4\n1 1\n4 6\n2 6\n6 4\n3 2\n4 1\n1 2\n6 4\n5 6\n1 4\n2 2\n5 4\n3 2\n1 2\n2 4\n2 5\n2 1\n3 6\n3 3\n1 1\n2 2\n4 4\n4 5\n3 3\n5 3\n6 2\n4 5\n6 5\n2 5\n5 6\n2 2",
"output": "Chris"
},
{
"input": "65\n1 1\n5 1\n2 2\n5 4\n4 5\n2 5\n3 2\n5 6\n6 3\n1 1\n6 1\n1 5\n1 1\n5 2\n6 4\n1 6\n1 1\n4 3\n2 3\n5 6\n4 4\n6 2\n1 3\n4 3\n1 3\n6 3\n3 5\n4 2\n4 1\n6 1\n3 2\n2 6\n3 2\n3 5\n6 3\n4 3\n1 5\n2 6\n1 3\n4 1\n4 1\n2 5\n2 5\n6 2\n5 3\n3 1\n3 3\n5 1\n2 4\n5 3\n3 3\n1 1\n6 3\n3 3\n5 1\n1 6\n4 5\n6 6\n5 5\n2 5\n4 1\n2 2\n1 4\n1 6\n6 5",
"output": "Mishka"
},
{
"input": "1\n1 1",
"output": "Friendship is magic!^^"
}
] | 1,674,590,852
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 69
| 46
| 0
|
n = int(input())
m = 0
c = 0
for i in range(n):
x,y = map(int,input().split())
if x>y:
m = m+1
elif x<y:
c = c+1
if (m>c):
print("Mishka")
elif m<c:
print("Chris")
else:
print("Friendship is magic!^^")
|
Title: Mishka and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
Input Specification:
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
Output Specification:
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
Demo Input:
['3\n3 5\n2 1\n4 2\n', '2\n6 1\n1 6\n', '3\n1 5\n3 3\n2 2\n']
Demo Output:
['Mishka', 'Friendship is magic!^^', 'Chris']
Note:
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.
|
```python
n = int(input())
m = 0
c = 0
for i in range(n):
x,y = map(int,input().split())
if x>y:
m = m+1
elif x<y:
c = c+1
if (m>c):
print("Mishka")
elif m<c:
print("Chris")
else:
print("Friendship is magic!^^")
```
| 3
|
|
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"
}
] | 1,532,103,495
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 109
| 0
|
x = int(input())
z = list(map(int,input().split()))
print((x-1)*max(z))
|
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
x = int(input())
z = list(map(int,input().split()))
print((x-1)*max(z))
```
| 0
|
|
469
|
A
|
I Wanna Be the Guy
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation"
] | null | null |
There is a game called "I Wanna Be the Guy", consisting of *n* levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only *p* levels of the game. And Little Y can pass only *q* levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
|
The first line contains a single integer *n* (1<=≤<=<=*n*<=≤<=100).
The next line contains an integer *p* (0<=≤<=*p*<=≤<=*n*) at first, then follows *p* distinct integers *a*1,<=*a*2,<=...,<=*a**p* (1<=≤<=*a**i*<=≤<=*n*). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to *n*.
|
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
|
[
"4\n3 1 2 3\n2 2 4\n",
"4\n3 1 2 3\n2 2 3\n"
] |
[
"I become the guy.\n",
"Oh, my keyboard!\n"
] |
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
| 500
|
[
{
"input": "4\n3 1 2 3\n2 2 4",
"output": "I become the guy."
},
{
"input": "4\n3 1 2 3\n2 2 3",
"output": "Oh, my keyboard!"
},
{
"input": "10\n5 8 6 1 5 4\n6 1 3 2 9 4 6",
"output": "Oh, my keyboard!"
},
{
"input": "10\n8 8 10 7 3 1 4 2 6\n8 9 5 10 3 7 2 4 8",
"output": "I become the guy."
},
{
"input": "10\n9 6 1 8 3 9 7 5 10 4\n7 1 3 2 7 6 9 5",
"output": "I become the guy."
},
{
"input": "100\n75 83 69 73 30 76 37 48 14 41 42 21 35 15 50 61 86 85 46 3 31 13 78 10 2 44 80 95 56 82 38 75 77 4 99 9 84 53 12 11 36 74 39 72 43 89 57 28 54 1 51 66 27 22 93 59 68 88 91 29 7 20 63 8 52 23 64 58 100 79 65 49 96 71 33 45\n83 50 89 73 34 28 99 67 77 44 19 60 68 42 8 27 94 85 14 39 17 78 24 21 29 63 92 32 86 22 71 81 31 82 65 48 80 59 98 3 70 55 37 12 15 72 47 9 11 33 16 7 91 74 13 64 38 84 6 61 93 90 45 69 1 54 52 100 57 10 35 49 53 75 76 43 62 5 4 18 36 96 79 23",
"output": "Oh, my keyboard!"
},
{
"input": "1\n1 1\n1 1",
"output": "I become the guy."
},
{
"input": "1\n0\n1 1",
"output": "I become the guy."
},
{
"input": "1\n1 1\n0",
"output": "I become the guy."
},
{
"input": "1\n0\n0",
"output": "Oh, my keyboard!"
},
{
"input": "100\n0\n0",
"output": "Oh, my keyboard!"
},
{
"input": "100\n44 71 70 55 49 43 16 53 7 95 58 56 38 76 67 94 20 73 29 90 25 30 8 84 5 14 77 52 99 91 66 24 39 37 22 44 78 12 63 59 32 51 15 82 34\n56 17 10 96 80 69 13 81 31 57 4 48 68 89 50 45 3 33 36 2 72 100 64 87 21 75 54 74 92 65 23 40 97 61 18 28 98 93 35 83 9 79 46 27 41 62 88 6 47 60 86 26 42 85 19 1 11",
"output": "I become the guy."
},
{
"input": "100\n78 63 59 39 11 58 4 2 80 69 22 95 90 26 65 16 30 100 66 99 67 79 54 12 23 28 45 56 70 74 60 82 73 91 68 43 92 75 51 21 17 97 86 44 62 47 85 78 72 64 50 81 71 5 57 13 31 76 87 9 49 96 25 42 19 35 88 53 7 83 38 27 29 41 89 93 10 84 18\n78 1 16 53 72 99 9 36 59 49 75 77 94 79 35 4 92 42 82 83 76 97 20 68 55 47 65 50 14 30 13 67 98 8 7 40 64 32 87 10 33 90 93 18 26 71 17 46 24 28 89 58 37 91 39 34 25 48 84 31 96 95 80 88 3 51 62 52 85 61 12 15 27 6 45 38 2 22 60",
"output": "I become the guy."
},
{
"input": "2\n2 2 1\n0",
"output": "I become the guy."
},
{
"input": "2\n1 2\n2 1 2",
"output": "I become the guy."
},
{
"input": "80\n57 40 1 47 36 69 24 76 5 72 26 4 29 62 6 60 3 70 8 64 18 37 16 14 13 21 25 7 66 68 44 74 61 39 38 33 15 63 34 65 10 23 56 51 80 58 49 75 71 12 50 57 2 30 54 27 17 52\n61 22 67 15 28 41 26 1 80 44 3 38 18 37 79 57 11 7 65 34 9 36 40 5 48 29 64 31 51 63 27 4 50 13 24 32 58 23 19 46 8 73 39 2 21 56 77 53 59 78 43 12 55 45 30 74 33 68 42 47 17 54",
"output": "Oh, my keyboard!"
},
{
"input": "100\n78 87 96 18 73 32 38 44 29 64 40 70 47 91 60 69 24 1 5 34 92 94 99 22 83 65 14 68 15 20 74 31 39 100 42 4 97 46 25 6 8 56 79 9 71 35 54 19 59 93 58 62 10 85 57 45 33 7 86 81 30 98 26 61 84 41 23 28 88 36 66 51 80 53 37 63 43 95 75\n76 81 53 15 26 37 31 62 24 87 41 39 75 86 46 76 34 4 51 5 45 65 67 48 68 23 71 27 94 47 16 17 9 96 84 89 88 100 18 52 69 42 6 92 7 64 49 12 98 28 21 99 25 55 44 40 82 19 36 30 77 90 14 43 50 3 13 95 78 35 20 54 58 11 2 1 33",
"output": "Oh, my keyboard!"
},
{
"input": "100\n77 55 26 98 13 91 78 60 23 76 12 11 36 62 84 80 18 1 68 92 81 67 19 4 2 10 17 77 96 63 15 69 46 97 82 42 83 59 50 72 14 40 89 9 52 29 56 31 74 39 45 85 22 99 44 65 95 6 90 38 54 32 49 34 3 70 75 33 94 53 21 71 5 66 73 41 100 24\n69 76 93 5 24 57 59 6 81 4 30 12 44 15 67 45 73 3 16 8 47 95 20 64 68 85 54 17 90 86 66 58 13 37 42 51 35 32 1 28 43 80 7 14 48 19 62 55 2 91 25 49 27 26 38 79 89 99 22 60 75 53 88 82 34 21 87 71 72 61",
"output": "I become the guy."
},
{
"input": "100\n74 96 32 63 12 69 72 99 15 22 1 41 79 77 71 31 20 28 75 73 85 37 38 59 42 100 86 89 55 87 68 4 24 57 52 8 92 27 56 98 95 58 34 9 45 14 11 36 66 76 61 19 25 23 78 49 90 26 80 43 70 13 65 10 5 74 81 21 44 60 97 3 47 93 6\n64 68 21 27 16 91 23 22 33 12 71 88 90 50 62 43 28 29 57 59 5 74 10 95 35 1 67 93 36 32 86 40 6 64 78 46 89 15 84 53 18 30 17 85 2 3 47 92 25 48 76 51 20 82 52 83 99 63 80 11 94 54 39 7 58",
"output": "I become the guy."
},
{
"input": "100\n75 11 98 44 47 88 94 23 78 59 70 2 43 39 34 63 71 19 42 61 30 74 14 77 97 53 92 60 67 36 37 13 6 86 62 46 41 3 25 93 7 12 27 48 55 49 31 35 51 10 57 54 95 82 28 90 73 26 17 50 81 56 20 87 40 85 72 64 99 29 91 5 80 18 24 52\n72 93 59 5 88 47 9 58 48 1 43 50 100 87 61 91 45 98 99 56 25 84 53 73 78 54 63 38 37 2 77 95 89 85 4 90 10 33 12 22 74 32 34 70 71 52 96 57 15 66 31 27 75 8 21 39 62 44 67 94 81 68 14 19 36 28 11 79 16 65 46 83 76",
"output": "Oh, my keyboard!"
},
{
"input": "3\n1 2\n2 2 3",
"output": "Oh, my keyboard!"
},
{
"input": "4\n1 2\n3 1 3 4",
"output": "I become the guy."
},
{
"input": "6\n2 1 2\n3 4 5 6",
"output": "Oh, my keyboard!"
}
] | 1,696,427,365
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 29
| 62
| 0
|
a,b,c=int(input()),list(map(int,input().split())),list(map(int,input().split()))
print("I become the guy.") if len(set(tuple(b[1::]+c[1::])))==a else print("Oh, my keyboard!")
|
Title: I Wanna Be the Guy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a game called "I Wanna Be the Guy", consisting of *n* levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.
Little X can pass only *p* levels of the game. And Little Y can pass only *q* levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?
Input Specification:
The first line contains a single integer *n* (1<=≤<=<=*n*<=≤<=100).
The next line contains an integer *p* (0<=≤<=*p*<=≤<=*n*) at first, then follows *p* distinct integers *a*1,<=*a*2,<=...,<=*a**p* (1<=≤<=*a**i*<=≤<=*n*). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to *n*.
Output Specification:
If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).
Demo Input:
['4\n3 1 2 3\n2 2 4\n', '4\n3 1 2 3\n2 2 3\n']
Demo Output:
['I become the guy.\n', 'Oh, my keyboard!\n']
Note:
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.
In the second sample, no one can pass level 4.
|
```python
a,b,c=int(input()),list(map(int,input().split())),list(map(int,input().split()))
print("I become the guy.") if len(set(tuple(b[1::]+c[1::])))==a else print("Oh, my keyboard!")
```
| 3
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,672,155,071
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 20
| 62
| 0
|
t=int(input())
for i in range(0,t):
w=str(input())
if len(w)>10:
print(w[0],end="")
x=len(w)-2
print(x,end="")
print(w[len(w)-1])
else:
print(w)
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
t=int(input())
for i in range(0,t):
w=str(input())
if len(w)>10:
print(w[0],end="")
x=len(w)-2
print(x,end="")
print(w[len(w)-1])
else:
print(w)
```
| 3.969
|
20
|
C
|
Dijkstra?
|
PROGRAMMING
| 1,900
|
[
"graphs",
"shortest paths"
] |
C. Dijkstra?
|
1
|
64
|
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
|
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
|
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
|
[
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n",
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n"
] |
[
"1 4 3 5 ",
"1 4 3 5 "
] |
none
| 1,500
|
[
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "2 1\n1 2 1",
"output": "1 2 "
},
{
"input": "3 1\n1 2 1",
"output": "-1"
},
{
"input": "3 3\n1 2 1\n1 3 2\n2 3 1",
"output": "1 3 "
},
{
"input": "10 10\n1 5 12\n2 4 140\n2 10 149\n3 6 154\n3 7 9\n3 8 226\n3 10 132\n4 10 55\n5 8 33\n7 8 173",
"output": "1 5 8 7 3 10 "
},
{
"input": "10 10\n1 5 178\n1 8 221\n2 7 92\n2 8 159\n3 5 55\n3 6 179\n3 10 237\n4 8 205\n5 6 191\n8 10 157",
"output": "1 8 10 "
},
{
"input": "10 10\n1 4 200\n1 9 197\n3 4 79\n3 5 213\n3 6 149\n5 8 3\n5 9 189\n6 7 130\n6 9 51\n8 10 135",
"output": "1 9 5 8 10 "
},
{
"input": "10 10\n1 4 201\n2 3 238\n3 4 40\n3 6 231\n3 8 45\n4 5 227\n4 6 58\n4 9 55\n5 7 14\n6 10 242",
"output": "1 4 6 10 "
}
] | 1,575,420,998
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 109
| 204,800
|
number_vertices, edge_number = input().split(" ")
edge_number = int(edge_number)
graph = {}
n_vertice = '1'
for i in range(edge_number):
node_from, node_to, edge_weight = input().split(" ")
n_vertice = str(max(int(n_vertice), int(node_from)))
n_vertice = str(max(int(n_vertice), int(node_to)))
graph[node_from] = {}
graph[node_to] = {}
if(node_from in graph):
value = graph[node_from]
value[node_to] = int(edge_weight)
graph[node_from] = value
else:
graph[node_from] = {node_to:int(edge_weight)}
initial = '1'
path = {}
adj_node = {}
queue = []
for node in graph:
path[node] = float("inf")
adj_node[node] = None
queue.append(node)
path[initial] = 0
while queue:
# find min distance which wasn't marked as current
key_min = queue[0]
min_val = path[key_min]
for n in range(1, len(queue)):
if path[queue[n]] < min_val:
key_min = queue[n]
min_val = path[key_min]
cur = key_min
queue.remove(cur)
for i in graph[cur]:
alternate = graph[cur][i] + path[cur]
if path[i] > alternate:
path[i] = alternate
adj_node[i] = cur
x = n_vertice
str_result = ""
str_result_end = ""
str_result_end += x
while True:
x = adj_node[x]
if x is None:
break
str_result += x
str_result = str_result[::-1]
if(len(str_result) == 0):
print(-1)
else:
final_str = ""
for i in range(len(str_result)):
final_str += str_result[i] + " "
final_str = final_str + str_result_end
print(final_str)
|
Title: Dijkstra?
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
Output Specification:
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
Demo Input:
['5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n', '5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n']
Demo Output:
['1 4 3 5 ', '1 4 3 5 ']
Note:
none
|
```python
number_vertices, edge_number = input().split(" ")
edge_number = int(edge_number)
graph = {}
n_vertice = '1'
for i in range(edge_number):
node_from, node_to, edge_weight = input().split(" ")
n_vertice = str(max(int(n_vertice), int(node_from)))
n_vertice = str(max(int(n_vertice), int(node_to)))
graph[node_from] = {}
graph[node_to] = {}
if(node_from in graph):
value = graph[node_from]
value[node_to] = int(edge_weight)
graph[node_from] = value
else:
graph[node_from] = {node_to:int(edge_weight)}
initial = '1'
path = {}
adj_node = {}
queue = []
for node in graph:
path[node] = float("inf")
adj_node[node] = None
queue.append(node)
path[initial] = 0
while queue:
# find min distance which wasn't marked as current
key_min = queue[0]
min_val = path[key_min]
for n in range(1, len(queue)):
if path[queue[n]] < min_val:
key_min = queue[n]
min_val = path[key_min]
cur = key_min
queue.remove(cur)
for i in graph[cur]:
alternate = graph[cur][i] + path[cur]
if path[i] > alternate:
path[i] = alternate
adj_node[i] = cur
x = n_vertice
str_result = ""
str_result_end = ""
str_result_end += x
while True:
x = adj_node[x]
if x is None:
break
str_result += x
str_result = str_result[::-1]
if(len(str_result) == 0):
print(-1)
else:
final_str = ""
for i in range(len(str_result)):
final_str += str_result[i] + " "
final_str = final_str + str_result_end
print(final_str)
```
| 0
|
302
|
A
|
Eugeny and Array
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Eugeny has array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* integers. Each integer *a**i* equals to -1, or to 1. Also, he has *m* queries:
- Query number *i* is given as a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). - The response to the query will be integer 1, if the elements of array *a* can be rearranged so as the sum *a**l**i*<=+<=*a**l**i*<=+<=1<=+<=...<=+<=*a**r**i*<==<=0, otherwise the response to the query will be integer 0.
Help Eugeny, answer all his queries.
|
The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2·105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (*a**i*<==<=-1,<=1). Next *m* lines contain Eugene's queries. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*).
|
Print *m* integers — the responses to Eugene's queries in the order they occur in the input.
|
[
"2 3\n1 -1\n1 1\n1 2\n2 2\n",
"5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5\n"
] |
[
"0\n1\n0\n",
"0\n1\n0\n1\n0\n"
] |
none
| 500
|
[
{
"input": "2 3\n1 -1\n1 1\n1 2\n2 2",
"output": "0\n1\n0"
},
{
"input": "5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5",
"output": "0\n1\n0\n1\n0"
},
{
"input": "3 3\n1 1 1\n2 2\n1 1\n1 1",
"output": "0\n0\n0"
},
{
"input": "4 4\n-1 -1 -1 -1\n1 3\n1 2\n1 2\n1 1",
"output": "0\n0\n0\n0"
},
{
"input": "5 5\n-1 -1 -1 -1 -1\n1 1\n1 1\n3 4\n1 1\n1 4",
"output": "0\n0\n0\n0\n0"
},
{
"input": "6 6\n-1 -1 1 -1 -1 1\n1 1\n3 4\n1 1\n1 1\n1 3\n1 4",
"output": "0\n1\n0\n0\n0\n1"
},
{
"input": "7 7\n-1 -1 -1 1 -1 -1 -1\n1 1\n2 7\n1 3\n1 5\n4 7\n1 7\n6 7",
"output": "0\n0\n0\n0\n0\n0\n1"
},
{
"input": "8 8\n1 1 1 1 1 1 1 1\n5 8\n2 6\n2 3\n1 7\n7 7\n1 6\n1 8\n1 3",
"output": "0\n0\n0\n0\n0\n0\n0\n0"
},
{
"input": "9 9\n-1 1 1 1 1 1 1 1 1\n1 7\n5 6\n1 4\n1 1\n1 1\n6 8\n1 1\n6 7\n3 5",
"output": "0\n1\n0\n0\n0\n0\n0\n1\n0"
},
{
"input": "10 10\n-1 1 -1 1 -1 -1 -1 -1 -1 -1\n6 7\n2 5\n3 6\n1 3\n3 5\n4 5\n3 4\n1 6\n1 1\n1 1",
"output": "1\n1\n1\n0\n0\n1\n1\n0\n0\n0"
},
{
"input": "1 1\n-1\n1 1",
"output": "0"
},
{
"input": "1 1\n1\n1 1",
"output": "0"
}
] | 1,628,991,116
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 10
| 1,000
| 11,571,200
|
from typing import Counter
n, m = map(int, input().split())
a = list(map(int, input().split()))
c = Counter(a)
for _ in range(m):
l, r = map(int, input().split())
d = r - l
if d & 1:
h = d >> 1
if c[1] > h and c[-1] > h:
print('1')
else:
print('0')
else:
print('0')
|
Title: Eugeny and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eugeny has array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* integers. Each integer *a**i* equals to -1, or to 1. Also, he has *m* queries:
- Query number *i* is given as a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). - The response to the query will be integer 1, if the elements of array *a* can be rearranged so as the sum *a**l**i*<=+<=*a**l**i*<=+<=1<=+<=...<=+<=*a**r**i*<==<=0, otherwise the response to the query will be integer 0.
Help Eugeny, answer all his queries.
Input Specification:
The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2·105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (*a**i*<==<=-1,<=1). Next *m* lines contain Eugene's queries. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*).
Output Specification:
Print *m* integers — the responses to Eugene's queries in the order they occur in the input.
Demo Input:
['2 3\n1 -1\n1 1\n1 2\n2 2\n', '5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5\n']
Demo Output:
['0\n1\n0\n', '0\n1\n0\n1\n0\n']
Note:
none
|
```python
from typing import Counter
n, m = map(int, input().split())
a = list(map(int, input().split()))
c = Counter(a)
for _ in range(m):
l, r = map(int, input().split())
d = r - l
if d & 1:
h = d >> 1
if c[1] > h and c[-1] > h:
print('1')
else:
print('0')
else:
print('0')
```
| 0
|
|
208
|
A
|
Dubstep
|
PROGRAMMING
| 900
|
[
"strings"
] | null | null |
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
|
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
|
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
|
[
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] |
[
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] |
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
| 500
|
[
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "ZJWUBWUBWUBJWUBWUBWUBL",
"output": "ZJ J L "
},
{
"input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB",
"output": "C B E Q "
},
{
"input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB",
"output": "JKD WBIRAQKF YE WV "
},
{
"input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB",
"output": "KSDHEMIXUJ R S H "
},
{
"input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB",
"output": "OG X I KO "
},
{
"input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH",
"output": "Q QQ I WW JOPJPBRH "
},
{
"input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB",
"output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C "
},
{
"input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV",
"output": "E IQMJNIQ GZZBQZAUHYP PMR DCV "
},
{
"input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB",
"output": "FV BPS RXNETCJ JDMBH B V B "
},
{
"input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL",
"output": "FBQ IDFSY CTWDM SXO QI L "
},
{
"input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL",
"output": "I QLHD YIIKZDFQ CX U K NL "
},
{
"input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE",
"output": "K UPDYXGOKU AGOAH IZD IY V P E "
},
{
"input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB",
"output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ "
},
{
"input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB",
"output": "PAMJGY XGPQM TKGSXUY E N H E "
},
{
"input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB",
"output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB "
},
{
"input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM",
"output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M "
},
{
"input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW",
"output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W "
},
{
"input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG",
"output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G "
},
{
"input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN",
"output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N "
},
{
"input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG",
"output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG "
},
{
"input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB",
"output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L "
},
{
"input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB",
"output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U "
},
{
"input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB",
"output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ "
},
{
"input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB",
"output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J "
},
{
"input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO",
"output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O "
},
{
"input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR",
"output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R "
},
{
"input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB",
"output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM "
},
{
"input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB",
"output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H "
},
{
"input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB",
"output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S "
},
{
"input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB",
"output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J "
},
{
"input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF",
"output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F "
},
{
"input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY",
"output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y "
},
{
"input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB",
"output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO "
},
{
"input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW",
"output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W "
},
{
"input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD",
"output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D "
},
{
"input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB",
"output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A "
},
{
"input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB",
"output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y "
},
{
"input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB",
"output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R "
},
{
"input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY",
"output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y "
},
{
"input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE",
"output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E "
},
{
"input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS",
"output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S "
},
{
"input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP",
"output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP "
},
{
"input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB",
"output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y "
},
{
"input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL",
"output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL "
},
{
"input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO",
"output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O "
},
{
"input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU",
"output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU "
},
{
"input": "A",
"output": "A "
},
{
"input": "WUBA",
"output": "A "
},
{
"input": "WUBWUBA",
"output": "A "
},
{
"input": "AWUBWUBWUB",
"output": "A "
},
{
"input": "AWUBBWUBCWUBD",
"output": "A B C D "
},
{
"input": "WUBWWUBWUBWUBUWUBWUBBWUB",
"output": "W U B "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "WUWUBBWWUBUB",
"output": "WU BW UB "
},
{
"input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB",
"output": "WUAB "
},
{
"input": "ZWUB",
"output": "Z "
},
{
"input": "WU",
"output": "WU "
},
{
"input": "UB",
"output": "UB "
},
{
"input": "U",
"output": "U "
},
{
"input": "WUBW",
"output": "W "
},
{
"input": "WUBWU",
"output": "WU "
},
{
"input": "WUWUB",
"output": "WU "
},
{
"input": "UBWUB",
"output": "UB "
},
{
"input": "WUWUBUBWUBUWUB",
"output": "WU UB U "
},
{
"input": "WUBWWUBAWUB",
"output": "W A "
},
{
"input": "WUUUUU",
"output": "WUUUUU "
}
] | 1,691,658,844
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 0
|
s=str(input())
a=s.replace("WUB","")
b=a.strip()
print(b)
|
Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
|
```python
s=str(input())
a=s.replace("WUB","")
b=a.strip()
print(b)
```
| 0
|
|
9
|
A
|
Die Roll
|
PROGRAMMING
| 800
|
[
"math",
"probabilities"
] |
A. Die Roll
|
1
|
64
|
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
|
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
|
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
|
[
"4 2\n"
] |
[
"1/2\n"
] |
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
| 0
|
[
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,620,622,118
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 154
| 20,172,800
|
y,w=list(map(int ,input().split()))
sample=[1,2,3,4,5,6]
max_ele=max(y,w)
index_max=sample.index(max_ele)
final=sample[index_max:]
len_final=len(final)
if(len_final%2==0):
print(str(len_final//2)+"/"+str(3))
elif(len_final%3==0):
print(str(len_final//3)+"/"+str(2))
else:
print(str(len_final)+"/"+str(6))
|
Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
|
```python
y,w=list(map(int ,input().split()))
sample=[1,2,3,4,5,6]
max_ele=max(y,w)
index_max=sample.index(max_ele)
final=sample[index_max:]
len_final=len(final)
if(len_final%2==0):
print(str(len_final//2)+"/"+str(3))
elif(len_final%3==0):
print(str(len_final//3)+"/"+str(2))
else:
print(str(len_final)+"/"+str(6))
```
| 0
|
462
|
B
|
Appleman and Card Game
|
PROGRAMMING
| 1,300
|
[
"greedy"
] | null | null |
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
|
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
|
Print a single integer – the answer to the problem.
|
[
"15 10\nDZFDFZDFDDDDDDF\n",
"6 4\nYJSNPI\n"
] |
[
"82\n",
"4\n"
] |
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
| 1,000
|
[
{
"input": "15 10\nDZFDFZDFDDDDDDF",
"output": "82"
},
{
"input": "6 4\nYJSNPI",
"output": "4"
},
{
"input": "5 3\nAOWBY",
"output": "3"
},
{
"input": "1 1\nV",
"output": "1"
},
{
"input": "2 1\nWT",
"output": "1"
},
{
"input": "2 2\nBL",
"output": "2"
},
{
"input": "5 1\nFACJT",
"output": "1"
},
{
"input": "5 5\nMJDIJ",
"output": "7"
},
{
"input": "15 5\nAZBIPTOFTJCJJIK",
"output": "13"
},
{
"input": "100 1\nEVEEVEEEGGECFEHEFVFVFHVHEEEEEFCVEEEEEEVFVEEVEEHEEVEFEVVEFEEEFEVECEHGHEEFGEEVCEECCECEFHEVEEEEEEGEEHVH",
"output": "1"
},
{
"input": "100 15\nKKTFFUTFCKUIKKKKFIFFKTUKUUKUKKIKKKTIFKTKUCFFKKKIIKKKKKKTFKFKKIRKKKFKUUKIKUUUFFKKKKTUZKITUIKKIKUKKTIK",
"output": "225"
},
{
"input": "100 50\nYYIYYAAAIEAAYAYAEAIIIAAEAAYEAEYYYIAEYAYAYYAAAIAYAEAAYAYYIYAAYYAAAAAAIYYYAAYAAEAAYAIEIYIYAYAYAYIIAAEY",
"output": "1972"
},
{
"input": "100 90\nFAFAOOAOOAFAOTFAFAFFATAAAOFAAOAFBAAAFBOAOFFFOAOAFAPFOFAOFAAFOAAAAFAAFOFAAOFPPAAOOAAOOFFOFFFOFAOTOFAF",
"output": "2828"
},
{
"input": "100 99\nBFFBBFBFBQFFFFFQBFFBFFBQFBFQFBBFQFFFBFFFBFQFQFBFFBBFYQFBFFFFFFFBQQFQBFBQBQFFFBQQFFFBQFYFBFBFFFBBBQQY",
"output": "3713"
},
{
"input": "100 100\nMQSBDAJABILIBCUEOWGWCEXMUTEYQKAIWGINXVQEOFDUBSVULROQHQRZZAALVQFEFRAAAYUIMGCAFQGIAEFBETRECGSFQJNXHHDN",
"output": "514"
},
{
"input": "100 50\nBMYIXQSJNHGFVFPJBIOBXIKSFNUFPVODCUBQYSIIQNVNXXCWXWRHKFEUPPIIDDGRDBJLZDCBMNJMYRMWFIHOSTDJJHXHPNRKWNFD",
"output": "328"
},
{
"input": "100 50\nENFNEMLJEMDMFMNNGNIMNINALGLLLAEMENEMNLMMIEIJNAINBJEJMFJLLIMINELGFLAIAMJMHMGNLIEFJIEEFEFGLLLDLMEAEIMM",
"output": "748"
}
] | 1,619,595,860
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 9
| 62
| 0
|
n, k = map(int, input().split())
counter = {}
for ch in input():
counter[ch] = counter.get(ch, 0) + 1
sorted_counts = sorted(counter.values(), reverse=True)
tmp_cards = coins = 0
for card_count in sorted_counts:
tmp_sum = tmp_cards + card_count
if tmp_sum < k:
tmp_cards = tmp_sum
coins += card_count * card_count
else:
partial = card_count-(tmp_sum%k)
coins += partial * partial
print(coins)
break
|
Title: Appleman and Card Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
Output Specification:
Print a single integer – the answer to the problem.
Demo Input:
['15 10\nDZFDFZDFDDDDDDF\n', '6 4\nYJSNPI\n']
Demo Output:
['82\n', '4\n']
Note:
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
|
```python
n, k = map(int, input().split())
counter = {}
for ch in input():
counter[ch] = counter.get(ch, 0) + 1
sorted_counts = sorted(counter.values(), reverse=True)
tmp_cards = coins = 0
for card_count in sorted_counts:
tmp_sum = tmp_cards + card_count
if tmp_sum < k:
tmp_cards = tmp_sum
coins += card_count * card_count
else:
partial = card_count-(tmp_sum%k)
coins += partial * partial
print(coins)
break
```
| 0
|
|
2
|
A
|
Winner
|
PROGRAMMING
| 1,500
|
[
"hashing",
"implementation"
] |
A. Winner
|
1
|
64
|
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
|
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
|
Print the name of the winner.
|
[
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] |
[
"andrew\n",
"andrew\n"
] |
none
| 0
|
[
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},
{
"input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734",
"output": "ksjuuerbnlklcfdjeyq"
},
{
"input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879",
"output": "fcgslzkicjrpbqaifgweyzreajjfdo"
},
{
"input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735",
"output": "aawtvezfntstrcpgbzjbf"
},
{
"input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158",
"output": "ivhgbxiv"
}
] | 1,409,492,622
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 124
| 0
|
##__________________________________________________________________
##
## Author: GogolGrind
##__________________________________________________________________
from sys import *
from math import *
def main ():
k = int(input())
d = {}
m = -1
for i in range(k):
n,s = list(map(str,input().split()))
s = int(s)
if (d.get(n) == None):
d[n] = s
else:
d[n] += s
t = []
for k,v in d.items():
m = max(m,v)
for k,v in d.items():
if (v == m):
t.append([k,v])
d = {}
for n,s in t:
if (d.get(n) == None):
d[n] = s
else:
d[n] += s
if (d[n] >= m):
print(n)
return 0
if __name__ == '__main__':
main()
|
Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none
|
```python
##__________________________________________________________________
##
## Author: GogolGrind
##__________________________________________________________________
from sys import *
from math import *
def main ():
k = int(input())
d = {}
m = -1
for i in range(k):
n,s = list(map(str,input().split()))
s = int(s)
if (d.get(n) == None):
d[n] = s
else:
d[n] += s
t = []
for k,v in d.items():
m = max(m,v)
for k,v in d.items():
if (v == m):
t.append([k,v])
d = {}
for n,s in t:
if (d.get(n) == None):
d[n] = s
else:
d[n] += s
if (d[n] >= m):
print(n)
return 0
if __name__ == '__main__':
main()
```
| 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"
},
{
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{
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},
{
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"output": "1"
}
] | 1,648,466,533
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 46
| 0
|
n=input()
n=list(n)
m=input()
m=list(m)
s=len(n)
for i in range(0,s):
if n[i]!=m[i]:
print(1,end="")
else:
print(0,end="")
|
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
n=input()
n=list(n)
m=input()
m=list(m)
s=len(n)
for i in range(0,s):
if n[i]!=m[i]:
print(1,end="")
else:
print(0,end="")
```
| 3.9885
|
Subsets and Splits
Successful Python Submissions
Retrieves all records from the train dataset where the verdict is 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Retrieves records of users with a rating of 1600 or higher and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a rating above 2000 and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a 'OK' verdict, providing a basic overview of a specific category within the dataset.