MatrixStudio/Codeforces-Python-Submissions

903

A

Hungry Student Problem

PROGRAMMING

900

[ "greedy", "implementation" ]

null

Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken. CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks. Ivan wants to eat exactly *x* chunks. Now he wonders whether he can buy exactly this amount of chicken. Formally, Ivan wants to know if he can choose two non-negative integers *a* and *b* in such a way that *a* small portions and *b* large ones contain exactly *x* chunks. Help Ivan to answer this question for several values of *x*!

The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of testcases. The *i*-th of the following *n* lines contains one integer *x**i* (1<=≤<=*x**i*<=≤<=100) — the number of chicken chunks Ivan wants to eat.

Print *n* lines, in *i*-th line output YES if Ivan can buy exactly *x**i* chunks. Otherwise, print NO.

[ "2\n6\n5\n" ]

[ "YES\nNO\n" ]

In the first example Ivan can buy two small portions. In the second example Ivan cannot buy exactly 5 chunks, since one small portion is not enough, but two small portions or one large is too much.

0

[ { "input": "2\n6\n5", "output": "YES\nNO" }, { "input": "100\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\n19\n20\n21\n22\n23\n24\n25\n26\n27\n28\n29\n30\n31\n32\n33\n34\n35\n36\n37\n38\n39\n40\n41\n42\n43\n44\n45\n46\n47\n48\n49\n50\n51\n52\n53\n54\n55\n56\n57\n58\n59\n60\n61\n62\n63\n64\n65\n66\n67\n68\n69\n70\n71\n72\n73\n74\n75\n76\n77\n78\n79\n80\n81\n82\n83\n84\n85\n86\n87\n88\n89\n90\n91\n92\n93\n94\n95\n96\n97\n98\n99\n100", "output": "NO\nNO\nYES\nNO\nNO\nYES\nYES\nNO\nYES\nYES\nNO\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES" }, { "input": "3\n6\n6\n6", "output": "YES\nYES\nYES" }, { "input": "47\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1", "output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO" }, { "input": "3\n1\n52\n76", "output": "NO\nYES\nYES" }, { "input": "87\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100", "output": "YES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES" }, { "input": "3\n3\n2\n1", "output": "YES\nNO\nNO" }, { "input": "100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100", "output": "YES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES" } ]

1,645,027,334

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

1

31

0

for _ in range(int(input())): n=int(input()) if n%3==0 or n%7==0 or (n//3+n//7)%3==0 or (n//3+n//7)%7==0: print('YES') else: print('NO')

Title: Hungry Student Problem Time Limit: None seconds Memory Limit: None megabytes Problem Description: Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken. CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks. Ivan wants to eat exactly *x* chunks. Now he wonders whether he can buy exactly this amount of chicken. Formally, Ivan wants to know if he can choose two non-negative integers *a* and *b* in such a way that *a* small portions and *b* large ones contain exactly *x* chunks. Help Ivan to answer this question for several values of *x*! Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of testcases. The *i*-th of the following *n* lines contains one integer *x**i* (1<=≤<=*x**i*<=≤<=100) — the number of chicken chunks Ivan wants to eat. Output Specification: Print *n* lines, in *i*-th line output YES if Ivan can buy exactly *x**i* chunks. Otherwise, print NO. Demo Input: ['2\n6\n5\n'] Demo Output: ['YES\nNO\n'] Note: In the first example Ivan can buy two small portions. In the second example Ivan cannot buy exactly 5 chunks, since one small portion is not enough, but two small portions or one large is too much.

```python for _ in range(int(input())): n=int(input()) if n%3==0 or n%7==0 or (n//3+n//7)%3==0 or (n//3+n//7)%7==0: print('YES') else: print('NO') ```

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" } ]

1,674,059,006

2,147,483,647

Python 3

OK

TESTS

32

92

0

n = int(input()) vals = list(map(int, input().split(" "))) dic = {0:0,1:0} for i in range (3): dic[vals[i]%2]+=1 if dic[0]>dic[1]: r = 1 else: r=0 for i in range (len(vals)): if vals[i]%2 == r: print(i+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()) vals = list(map(int, input().split(" "))) dic = {0:0,1:0} for i in range (3): dic[vals[i]%2]+=1 if dic[0]>dic[1]: r = 1 else: r=0 for i in range (len(vals)): if vals[i]%2 == r: print(i+1) ```

3.977

0

none

0

[ "none" ]

null

Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture). The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right. Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!

The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.

Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.

[ "4 3 9\n", "4 3 24\n", "2 4 4\n" ]

[ "2 2 L\n", "4 3 R\n", "1 2 R\n" ]

The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example. In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.

0

[ { "input": "4 3 9", "output": "2 2 L" }, { "input": "4 3 24", "output": "4 3 R" }, { "input": "2 4 4", "output": "1 2 R" }, { "input": "3 10 24", "output": "2 2 R" }, { "input": "10 3 59", "output": "10 3 L" }, { "input": "10000 10000 160845880", "output": "8043 2940 R" }, { "input": "1 1 1", "output": "1 1 L" }, { "input": "1 1 2", "output": "1 1 R" }, { "input": "1 10000 1", "output": "1 1 L" }, { "input": "1 10000 20000", "output": "1 10000 R" }, { "input": "10000 1 1", "output": "1 1 L" }, { "input": "10000 1 10000", "output": "5000 1 R" }, { "input": "10000 1 20000", "output": "10000 1 R" }, { "input": "3 2 1", "output": "1 1 L" }, { "input": "3 2 2", "output": "1 1 R" }, { "input": "3 2 3", "output": "1 2 L" }, { "input": "3 2 4", "output": "1 2 R" }, { "input": "3 2 5", "output": "2 1 L" }, { "input": "3 2 6", "output": "2 1 R" }, { "input": "3 2 7", "output": "2 2 L" }, { "input": "3 2 8", "output": "2 2 R" }, { "input": "3 2 9", "output": "3 1 L" }, { "input": "3 2 10", "output": "3 1 R" }, { "input": "3 2 11", "output": "3 2 L" }, { "input": "3 2 12", "output": "3 2 R" }, { "input": "300 2000 1068628", "output": "268 314 R" }, { "input": "300 2000 584756", "output": "147 378 R" }, { "input": "300 2000 268181", "output": "68 91 L" }, { "input": "10000 9999 186450844", "output": "9324 4745 R" }, { "input": "10000 9999 197114268", "output": "9857 6990 R" }, { "input": "10000 9999 112390396", "output": "5621 818 R" }, { "input": "10000 10000 1", "output": "1 1 L" }, { "input": "10000 10000 2", "output": "1 1 R" }, { "input": "10000 10000 100000001", "output": "5001 1 L" }, { "input": "10000 10000 199999999", "output": "10000 10000 L" }, { "input": "10000 10000 200000000", "output": "10000 10000 R" }, { "input": "1 2 1", "output": "1 1 L" }, { "input": "1 2 2", "output": "1 1 R" }, { "input": "1 2 3", "output": "1 2 L" }, { "input": "1 2 4", "output": "1 2 R" }, { "input": "2 1 1", "output": "1 1 L" }, { "input": "2 1 2", "output": "1 1 R" }, { "input": "2 1 3", "output": "2 1 L" }, { "input": "2 1 4", "output": "2 1 R" }, { "input": "4 3 7", "output": "2 1 L" }, { "input": "1 1 1", "output": "1 1 L" } ]

1,482,816,340

2,147,483,647

Python 3

OK

TESTS

46

77

4,608,000

n, m, k = map(int, input().split()) if k&1: lr = "L" else: lr = "R" k -= 1 k //= 2 print(k // m + 1, k % m + 1, lr)

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture). The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right. Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right! Input Specification: The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place. Output Specification: Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right. Demo Input: ['4 3 9\n', '4 3 24\n', '2 4 4\n'] Demo Output: ['2 2 L\n', '4 3 R\n', '1 2 R\n'] Note: The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example. In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.

```python n, m, k = map(int, input().split()) if k&1: lr = "L" else: lr = "R" k -= 1 k //= 2 print(k // m + 1, k % m + 1, lr) ```

3

469

A

I Wanna Be the Guy

PROGRAMMING

800

[ "greedy", "implementation" ]

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!" } ]

1,694,137,619

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

26

46

0

n=int(input()) l=list(map(int,input().split())) m=list(map(int,input().split())) l=list(set(l)) m=list(set(m)) a=list(set(l+m)) c=0 for i in range(1,n+1): if i in a: c+=1 # print(c) if n==c: print("I become the guy.") else: print("Oh, my keyboard!") # print(a)

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 n=int(input()) l=list(map(int,input().split())) m=list(map(int,input().split())) l=list(set(l)) m=list(set(m)) a=list(set(l+m)) c=0 for i in range(1,n+1): if i in a: c+=1 # print(c) if n==c: print("I become the guy.") else: print("Oh, my keyboard!") # print(a) ```

0

549

A

Face Detection

PROGRAMMING

900

[ "implementation", "strings" ]

null

The developers of Looksery have to write an efficient algorithm that detects faces on a picture. Unfortunately, they are currently busy preparing a contest for you, so you will have to do it for them. In this problem an image is a rectangular table that consists of lowercase Latin letters. A face on the image is a 2<=×<=2 square, such that from the four letters of this square you can make word "face". You need to write a program that determines the number of faces on the image. The squares that correspond to the faces can overlap.

The first line contains two space-separated integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the height and the width of the image, respectively. Next *n* lines define the image. Each line contains *m* lowercase Latin letters.

In the single line print the number of faces on the image.

[ "4 4\nxxxx\nxfax\nxcex\nxxxx\n", "4 2\nxx\ncf\nae\nxx\n", "2 3\nfac\ncef\n", "1 4\nface\n" ]

[ "1\n", "1\n", "2\n", "0\n" ]

In the first sample the image contains a single face, located in a square with the upper left corner at the second line and the second column: In the second sample the image also contains exactly one face, its upper left corner is at the second row and the first column. In the third sample two faces are shown: In the fourth sample the image has no faces on it.

250

[ { "input": "4 4\nxxxx\nxfax\nxcex\nxxxx", "output": "1" }, { "input": "4 2\nxx\ncf\nae\nxx", "output": "1" }, { "input": "2 3\nfac\ncef", "output": "2" }, { "input": "1 4\nface", "output": "0" }, { "input": "5 5\nwmmwn\nlurcm\nkeetd\nfokon\ncxxgx", "output": "0" }, { "input": "5 5\nkjxbw\neacra\nxefhx\nucmcz\npgtjk", "output": "1" }, { "input": "1 1\np", "output": "0" }, { "input": "2 5\nacdmw\nefazb", "output": "1" }, { "input": "5 2\ndz\nda\nsx\nyu\nzz", "output": "0" }, { "input": "5 5\nxeljd\nwriac\nveief\nlcacf\nbqefn", "output": "2" }, { "input": "5 5\nacnbx\nefacp\nlrefa\norqce\nzvbay", "output": "3" }, { "input": "5 5\nbyjvu\nkmaca\nalefe\nwcacg\nrefez", "output": "5" }, { "input": "5 5\npuxac\nbbaef\naccfa\nefaec\nligsr", "output": "5" }, { "input": "37 4\nacjo\nefac\nacef\nefac\nwpef\nicac\naefe\ncfac\naece\ncfaf\nyqce\nmiaf\nirce\nycaf\naefc\ncfae\nrsnc\nbacz\nqefb\npdhs\nffac\nfaef\nacfd\nacmi\nefvm\nacaz\nefpn\nacao\nefer\nacap\nefec\nacaf\nefef\nacbj\nefac\nacef\nefoz", "output": "49" }, { "input": "7 3\njac\naef\ncfa\naec\ncfq\ndig\nxyq", "output": "5" }, { "input": "35 1\ny\na\nk\ng\ni\nd\nv\nn\nl\nx\nu\nx\nu\no\nd\nf\nk\nj\nr\nm\nq\ns\nc\nd\nc\nm\nv\nh\nn\ne\nl\nt\nz\ny\no", "output": "0" }, { "input": "9 46\nuuexbaacesjclggslacermcbkxlcxhdgqtacdwfryxzuxc\naclrsaefakndbnzlkefenuphgcgoedhkaxefjtnkgfeaca\nefuqunpmfxdyyffyhvracozzrxlpekhtsrfhlilfmyhefg\numyacfzffvicqtdpiulefnwcojuwtfbvlxkfsiapdnzpqo\nactefvuxqptremlqjhdbdwacjxdxitxjktecvefacamjcz\neflarseklqrkayhosverpfefzirqigzlxezabhzeferkwm\nztpypwxyohtacsnnvoigqwcachtaceyxlkhfhgwbnfmklb\nfzpqeymhljiefctvqtbgcdjefkxefarbciyosmktmoziac\nbssxizyrzyhacszttwhrzspbfkgfcejybdjcpuczhrkqef", "output": "20" }, { "input": "7 30\nmjfracgaacacctacrreyrlkacuacay\nrlacefacefeftaeftkacacaefcefev\nacefacefacraccfaeaefefecaeacaf\nefacefacefaefaecfcfacacaecfefa\nncefacefacecacfaeaecefefcaeace\nfafaceacuafaefadcfcafacaefcfea\nzsvefafukcecfarkaeaecefecailgu", "output": "95" }, { "input": "2 2\nff\nee", "output": "0" }, { "input": "2 2\nfa\ncc", "output": "0" }, { "input": "2 2\nfa\naa", "output": "0" }, { "input": "2 2\nfa\nea", "output": "0" }, { "input": "2 2\nfa\ndd", "output": "0" }, { "input": "2 2\nee\nff", "output": "0" }, { "input": "2 2\ncc\ncf", "output": "0" }, { "input": "2 2\naa\nae", "output": "0" }, { "input": "2 2\nef\nac", "output": "1" }, { "input": "2 2\nfe\nca", "output": "1" } ]

1,485,940,343

2,147,483,647

Python 3

OK

TESTS

37

78

4,608,000

n,m=map(int,input().split()) M=[input() for i in range(n)] k=0 L=sorted('face') for i in range(n) : for j in range(m) : F='' if i+2<=n and j+2<=m : F=F+M[i][j]+M[i][j+1]+M[i+1][j]+M[i+1][j+1] if sorted(F)==L : k=k+1 print(k)

Title: Face Detection Time Limit: None seconds Memory Limit: None megabytes Problem Description: The developers of Looksery have to write an efficient algorithm that detects faces on a picture. Unfortunately, they are currently busy preparing a contest for you, so you will have to do it for them. In this problem an image is a rectangular table that consists of lowercase Latin letters. A face on the image is a 2<=×<=2 square, such that from the four letters of this square you can make word "face". You need to write a program that determines the number of faces on the image. The squares that correspond to the faces can overlap. Input Specification: The first line contains two space-separated integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the height and the width of the image, respectively. Next *n* lines define the image. Each line contains *m* lowercase Latin letters. Output Specification: In the single line print the number of faces on the image. Demo Input: ['4 4\nxxxx\nxfax\nxcex\nxxxx\n', '4 2\nxx\ncf\nae\nxx\n', '2 3\nfac\ncef\n', '1 4\nface\n'] Demo Output: ['1\n', '1\n', '2\n', '0\n'] Note: In the first sample the image contains a single face, located in a square with the upper left corner at the second line and the second column: In the second sample the image also contains exactly one face, its upper left corner is at the second row and the first column. In the third sample two faces are shown: In the fourth sample the image has no faces on it.

```python n,m=map(int,input().split()) M=[input() for i in range(n)] k=0 L=sorted('face') for i in range(n) : for j in range(m) : F='' if i+2<=n and j+2<=m : F=F+M[i][j]+M[i][j+1]+M[i+1][j]+M[i+1][j+1] if sorted(F)==L : k=k+1 print(k) ```

3

371

C

Hamburgers

PROGRAMMING

1,600

[ "binary search", "brute force" ]

null

Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again. Polycarpus has *n**b* pieces of bread, *n**s* pieces of sausage and *n**c* pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are *p**b* rubles for a piece of bread, *p**s* for a piece of sausage and *p**c* for a piece of cheese. Polycarpus has *r* rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient.

The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C). The second line contains three integers *n**b*, *n**s*, *n**c* (1<=≤<=*n**b*,<=*n**s*,<=*n**c*<=≤<=100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers *p**b*, *p**s*, *p**c* (1<=≤<=*p**b*,<=*p**s*,<=*p**c*<=≤<=100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer *r* (1<=≤<=*r*<=≤<=1012) — the number of rubles Polycarpus has. Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0.

[ "BBBSSC\n6 4 1\n1 2 3\n4\n", "BBC\n1 10 1\n1 10 1\n21\n", "BSC\n1 1 1\n1 1 3\n1000000000000\n" ]

[ "2\n", "7\n", "200000000001\n" ]

none

1,500

[ { "input": "BBBSSC\n6 4 1\n1 2 3\n4", "output": "2" }, { "input": "BBC\n1 10 1\n1 10 1\n21", "output": "7" }, { "input": "BSC\n1 1 1\n1 1 3\n1000000000000", "output": "200000000001" }, { "input": "B\n1 1 1\n1 1 1\n381", "output": "382" }, { "input": "BSC\n3 5 6\n7 3 9\n100", "output": "10" }, { "input": "BSC\n100 1 1\n100 1 1\n100", "output": "51" }, { "input": "SBBCCSBB\n1 50 100\n31 59 21\n100000", "output": "370" }, { "input": "BBBBCCCCCCCCCCCCCCCCCCCCSSSSBBBBBBBBSS\n100 100 100\n1 1 1\n3628800", "output": "95502" }, { "input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n200", "output": "0" }, { "input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n2000", "output": "1" }, { "input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n300", "output": "0" }, { "input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n300000000", "output": "42858" }, { "input": "BBBBBBBBBBCCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n10 20 40\n100 100 100\n914159265358", "output": "130594181" }, { "input": "SSSSSSSSSSBBBBBBBBBCCCCCCCCCCCCCCCCCCCSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSBB\n31 53 97\n13 17 31\n914159265358", "output": "647421579" }, { "input": "BBBCSBSBBSSSSCCCCBBCSBBBBSSBBBCBSCCSSCSSCSBSSSCCCCBSCSSBSSSCCCBBCCCSCBCBBCCSCCCCSBBCCBBBBCCCCCCBSSCB\n91 87 17\n64 44 43\n958532915587", "output": "191668251" }, { "input": "CSSCBBCCCSBSCBBBCSBBBCBSBCSCBCSCBCBSBCBCSSBBSBBCBBBBSCSBBCCBCCBCBBSBSBCSCSBBSSBBCSSBCSCSCCSSBCBBCBSB\n56 34 48\n78 6 96\n904174875419", "output": "140968956" }, { "input": "CCSCCCSBBBSCBSCSCCSSBBBSSBBBSBBBCBCSSBCSCBBCCCBCBCBCCCSSBSBBCCCCCBBSCBSCBCBBCBBCSSBCSBSSCCSCCSCCBBBS\n33 73 67\n4 56 42\n886653164314", "output": "277425898" }, { "input": "SBCSSCBBSSBCSSBBBSSBSCBSSSCBBSBBBBCSBCSBSCBSCBSCBSBSSCCCCBSBCCBCBSCCCBSCCBSBBCBSSCCCCSBSBBBSSSBCSCBC\n94 16 85\n14 18 91\n836590091442", "output": "217522127" }, { "input": "BSCSBSCCSCSSCCCSBCSSBCBBSCCBSCCSSSSSSSSSCCSBSCCBBCBBSBSCCCCBCSBSBSSBBBBBSSBSSCBCCSSBSSSCBBCSBBSBCCCB\n67 54 8\n36 73 37\n782232051273", "output": "154164772" }, { "input": "CBBCBSBCCSCBSSCCBCSBCSBBSCBBCSCCBSCCSCSBBSSBSBSCBBSBBCSSSSBBBBSBBCBCSBBCBCSSBBCSBSCCSCSBCSCBSCCBBCSC\n71 71 52\n52 88 3\n654400055575", "output": "137826467" }, { "input": "CBBCBSBCCSCBSSCCBCSBCSBBSCBBCSCCBSCCSCSBBSBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBCBBCSC\n100 1 1\n1 17 23\n954400055575", "output": "1355681897" }, { "input": "C\n100 100 100\n1 1 1\n1000000000000", "output": "1000000000100" }, { "input": "SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\n100 100 100\n100 100 100\n1000000000000", "output": "100000001" }, { "input": "B\n100 100 100\n1 1 1\n1", "output": "101" }, { "input": "SC\n2 1 1\n1 1 1\n100000000000", "output": "50000000001" }, { "input": "B\n100 1 1\n1 1 1\n1000000000000", "output": "1000000000100" }, { "input": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\n1 1 1\n100 100 100\n1000000000000", "output": "100000000" }, { "input": "CC\n1 1 1\n100 100 100\n1", "output": "0" }, { "input": "B\n100 100 100\n1 1 1\n1000000000000", "output": "1000000000100" }, { "input": "BSC\n100 100 100\n1 1 1\n1000000000000", "output": "333333333433" }, { "input": "BSC\n100 100 100\n1 1 1\n1", "output": "100" } ]

1,659,949,531

2,147,483,647

PyPy 3-64

OK

TESTS

31

62

0

a = input() n = list(int(x) for x in input().split()) p = list(int(x) for x in input().split()) rub = int(input()) bread = a.count("B") cheese = a.count("C") sausage = a.count("S") l = 0 r = 1e15 while (r - l > 1): m = (r + l) // 2 if max(0, (m* bread - n[0])) * p[0] + max(0, (m * cheese - n[2])) * p[2] + max(0, (m * sausage - n[1]) * p[1]) > rub: r = m else: l = m print(int(l))

Title: Hamburgers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again. Polycarpus has *n**b* pieces of bread, *n**s* pieces of sausage and *n**c* pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are *p**b* rubles for a piece of bread, *p**s* for a piece of sausage and *p**c* for a piece of cheese. Polycarpus has *r* rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient. Input Specification: The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C). The second line contains three integers *n**b*, *n**s*, *n**c* (1<=≤<=*n**b*,<=*n**s*,<=*n**c*<=≤<=100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers *p**b*, *p**s*, *p**c* (1<=≤<=*p**b*,<=*p**s*,<=*p**c*<=≤<=100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer *r* (1<=≤<=*r*<=≤<=1012) — the number of rubles Polycarpus has. Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. Output Specification: Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0. Demo Input: ['BBBSSC\n6 4 1\n1 2 3\n4\n', 'BBC\n1 10 1\n1 10 1\n21\n', 'BSC\n1 1 1\n1 1 3\n1000000000000\n'] Demo Output: ['2\n', '7\n', '200000000001\n'] Note: none

```python a = input() n = list(int(x) for x in input().split()) p = list(int(x) for x in input().split()) rub = int(input()) bread = a.count("B") cheese = a.count("C") sausage = a.count("S") l = 0 r = 1e15 while (r - l > 1): m = (r + l) // 2 if max(0, (m* bread - n[0])) * p[0] + max(0, (m * cheese - n[2])) * p[2] + max(0, (m * sausage - n[1]) * p[1]) > rub: r = m else: l = m print(int(l)) ```

3

994

A

Fingerprints

PROGRAMMING

800

[ "implementation" ]

null

You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits. Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code.

The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints. The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence. The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints.

In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable.

[ "7 3\n3 5 7 1 6 2 8\n1 2 7\n", "4 4\n3 4 1 0\n0 1 7 9\n" ]

[ "7 1 2\n", "1 0\n" ]

In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence. In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important.

500

[ { "input": "7 3\n3 5 7 1 6 2 8\n1 2 7", "output": "7 1 2" }, { "input": "4 4\n3 4 1 0\n0 1 7 9", "output": "1 0" }, { "input": "9 4\n9 8 7 6 5 4 3 2 1\n2 4 6 8", "output": "8 6 4 2" }, { "input": "10 5\n3 7 1 2 4 6 9 0 5 8\n4 3 0 7 9", "output": "3 7 4 9 0" }, { "input": "10 10\n1 2 3 4 5 6 7 8 9 0\n4 5 6 7 1 2 3 0 9 8", "output": "1 2 3 4 5 6 7 8 9 0" }, { "input": "1 1\n4\n4", "output": "4" }, { "input": "3 7\n6 3 4\n4 9 0 1 7 8 6", "output": "6 4" }, { "input": "10 1\n9 0 8 1 7 4 6 5 2 3\n0", "output": "0" }, { "input": "5 10\n6 0 3 8 1\n3 1 0 5 4 7 2 8 9 6", "output": "6 0 3 8 1" }, { "input": "8 2\n7 2 9 6 1 0 3 4\n6 3", "output": "6 3" }, { "input": "5 4\n7 0 1 4 9\n0 9 5 3", "output": "0 9" }, { "input": "10 1\n9 6 2 0 1 8 3 4 7 5\n6", "output": "6" }, { "input": "10 2\n7 1 0 2 4 6 5 9 3 8\n3 2", "output": "2 3" }, { "input": "5 9\n3 7 9 2 4\n3 8 4 5 9 6 1 0 2", "output": "3 9 2 4" }, { "input": "10 6\n7 1 2 3 8 0 6 4 5 9\n1 5 8 2 3 6", "output": "1 2 3 8 6 5" }, { "input": "8 2\n7 4 8 9 2 5 6 1\n6 4", "output": "4 6" }, { "input": "10 2\n1 0 3 5 8 9 4 7 6 2\n0 3", "output": "0 3" }, { "input": "7 6\n9 2 8 6 1 3 7\n4 2 0 3 1 8", "output": "2 8 1 3" }, { "input": "1 6\n3\n6 8 2 4 5 3", "output": "3" }, { "input": "1 8\n0\n9 2 4 8 1 5 0 7", "output": "0" }, { "input": "6 9\n7 3 9 4 1 0\n9 1 5 8 0 6 2 7 4", "output": "7 9 4 1 0" }, { "input": "10 2\n4 9 6 8 3 0 1 5 7 2\n0 1", "output": "0 1" }, { "input": "10 5\n5 2 8 0 9 7 6 1 4 3\n9 6 4 1 2", "output": "2 9 6 1 4" }, { "input": "6 3\n8 3 9 2 7 6\n5 4 3", "output": "3" }, { "input": "4 10\n8 3 9 6\n4 9 6 2 7 0 8 1 3 5", "output": "8 3 9 6" }, { "input": "1 2\n1\n1 0", "output": "1" }, { "input": "3 6\n1 2 3\n4 5 6 1 2 3", "output": "1 2 3" }, { "input": "1 2\n2\n1 2", "output": "2" }, { "input": "1 10\n9\n0 1 2 3 4 5 6 7 8 9", "output": "9" } ]

1,557,325,750

2,147,483,647

Python 3

OK

TESTS

31

108

0

n, m = input().split() n, m = int(n), int(m) a = [int(s) for s in input().split()] b = [int(s) for s in input().split()] c = [] for i in range(n): if a[i] in b: c.append(a[i]) print(*c)

Title: Fingerprints Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits. Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code. Input Specification: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints. The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence. The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints. Output Specification: In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable. Demo Input: ['7 3\n3 5 7 1 6 2 8\n1 2 7\n', '4 4\n3 4 1 0\n0 1 7 9\n'] Demo Output: ['7 1 2\n', '1 0\n'] Note: In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence. In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important.

```python n, m = input().split() n, m = int(n), int(m) a = [int(s) for s in input().split()] b = [int(s) for s in input().split()] c = [] for i in range(n): if a[i] in b: c.append(a[i]) print(*c) ```

3

294

A

Shaass and Oskols

PROGRAMMING

800

[ "implementation", "math" ]

null

Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire. Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away. Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.

The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100). The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.

On the *i*-th line of the output print the number of birds on the *i*-th wire.

[ "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n", "3\n2 4 1\n1\n2 2\n" ]

[ "0\n12\n5\n0\n16\n", "3\n0\n3\n" ]

none

500

[ { "input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6", "output": "0\n12\n5\n0\n16" }, { "input": "3\n2 4 1\n1\n2 2", "output": "3\n0\n3" }, { "input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43", "output": "0\n66\n57\n7\n0" }, { "input": "10\n48 53 10 28 91 56 81 2 67 52\n2\n2 40\n6 51", "output": "87\n0\n23\n28\n141\n0\n86\n2\n67\n52" }, { "input": "2\n72 45\n6\n1 69\n2 41\n1 19\n2 7\n1 5\n2 1", "output": "0\n0" }, { "input": "10\n95 54 36 39 98 30 19 24 14 12\n3\n9 5\n8 15\n7 5", "output": "95\n54\n36\n39\n98\n34\n0\n28\n13\n21" }, { "input": "100\n95 15 25 18 64 62 23 59 70 84 50 26 87 35 75 86 0 22 77 60 66 41 21 9 75 50 25 3 69 14 39 68 64 46 59 99 2 0 21 76 90 12 61 42 6 91 36 39 47 41 93 81 66 57 70 36 68 89 52 1 19 93 67 22 76 20 8 81 98 18 100 73 61 93 75 80 53 72 40 20 2 86 33 59 27 16 11 26 55 44 47 55 94 58 2 55 82 22 9 95\n0", "output": "95\n15\n25\n18\n64\n62\n23\n59\n70\n84\n50\n26\n87\n35\n75\n86\n0\n22\n77\n60\n66\n41\n21\n9\n75\n50\n25\n3\n69\n14\n39\n68\n64\n46\n59\n99\n2\n0\n21\n76\n90\n12\n61\n42\n6\n91\n36\n39\n47\n41\n93\n81\n66\n57\n70\n36\n68\n89\n52\n1\n19\n93\n67\n22\n76\n20\n8\n81\n98\n18\n100\n73\n61\n93\n75\n80\n53\n72\n40\n20\n2\n86\n33\n59\n27\n16\n11\n26\n55\n44\n47\n55\n94\n58\n2\n55\n82\n22\n9\n95" }, { "input": "1\n100\n1\n1 100", "output": "0" }, { "input": "1\n100\n1\n1 1", "output": "0" }, { "input": "1\n50\n1\n1 25", "output": "0" }, { "input": "2\n50 0\n1\n1 1", "output": "0\n49" }, { "input": "1\n10\n0", "output": "10" } ]

1,693,331,827

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

30

0

def eratosthene_sieve(n: int) -> list: prime_numbers = [0 for _ in range(n+1)] for i in range(2, n + 1): if prime_numbers[j] == 0: for j in range(2 * i, n + 1, i): prime_numbers[j] = -1 prime_numbers = prime_numbers[1:] for i, v in enumerate(prime_numbers): if v == 0: prime_numbers[i] = i + 1 prime_numbers = list(filter(lambda x: x > 0, prime_numbers)) return prime_numbers n,m = list(map(int, input().split())) max_value = max(n, m) prime_numbers = eratosthene_sieve(max_value) n_index = prime_numbers.index(n) answer = "YES" if m in prime_numbers and prime_numbers[n_index + 1] == m else "NO" print(answer)

Title: Shaass and Oskols Time Limit: None seconds Memory Limit: None megabytes Problem Description: Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire. Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away. Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots. Input Specification: The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100). The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment. Output Specification: On the *i*-th line of the output print the number of birds on the *i*-th wire. Demo Input: ['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n'] Demo Output: ['0\n12\n5\n0\n16\n', '3\n0\n3\n'] Note: none

```python def eratosthene_sieve(n: int) -> list: prime_numbers = [0 for _ in range(n+1)] for i in range(2, n + 1): if prime_numbers[j] == 0: for j in range(2 * i, n + 1, i): prime_numbers[j] = -1 prime_numbers = prime_numbers[1:] for i, v in enumerate(prime_numbers): if v == 0: prime_numbers[i] = i + 1 prime_numbers = list(filter(lambda x: x > 0, prime_numbers)) return prime_numbers n,m = list(map(int, input().split())) max_value = max(n, m) prime_numbers = eratosthene_sieve(max_value) n_index = prime_numbers.index(n) answer = "YES" if m in prime_numbers and prime_numbers[n_index + 1] == m else "NO" print(answer) ```

-1

478

B

Random Teams

PROGRAMMING

1,300

[ "combinatorics", "constructive algorithms", "greedy", "math" ]

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" } ]

1,645,867,924

2,147,483,647

PyPy 3-64

OK

TESTS

26

46

0

n, m = map(int,input().split()) def sum_prog(k): return k * (k - 1) // 2 ans1 = 0 ans2 = 0 temp = n % m ans1 += sum_prog( n // m ) * (m - temp) ans1 += sum_prog( n // m + 1) * temp ans2 += sum_prog( n - m + 1 ) print(int(min(ans1, ans2)), int(max(ans1, ans2)) )

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()) def sum_prog(k): return k * (k - 1) // 2 ans1 = 0 ans2 = 0 temp = n % m ans1 += sum_prog( n // m ) * (m - temp) ans1 += sum_prog( n // m + 1) * temp ans2 += sum_prog( n - m + 1 ) print(int(min(ans1, ans2)), int(max(ans1, ans2)) ) ```

3

275

A

Lights Out

PROGRAMMING

900

[ "implementation" ]

null

Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on. Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.

The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.

Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".

[ "1 0 0\n0 0 0\n0 0 1\n", "1 0 1\n8 8 8\n2 0 3\n" ]

[ "001\n010\n100\n", "010\n011\n100\n" ]

none

500

[ { "input": "1 0 0\n0 0 0\n0 0 1", "output": "001\n010\n100" }, { "input": "1 0 1\n8 8 8\n2 0 3", "output": "010\n011\n100" }, { "input": "13 85 77\n25 50 45\n65 79 9", "output": "000\n010\n000" }, { "input": "96 95 5\n8 84 74\n67 31 61", "output": "011\n011\n101" }, { "input": "24 54 37\n60 63 6\n1 84 26", "output": "110\n101\n011" }, { "input": "23 10 40\n15 6 40\n92 80 77", "output": "101\n100\n000" }, { "input": "62 74 80\n95 74 93\n2 47 95", "output": "010\n001\n110" }, { "input": "80 83 48\n26 0 66\n47 76 37", "output": "000\n000\n010" }, { "input": "32 15 65\n7 54 36\n5 51 3", "output": "111\n101\n001" }, { "input": "22 97 12\n71 8 24\n100 21 64", "output": "100\n001\n100" }, { "input": "46 37 13\n87 0 50\n90 8 55", "output": "111\n011\n000" }, { "input": "57 43 58\n20 82 83\n66 16 52", "output": "111\n010\n110" }, { "input": "45 56 93\n47 51 59\n18 51 63", "output": "101\n011\n100" }, { "input": "47 66 67\n14 1 37\n27 81 69", "output": "001\n001\n110" }, { "input": "26 69 69\n85 18 23\n14 22 74", "output": "110\n001\n010" }, { "input": "10 70 65\n94 27 25\n74 66 30", "output": "111\n010\n100" }, { "input": "97 1 74\n15 99 1\n88 68 86", "output": "001\n011\n000" }, { "input": "36 48 42\n45 41 66\n26 64 1", "output": "001\n111\n010" }, { "input": "52 81 97\n29 77 71\n66 11 2", "output": "100\n100\n111" }, { "input": "18 66 33\n19 49 49\n48 46 26", "output": "011\n100\n000" }, { "input": "68 79 52\n51 39 100\n29 14 26", "output": "110\n000\n111" }, { "input": "91 69 77\n91 26 64\n91 88 57", "output": "001\n011\n110" }, { "input": "16 69 64\n48 21 80\n81 51 51", "output": "010\n101\n111" }, { "input": "96 14 2\n100 18 12\n65 34 89", "output": "111\n010\n010" }, { "input": "93 95 90\n8 59 42\n53 13 19", "output": "100\n001\n111" }, { "input": "71 84 18\n100 19 67\n9 76 15", "output": "010\n010\n001" }, { "input": "38 93 85\n21 88 64\n4 96 25", "output": "111\n011\n000" }, { "input": "75 20 20\n60 5 78\n77 4 69", "output": "011\n001\n000" }, { "input": "65 70 96\n19 6 83\n33 37 82", "output": "100\n000\n011" }, { "input": "11 13 60\n17 13 46\n42 21 39", "output": "000\n011\n101" }, { "input": "0 0 0\n0 0 0\n0 0 0", "output": "111\n111\n111" }, { "input": "0 0 0\n0 1 0\n0 0 0", "output": "101\n000\n101" }, { "input": "0 0 0\n0 0 0\n0 0 1", "output": "111\n110\n100" } ]

1,692,924,580

2,147,483,647

PyPy 3-64

OK

TESTS

33

62

0

odd_count = [[0, 0, 0], [0, 0, 0], [0, 0, 0]] for i in range(3): row = [int(x) for x in input().split()] for j in range(3): if row[j] % 2: odd_count[i][j] += 1 if i < 2: odd_count[i+1][j] += 1 if i > 0: odd_count[i-1][j] += 1 if j < 2: odd_count[i][j+1] += 1 if j > 0: odd_count[i][j-1] += 1 for i in range(3): print("".join([str((x + 1) % 2) for x in odd_count[i]]))

Title: Lights Out Time Limit: None seconds Memory Limit: None megabytes Problem Description: Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on. Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light. Input Specification: The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed. Output Specification: Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0". Demo Input: ['1 0 0\n0 0 0\n0 0 1\n', '1 0 1\n8 8 8\n2 0 3\n'] Demo Output: ['001\n010\n100\n', '010\n011\n100\n'] Note: none

```python odd_count = [[0, 0, 0], [0, 0, 0], [0, 0, 0]] for i in range(3): row = [int(x) for x in input().split()] for j in range(3): if row[j] % 2: odd_count[i][j] += 1 if i < 2: odd_count[i+1][j] += 1 if i > 0: odd_count[i-1][j] += 1 if j < 2: odd_count[i][j+1] += 1 if j > 0: odd_count[i][j-1] += 1 for i in range(3): print("".join([str((x + 1) % 2) for x in odd_count[i]])) ```

3

887

A

Div. 64

PROGRAMMING

1,000

[ "implementation" ]

null

Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills. Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system.

In the only line given a non-empty binary string *s* with length up to 100.

Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise.

[ "100010001\n", "100\n" ]

[ "yes", "no" ]

In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system. You can read more about binary numeral system representation here: [https://en.wikipedia.org/wiki/Binary_system](https://en.wikipedia.org/wiki/Binary_system)

500

[ { "input": "100010001", "output": "yes" }, { "input": "100", "output": "no" }, { "input": "0000001000000", "output": "yes" }, { "input": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "no" }, { "input": "1111111111111111111111111111111111111111111111111111111111111111111111110111111111111111111111111111", "output": "no" }, { "input": "0111111101111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "no" }, { "input": "1111011111111111111111111111110111110111111111111111111111011111111111111110111111111111111111111111", "output": "no" }, { "input": "1111111111101111111111111111111111111011111111111111111111111101111011111101111111111101111111111111", "output": "yes" }, { "input": "0110111111111111111111011111111110110111110111111111111111111111111111111111111110111111111111111111", "output": "yes" }, { "input": "1100110001111011001101101000001110111110011110111110010100011000100101000010010111100000010001001101", "output": "yes" }, { "input": "000000", "output": "no" }, { "input": "0001000", "output": "no" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "no" }, { "input": "1000000", "output": "yes" }, { "input": "0", "output": "no" }, { "input": "1", "output": "no" }, { "input": "10000000000", "output": "yes" }, { "input": "0000000000", "output": "no" }, { "input": "0010000", "output": "no" }, { "input": "000000011", "output": "no" }, { "input": "000000000", "output": "no" }, { "input": "00000000", "output": "no" }, { "input": "000000000011", "output": "no" }, { "input": "0000000", "output": "no" }, { "input": "00000000011", "output": "no" }, { "input": "000000001", "output": "no" }, { "input": "000000000000000000000000000", "output": "no" }, { "input": "0000001", "output": "no" }, { "input": "00000001", "output": "no" }, { "input": "00000000100", "output": "no" }, { "input": "00000000000000000000", "output": "no" }, { "input": "0000000000000000000", "output": "no" }, { "input": "00001000", "output": "no" }, { "input": "0000000000010", "output": "no" }, { "input": "000000000010", "output": "no" }, { "input": "000000000000010", "output": "no" }, { "input": "0100000", "output": "no" }, { "input": "00010000", "output": "no" }, { "input": "00000000000000000", "output": "no" }, { "input": "00000000000", "output": "no" }, { "input": "000001000", "output": "no" }, { "input": "000000000000", "output": "no" }, { "input": "100000000000000", "output": "yes" }, { "input": "000010000", "output": "no" }, { "input": "00000100", "output": "no" }, { "input": "0001100000", "output": "no" }, { "input": "000000000000000000000000001", "output": "no" }, { "input": "000000100", "output": "no" }, { "input": "0000000000001111111111", "output": "no" }, { "input": "00000010", "output": "no" }, { "input": "0001110000", "output": "no" }, { "input": "0000000000000000000000", "output": "no" }, { "input": "000000010010", "output": "no" }, { "input": "0000100", "output": "no" }, { "input": "0000000001", "output": "no" }, { "input": "000000111", "output": "no" }, { "input": "0000000000000", "output": "no" }, { "input": "000000000000000000", "output": "no" }, { "input": "0000000000000000000000000", "output": "no" }, { "input": "000000000000000", "output": "no" }, { "input": "0010000000000100", "output": "yes" }, { "input": "0000001000", "output": "no" }, { "input": "00000000000000000001", "output": "no" }, { "input": "100000000", "output": "yes" }, { "input": "000000000001", "output": "no" }, { "input": "0000011001", "output": "no" }, { "input": "000", "output": "no" }, { "input": "000000000000000000000", "output": "no" }, { "input": "0000000000011", "output": "no" }, { "input": "0000000000000000", "output": "no" }, { "input": "00000000000000001", "output": "no" }, { "input": "00000000000000", "output": "no" }, { "input": "0000000000000000010", "output": "no" }, { "input": "00000000000000000000000000000000000000000000000000000000", "output": "no" }, { "input": "000011000", "output": "no" }, { "input": "00000011", "output": "no" }, { "input": "0000000000001100", "output": "no" }, { "input": "00000", "output": "no" }, { "input": "000000000000000000000000000111111111111111", "output": "no" }, { "input": "000000010", "output": "no" }, { "input": "00000000111", "output": "no" }, { "input": "000000000000001", "output": "no" }, { "input": "0000000000000011111111111111111", "output": "no" }, { "input": "0000000010", "output": "no" }, { "input": "0000000000000000000000000000000000000000000000000", "output": "no" }, { "input": "00000000010", "output": "no" }, { "input": "101000000000", "output": "yes" }, { "input": "00100000", "output": "no" }, { "input": "00000000000001", "output": "no" }, { "input": "0000000000100", "output": "no" }, { "input": "0000", "output": "no" }, { "input": "00000000000111", "output": "no" }, { "input": "0000000000000011", "output": "no" }, { "input": "0000000000000000000000000000000000000000", "output": "no" }, { "input": "0000000000000010", "output": "no" }, { "input": "0010101010", "output": "no" }, { "input": "0000000000000001", "output": "no" }, { "input": "1010101", "output": "no" } ]

1,509,877,852

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

10

62

0

n=input() flag=0 c=0 for i in range(len(n)-1,-1,-1): if n[i]=='0': c+=1 if c==6: flag=1 print ("Yes") break if flag==0: print ("No")

Title: Div. 64 Time Limit: None seconds Memory Limit: None megabytes Problem Description: Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills. Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system. Input Specification: In the only line given a non-empty binary string *s* with length up to 100. Output Specification: Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise. Demo Input: ['100010001\n', '100\n'] Demo Output: ['yes', 'no'] Note: In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system. You can read more about binary numeral system representation here: [https://en.wikipedia.org/wiki/Binary_system](https://en.wikipedia.org/wiki/Binary_system)

```python n=input() flag=0 c=0 for i in range(len(n)-1,-1,-1): if n[i]=='0': c+=1 if c==6: flag=1 print ("Yes") break if flag==0: print ("No") ```

0

19

A

World Football Cup

PROGRAMMING

1,400

[ "implementation" ]

A. World Football Cup

2

64

Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations: - the final tournament features *n* teams (*n* is always even) - the first *n*<=/<=2 teams (according to the standings) come through to the knockout stage - the standings are made on the following principle: for a victory a team gets 3 points, for a draw — 1 point, for a defeat — 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place — in decreasing order of the difference between scored and missed goals; in the third place — in the decreasing order of scored goals - it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity. You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.

The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=50) — amount of the teams, taking part in the final tournament of World Cup. The following *n* lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following *n*·(*n*<=-<=1)<=/<=2 lines describe the held matches in the format name1-name2 num1:num2, where *name*1, *name*2 — names of the teams; *num*1, *num*2 (0<=≤<=*num*1,<=*num*2<=≤<=100) — amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.

Output *n*<=/<=2 lines — names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.

[ "4\nA\nB\nC\nD\nA-B 1:1\nA-C 2:2\nA-D 1:0\nB-C 1:0\nB-D 0:3\nC-D 0:3\n", "2\na\nA\na-A 2:1\n" ]

[ "A\nD\n", "a\n" ]

none

0

[ { "input": "4\nA\nB\nC\nD\nA-B 1:1\nA-C 2:2\nA-D 1:0\nB-C 1:0\nB-D 0:3\nC-D 0:3", "output": "A\nD" }, { "input": "2\na\nA\na-A 2:1", "output": "a" }, { "input": "2\nEULEUbCmfrmqxtzvg\nuHGRmKUhDcxcfqyruwzen\nuHGRmKUhDcxcfqyruwzen-EULEUbCmfrmqxtzvg 13:92", "output": "EULEUbCmfrmqxtzvg" }, { "input": "4\nTeMnHVvWKpwlpubwyhzqvc\nAWJwc\nbhbxErlydiwtoxy\nEVASMeLpfqwjkke\nAWJwc-TeMnHVvWKpwlpubwyhzqvc 37:34\nbhbxErlydiwtoxy-TeMnHVvWKpwlpubwyhzqvc 38:99\nbhbxErlydiwtoxy-AWJwc 33:84\nEVASMeLpfqwjkke-TeMnHVvWKpwlpubwyhzqvc 79:34\nEVASMeLpfqwjkke-AWJwc 24:37\nEVASMeLpfqwjkke-bhbxErlydiwtoxy 3:6", "output": "AWJwc\nEVASMeLpfqwjkke" }, { "input": "6\nA\nB\nC\nD\nE\nF\nA-B 1:0\nA-C 0:0\nA-D 1:0\nA-E 5:5\nA-F 0:1\nB-C 1:0\nB-D 1:0\nB-E 1:0\nB-F 0:2\nC-D 2:2\nC-E 1:0\nC-F 1:0\nD-E 1:0\nD-F 1:0\nE-F 0:1", "output": "A\nB\nF" }, { "input": "6\nA\nB\nC\nD\nE\nF\nA-B 1:0\nA-C 0:0\nA-D 1:0\nA-E 5:5\nA-F 0:1\nB-C 1:0\nB-D 1:0\nB-E 1:0\nB-F 0:2\nC-D 7:7\nC-E 1:0\nC-F 1:0\nD-E 1:0\nD-F 1:0\nE-F 0:1", "output": "B\nC\nF" } ]

1,683,829,443

2,147,483,647

Python 3

OK

TESTS

29

92

102,400

class Info: def __init__(self, newTeamName, newPoints, newGoalDiff, newScoredGoals): self.teamName = newTeamName self.points = newPoints self.goalDiff = newGoalDiff self.scoredGoals = newScoredGoals def __str__(self): return f'teamName: \'{self.teamName}\', points: {self.points}, goalDiff: {self.goalDiff}, scoredGoals: {self.scoredGoals}' def out(cont): for item in cont: print(item) def findIndexByName(cont, searchName): for i in range(len(cont)): if cont[i].teamName == searchName: return i return -1 n, cont = int(input()), [] for i in range(n): cont.append(Info(input(), 0, 0, 0)) ''' 01234567890123456789 line = 'Barcelona-Real 15:10' ''' for i in range(n * (n - 1) // 2): line = input() dashInd = line.index('-') spaceInd = line.index(' ') colonInd = line.index(':') team1Name = line[:dashInd] team2Name = line[dashInd + 1:spaceInd] goal1 = int(line[spaceInd + 1: colonInd]) goal2 = int(line[colonInd + 1:]) team1Index = findIndexByName(cont, team1Name) team2Index = findIndexByName(cont, team2Name) # update points if goal1 > goal2: cont[team1Index].points += 3 elif goal1 < goal2: cont[team2Index].points += 3 else: cont[team1Index].points += 1 cont[team2Index].points += 1 # update goal diff cont[team1Index].goalDiff += goal1 - goal2 cont[team2Index].goalDiff += goal2 - goal1 # update scored goals cont[team1Index].scoredGoals += goal1 cont[team2Index].scoredGoals += goal2 cont.sort(key=lambda obj: (obj.points, obj.goalDiff, obj.scoredGoals), reverse=True) del cont[n // 2:] cont.sort(key=lambda obj: obj.teamName) # out(cont) for obj in cont: print(obj.teamName)

Title: World Football Cup Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations: - the final tournament features *n* teams (*n* is always even) - the first *n*<=/<=2 teams (according to the standings) come through to the knockout stage - the standings are made on the following principle: for a victory a team gets 3 points, for a draw — 1 point, for a defeat — 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place — in decreasing order of the difference between scored and missed goals; in the third place — in the decreasing order of scored goals - it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity. You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage. Input Specification: The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=50) — amount of the teams, taking part in the final tournament of World Cup. The following *n* lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following *n*·(*n*<=-<=1)<=/<=2 lines describe the held matches in the format name1-name2 num1:num2, where *name*1, *name*2 — names of the teams; *num*1, *num*2 (0<=≤<=*num*1,<=*num*2<=≤<=100) — amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once. Output Specification: Output *n*<=/<=2 lines — names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity. Demo Input: ['4\nA\nB\nC\nD\nA-B 1:1\nA-C 2:2\nA-D 1:0\nB-C 1:0\nB-D 0:3\nC-D 0:3\n', '2\na\nA\na-A 2:1\n'] Demo Output: ['A\nD\n', 'a\n'] Note: none

```python class Info: def __init__(self, newTeamName, newPoints, newGoalDiff, newScoredGoals): self.teamName = newTeamName self.points = newPoints self.goalDiff = newGoalDiff self.scoredGoals = newScoredGoals def __str__(self): return f'teamName: \'{self.teamName}\', points: {self.points}, goalDiff: {self.goalDiff}, scoredGoals: {self.scoredGoals}' def out(cont): for item in cont: print(item) def findIndexByName(cont, searchName): for i in range(len(cont)): if cont[i].teamName == searchName: return i return -1 n, cont = int(input()), [] for i in range(n): cont.append(Info(input(), 0, 0, 0)) ''' 01234567890123456789 line = 'Barcelona-Real 15:10' ''' for i in range(n * (n - 1) // 2): line = input() dashInd = line.index('-') spaceInd = line.index(' ') colonInd = line.index(':') team1Name = line[:dashInd] team2Name = line[dashInd + 1:spaceInd] goal1 = int(line[spaceInd + 1: colonInd]) goal2 = int(line[colonInd + 1:]) team1Index = findIndexByName(cont, team1Name) team2Index = findIndexByName(cont, team2Name) # update points if goal1 > goal2: cont[team1Index].points += 3 elif goal1 < goal2: cont[team2Index].points += 3 else: cont[team1Index].points += 1 cont[team2Index].points += 1 # update goal diff cont[team1Index].goalDiff += goal1 - goal2 cont[team2Index].goalDiff += goal2 - goal1 # update scored goals cont[team1Index].scoredGoals += goal1 cont[team2Index].scoredGoals += goal2 cont.sort(key=lambda obj: (obj.points, obj.goalDiff, obj.scoredGoals), reverse=True) del cont[n // 2:] cont.sort(key=lambda obj: obj.teamName) # out(cont) for obj in cont: print(obj.teamName) ```

3.976237

227

B

Effective Approach

PROGRAMMING

1,100

[ "implementation" ]

null

Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array. According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is. Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent. To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand. But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.

The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array. The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.

Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

[ "2\n1 2\n1\n1\n", "2\n2 1\n1\n1\n", "3\n3 1 2\n3\n1 2 3\n" ]

[ "1 2\n", "2 1\n", "6 6\n" ]

In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element). In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).

1,000

[ { "input": "2\n1 2\n1\n1", "output": "1 2" }, { "input": "2\n2 1\n1\n1", "output": "2 1" }, { "input": "3\n3 1 2\n3\n1 2 3", "output": "6 6" }, { "input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5", "output": "58 32" }, { "input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4", "output": "9 2" }, { "input": "10\n5 2 10 8 3 1 9 7 6 4\n9\n2 5 9 2 3 2 5 9 8", "output": "31 68" }, { "input": "9\n3 8 4 7 1 2 5 6 9\n3\n2 7 1", "output": "15 15" }, { "input": "9\n5 3 8 4 2 6 1 7 9\n4\n6 1 9 2", "output": "27 13" }, { "input": "4\n1 3 2 4\n4\n3 1 2 3", "output": "8 12" }, { "input": "3\n1 2 3\n8\n3 2 1 1 2 3 1 2", "output": "15 17" } ]

1,684,655,992

2,147,483,647

Python 3

COMPILATION_ERROR

TESTS

0

#include <iostream> #include <bits/stdc++.h> using namespace std; int main() { long int n,i,m,s1=0,j,s2; cin>>n; long int a[n]; for(i=0;i<n;i++) { cin>>a[i]; } cin>>m; long int b[m]; for(i=0;i<m;i++) { cin>>b[i]; } for(j=0;j<m;j++) for(i=0;i<n;i++) { if(a[i]==b[j]) { s1+=i+1; } } s2=(m*n)+m-(m*s1); cout<<s1<<" "<<abs(s2); }

Title: Effective Approach Time Limit: None seconds Memory Limit: None megabytes Problem Description: Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array. According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is. Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent. To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand. But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array. The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. Output Specification: Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. Demo Input: ['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n'] Demo Output: ['1 2\n', '2 1\n', '6 6\n'] Note: In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element). In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).

```python #include <iostream> #include <bits/stdc++.h> using namespace std; int main() { long int n,i,m,s1=0,j,s2; cin>>n; long int a[n]; for(i=0;i<n;i++) { cin>>a[i]; } cin>>m; long int b[m]; for(i=0;i<m;i++) { cin>>b[i]; } for(j=0;j<m;j++) for(i=0;i<n;i++) { if(a[i]==b[j]) { s1+=i+1; } } s2=(m*n)+m-(m*s1); cout<<s1<<" "<<abs(s2); } ```

-1

999

A

Mishka and Contest

PROGRAMMING

800

[ "brute force", "implementation" ]

null

Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$. Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list. Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list. How many problems can Mishka solve?

The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill. The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.

Print one integer — the maximum number of problems Mishka can solve.

[ "8 4\n4 2 3 1 5 1 6 4\n", "5 2\n3 1 2 1 3\n", "5 100\n12 34 55 43 21\n" ]

[ "5\n", "0\n", "5\n" ]

In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$. In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$. In the third example, Mishka's solving skill is so amazing that he can solve all the problems.

0

[ { "input": "8 4\n4 2 3 1 5 1 6 4", "output": "5" }, { "input": "5 2\n3 1 2 1 3", "output": "0" }, { "input": "5 100\n12 34 55 43 21", "output": "5" }, { "input": "100 100\n44 47 36 83 76 94 86 69 31 2 22 77 37 51 10 19 25 78 53 25 1 29 48 95 35 53 22 72 49 86 60 38 13 91 89 18 54 19 71 2 25 33 65 49 53 5 95 90 100 68 25 5 87 48 45 72 34 14 100 44 94 75 80 26 25 7 57 82 49 73 55 43 42 60 34 8 51 11 71 41 81 23 20 89 12 72 68 26 96 92 32 63 13 47 19 9 35 56 79 62", "output": "100" }, { "input": "100 99\n84 82 43 4 71 3 30 92 15 47 76 43 2 17 76 4 1 33 24 96 44 98 75 99 59 11 73 27 67 17 8 88 69 41 44 22 91 48 4 46 42 21 21 67 85 51 57 84 11 100 100 59 39 72 89 82 74 19 98 14 37 97 20 78 38 52 44 83 19 83 69 32 56 6 93 13 98 80 80 2 33 71 11 15 55 51 98 58 16 91 39 32 83 58 77 79 88 81 17 98", "output": "98" }, { "input": "100 69\n80 31 12 89 16 35 8 28 39 12 32 51 42 67 64 53 17 88 63 97 29 41 57 28 51 33 82 75 93 79 57 86 32 100 83 82 99 33 1 27 86 22 65 15 60 100 42 37 38 85 26 43 90 62 91 13 1 92 16 20 100 19 28 30 23 6 5 69 24 22 9 1 10 14 28 14 25 9 32 8 67 4 39 7 10 57 15 7 8 35 62 6 53 59 62 13 24 7 53 2", "output": "39" }, { "input": "100 2\n2 2 2 2 1 1 1 2 1 2 2 2 1 2 2 2 2 1 2 1 2 1 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 2 1 2 1 2 1 1 2 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 2 1 1 1 2 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 2 2 16", "output": "99" }, { "input": "100 3\n86 53 82 40 2 20 59 2 46 63 75 49 24 81 70 22 9 9 93 72 47 23 29 77 78 51 17 59 19 71 35 3 20 60 70 9 11 96 71 94 91 19 88 93 50 49 72 19 53 30 38 67 62 71 81 86 5 26 5 32 63 98 1 97 22 32 87 65 96 55 43 85 56 37 56 67 12 100 98 58 77 54 18 20 33 53 21 66 24 64 42 71 59 32 51 69 49 79 10 1", "output": "1" }, { "input": "13 7\n1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "13" }, { "input": "1 5\n4", "output": "1" }, { "input": "3 2\n1 4 1", "output": "2" }, { "input": "1 2\n100", "output": "0" }, { "input": "7 4\n4 2 3 4 4 2 3", "output": "7" }, { "input": "1 2\n1", "output": "1" }, { "input": "1 2\n15", "output": "0" }, { "input": "2 1\n1 1", "output": "2" }, { "input": "5 3\n3 4 3 2 1", "output": "4" }, { "input": "1 1\n2", "output": "0" }, { "input": "1 5\n1", "output": "1" }, { "input": "6 6\n7 1 1 1 1 1", "output": "5" }, { "input": "5 5\n6 5 5 5 5", "output": "4" }, { "input": "1 4\n2", "output": "1" }, { "input": "9 4\n1 2 1 2 4 2 1 2 1", "output": "9" }, { "input": "1 1\n1", "output": "1" }, { "input": "1 10\n5", "output": "1" }, { "input": "5 5\n1 1 1 1 1", "output": "5" }, { "input": "100 10\n2 5 1 10 10 2 7 7 9 4 1 8 1 1 8 4 7 9 10 5 7 9 5 6 7 2 7 5 3 2 1 82 4 80 9 8 6 1 10 7 5 7 1 5 6 7 19 4 2 4 6 2 1 8 31 6 2 2 57 42 3 2 7 1 9 5 10 8 5 4 10 8 3 5 8 7 2 7 6 5 3 3 4 10 6 7 10 8 7 10 7 2 4 6 8 10 10 2 6 4", "output": "71" }, { "input": "100 90\n17 16 5 51 17 62 24 45 49 41 90 30 19 78 67 66 59 34 28 47 42 8 33 77 90 41 61 16 86 33 43 71 90 95 23 9 56 41 24 90 31 12 77 36 90 67 47 15 92 50 79 88 42 19 21 79 86 60 41 26 47 4 70 62 44 90 82 89 84 91 54 16 90 53 29 69 21 44 18 28 88 74 56 43 12 76 10 22 34 24 27 52 28 76 90 75 5 29 50 90", "output": "63" }, { "input": "100 10\n6 4 8 4 1 9 4 8 5 2 2 5 2 6 10 2 2 5 3 5 2 3 10 5 2 9 1 1 6 1 5 9 16 42 33 49 26 31 81 27 53 63 81 90 55 97 70 51 87 21 79 62 60 91 54 95 26 26 30 61 87 79 47 11 59 34 40 82 37 40 81 2 7 1 8 4 10 7 1 10 8 7 3 5 2 8 3 3 9 2 1 1 5 7 8 7 1 10 9 8", "output": "61" }, { "input": "100 90\n45 57 52 69 17 81 85 60 59 39 55 14 87 90 90 31 41 57 35 89 74 20 53 4 33 49 71 11 46 90 71 41 71 90 63 74 51 13 99 92 99 91 100 97 93 40 93 96 100 99 100 92 98 96 78 91 91 91 91 100 94 97 95 97 96 95 17 13 45 35 54 26 2 74 6 51 20 3 73 90 90 42 66 43 86 28 84 70 37 27 90 30 55 80 6 58 57 51 10 22", "output": "72" }, { "input": "100 10\n10 2 10 10 10 10 10 10 10 7 10 10 10 10 10 10 9 10 10 10 10 10 10 10 10 7 9 10 10 10 37 10 4 10 10 10 59 5 95 10 10 10 10 39 10 10 10 10 10 10 10 5 10 10 10 10 10 10 10 10 10 10 10 10 66 10 10 10 10 10 5 10 10 10 10 10 10 44 10 10 10 10 10 10 10 10 10 10 10 7 10 10 10 10 10 10 10 10 10 2", "output": "52" }, { "input": "100 90\n57 90 90 90 90 90 90 90 81 90 3 90 39 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 92 90 90 90 90 90 90 90 90 98 90 90 90 90 90 90 90 90 90 90 90 90 90 54 90 90 90 90 90 62 90 90 91 90 90 90 90 90 90 91 90 90 90 90 90 90 90 3 90 90 90 90 90 90 90 2 90 90 90 90 90 90 90 90 90 2 90 90 90 90 90", "output": "60" }, { "input": "100 10\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 78 90 61 40 87 39 91 50 64 30 10 24 10 55 28 11 28 35 26 26 10 57 45 67 14 99 96 51 67 79 59 11 21 55 70 33 10 16 92 70 38 50 66 52 5 10 10 10 2 4 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 10 10 10 10 8 10 10 10 10 10", "output": "56" }, { "input": "100 90\n90 90 90 90 90 90 55 21 90 90 90 90 90 90 90 90 90 90 69 83 90 90 90 90 90 90 90 90 93 95 92 98 92 97 91 92 92 91 91 95 94 95 100 100 96 97 94 93 90 90 95 95 97 99 90 95 98 91 94 96 99 99 94 95 95 97 99 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 12 90 3 90 90 90 90 90 90 90", "output": "61" }, { "input": "100 49\n71 25 14 36 36 48 36 49 28 40 49 49 49 38 40 49 33 22 49 49 14 46 8 44 49 11 37 49 40 49 2 49 3 49 37 49 49 11 25 49 49 32 49 11 49 30 16 21 49 49 23 24 30 49 49 49 49 49 49 27 49 42 49 49 20 32 30 29 35 49 30 49 9 49 27 25 5 49 49 42 49 20 49 35 49 22 15 49 49 49 19 49 29 28 13 49 22 7 6 24", "output": "99" }, { "input": "100 50\n38 68 9 6 50 18 19 50 50 20 33 34 43 50 24 50 50 2 50 50 50 50 50 21 30 50 41 40 50 50 50 50 50 7 50 21 19 23 1 50 24 50 50 50 25 50 50 50 50 50 50 50 7 24 28 18 50 5 43 50 20 50 13 50 50 16 50 3 2 24 50 50 18 5 50 4 50 50 38 50 33 49 12 33 11 14 50 50 50 33 50 50 50 50 50 50 7 4 50 50", "output": "99" }, { "input": "100 48\n8 6 23 47 29 48 48 48 48 48 48 26 24 48 48 48 3 48 27 28 41 45 9 29 48 48 48 48 48 48 48 48 48 48 47 23 48 48 48 5 48 22 40 48 48 48 20 48 48 57 48 32 19 48 33 2 4 19 48 48 39 48 16 48 48 44 48 48 48 48 29 14 25 43 46 7 48 19 30 48 18 8 39 48 30 47 35 18 48 45 48 48 30 13 48 48 48 17 9 48", "output": "99" }, { "input": "100 57\n57 9 57 4 43 57 57 57 57 26 57 18 57 57 57 57 57 57 57 47 33 57 57 43 57 57 55 57 14 57 57 4 1 57 57 57 57 57 46 26 57 57 57 57 57 57 57 39 57 57 57 5 57 12 11 57 57 57 25 37 34 57 54 18 29 57 39 57 5 57 56 34 57 24 7 57 57 57 2 57 57 57 57 1 55 39 19 57 57 57 57 21 3 40 13 3 57 57 62 57", "output": "99" }, { "input": "100 51\n51 51 38 51 51 45 51 51 51 18 51 36 51 19 51 26 37 51 11 51 45 34 51 21 51 51 33 51 6 51 51 51 21 47 51 13 51 51 30 29 50 51 51 51 51 51 51 45 14 51 2 51 51 23 9 51 50 23 51 29 34 51 40 32 1 36 31 51 11 51 51 47 51 51 51 51 51 51 51 50 39 51 14 4 4 12 3 11 51 51 51 51 41 51 51 51 49 37 5 93", "output": "99" }, { "input": "100 50\n87 91 95 73 50 50 16 97 39 24 58 50 33 89 42 37 50 50 12 71 3 55 50 50 80 10 76 50 52 36 88 44 66 69 86 71 77 50 72 50 21 55 50 50 78 61 75 89 65 2 50 69 62 47 11 92 97 77 41 31 55 29 35 51 36 48 50 91 92 86 50 36 50 94 51 74 4 27 55 63 50 36 87 50 67 7 65 75 20 96 88 50 41 73 35 51 66 21 29 33", "output": "3" }, { "input": "100 50\n50 37 28 92 7 76 50 50 50 76 100 57 50 50 50 32 76 50 8 72 14 8 50 91 67 50 55 82 50 50 24 97 88 50 59 61 68 86 44 15 61 67 88 50 40 50 36 99 1 23 63 50 88 59 76 82 99 76 68 50 50 30 31 68 57 98 71 12 15 60 35 79 90 6 67 50 50 50 50 68 13 6 50 50 16 87 84 50 67 67 50 64 50 58 50 50 77 51 50 51", "output": "3" }, { "input": "100 50\n43 50 50 91 97 67 6 50 86 50 76 60 50 59 4 56 11 38 49 50 37 50 50 20 60 47 33 54 95 58 22 50 77 77 72 9 57 40 81 57 95 50 81 63 62 76 13 87 50 39 74 69 50 99 63 1 11 62 84 31 97 99 56 73 70 36 45 100 28 91 93 9 19 52 73 50 83 58 84 52 86 12 50 44 64 52 97 50 12 71 97 52 87 66 83 66 86 50 9 49", "output": "6" }, { "input": "88 10\n10 8 1 10 10 1 3 7 10 5 8 8 10 2 7 10 10 10 10 10 1 10 10 10 10 1 2 9 10 9 10 10 10 64 100 25 10 12 9 52 13 8 10 56 10 4 10 7 10 3 10 79 74 8 73 10 10 10 9 10 3 5 10 10 10 5 1 10 10 4 3 10 10 10 4 10 6 4 10 10 10 10 3 3 8 5 6 8", "output": "66" }, { "input": "100 50\n80 39 33 69 75 50 23 88 50 50 67 90 87 50 29 15 55 32 60 50 50 50 38 95 62 50 50 88 8 97 45 50 42 12 22 93 49 50 24 50 50 71 60 4 50 72 57 57 50 50 50 83 69 17 1 31 72 55 50 11 50 80 93 41 91 94 20 60 50 50 51 48 53 56 76 73 50 72 19 98 50 50 50 50 50 28 48 45 62 11 16 67 93 88 63 50 50 66 48 95", "output": "0" }, { "input": "100 50\n70 50 38 50 38 50 32 30 50 31 26 42 50 33 34 50 50 50 28 21 50 44 50 47 50 50 9 40 50 50 50 50 50 42 50 50 16 50 50 3 24 50 50 50 4 26 50 2 50 50 33 1 27 50 50 50 8 29 50 23 33 50 6 29 50 50 15 50 50 50 32 50 43 50 50 50 31 50 4 50 50 31 50 50 31 16 50 17 50 17 31 13 25 16 50 10 50 47 50 66", "output": "0" }, { "input": "2 8\n8 8", "output": "2" }, { "input": "1 6\n3", "output": "1" }, { "input": "1 5\n5", "output": "1" } ]

1,616,070,378

2,147,483,647

Python 3

OK

TESTS

48

77

0

n, k = map(int, input().split()) lst = [int(i) for i in input().split()] count = 0 for i in lst: if i > k: break else: count += 1 for i in lst[::-1]: if i > k: break else: count += 1 print(min(n, count))

Title: Mishka and Contest Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$. Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list. Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list. How many problems can Mishka solve? Input Specification: The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill. The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list. Output Specification: Print one integer — the maximum number of problems Mishka can solve. Demo Input: ['8 4\n4 2 3 1 5 1 6 4\n', '5 2\n3 1 2 1 3\n', '5 100\n12 34 55 43 21\n'] Demo Output: ['5\n', '0\n', '5\n'] Note: In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$. In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$. In the third example, Mishka's solving skill is so amazing that he can solve all the problems.

```python n, k = map(int, input().split()) lst = [int(i) for i in input().split()] count = 0 for i in lst: if i > k: break else: count += 1 for i in lst[::-1]: if i > k: break else: count += 1 print(min(n, count)) ```

3

841

B

Godsend

PROGRAMMING

1,100

[ "games", "math" ]

null

Leha somehow found an array consisting of *n* integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?

First line of input data contains single integer *n* (1<=≤<=*n*<=≤<=106) — length of the array. Next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109).

Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).

[ "4\n1 3 2 3\n", "2\n2 2\n" ]

[ "First\n", "Second\n" ]

In first sample first player remove whole array in one move and win. In second sample first player can't make a move and lose.

1,000

[ { "input": "4\n1 3 2 3", "output": "First" }, { "input": "2\n2 2", "output": "Second" }, { "input": "4\n2 4 6 8", "output": "Second" }, { "input": "5\n1 1 1 1 1", "output": "First" }, { "input": "4\n720074544 345031254 849487632 80870826", "output": "Second" }, { "input": "1\n0", "output": "Second" }, { "input": "1\n999999999", "output": "First" }, { "input": "2\n1 999999999", "output": "First" }, { "input": "4\n3 3 4 4", "output": "First" }, { "input": "2\n1 2", "output": "First" }, { "input": "8\n2 2 2 1 1 2 2 2", "output": "First" }, { "input": "5\n3 3 2 2 2", "output": "First" }, { "input": "4\n0 1 1 0", "output": "First" }, { "input": "3\n1 2 2", "output": "First" }, { "input": "6\n2 2 1 1 4 2", "output": "First" }, { "input": "8\n2 2 2 3 3 2 2 2", "output": "First" }, { "input": "4\n2 3 3 4", "output": "First" }, { "input": "10\n2 2 2 2 3 1 2 2 2 2", "output": "First" }, { "input": "6\n2 2 1 1 2 2", "output": "First" }, { "input": "3\n1 1 2", "output": "First" }, { "input": "6\n2 4 3 3 4 6", "output": "First" }, { "input": "6\n4 4 3 3 4 4", "output": "First" }, { "input": "4\n1 1 2 2", "output": "First" }, { "input": "4\n1 3 5 7", "output": "First" }, { "input": "4\n2 1 1 2", "output": "First" }, { "input": "4\n1 3 3 2", "output": "First" }, { "input": "5\n3 2 2 2 2", "output": "First" }, { "input": "3\n2 1 1", "output": "First" }, { "input": "4\n1000000000 1000000000 1000000000 99999999", "output": "First" }, { "input": "4\n2 2 1 1", "output": "First" }, { "input": "5\n2 3 2 3 2", "output": "First" }, { "input": "1\n1", "output": "First" }, { "input": "4\n1000000000 1000000000 1000000000 1", "output": "First" }, { "input": "5\n2 2 2 1 1", "output": "First" }, { "input": "6\n2 1 1 1 1 2", "output": "First" }, { "input": "6\n1 2 2 2 2 1", "output": "First" }, { "input": "11\n2 2 2 2 2 1 2 2 2 2 2", "output": "First" }, { "input": "5\n1 3 2 2 2", "output": "First" }, { "input": "3\n2 3 2", "output": "First" }, { "input": "2\n1 1", "output": "First" }, { "input": "5\n4 4 4 3 3", "output": "First" }, { "input": "5\n3 3 4 4 4", "output": "First" }, { "input": "1\n2", "output": "Second" } ]

1,504,242,625

2,147,483,647

Python 3

OK

TESTS

88

529

57,241,600

n = input() print("First" if any(int(i) % 2 for i in list(input().split())) else "Second")

Title: Godsend Time Limit: None seconds Memory Limit: None megabytes Problem Description: Leha somehow found an array consisting of *n* integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally? Input Specification: First line of input data contains single integer *n* (1<=≤<=*n*<=≤<=106) — length of the array. Next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109). Output Specification: Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes). Demo Input: ['4\n1 3 2 3\n', '2\n2 2\n'] Demo Output: ['First\n', 'Second\n'] Note: In first sample first player remove whole array in one move and win. In second sample first player can't make a move and lose.

```python n = input() print("First" if any(int(i) % 2 for i in list(input().split())) else "Second") ```

3

884

D

Boxes And Balls

PROGRAMMING

2,300

[ "data structures", "greedy" ]

null

Ivan has *n* different boxes. The first of them contains some balls of *n* different colors. Ivan wants to play a strange game. He wants to distribute the balls into boxes in such a way that for every *i* (1<=≤<=*i*<=≤<=*n*) *i*-th box will contain all balls with color *i*. In order to do this, Ivan will make some turns. Each turn he does the following: 1. Ivan chooses any non-empty box and takes all balls from this box; 1. Then Ivan chooses any *k* empty boxes (the box from the first step becomes empty, and Ivan is allowed to choose it), separates the balls he took on the previous step into *k* non-empty groups and puts each group into one of the boxes. He should put each group into a separate box. He can choose either *k*<==<=2 or *k*<==<=3. The penalty of the turn is the number of balls Ivan takes from the box during the first step of the turn. And penalty of the game is the total penalty of turns made by Ivan until he distributes all balls to corresponding boxes. Help Ivan to determine the minimum possible penalty of the game!

The first line contains one integer number *n* (1<=≤<=*n*<=≤<=200000) — the number of boxes and colors. The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is the number of balls with color *i*.

Print one number — the minimum possible penalty of the game.

[ "3\n1 2 3\n", "4\n2 3 4 5\n" ]

[ "6\n", "19\n" ]

In the first example you take all the balls from the first box, choose *k* = 3 and sort all colors to corresponding boxes. Penalty is 6. In the second example you make two turns: 1. Take all the balls from the first box, choose *k* = 3, put balls of color 3 to the third box, of color 4 — to the fourth box and the rest put back into the first box. Penalty is 14; 1. Take all the balls from the first box, choose *k* = 2, put balls of color 1 to the first box, of color 2 — to the second box. Penalty is 5. Total penalty is 19.

0

[ { "input": "3\n1 2 3", "output": "6" }, { "input": "4\n2 3 4 5", "output": "19" }, { "input": "6\n1 4 4 4 4 4", "output": "38" }, { "input": "8\n821407370 380061316 428719552 90851747 825473738 704702117 845629927 245820158", "output": "8176373828" }, { "input": "1\n10", "output": "0" }, { "input": "1\n4", "output": "0" }, { "input": "1\n12312", "output": "0" }, { "input": "1\n1", "output": "0" }, { "input": "2\n3 4", "output": "7" } ]

1,509,472,543

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

46

0

n=int(input()) a=sorted(list(map(int,input().split())),reverse=True) a=a+[0]*(n%2) straf=sum(map(lambda x,y:sum(x)*y,zip(a[::2],a[1::2]),[i for i in range (1,n+1)])) print(straf if a[-1] != 0 else straf-a[-2])

Title: Boxes And Balls Time Limit: None seconds Memory Limit: None megabytes Problem Description: Ivan has *n* different boxes. The first of them contains some balls of *n* different colors. Ivan wants to play a strange game. He wants to distribute the balls into boxes in such a way that for every *i* (1<=≤<=*i*<=≤<=*n*) *i*-th box will contain all balls with color *i*. In order to do this, Ivan will make some turns. Each turn he does the following: 1. Ivan chooses any non-empty box and takes all balls from this box; 1. Then Ivan chooses any *k* empty boxes (the box from the first step becomes empty, and Ivan is allowed to choose it), separates the balls he took on the previous step into *k* non-empty groups and puts each group into one of the boxes. He should put each group into a separate box. He can choose either *k*<==<=2 or *k*<==<=3. The penalty of the turn is the number of balls Ivan takes from the box during the first step of the turn. And penalty of the game is the total penalty of turns made by Ivan until he distributes all balls to corresponding boxes. Help Ivan to determine the minimum possible penalty of the game! Input Specification: The first line contains one integer number *n* (1<=≤<=*n*<=≤<=200000) — the number of boxes and colors. The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is the number of balls with color *i*. Output Specification: Print one number — the minimum possible penalty of the game. Demo Input: ['3\n1 2 3\n', '4\n2 3 4 5\n'] Demo Output: ['6\n', '19\n'] Note: In the first example you take all the balls from the first box, choose *k* = 3 and sort all colors to corresponding boxes. Penalty is 6. In the second example you make two turns: 1. Take all the balls from the first box, choose *k* = 3, put balls of color 3 to the third box, of color 4 — to the fourth box and the rest put back into the first box. Penalty is 14; 1. Take all the balls from the first box, choose *k* = 2, put balls of color 1 to the first box, of color 2 — to the second box. Penalty is 5. Total penalty is 19.

```python n=int(input()) a=sorted(list(map(int,input().split())),reverse=True) a=a+[0]*(n%2) straf=sum(map(lambda x,y:sum(x)*y,zip(a[::2],a[1::2]),[i for i in range (1,n+1)])) print(straf if a[-1] != 0 else straf-a[-2]) ```

0

631

D

Messenger

PROGRAMMING

2,100

[ "data structures", "hashing", "implementation", "string suffix structures", "strings" ]

null

Each employee of the "Blake Techologies" company uses a special messaging app "Blake Messenger". All the stuff likes this app and uses it constantly. However, some important futures are missing. For example, many users want to be able to search through the message history. It was already announced that the new feature will appear in the nearest update, when developers faced some troubles that only you may help them to solve. All the messages are represented as a strings consisting of only lowercase English letters. In order to reduce the network load strings are represented in the special compressed form. Compression algorithm works as follows: string is represented as a concatenation of *n* blocks, each block containing only equal characters. One block may be described as a pair (*l**i*,<=*c**i*), where *l**i* is the length of the *i*-th block and *c**i* is the corresponding letter. Thus, the string *s* may be written as the sequence of pairs . Your task is to write the program, that given two compressed string *t* and *s* finds all occurrences of *s* in *t*. Developers know that there may be many such occurrences, so they only ask you to find the number of them. Note that *p* is the starting position of some occurrence of *s* in *t* if and only if *t**p**t**p*<=+<=1...*t**p*<=+<=|*s*|<=-<=1<==<=*s*, where *t**i* is the *i*-th character of string *t*. Note that the way to represent the string in compressed form may not be unique. For example string "aaaa" may be given as , , ...

The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=200<=000) — the number of blocks in the strings *t* and *s*, respectively. The second line contains the descriptions of *n* parts of string *t* in the format "*l**i*-*c**i*" (1<=≤<=*l**i*<=≤<=1<=000<=000) — the length of the *i*-th part and the corresponding lowercase English letter. The second line contains the descriptions of *m* parts of string *s* in the format "*l**i*-*c**i*" (1<=≤<=*l**i*<=≤<=1<=000<=000) — the length of the *i*-th part and the corresponding lowercase English letter.

Print a single integer — the number of occurrences of *s* in *t*.

[ "5 3\n3-a 2-b 4-c 3-a 2-c\n2-a 2-b 1-c\n", "6 1\n3-a 6-b 7-a 4-c 8-e 2-a\n3-a\n", "5 5\n1-h 1-e 1-l 1-l 1-o\n1-w 1-o 1-r 1-l 1-d\n" ]

[ "1", "6", "0" ]

In the first sample, *t* = "aaabbccccaaacc", and string *s* = "aabbc". The only occurrence of string *s* in string *t* starts at position *p* = 2. In the second sample, *t* = "aaabbbbbbaaaaaaacccceeeeeeeeaa", and *s* = "aaa". The occurrences of *s* in *t* start at positions *p* = 1, *p* = 10, *p* = 11, *p* = 12, *p* = 13 and *p* = 14.

2,000

[ { "input": "5 3\n3-a 2-b 4-c 3-a 2-c\n2-a 2-b 1-c", "output": "1" }, { "input": "6 1\n3-a 6-b 7-a 4-c 8-e 2-a\n3-a", "output": "6" }, { "input": "5 5\n1-h 1-e 1-l 1-l 1-o\n1-w 1-o 1-r 1-l 1-d", "output": "0" }, { "input": "9 3\n1-h 1-e 2-l 1-o 1-w 1-o 1-r 1-l 1-d\n2-l 1-o 1-w", "output": "1" }, { "input": "5 3\n1-m 1-i 2-r 1-o 1-r\n1-m 1-i 1-r", "output": "1" }, { "input": "9 2\n1-a 2-b 1-o 1-k 1-l 1-m 1-a 3-b 3-z\n1-a 2-b", "output": "2" }, { "input": "10 3\n1-b 1-a 2-b 1-a 1-b 1-a 4-b 1-a 1-a 2-b\n1-b 1-a 1-b", "output": "3" }, { "input": "4 2\n7-a 3-b 2-c 11-a\n3-a 4-a", "output": "6" }, { "input": "4 3\n8-b 2-a 7-b 3-a\n3-b 2-b 1-a", "output": "2" }, { "input": "1 1\n12344-a\n12345-a", "output": "0" }, { "input": "1 1\n5352-k\n5234-j", "output": "0" }, { "input": "1 1\n6543-o\n34-o", "output": "6510" }, { "input": "1 1\n1-z\n1-z", "output": "1" }, { "input": "5 2\n7-a 6-b 6-a 5-b 2-b\n6-a 7-b", "output": "1" }, { "input": "10 3\n7-a 1-c 6-b 1-c 8-a 1-c 8-b 6-a 2-c 5-b\n5-a 1-c 4-b", "output": "2" }, { "input": "4 2\n10-c 3-c 2-d 8-a\n6-a 1-b", "output": "0" }, { "input": "4 1\n10-a 2-b 8-d 11-e\n1-c", "output": "0" }, { "input": "28 7\n1-a 1-b 1-c 1-d 1-e 1-f 1-t 1-a 1-b 1-c 1-d 1-e 1-f 1-j 1-a 1-b 1-c 1-d 1-e 1-f 1-g 1-a 1-b 1-c 1-d 1-e 1-f 2-g\n1-a 1-b 1-c 1-d 1-e 1-f 1-g", "output": "2" }, { "input": "10 3\n2-w 4-l 2-w 4-l 2-w 5-l 2-w 6-l 3-w 3-l\n2-l 2-w 2-l", "output": "3" }, { "input": "15 7\n1-b 1-a 1-b 1-c 1-b 1-a 1-b 1-c 1-b 1-a 1-b 1-c 1-b 1-a 1-b\n1-b 1-a 1-b 1-c 1-b 1-a 1-b", "output": "3" }, { "input": "15 7\n1-b 2-a 1-b 1-c 1-b 1-a 1-b 1-c 1-b 2-a 1-b 1-c 1-b 1-a 1-b\n1-b 2-a 1-b 1-c 1-b 1-a 1-b", "output": "2" }, { "input": "2 2\n1-a 1-b\n2-a 1-b", "output": "0" }, { "input": "8 5\n1-a 1-b 1-c 1-a 2-b 1-c 1-a 1-b\n1-a 1-b 1-c 1-a 1-b", "output": "1" }, { "input": "9 5\n7-a 6-b 7-a 6-b 7-a 6-b 8-a 6-b 7-a\n7-a 6-b 7-a 6-b 7-a", "output": "2" } ]

1,689,645,964

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

0

46

0

print("_RANDOM_GUESS_1689645962.2429454")# 1689645962.2429676

Title: Messenger Time Limit: None seconds Memory Limit: None megabytes Problem Description: Each employee of the "Blake Techologies" company uses a special messaging app "Blake Messenger". All the stuff likes this app and uses it constantly. However, some important futures are missing. For example, many users want to be able to search through the message history. It was already announced that the new feature will appear in the nearest update, when developers faced some troubles that only you may help them to solve. All the messages are represented as a strings consisting of only lowercase English letters. In order to reduce the network load strings are represented in the special compressed form. Compression algorithm works as follows: string is represented as a concatenation of *n* blocks, each block containing only equal characters. One block may be described as a pair (*l**i*,<=*c**i*), where *l**i* is the length of the *i*-th block and *c**i* is the corresponding letter. Thus, the string *s* may be written as the sequence of pairs . Your task is to write the program, that given two compressed string *t* and *s* finds all occurrences of *s* in *t*. Developers know that there may be many such occurrences, so they only ask you to find the number of them. Note that *p* is the starting position of some occurrence of *s* in *t* if and only if *t**p**t**p*<=+<=1...*t**p*<=+<=|*s*|<=-<=1<==<=*s*, where *t**i* is the *i*-th character of string *t*. Note that the way to represent the string in compressed form may not be unique. For example string "aaaa" may be given as , , ... Input Specification: The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=200<=000) — the number of blocks in the strings *t* and *s*, respectively. The second line contains the descriptions of *n* parts of string *t* in the format "*l**i*-*c**i*" (1<=≤<=*l**i*<=≤<=1<=000<=000) — the length of the *i*-th part and the corresponding lowercase English letter. The second line contains the descriptions of *m* parts of string *s* in the format "*l**i*-*c**i*" (1<=≤<=*l**i*<=≤<=1<=000<=000) — the length of the *i*-th part and the corresponding lowercase English letter. Output Specification: Print a single integer — the number of occurrences of *s* in *t*. Demo Input: ['5 3\n3-a 2-b 4-c 3-a 2-c\n2-a 2-b 1-c\n', '6 1\n3-a 6-b 7-a 4-c 8-e 2-a\n3-a\n', '5 5\n1-h 1-e 1-l 1-l 1-o\n1-w 1-o 1-r 1-l 1-d\n'] Demo Output: ['1', '6', '0'] Note: In the first sample, *t* = "aaabbccccaaacc", and string *s* = "aabbc". The only occurrence of string *s* in string *t* starts at position *p* = 2. In the second sample, *t* = "aaabbbbbbaaaaaaacccceeeeeeeeaa", and *s* = "aaa". The occurrences of *s* in *t* start at positions *p* = 1, *p* = 10, *p* = 11, *p* = 12, *p* = 13 and *p* = 14.

```python print("_RANDOM_GUESS_1689645962.2429454")# 1689645962.2429676 ```

0

none

0

[ "none" ]

null

Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture). The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right. Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!

The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.

Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.

[ "4 3 9\n", "4 3 24\n", "2 4 4\n" ]

[ "2 2 L\n", "4 3 R\n", "1 2 R\n" ]

The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example. In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.

0

[ { "input": "4 3 9", "output": "2 2 L" }, { "input": "4 3 24", "output": "4 3 R" }, { "input": "2 4 4", "output": "1 2 R" }, { "input": "3 10 24", "output": "2 2 R" }, { "input": "10 3 59", "output": "10 3 L" }, { "input": "10000 10000 160845880", "output": "8043 2940 R" }, { "input": "1 1 1", "output": "1 1 L" }, { "input": "1 1 2", "output": "1 1 R" }, { "input": "1 10000 1", "output": "1 1 L" }, { "input": "1 10000 20000", "output": "1 10000 R" }, { "input": "10000 1 1", "output": "1 1 L" }, { "input": "10000 1 10000", "output": "5000 1 R" }, { "input": "10000 1 20000", "output": "10000 1 R" }, { "input": "3 2 1", "output": "1 1 L" }, { "input": "3 2 2", "output": "1 1 R" }, { "input": "3 2 3", "output": "1 2 L" }, { "input": "3 2 4", "output": "1 2 R" }, { "input": "3 2 5", "output": "2 1 L" }, { "input": "3 2 6", "output": "2 1 R" }, { "input": "3 2 7", "output": "2 2 L" }, { "input": "3 2 8", "output": "2 2 R" }, { "input": "3 2 9", "output": "3 1 L" }, { "input": "3 2 10", "output": "3 1 R" }, { "input": "3 2 11", "output": "3 2 L" }, { "input": "3 2 12", "output": "3 2 R" }, { "input": "300 2000 1068628", "output": "268 314 R" }, { "input": "300 2000 584756", "output": "147 378 R" }, { "input": "300 2000 268181", "output": "68 91 L" }, { "input": "10000 9999 186450844", "output": "9324 4745 R" }, { "input": "10000 9999 197114268", "output": "9857 6990 R" }, { "input": "10000 9999 112390396", "output": "5621 818 R" }, { "input": "10000 10000 1", "output": "1 1 L" }, { "input": "10000 10000 2", "output": "1 1 R" }, { "input": "10000 10000 100000001", "output": "5001 1 L" }, { "input": "10000 10000 199999999", "output": "10000 10000 L" }, { "input": "10000 10000 200000000", "output": "10000 10000 R" }, { "input": "1 2 1", "output": "1 1 L" }, { "input": "1 2 2", "output": "1 1 R" }, { "input": "1 2 3", "output": "1 2 L" }, { "input": "1 2 4", "output": "1 2 R" }, { "input": "2 1 1", "output": "1 1 L" }, { "input": "2 1 2", "output": "1 1 R" }, { "input": "2 1 3", "output": "2 1 L" }, { "input": "2 1 4", "output": "2 1 R" }, { "input": "4 3 7", "output": "2 1 L" }, { "input": "1 1 1", "output": "1 1 L" } ]

1,556,671,904

1,304

Python 3

OK

TESTS

46

109

0

n, m, k = map(int, input().split()) r = k // (2 * m) + int(k % (2 * m) != 0) if k % (2 * m) == 0: print(r, m, 'R') else: mod = k % (2 * m) d = mod // 2 + int(mod % 2 != 0) if mod % 2 == 0: print(r, d, 'R') else: print(r, d, 'L')

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture). The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right. Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right! Input Specification: The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place. Output Specification: Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right. Demo Input: ['4 3 9\n', '4 3 24\n', '2 4 4\n'] Demo Output: ['2 2 L\n', '4 3 R\n', '1 2 R\n'] Note: The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example. In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.

```python n, m, k = map(int, input().split()) r = k // (2 * m) + int(k % (2 * m) != 0) if k % (2 * m) == 0: print(r, m, 'R') else: mod = k % (2 * m) d = mod // 2 + int(mod % 2 != 0) if mod % 2 == 0: print(r, d, 'R') else: print(r, d, 'L') ```

3

908

A

New Year and Counting Cards

PROGRAMMING

800

[ "brute force", "implementation" ]

null

Your friend has *n* cards. You know that each card has a lowercase English letter on one side and a digit on the other. Currently, your friend has laid out the cards on a table so only one side of each card is visible. You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'. For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true. To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.

The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.

Print a single integer, the minimum number of cards you must turn over to verify your claim.

[ "ee\n", "z\n", "0ay1\n" ]

[ "2\n", "0\n", "2\n" ]

In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side. In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them. In the third sample, we need to flip the second and fourth cards.

500

[ { "input": "ee", "output": "2" }, { "input": "z", "output": "0" }, { "input": "0ay1", "output": "2" }, { "input": "0abcdefghijklmnopqrstuvwxyz1234567896", "output": "10" }, { "input": "0a0a9e9e2i2i9o9o6u6u9z9z4x4x9b9b", "output": "18" }, { "input": "01234567890123456789012345678901234567890123456789", "output": "25" }, { "input": "qwertyuioplkjhgfdsazxcvbnmqwertyuioplkjhgfdsazxcvb", "output": "10" }, { "input": "cjw2dwmr10pku4yxohe0wglktd", "output": "4" }, { "input": "6z2tx805jie8cfybwtfqvmlveec3iak5z5u3lu62vbxyqht6", "output": "13" }, { "input": "kaq7jyialrfp4ilkni90eq8v3amcbygon7py0hb8z26fbl8ss1", "output": "13" }, { "input": "hpwn50zgbmct80k9rizjqg40nycgs0acwikjqt11nr6m61krfs", "output": "8" }, { "input": "l3rw91a4m25l8iytxyeuixsegzcbm4h41ornf3pixkrmwznrzc", "output": "14" }, { "input": "2222", "output": "0" }, { "input": "13579", "output": "5" }, { "input": "1", "output": "1" }, { "input": "0", "output": "0" }, { "input": "a", "output": "1" }, { "input": "y", "output": "0" }, { "input": "w", "output": "0" }, { "input": "oo", "output": "2" }, { "input": "oy", "output": "1" }, { "input": "yo", "output": "1" }, { "input": "yy", "output": "0" }, { "input": "a0", "output": "1" }, { "input": "a9", "output": "2" }, { "input": "y0", "output": "0" }, { "input": "y7", "output": "1" }, { "input": "0a", "output": "1" }, { "input": "3a", "output": "2" }, { "input": "06", "output": "0" }, { "input": "07", "output": "1" }, { "input": "70", "output": "1" }, { "input": "77", "output": "2" }, { "input": "13570", "output": "4" }, { "input": "0000000000011111", "output": "5" }, { "input": "1357", "output": "4" }, { "input": "uuuuuuuuuuuuuuuuuu", "output": "18" }, { "input": "gabieurat", "output": "5" } ]

1,602,241,608

2,147,483,647

Python 3

OK

TESTS

38

108

307,200

def needFlip(card): verifications = set({ 97,101,105,111,117, 49,51,53,55,57 }) return ord(card) in verifications string = input() i = 0 j = len(string) - 1 flips = 0 while i <= j: if i==j and needFlip(string[i]): flips += 1 break if needFlip(string[i]): flips += 1 if needFlip(string[j]): flips += 1 i += 1 j -= 1 print(flips)

Title: New Year and Counting Cards Time Limit: None seconds Memory Limit: None megabytes Problem Description: Your friend has *n* cards. You know that each card has a lowercase English letter on one side and a digit on the other. Currently, your friend has laid out the cards on a table so only one side of each card is visible. You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'. For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true. To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true. Input Specification: The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit. Output Specification: Print a single integer, the minimum number of cards you must turn over to verify your claim. Demo Input: ['ee\n', 'z\n', '0ay1\n'] Demo Output: ['2\n', '0\n', '2\n'] Note: In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side. In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them. In the third sample, we need to flip the second and fourth cards.

```python def needFlip(card): verifications = set({ 97,101,105,111,117, 49,51,53,55,57 }) return ord(card) in verifications string = input() i = 0 j = len(string) - 1 flips = 0 while i <= j: if i==j and needFlip(string[i]): flips += 1 break if needFlip(string[i]): flips += 1 if needFlip(string[j]): flips += 1 i += 1 j -= 1 print(flips) ```

3

476

B

Dreamoon and WiFi

PROGRAMMING

1,300

[ "bitmasks", "brute force", "combinatorics", "dp", "math", "probabilities" ]

null

Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them. Each command is one of the following two types: 1. Go 1 unit towards the positive direction, denoted as '+' 1. Go 1 unit towards the negative direction, denoted as '-' But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the 1 unit to the negative or positive direction with the same probability 0.5). You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands?

The first line contains a string *s*1 — the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}. The second line contains a string *s*2 — the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command. Lengths of two strings are equal and do not exceed 10.

Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=9.

[ "++-+-\n+-+-+\n", "+-+-\n+-??\n", "+++\n??-\n" ]

[ "1.000000000000\n", "0.500000000000\n", "0.000000000000\n" ]

For the first sample, both *s*1 and *s*2 will lead Dreamoon to finish at the same position + 1. For the second sample, *s*1 will lead Dreamoon to finish at position 0, while there are four possibilites for *s*2: {"+-++", "+-+-", "+--+", "+---"} with ending position {+2, 0, 0, -2} respectively. So there are 2 correct cases out of 4, so the probability of finishing at the correct position is 0.5. For the third sample, *s*2 could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position + 3 is 0.

1,500

[ { "input": "++-+-\n+-+-+", "output": "1.000000000000" }, { "input": "+-+-\n+-??", "output": "0.500000000000" }, { "input": "+++\n??-", "output": "0.000000000000" }, { "input": "++++++++++\n+++??++?++", "output": "0.125000000000" }, { "input": "--+++---+-\n??????????", "output": "0.205078125000" }, { "input": "+--+++--+-\n??????????", "output": "0.246093750000" }, { "input": "+\n+", "output": "1.000000000000" }, { "input": "-\n?", "output": "0.500000000000" }, { "input": "+\n-", "output": "0.000000000000" }, { "input": "-\n-", "output": "1.000000000000" }, { "input": "-\n+", "output": "0.000000000000" }, { "input": "+\n?", "output": "0.500000000000" }, { "input": "++++++++++\n++++++++++", "output": "1.000000000000" }, { "input": "++++++++++\n++++-+++++", "output": "0.000000000000" }, { "input": "----------\n++++++++++", "output": "0.000000000000" }, { "input": "++++++++++\n++++??++++", "output": "0.250000000000" }, { "input": "----------\n+++?++++-+", "output": "0.000000000000" }, { "input": "++++++++++\n++++++++?+", "output": "0.500000000000" }, { "input": "--++++--+\n?-+?-??+-", "output": "0.250000000000" }, { "input": "----------\n??????????", "output": "0.000976562500" }, { "input": "+--++\n+--+-", "output": "0.000000000000" }, { "input": "-----++---\n????????+?", "output": "0.017578125000" }, { "input": "------+--+\n??????????", "output": "0.043945312500" }, { "input": "---++--\n???????", "output": "0.164062500000" }, { "input": "-----++\n???????", "output": "0.164062500000" }, { "input": "+---+--\n???????", "output": "0.164062500000" }, { "input": "---+-+\n??????", "output": "0.234375000000" }, { "input": "+++-+\n---++", "output": "0.000000000000" }, { "input": "++-+--+\n?-?+??+", "output": "0.375000000000" }, { "input": "----+++--\n-+?+++?--", "output": "0.000000000000" }, { "input": "+-----\n+?----", "output": "0.500000000000" } ]

1,696,877,955

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

1

77

1,536,000

#python 3.7.1 import operator as op from functools import reduce def comb(n, r): r = min(r, n-r) numer = reduce(op.mul, range(n, n-r, -1), 1) denom = reduce(op.mul, range(1, r+1), 1) return numer // denom # or / in Python 2 first= input() second=input() length= len(first) diff=0 avlb =0 unknown=0 plus =0 for i in range(length): if first[i] == '+' : diff+=1 plus+=1 else: diff -=1 if second[i] == '?' : unknown+=1 continue if second[i] == '+' : avlb+=1 else : avlb-=1 if(diff == avlb and unknown == 0 ): print(1.00) elif((diff == avlb and unknown == 1) or (diff != avlb and unknown == 0) ): print(0.00) else: total = 2 ** unknown r = abs(diff-avlb) if r > unknown or (r==0 and unknown %2 !=0): print(0.00) else: if r==0 or unknown == length : r= plus comb = comb(unknown , r) print(comb/total)

Title: Dreamoon and WiFi Time Limit: None seconds Memory Limit: None megabytes Problem Description: Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them. Each command is one of the following two types: 1. Go 1 unit towards the positive direction, denoted as '+' 1. Go 1 unit towards the negative direction, denoted as '-' But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the 1 unit to the negative or positive direction with the same probability 0.5). You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands? Input Specification: The first line contains a string *s*1 — the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}. The second line contains a string *s*2 — the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command. Lengths of two strings are equal and do not exceed 10. Output Specification: Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=9. Demo Input: ['++-+-\n+-+-+\n', '+-+-\n+-??\n', '+++\n??-\n'] Demo Output: ['1.000000000000\n', '0.500000000000\n', '0.000000000000\n'] Note: For the first sample, both *s*1 and *s*2 will lead Dreamoon to finish at the same position + 1. For the second sample, *s*1 will lead Dreamoon to finish at position 0, while there are four possibilites for *s*2: {"+-++", "+-+-", "+--+", "+---"} with ending position {+2, 0, 0, -2} respectively. So there are 2 correct cases out of 4, so the probability of finishing at the correct position is 0.5. For the third sample, *s*2 could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position + 3 is 0.

```python #python 3.7.1 import operator as op from functools import reduce def comb(n, r): r = min(r, n-r) numer = reduce(op.mul, range(n, n-r, -1), 1) denom = reduce(op.mul, range(1, r+1), 1) return numer // denom # or / in Python 2 first= input() second=input() length= len(first) diff=0 avlb =0 unknown=0 plus =0 for i in range(length): if first[i] == '+' : diff+=1 plus+=1 else: diff -=1 if second[i] == '?' : unknown+=1 continue if second[i] == '+' : avlb+=1 else : avlb-=1 if(diff == avlb and unknown == 0 ): print(1.00) elif((diff == avlb and unknown == 1) or (diff != avlb and unknown == 0) ): print(0.00) else: total = 2 ** unknown r = abs(diff-avlb) if r > unknown or (r==0 and unknown %2 !=0): print(0.00) else: if r==0 or unknown == length : r= plus comb = comb(unknown , r) print(comb/total) ```

0

226

B

Naughty Stone Piles

PROGRAMMING

1,900

[ "greedy" ]

null

There are *n* piles of stones of sizes *a*1,<=*a*2,<=...,<=*a**n* lying on the table in front of you. During one move you can take one pile and add it to the other. As you add pile *i* to pile *j*, the size of pile *j* increases by the current size of pile *i*, and pile *i* stops existing. The cost of the adding operation equals the size of the added pile. Your task is to determine the minimum cost at which you can gather all stones in one pile. To add some challenge, the stone piles built up conspiracy and decided that each pile will let you add to it not more than *k* times (after that it can only be added to another pile). Moreover, the piles decided to puzzle you completely and told you *q* variants (not necessarily distinct) of what *k* might equal. Your task is to find the minimum cost for each of *q* variants.

The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of stone piles. The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the initial sizes of the stone piles. The third line contains integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries. The last line contains *q* space-separated integers *k*1,<=*k*2,<=...,<=*k**q* (1<=≤<=*k**i*<=≤<=105) — the values of number *k* for distinct queries. Note that numbers *k**i* can repeat.

Print *q* whitespace-separated integers — the answers to the queries in the order, in which the queries are given in the input. Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

[ "5\n2 3 4 1 1\n2\n2 3\n" ]

[ "9 8 " ]

In the first sample one way to get the optimal answer goes like this: we add in turns the 4-th and the 5-th piles to the 2-nd one; then we add the 1-st pile to the 3-rd one; we add the 2-nd pile to the 3-rd one. The first two operations cost 1 each; the third one costs 2, the fourth one costs 5 (the size of the 2-nd pile after the first two operations is not 3, it already is 5). In the second sample you can add the 2-nd pile to the 3-rd one (the operations costs 3); then the 1-st one to the 3-th one (the cost is 2); then the 5-th one to the 4-th one (the costs is 1); and at last, the 4-th one to the 3-rd one (the cost is 2).

1,000

[ { "input": "5\n2 3 4 1 1\n2\n2 3", "output": "9 8 " }, { "input": "2\n2 9\n5\n4 10 7 3 4", "output": "2 2 2 2 2 " }, { "input": "1\n7\n4\n6 2 3 3", "output": "0 0 0 0 " }, { "input": "2\n7 10\n2\n2 4", "output": "7 7 " }, { "input": "1\n10\n5\n5 3 7 7 1", "output": "0 0 0 0 0 " }, { "input": "1\n2\n5\n7 3 9 8 1", "output": "0 0 0 0 0 " }, { "input": "4\n8 10 4 4\n3\n7 8 1", "output": "16 16 28 " }, { "input": "2\n7 9\n1\n9", "output": "7 " }, { "input": "3\n4 5 4\n2\n10 2", "output": "8 8 " }, { "input": "3\n1 6 8\n1\n6", "output": "7 " }, { "input": "2\n9 3\n1\n6", "output": "3 " }, { "input": "5\n9 5 7 3 3\n1\n3", "output": "21 " }, { "input": "2\n7 4\n1\n7", "output": "4 " }, { "input": "4\n7 4 1 7\n3\n6 8 3", "output": "12 12 12 " }, { "input": "3\n3 7 3\n1\n5", "output": "6 " }, { "input": "1\n3\n1\n2", "output": "0 " }, { "input": "1\n1\n3\n2 1 10", "output": "0 0 0 " } ]

1,544,720,350

2,147,483,647

Python 3

OK

TESTS

69

717

12,492,800

n = int(input()) stones = list(map(lambda t : int(t), input().split())) q = int(input()) queries = list(map(lambda t : int(t), input().split())) stones.sort() added_stones = [] added_stones.append(stones[0]) for i in range(1, n, 1): added_stones.append(stones[i] + added_stones[i - 1]) computed_queries = {} for qidx, qq in enumerate(queries): if qq in computed_queries: queries[qidx] = computed_queries[qq] continue i = n - 2 multiplier = 1 cost = 0 while i >= 0: pp = pow(qq, multiplier) nexti = i - pp if nexti < 0: cost += added_stones[i] * multiplier break cost += (added_stones[i] - added_stones[nexti]) * multiplier multiplier += 1 i = nexti queries[qidx] = cost computed_queries[qq] = cost print(*queries, sep=' ')

Title: Naughty Stone Piles Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* piles of stones of sizes *a*1,<=*a*2,<=...,<=*a**n* lying on the table in front of you. During one move you can take one pile and add it to the other. As you add pile *i* to pile *j*, the size of pile *j* increases by the current size of pile *i*, and pile *i* stops existing. The cost of the adding operation equals the size of the added pile. Your task is to determine the minimum cost at which you can gather all stones in one pile. To add some challenge, the stone piles built up conspiracy and decided that each pile will let you add to it not more than *k* times (after that it can only be added to another pile). Moreover, the piles decided to puzzle you completely and told you *q* variants (not necessarily distinct) of what *k* might equal. Your task is to find the minimum cost for each of *q* variants. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of stone piles. The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the initial sizes of the stone piles. The third line contains integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries. The last line contains *q* space-separated integers *k*1,<=*k*2,<=...,<=*k**q* (1<=≤<=*k**i*<=≤<=105) — the values of number *k* for distinct queries. Note that numbers *k**i* can repeat. Output Specification: Print *q* whitespace-separated integers — the answers to the queries in the order, in which the queries are given in the input. Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. Demo Input: ['5\n2 3 4 1 1\n2\n2 3\n'] Demo Output: ['9 8 '] Note: In the first sample one way to get the optimal answer goes like this: we add in turns the 4-th and the 5-th piles to the 2-nd one; then we add the 1-st pile to the 3-rd one; we add the 2-nd pile to the 3-rd one. The first two operations cost 1 each; the third one costs 2, the fourth one costs 5 (the size of the 2-nd pile after the first two operations is not 3, it already is 5). In the second sample you can add the 2-nd pile to the 3-rd one (the operations costs 3); then the 1-st one to the 3-th one (the cost is 2); then the 5-th one to the 4-th one (the costs is 1); and at last, the 4-th one to the 3-rd one (the cost is 2).

```python n = int(input()) stones = list(map(lambda t : int(t), input().split())) q = int(input()) queries = list(map(lambda t : int(t), input().split())) stones.sort() added_stones = [] added_stones.append(stones[0]) for i in range(1, n, 1): added_stones.append(stones[i] + added_stones[i - 1]) computed_queries = {} for qidx, qq in enumerate(queries): if qq in computed_queries: queries[qidx] = computed_queries[qq] continue i = n - 2 multiplier = 1 cost = 0 while i >= 0: pp = pow(qq, multiplier) nexti = i - pp if nexti < 0: cost += added_stones[i] * multiplier break cost += (added_stones[i] - added_stones[nexti]) * multiplier multiplier += 1 i = nexti queries[qidx] = cost computed_queries[qq] = cost print(*queries, sep=' ') ```

3

157

B

Trace

PROGRAMMING

1,000

[ "geometry", "sortings" ]

null

One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall. Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.

The first line contains the single integer *n* (1<=≤<=*n*<=≤<=100). The second line contains *n* space-separated integers *r**i* (1<=≤<=*r**i*<=≤<=1000) — the circles' radii. It is guaranteed that all circles are different.

Print the single real number — total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10<=-<=4.

[ "1\n1\n", "3\n1 4 2\n" ]

[ "3.1415926536\n", "40.8407044967\n" ]

In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals π × 12 = π. In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (π × 42 - π × 22) + π × 12 = π × 12 + π = 13π

1,000

[ { "input": "1\n1", "output": "3.1415926536" }, { "input": "3\n1 4 2", "output": "40.8407044967" }, { "input": "4\n4 1 3 2", "output": "31.4159265359" }, { "input": "4\n100 10 2 1", "output": "31111.1920484997" }, { "input": "10\n10 9 8 7 6 5 4 3 2 1", "output": "172.7875959474" }, { "input": "1\n1000", "output": "3141592.6535897931" }, { "input": "8\n8 1 7 2 6 3 5 4", "output": "113.0973355292" }, { "input": "100\n1000 999 998 997 996 995 994 993 992 991 990 989 988 987 986 985 984 983 982 981 980 979 978 977 976 975 974 973 972 971 970 969 968 967 966 965 964 963 962 961 960 959 958 957 956 955 954 953 952 951 950 949 948 947 946 945 944 943 942 941 940 939 938 937 936 935 934 933 932 931 930 929 928 927 926 925 924 923 922 921 920 919 918 917 916 915 914 913 912 911 910 909 908 907 906 905 904 903 902 901", "output": "298608.3817237098" }, { "input": "6\n109 683 214 392 678 10", "output": "397266.9574170437" }, { "input": "2\n151 400", "output": "431023.3704798660" }, { "input": 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323 386 348 410 468 875 107 50 960 82 834 234 663 651 422 794 633 294 771 945 607 146 913 950 858 297 88 882 725 247 872 645 749 799 987 115 394 380 382 971 429 593 426 652 353 351 233 868 598 889 116 71 376 916 464 414 976 138 903", "output": "1528494.7817143100" }, { "input": "70\n12 347 748 962 514 686 192 159 990 4 10 788 602 542 946 215 523 727 799 717 955 796 529 465 897 103 181 515 495 153 710 179 747 145 16 585 943 998 923 708 156 399 770 547 775 285 9 68 713 722 570 143 913 416 663 624 925 218 64 237 797 138 942 213 188 818 780 840 480 758", "output": "1741821.4892636713" }, { "input": "26\n656 508 45 189 561 366 96 486 547 386 703 570 780 689 264 26 11 74 466 76 421 48 982 886 215 650", "output": "1818821.9252031571" }, { "input": "52\n270 658 808 249 293 707 700 78 791 167 92 772 807 502 830 991 945 102 968 376 556 578 326 980 688 368 280 853 646 256 666 638 424 737 321 996 925 405 199 680 953 541 716 481 727 143 577 919 892 355 346 298", "output": "1272941.9273080483" }, { 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"1258248.6984672088" }, { "input": "37\n280 281 169 68 249 389 977 101 360 43 448 447 368 496 125 507 747 392 338 270 916 150 929 428 118 266 589 470 774 852 263 644 187 817 808 58 637", "output": "1495219.0323274869" }, { "input": "97\n768 569 306 968 437 779 227 561 412 60 44 807 234 645 169 858 580 396 343 145 842 723 416 80 456 247 81 150 297 116 760 964 312 558 101 850 549 650 299 868 121 435 579 705 118 424 302 812 970 397 659 565 916 183 933 459 6 593 518 717 326 305 744 470 75 981 824 221 294 324 194 293 251 446 481 215 338 861 528 829 921 945 540 89 450 178 24 460 990 392 148 219 934 615 932 340 937", "output": "1577239.7333274092" }, { "input": "94\n145 703 874 425 277 652 239 496 458 658 339 842 564 699 893 352 625 980 432 121 798 872 499 859 850 721 414 825 543 843 304 111 342 45 219 311 50 748 465 902 781 822 504 985 919 656 280 310 917 438 464 527 491 713 906 329 635 777 223 810 501 535 156 252 806 112 971 719 103 443 165 98 579 554 244 996 221 560 301 51 977 422 314 858 528 772 448 626 185 194 536 66 577 677", "output": "1624269.3753516484" }, { "input": "97\n976 166 649 81 611 927 480 231 998 711 874 91 969 521 531 414 993 790 317 981 9 261 437 332 173 573 904 777 882 990 658 878 965 64 870 896 271 732 431 53 761 943 418 602 708 949 930 130 512 240 363 458 673 319 131 784 224 48 919 126 208 212 911 59 677 535 450 273 479 423 79 807 336 18 72 290 724 28 123 605 287 228 350 897 250 392 885 655 746 417 643 114 813 378 355 635 905", "output": "1615601.7212203942" }, { "input": "91\n493 996 842 9 748 178 1 807 841 519 796 998 84 670 778 143 707 208 165 893 154 943 336 150 761 881 434 112 833 55 412 682 552 945 758 189 209 600 354 325 440 844 410 20 136 665 88 791 688 17 539 821 133 236 94 606 483 446 429 60 960 476 915 134 137 852 754 908 276 482 117 252 297 903 981 203 829 811 471 135 188 667 710 393 370 302 874 872 551 457 692", "output": "1806742.5014501044" }, { "input": "95\n936 736 17 967 229 607 589 291 242 244 29 698 800 566 630 667 90 416 11 94 812 838 668 520 678 111 490 823 199 973 681 676 683 721 262 896 682 713 402 691 874 44 95 704 56 322 822 887 639 433 406 35 988 61 176 496 501 947 440 384 372 959 577 370 754 802 1 945 427 116 746 408 308 391 397 730 493 183 203 871 831 862 461 565 310 344 504 378 785 137 279 123 475 138 415", "output": "1611115.5269110680" }, { "input": "90\n643 197 42 218 582 27 66 704 195 445 641 675 285 639 503 686 242 327 57 955 848 287 819 992 756 749 363 48 648 736 580 117 752 921 923 372 114 313 202 337 64 497 399 25 883 331 24 871 917 8 517 486 323 529 325 92 891 406 864 402 263 773 931 253 625 31 17 271 140 131 232 586 893 525 846 54 294 562 600 801 214 55 768 683 389 738 314 284 328 804", "output": "1569819.2914796301" }, { "input": "98\n29 211 984 75 333 96 840 21 352 168 332 433 130 944 215 210 620 442 363 877 91 491 513 955 53 82 351 19 998 706 702 738 770 453 344 117 893 590 723 662 757 16 87 546 312 669 568 931 224 374 927 225 751 962 651 587 361 250 256 240 282 600 95 64 384 589 813 783 39 918 412 648 506 283 886 926 443 173 946 241 310 33 622 565 261 360 547 339 943 367 354 25 479 743 385 485 896 741", "output": "2042921.1539616778" }, { "input": "93\n957 395 826 67 185 4 455 880 683 654 463 84 258 878 553 592 124 585 9 133 20 609 43 452 725 125 801 537 700 685 771 155 566 376 19 690 383 352 174 208 177 416 304 1000 533 481 87 509 358 233 681 22 507 659 36 859 952 259 138 271 594 779 576 782 119 69 608 758 283 616 640 523 710 751 34 106 774 92 874 568 864 660 998 992 474 679 180 409 15 297 990 689 501", "output": "1310703.8710041976" }, { "input": "97\n70 611 20 30 904 636 583 262 255 501 604 660 212 128 199 138 545 576 506 528 12 410 77 888 783 972 431 188 338 485 148 793 907 678 281 922 976 680 252 724 253 920 177 361 721 798 960 572 99 622 712 466 608 49 612 345 266 751 63 594 40 695 532 789 520 930 825 929 48 59 405 135 109 735 508 186 495 772 375 587 201 324 447 610 230 947 855 318 856 956 313 810 931 175 668 183 688", "output": "1686117.9099228707" }, { "input": "96\n292 235 391 180 840 172 218 997 166 287 329 20 886 325 400 471 182 356 448 337 417 319 58 106 366 764 393 614 90 831 924 314 667 532 64 874 3 434 350 352 733 795 78 640 967 63 47 879 635 272 145 569 468 792 153 761 770 878 281 467 209 208 298 37 700 18 334 93 5 750 412 779 523 517 360 649 447 328 311 653 57 578 767 460 647 663 50 670 151 13 511 580 625 907 227 89", "output": "1419726.5608617242" }, { "input": "100\n469 399 735 925 62 153 707 723 819 529 200 624 57 708 245 384 889 11 639 638 260 419 8 142 403 298 204 169 887 388 241 983 885 267 643 943 417 237 452 562 6 839 149 742 832 896 100 831 712 754 679 743 135 222 445 680 210 955 220 63 960 487 514 824 481 584 441 997 795 290 10 45 510 678 844 503 407 945 850 84 858 934 500 320 936 663 736 592 161 670 606 465 864 969 293 863 868 393 899 744", "output": "1556458.0979239127" }, { "input": "100\n321 200 758 415 190 710 920 992 873 898 814 259 359 66 971 210 838 545 663 652 684 277 36 756 963 459 335 484 462 982 532 423 131 703 307 229 391 938 253 847 542 975 635 928 220 980 222 567 557 181 366 824 900 180 107 979 112 564 525 413 300 422 876 615 737 343 902 8 654 628 469 913 967 785 893 314 909 215 912 262 20 709 363 915 997 954 986 454 596 124 74 159 660 550 787 418 895 786 293 50", "output": "1775109.8050211088" }, { "input": "100\n859 113 290 762 701 63 188 431 810 485 671 673 99 658 194 227 511 435 941 212 551 124 89 222 42 321 657 815 898 171 216 482 707 567 724 491 414 942 820 351 48 653 685 312 586 24 20 627 602 498 533 173 463 262 621 466 119 299 580 964 510 987 40 698 521 998 847 651 746 215 808 563 785 837 631 772 404 923 682 244 232 214 390 350 968 771 517 900 70 543 934 554 681 368 642 575 891 728 478 317", "output": "1447969.4788174964" }, { "input": "100\n941 283 349 457 52 837 299 284 796 305 893 624 101 972 738 204 121 70 17 704 836 791 95 111 162 952 472 724 733 580 878 177 705 804 11 211 463 417 288 409 410 485 896 755 921 267 164 656 505 765 539 439 535 19 991 689 220 474 114 944 884 144 926 849 486 566 117 35 749 499 797 303 362 905 690 890 976 66 590 183 234 683 39 297 769 787 376 541 571 759 495 200 261 352 73 493 831 442 273 339", "output": "1597889.4218394549" }, { "input": "100\n110 868 147 888 291 282 916 542 917 337 235 595 498 621 814 249 261 713 445 666 142 275 319 49 900 543 333 606 487 670 620 769 852 258 230 867 603 491 714 839 879 37 21 74 590 190 397 506 499 967 152 862 200 107 145 23 614 167 857 475 79 598 270 906 626 116 89 59 651 846 723 201 50 699 522 408 634 132 226 414 824 764 513 586 526 238 556 162 667 173 684 415 527 743 10 442 552 274 350 822", "output": "1567230.6191330721" }, { "input": "100\n901 902 903 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 937 938 939 940 941 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000", "output": "298608.3817237098" } ]

1,619,833,609

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

3

184

0

from math import pi n = int(input()) r = input().split(' ', n) r.sort() r.reverse() r = [int(i) ** 2 for i in r] pi_sum = 0 red = True for num in r: if red is True: pi_sum += num red = False elif red is False: pi_sum -= num red = True print(pi_sum * pi)

Title: Trace Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall. Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric. Input Specification: The first line contains the single integer *n* (1<=≤<=*n*<=≤<=100). The second line contains *n* space-separated integers *r**i* (1<=≤<=*r**i*<=≤<=1000) — the circles' radii. It is guaranteed that all circles are different. Output Specification: Print the single real number — total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10<=-<=4. Demo Input: ['1\n1\n', '3\n1 4 2\n'] Demo Output: ['3.1415926536\n', '40.8407044967\n'] Note: In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals π × 12 = π. In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (π × 42 - π × 22) + π × 12 = π × 12 + π = 13π

```python from math import pi n = int(input()) r = input().split(' ', n) r.sort() r.reverse() r = [int(i) ** 2 for i in r] pi_sum = 0 red = True for num in r: if red is True: pi_sum += num red = False elif red is False: pi_sum -= num red = True print(pi_sum * pi) ```

0

448

A

Rewards

PROGRAMMING

800

[ "implementation" ]

null

Bizon the Champion is called the Champion for a reason. Bizon the Champion has recently got a present — a new glass cupboard with *n* shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has *a*1 first prize cups, *a*2 second prize cups and *a*3 third prize cups. Besides, he has *b*1 first prize medals, *b*2 second prize medals and *b*3 third prize medals. Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules: - any shelf cannot contain both cups and medals at the same time; - no shelf can contain more than five cups; - no shelf can have more than ten medals. Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled.

The first line contains integers *a*1, *a*2 and *a*3 (0<=≤<=*a*1,<=*a*2,<=*a*3<=≤<=100). The second line contains integers *b*1, *b*2 and *b*3 (0<=≤<=*b*1,<=*b*2,<=*b*3<=≤<=100). The third line contains integer *n* (1<=≤<=*n*<=≤<=100). The numbers in the lines are separated by single spaces.

Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes).

[ "1 1 1\n1 1 1\n4\n", "1 1 3\n2 3 4\n2\n", "1 0 0\n1 0 0\n1\n" ]

[ "YES\n", "YES\n", "NO\n" ]

none

500

[ { "input": "1 1 1\n1 1 1\n4", "output": "YES" }, { "input": "1 1 3\n2 3 4\n2", "output": "YES" }, { "input": "1 0 0\n1 0 0\n1", "output": "NO" }, { "input": "0 0 0\n0 0 0\n1", "output": "YES" }, { "input": "100 100 100\n100 100 100\n100", "output": "YES" }, { "input": "100 100 100\n100 100 100\n1", "output": "NO" }, { "input": "1 10 100\n100 10 1\n20", "output": "NO" }, { "input": "1 1 1\n0 0 0\n1", "output": "YES" }, { "input": "0 0 0\n1 1 1\n1", "output": "YES" }, { "input": "5 5 5\n0 0 0\n2", "output": "NO" }, { "input": "0 0 0\n10 10 10\n2", "output": "NO" }, { "input": "21 61 39\n63 58 69\n44", "output": "YES" }, { "input": "18 95 4\n7 1 75\n46", "output": "YES" }, { "input": "64 27 81\n72 35 23\n48", "output": "YES" }, { "input": "6 6 6\n11 11 11\n7", "output": "NO" }, { "input": "1 2 3\n2 4 6\n3", "output": "NO" }, { "input": "1 2 3\n2 4 6\n4", "output": "YES" }, { "input": "99 99 99\n99 99 99\n89", "output": "NO" }, { "input": "5 0 0\n15 0 0\n2", "output": "NO" }, { "input": "10 10 10\n0 0 0\n1", "output": "NO" }, { "input": "1 1 1\n1 1 1\n15", "output": "YES" }, { "input": "2 3 5\n2 3 5\n2", "output": "NO" }, { "input": "2 2 2\n3 3 5\n3", "output": "NO" }, { "input": "1 2 2\n2 4 4\n1", "output": "NO" }, { "input": "1 2 3\n1 5 5\n2", "output": "NO" } ]

1,643,965,760

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

31

0

a=sum(list(map(int,input().split()))) b=sum(list(map(int,input().split()))) print("YES" if (a/5)+(b/10)<=int(input()) else"NO")

Title: Rewards Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bizon the Champion is called the Champion for a reason. Bizon the Champion has recently got a present — a new glass cupboard with *n* shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has *a*1 first prize cups, *a*2 second prize cups and *a*3 third prize cups. Besides, he has *b*1 first prize medals, *b*2 second prize medals and *b*3 third prize medals. Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules: - any shelf cannot contain both cups and medals at the same time; - no shelf can contain more than five cups; - no shelf can have more than ten medals. Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled. Input Specification: The first line contains integers *a*1, *a*2 and *a*3 (0<=≤<=*a*1,<=*a*2,<=*a*3<=≤<=100). The second line contains integers *b*1, *b*2 and *b*3 (0<=≤<=*b*1,<=*b*2,<=*b*3<=≤<=100). The third line contains integer *n* (1<=≤<=*n*<=≤<=100). The numbers in the lines are separated by single spaces. Output Specification: Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes). Demo Input: ['1 1 1\n1 1 1\n4\n', '1 1 3\n2 3 4\n2\n', '1 0 0\n1 0 0\n1\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: none

```python a=sum(list(map(int,input().split()))) b=sum(list(map(int,input().split()))) print("YES" if (a/5)+(b/10)<=int(input()) else"NO") ```

0

61

A

Ultra-Fast Mathematician

PROGRAMMING

800

[ "implementation" ]

A. Ultra-Fast Mathematician

2

256

Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate.

There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.

Write one line — the corresponding answer. Do not omit the leading 0s.

[ "1010100\n0100101\n", "000\n111\n", "1110\n1010\n", "01110\n01100\n" ]

[ "1110001\n", "111\n", "0100\n", "00010\n" ]

none

500

[ { "input": "1010100\n0100101", "output": "1110001" }, { "input": "000\n111", "output": "111" }, { "input": "1110\n1010", "output": "0100" }, { "input": "01110\n01100", "output": "00010" }, { "input": "011101\n000001", "output": "011100" }, { "input": "10\n01", "output": "11" }, { "input": "00111111\n11011101", "output": "11100010" }, { "input": "011001100\n101001010", "output": "110000110" }, { "input": "1100100001\n0110101100", "output": "1010001101" }, { "input": "00011101010\n10010100101", "output": "10001001111" }, { "input": "100000101101\n111010100011", "output": "011010001110" }, { "input": "1000001111010\n1101100110001", "output": "0101101001011" }, { "input": "01011111010111\n10001110111010", "output": "11010001101101" }, { "input": "110010000111100\n001100101011010", "output": "111110101100110" }, { "input": "0010010111110000\n0000000011010110", "output": "0010010100100110" }, { "input": "00111110111110000\n01111100001100000", "output": "01000010110010000" }, { "input": "101010101111010001\n001001111101111101", "output": "100011010010101100" }, { "input": "0110010101111100000\n0011000101000000110", "output": "0101010000111100110" }, { "input": "11110100011101010111\n00001000011011000000", "output": "11111100000110010111" }, { "input": "101010101111101101001\n111010010010000011111", "output": "010000111101101110110" }, { "input": "0000111111100011000010\n1110110110110000001010", "output": "1110001001010011001000" }, { "input": "10010010101000110111000\n00101110100110111000111", "output": "10111100001110001111111" }, { "input": "010010010010111100000111\n100100111111100011001110", "output": "110110101101011111001001" }, { "input": "0101110100100111011010010\n0101100011010111001010001", "output": "0000010111110000010000011" }, { "input": "10010010100011110111111011\n10000110101100000001000100", "output": "00010100001111110110111111" }, { "input": "000001111000000100001000000\n011100111101111001110110001", "output": "011101000101111101111110001" }, { "input": "0011110010001001011001011100\n0000101101000011101011001010", "output": "0011011111001010110010010110" }, { "input": "11111000000000010011001101111\n11101110011001010100010000000", "output": "00010110011001000111011101111" }, { "input": "011001110000110100001100101100\n001010000011110000001000101001", "output": "010011110011000100000100000101" }, { "input": "1011111010001100011010110101111\n1011001110010000000101100010101", "output": "0000110100011100011111010111010" }, { "input": "10111000100001000001010110000001\n10111000001100101011011001011000", "output": "00000000101101101010001111011001" }, { "input": "000001010000100001000000011011100\n111111111001010100100001100000111", "output": "111110101001110101100001111011011" }, { "input": "1101000000000010011011101100000110\n1110000001100010011010000011011110", "output": "0011000001100000000001101111011000" }, { "input": "01011011000010100001100100011110001\n01011010111000001010010100001110000", "output": "00000001111010101011110000010000001" }, { "input": "000011111000011001000110111100000100\n011011000110000111101011100111000111", "output": "011000111110011110101101011011000011" }, { "input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000", "output": "1011001001111001001011101010101000010" }, { "input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011", "output": "10001110000010101110000111000011111110" }, { "input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100", "output": "000100001011110000011101110111010001110" }, { "input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001", "output": "1101110101010110000011000000101011110011" }, { "input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100", "output": "11001011110010010000010111001100001001110" }, { "input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110", "output": "001100101000011111111101111011101010111001" }, { "input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001", "output": "0111010010100110110101100010000100010100000" }, { "input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100", "output": "11111110000000100101000100110111001100011001" }, { "input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011", "output": "101011011100100010100011011001101010100100010" }, { "input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001", "output": "1101001100111011010111110110101111001011110111" }, { "input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001", "output": "10010101000101000000011010011110011110011110001" }, { "input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100", "output": "011011011100000000010101110010000000101000111101" }, { "input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100", "output": "0101010111101001011011110110011101010101010100011" }, { "input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011", "output": "11001011010010111000010110011101100100001110111111" }, { "input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011", "output": "111011101010011100001111101001101011110010010110001" }, { "input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001", "output": "0100111110110011111110010010010000110111100101101101" }, { "input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100", "output": "01011001110111010111001100010011010100010000111011000" }, { "input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111", "output": "100011101001001000011011011001111000100000010100100100" }, { "input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110", "output": "1100110010000101101010111111101001001001110101110010110" }, { "input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110", "output": "01000111100111001011110010100011111111110010101100001101" }, { "input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010", "output": "110001010001000011000101110101000100001011111001011001001" }, { "input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111", "output": "1110100010111000101001001011101110011111100111000011011011" }, { "input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110", "output": "01110110101110100100110011010000001000101100101111000111011" }, { "input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011", "output": "111100101000000011101011011001110010101111000110010010000000" }, { "input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111", "output": "0100100010111110010011101010000011111110001110010110010111001" }, { "input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111", "output": "00110100000011001101101100100010110010001100000001100110011101" }, { "input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011", "output": "000000011000111011110011101000010000010100101000000011010110010" }, { "input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010", "output": "0010100110110100111100100100101101010100100111011010001001010101" }, { "input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111", "output": "11010110111100101111101001100001110100010110010110110111100110100" }, { "input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111", "output": "111111010011011100101110100110111111111001111110011010111111110000" }, { "input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110", "output": "1010101010100010001001001001100000111000010010010100010011000100000" }, { "input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000", "output": "00011111011111001000011100010011100011010100101011011000001001111110" }, { "input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111", "output": "001111000011001110100111010101111111011100110011001010010010000111011" }, { "input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101", "output": "0110001100110100010000110111000010011010011000011001010011010100010100" }, { "input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010", "output": "00010000000110110101000011001000000100100110111010011111101010001010000" }, { "input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001", "output": "000100100000000110011100100001010110101001100101110010010011111001110111" }, { "input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000", "output": "1000111100010011010110011101000000101010101100011111100001101111001010010" }, { "input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011", "output": "01000010011110111001011011110000001011000111101101101010010110001010100100" }, { "input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101", "output": "101101110110110010011100001011111100100001110000101100110000100011011100110" }, { "input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001", "output": "1110111111110010111000001100101010101011010100101010100101100011001001111111" }, { "input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111", "output": "10111011000111000101110100101000100111011011100011110110000101010001111010111" }, { "input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110", "output": "110100100110000101010010011010011001100110000111010000010100001011110111111101" }, { "input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111", "output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111" }, { "input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001", "output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001" }, { "input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110", "output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011" }, { "input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111", "output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101" }, { "input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010", "output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101" }, { "input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000", "output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001" }, { "input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011", "output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110" }, { "input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011", "output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011" }, { "input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011", "output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011" }, { "input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010", "output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011" }, { "input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001", "output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110" }, { "input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111", "output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010" }, { "input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011", "output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011" }, { "input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100", "output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000" }, { "input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001", "output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101" }, { "input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110", "output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101" }, { "input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011", "output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110" }, { "input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111", "output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110" }, { "input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000", "output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001" }, { "input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010", "output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101" }, { "input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101", "output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011" }, { "input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011", "output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010", "output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011" }, { "input": "0\n0", "output": "0" }, { "input": "0\n1", "output": "1" } ]

1,619,522,875

2,147,483,647

Python 3

OK

TESTS

102

62

0

s = input() s2 = input() result = '' for i in range(len(s)): if s[i] == s2[i]: result+='0' else: result+='1' print(result)

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 s = input() s2 = input() result = '' for i in range(len(s)): if s[i] == s2[i]: result+='0' else: result+='1' print(result) ```

3.9845

245

A

System Administrator

PROGRAMMING

800

[ "implementation" ]

null

Polycarpus is a system administrator. There are two servers under his strict guidance — *a* and *b*. To stay informed about the servers' performance, Polycarpus executes commands "ping a" and "ping b". Each ping command sends exactly ten packets to the server specified in the argument of the command. Executing a program results in two integers *x* and *y* (*x*<=+<=*y*<==<=10; *x*,<=*y*<=≥<=0). These numbers mean that *x* packets successfully reached the corresponding server through the network and *y* packets were lost. Today Polycarpus has performed overall *n* ping commands during his workday. Now for each server Polycarpus wants to know whether the server is "alive" or not. Polycarpus thinks that the server is "alive", if at least half of the packets that we send to this server reached it successfully along the network. Help Polycarpus, determine for each server, whether it is "alive" or not by the given commands and their results.

The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of commands Polycarpus has fulfilled. Each of the following *n* lines contains three integers — the description of the commands. The *i*-th of these lines contains three space-separated integers *t**i*, *x**i*, *y**i* (1<=≤<=*t**i*<=≤<=2; *x**i*,<=*y**i*<=≥<=0; *x**i*<=+<=*y**i*<==<=10). If *t**i*<==<=1, then the *i*-th command is "ping a", otherwise the *i*-th command is "ping b". Numbers *x**i*, *y**i* represent the result of executing this command, that is, *x**i* packets reached the corresponding server successfully and *y**i* packets were lost. It is guaranteed that the input has at least one "ping a" command and at least one "ping b" command.

In the first line print string "LIVE" (without the quotes) if server *a* is "alive", otherwise print "DEAD" (without the quotes). In the second line print the state of server *b* in the similar format.

[ "2\n1 5 5\n2 6 4\n", "3\n1 0 10\n2 0 10\n1 10 0\n" ]

[ "LIVE\nLIVE\n", "LIVE\nDEAD\n" ]

Consider the first test case. There 10 packets were sent to server *a*, 5 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall there were 10 packets sent to server *b*, 6 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Consider the second test case. There were overall 20 packages sent to server *a*, 10 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall 10 packets were sent to server *b*, 0 of them reached it. Therefore, less than half of all packets sent to this server successfully reached it through the network.

0

[ { "input": "2\n1 5 5\n2 6 4", "output": "LIVE\nLIVE" }, { "input": "3\n1 0 10\n2 0 10\n1 10 0", "output": "LIVE\nDEAD" }, { "input": "10\n1 3 7\n2 4 6\n1 2 8\n2 5 5\n2 10 0\n2 10 0\n1 8 2\n2 2 8\n2 10 0\n1 1 9", "output": "DEAD\nLIVE" }, { "input": "11\n1 8 2\n1 6 4\n1 9 1\n1 7 3\n2 0 10\n2 0 10\n1 8 2\n2 2 8\n2 6 4\n2 7 3\n2 9 1", "output": "LIVE\nDEAD" }, { "input": "12\n1 5 5\n1 0 10\n1 4 6\n1 2 8\n1 2 8\n1 5 5\n1 9 1\n2 9 1\n1 5 5\n1 1 9\n2 9 1\n2 7 3", "output": "DEAD\nLIVE" }, { "input": "13\n1 8 2\n1 4 6\n1 5 5\n1 5 5\n2 10 0\n2 9 1\n1 3 7\n2 6 4\n2 6 4\n2 5 5\n1 7 3\n2 3 7\n2 9 1", "output": "LIVE\nLIVE" }, { "input": "14\n1 7 3\n1 0 10\n1 7 3\n1 1 9\n2 2 8\n2 0 10\n1 1 9\n2 8 2\n2 6 4\n1 3 7\n1 3 7\n2 6 4\n2 1 9\n2 7 3", "output": "DEAD\nDEAD" } ]

1,547,280,987

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

154

0

n = int(input()) a = [0, 0] b = [0, 0] for i in range(n): x = list(map(int,input().split())) if x[0] == 1: a[0], a[1] = a[0] + x[1], a[1] + x[2] else: b[0], b[1] = b[0] + x[1], b[1] + x[2] print('LIVE' if a[0] > a[1] else 'DEAD') print('LIVE' if b[0] > b[1] else 'DAED')

Title: System Administrator Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarpus is a system administrator. There are two servers under his strict guidance — *a* and *b*. To stay informed about the servers' performance, Polycarpus executes commands "ping a" and "ping b". Each ping command sends exactly ten packets to the server specified in the argument of the command. Executing a program results in two integers *x* and *y* (*x*<=+<=*y*<==<=10; *x*,<=*y*<=≥<=0). These numbers mean that *x* packets successfully reached the corresponding server through the network and *y* packets were lost. Today Polycarpus has performed overall *n* ping commands during his workday. Now for each server Polycarpus wants to know whether the server is "alive" or not. Polycarpus thinks that the server is "alive", if at least half of the packets that we send to this server reached it successfully along the network. Help Polycarpus, determine for each server, whether it is "alive" or not by the given commands and their results. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of commands Polycarpus has fulfilled. Each of the following *n* lines contains three integers — the description of the commands. The *i*-th of these lines contains three space-separated integers *t**i*, *x**i*, *y**i* (1<=≤<=*t**i*<=≤<=2; *x**i*,<=*y**i*<=≥<=0; *x**i*<=+<=*y**i*<==<=10). If *t**i*<==<=1, then the *i*-th command is "ping a", otherwise the *i*-th command is "ping b". Numbers *x**i*, *y**i* represent the result of executing this command, that is, *x**i* packets reached the corresponding server successfully and *y**i* packets were lost. It is guaranteed that the input has at least one "ping a" command and at least one "ping b" command. Output Specification: In the first line print string "LIVE" (without the quotes) if server *a* is "alive", otherwise print "DEAD" (without the quotes). In the second line print the state of server *b* in the similar format. Demo Input: ['2\n1 5 5\n2 6 4\n', '3\n1 0 10\n2 0 10\n1 10 0\n'] Demo Output: ['LIVE\nLIVE\n', 'LIVE\nDEAD\n'] Note: Consider the first test case. There 10 packets were sent to server *a*, 5 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall there were 10 packets sent to server *b*, 6 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Consider the second test case. There were overall 20 packages sent to server *a*, 10 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall 10 packets were sent to server *b*, 0 of them reached it. Therefore, less than half of all packets sent to this server successfully reached it through the network.

```python n = int(input()) a = [0, 0] b = [0, 0] for i in range(n): x = list(map(int,input().split())) if x[0] == 1: a[0], a[1] = a[0] + x[1], a[1] + x[2] else: b[0], b[1] = b[0] + x[1], b[1] + x[2] print('LIVE' if a[0] > a[1] else 'DEAD') print('LIVE' if b[0] > b[1] else 'DAED') ```

0

988

A

Diverse Team

PROGRAMMING

800

[ "brute force", "implementation" ]

null

There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct. If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them.

The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the number of students and the size of the team you have to form. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student.

If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them. Assume that the students are numbered from $1$ to $n$.

[ "5 3\n15 13 15 15 12\n", "5 4\n15 13 15 15 12\n", "4 4\n20 10 40 30\n" ]

[ "YES\n1 2 5 \n", "NO\n", "YES\n1 2 3 4 \n" ]

All possible answers for the first example: - {1 2 5} - {2 3 5} - {2 4 5} Note that the order does not matter.

0

[ { "input": "5 3\n15 13 15 15 12", "output": "YES\n1 2 5 " }, { "input": "5 4\n15 13 15 15 12", "output": "NO" }, { "input": "4 4\n20 10 40 30", "output": "YES\n1 2 3 4 " }, { "input": "1 1\n1", "output": "YES\n1 " }, { "input": "100 53\n16 17 1 2 27 5 9 9 53 24 17 33 35 24 20 48 56 73 12 14 39 55 58 13 59 73 29 26 40 33 22 29 34 22 55 38 63 66 36 13 60 42 10 15 21 9 11 5 23 37 79 47 26 3 79 53 44 8 71 75 42 11 34 39 79 33 10 26 23 23 17 14 54 41 60 31 83 5 45 4 14 35 6 60 28 48 23 18 60 36 21 28 7 34 9 25 52 43 54 19", "output": "YES\n1 2 3 4 5 6 7 9 10 12 13 15 16 17 18 19 20 21 22 23 24 25 27 28 29 31 33 36 37 38 39 41 42 43 44 45 47 49 50 51 52 54 57 58 59 60 73 74 76 77 79 80 83 " }, { "input": "2 2\n100 100", "output": "NO" }, { "input": "2 2\n100 99", "output": "YES\n1 2 " }, { "input": "100 100\n63 100 75 32 53 24 73 98 76 15 70 48 8 81 88 58 95 78 27 92 14 16 72 43 46 39 66 38 64 42 59 9 22 51 4 6 10 94 28 99 68 80 35 50 45 20 47 7 30 26 49 91 77 19 96 57 65 1 11 13 31 12 82 87 93 34 62 3 21 79 56 41 89 18 44 23 74 86 2 33 69 36 61 67 25 83 5 84 90 37 40 29 97 60 52 55 54 71 17 85", "output": "YES\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 " }, { "input": "100 41\n54 16 42 3 45 6 9 72 100 13 24 57 35 5 89 13 97 27 43 9 73 89 48 16 48 55 18 15 55 28 30 6 18 41 100 61 9 42 35 54 57 25 73 15 42 54 49 5 72 48 30 55 4 43 94 5 60 92 93 23 89 75 53 92 74 93 89 28 69 6 3 49 15 28 49 57 54 55 30 57 69 18 89 6 25 23 93 74 30 13 87 53 6 42 4 54 60 30 4 35", "output": "NO" }, { "input": "100 2\n70 64 70 32 70 64 32 70 64 32 32 64 70 64 64 32 64 64 64 70 70 64 64 64 64 70 32 64 70 64 32 70 70 70 64 70 64 70 64 32 70 32 70 64 64 64 32 70 64 70 70 32 70 32 32 32 70 32 70 32 64 64 70 32 32 64 70 64 32 32 64 64 32 32 70 70 32 70 32 64 32 70 64 64 32 64 32 64 70 32 70 32 70 64 64 64 70 70 64 70", "output": "YES\n1 2 " } ]

1,659,342,972

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

4

61

0

a = [int(x) for x in input().split()] b = [int(x) for x in input().split()] n , k = a[0] , a[1] dic = {} for i in range(len(b)): if b[i] not in dic.keys(): dic[b[i]] = i + 1 if len(dic) == k: print("YES") for i in dic.keys(): print(dic[i] , end = " ") else: print("NO")

Title: Diverse Team Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct. If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them. Input Specification: The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the number of students and the size of the team you have to form. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student. Output Specification: If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them. Assume that the students are numbered from $1$ to $n$. Demo Input: ['5 3\n15 13 15 15 12\n', '5 4\n15 13 15 15 12\n', '4 4\n20 10 40 30\n'] Demo Output: ['YES\n1 2 5 \n', 'NO\n', 'YES\n1 2 3 4 \n'] Note: All possible answers for the first example: - {1 2 5} - {2 3 5} - {2 4 5} Note that the order does not matter.

```python a = [int(x) for x in input().split()] b = [int(x) for x in input().split()] n , k = a[0] , a[1] dic = {} for i in range(len(b)): if b[i] not in dic.keys(): dic[b[i]] = i + 1 if len(dic) == k: print("YES") for i in dic.keys(): print(dic[i] , end = " ") else: print("NO") ```

0

606

A

Magic Spheres

PROGRAMMING

1,200

[ "implementation" ]

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" } ]

1,449,679,232

2,132

Python 3

WRONG_ANSWER

PRETESTS

10

62

0

abc = input().split() abc = [int(abc[i]) for i in range(3)] xyz = input().split() xyz = [int(xyz[i]) for i in range(3)] out = 'None' if abc[0] >= xyz[0] and abc[1] >= xyz[1] and abc[2] >= xyz[2]: print('Yes') elif sum(abc) < sum(xyz): print('No') else: while out == 'None': for i in range(3): b = [] if abc[i] >= xyz[i]+2: l = [] for x in range(3): l.append(xyz[x]-abc[x]) abc[l.index(max(l))] += 1 abc[i] -= 2 b.append(True) else: b.append(False) if not True in b: out = 'No' if abc[0] >= xyz[0] and abc[1] >= xyz[1] and abc[2] >= xyz[2]: out = 'Yes' print(out)

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 abc = input().split() abc = [int(abc[i]) for i in range(3)] xyz = input().split() xyz = [int(xyz[i]) for i in range(3)] out = 'None' if abc[0] >= xyz[0] and abc[1] >= xyz[1] and abc[2] >= xyz[2]: print('Yes') elif sum(abc) < sum(xyz): print('No') else: while out == 'None': for i in range(3): b = [] if abc[i] >= xyz[i]+2: l = [] for x in range(3): l.append(xyz[x]-abc[x]) abc[l.index(max(l))] += 1 abc[i] -= 2 b.append(True) else: b.append(False) if not True in b: out = 'No' if abc[0] >= xyz[0] and abc[1] >= xyz[1] and abc[2] >= xyz[2]: out = 'Yes' print(out) ```

0

709

A

Juicer

PROGRAMMING

900

[ "implementation" ]

null

Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one. The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section?

The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer.

Print one integer — the number of times Kolya will have to empty the waste section.

[ "2 7 10\n5 6\n", "1 5 10\n7\n", "3 10 10\n5 7 7\n", "1 1 1\n1\n" ]

[ "1\n", "0\n", "1\n", "0\n" ]

In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards. In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all.

500

[ { "input": "2 7 10\n5 6", "output": "1" }, { "input": "1 5 10\n7", "output": "0" }, { "input": "3 10 10\n5 7 7", "output": "1" }, { "input": "1 1 1\n1", "output": "0" }, { "input": "2 951637 951638\n44069 951637", "output": "1" }, { "input": "50 100 129\n55 130 91 19 116 3 63 52 104 76 75 27 151 99 149 147 39 148 84 9 132 49 40 112 124 141 144 93 36 32 146 74 48 38 150 55 94 32 107 69 77 81 33 57 62 98 78 127 154 126", "output": "12" }, { "input": "100 1000 1083\n992 616 818 359 609 783 263 989 501 929 362 394 919 1081 870 830 1097 975 62 346 531 367 323 457 707 360 949 334 867 116 478 417 961 963 1029 114 867 1008 988 916 983 1077 959 942 572 961 579 318 721 337 488 717 111 70 416 685 987 130 353 107 61 191 827 849 106 815 211 953 111 398 889 860 801 71 375 320 395 1059 116 222 931 444 582 74 677 655 88 173 686 491 661 186 114 832 615 814 791 464 517 850", "output": "36" }, { "input": "2 6 8\n2 1", "output": "0" }, { "input": "5 15 16\n7 11 5 12 8", "output": "2" }, { "input": "15 759966 759967\n890397 182209 878577 548548 759966 812923 759966 860479 200595 381358 299175 339368 759966 907668 69574", "output": "4" }, { "input": "5 234613 716125\n642626 494941 234613 234613 234613", "output": "0" }, { "input": "50 48547 567054\n529808 597004 242355 559114 78865 537318 631455 733020 655072 645093 309010 855034 306058 625046 524574 834944 27330 664392 443637 821584 338013 490702 289520 675471 885846 258814 134220 571301 84875 94132 200425 928833 375166 521232 317961 175315 947093 89971 322071 174033 48547 998535 954205 704114 943163 438900 48547 538422 48547 48547", "output": "0" }, { "input": "5 10 20\n10 10 10 10 1", "output": "1" }, { "input": "5 10 11\n10 10 10 10 1", "output": "2" }, { "input": "3 10 10\n4 3 3", "output": "0" }, { "input": "3 5 5\n5 5 5", "output": "1" }, { "input": "3 4 14\n5 5 5", "output": "0" }, { "input": "2 7 10\n1234 1234", "output": "0" }, { "input": "1 5 6\n10", "output": "0" }, { "input": "3 4 6\n1 2 3", "output": "0" }, { "input": "5 10 12\n13 13 13 13 13", "output": "0" }, { "input": "3 4 5\n5 7 9", "output": "0" }, { "input": "3 10 100\n5 5 5", "output": "0" }, { "input": "5 1 2\n2 2 2 2 2", "output": "0" }, { "input": "5 5 5\n5 5 5 5 5", "output": "2" }, { "input": "2 2 5\n5 5", "output": "0" }, { "input": "3 1 4\n2 2 2", "output": "0" } ]

1,674,978,319

2,147,483,647

Python 3

OK

TESTS

58

93

7,065,600

def main(list): n = list[0] b = list[1] d = list[2] inputer = [int(i) for i in input().split()] res = 0 trash = 0 for i in range(n): if inputer[i]<=b: trash+=inputer[i] if trash>d: trash = 0 res+=1 return res print(main([int(i) for i in input().split()]))

Title: Juicer Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one. The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section? Input Specification: The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer. Output Specification: Print one integer — the number of times Kolya will have to empty the waste section. Demo Input: ['2 7 10\n5 6\n', '1 5 10\n7\n', '3 10 10\n5 7 7\n', '1 1 1\n1\n'] Demo Output: ['1\n', '0\n', '1\n', '0\n'] Note: In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards. In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all.

```python def main(list): n = list[0] b = list[1] d = list[2] inputer = [int(i) for i in input().split()] res = 0 trash = 0 for i in range(n): if inputer[i]<=b: trash+=inputer[i] if trash>d: trash = 0 res+=1 return res print(main([int(i) for i in input().split()])) ```

3

888

B

Buggy Robot

PROGRAMMING

1,000

[ "greedy" ]

null

Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell (0,<=0). The robot can process commands. There are four types of commands it can perform: - U — move from the cell (*x*,<=*y*) to (*x*,<=*y*<=+<=1); - D — move from (*x*,<=*y*) to (*x*,<=*y*<=-<=1); - L — move from (*x*,<=*y*) to (*x*<=-<=1,<=*y*); - R — move from (*x*,<=*y*) to (*x*<=+<=1,<=*y*). Ivan entered a sequence of *n* commands, and the robot processed it. After this sequence the robot ended up in the starting cell (0,<=0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations!

The first line contains one number *n* — the length of sequence of commands entered by Ivan (1<=≤<=*n*<=≤<=100). The second line contains the sequence itself — a string consisting of *n* characters. Each character can be U, D, L or R.

Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell.

[ "4\nLDUR\n", "5\nRRRUU\n", "6\nLLRRRR\n" ]

[ "4\n", "0\n", "4\n" ]

none

0

[ { "input": "4\nLDUR", "output": "4" }, { "input": "5\nRRRUU", "output": "0" }, { "input": "6\nLLRRRR", "output": "4" }, { "input": "88\nLLUUULRDRRURDDLURRLRDRLLRULRUUDDLLLLRRDDURDURRLDURRLDRRRUULDDLRRRDDRRLUULLURDURUDDDDDLDR", "output": "76" }, { "input": "89\nLDLLLDRDUDURRRRRUDULDDDLLUDLRLRLRLDLDUULRDUDLRRDLUDLURRDDRRDLDUDUUURUUUDRLUDUDLURDLDLLDDU", "output": "80" }, { "input": "90\nRRRDUULLLRDUUDDRLDLRLUDURDRDUUURUURDDRRRURLDDDUUDRLLLULURDRDRURLDRRRRUULDULDDLLLRRLRDLLLLR", "output": "84" }, { "input": "91\nRLDRLRRLLDLULULLURULLRRULUDUULLUDULDUULURUDRUDUURDULDUDDUUUDRRUUDLLRULRULURLDRDLDRURLLLRDDD", "output": "76" }, { "input": "92\nRLRDDLULRLLUURRDDDLDDDLDDUURRRULLRDULDULLLUUULDUDLRLRRDRDRDDULDRLUDRDULDRURUDUULLRDRRLLDRLRR", "output": "86" }, { "input": "93\nRLLURLULRURDDLUURLUDDRDLUURLRDLRRRDUULLRDRRLRLDURRDLLRDDLLLDDDLDRRURLLDRUDULDDRRULRRULRLDRDLR", "output": "84" }, { "input": "94\nRDULDDDLULRDRUDRUUDUUDRRRULDRRUDURUULRDUUDLULLLUDURRDRDLUDRULRRRULUURUDDDDDUDLLRDLDRLLRUUURLUL", "output": "86" }, { "input": "95\nRDLUUULLUURDDRLDLLRRRULRLRDULULRULRUDURLULDDDRLURLDRULDUDUUULLRDDURUULULLDDLDRDRLLLURLRDLLDDDDU", "output": "86" }, { "input": "96\nRDDRLRLLDDULRLRURUDLRLDUDRURLLUUDLLURDLRRUURDRRUDRURLLDLLRDURDURLRLUDURULLLRDUURULUUULRRURRDLURL", "output": "84" }, { "input": "97\nRURDDLRLLRULUDURDLRLLUUDURRLLUDLLLDUDRUULDRUUURURULRDLDRRLLUUUDLLLDDLLLLRLLDUDRRDLLUDLRURUDULRLUR", "output": "82" }, { "input": "98\nRUDURLULLDDLLRDLLRDDLLLLRLDDDDRRRDDRRURLDRLLRUUUDLUUUDDDUDRUURLURUUDUUDRULRRULLRRLRULLULDLUURLULRD", "output": "92" }, { "input": "99\nRRULLDULRRDRULLDUDRUDDDRLLUUDRDDUDURLDDRUUDRRUUURRRURDDLDUURDLRLURRDDLUDDLUDURDRUDDURLURURLRUDRURLD", "output": "86" }, { "input": "100\nUDRLRRLLRRLRRRDDLLDDDLULLDDLURUURUULUDDDRDDLLRDLLUURLRDRLRRLRLLLULDUDDUURRLRDULDRDURRRRRRULDRRDLDRRL", "output": "88" }, { "input": "1\nU", "output": "0" }, { "input": "5\nUUULD", "output": "2" }, { "input": "1\nD", "output": "0" }, { "input": "5\nURLUL", "output": "2" }, { "input": "5\nDDDRU", "output": "2" }, { "input": "2\nLR", "output": "2" }, { "input": "8\nDDRDLDUR", "output": "4" }, { "input": "6\nLLLLUD", "output": "2" }, { "input": "13\nRRRLLLLLLLLLL", "output": "6" } ]

1,625,490,062

2,147,483,647

Python 3

OK

TESTS

25

62

0

n = int(input()) s = input() h, v = 0, 0 for i in range(len(s)): if s[i] == 'L': h -= 1 elif s[i] == 'R': h += 1 elif s[i] == 'D': v -= 1 else: v += 1 h = abs(h) v = abs(v) delta = len(s) - h - v print(delta)

Title: Buggy Robot Time Limit: None seconds Memory Limit: None megabytes Problem Description: Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell (0,<=0). The robot can process commands. There are four types of commands it can perform: - U — move from the cell (*x*,<=*y*) to (*x*,<=*y*<=+<=1); - D — move from (*x*,<=*y*) to (*x*,<=*y*<=-<=1); - L — move from (*x*,<=*y*) to (*x*<=-<=1,<=*y*); - R — move from (*x*,<=*y*) to (*x*<=+<=1,<=*y*). Ivan entered a sequence of *n* commands, and the robot processed it. After this sequence the robot ended up in the starting cell (0,<=0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations! Input Specification: The first line contains one number *n* — the length of sequence of commands entered by Ivan (1<=≤<=*n*<=≤<=100). The second line contains the sequence itself — a string consisting of *n* characters. Each character can be U, D, L or R. Output Specification: Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell. Demo Input: ['4\nLDUR\n', '5\nRRRUU\n', '6\nLLRRRR\n'] Demo Output: ['4\n', '0\n', '4\n'] Note: none

```python n = int(input()) s = input() h, v = 0, 0 for i in range(len(s)): if s[i] == 'L': h -= 1 elif s[i] == 'R': h += 1 elif s[i] == 'D': v -= 1 else: v += 1 h = abs(h) v = abs(v) delta = len(s) - h - v print(delta) ```

3

152

A

Marks

PROGRAMMING

900

[ "implementation" ]

null

Vasya, or Mr. Vasily Petrov is a dean of a department in a local university. After the winter exams he got his hands on a group's gradebook. Overall the group has *n* students. They received marks for *m* subjects. Each student got a mark from 1 to 9 (inclusive) for each subject. Let's consider a student the best at some subject, if there is no student who got a higher mark for this subject. Let's consider a student successful, if there exists a subject he is the best at. Your task is to find the number of successful students in the group.

The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of students and the number of subjects, correspondingly. Next *n* lines each containing *m* characters describe the gradebook. Each character in the gradebook is a number from 1 to 9. Note that the marks in a rows are not sepatated by spaces.

Print the single number — the number of successful students in the given group.

[ "3 3\n223\n232\n112\n", "3 5\n91728\n11828\n11111\n" ]

[ "2\n", "3\n" ]

In the first sample test the student number 1 is the best at subjects 1 and 3, student 2 is the best at subjects 1 and 2, but student 3 isn't the best at any subject. In the second sample test each student is the best at at least one subject.

500

[ { "input": "3 3\n223\n232\n112", "output": "2" }, { "input": "3 5\n91728\n11828\n11111", "output": "3" }, { "input": "2 2\n48\n27", "output": "1" }, { "input": "2 1\n4\n6", "output": "1" }, { "input": "1 2\n57", "output": "1" }, { "input": "1 1\n5", "output": "1" }, { "input": "3 4\n2553\n6856\n5133", "output": "2" }, { "input": "8 7\n6264676\n7854895\n3244128\n2465944\n8958761\n1378945\n3859353\n6615285", "output": "6" }, { "input": "9 8\n61531121\n43529859\n18841327\n88683622\n98995641\n62741632\n57441743\n49396792\n63381994", "output": "4" }, { "input": "10 20\n26855662887514171367\n48525577498621511535\n47683778377545341138\n47331616748732562762\n44876938191354974293\n24577238399664382695\n42724955594463126746\n79187344479926159359\n48349683283914388185\n82157191115518781898", "output": "9" }, { "input": "20 15\n471187383859588\n652657222494199\n245695867594992\n726154672861295\n614617827782772\n862889444974692\n373977167653235\n645434268565473\n785993468314573\n722176861496755\n518276853323939\n723712762593348\n728935312568886\n373898548522463\n769777587165681\n247592995114377\n182375946483965\n497496542536127\n988239919677856\n859844339819143", "output": "18" }, { "input": "13 9\n514562255\n322655246\n135162979\n733845982\n473117129\n513967187\n965649829\n799122777\n661249521\n298618978\n659352422\n747778378\n723261619", "output": "11" }, { "input": "75 1\n2\n3\n8\n3\n2\n1\n3\n1\n5\n1\n5\n4\n8\n8\n4\n2\n5\n1\n7\n6\n3\n2\n2\n3\n5\n5\n2\n4\n7\n7\n9\n2\n9\n5\n1\n4\n9\n5\n2\n4\n6\n6\n3\n3\n9\n3\n3\n2\n3\n4\n2\n6\n9\n1\n1\n1\n1\n7\n2\n3\n2\n9\n7\n4\n9\n1\n7\n5\n6\n8\n3\n4\n3\n4\n6", "output": "7" }, { "input": "92 3\n418\n665\n861\n766\n529\n416\n476\n676\n561\n995\n415\n185\n291\n176\n776\n631\n556\n488\n118\n188\n437\n496\n466\n131\n914\n118\n766\n365\n113\n897\n386\n639\n276\n946\n759\n169\n494\n837\n338\n351\n783\n311\n261\n862\n598\n132\n246\n982\n575\n364\n615\n347\n374\n368\n523\n132\n774\n161\n552\n492\n598\n474\n639\n681\n635\n342\n516\n483\n141\n197\n571\n336\n175\n596\n481\n327\n841\n133\n142\n146\n246\n396\n287\n582\n556\n996\n479\n814\n497\n363\n963\n162", "output": "23" }, { "input": "100 1\n1\n6\n9\n1\n1\n5\n5\n4\n6\n9\n6\n1\n7\n8\n7\n3\n8\n8\n7\n6\n2\n1\n5\n8\n7\n3\n5\n4\n9\n7\n1\n2\n4\n1\n6\n5\n1\n3\n9\n4\n5\n8\n1\n2\n1\n9\n7\n3\n7\n1\n2\n2\n2\n2\n3\n9\n7\n2\n4\n7\n1\n6\n8\n1\n5\n6\n1\n1\n2\n9\n7\n4\n9\n1\n9\n4\n1\n3\n5\n2\n4\n4\n6\n5\n1\n4\n5\n8\n4\n7\n6\n5\n6\n9\n5\n8\n1\n5\n1\n6", "output": "10" }, { "input": "100 2\n71\n87\n99\n47\n22\n87\n49\n73\n21\n12\n77\n43\n18\n41\n78\n62\n61\n16\n64\n89\n81\n54\n53\n92\n93\n94\n68\n93\n15\n68\n42\n93\n28\n19\n86\n16\n97\n17\n11\n43\n72\n76\n54\n95\n58\n53\n48\n45\n85\n85\n74\n21\n44\n51\n89\n75\n76\n17\n38\n62\n81\n22\n66\n59\n89\n85\n91\n87\n12\n97\n52\n87\n43\n89\n51\n58\n57\n98\n78\n68\n82\n41\n87\n29\n75\n72\n48\n14\n35\n71\n74\n91\n66\n67\n42\n98\n52\n54\n22\n41", "output": "21" }, { "input": "5 20\n11111111111111111111\n11111111111111111111\n11111111111111111111\n99999999999999999999\n11111111111111111119", "output": "2" }, { "input": "3 3\n111\n111\n999", "output": "1" }, { "input": "3 3\n119\n181\n711", "output": "3" }, { "input": "15 5\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111", "output": "15" }, { "input": "2 20\n22222222222222222222\n11111111111111111111", "output": "1" }, { "input": "3 3\n233\n222\n111", "output": "2" }, { "input": "4 15\n222222222222222\n111111111111119\n111111111111119\n111111111111111", "output": "3" }, { "input": "4 1\n1\n9\n9\n9", "output": "3" }, { "input": "3 3\n123\n321\n132", "output": "3" }, { "input": "3 3\n113\n332\n322", "output": "3" }, { "input": "2 100\n2222222222222222222222222222222222222222222222222222222222222222222222221222222222222222222222222222\n1111111111111111111111111111111111111111111111111111111111111111111111119111111111111111111111111111", "output": "2" }, { "input": "3 3\n321\n231\n123", "output": "3" }, { "input": "2 100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222", "output": "1" }, { "input": "3 3\n221\n111\n111", "output": "3" } ]

1,589,124,696

2,147,483,647

PyPy 3

OK

TESTS

44

140

22,016,000

n, m = map(int, input().split()) scores=[] for i in range(n): scores.append(str(input())) answer = [] for i in range(m): temp=[] for j in scores: temp.append(int(j[i])) highest = max(temp) for i in range(len(temp)): if temp[i]==highest: answer.append(i) print(len(set(answer)))

Title: Marks Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya, or Mr. Vasily Petrov is a dean of a department in a local university. After the winter exams he got his hands on a group's gradebook. Overall the group has *n* students. They received marks for *m* subjects. Each student got a mark from 1 to 9 (inclusive) for each subject. Let's consider a student the best at some subject, if there is no student who got a higher mark for this subject. Let's consider a student successful, if there exists a subject he is the best at. Your task is to find the number of successful students in the group. Input Specification: The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of students and the number of subjects, correspondingly. Next *n* lines each containing *m* characters describe the gradebook. Each character in the gradebook is a number from 1 to 9. Note that the marks in a rows are not sepatated by spaces. Output Specification: Print the single number — the number of successful students in the given group. Demo Input: ['3 3\n223\n232\n112\n', '3 5\n91728\n11828\n11111\n'] Demo Output: ['2\n', '3\n'] Note: In the first sample test the student number 1 is the best at subjects 1 and 3, student 2 is the best at subjects 1 and 2, but student 3 isn't the best at any subject. In the second sample test each student is the best at at least one subject.

```python n, m = map(int, input().split()) scores=[] for i in range(n): scores.append(str(input())) answer = [] for i in range(m): temp=[] for j in scores: temp.append(int(j[i])) highest = max(temp) for i in range(len(temp)): if temp[i]==highest: answer.append(i) print(len(set(answer))) ```

3

463

B

Caisa and Pylons

PROGRAMMING

1,100

[ "brute force", "implementation", "math" ]

null

Caisa solved the problem with the sugar and now he is on the way back to home. Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=><=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time. Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?

The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons.

Print a single number representing the minimum number of dollars paid by Caisa.

[ "5\n3 4 3 2 4\n", "3\n4 4 4\n" ]

[ "4\n", "4\n" ]

In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.

1,000

[ { "input": "5\n3 4 3 2 4", "output": "4" }, { "input": "3\n4 4 4", "output": "4" }, { "input": "99\n1401 2019 1748 3785 3236 3177 3443 3772 2138 1049 353 908 310 2388 1322 88 2160 2783 435 2248 1471 706 2468 2319 3156 3506 2794 1999 1983 2519 2597 3735 537 344 3519 3772 3872 2961 3895 2010 10 247 3269 671 2986 942 758 1146 77 1545 3745 1547 2250 2565 217 1406 2070 3010 3404 404 1528 2352 138 2065 3047 3656 2188 2919 2616 2083 1280 2977 2681 548 4000 1667 1489 1109 3164 1565 2653 3260 3463 903 1824 3679 2308 245 2689 2063 648 568 766 785 2984 3812 440 1172 2730", "output": "4000" }, { "input": "68\n477 1931 3738 3921 2306 1823 3328 2057 661 3993 2967 3520 171 1739 1525 1817 209 3475 1902 2666 518 3283 3412 3040 3383 2331 1147 1460 1452 1800 1327 2280 82 1416 2200 2388 3238 1879 796 250 1872 114 121 2042 1853 1645 211 2061 1472 2464 726 1989 1746 489 1380 1128 2819 2527 2939 622 678 265 2902 1111 2032 1453 3850 1621", "output": "3993" }, { "input": "30\n30 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": "30" }, { "input": "3\n3 2 1", "output": "3" }, { "input": "1\n69", "output": "69" } ]

1,638,172,537

2,147,483,647

Python 3

OK

TESTS

49

62

7,270,400

import math #s = input() #n= (map(int, input().split())) #(map(int, input().split())) #a, b = (map(int, input().split())) #for i in range(0, int(input())): n = int(input()) mass = list(map(int, input().split())) print(max(mass))

Title: Caisa and Pylons Time Limit: None seconds Memory Limit: None megabytes Problem Description: Caisa solved the problem with the sugar and now he is on the way back to home. Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=><=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time. Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game? Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons. Output Specification: Print a single number representing the minimum number of dollars paid by Caisa. Demo Input: ['5\n3 4 3 2 4\n', '3\n4 4 4\n'] Demo Output: ['4\n', '4\n'] Note: In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.

```python import math #s = input() #n= (map(int, input().split())) #(map(int, input().split())) #a, b = (map(int, input().split())) #for i in range(0, int(input())): n = int(input()) mass = list(map(int, input().split())) print(max(mass)) ```

3

950

A

Left-handers, Right-handers and Ambidexters

PROGRAMMING

800

[ "implementation", "math" ]

null

You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand. The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands. Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand. Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.

The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.

Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.

[ "1 4 2\n", "5 5 5\n", "0 2 0\n" ]

[ "6\n", "14\n", "0\n" ]

In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team. In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.

500

[ { "input": "1 4 2", "output": "6" }, { "input": "5 5 5", "output": "14" }, { "input": "0 2 0", "output": "0" }, { "input": "30 70 34", "output": "128" }, { "input": "89 32 24", "output": "112" }, { "input": "89 44 77", "output": "210" }, { "input": "0 0 0", "output": "0" }, { "input": "100 100 100", "output": "300" }, { "input": "1 1 1", "output": "2" }, { "input": "30 70 35", "output": "130" }, { "input": "89 44 76", "output": "208" }, { "input": "0 100 100", "output": "200" }, { "input": "100 0 100", "output": "200" }, { "input": "100 1 100", "output": "200" }, { "input": "1 100 100", "output": "200" }, { "input": "100 100 0", "output": "200" }, { "input": "100 100 1", "output": "200" }, { "input": "1 2 1", "output": "4" }, { "input": "0 0 100", "output": "100" }, { "input": "0 100 0", "output": "0" }, { "input": "100 0 0", "output": "0" }, { "input": "10 8 7", "output": "24" }, { "input": "45 47 16", "output": "108" }, { "input": "59 43 100", "output": "202" }, { "input": "34 1 30", "output": "62" }, { "input": "14 81 1", "output": "30" }, { "input": "53 96 94", "output": "242" }, { "input": "62 81 75", "output": "218" }, { "input": "21 71 97", "output": "188" }, { "input": "49 82 73", "output": "204" }, { "input": "88 19 29", "output": "96" }, { "input": "89 4 62", "output": "132" }, { "input": "58 3 65", "output": "126" }, { "input": "27 86 11", "output": "76" }, { "input": "35 19 80", "output": "134" }, { "input": "4 86 74", "output": "156" }, { "input": "32 61 89", "output": "182" }, { "input": "68 60 98", "output": "226" }, { "input": "37 89 34", "output": "142" }, { "input": "92 9 28", "output": "74" }, { "input": "79 58 98", "output": "234" }, { "input": "35 44 88", "output": "166" }, { "input": "16 24 19", "output": "58" }, { "input": "74 71 75", "output": "220" }, { "input": "83 86 99", "output": "268" }, { "input": "97 73 15", "output": "176" }, { "input": "77 76 73", "output": "226" }, { "input": "48 85 55", "output": "188" }, { "input": "1 2 2", "output": "4" }, { "input": "2 2 2", "output": "6" }, { "input": "2 1 2", "output": "4" }, { "input": "2 2 1", "output": "4" }, { "input": "3 2 1", "output": "6" }, { "input": "1 2 3", "output": "6" }, { "input": "1 3 2", "output": "6" }, { "input": "2 1 3", "output": "6" }, { "input": "2 3 1", "output": "6" }, { "input": "3 1 2", "output": "6" }, { "input": "99 99 99", "output": "296" }, { "input": "99 99 100", "output": "298" }, { "input": "99 100 99", "output": "298" }, { "input": "99 100 100", "output": "298" }, { "input": "100 99 99", "output": "298" }, { "input": "100 99 100", "output": "298" }, { "input": "100 100 99", "output": "298" }, { "input": "89 32 23", "output": "110" }, { "input": "4 5 0", "output": "8" }, { "input": "3 0 3", "output": "6" }, { "input": "0 0 2", "output": "2" }, { "input": "97 97 0", "output": "194" }, { "input": "1 4 0", "output": "2" }, { "input": "5 2 0", "output": "4" }, { "input": "0 5 10", "output": "14" }, { "input": "0 1 2", "output": "2" }, { "input": "5 2 3", "output": "10" }, { "input": "5 5 0", "output": "10" }, { "input": "0 0 10", "output": "10" }, { "input": "0 1 1", "output": "2" }, { "input": "0 0 1", "output": "0" } ]

1,631,174,570

2,147,483,647

Python 3

OK

TESTS

79

62

6,758,400

l, r, a = map(int, input().split()) print(2 * min(l + a, r + a, (l + r + a) // 2))

Title: Left-handers, Right-handers and Ambidexters Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand. The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands. Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand. Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively. Input Specification: The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training. Output Specification: Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players. Demo Input: ['1 4 2\n', '5 5 5\n', '0 2 0\n'] Demo Output: ['6\n', '14\n', '0\n'] Note: In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team. In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.

```python l, r, a = map(int, input().split()) print(2 * min(l + a, r + a, (l + r + a) // 2)) ```

3

600

B

Queries about less or equal elements

PROGRAMMING

1,300

[ "binary search", "data structures", "sortings", "two pointers" ]

null

You are given two arrays of integers *a* and *b*. For each element of the second array *b**j* you should find the number of elements in array *a* that are less than or equal to the value *b**j*.

The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=2·105) — the sizes of arrays *a* and *b*. The second line contains *n* integers — the elements of array *a* (<=-<=109<=≤<=*a**i*<=≤<=109). The third line contains *m* integers — the elements of array *b* (<=-<=109<=≤<=*b**j*<=≤<=109).

Print *m* integers, separated by spaces: the *j*-th of which is equal to the number of such elements in array *a* that are less than or equal to the value *b**j*.

[ "5 4\n1 3 5 7 9\n6 4 2 8\n", "5 5\n1 2 1 2 5\n3 1 4 1 5\n" ]

[ "3 2 1 4\n", "4 2 4 2 5\n" ]

none

0

[ { "input": "5 4\n1 3 5 7 9\n6 4 2 8", "output": "3 2 1 4" }, { "input": "5 5\n1 2 1 2 5\n3 1 4 1 5", "output": "4 2 4 2 5" }, { "input": "1 1\n-1\n-2", "output": "0" }, { "input": "1 1\n-80890826\n686519510", "output": "1" }, { "input": "11 11\n237468511 -779187544 -174606592 193890085 404563196 -71722998 -617934776 170102710 -442808289 109833389 953091341\n994454001 322957429 216874735 -606986750 -455806318 -663190696 3793295 41395397 -929612742 -787653860 -684738874", "output": "11 9 8 2 2 1 5 5 0 0 1" }, { "input": "20 22\n858276994 -568758442 -918490847 -983345984 -172435358 389604931 200224783 486556113 413281867 -258259500 -627945379 -584563643 444685477 -602481243 -370745158 965672503 630955806 -626138773 -997221880 633102929\n-61330638 -977252080 -212144219 385501731 669589742 954357160 563935906 584468977 -895883477 405774444 853372186 186056475 -964575261 -952431965 632332084 -388829939 -23011650 310957048 -770695392 977376693 321435214 199223897", "output": "11 2 10 12 18 19 16 16 3 13 18 11 2 2 17 8 11 12 3 20 12 11" }, { "input": "5 9\n1 3 5 7 9\n1 2 3 4 5 6 7 8 9", "output": "1 1 2 2 3 3 4 4 5" }, { "input": "22 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22\n1", "output": "1" }, { "input": "5 1\n1 3 3 3 5\n3", "output": "4" }, { "input": "4 5\n1 1 1 4\n1 5 5 4 3", "output": "3 4 4 4 3" }, { "input": "5 4\n0 5 5 5 6\n5 1 6 3", "output": "4 1 5 1" }, { "input": "1 3\n0\n-1 0 1", "output": "0 1 1" }, { "input": "96 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1", "output": "96" }, { "input": "7 1\n1 2 3 4 5 6 7\n1", "output": "1" }, { "input": "13 13\n-1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000\n-1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000", "output": "7 13 7 13 7 13 7 13 7 13 7 13 7" }, { "input": "9 5\n1 2 3 4 5 6 7 8 9\n1 2 3 4 5", "output": "1 2 3 4 5" }, { "input": "3 8\n1 1 1\n1 1 1 1 1 1 1 1", "output": "3 3 3 3 3 3 3 3" }, { "input": "1 1\n-11111\n-5938", "output": "1" }, { "input": "1 1\n1\n400000009", "output": "1" }, { "input": "1 1\n1\n300000009", "output": "1" }, { "input": "1 1\n1\n200000009", "output": "1" }, { "input": "1 1\n1\n200000003", "output": "1" } ]

1,684,921,174

2,147,483,647

PyPy 3-64

OK

TESTS

60

545

39,833,600

import sys,random,bisect from collections import deque,defaultdict,Counter from heapq import heapify,heappop,heappush from math import gcd from types import GeneratorType #from functools import cache 3.9 mod = int(1e9 + 7) #998244353 inf = int(1e20) input = lambda :sys.stdin.readline().rstrip() mi = lambda :map(int,input().split()) li = lambda :list(mi()) ii = lambda :int(input()) py = lambda :print("YES") pn = lambda :print("NO") class BIT1: """单点修改,区间和查询""" __slots__ = "size", "bit", "tree" def __init__(self, n: int): self.size = n self.bit = n.bit_length() self.tree = dict() def add(self, index: int, delta: int) -> None: # index 必须大于0 while index <= self.size: self.tree[index] = self.tree.get(index, 0) + delta index += index & -index def _query(self, index: int) -> int: #外部不要调用，这个不是单点值，是前缀和，查单点值还是用queryRange if index > self.size: index = self.size res = 0 while index > 0: res += self.tree.get(index, 0) index -= index & -index return res def queryRange(self, left: int, right: int) -> int: return self._query(right) - self._query(left - 1) n,m=li() arr=li() brr=li() arr.sort(reverse=True) res=[0]*m for i,b in sorted(enumerate(brr),key=lambda x:x[1]): while arr and arr[-1]<=b: arr.pop() res[i]=n-len(arr) print(" ".join([str(i) for i in res]))

Title: Queries about less or equal elements Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given two arrays of integers *a* and *b*. For each element of the second array *b**j* you should find the number of elements in array *a* that are less than or equal to the value *b**j*. Input Specification: The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=2·105) — the sizes of arrays *a* and *b*. The second line contains *n* integers — the elements of array *a* (<=-<=109<=≤<=*a**i*<=≤<=109). The third line contains *m* integers — the elements of array *b* (<=-<=109<=≤<=*b**j*<=≤<=109). Output Specification: Print *m* integers, separated by spaces: the *j*-th of which is equal to the number of such elements in array *a* that are less than or equal to the value *b**j*. Demo Input: ['5 4\n1 3 5 7 9\n6 4 2 8\n', '5 5\n1 2 1 2 5\n3 1 4 1 5\n'] Demo Output: ['3 2 1 4\n', '4 2 4 2 5\n'] Note: none

```python import sys,random,bisect from collections import deque,defaultdict,Counter from heapq import heapify,heappop,heappush from math import gcd from types import GeneratorType #from functools import cache 3.9 mod = int(1e9 + 7) #998244353 inf = int(1e20) input = lambda :sys.stdin.readline().rstrip() mi = lambda :map(int,input().split()) li = lambda :list(mi()) ii = lambda :int(input()) py = lambda :print("YES") pn = lambda :print("NO") class BIT1: """单点修改,区间和查询""" __slots__ = "size", "bit", "tree" def __init__(self, n: int): self.size = n self.bit = n.bit_length() self.tree = dict() def add(self, index: int, delta: int) -> None: # index 必须大于0 while index <= self.size: self.tree[index] = self.tree.get(index, 0) + delta index += index & -index def _query(self, index: int) -> int: #外部不要调用，这个不是单点值，是前缀和，查单点值还是用queryRange if index > self.size: index = self.size res = 0 while index > 0: res += self.tree.get(index, 0) index -= index & -index return res def queryRange(self, left: int, right: int) -> int: return self._query(right) - self._query(left - 1) n,m=li() arr=li() brr=li() arr.sort(reverse=True) res=[0]*m for i,b in sorted(enumerate(brr),key=lambda x:x[1]): while arr and arr[-1]<=b: arr.pop() res[i]=n-len(arr) print(" ".join([str(i) for i in res])) ```

3

144

A

Arrival of the General

PROGRAMMING

800

[ "implementation" ]

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" } ]

1,689,361,258

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

62

0

num1 = int(input()) list1 = list(map(int,input().split())) num2 = min(list1) num3 = max(list1) num4 = list1.index(num3) for x in range(num1): if list1[-(x+1)] == num2: if num4 > num1-x: num4 -= 1 num4 += x break print(num4)

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 num1 = int(input()) list1 = list(map(int,input().split())) num2 = min(list1) num3 = max(list1) num4 = list1.index(num3) for x in range(num1): if list1[-(x+1)] == num2: if num4 > num1-x: num4 -= 1 num4 += x break print(num4) ```

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" } ]

1,643,631,825

2,147,483,647

Python 3

OK

TESTS

32

92

0

n = int(input()) s = input().split() chet = [] nechet = [] for i in s: if int(i) % 2 == 0: chet.append(i) else: nechet.append(i) if len(chet) > len(nechet): print(s.index(nechet[0])+1) else: print(s.index(chet[0])+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()) s = input().split() chet = [] nechet = [] for i in s: if int(i) % 2 == 0: chet.append(i) else: nechet.append(i) if len(chet) > len(nechet): print(s.index(nechet[0])+1) else: print(s.index(chet[0])+1) ```

3.977

478

B

Random Teams

PROGRAMMING

1,300

[ "combinatorics", "constructive algorithms", "greedy", "math" ]

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" } ]

1,681,107,492

2,147,483,647

PyPy 3

OK

TESTS

26

93

0

def nc2_(n): if n <= 1: return 0 return n*(n-1)//2 n, m = [int(i) for i in input().split()] mini = nc2_(n//m+1)*(n%m) + nc2_(n//m) * (m - (n%m)) print(mini, end = ' ') maxi = n-m+1 print(maxi*(maxi-1)//2)

Title: Random Teams Time Limit: None seconds Memory Limit: None megabytes Problem Description: *n* participants of the competition were split into *m* teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends. Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition. Input Specification: The only line of input contains two integers *n* and *m*, separated by a single space (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the number of participants and the number of teams respectively. Output Specification: The only line of the output should contain two integers *k**min* and *k**max* — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively. Demo Input: ['5 1\n', '3 2\n', '6 3\n'] Demo Output: ['10 10\n', '1 1\n', '3 6\n'] Note: In the first sample all the participants get into one team, so there will be exactly ten pairs of friends. In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one. In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.

```python def nc2_(n): if n <= 1: return 0 return n*(n-1)//2 n, m = [int(i) for i in input().split()] mini = nc2_(n//m+1)*(n%m) + nc2_(n//m) * (m - (n%m)) print(mini, end = ' ') maxi = n-m+1 print(maxi*(maxi-1)//2) ```

3

577

A

Multiplication Table

PROGRAMMING

1,000

[ "implementation", "number theory" ]

null

Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1. You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.

The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table.

Print a single number: the number of times *x* occurs in the table.

[ "10 5\n", "6 12\n", "5 13\n" ]

[ "2\n", "4\n", "0\n" ]

A table for the second sample test is given below. The occurrences of number 12 are marked bold.

500

[ { "input": "10 5", "output": "2" }, { "input": "6 12", "output": "4" }, { "input": "5 13", "output": "0" }, { "input": "1 1", "output": "1" }, { "input": "2 1", "output": "1" }, { "input": "100000 1", "output": "1" }, { "input": "1 1000000000", "output": "0" }, { "input": "100000 1000000000", "output": "16" }, { "input": "100000 362880", "output": "154" }, { "input": "1 4", "output": "0" }, { "input": "9 12", "output": "4" }, { "input": "10 123", "output": "0" }, { "input": "9551 975275379", "output": "0" }, { "input": "17286 948615687", "output": "0" }, { "input": "58942 936593001", "output": "0" }, { "input": "50000 989460910", "output": "4" }, { "input": "22741 989460910", "output": "0" }, { "input": "22740 989460910", "output": "0" }, { "input": "100000 989460910", "output": "4" }, { "input": "100000 98280", "output": "128" }, { "input": "100000 997920", "output": "222" }, { "input": "100000 720720", "output": "226" }, { "input": "100000 2162160", "output": "282" }, { "input": "100000 4324320", "output": "320" }, { "input": "100000 8648640", "output": "348" }, { "input": "100000 183783600", "output": "438" }, { "input": "100000 551350800", "output": "392" }, { "input": "40000 551350800", "output": "150" }, { "input": "20000 400000000", "output": "1" }, { "input": "19999 400000000", "output": "0" }, { "input": "19999 399960001", "output": "1" }, { "input": "31621 999887641", "output": "1" }, { "input": "31622 999887641", "output": "1" }, { "input": "31620 999887641", "output": "0" }, { "input": "100000 999887641", "output": "3" }, { "input": "100000 25", "output": "3" }, { "input": "100000 1", "output": "1" }, { "input": "100000 3628800", "output": "220" }, { "input": "100000 39916800", "output": "328" }, { "input": "100000 479001600", "output": "254" }, { "input": "4 9", "output": "1" }, { "input": "2 6", "output": "0" }, { "input": "20 100", "output": "3" }, { "input": "10 3", "output": "2" }, { "input": "4 4", "output": "3" }, { "input": "2 4", "output": "1" }, { "input": "89874 1", "output": "1" } ]

1,658,258,067

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

30

0

[n,x]=[int(i) for i in input().split()][:2] i=1 j=0 while i<n and i<=x: if x%i==0 and (x/i)<=n: j+=1 i+=1 else: i+=1 print(j)

Title: Multiplication Table Time Limit: None seconds Memory Limit: None megabytes Problem Description: Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1. You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*. Input Specification: The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table. Output Specification: Print a single number: the number of times *x* occurs in the table. Demo Input: ['10 5\n', '6 12\n', '5 13\n'] Demo Output: ['2\n', '4\n', '0\n'] Note: A table for the second sample test is given below. The occurrences of number 12 are marked bold.

```python [n,x]=[int(i) for i in input().split()][:2] i=1 j=0 while i<n and i<=x: if x%i==0 and (x/i)<=n: j+=1 i+=1 else: i+=1 print(j) ```

0

189

A

Cut Ribbon

PROGRAMMING

1,300

[ "brute force", "dp" ]

null

Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions: - After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum. Help Polycarpus and find the number of ribbon pieces after the required cutting.

The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide.

Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.

[ "5 5 3 2\n", "7 5 5 2\n" ]

[ "2\n", "2\n" ]

In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3. In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.

500

[ { "input": "5 5 3 2", "output": "2" }, { "input": "7 5 5 2", "output": "2" }, { "input": "4 4 4 4", "output": "1" }, { "input": "1 1 1 1", "output": "1" }, { "input": "4000 1 2 3", "output": "4000" }, { "input": "4000 3 4 5", "output": "1333" }, { "input": "10 3 4 5", "output": "3" }, { "input": "100 23 15 50", "output": "2" }, { "input": "3119 3515 1021 7", "output": "11" }, { "input": "918 102 1327 1733", "output": "9" }, { "input": "3164 42 430 1309", "output": "15" }, { "input": "3043 317 1141 2438", "output": "7" }, { "input": "26 1 772 2683", "output": "26" }, { "input": "370 2 1 15", "output": "370" }, { "input": "734 12 6 2", "output": "367" }, { "input": "418 18 14 17", "output": "29" }, { "input": "18 16 28 9", "output": "2" }, { "input": "14 6 2 17", "output": "7" }, { "input": "29 27 18 2", "output": "2" }, { "input": "29 12 7 10", "output": "3" }, { "input": "27 23 4 3", "output": "9" }, { "input": "5 14 5 2", "output": "1" }, { "input": "5 17 26 5", "output": "1" }, { "input": "9 1 10 3", "output": "9" }, { "input": "2 19 15 1", "output": "2" }, { "input": "4 6 4 9", "output": "1" }, { "input": "10 6 2 9", "output": "5" }, { "input": "2 2 9 6", "output": "1" }, { "input": "6 2 4 1", "output": "6" }, { "input": "27 24 5 27", "output": "1" }, { "input": "2683 83 26 2709", "output": "101" }, { "input": "728 412 789 158", "output": "3" }, { "input": "3964 4 2916 176", "output": "991" }, { "input": "3399 2035 2 3334", "output": "683" }, { "input": "3455 244 3301 3", "output": "991" }, { "input": "595 2263 3625 1", "output": "595" }, { "input": "4000 1 1 1", "output": "4000" }, { "input": "3999 2 2 3999", "output": "1" }, { "input": "25 6 8 11", "output": "3" }, { "input": "4000 500 1000 2000", "output": "8" }, { "input": "53 10 11 23", "output": "5" }, { "input": "100 100 1 1", "output": "100" }, { "input": "17 3 4 10", "output": "5" }, { "input": "413 101 102 105", "output": "4" }, { "input": "490 4 49 50", "output": "111" }, { "input": "3999 2 3 3", "output": "1999" }, { "input": "8 3 8 4", "output": "2" }, { "input": "5 1 3 3", "output": "5" }, { "input": "100 3 17 22", "output": "27" }, { "input": "4000 2 3 4", "output": "2000" }, { "input": "4000 3 3 5", "output": "1332" }, { "input": "13 4 6 7", "output": "2" }, { "input": "4000 5 2 2", "output": "2000" }, { "input": "3999 2 2 3", "output": "1999" }, { "input": "4000 33 7 3333", "output": "564" }, { "input": "60 33 20 9", "output": "4" }, { "input": "100 9 11 99", "output": "10" }, { "input": "2009 6 8 9", "output": "334" } ]

1,697,619,724

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

30

0

n, a, b, c = map(int, input().split()) # Initialize an array to store the maximum number of pieces for each possible length dp = [0] * (n + 1) # Initialize the first element to -1 (impossible to cut) dp[0] = -1 # Iterate through all possible lengths from a to n for i in range(a, n + 1): # Update the maximum number of pieces dp[i] = max(dp[i], dp[i - a] + 1) for i in range(b, n + 1): dp[i] = max(dp[i], dp[i - b] + 1) for i in range(c, n + 1): dp[i] = max(dp[i], dp[i - c] + 1) # The maximum number of pieces is stored in dp[n] print(dp[n])

Title: Cut Ribbon Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions: - After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum. Help Polycarpus and find the number of ribbon pieces after the required cutting. Input Specification: The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide. Output Specification: Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists. Demo Input: ['5 5 3 2\n', '7 5 5 2\n'] Demo Output: ['2\n', '2\n'] Note: In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3. In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.

```python n, a, b, c = map(int, input().split()) # Initialize an array to store the maximum number of pieces for each possible length dp = [0] * (n + 1) # Initialize the first element to -1 (impossible to cut) dp[0] = -1 # Iterate through all possible lengths from a to n for i in range(a, n + 1): # Update the maximum number of pieces dp[i] = max(dp[i], dp[i - a] + 1) for i in range(b, n + 1): dp[i] = max(dp[i], dp[i - b] + 1) for i in range(c, n + 1): dp[i] = max(dp[i], dp[i - c] + 1) # The maximum number of pieces is stored in dp[n] print(dp[n]) ```

0

361

B

Levko and Permutation

PROGRAMMING

1,200

[ "constructive algorithms", "math", "number theory" ]

null

Levko loves permutations very much. A permutation of length *n* is a sequence of distinct positive integers, each is at most *n*. Let’s assume that value *gcd*(*a*,<=*b*) shows the greatest common divisor of numbers *a* and *b*. Levko assumes that element *p**i* of permutation *p*1,<=*p*2,<=... ,<=*p**n* is good if *gcd*(*i*,<=*p**i*)<=><=1. Levko considers a permutation beautiful, if it has exactly *k* good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them.

The single line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*).

In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist. If there are multiple suitable permutations, you are allowed to print any of them.

[ "4 2\n", "1 1\n" ]

[ "2 4 3 1", "-1\n" ]

In the first sample elements 4 and 3 are good because *gcd*(2, 4) = 2 > 1 and *gcd*(3, 3) = 3 > 1. Elements 2 and 1 are not good because *gcd*(1, 2) = 1 and *gcd*(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful. The second sample has no beautiful permutations.

1,000

[ { "input": "4 2", "output": "2 1 3 4 " }, { "input": "1 1", "output": "-1" }, { "input": "7 4", "output": "3 1 2 4 5 6 7 " }, { "input": "10 9", "output": "1 2 3 4 5 6 7 8 9 10 " }, { "input": "10000 5000", "output": "5000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "7 0", "output": "7 1 2 3 4 5 6 " }, { "input": "1 0", "output": "1 " }, { "input": "7 7", "output": "-1" }, { "input": "100000 47", "output": "99953 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "100000 100000", "output": "-1" }, { "input": "100000 43425", "output": "56575 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "7 6", "output": "1 2 3 4 5 6 7 " }, { "input": "100000 99999", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "47 46", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 " }, { "input": "5 0", "output": "5 1 2 3 4 " }, { "input": "4 2", "output": "2 1 3 4 " }, { "input": "1533 1052", "output": "481 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "81314 52747", "output": "28567 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "17767 145", "output": "17622 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "18168 7942", "output": "10226 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "26593 15915", "output": "10678 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "26593 8877", "output": "17716 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "13852 12727", "output": "1125 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "4 1", "output": "3 1 2 4 " }, { "input": "8834 8834", "output": "-1" }, { "input": "8485 8484", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "14564 14564", "output": "-1" }, { "input": "8254 8253", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "81314 81312", "output": "2 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "5795 5792", "output": "3 1 2 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "6417 3", "output": "6414 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "6896 0", "output": "6896 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "6778 1", "output": "6777 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "9448 1", "output": "9447 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "5938 2", "output": "5936 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "3072 0", "output": "3072 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "8576 0", "output": "8576 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "2 1", "output": "1 2 " }, { "input": "4 4", "output": "-1" }, { "input": "5 5", "output": "-1" }, { "input": "2 2", "output": "-1" }, { "input": "100000 1", "output": "99999 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "100000 50000", "output": "50000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "4 1", "output": "3 1 2 4 " }, { "input": "100000 9999", "output": "90001 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "100000 99000", "output": "1000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "100000 12347", "output": "87653 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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..." } ]

1,601,530,736

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

93

307,200

entrada = input().split() n = int(entrada[0]) k = int(entrada[1]) aux = [] x = 0 if((n - k) % 2 == 0): x = 1 for i in range(1,n+1): aux.append(i) for j in range(x,n-1,k): aux[j],aux[j+1] = aux[j+1],aux[j] print(*aux)

Title: Levko and Permutation Time Limit: None seconds Memory Limit: None megabytes Problem Description: Levko loves permutations very much. A permutation of length *n* is a sequence of distinct positive integers, each is at most *n*. Let’s assume that value *gcd*(*a*,<=*b*) shows the greatest common divisor of numbers *a* and *b*. Levko assumes that element *p**i* of permutation *p*1,<=*p*2,<=... ,<=*p**n* is good if *gcd*(*i*,<=*p**i*)<=><=1. Levko considers a permutation beautiful, if it has exactly *k* good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them. Input Specification: The single line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*). Output Specification: In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist. If there are multiple suitable permutations, you are allowed to print any of them. Demo Input: ['4 2\n', '1 1\n'] Demo Output: ['2 4 3 1', '-1\n'] Note: In the first sample elements 4 and 3 are good because *gcd*(2, 4) = 2 > 1 and *gcd*(3, 3) = 3 > 1. Elements 2 and 1 are not good because *gcd*(1, 2) = 1 and *gcd*(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful. The second sample has no beautiful permutations.

```python entrada = input().split() n = int(entrada[0]) k = int(entrada[1]) aux = [] x = 0 if((n - k) % 2 == 0): x = 1 for i in range(1,n+1): aux.append(i) for j in range(x,n-1,k): aux[j],aux[j+1] = aux[j+1],aux[j] print(*aux) ```

0

445

A

DZY Loves Chessboard

PROGRAMMING

1,200

[ "dfs and similar", "implementation" ]

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" } ]

1,678,473,696

2,147,483,647

Python 3

OK

TESTS

37

46

0

# Read in the dimensions of the chessboard n, m = map(int, input().split()) # Read in the chessboard as a list of strings board = [input() for _ in range(n)] # Iterate over each cell in the chessboard and assign a chessman to it for i in range(n): for j in range(m): # If the cell is good if board[i][j] == ".": # Assign a black or white chessman based on the position of the cell if (i+j) % 2 == 0: board[i] = board[i][:j] + "B" + board[i][j+1:] else: board[i] = board[i][:j] + "W" + board[i][j+1:] # Output for row in board: print(row)

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 # Read in the dimensions of the chessboard n, m = map(int, input().split()) # Read in the chessboard as a list of strings board = [input() for _ in range(n)] # Iterate over each cell in the chessboard and assign a chessman to it for i in range(n): for j in range(m): # If the cell is good if board[i][j] == ".": # Assign a black or white chessman based on the position of the cell if (i+j) % 2 == 0: board[i] = board[i][:j] + "B" + board[i][j+1:] else: board[i] = board[i][:j] + "W" + board[i][j+1:] # Output for row in board: print(row) ```

3

431

C

k-Tree

PROGRAMMING

1,600

[ "dp", "implementation", "trees" ]

null

Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a *k*-tree. A *k*-tree is an infinite rooted tree where: - each vertex has exactly *k* children; - each edge has some weight; - if we look at the edges that goes from some vertex to its children (exactly *k* edges), then their weights will equal 1,<=2,<=3,<=...,<=*k*. The picture below shows a part of a 3-tree. Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (109<=+<=7).

A single line contains three space-separated integers: *n*, *k* and *d* (1<=≤<=*n*,<=*k*<=≤<=100; 1<=≤<=*d*<=≤<=*k*).

Print a single integer — the answer to the problem modulo 1000000007 (109<=+<=7).

[ "3 3 2\n", "3 3 3\n", "4 3 2\n", "4 5 2\n" ]

[ "3\n", "1\n", "6\n", "7\n" ]

none

1,500

[ { "input": "3 3 2", "output": "3" }, { "input": "3 3 3", "output": "1" }, { "input": "4 3 2", "output": "6" }, { "input": "4 5 2", "output": "7" }, { "input": "28 6 3", "output": "110682188" }, { "input": "5 100 1", "output": "16" }, { "input": "50 6 3", "output": "295630102" }, { "input": "10 13 6", "output": "48" }, { "input": "20 16 14", "output": "236" }, { "input": "1 10 1", "output": "1" }, { "input": "8 11 4", "output": "47" }, { "input": "16 5 4", "output": "16175" }, { "input": "5 26 17", "output": "0" }, { "input": "35 47 42", "output": "0" }, { "input": "11 6 2", "output": "975" }, { "input": "54 60 16", "output": "931055544" }, { "input": "47 5 1", "output": "164058640" }, { "input": "70 6 1", "output": "592826579" }, { "input": "40 77 77", "output": "0" }, { "input": "96 9 6", "output": "362487247" }, { "input": "52 46 4", "output": "27907693" }, { "input": "74 41 28", "output": "806604424" }, { "input": "100 100 100", "output": "1" }, { "input": "99 99 5", "output": "700732369" }, { "input": "100 100 1", "output": "988185646" }, { "input": "98 98 64", "output": "237643149" }, { "input": "28 74 2", "output": "134217727" }, { "input": "86 69 62", "output": "217513984" }, { "input": "9 17 14", "output": "0" }, { "input": "73 72 12", "output": "426374014" }, { "input": "41 98 76", "output": "0" }, { "input": "1 1 1", "output": "1" }, { "input": "1 100 100", "output": "0" }, { "input": "1 100 1", "output": "1" }, { "input": "1 100 2", "output": "0" }, { "input": "2 100 2", "output": "1" }, { "input": "2 100 1", "output": "2" }, { "input": "50 50 1", "output": "949480669" }, { "input": "100 50 50", "output": "661237556" }, { "input": "3 2 2", "output": "2" }, { "input": "100 50 3", "output": "494224664" }, { "input": "90 97 24", "output": "413496885" }, { "input": "31 8 8", "output": "52532592" }, { "input": "78 90 38", "output": "744021655" }, { "input": "100 13 11", "output": "883875774" }, { "input": "100 45 8", "output": "367847193" }, { "input": "31 8 4", "output": "924947104" }, { "input": "35 95 9", "output": "927164672" }, { "input": "45 5 3", "output": "252804490" }, { "input": "1 5 5", "output": "0" }, { "input": "89 75 59", "output": "179807625" }, { "input": "90 100 30", "output": "697322870" }, { "input": "89 73 32", "output": "152673180" }, { "input": "100 90 80", "output": "11531520" } ]

1,685,465,128

2,147,483,647

PyPy 3-64

TIME_LIMIT_EXCEEDED

TESTS

4

1,000

12,800,000

from functools import lru_cache def numI(): return int(input()) def numII(): return list(input()) def numIII(): return list(map(int, input().split(' '))) def strInp(): return input() def strInpI(): return list(input()) @lru_cache def solve(): n, k, d = numIII() c = 0 memo = {} def dfs(num, maxi): nonlocal c if num == n and maxi >= d: c+=1 return 1 # if num in memo: # return memo[num] temp = 0 for x in range(1, k+1): if x + num <= n: ret = dfs(x + num, max(maxi, x)) temp+=ret memo[num] = temp return temp dfs(0, -1) print(c) solve()

Title: k-Tree Time Limit: None seconds Memory Limit: None megabytes Problem Description: Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a *k*-tree. A *k*-tree is an infinite rooted tree where: - each vertex has exactly *k* children; - each edge has some weight; - if we look at the edges that goes from some vertex to its children (exactly *k* edges), then their weights will equal 1,<=2,<=3,<=...,<=*k*. The picture below shows a part of a 3-tree. Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (109<=+<=7). Input Specification: A single line contains three space-separated integers: *n*, *k* and *d* (1<=≤<=*n*,<=*k*<=≤<=100; 1<=≤<=*d*<=≤<=*k*). Output Specification: Print a single integer — the answer to the problem modulo 1000000007 (109<=+<=7). Demo Input: ['3 3 2\n', '3 3 3\n', '4 3 2\n', '4 5 2\n'] Demo Output: ['3\n', '1\n', '6\n', '7\n'] Note: none

```python from functools import lru_cache def numI(): return int(input()) def numII(): return list(input()) def numIII(): return list(map(int, input().split(' '))) def strInp(): return input() def strInpI(): return list(input()) @lru_cache def solve(): n, k, d = numIII() c = 0 memo = {} def dfs(num, maxi): nonlocal c if num == n and maxi >= d: c+=1 return 1 # if num in memo: # return memo[num] temp = 0 for x in range(1, k+1): if x + num <= n: ret = dfs(x + num, max(maxi, x)) temp+=ret memo[num] = temp return temp dfs(0, -1) print(c) solve() ```

0

122

A

Lucky Division

PROGRAMMING

1,000

[ "brute force", "number theory" ]

null

Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.

The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.

In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).

[ "47\n", "16\n", "78\n" ]

[ "YES\n", "YES\n", "NO\n" ]

Note that all lucky numbers are almost lucky as any number is evenly divisible by itself. In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.

500

[ { "input": "47", "output": "YES" }, { "input": "16", "output": "YES" }, { "input": "78", "output": "NO" }, { "input": "48", "output": "YES" }, { "input": "100", "output": "YES" }, { "input": "107", "output": "NO" }, { "input": "77", "output": "YES" }, { "input": "477", "output": "YES" }, { "input": "480", "output": "YES" }, { "input": "1", "output": "NO" }, { "input": "3", "output": "NO" }, { "input": "4", "output": "YES" }, { "input": "49", "output": "YES" }, { "input": "56", "output": "YES" }, { "input": "124", "output": "YES" }, { "input": "1000", "output": "YES" }, { "input": "999", "output": "NO" }, { "input": "298", "output": "NO" }, { "input": "274", "output": "NO" }, { "input": "998", "output": "NO" }, { "input": "42", "output": "YES" }, { "input": "788", "output": "YES" }, { "input": "70", "output": "YES" }, { "input": "444", "output": "YES" }, { "input": "777", "output": "YES" }, { "input": "799", "output": "YES" }, { "input": "25", "output": "NO" }, { "input": "882", "output": "YES" }, { "input": "88", "output": "YES" }, { "input": "11", "output": "NO" }, { "input": "2", "output": "NO" }, { "input": "7", "output": "YES" }, { "input": "8", "output": "YES" }, { "input": "94", "output": "YES" }, { "input": "477", "output": "YES" }, { "input": "141", "output": "YES" } ]

1,691,814,490

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

32

92

0

from sys import stdin a = stdin.readline().strip() b = a.count("4");c = a.count("7") if b+c == len(list(a)): print("YES") else: for i in range(4,int(a)//2): if int(a)%i == 0 : c = str(i) b = c.count("4");c = c.count("7") if b+c == len(list(str(i))): print("YES") exit() print("NO")

Title: Lucky Division Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky. Input Specification: The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked. Output Specification: In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes). Demo Input: ['47\n', '16\n', '78\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: Note that all lucky numbers are almost lucky as any number is evenly divisible by itself. In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.

```python from sys import stdin a = stdin.readline().strip() b = a.count("4");c = a.count("7") if b+c == len(list(a)): print("YES") else: for i in range(4,int(a)//2): if int(a)%i == 0 : c = str(i) b = c.count("4");c = c.count("7") if b+c == len(list(str(i))): print("YES") exit() print("NO") ```

0

501

A

Contest

PROGRAMMING

900

[ "implementation" ]

null

Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points. Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.

The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180). It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round).

Output on a single line: "Misha" (without the quotes), if Misha got more points than Vasya. "Vasya" (without the quotes), if Vasya got more points than Misha. "Tie" (without the quotes), if both of them got the same number of points.

[ "500 1000 20 30\n", "1000 1000 1 1\n", "1500 1000 176 177\n" ]

[ "Vasya\n", "Tie\n", "Misha\n" ]

none

500

[ { "input": "500 1000 20 30", "output": "Vasya" }, { "input": "1000 1000 1 1", "output": "Tie" }, { "input": "1500 1000 176 177", "output": "Misha" }, { "input": "1500 1000 74 177", "output": "Misha" }, { "input": "750 2500 175 178", "output": "Vasya" }, { "input": "750 1000 54 103", "output": "Tie" }, { "input": "2000 1250 176 130", "output": "Tie" }, { "input": "1250 1750 145 179", "output": "Tie" }, { "input": "2000 2000 176 179", "output": "Tie" }, { "input": "1500 1500 148 148", "output": "Tie" }, { "input": "2750 1750 134 147", "output": "Misha" }, { "input": "3250 250 175 173", "output": "Misha" }, { "input": "500 500 170 176", "output": "Misha" }, { "input": "250 1000 179 178", "output": "Vasya" }, { "input": "3250 1000 160 138", "output": "Misha" }, { "input": "3000 2000 162 118", "output": "Tie" }, { "input": "1500 1250 180 160", "output": "Tie" }, { "input": "1250 2500 100 176", "output": "Tie" }, { "input": "3500 3500 177 178", "output": "Tie" }, { "input": "3000 3250 16 34", "output": "Tie" }, { "input": "1750 3000 137 49", "output": "Vasya" }, { "input": "500 1500 179 71", "output": "Vasya" }, { "input": "1250 2000 101 180", "output": "Misha" }, { "input": "250 750 180 176", "output": "Vasya" }, { "input": "2250 2250 163 145", "output": "Vasya" }, { "input": "3000 3000 176 78", "output": "Vasya" }, { "input": "250 3500 8 178", "output": "Vasya" }, { "input": "1750 1250 179 180", "output": "Misha" }, { "input": "2750 1750 13 164", "output": "Misha" }, { "input": "1750 2250 178 53", "output": "Vasya" }, { "input": "2500 2750 73 179", "output": "Misha" }, { "input": "1000 3500 178 175", "output": "Vasya" }, { "input": "1000 500 7 162", "output": "Misha" }, { "input": "1000 250 175 48", "output": "Misha" }, { "input": "1750 500 166 177", "output": "Misha" }, { "input": "250 250 0 0", "output": "Tie" }, { "input": "250 3500 0 0", "output": "Vasya" }, { "input": "250 3500 0 180", "output": "Vasya" }, { "input": "3500 3500 180 180", "output": "Tie" }, { "input": "3500 250 0 180", "output": "Misha" } ]

1,515,801,697

2,147,483,647

Python 3

OK

TESTS

40

77

5,632,000

a, b, c, d = list(map(int, input().split())) if max(3*a/10, a-a/250*c)> max(3*b/10, b-b/250*d): print("Misha") elif max(3*a/10, a-a/250*c)< max(3*b/10, b-b/250*d): print("Vasya") else: print("Tie")

Title: Contest Time Limit: None seconds Memory Limit: None megabytes Problem Description: Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points. Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth. Input Specification: The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180). It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round). Output Specification: Output on a single line: "Misha" (without the quotes), if Misha got more points than Vasya. "Vasya" (without the quotes), if Vasya got more points than Misha. "Tie" (without the quotes), if both of them got the same number of points. Demo Input: ['500 1000 20 30\n', '1000 1000 1 1\n', '1500 1000 176 177\n'] Demo Output: ['Vasya\n', 'Tie\n', 'Misha\n'] Note: none

```python a, b, c, d = list(map(int, input().split())) if max(3*a/10, a-a/250*c)> max(3*b/10, b-b/250*d): print("Misha") elif max(3*a/10, a-a/250*c)< max(3*b/10, b-b/250*d): print("Vasya") else: print("Tie") ```

3

344

A

Magnets

PROGRAMMING

800

[ "implementation" ]

null

Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other. Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.

The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.

On the single line of the output print the number of groups of magnets.

[ "6\n10\n10\n10\n01\n10\n10\n", "4\n01\n01\n10\n10\n" ]

[ "3\n", "2\n" ]

The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets. The second testcase has two groups, each consisting of two magnets.

500

[ { "input": "6\n10\n10\n10\n01\n10\n10", "output": "3" }, { "input": "4\n01\n01\n10\n10", "output": "2" }, { "input": "1\n10", "output": "1" }, { "input": "2\n01\n10", "output": "2" }, { "input": "2\n10\n10", "output": "1" }, { "input": "3\n10\n01\n10", "output": "3" }, { "input": "1\n01", "output": "1" }, { "input": "2\n01\n01", "output": "1" }, { "input": "2\n10\n01", "output": "2" }, { "input": "3\n01\n01\n01", "output": "1" }, { "input": "3\n10\n10\n01", "output": "2" }, { "input": "3\n01\n10\n10", "output": "2" }, { "input": "115\n10\n10\n10\n10\n01\n01\n10\n10\n10\n01\n01\n10\n01\n01\n10\n10\n10\n01\n10\n01\n10\n10\n01\n01\n10\n10\n10\n10\n01\n10\n01\n01\n10\n10\n10\n10\n01\n10\n10\n10\n01\n10\n01\n10\n10\n10\n10\n01\n01\n01\n10\n10\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n01\n10\n01\n10\n10\n01\n01\n01\n10\n01\n01\n10\n10\n01\n01\n01\n01\n01\n10\n01\n10\n01\n10\n01\n01\n01\n10\n01\n10\n10\n01\n10\n10\n01\n01\n01\n10\n10\n10\n10\n10\n10\n10\n10", "output": "55" } ]

1,691,051,959

2,147,483,647

PyPy 3-64

TIME_LIMIT_EXCEEDED

TESTS

6

1,000

8,806,400

c=1 t=int(input()) l=[] for i in range(t): l.append(int(input())) for i in range(1,t): if l[i-1]!=l[i]: c+=1 print(c)

Title: Magnets Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other. Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. Input Specification: The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. Output Specification: On the single line of the output print the number of groups of magnets. Demo Input: ['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n'] Demo Output: ['3\n', '2\n'] Note: The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets. The second testcase has two groups, each consisting of two magnets.

```python c=1 t=int(input()) l=[] for i in range(t): l.append(int(input())) for i in range(1,t): if l[i-1]!=l[i]: c+=1 print(c) ```

0

802

G

Fake News (easy)

PROGRAMMING

800

[ "implementation", "strings" ]

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" } ]

1,633,717,459

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

30

6,963,200

words = list(input()) subsequence = 'heidi' count = 0 fh = 0 fe = 0 fi = 0 fd = 0 for word in words: if word == 'h' and fh == 0: fh = 1 count += 1 elif fh == 1 and word == 'e': fe = 1 count += 1 elif fe == 1 and word == 'i': fi = 1 count += 1 elif fi == 1 and word == 'd': fd = 1 count += 1 elif fd == 1 and word == 'i': count += 1 break if count == 5: print('YES') else: print('NO')

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 words = list(input()) subsequence = 'heidi' count = 0 fh = 0 fe = 0 fi = 0 fd = 0 for word in words: if word == 'h' and fh == 0: fh = 1 count += 1 elif fh == 1 and word == 'e': fe = 1 count += 1 elif fe == 1 and word == 'i': fi = 1 count += 1 elif fi == 1 and word == 'd': fd = 1 count += 1 elif fd == 1 and word == 'i': count += 1 break if count == 5: print('YES') else: print('NO') ```

0

688

B

Lovely Palindromes

PROGRAMMING

1,000

[ "constructive algorithms", "math" ]

null

Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not. Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them. Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number?

The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000).

Print the *n*-th even-length palindrome number.

[ "1\n", "10\n" ]

[ "11\n", "1001\n" ]

The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.

1,000

[ { "input": "1", "output": "11" }, { "input": "10", "output": "1001" }, { "input": "11", "output": "1111" }, { "input": "12", "output": "1221" }, { "input": "100", "output": "100001" }, { "input": "1321", "output": "13211231" }, { "input": "2", "output": "22" }, { "input": "3", "output": "33" }, { "input": "4", "output": "44" }, { "input": "5", "output": "55" }, { "input": "6", "output": "66" }, { "input": "7", "output": "77" }, { "input": "8", "output": "88" }, { "input": "9", "output": "99" }, { "input": "13", "output": "1331" }, { "input": "14", "output": "1441" }, { "input": "15", "output": "1551" }, { "input": "16", "output": "1661" }, { "input": "17", "output": "1771" }, { "input": "18", "output": "1881" }, { "input": "19", "output": "1991" }, { "input": "20", "output": "2002" }, { "input": "26550", "output": "2655005562" }, { "input": "16137", "output": "1613773161" }, { "input": "91471", "output": "9147117419" }, { "input": "41242", "output": "4124224214" }, { "input": "30866", "output": "3086666803" }, { "input": "4442231232741003009964183199713851566466079882929007253443848371978048610682180431295671905105980116", "output": "44422312327410030099641831997138515664660798829290072534438483719780486106821804312956719051059801166110895015091765921340812860168408791738483443527009292889706646651583179913814699003001472321322444" }, { "input": "7806526084732010742850470902979267046151204705563201965818390803445852115032119297648697933316807252", "output": "78065260847320107428504709029792670461512047055632019658183908034458521150321192976486979333168072522527086133397968467929112305112585443080938185691023655074021516407629792090740582470102374806256087" }, { "input": "2244399823612183124978103773977739948197709395321563148611877999503840818596183803323258317041875913", "output": "22443998236121831249781037739777399481977093953215631486118779995038408185961838033232583170418759133195781407138523233083816958180483059997781168413651235939077918499377793773018794213812163289934422" }, { "input": "123451", "output": "123451154321" } ]

1,596,877,756

2,147,483,647

PyPy 3

OK

TESTS

40

155

21,606,400

import sys import math input = sys.stdin.readline s = input().strip() print(s + s[::-1])

Title: Lovely Palindromes Time Limit: None seconds Memory Limit: None megabytes Problem Description: Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not. Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them. Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number? Input Specification: The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000). Output Specification: Print the *n*-th even-length palindrome number. Demo Input: ['1\n', '10\n'] Demo Output: ['11\n', '1001\n'] Note: The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.

```python import sys import math input = sys.stdin.readline s = input().strip() print(s + s[::-1]) ```

3

926

G

Large Bouquets

PROGRAMMING

1,500

[]

null

A flower shop has got *n* bouquets, and the *i*-th bouquet consists of *a**i* flowers. Vasya, the manager of the shop, decided to make large bouquets from these bouquets. Vasya thinks that a bouquet is large if it is made of two or more initial bouquets, and there is a constraint: the total number of flowers in a large bouquet should be odd. Each of the initial bouquets can be a part of at most one large bouquet. If an initial bouquet becomes a part of a large bouquet, all its flowers are included in the large bouquet. Determine the maximum possible number of large bouquets Vasya can make.

The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of initial bouquets. The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the number of flowers in each of the initial bouquets.

Print the maximum number of large bouquets Vasya can make.

[ "5\n2 3 4 2 7\n", "6\n2 2 6 8 6 12\n", "3\n11 4 10\n" ]

[ "2\n", "0\n", "1\n" ]

In the first example Vasya can make 2 large bouquets. For example, the first bouquet can contain the first and the fifth initial bouquets (the total number of flowers is then equal to 9), and the second bouquet can consist of the second and the third initial bouquets (the total number of flowers is then equal to 7). The fourth initial bouquet is unused in this scheme. In the second example it is not possible to form a single bouquet with odd number of flowers. In the third example Vasya can make one large bouquet. For example, he can make it using all three initial bouquets. The size of the large bouquet is then equal to 11 + 4 + 10 = 25.

0

[ { "input": "5\n2 3 4 2 7", "output": "2" }, { "input": "6\n2 2 6 8 6 12", "output": "0" }, { "input": "3\n11 4 10", "output": "1" }, { "input": "1\n1", "output": "0" }, { "input": "1\n2", "output": "0" }, { "input": "1\n999999", "output": "0" }, { "input": "1\n1000000", "output": "0" }, { "input": "4\n943543 151729 379602 589828", "output": "2" }, { "input": "2\n468463 62253", "output": "0" }, { "input": "3\n352987 849349 967007", "output": "1" }, { "input": "20\n274039 899325 798709 157662 963297 276599 529230 80095 252956 980560 358150 82383 29856 901568 123794 275349 512273 508369 120076 170206", "output": "10" }, { "input": "25\n742168 377547 485672 437223 96307 902863 759104 747933 512899 410317 588598 666688 823202 257684 520631 910066 168864 71499 899972 565350 764848 754913 929040 864132 289976", "output": "10" } ]

1,521,303,699

2,799

Python 3

OK

TESTS

49

140

14,438,400

n = int(input()) s = list(map(int, input().split())) oddd = 0 evend = 0 for i in range(n): if s[i] % 2 == 0: oddd += 1 else: evend += 1 if evend == 0: print(0) else: if oddd == 0: print(evend // 3) else: if oddd < evend: print(oddd + (evend - oddd) // 3) else: print(evend)

Title: Large Bouquets Time Limit: None seconds Memory Limit: None megabytes Problem Description: A flower shop has got *n* bouquets, and the *i*-th bouquet consists of *a**i* flowers. Vasya, the manager of the shop, decided to make large bouquets from these bouquets. Vasya thinks that a bouquet is large if it is made of two or more initial bouquets, and there is a constraint: the total number of flowers in a large bouquet should be odd. Each of the initial bouquets can be a part of at most one large bouquet. If an initial bouquet becomes a part of a large bouquet, all its flowers are included in the large bouquet. Determine the maximum possible number of large bouquets Vasya can make. Input Specification: The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of initial bouquets. The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the number of flowers in each of the initial bouquets. Output Specification: Print the maximum number of large bouquets Vasya can make. Demo Input: ['5\n2 3 4 2 7\n', '6\n2 2 6 8 6 12\n', '3\n11 4 10\n'] Demo Output: ['2\n', '0\n', '1\n'] Note: In the first example Vasya can make 2 large bouquets. For example, the first bouquet can contain the first and the fifth initial bouquets (the total number of flowers is then equal to 9), and the second bouquet can consist of the second and the third initial bouquets (the total number of flowers is then equal to 7). The fourth initial bouquet is unused in this scheme. In the second example it is not possible to form a single bouquet with odd number of flowers. In the third example Vasya can make one large bouquet. For example, he can make it using all three initial bouquets. The size of the large bouquet is then equal to 11 + 4 + 10 = 25.

```python n = int(input()) s = list(map(int, input().split())) oddd = 0 evend = 0 for i in range(n): if s[i] % 2 == 0: oddd += 1 else: evend += 1 if evend == 0: print(0) else: if oddd == 0: print(evend // 3) else: if oddd < evend: print(oddd + (evend - oddd) // 3) else: print(evend) ```

3

454

B

Little Pony and Sort by Shift

PROGRAMMING

1,200

[ "implementation" ]

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" } ]

1,604,692,763

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

6

187

9,216,000

import sys input = sys.stdin.readline n = int(input()) a = list(map(int, input().split())) _min = [(10**5) + 1, 0] for i in range(n): if a[i] < _min[0]: _min = [a[i], i] a += a[:_min[1]] a = a[_min[1]:] ct = 0 for i in range(n-1): if a[i+1] < a[i] and a[i+1] != _min[0]: print(-1) break if a[i+1] == _min[0]: ct += 1 else: print((n-_min[1]+ct) % n)

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 import sys input = sys.stdin.readline n = int(input()) a = list(map(int, input().split())) _min = [(10**5) + 1, 0] for i in range(n): if a[i] < _min[0]: _min = [a[i], i] a += a[:_min[1]] a = a[_min[1]:] ct = 0 for i in range(n-1): if a[i+1] < a[i] and a[i+1] != _min[0]: print(-1) break if a[i+1] == _min[0]: ct += 1 else: print((n-_min[1]+ct) % n) ```

0

1,008

A

Romaji

PROGRAMMING

900

[ "implementation", "strings" ]

null

Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant. In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not. Help Vitya find out if a word $s$ is Berlanese.

The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.

Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO". You can print each letter in any case (upper or lower).

[ "sumimasen\n", "ninja\n", "codeforces\n" ]

[ "YES\n", "YES\n", "NO\n" ]

In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese. In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.

500

[ { "input": "sumimasen", "output": "YES" }, { "input": "ninja", "output": "YES" }, { "input": "codeforces", "output": "NO" }, { "input": "auuaoonntanonnuewannnnpuuinniwoonennyolonnnvienonpoujinndinunnenannmuveoiuuhikucuziuhunnnmunzancenen", "output": "YES" }, { "input": "n", "output": "YES" }, { "input": "necnei", "output": "NO" }, { "input": "nternn", "output": "NO" }, { "input": "aucunuohja", "output": "NO" }, { "input": "a", "output": "YES" }, { "input": "b", "output": "NO" }, { "input": "nn", "output": "YES" }, { "input": "nnnzaaa", "output": "YES" }, { "input": "zn", "output": "NO" }, { "input": "ab", "output": "NO" }, { "input": "aaaaaaaaaa", "output": "YES" }, { "input": "aaaaaaaaab", "output": "NO" }, { "input": "aaaaaaaaan", "output": "YES" }, { "input": "baaaaaaaaa", "output": "YES" }, { "input": "naaaaaaaaa", "output": "YES" }, { "input": "nbaaaaaaaa", "output": "YES" }, { "input": "bbaaaaaaaa", "output": "NO" }, { "input": "bnaaaaaaaa", "output": "NO" }, { "input": "eonwonojannonnufimiiniewuqaienokacevecinfuqihatenhunliquuyebayiaenifuexuanenuaounnboancaeowonu", "output": "YES" }, { "input": "uixinnepnlinqaingieianndeakuniooudidonnnqeaituioeneiroionxuowudiooonayenfeonuino", "output": "NO" }, { "input": "nnnnnyigaveteononnnnxaalenxuiiwannntoxonyoqonlejuoxuoconnnentoinnul", "output": "NO" }, { "input": "ndonneasoiunhomuunnhuitonnntunntoanerekonoupunanuauenu", "output": "YES" }, { "input": "anujemogawautiedoneobninnibonuunaoennnyoorufonxionntinimiboonununnnnnleenqunminzayoutanlalo", "output": "NO" }, { "input": "y", "output": "NO" }, { "input": "by", "output": "NO" }, { "input": "yy", "output": "NO" }, { "input": "nbn", "output": "NO" }, { "input": "nz", "output": "NO" }, { "input": "king", "output": "NO" }, { "input": "g", "output": "NO" }, { "input": "az", "output": "NO" }, { "input": "x", "output": "NO" }, { "input": "z", "output": "NO" }, { "input": "yn", "output": "NO" }, { "input": "aeo", "output": "YES" }, { "input": "nb", "output": "NO" }, { "input": "npn", "output": "NO" }, { "input": "kini", "output": "YES" }, { "input": "pya", "output": "NO" }, { "input": "m", "output": "NO" }, { "input": "p", "output": "NO" }, { "input": "aaaaaaaak", "output": "NO" }, { "input": "aab", "output": "NO" }, { "input": "d", "output": "NO" }, { "input": "at", "output": "NO" }, { "input": "aaaaaak", "output": "NO" }, { "input": "aaz", "output": "NO" }, { "input": "aaab", "output": "NO" }, { "input": "s", "output": "NO" }, { "input": "nzzen", "output": "NO" }, { "input": "aeionnhhhn", "output": "NO" }, { "input": "h", "output": "NO" }, { "input": "ny", "output": "NO" } ]

1,670,991,299

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

1

46

0

x = ['a', 'e', 'i', 'o', 'u', 'n'] belanese = input() n = len(belanese) i = 0 if n % 2 == 1: while i < n: if ((belanese[i] not in x) and (belanese[i+1] in x)): #print(belanese[i], belanese[i+1]) i+=2 if i == n -1: print("Yes") else: i+=1 elif n % 2 == 0: while i < n-1: #print(belanese[i], belanese[i+1]) i+=1 if i == n-1: print("No")

Title: Romaji Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant. In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not. Help Vitya find out if a word $s$ is Berlanese. Input Specification: The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters. Output Specification: Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO". You can print each letter in any case (upper or lower). Demo Input: ['sumimasen\n', 'ninja\n', 'codeforces\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese. In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.

```python x = ['a', 'e', 'i', 'o', 'u', 'n'] belanese = input() n = len(belanese) i = 0 if n % 2 == 1: while i < n: if ((belanese[i] not in x) and (belanese[i+1] in x)): #print(belanese[i], belanese[i+1]) i+=2 if i == n -1: print("Yes") else: i+=1 elif n % 2 == 0: while i < n-1: #print(belanese[i], belanese[i+1]) i+=1 if i == n-1: print("No") ```

0

849

B

Tell Your World

PROGRAMMING

1,600

[ "brute force", "geometry" ]

null

Connect the countless points with lines, till we reach the faraway yonder. There are *n* points on a coordinate plane, the *i*-th of which being (*i*,<=*y**i*). Determine whether it's possible to draw two parallel and non-overlapping lines, such that every point in the set lies on exactly one of them, and each of them passes through at least one point in the set.

The first line of input contains a positive integer *n* (3<=≤<=*n*<=≤<=1<=000) — the number of points. The second line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (<=-<=109<=≤<=*y**i*<=≤<=109) — the vertical coordinates of each point.

Output "Yes" (without quotes) if it's possible to fulfill the requirements, and "No" otherwise. You can print each letter in any case (upper or lower).

[ "5\n7 5 8 6 9\n", "5\n-1 -2 0 0 -5\n", "5\n5 4 3 2 1\n", "5\n1000000000 0 0 0 0\n" ]

[ "Yes\n", "No\n", "No\n", "Yes\n" ]

In the first example, there are five points: (1, 7), (2, 5), (3, 8), (4, 6) and (5, 9). It's possible to draw a line that passes through points 1, 3, 5, and another one that passes through points 2, 4 and is parallel to the first one. In the second example, while it's possible to draw two lines that cover all points, they cannot be made parallel. In the third example, it's impossible to satisfy both requirements at the same time.

1,000

[ { "input": "5\n7 5 8 6 9", "output": "Yes" }, { "input": "5\n-1 -2 0 0 -5", "output": "No" }, { "input": "5\n5 4 3 2 1", "output": "No" }, { "input": "5\n1000000000 0 0 0 0", "output": "Yes" }, { "input": "5\n1000000000 1 0 -999999999 -1000000000", "output": "Yes" }, { "input": "3\n998 244 353", "output": "Yes" }, { "input": "3\n-1000000000 0 1000000000", "output": "No" }, { "input": "5\n-1 -1 -1 -1 1", "output": "Yes" }, { "input": "4\n-9763 530 3595 6660", "output": "Yes" }, { "input": "4\n-253090305 36298498 374072642 711846786", "output": "Yes" }, { "input": "5\n-186772848 -235864239 -191561068 -193955178 -243046569", "output": "Yes" }, { "input": "5\n-954618456 -522919664 -248330428 -130850748 300848044", "output": "Yes" }, { "input": "10\n4846 6705 2530 5757 5283 -944 -2102 -3260 -4418 2913", "output": "No" }, { "input": "10\n-6568 -5920 -5272 -4624 -2435 -635 -2680 -2032 -1384 6565", "output": "No" }, { "input": "20\n319410377 286827025 254243673 221660321 189076969 156493617 123910265 91326913 58743561 26160209 -6423143 -39006495 -71589847 -104173199 -136756551 -169339903 -201923255 -234506607 -267089959 -299673311", "output": "No" }, { "input": "20\n-975467170 758268840 -975467171 758268839 -975467172 758268838 -975467173 758268837 -975467174 758268836 -975467175 758268835 -975467176 758268834 -975467177 758268833 -975467178 758268832 -975467179 758268831", "output": "Yes" }, { "input": "4\n1 0 3 0", "output": "No" }, { "input": "4\n100 2 3 4", "output": "Yes" }, { "input": "5\n7 5 8 6 3", "output": "No" }, { "input": "3\n1000000000 1000000000 -1000000000", "output": "Yes" }, { "input": "4\n1 0 1 4", "output": "Yes" }, { "input": "7\n1 2 -1 0 1 6 7", "output": "Yes" }, { "input": "4\n0 0 4 0", "output": "Yes" }, { "input": "7\n0 0 2 3 4 5 5", "output": "Yes" }, { "input": "5\n7 5 8 6 8", "output": "No" }, { "input": "5\n1 2 9 4 5", "output": "Yes" }, { "input": "8\n1 12 3 14 5 16 7 8", "output": "Yes" }, { "input": "5\n1 6 7 4 9", "output": "Yes" }, { "input": "5\n2 1 0 1 2", "output": "No" }, { "input": "4\n0 0 1 3", "output": "Yes" }, { "input": "4\n100 50 50 10000000", "output": "No" }, { "input": "5\n1 2 3 3 3", "output": "No" }, { "input": "5\n1 2 6 10 17", "output": "Yes" }, { "input": "4\n1 3 4 4", "output": "Yes" }, { "input": "4\n100 50 50 1000000", "output": "No" }, { "input": "6\n1 2 4 5 7 9", "output": "No" }, { "input": "6\n0 0 1 2 3 4", "output": "Yes" }, { "input": "5\n7 5 9 10 8", "output": "Yes" }, { "input": "7\n1 2 2 1 2 2 1", "output": "Yes" }, { "input": "4\n2 2 4 5", "output": "Yes" }, { "input": "6\n1 2 1 3 4 5", "output": "No" }, { "input": "4\n1 3 3 6", "output": "No" }, { "input": "5\n1 2 -3 4 -1", "output": "Yes" } ]

1,504,275,779

2,879

Python 3

WRONG_ANSWER

TESTS

14

358

512,000

from collections import defaultdict n=int(input()) ip=list(map(int,input().split())) b=0 slopes=defaultdict(lambda: 0) for i in range(n): for j in range(i+1,n): slope=((ip[j]-ip[i])/(j-i)) slopes[slope] = slopes.get(slope, 0) + 1 #print(slopes) for slope in slopes: y=slopes[slope] if (2*n+4*y-n*n)>=0 and y!=((n*n-n)/2): if (n+(2*n+4*y-n*n)**0.5)%2==0 or (n-(2*n+4*y-n*n)**0.5)%2==0: print('Yes') b=1 break if b==0: print('No')

Title: Tell Your World Time Limit: None seconds Memory Limit: None megabytes Problem Description: Connect the countless points with lines, till we reach the faraway yonder. There are *n* points on a coordinate plane, the *i*-th of which being (*i*,<=*y**i*). Determine whether it's possible to draw two parallel and non-overlapping lines, such that every point in the set lies on exactly one of them, and each of them passes through at least one point in the set. Input Specification: The first line of input contains a positive integer *n* (3<=≤<=*n*<=≤<=1<=000) — the number of points. The second line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (<=-<=109<=≤<=*y**i*<=≤<=109) — the vertical coordinates of each point. Output Specification: Output "Yes" (without quotes) if it's possible to fulfill the requirements, and "No" otherwise. You can print each letter in any case (upper or lower). Demo Input: ['5\n7 5 8 6 9\n', '5\n-1 -2 0 0 -5\n', '5\n5 4 3 2 1\n', '5\n1000000000 0 0 0 0\n'] Demo Output: ['Yes\n', 'No\n', 'No\n', 'Yes\n'] Note: In the first example, there are five points: (1, 7), (2, 5), (3, 8), (4, 6) and (5, 9). It's possible to draw a line that passes through points 1, 3, 5, and another one that passes through points 2, 4 and is parallel to the first one. In the second example, while it's possible to draw two lines that cover all points, they cannot be made parallel. In the third example, it's impossible to satisfy both requirements at the same time.

```python from collections import defaultdict n=int(input()) ip=list(map(int,input().split())) b=0 slopes=defaultdict(lambda: 0) for i in range(n): for j in range(i+1,n): slope=((ip[j]-ip[i])/(j-i)) slopes[slope] = slopes.get(slope, 0) + 1 #print(slopes) for slope in slopes: y=slopes[slope] if (2*n+4*y-n*n)>=0 and y!=((n*n-n)/2): if (n+(2*n+4*y-n*n)**0.5)%2==0 or (n-(2*n+4*y-n*n)**0.5)%2==0: print('Yes') b=1 break if b==0: print('No') ```

0

883

M

Quadcopter Competition

PROGRAMMING

1,100

[ "greedy", "math" ]

null

Polycarp takes part in a quadcopter competition. According to the rules a flying robot should: - start the race from some point of a field, - go around the flag, - close cycle returning back to the starting point. Polycarp knows the coordinates of the starting point (*x*1,<=*y*1) and the coordinates of the point where the flag is situated (*x*2,<=*y*2). Polycarp’s quadcopter can fly only parallel to the sides of the field each tick changing exactly one coordinate by 1. It means that in one tick the quadcopter can fly from the point (*x*,<=*y*) to any of four points: (*x*<=-<=1,<=*y*), (*x*<=+<=1,<=*y*), (*x*,<=*y*<=-<=1) or (*x*,<=*y*<=+<=1). Thus the quadcopter path is a closed cycle starting and finishing in (*x*1,<=*y*1) and containing the point (*x*2,<=*y*2) strictly inside. What is the minimal length of the quadcopter path?

The first line contains two integer numbers *x*1 and *y*1 (<=-<=100<=≤<=*x*1,<=*y*1<=≤<=100) — coordinates of the quadcopter starting (and finishing) point. The second line contains two integer numbers *x*2 and *y*2 (<=-<=100<=≤<=*x*2,<=*y*2<=≤<=100) — coordinates of the flag. It is guaranteed that the quadcopter starting point and the flag do not coincide.

Print the length of minimal path of the quadcopter to surround the flag and return back.

[ "1 5\n5 2\n", "0 1\n0 0\n" ]

[ "18\n", "8\n" ]

none

0

[ { "input": "1 5\n5 2", "output": "18" }, { "input": "0 1\n0 0", "output": "8" }, { "input": "-100 -100\n100 100", "output": "804" }, { "input": "-100 -100\n-100 100", "output": "406" }, { "input": "-100 -100\n100 -100", "output": "406" }, { "input": "100 -100\n-100 -100", "output": "406" }, { "input": "100 -100\n-100 100", "output": "804" }, { "input": "100 -100\n100 100", "output": "406" }, { "input": "-100 100\n-100 -100", "output": "406" }, { "input": "-100 100\n100 -100", "output": "804" }, { "input": "-100 100\n100 100", "output": "406" }, { "input": "100 100\n-100 -100", "output": "804" }, { "input": "100 100\n-100 100", "output": "406" }, { "input": "100 100\n100 -100", "output": "406" }, { "input": "45 -43\n45 -44", "output": "8" }, { "input": "76 76\n75 75", "output": "8" }, { "input": "-34 -56\n-35 -56", "output": "8" }, { "input": "56 -7\n55 -6", "output": "8" }, { "input": "43 -11\n43 -10", "output": "8" }, { "input": "1 -3\n2 -2", "output": "8" }, { "input": "55 71\n56 71", "output": "8" }, { "input": "54 -87\n55 -88", "output": "8" }, { "input": "22 98\n100 33", "output": "290" }, { "input": "37 84\n-83 5", "output": "402" }, { "input": "52 74\n-73 -39", "output": "480" }, { "input": "66 51\n51 -71", "output": "278" }, { "input": "-31 44\n73 86", "output": "296" }, { "input": "-20 34\n-9 55", "output": "68" }, { "input": "-5 19\n-91 -86", "output": "386" }, { "input": "-82 5\n28 -17", "output": "268" }, { "input": "-90 -100\n55 48", "output": "590" }, { "input": "-75 -14\n-32 8", "output": "134" }, { "input": "-53 -28\n-13 -28", "output": "86" }, { "input": "-42 -46\n10 -64", "output": "144" }, { "input": "55 -42\n25 2", "output": "152" }, { "input": "70 -64\n-54 70", "output": "520" }, { "input": "93 -78\n-32 -75", "output": "260" }, { "input": "8 -93\n79 -6", "output": "320" }, { "input": "50 43\n54 10", "output": "78" }, { "input": "65 32\n-37 71", "output": "286" }, { "input": "80 18\n-15 -58", "output": "346" }, { "input": "94 92\n4 -1", "output": "370" }, { "input": "-10 96\n27 64", "output": "142" }, { "input": "-96 78\n-56 32", "output": "176" }, { "input": "-81 64\n-37 -8", "output": "236" }, { "input": "-58 49\n74 -40", "output": "446" }, { "input": "-62 -55\n1 18", "output": "276" }, { "input": "-51 -69\n-78 86", "output": "368" }, { "input": "-29 -80\n-56 -47", "output": "124" }, { "input": "-14 -94\n55 -90", "output": "150" }, { "input": "83 -2\n82 83", "output": "176" }, { "input": "98 -16\n-96 40", "output": "504" }, { "input": "17 -34\n-86 -93", "output": "328" }, { "input": "32 -48\n33 -37", "output": "28" }, { "input": "74 87\n3 92", "output": "156" }, { "input": "89 73\n-80 49", "output": "390" }, { "input": "4 58\n-61 -80", "output": "410" }, { "input": "15 48\n50 -20", "output": "210" }, { "input": "-82 45\n81 46", "output": "332" }, { "input": "-68 26\n-2 6", "output": "176" }, { "input": "-53 4\n-92 -31", "output": "152" }, { "input": "-30 94\n31 -58", "output": "430" }, { "input": "-38 -11\n58 99", "output": "416" }, { "input": "-27 -25\n-28 68", "output": "192" }, { "input": "-5 -39\n-10 -77", "output": "90" }, { "input": "-90 -54\n9 -9", "output": "292" }, { "input": "7 -57\n28 61", "output": "282" }, { "input": "18 -67\n-51 21", "output": "318" }, { "input": "41 -82\n-33 -15", "output": "286" }, { "input": "56 -8\n91 -55", "output": "168" }, { "input": "-23 -13\n-24 -12", "output": "8" }, { "input": "1 32\n1 33", "output": "8" }, { "input": "25 76\n24 76", "output": "8" }, { "input": "-29 -78\n-28 -79", "output": "8" }, { "input": "-77 19\n-76 19", "output": "8" }, { "input": "-53 63\n-53 62", "output": "8" }, { "input": "86 12\n86 11", "output": "8" }, { "input": "14 56\n13 56", "output": "8" }, { "input": "63 41\n62 40", "output": "8" }, { "input": "0 -4\n1 -4", "output": "8" }, { "input": "24 41\n24 42", "output": "8" }, { "input": "48 85\n49 86", "output": "8" }, { "input": "0 0\n0 1", "output": "8" }, { "input": "0 0\n1 0", "output": "8" }, { "input": "0 0\n1 1", "output": "8" }, { "input": "0 1\n0 0", "output": "8" }, { "input": "0 1\n1 0", "output": "8" }, { "input": "0 1\n1 1", "output": "8" }, { "input": "1 0\n0 0", "output": "8" }, { "input": "1 0\n0 1", "output": "8" }, { "input": "1 0\n1 1", "output": "8" }, { "input": "1 1\n0 0", "output": "8" }, { "input": "1 1\n0 1", "output": "8" }, { "input": "1 1\n1 0", "output": "8" }, { "input": "100 100\n99 -100", "output": "406" }, { "input": "100 100\n-100 99", "output": "406" }, { "input": "-100 -100\n-99 100", "output": "406" }, { "input": "-100 -100\n100 -99", "output": "406" }, { "input": "0 0\n1 2", "output": "10" }, { "input": "0 0\n2 1", "output": "10" } ]

1,594,891,387

2,147,483,647

PyPy 3

OK

TESTS

100

140

20,172,800

x1,y1 = map(int,input().split()) x2,y2= map(int,input().split()) a = abs(x1-x2)+1 b = abs(y1-y2)+1 if a==1: a+=1 if b==1: b+=1 print((a+b)*2)

Title: Quadcopter Competition Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp takes part in a quadcopter competition. According to the rules a flying robot should: - start the race from some point of a field, - go around the flag, - close cycle returning back to the starting point. Polycarp knows the coordinates of the starting point (*x*1,<=*y*1) and the coordinates of the point where the flag is situated (*x*2,<=*y*2). Polycarp’s quadcopter can fly only parallel to the sides of the field each tick changing exactly one coordinate by 1. It means that in one tick the quadcopter can fly from the point (*x*,<=*y*) to any of four points: (*x*<=-<=1,<=*y*), (*x*<=+<=1,<=*y*), (*x*,<=*y*<=-<=1) or (*x*,<=*y*<=+<=1). Thus the quadcopter path is a closed cycle starting and finishing in (*x*1,<=*y*1) and containing the point (*x*2,<=*y*2) strictly inside. What is the minimal length of the quadcopter path? Input Specification: The first line contains two integer numbers *x*1 and *y*1 (<=-<=100<=≤<=*x*1,<=*y*1<=≤<=100) — coordinates of the quadcopter starting (and finishing) point. The second line contains two integer numbers *x*2 and *y*2 (<=-<=100<=≤<=*x*2,<=*y*2<=≤<=100) — coordinates of the flag. It is guaranteed that the quadcopter starting point and the flag do not coincide. Output Specification: Print the length of minimal path of the quadcopter to surround the flag and return back. Demo Input: ['1 5\n5 2\n', '0 1\n0 0\n'] Demo Output: ['18\n', '8\n'] Note: none

```python x1,y1 = map(int,input().split()) x2,y2= map(int,input().split()) a = abs(x1-x2)+1 b = abs(y1-y2)+1 if a==1: a+=1 if b==1: b+=1 print((a+b)*2) ```

3

445

A

DZY Loves Chessboard

PROGRAMMING

1,200

[ "dfs and similar", "implementation" ]

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" } ]

1,672,074,496

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

30

0

#https://codeforces.com/problemset/problem/445/A #creating a bipartite graph #starting with 'B' import heapq from collections import defaultdict s=set() #storig valid cells (good cells) n , m =map(int,input().split()) l=[] for i in range(n): x =list(input()) #print(x) l.append(x) #print(l) for i in range(n): for j in range(m): #print(i,j) #print(l[i][j]) if l[i][j]=='.': s.add((i,j)) di=[[0,1],[0,-1],[1,0],[-1,0]] vis=set() for i in range(n): for j in range(m): if l[i][j]=='.': q=[(i,j,'B')] l[i][j]='B' while(q): #print(l) x,y,col=q.pop(0) for dire in di: a=x+dire[0] b=y+dire[1] if a>=0 and a<n and b>=0 and b<m and l[a][b]=='.': if col=='B': l[a][b]='W' else: l[a][b]='B' q.append((a,b,l[a][b])) for i in range(n): for j in range(m): print(l[i][j],end=" ") print()

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 #https://codeforces.com/problemset/problem/445/A #creating a bipartite graph #starting with 'B' import heapq from collections import defaultdict s=set() #storig valid cells (good cells) n , m =map(int,input().split()) l=[] for i in range(n): x =list(input()) #print(x) l.append(x) #print(l) for i in range(n): for j in range(m): #print(i,j) #print(l[i][j]) if l[i][j]=='.': s.add((i,j)) di=[[0,1],[0,-1],[1,0],[-1,0]] vis=set() for i in range(n): for j in range(m): if l[i][j]=='.': q=[(i,j,'B')] l[i][j]='B' while(q): #print(l) x,y,col=q.pop(0) for dire in di: a=x+dire[0] b=y+dire[1] if a>=0 and a<n and b>=0 and b<m and l[a][b]=='.': if col=='B': l[a][b]='W' else: l[a][b]='B' q.append((a,b,l[a][b])) for i in range(n): for j in range(m): print(l[i][j],end=" ") print() ```

0

115

A

Party

PROGRAMMING

900

[ "dfs and similar", "graphs", "trees" ]

null

A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true: - Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*. The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager. Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*. What is the minimum number of groups that must be formed?

The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees. The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager. It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.

Print a single integer denoting the minimum number of groups that will be formed in the party.

[ "5\n-1\n1\n2\n1\n-1\n" ]

[ "3\n" ]

For the first example, three groups are sufficient, for example: - Employee 1 - Employees 2 and 4 - Employees 3 and 5

500

[ { "input": "5\n-1\n1\n2\n1\n-1", "output": "3" }, { "input": "4\n-1\n1\n2\n3", "output": "4" }, { "input": "12\n-1\n1\n2\n3\n-1\n5\n6\n7\n-1\n9\n10\n11", "output": "4" }, { "input": "6\n-1\n-1\n2\n3\n1\n1", "output": "3" }, { "input": "3\n-1\n1\n1", "output": "2" }, { "input": "1\n-1", "output": "1" }, { "input": "2\n2\n-1", "output": "2" }, { "input": "2\n-1\n-1", "output": "1" }, { "input": "3\n2\n-1\n1", "output": "3" }, { "input": "3\n-1\n-1\n-1", "output": "1" }, { "input": "5\n4\n5\n1\n-1\n4", "output": "3" }, { "input": "12\n-1\n1\n1\n1\n1\n1\n3\n4\n3\n3\n4\n7", "output": "4" }, { "input": "12\n-1\n-1\n1\n-1\n1\n1\n5\n11\n8\n6\n6\n4", "output": "5" }, { "input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n2\n-1\n-1\n-1", "output": "2" }, { "input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1", "output": "1" }, { "input": "12\n3\n4\n2\n8\n7\n1\n10\n12\n5\n-1\n9\n11", "output": "12" }, { "input": "12\n5\n6\n7\n1\n-1\n9\n12\n4\n8\n-1\n3\n2", "output": "11" }, { "input": "12\n-1\n9\n11\n6\n6\n-1\n6\n3\n8\n6\n1\n6", "output": "6" }, { "input": "12\n7\n8\n4\n12\n7\n9\n-1\n-1\n-1\n8\n6\n-1", "output": "3" }, { "input": "12\n-1\n10\n-1\n1\n-1\n5\n9\n12\n-1\n-1\n3\n-1", "output": "2" }, { "input": "12\n-1\n7\n9\n12\n1\n7\n-1\n-1\n8\n5\n4\n-1", "output": "3" }, { "input": "12\n11\n11\n8\n9\n1\n1\n2\n-1\n10\n3\n-1\n8", "output": "5" }, { "input": "12\n-1\n8\n9\n-1\n4\n2\n11\n1\n-1\n6\n-1\n10", "output": "6" }, { "input": "12\n7\n4\n4\n-1\n6\n7\n9\n7\n-1\n7\n4\n1", "output": "4" }, { "input": "12\n-1\n6\n10\n6\n1\n11\n5\n9\n11\n5\n1\n9", "output": "4" }, { "input": "12\n8\n6\n6\n8\n6\n-1\n6\n6\n8\n6\n8\n8", "output": "3" }, { "input": "12\n3\n8\n9\n12\n2\n8\n11\n12\n2\n9\n-1\n11", "output": "7" }, { "input": "7\n7\n6\n-1\n-1\n4\n4\n-1", "output": "3" }, { "input": "77\n53\n-1\n48\n-1\n51\n-1\n54\n43\n70\n19\n64\n31\n45\n58\n6\n61\n41\n56\n34\n-1\n54\n-1\n66\n65\n74\n16\n38\n1\n2\n61\n-1\n34\n-1\n-1\n7\n19\n57\n13\n61\n42\n-1\n34\n65\n16\n31\n61\n6\n5\n34\n15\n-1\n13\n4\n22\n74\n7\n-1\n70\n38\n61\n45\n53\n46\n22\n-1\n65\n34\n19\n43\n48\n33\n58\n36\n-1\n47\n24\n34", "output": "6" }, { "input": "5\n2\n3\n4\n5\n-1", "output": "5" } ]

1,689,769,824

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

6

92

204,800

# 5 # # # -1 # 1 # 2 # 1 # -1 # adj[-1] = # id = 1 # boss = -1 # adj[1] = [] # adj[1] # 1 - no boss # 2 - 1 is a manger # 3 - 2 is a manager # 4 - 1 is a manger # 5 - no boss # # // put each root of each componenyt # // each level can be in the same group: 2 and 4 # // 3 and # // find depth of each node # [1: 1, 5:1, 2 # # # 1 5 # / \ # 2 4 # / # 3 n = int(input()) adj = {} roots = [] result = 0 #glabalVar = 0 def buildTree(n): for i in range(n): eId = i + 1 boss = int(input()) if boss == -1: adj[eId] = [] roots.append(eId) else: if boss in adj: adj[boss].append(eId) else: adj[boss] = [eId] def dfs(root, depth): global result result = max(result, depth) # temp = glabalVar #glabalVar = temp #result1 = temp if not root in adj or adj[root] == []: return for child in adj[root]: dfs(child, depth + 1) buildTree(n) resultTask = 0 for root in roots: result = 0 dfs(root, 0) resultTask = max(resultTask, result) if resultTask == 0: print(resultTask) else: print(resultTask + 1) # print(roots) # print(adj) # print(depth) # if depth: # print(max(list(depth.keys())) + 1) # else: # print(1)

Title: Party Time Limit: None seconds Memory Limit: None megabytes Problem Description: A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true: - Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*. The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager. Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*. What is the minimum number of groups that must be formed? Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees. The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager. It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles. Output Specification: Print a single integer denoting the minimum number of groups that will be formed in the party. Demo Input: ['5\n-1\n1\n2\n1\n-1\n'] Demo Output: ['3\n'] Note: For the first example, three groups are sufficient, for example: - Employee 1 - Employees 2 and 4 - Employees 3 and 5

```python # 5 # # # -1 # 1 # 2 # 1 # -1 # adj[-1] = # id = 1 # boss = -1 # adj[1] = [] # adj[1] # 1 - no boss # 2 - 1 is a manger # 3 - 2 is a manager # 4 - 1 is a manger # 5 - no boss # # // put each root of each componenyt # // each level can be in the same group: 2 and 4 # // 3 and # // find depth of each node # [1: 1, 5:1, 2 # # # 1 5 # / \ # 2 4 # / # 3 n = int(input()) adj = {} roots = [] result = 0 #glabalVar = 0 def buildTree(n): for i in range(n): eId = i + 1 boss = int(input()) if boss == -1: adj[eId] = [] roots.append(eId) else: if boss in adj: adj[boss].append(eId) else: adj[boss] = [eId] def dfs(root, depth): global result result = max(result, depth) # temp = glabalVar #glabalVar = temp #result1 = temp if not root in adj or adj[root] == []: return for child in adj[root]: dfs(child, depth + 1) buildTree(n) resultTask = 0 for root in roots: result = 0 dfs(root, 0) resultTask = max(resultTask, result) if resultTask == 0: print(resultTask) else: print(resultTask + 1) # print(roots) # print(adj) # print(depth) # if depth: # print(max(list(depth.keys())) + 1) # else: # print(1) ```

0

718

C

Sasha and Array

PROGRAMMING

2,300

[ "data structures", "math", "matrices" ]

null

Sasha has an array of integers *a*1,<=*a*2,<=...,<=*a**n*. You have to perform *m* queries. There might be queries of two types: 1. 1 l r x — increase all integers on the segment from *l* to *r* by values *x*; 1. 2 l r — find , where *f*(*x*) is the *x*-th Fibonacci number. As this number may be large, you only have to find it modulo 109<=+<=7. In this problem we define Fibonacci numbers as follows: *f*(1)<==<=1, *f*(2)<==<=1, *f*(*x*)<==<=*f*(*x*<=-<=1)<=+<=*f*(*x*<=-<=2) for all *x*<=><=2. Sasha is a very talented boy and he managed to perform all queries in five seconds. Will you be able to write the program that performs as well as Sasha?

The first line of the input contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000) — the number of elements in the array and the number of queries respectively. The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). Then follow *m* lines with queries descriptions. Each of them contains integers *tp**i*, *l**i*, *r**i* and may be *x**i* (1<=≤<=*tp**i*<=≤<=2, 1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*, 1<=≤<=*x**i*<=≤<=109). Here *tp**i*<==<=1 corresponds to the queries of the first type and *tp**i* corresponds to the queries of the second type. It's guaranteed that the input will contains at least one query of the second type.

For each query of the second type print the answer modulo 109<=+<=7.

[ "5 4\n1 1 2 1 1\n2 1 5\n1 2 4 2\n2 2 4\n2 1 5\n" ]

[ "5\n7\n9\n" ]

Initially, array *a* is equal to 1, 1, 2, 1, 1. The answer for the first query of the second type is *f*(1) + *f*(1) + *f*(2) + *f*(1) + *f*(1) = 1 + 1 + 1 + 1 + 1 = 5. After the query 1 2 4 2 array *a* is equal to 1, 3, 4, 3, 1. The answer for the second query of the second type is *f*(3) + *f*(4) + *f*(3) = 2 + 3 + 2 = 7. The answer for the third query of the second type is *f*(1) + *f*(3) + *f*(4) + *f*(3) + *f*(1) = 1 + 2 + 3 + 2 + 1 = 9.

1,250

[ { "input": "5 4\n1 1 2 1 1\n2 1 5\n1 2 4 2\n2 2 4\n2 1 5", "output": "5\n7\n9" }, { "input": "2 3\n1 3\n2 1 1\n1 1 2 3\n1 1 2 2", "output": "1" }, { "input": "7 4\n2 2 1 1 3 3 2\n2 1 5\n2 6 7\n1 3 4 3\n2 6 6", "output": "6\n3\n2" }, { "input": "9 4\n2 1 2 3 3 3 2 1 3\n2 1 8\n1 7 7 3\n1 1 3 1\n1 3 5 2", "output": "11" }, { "input": "18 19\n2 2 2 2 1 1 1 1 1 2 1 2 2 1 2 1 2 2\n2 4 14\n2 16 16\n2 5 8\n1 1 6 2\n2 14 14\n2 1 3\n2 4 5\n2 11 13\n2 18 18\n1 8 15 2\n2 17 18\n2 14 16\n1 3 10 2\n2 1 9\n1 3 18 1\n1 17 18 1\n2 17 17\n1 15 15 1\n1 3 18 2", "output": "11\n1\n4\n1\n9\n5\n3\n1\n2\n6\n44\n3" }, { "input": "18 31\n1 1 2 2 2 1 2 1 2 2 2 2 1 1 1 2 2 1\n1 4 7 2\n1 9 10 1\n1 2 4 2\n1 15 18 2\n1 3 10 2\n1 7 8 1\n1 13 18 2\n1 15 17 2\n1 16 17 1\n1 16 16 2\n1 10 12 1\n1 4 13 2\n1 3 8 1\n1 7 9 1\n1 13 18 2\n1 3 14 2\n1 2 2 1\n1 14 17 1\n1 9 18 1\n1 7 17 1\n1 2 5 1\n1 8 13 2\n1 12 18 1\n2 13 14\n2 15 18\n1 14 14 2\n1 8 14 2\n1 3 11 2\n1 17 17 2\n1 7 10 2\n2 3 12", "output": "466\n2474\n22934" }, { "input": "17 30\n8 10 7 8 7 9 10 6 7 3 7 2 8 8 10 9 8\n1 16 16 5\n2 13 13\n1 16 17 4\n1 1 17 2\n2 17 17\n1 2 10 10\n1 4 8 3\n1 17 17 4\n1 13 15 6\n1 10 16 2\n2 13 14\n2 12 14\n1 6 11 5\n2 4 14\n1 12 16 7\n1 8 8 4\n1 1 16 9\n1 17 17 6\n1 7 8 7\n1 7 10 1\n1 11 14 6\n1 6 10 8\n1 9 14 8\n1 16 16 9\n1 6 13 10\n1 9 13 10\n2 8 8\n1 5 12 6\n1 8 11 10\n1 6 15 7", "output": "21\n377\n5168\n5176\n1584272\n680057396" }, { "input": "1 1\n1\n2 1 1", "output": "1" }, { "input": "1 2\n1000000000\n1 1 1 1000000000\n2 1 1", "output": "999999020" } ]

1,600,684,360

2,147,483,647

PyPy 3

TIME_LIMIT_EXCEEDED

TESTS

9

5,000

70,553,600

from math import inf, log2 class LazySegmentTree: def __init__(self, array): self.n = len(array) self.mod = 10**9 + 7 self.size = 2 ** (int(log2(self.n - 1)) + 1) if self.n != 1 else 1 self.func = lambda a,b:[(a[i] + b[i]) % self.mod for i in range(len(a))] self.func2 = lambda a, b: [(a[0] * b[0] + a[1] * b[2]) % self.mod, (a[0] * b[1] + a[1] * b[3]) % self.mod, (a[2] * b[0] + a[3] * b[2]) % self.mod, (a[2] * b[1] + a[3] * b[3]) % self.mod] self.func3 = lambda a,b: [(a[0]*b[0] + a[1]*b[2]) % self.mod, (a[0]*b[2] + a[1]*b[3]) % self.mod] self.default = [0,1] self.data = [self.default for _ in range(2 * self.size)] self.lazy = [0] * (2 * self.size) self.process(array) def nth(self, n): m = [0, 1, 1, 1] b = list(bin(n)[2:]) ans = [1, 0, 0, 1] while b: x = int(b.pop()) if x == 1: ans = self.func2(ans, m) m = self.func2(m, m) return ans def process(self, array): for i in range(self.n): ans = self.nth(array[i]) self.data[self.size+i] = self.func3(self.data[self.size+i], ans) for i in range(self.size - 1, -1, -1): self.data[i] = self.func(self.data[2 * i], self.data[2 * i + 1]) def push(self, index): """Push the information of the root to it's children!""" if not self.lazy[index]:return self.lazy[2 * index] += self.lazy[index] self.lazy[2 * index + 1] += self.lazy[index] self.data[2 * index] = self.func3(self.data[2*index], self.nth(self.lazy[index])) self.data[2 * index + 1] = self.func3(self.data[2*index+1], self.nth(self.lazy[index])) self.lazy[index] = 0 def build(self, index): """Build data with the new changes!""" index >>= 1 while index: self.data[index] = self.func3(self.func(self.data[2 * index], self.data[2 * index + 1]), self.nth(self.lazy[index])) index >>= 1 def query(self, alpha, omega): """Returns the result of function over the range (inclusive)!""" res = [0, 0] alpha += self.size omega += self.size + 1 for i in range(len(bin(alpha)[2:]) - 1, 0, -1): self.push(alpha >> i) for i in range(len(bin(omega - 1)[2:]) - 1, 0, -1): self.push((omega - 1) >> i) while alpha < omega: if alpha & 1: res = self.func(res, self.data[alpha]) alpha += 1 if omega & 1: omega -= 1 res = self.func(res, self.data[omega]) alpha >>= 1 omega >>= 1 return res def update(self, alpha, omega, value): """Increases all elements in the range (inclusive) by given value!""" alpha += self.size omega += self.size + 1 l, r = alpha, omega ans = self.nth(value) while alpha < omega: if alpha & 1: self.data[alpha] = self.func3(self.data[alpha], ans) self.lazy[alpha] += value alpha += 1 if omega & 1: omega -= 1 self.data[omega] = self.func3(self.data[omega], ans) self.lazy[omega] += value alpha >>= 1 omega >>= 1 self.build(l) self.build(r - 1) # ------------------- fast io -------------------- import os import sys from io import BytesIO, IOBase BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") # ------------------- fast io -------------------- from math import gcd, ceil def pre(s): n = len(s) pi = [0] * n for i in range(1, n): j = pi[i - 1] while j and s[i] != s[j]: j = pi[j - 1] if s[i] == s[j]: j += 1 pi[i] = j return pi def prod(a): ans = 1 for each in a: ans = (ans * each) return ans def lcm(a, b): return a * b // gcd(a, b) def binary(x, length=16): y = bin(x)[2:] return y if len(y) >= length else "0" * (length - len(y)) + y n, m = map(int, input().split()) a = list(map(int, input().split())) st = LazySegmentTree(a) for _ in range(m): a = list(map(int, input().split())) if len(a) == 4: st.update(a[1] - 1, a[2]-1, a[3]) else: print(st.query(a[1] - 1, a[2] - 1)[0])

Title: Sasha and Array Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sasha has an array of integers *a*1,<=*a*2,<=...,<=*a**n*. You have to perform *m* queries. There might be queries of two types: 1. 1 l r x — increase all integers on the segment from *l* to *r* by values *x*; 1. 2 l r — find , where *f*(*x*) is the *x*-th Fibonacci number. As this number may be large, you only have to find it modulo 109<=+<=7. In this problem we define Fibonacci numbers as follows: *f*(1)<==<=1, *f*(2)<==<=1, *f*(*x*)<==<=*f*(*x*<=-<=1)<=+<=*f*(*x*<=-<=2) for all *x*<=><=2. Sasha is a very talented boy and he managed to perform all queries in five seconds. Will you be able to write the program that performs as well as Sasha? Input Specification: The first line of the input contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000) — the number of elements in the array and the number of queries respectively. The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). Then follow *m* lines with queries descriptions. Each of them contains integers *tp**i*, *l**i*, *r**i* and may be *x**i* (1<=≤<=*tp**i*<=≤<=2, 1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*, 1<=≤<=*x**i*<=≤<=109). Here *tp**i*<==<=1 corresponds to the queries of the first type and *tp**i* corresponds to the queries of the second type. It's guaranteed that the input will contains at least one query of the second type. Output Specification: For each query of the second type print the answer modulo 109<=+<=7. Demo Input: ['5 4\n1 1 2 1 1\n2 1 5\n1 2 4 2\n2 2 4\n2 1 5\n'] Demo Output: ['5\n7\n9\n'] Note: Initially, array *a* is equal to 1, 1, 2, 1, 1. The answer for the first query of the second type is *f*(1) + *f*(1) + *f*(2) + *f*(1) + *f*(1) = 1 + 1 + 1 + 1 + 1 = 5. After the query 1 2 4 2 array *a* is equal to 1, 3, 4, 3, 1. The answer for the second query of the second type is *f*(3) + *f*(4) + *f*(3) = 2 + 3 + 2 = 7. The answer for the third query of the second type is *f*(1) + *f*(3) + *f*(4) + *f*(3) + *f*(1) = 1 + 2 + 3 + 2 + 1 = 9.

```python from math import inf, log2 class LazySegmentTree: def __init__(self, array): self.n = len(array) self.mod = 10**9 + 7 self.size = 2 ** (int(log2(self.n - 1)) + 1) if self.n != 1 else 1 self.func = lambda a,b:[(a[i] + b[i]) % self.mod for i in range(len(a))] self.func2 = lambda a, b: [(a[0] * b[0] + a[1] * b[2]) % self.mod, (a[0] * b[1] + a[1] * b[3]) % self.mod, (a[2] * b[0] + a[3] * b[2]) % self.mod, (a[2] * b[1] + a[3] * b[3]) % self.mod] self.func3 = lambda a,b: [(a[0]*b[0] + a[1]*b[2]) % self.mod, (a[0]*b[2] + a[1]*b[3]) % self.mod] self.default = [0,1] self.data = [self.default for _ in range(2 * self.size)] self.lazy = [0] * (2 * self.size) self.process(array) def nth(self, n): m = [0, 1, 1, 1] b = list(bin(n)[2:]) ans = [1, 0, 0, 1] while b: x = int(b.pop()) if x == 1: ans = self.func2(ans, m) m = self.func2(m, m) return ans def process(self, array): for i in range(self.n): ans = self.nth(array[i]) self.data[self.size+i] = self.func3(self.data[self.size+i], ans) for i in range(self.size - 1, -1, -1): self.data[i] = self.func(self.data[2 * i], self.data[2 * i + 1]) def push(self, index): """Push the information of the root to it's children!""" if not self.lazy[index]:return self.lazy[2 * index] += self.lazy[index] self.lazy[2 * index + 1] += self.lazy[index] self.data[2 * index] = self.func3(self.data[2*index], self.nth(self.lazy[index])) self.data[2 * index + 1] = self.func3(self.data[2*index+1], self.nth(self.lazy[index])) self.lazy[index] = 0 def build(self, index): """Build data with the new changes!""" index >>= 1 while index: self.data[index] = self.func3(self.func(self.data[2 * index], self.data[2 * index + 1]), self.nth(self.lazy[index])) index >>= 1 def query(self, alpha, omega): """Returns the result of function over the range (inclusive)!""" res = [0, 0] alpha += self.size omega += self.size + 1 for i in range(len(bin(alpha)[2:]) - 1, 0, -1): self.push(alpha >> i) for i in range(len(bin(omega - 1)[2:]) - 1, 0, -1): self.push((omega - 1) >> i) while alpha < omega: if alpha & 1: res = self.func(res, self.data[alpha]) alpha += 1 if omega & 1: omega -= 1 res = self.func(res, self.data[omega]) alpha >>= 1 omega >>= 1 return res def update(self, alpha, omega, value): """Increases all elements in the range (inclusive) by given value!""" alpha += self.size omega += self.size + 1 l, r = alpha, omega ans = self.nth(value) while alpha < omega: if alpha & 1: self.data[alpha] = self.func3(self.data[alpha], ans) self.lazy[alpha] += value alpha += 1 if omega & 1: omega -= 1 self.data[omega] = self.func3(self.data[omega], ans) self.lazy[omega] += value alpha >>= 1 omega >>= 1 self.build(l) self.build(r - 1) # ------------------- fast io -------------------- import os import sys from io import BytesIO, IOBase BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") # ------------------- fast io -------------------- from math import gcd, ceil def pre(s): n = len(s) pi = [0] * n for i in range(1, n): j = pi[i - 1] while j and s[i] != s[j]: j = pi[j - 1] if s[i] == s[j]: j += 1 pi[i] = j return pi def prod(a): ans = 1 for each in a: ans = (ans * each) return ans def lcm(a, b): return a * b // gcd(a, b) def binary(x, length=16): y = bin(x)[2:] return y if len(y) >= length else "0" * (length - len(y)) + y n, m = map(int, input().split()) a = list(map(int, input().split())) st = LazySegmentTree(a) for _ in range(m): a = list(map(int, input().split())) if len(a) == 4: st.update(a[1] - 1, a[2]-1, a[3]) else: print(st.query(a[1] - 1, a[2] - 1)[0]) ```

0

1,005

A

Tanya and Stairways

PROGRAMMING

800

[ "implementation" ]

null

Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$. You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway. The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.

The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya. The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order. The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.

In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways.

[ "7\n1 2 3 1 2 3 4\n", "4\n1 1 1 1\n", "5\n1 2 3 4 5\n", "5\n1 2 1 2 1\n" ]

[ "2\n3 4 ", "4\n1 1 1 1 ", "1\n5 ", "3\n2 2 1 " ]

none

0

[ { "input": "7\n1 2 3 1 2 3 4", "output": "2\n3 4 " }, { "input": "4\n1 1 1 1", "output": "4\n1 1 1 1 " }, { "input": "5\n1 2 3 4 5", "output": "1\n5 " }, { "input": "5\n1 2 1 2 1", "output": "3\n2 2 1 " }, { "input": "1\n1", "output": "1\n1 " }, { "input": "48\n1 2 3 4 1 2 3 1 1 2 3 1 2 3 4 1 1 2 3 4 1 2 3 4 1 2 3 4 1 1 2 1 2 1 2 1 1 2 1 2 1 2 3 1 2 1 2 1", "output": "20\n4 3 1 3 4 1 4 4 4 1 2 2 2 1 2 2 3 2 2 1 " }, { "input": "2\n1 2", "output": "1\n2 " }, { "input": "3\n1 1 2", "output": "2\n1 2 " }, { "input": "4\n1 1 2 3", "output": "2\n1 3 " }, { "input": "8\n1 2 3 1 2 3 4 5", "output": "2\n3 5 " }, { "input": "5\n1 1 1 2 3", "output": "3\n1 1 3 " } ]

1,671,337,679

2,147,483,647

PyPy 3-64

OK

TESTS

16

61

1,638,400

from itertools import groupby input() ans = [ sum(1 for _ in v) for _, v in groupby(x - i for i, x in enumerate(map(int, input().split()))) ] print(len(ans)) print(*ans)

Title: Tanya and Stairways Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$. You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway. The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways. Input Specification: The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya. The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order. The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways. Output Specification: In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways. Demo Input: ['7\n1 2 3 1 2 3 4\n', '4\n1 1 1 1\n', '5\n1 2 3 4 5\n', '5\n1 2 1 2 1\n'] Demo Output: ['2\n3 4 ', '4\n1 1 1 1 ', '1\n5 ', '3\n2 2 1 '] Note: none

```python from itertools import groupby input() ans = [ sum(1 for _ in v) for _, v in groupby(x - i for i, x in enumerate(map(int, input().split()))) ] print(len(ans)) print(*ans) ```

3

433

B

Kuriyama Mirai's Stones

PROGRAMMING

1,200

[ "dp", "implementation", "sortings" ]

null

Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions: 1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones. The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.

Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.

[ "6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n", "4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n" ]

[ "24\n9\n28\n", "10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n" ]

Please note that the answers to the questions may overflow 32-bit integer type.

1,500

[ { "input": "6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6", "output": "24\n9\n28" }, { "input": "4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2", "output": "10\n15\n5\n15\n5\n5\n2\n12\n3\n5" }, { "input": "4\n2 2 3 6\n9\n2 2 3\n1 1 3\n2 2 3\n2 2 3\n2 2 2\n1 1 3\n1 1 3\n2 1 4\n1 1 2", "output": "5\n7\n5\n5\n2\n7\n7\n13\n4" }, { "input": "18\n26 46 56 18 78 88 86 93 13 77 21 84 59 61 5 74 72 52\n25\n1 10 10\n1 9 13\n2 13 17\n1 8 14\n2 2 6\n1 12 16\n2 15 17\n2 3 6\n1 3 13\n2 8 9\n2 17 17\n1 17 17\n2 5 10\n2 1 18\n1 4 16\n1 1 13\n1 1 8\n2 7 11\n2 6 12\n1 5 9\n1 4 5\n2 7 15\n1 8 8\n1 8 14\n1 3 7", "output": "77\n254\n413\n408\n124\n283\n258\n111\n673\n115\n88\n72\n300\n1009\n757\n745\n491\n300\n420\n358\n96\n613\n93\n408\n326" }, { "input": "56\n43 100 44 66 65 11 26 75 96 77 5 15 75 96 11 44 11 97 75 53 33 26 32 33 90 26 68 72 5 44 53 26 33 88 68 25 84 21 25 92 1 84 21 66 94 35 76 51 11 95 67 4 61 3 34 18\n27\n1 20 38\n1 11 46\n2 42 53\n1 8 11\n2 11 42\n2 35 39\n2 37 41\n1 48 51\n1 32 51\n1 36 40\n1 31 56\n1 18 38\n2 9 51\n1 7 48\n1 15 52\n1 27 31\n2 5 19\n2 35 50\n1 31 34\n1 2 7\n2 15 33\n2 46 47\n1 26 28\n2 3 29\n1 23 45\n2 29 55\n1 14 29", "output": "880\n1727\n1026\n253\n1429\n335\n350\n224\n1063\n247\n1236\n1052\n2215\n2128\n1840\n242\n278\n1223\n200\n312\n722\n168\n166\n662\n1151\n2028\n772" }, { "input": "18\n38 93 48 14 69 85 26 47 71 11 57 9 38 65 72 78 52 47\n38\n2 10 12\n1 6 18\n2 2 2\n1 3 15\n2 1 16\n2 5 13\n1 9 17\n1 2 15\n2 5 17\n1 15 15\n2 4 11\n2 3 4\n2 2 5\n2 1 17\n2 6 16\n2 8 16\n2 8 14\n1 9 12\n2 8 13\n2 1 14\n2 5 13\n1 2 3\n1 9 14\n2 12 15\n2 3 3\n2 9 13\n2 4 12\n2 11 14\n2 6 16\n1 8 14\n1 12 15\n2 3 4\n1 3 5\n2 4 14\n1 6 6\n2 7 14\n2 7 18\n1 8 12", "output": "174\n658\n11\n612\n742\n461\n453\n705\n767\n72\n353\n40\n89\n827\n644\n559\n409\n148\n338\n592\n461\n141\n251\n277\n14\n291\n418\n262\n644\n298\n184\n40\n131\n558\n85\n456\n784\n195" }, { "input": "1\n2\n10\n1 1 1\n1 1 1\n2 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n2 1 1\n1 1 1\n1 1 1", "output": "2\n2\n2\n2\n2\n2\n2\n2\n2\n2" }, { "input": "2\n1 5\n8\n2 1 2\n1 1 1\n1 1 2\n1 1 1\n2 2 2\n2 1 2\n1 1 1\n1 2 2", "output": "6\n1\n6\n1\n5\n6\n1\n5" }, { "input": "8\n2 6 4 6 8 4 7 7\n9\n2 6 8\n1 2 3\n2 3 4\n2 1 2\n2 4 5\n1 3 8\n2 4 4\n2 2 4\n1 2 4", "output": "22\n10\n10\n6\n12\n36\n6\n14\n16" }, { "input": "4\n1 1 2 8\n9\n1 2 4\n1 2 4\n1 1 2\n2 3 4\n2 1 1\n1 2 3\n2 3 4\n1 1 4\n1 1 4", "output": "11\n11\n2\n10\n1\n3\n10\n12\n12" }, { "input": "5\n8 8 1 10 7\n5\n1 2 2\n1 1 4\n1 3 3\n1 2 2\n1 2 5", "output": "8\n27\n1\n8\n26" }, { "input": "10\n10 3 3 2 1 2 7 1 1 1\n4\n2 10 10\n2 6 6\n1 5 10\n1 6 8", "output": "10\n2\n13\n10" }, { "input": "6\n6 6 3 8 5 4\n3\n2 2 3\n2 1 5\n1 1 5", "output": "9\n24\n28" }, { "input": "1\n1\n2\n1 1 1\n2 1 1", "output": "1\n1" } ]

1,631,996,095

2,147,483,647

PyPy 3

TIME_LIMIT_EXCEEDED

TESTS

4

2,000

32,256,000

input() lst = [] vn = list(map(int,input().split())) vs = sorted(vn) vs2 = []; vn2 = [] for i in range(0, len(vn), 2): vn2.append(vn[i] + vn[i + 1]) for i in range(0, len(vs), 2): vs2.append(vs[i] + vs[i + 1]) for i in range(int(input())): a,b,c = map(int,(input().split())) b -= 1 bOdd = b % 2; cOdd = c % 2 if a == 1: if not bOdd and not cOdd: lst.append(sum(vn2[b // 2 : c // 2])) elif bOdd and not cOdd: lst.append(sum(vn2[(b + 1) // 2 : c // 2]) + vn[b]) elif not bOdd and cOdd: lst.append(sum(vn2[b // 2 : (c - 1) // 2]) + vn[c - 1]) elif bOdd and cOdd: lst.append(sum(vn2[(b + 1) // 2 : (c - 1) // 2]) + vn[b] + vn[c - 1]) else: if not bOdd and not cOdd: lst.append(sum(vs2[b // 2 : c // 2])) elif bOdd and not cOdd: lst.append(sum(vs2[(b + 1) // 2 : c // 2]) + vs[b]) elif not bOdd and cOdd: lst.append(sum(vs2[b // 2 : (c - 1) // 2]) + vs[c - 1]) elif bOdd and cOdd: lst.append(sum(vs2[(b + 1) // 2 : (c - 1) // 2]) + vs[b] + vs[c - 1]) for i in lst: print(i)

Title: Kuriyama Mirai's Stones Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions: 1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones. The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one. Output Specification: Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input. Demo Input: ['6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n', '4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n'] Demo Output: ['24\n9\n28\n', '10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n'] Note: Please note that the answers to the questions may overflow 32-bit integer type.

```python input() lst = [] vn = list(map(int,input().split())) vs = sorted(vn) vs2 = []; vn2 = [] for i in range(0, len(vn), 2): vn2.append(vn[i] + vn[i + 1]) for i in range(0, len(vs), 2): vs2.append(vs[i] + vs[i + 1]) for i in range(int(input())): a,b,c = map(int,(input().split())) b -= 1 bOdd = b % 2; cOdd = c % 2 if a == 1: if not bOdd and not cOdd: lst.append(sum(vn2[b // 2 : c // 2])) elif bOdd and not cOdd: lst.append(sum(vn2[(b + 1) // 2 : c // 2]) + vn[b]) elif not bOdd and cOdd: lst.append(sum(vn2[b // 2 : (c - 1) // 2]) + vn[c - 1]) elif bOdd and cOdd: lst.append(sum(vn2[(b + 1) // 2 : (c - 1) // 2]) + vn[b] + vn[c - 1]) else: if not bOdd and not cOdd: lst.append(sum(vs2[b // 2 : c // 2])) elif bOdd and not cOdd: lst.append(sum(vs2[(b + 1) // 2 : c // 2]) + vs[b]) elif not bOdd and cOdd: lst.append(sum(vs2[b // 2 : (c - 1) // 2]) + vs[c - 1]) elif bOdd and cOdd: lst.append(sum(vs2[(b + 1) // 2 : (c - 1) // 2]) + vs[b] + vs[c - 1]) for i in lst: print(i) ```

0

618

B

Guess the Permutation

PROGRAMMING

1,100

[ "constructive algorithms" ]

null

Bob has a permutation of integers from 1 to *n*. Denote this permutation as *p*. The *i*-th element of *p* will be denoted as *p**i*. For all pairs of distinct integers *i*,<=*j* between 1 and *n*, he wrote the number *a**i*,<=*j*<==<=*min*(*p**i*,<=*p**j*). He writes *a**i*,<=*i*<==<=0 for all integer *i* from 1 to *n*. Bob gave you all the values of *a**i*,<=*j* that he wrote down. Your job is to reconstruct any permutation that could have generated these values. The input will be formed so that it is guaranteed that there is at least one solution that is consistent with the information given.

The first line of the input will contain a single integer *n* (2<=≤<=*n*<=≤<=50). The next *n* lines will contain the values of *a**i*,<=*j*. The *j*-th number on the *i*-th line will represent *a**i*,<=*j*. The *i*-th number on the *i*-th line will be 0. It's guaranteed that *a**i*,<=*j*<==<=*a**j*,<=*i* and there is at least one solution consistent with the information given.

Print *n* space separated integers, which represents a permutation that could have generated these values. If there are multiple possible solutions, print any of them.

[ "2\n0 1\n1 0\n", "5\n0 2 2 1 2\n2 0 4 1 3\n2 4 0 1 3\n1 1 1 0 1\n2 3 3 1 0\n" ]

[ "2 1\n", "2 5 4 1 3\n" ]

In the first case, the answer can be {1, 2} or {2, 1}. In the second case, another possible answer is {2, 4, 5, 1, 3}.

1,000

[ { "input": "2\n0 1\n1 0", "output": "2 1" }, { "input": "5\n0 2 2 1 2\n2 0 4 1 3\n2 4 0 1 3\n1 1 1 0 1\n2 3 3 1 0", "output": "2 5 4 1 3" }, { "input": "10\n0 1 5 2 5 3 4 5 5 5\n1 0 1 1 1 1 1 1 1 1\n5 1 0 2 6 3 4 6 6 6\n2 1 2 0 2 2 2 2 2 2\n5 1 6 2 0 3 4 8 8 7\n3 1 3 2 3 0 3 3 3 3\n4 1 4 2 4 3 0 4 4 4\n5 1 6 2 8 3 4 0 9 7\n5 1 6 2 8 3 4 9 0 7\n5 1 6 2 7 3 4 7 7 0", "output": "5 1 6 2 8 3 4 10 9 7" }, { "input": "4\n0 1 3 2\n1 0 1 1\n3 1 0 2\n2 1 2 0", "output": "4 1 3 2" }, { "input": "7\n0 3 2 4 1 4 4\n3 0 2 3 1 3 3\n2 2 0 2 1 2 2\n4 3 2 0 1 5 5\n1 1 1 1 0 1 1\n4 3 2 5 1 0 6\n4 3 2 5 1 6 0", "output": "4 3 2 5 1 7 6" }, { "input": "10\n0 4 4 1 4 4 4 2 3 4\n4 0 5 1 6 8 9 2 3 7\n4 5 0 1 5 5 5 2 3 5\n1 1 1 0 1 1 1 1 1 1\n4 6 5 1 0 6 6 2 3 6\n4 8 5 1 6 0 8 2 3 7\n4 9 5 1 6 8 0 2 3 7\n2 2 2 1 2 2 2 0 2 2\n3 3 3 1 3 3 3 2 0 3\n4 7 5 1 6 7 7 2 3 0", "output": "4 10 5 1 6 8 9 2 3 7" }, { "input": "13\n0 5 5 2 5 4 5 5 3 5 5 5 1\n5 0 6 2 6 4 6 6 3 6 6 6 1\n5 6 0 2 10 4 7 10 3 8 10 9 1\n2 2 2 0 2 2 2 2 2 2 2 2 1\n5 6 10 2 0 4 7 12 3 8 11 9 1\n4 4 4 2 4 0 4 4 3 4 4 4 1\n5 6 7 2 7 4 0 7 3 7 7 7 1\n5 6 10 2 12 4 7 0 3 8 11 9 1\n3 3 3 2 3 3 3 3 0 3 3 3 1\n5 6 8 2 8 4 7 8 3 0 8 8 1\n5 6 10 2 11 4 7 11 3 8 0 9 1\n5 6 9 2 9 4 7 9 3 8 9 0 1\n1 1 1 1 1 1 1 1 1 1 1 1 0", "output": "5 6 10 2 13 4 7 12 3 8 11 9 1" } ]

1,454,089,894

2,494

Python 3

TIME_LIMIT_EXCEEDED

PRETESTS

2

2,000

307,200

import sys from itertools import permutations inpt = [int(num) for num in sys.stdin.read().split()] a = [inpt[i:i+inpt[0]] for i in range(1, len(inpt), inpt[0])] for p in permutations([num for num in range(1, inpt[0]+1)]): satisfies = True for i in range(1, inpt[0]): for j in range(i): if min(p[i], p[j]) != a[i][j]: satisfies = False break if not satisfies: break if satisfies: print(' '.join(map(str, p))) break

Title: Guess the Permutation Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bob has a permutation of integers from 1 to *n*. Denote this permutation as *p*. The *i*-th element of *p* will be denoted as *p**i*. For all pairs of distinct integers *i*,<=*j* between 1 and *n*, he wrote the number *a**i*,<=*j*<==<=*min*(*p**i*,<=*p**j*). He writes *a**i*,<=*i*<==<=0 for all integer *i* from 1 to *n*. Bob gave you all the values of *a**i*,<=*j* that he wrote down. Your job is to reconstruct any permutation that could have generated these values. The input will be formed so that it is guaranteed that there is at least one solution that is consistent with the information given. Input Specification: The first line of the input will contain a single integer *n* (2<=≤<=*n*<=≤<=50). The next *n* lines will contain the values of *a**i*,<=*j*. The *j*-th number on the *i*-th line will represent *a**i*,<=*j*. The *i*-th number on the *i*-th line will be 0. It's guaranteed that *a**i*,<=*j*<==<=*a**j*,<=*i* and there is at least one solution consistent with the information given. Output Specification: Print *n* space separated integers, which represents a permutation that could have generated these values. If there are multiple possible solutions, print any of them. Demo Input: ['2\n0 1\n1 0\n', '5\n0 2 2 1 2\n2 0 4 1 3\n2 4 0 1 3\n1 1 1 0 1\n2 3 3 1 0\n'] Demo Output: ['2 1\n', '2 5 4 1 3\n'] Note: In the first case, the answer can be {1, 2} or {2, 1}. In the second case, another possible answer is {2, 4, 5, 1, 3}.

```python import sys from itertools import permutations inpt = [int(num) for num in sys.stdin.read().split()] a = [inpt[i:i+inpt[0]] for i in range(1, len(inpt), inpt[0])] for p in permutations([num for num in range(1, inpt[0]+1)]): satisfies = True for i in range(1, inpt[0]): for j in range(i): if min(p[i], p[j]) != a[i][j]: satisfies = False break if not satisfies: break if satisfies: print(' '.join(map(str, p))) break ```

0

898

B

Proper Nutrition

PROGRAMMING

1,100

[ "brute force", "implementation", "number theory" ]

null

Vasya has *n* burles. One bottle of Ber-Cola costs *a* burles and one Bars bar costs *b* burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars. Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly *n* burles. In other words, you should find two non-negative integers *x* and *y* such that Vasya can buy *x* bottles of Ber-Cola and *y* Bars bars and *x*·*a*<=+<=*y*·*b*<==<=*n* or tell that it's impossible.

First line contains single integer *n* (1<=≤<=*n*<=≤<=10<=000<=000) — amount of money, that Vasya has. Second line contains single integer *a* (1<=≤<=*a*<=≤<=10<=000<=000) — cost of one bottle of Ber-Cola. Third line contains single integer *b* (1<=≤<=*b*<=≤<=10<=000<=000) — cost of one Bars bar.

If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly *n* burles print «NO» (without quotes). Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers *x* and *y* — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly *n* burles, i.e. *x*·*a*<=+<=*y*·*b*<==<=*n*. If there are multiple answers print any of them. Any of numbers *x* and *y* can be equal 0.

[ "7\n2\n3\n", "100\n25\n10\n", "15\n4\n8\n", "9960594\n2551\n2557\n" ]

[ "YES\n2 1\n", "YES\n0 10\n", "NO\n", "YES\n1951 1949\n" ]

In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles. In second example Vasya can spend exactly *n* burles multiple ways: - buy two bottles of Ber-Cola and five Bars bars; - buy four bottles of Ber-Cola and don't buy Bars bars; - don't buy Ber-Cola and buy 10 Bars bars. In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly *n* burles.

750

[ { "input": "7\n2\n3", "output": "YES\n2 1" }, { "input": "100\n25\n10", "output": "YES\n0 10" }, { "input": "15\n4\n8", "output": "NO" }, { "input": "9960594\n2551\n2557", "output": "YES\n1951 1949" }, { "input": "10000000\n1\n1", "output": "YES\n0 10000000" }, { "input": "9999999\n9999\n9999", "output": "NO" }, { "input": "9963629\n2591\n2593", "output": "YES\n635 3208" }, { "input": "1\n7\n8", "output": "NO" }, { "input": "9963630\n2591\n2593", "output": "YES\n1931 1913" }, { "input": "7516066\n1601\n4793", "output": "YES\n4027 223" }, { "input": "6509546\n1607\n6221", "output": "YES\n617 887" }, { "input": "2756250\n8783\n29", "output": "YES\n21 88683" }, { "input": "7817510\n2377\n743", "output": "YES\n560 8730" }, { "input": "6087210\n1583\n1997", "output": "YES\n1070 2200" }, { "input": "4\n2\n2", "output": "YES\n0 2" }, { "input": "7996960\n4457\n5387", "output": "YES\n727 883" }, { "input": "7988988\n4021\n3169", "output": "YES\n1789 251" }, { "input": "4608528\n9059\n977", "output": "YES\n349 1481" }, { "input": "8069102\n2789\n47", "output": "YES\n3 171505" }, { "input": "3936174\n4783\n13", "output": "YES\n5 300943" }, { "input": "10000000\n9999999\n1", "output": "YES\n0 10000000" }, { "input": "10000000\n1\n9999999", "output": "YES\n1 1" }, { "input": "4\n1\n3", "output": "YES\n1 1" }, { "input": "4\n1\n2", "output": "YES\n0 2" }, { "input": "4\n3\n1", "output": "YES\n0 4" }, { "input": "4\n2\n1", "output": "YES\n0 4" }, { "input": "100\n10\n20", "output": "YES\n0 5" }, { "input": "101\n11\n11", "output": "NO" }, { "input": "121\n11\n11", "output": "YES\n0 11" }, { "input": "25\n5\n6", "output": "YES\n5 0" }, { "input": "1\n1\n1", "output": "YES\n0 1" }, { "input": "10000000\n2\n1", "output": "YES\n0 10000000" }, { "input": "10000000\n1234523\n1", "output": "YES\n0 10000000" }, { "input": "10000000\n5000000\n5000000", "output": "YES\n0 2" }, { "input": "10000000\n5000001\n5000000", "output": "YES\n0 2" }, { "input": "10000000\n5000000\n5000001", "output": "YES\n2 0" }, { "input": "9999999\n9999999\n9999999", "output": "YES\n0 1" }, { "input": "10000000\n10000000\n10000000", "output": "YES\n0 1" }, { "input": "10\n1\n3", "output": "YES\n1 3" }, { "input": "97374\n689\n893", "output": "NO" }, { "input": "100096\n791\n524", "output": "NO" }, { "input": "75916\n651\n880", "output": "NO" }, { "input": "110587\n623\n806", "output": "NO" }, { "input": "5600\n670\n778", "output": "NO" }, { "input": "81090\n527\n614", "output": "NO" }, { "input": "227718\n961\n865", "output": "NO" }, { "input": "10000000\n3\n999999", "output": "NO" }, { "input": "3\n4\n5", "output": "NO" }, { "input": "9999999\n2\n2", "output": "NO" }, { "input": "9999999\n2\n4", "output": "NO" }, { "input": "9999997\n2\n5", "output": "YES\n1 1999999" }, { "input": "9366189\n4326262\n8994187", "output": "NO" }, { "input": "1000000\n1\n10000000", "output": "YES\n1000000 0" }, { "input": "9999991\n2\n2", "output": "NO" }, { "input": "10000000\n7\n7", "output": "NO" }, { "input": "9999991\n2\n4", "output": "NO" }, { "input": "10000000\n3\n6", "output": "NO" }, { "input": "10000000\n11\n11", "output": "NO" }, { "input": "4\n7\n3", "output": "NO" }, { "input": "1000003\n2\n2", "output": "NO" }, { "input": "1000000\n7\n7", "output": "NO" }, { "input": "999999\n2\n2", "output": "NO" }, { "input": "8\n13\n5", "output": "NO" }, { "input": "1000003\n15\n3", "output": "NO" }, { "input": "7\n7\n2", "output": "YES\n1 0" }, { "input": "9999999\n2\n8", "output": "NO" }, { "input": "1000000\n3\n7", "output": "YES\n5 142855" }, { "input": "9999999\n1\n10000000", "output": "YES\n9999999 0" }, { "input": "100\n1\n1000000", "output": "YES\n100 0" }, { "input": "10000000\n9999999\n9999997", "output": "NO" }, { "input": "2\n1\n3", "output": "YES\n2 0" }, { "input": "3\n5\n2", "output": "NO" }, { "input": "5\n2\n3", "output": "YES\n1 1" }, { "input": "10000000\n7\n14", "output": "NO" }, { "input": "10000000\n2\n9999999", "output": "YES\n5000000 0" }, { "input": "10000000\n3\n3", "output": "NO" }, { "input": "1\n3\n2", "output": "NO" }, { "input": "25\n27\n2", "output": "NO" }, { "input": "3\n2\n17", "output": "NO" }, { "input": "999997\n4\n8", "output": "NO" }, { "input": "2000000\n1\n2000001", "output": "YES\n2000000 0" }, { "input": "8\n7\n3", "output": "NO" }, { "input": "7005920\n5705\n28145", "output": "NO" }, { "input": "2\n6\n4", "output": "NO" }, { "input": "10000000\n9999999\n3", "output": "NO" }, { "input": "10000000\n77\n99", "output": "NO" }, { "input": "100\n8\n70", "output": "NO" }, { "input": "99999\n2\n2", "output": "NO" }, { "input": "5\n7\n2", "output": "NO" }, { "input": "999999\n12\n14", "output": "NO" }, { "input": "100\n1\n1000", "output": "YES\n100 0" }, { "input": "10000000\n123\n321", "output": "NO" }, { "input": "9369319\n4\n2", "output": "NO" }, { "input": "9999998\n3\n3", "output": "NO" }, { "input": "85\n5\n85", "output": "YES\n0 1" }, { "input": "64549\n9999999\n2", "output": "NO" }, { "input": "10000000\n3\n7", "output": "YES\n1 1428571" }, { "input": "9999889\n2\n2", "output": "NO" }, { "input": "10000000\n9999999\n123", "output": "NO" }, { "input": "64549\n2\n9999999", "output": "NO" } ]

1,589,131,862

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

78

6,656,000

n = int(input()) b1 = int(input()) b2 = int(input()) f = False for r in range(int(n/b1)): if (n - (r*b1)) % b2 == 0: print('YES') print(str(r) + ' ' + str(int((n - (r*b1)) / b2))) f = True if not f: print('NO')

Title: Proper Nutrition Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya has *n* burles. One bottle of Ber-Cola costs *a* burles and one Bars bar costs *b* burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars. Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly *n* burles. In other words, you should find two non-negative integers *x* and *y* such that Vasya can buy *x* bottles of Ber-Cola and *y* Bars bars and *x*·*a*<=+<=*y*·*b*<==<=*n* or tell that it's impossible. Input Specification: First line contains single integer *n* (1<=≤<=*n*<=≤<=10<=000<=000) — amount of money, that Vasya has. Second line contains single integer *a* (1<=≤<=*a*<=≤<=10<=000<=000) — cost of one bottle of Ber-Cola. Third line contains single integer *b* (1<=≤<=*b*<=≤<=10<=000<=000) — cost of one Bars bar. Output Specification: If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly *n* burles print «NO» (without quotes). Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers *x* and *y* — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly *n* burles, i.e. *x*·*a*<=+<=*y*·*b*<==<=*n*. If there are multiple answers print any of them. Any of numbers *x* and *y* can be equal 0. Demo Input: ['7\n2\n3\n', '100\n25\n10\n', '15\n4\n8\n', '9960594\n2551\n2557\n'] Demo Output: ['YES\n2 1\n', 'YES\n0 10\n', 'NO\n', 'YES\n1951 1949\n'] Note: In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles. In second example Vasya can spend exactly *n* burles multiple ways: - buy two bottles of Ber-Cola and five Bars bars; - buy four bottles of Ber-Cola and don't buy Bars bars; - don't buy Ber-Cola and buy 10 Bars bars. In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly *n* burles.

```python n = int(input()) b1 = int(input()) b2 = int(input()) f = False for r in range(int(n/b1)): if (n - (r*b1)) % b2 == 0: print('YES') print(str(r) + ' ' + str(int((n - (r*b1)) / b2))) f = True if not f: print('NO') ```

0

764

A

Taymyr is calling you

PROGRAMMING

800

[ "brute force", "implementation", "math" ]

null

Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist. Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.

The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104).

Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.

[ "1 1 10\n", "1 2 5\n", "2 3 9\n" ]

[ "10\n", "2\n", "1\n" ]

Taymyr is a place in the north of Russia. In the first test the artists come each minute, as well as the calls, so we need to kill all of them. In the second test we need to kill artists which come on the second and the fourth minutes. In the third test — only the artist which comes on the sixth minute.

500

[ { "input": "1 1 10", "output": "10" }, { "input": "1 2 5", "output": "2" }, { "input": "2 3 9", "output": "1" }, { "input": "4 8 9", "output": "1" }, { "input": "7 9 2", "output": "0" }, { "input": "10000 10000 10000", "output": "1" }, { "input": "24 22 9235", "output": "34" }, { "input": "74 8 417", "output": "1" }, { "input": "972 1 203", "output": "0" }, { "input": "550 1 754", "output": "1" }, { "input": "860 1 884", "output": "1" }, { "input": "358 2 809", "output": "2" }, { "input": "33 27 216", "output": "0" }, { "input": "2940 1 9311", "output": "3" }, { "input": "4624 1 1953", "output": "0" }, { "input": "2696 2 7345", "output": "2" }, { "input": "3443 2 6701", "output": "0" }, { "input": "3 613 2275", "output": "1" }, { "input": "1 10000 10000", "output": "1" }, { "input": "10000 1 10000", "output": "1" }, { "input": "1 1 1", "output": "1" }, { "input": "1 1 10000", "output": "10000" }, { "input": "34 27 10000", "output": "10" }, { "input": "2 2 9999", "output": "4999" }, { "input": "2 2 1", "output": "0" }, { "input": "6 4 36", "output": "3" }, { "input": "33 6 3005", "output": "45" }, { "input": "5 1 20", "output": "4" }, { "input": "1 2 10", "output": "5" }, { "input": "2 1 100", "output": "50" }, { "input": "10 20 10000", "output": "500" }, { "input": "8 12 12", "output": "0" } ]

1,649,738,942

2,147,483,647

Python 3

OK

TESTS

32

46

0

n, m, z = list(map(int, input().split())) i = 1 while n*i % m != 0: i += 1 print(z//(n*i))

Title: Taymyr is calling you Time Limit: None seconds Memory Limit: None megabytes Problem Description: Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist. Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute. Input Specification: The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104). Output Specification: Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls. Demo Input: ['1 1 10\n', '1 2 5\n', '2 3 9\n'] Demo Output: ['10\n', '2\n', '1\n'] Note: Taymyr is a place in the north of Russia. In the first test the artists come each minute, as well as the calls, so we need to kill all of them. In the second test we need to kill artists which come on the second and the fourth minutes. In the third test — only the artist which comes on the sixth minute.

```python n, m, z = list(map(int, input().split())) i = 1 while n*i % m != 0: i += 1 print(z//(n*i)) ```

3

493

C

Vasya and Basketball

PROGRAMMING

1,600

[ "binary search", "brute force", "data structures", "implementation", "sortings", "two pointers" ]

null

Vasya follows a basketball game and marks the distances from which each team makes a throw. He knows that each successful throw has value of either 2 or 3 points. A throw is worth 2 points if the distance it was made from doesn't exceed some value of *d* meters, and a throw is worth 3 points if the distance is larger than *d* meters, where *d* is some non-negative integer. Vasya would like the advantage of the points scored by the first team (the points of the first team minus the points of the second team) to be maximum. For that he can mentally choose the value of *d*. Help him to do that.

The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of throws of the first team. Then follow *n* integer numbers — the distances of throws *a**i* (1<=≤<=*a**i*<=≤<=2·109). Then follows number *m* (1<=≤<=*m*<=≤<=2·105) — the number of the throws of the second team. Then follow *m* integer numbers — the distances of throws of *b**i* (1<=≤<=*b**i*<=≤<=2·109).

Print two numbers in the format a:b — the score that is possible considering the problem conditions where the result of subtraction *a*<=-<=*b* is maximum. If there are several such scores, find the one in which number *a* is maximum.

[ "3\n1 2 3\n2\n5 6\n", "5\n6 7 8 9 10\n5\n1 2 3 4 5\n" ]

[ "9:6\n", "15:10\n" ]

none

2,000

[ { "input": "3\n1 2 3\n2\n5 6", "output": "9:6" }, { "input": "5\n6 7 8 9 10\n5\n1 2 3 4 5", "output": "15:10" }, { "input": "5\n1 2 3 4 5\n5\n6 7 8 9 10", "output": "15:15" }, { "input": "3\n1 2 3\n3\n6 4 5", "output": "9:9" }, { "input": "10\n1 2 3 4 5 6 7 8 9 10\n1\n11", "output": "30:3" }, { "input": "10\n1 2 3 4 5 6 7 8 9 11\n1\n10", "output": "30:3" }, { "input": "3\n1 2 3\n3\n1 2 3", "output": "9:9" }, { "input": "3\n1 2 3\n3\n3 4 5", "output": "9:9" }, { "input": "4\n2 5 3 2\n4\n1 5 6 2", "output": "12:11" }, { "input": "2\n3 3\n3\n1 3 3", "output": "6:8" }, { "input": "3\n1 1 1\n4\n1 3 1 1", "output": "6:8" }, { "input": "4\n4 2 1 1\n4\n3 2 2 2", "output": "9:8" }, { "input": "3\n3 9 4\n2\n10 1", "output": "9:5" }, { "input": "14\n4336 24047 24846 25681 28597 30057 32421 34446 48670 67750 68185 69661 85721 89013\n30\n8751 10576 14401 22336 22689 35505 38649 43073 43176 44359 44777 50210 50408 51361 53181 60095 65554 68201 68285 68801 72501 75881 80251 80509 83306 93167 95365 95545 97201 97731", "output": "28:60" }, { "input": "1\n1\n2\n1 2", "output": "2:4" }, { "input": "18\n450 3726 12063 27630 29689 30626 33937 35015 45951 46217 53004 59541 75551 75836 78996 81297 93876 96211\n47\n3393 5779 6596 7935 9549 10330 11145 13121 14801 15578 24104 24125 25871 31280 35036 38969 40077 41342 42708 46033 47491 48451 49152 51905 55002 55689 56565 57901 59481 60017 66075 67081 68397 71122 74961 78501 84098 87083 87893 89281 89739 90321 92046 95821 96717 96921 96951", "output": "36:94" }, { "input": "3\n3 3 4\n6\n2 2 3 3 3 3", "output": "7:12" }, { "input": "3\n2 2 2\n3\n1 1 1", "output": "9:6" }, { "input": "2\n2 2\n2\n2 2", "output": "6:6" }, { "input": "1\n7\n6\n6 7 8 9 10 11", "output": "2:12" }, { "input": "1\n1\n2\n1 1", "output": "2:4" }, { "input": "3\n1 2 3\n1\n1", "output": "9:3" }, { "input": "3\n3 3 4\n6\n3 2 2 2 3 2", "output": "9:14" }, { "input": "1\n3\n1\n3", "output": "3:3" }, { "input": "1\n1\n5\n1 1 1 1 1", "output": "2:10" }, { "input": "2\n1 999999999\n2\n2 4", "output": "5:4" } ]

1,673,282,655

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

2

61

0

def check(mid,arr): v1=0 v2=0 for i in range(len(arr)): j=arr[i] # print(j,"'jj'",mid) if(j[0]<=mid): if(j[1]=='1'): v1+=2 else: v2+=2 else: if(j[1]=='1'): v1+=3 else: v2+=3 # print(v1,v2,mid,"---") return [v1-v2,v1,v2] n=int(input()) l1=list(map(int,input().split())) m=int(input()) l2=list(map(int,input().split())) p=[] for i in l1: p.append([i,'1']) for i in l2: p.append([i,'2']) p=sorted(p) max_=0 ans=[] for i in range(0,n+m+1): cal_= check(i,p) cal = cal_[0] if(cal>max_): max_=cal ans=[] ans.append([cal_[1],cal_[2]]) elif(cal==max_): ans.append([cal_[1],cal_[2]]) ans=sorted(ans) print(ans[0][0],end="") print(":",end="") print(ans[0][1])

Title: Vasya and Basketball Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya follows a basketball game and marks the distances from which each team makes a throw. He knows that each successful throw has value of either 2 or 3 points. A throw is worth 2 points if the distance it was made from doesn't exceed some value of *d* meters, and a throw is worth 3 points if the distance is larger than *d* meters, where *d* is some non-negative integer. Vasya would like the advantage of the points scored by the first team (the points of the first team minus the points of the second team) to be maximum. For that he can mentally choose the value of *d*. Help him to do that. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of throws of the first team. Then follow *n* integer numbers — the distances of throws *a**i* (1<=≤<=*a**i*<=≤<=2·109). Then follows number *m* (1<=≤<=*m*<=≤<=2·105) — the number of the throws of the second team. Then follow *m* integer numbers — the distances of throws of *b**i* (1<=≤<=*b**i*<=≤<=2·109). Output Specification: Print two numbers in the format a:b — the score that is possible considering the problem conditions where the result of subtraction *a*<=-<=*b* is maximum. If there are several such scores, find the one in which number *a* is maximum. Demo Input: ['3\n1 2 3\n2\n5 6\n', '5\n6 7 8 9 10\n5\n1 2 3 4 5\n'] Demo Output: ['9:6\n', '15:10\n'] Note: none

```python def check(mid,arr): v1=0 v2=0 for i in range(len(arr)): j=arr[i] # print(j,"'jj'",mid) if(j[0]<=mid): if(j[1]=='1'): v1+=2 else: v2+=2 else: if(j[1]=='1'): v1+=3 else: v2+=3 # print(v1,v2,mid,"---") return [v1-v2,v1,v2] n=int(input()) l1=list(map(int,input().split())) m=int(input()) l2=list(map(int,input().split())) p=[] for i in l1: p.append([i,'1']) for i in l2: p.append([i,'2']) p=sorted(p) max_=0 ans=[] for i in range(0,n+m+1): cal_= check(i,p) cal = cal_[0] if(cal>max_): max_=cal ans=[] ans.append([cal_[1],cal_[2]]) elif(cal==max_): ans.append([cal_[1],cal_[2]]) ans=sorted(ans) print(ans[0][0],end="") print(":",end="") print(ans[0][1]) ```

0

727

A

Transformation: from A to B

PROGRAMMING

1,000

[ "brute force", "dfs and similar", "math" ]

null

Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations: - multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1). You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible. Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*.

The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have.

If there is no way to get *b* from *a*, print "NO" (without quotes). Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where: - *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=<<=*i*<=≤<=*k*). If there are multiple answers, print any of them.

[ "2 162\n", "4 42\n", "100 40021\n" ]

[ "YES\n5\n2 4 8 81 162 \n", "NO\n", "YES\n5\n100 200 2001 4002 40021 \n" ]

none

1,000

[ { "input": "2 162", "output": "YES\n5\n2 4 8 81 162 " }, { "input": "4 42", "output": "NO" }, { "input": "100 40021", "output": "YES\n5\n100 200 2001 4002 40021 " }, { "input": "1 111111111", "output": "YES\n9\n1 11 111 1111 11111 111111 1111111 11111111 111111111 " }, { "input": "1 1000000000", "output": "NO" }, { "input": "999999999 1000000000", "output": "NO" }, { "input": "1 2", "output": "YES\n2\n1 2 " }, { "input": "1 536870912", "output": "YES\n30\n1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 " }, { "input": "11111 11111111", "output": "YES\n4\n11111 111111 1111111 11111111 " }, { "input": "59139 946224", "output": "YES\n5\n59139 118278 236556 473112 946224 " }, { "input": "9859 19718", "output": "YES\n2\n9859 19718 " }, { "input": "25987 51974222", "output": "YES\n5\n25987 259871 2598711 25987111 51974222 " }, { "input": "9411 188222222", "output": "YES\n6\n9411 94111 941111 9411111 94111111 188222222 " }, { "input": "25539 510782222", "output": "YES\n6\n25539 255391 2553911 25539111 255391111 510782222 " }, { "input": "76259 610072", "output": "YES\n4\n76259 152518 305036 610072 " }, { "input": "92387 184774", "output": "YES\n2\n92387 184774 " }, { "input": "8515 85151111", "output": "YES\n5\n8515 85151 851511 8515111 85151111 " }, { "input": "91939 9193911", "output": "YES\n3\n91939 919391 9193911 " }, { "input": "30518 610361", "output": "YES\n3\n30518 61036 610361 " }, { "input": "46646 373168844", "output": "YES\n7\n46646 466461 932922 9329221 93292211 186584422 373168844 " }, { "input": "30070 300701", "output": "YES\n2\n30070 300701 " }, { "input": "13494 1079528", "output": "YES\n5\n13494 134941 269882 539764 1079528 " }, { "input": "96918 775344422", "output": "YES\n7\n96918 193836 1938361 3876722 38767221 387672211 775344422 " }, { "input": "13046 260921", "output": "YES\n3\n13046 26092 260921 " }, { "input": "29174 5834811", "output": "YES\n4\n29174 58348 583481 5834811 " }, { "input": "79894 319576421", "output": "YES\n6\n79894 798941 1597882 15978821 31957642 319576421 " }, { "input": "96022 1920442", "output": "YES\n3\n96022 960221 1920442 " }, { "input": "79446 6355681", "output": "YES\n5\n79446 158892 317784 635568 6355681 " }, { "input": "5440 27853056", "output": "YES\n11\n5440 10880 108801 217602 435204 870408 1740816 3481632 6963264 13926528 27853056 " }, { "input": "250000000 705032705", "output": "NO" }, { "input": "17 35", "output": "NO" }, { "input": "1 3", "output": "NO" }, { "input": "2 11", "output": "NO" } ]

1,675,759,986

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

15

0

a,b=map(int,input().split()) lst=[b] flag=0 while(b!=a): if b%2==0: b=b/2 lst.append(b) else: b=b//10 lst.append(b) if b==1: flag=5 lst.sort(reverse=True) if flag==5: print("NO") else: print("YES") print(len(lst)) for i in range (len(lst)): print(int(lst[i]),end=" ")

Title: Transformation: from A to B Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations: - multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1). You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible. Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*. Input Specification: The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have. Output Specification: If there is no way to get *b* from *a*, print "NO" (without quotes). Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where: - *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=<<=*i*<=≤<=*k*). If there are multiple answers, print any of them. Demo Input: ['2 162\n', '4 42\n', '100 40021\n'] Demo Output: ['YES\n5\n2 4 8 81 162 \n', 'NO\n', 'YES\n5\n100 200 2001 4002 40021 \n'] Note: none

```python a,b=map(int,input().split()) lst=[b] flag=0 while(b!=a): if b%2==0: b=b/2 lst.append(b) else: b=b//10 lst.append(b) if b==1: flag=5 lst.sort(reverse=True) if flag==5: print("NO") else: print("YES") print(len(lst)) for i in range (len(lst)): print(int(lst[i]),end=" ") ```

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" } ]

1,643,392,418

2,147,483,647

PyPy 3-64

COMPILATION_ERROR

TESTS

0

n = int(input()) # n represents n of lines (forces) # input() alwayes returns str value x=0 y=0 z=0 for x in range(n): a, b, c = map(int, input().split()) # this the first line containing f_force, within each loop a, b, c will change therefore i will store keep summing x += a y += b z += c # sounds like i got the sum of x column if (x=0) and (y=0) and (z=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()) # n represents n of lines (forces) # input() alwayes returns str value x=0 y=0 z=0 for x in range(n): a, b, c = map(int, input().split()) # this the first line containing f_force, within each loop a, b, c will change therefore i will store keep summing x += a y += b z += c # sounds like i got the sum of x column if (x=0) and (y=0) and (z=0): print("YES") else: print("NO") ```

-1

11

A

Increasing Sequence

PROGRAMMING

900

[ "constructive algorithms", "implementation", "math" ]

A. Increasing Sequence

1

64

A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*. You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?

The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).

Output the minimal number of moves needed to make the sequence increasing.

[ "4 2\n1 3 3 2\n" ]

[ "3\n" ]

none

0

[ { "input": "4 2\n1 3 3 2", "output": "3" }, { "input": "2 1\n1 1", "output": "1" }, { "input": "2 1\n2 5", "output": "0" }, { "input": "2 1\n1 2", "output": "0" }, { "input": "2 1\n1 1", "output": "1" }, { "input": "2 7\n10 20", "output": "0" }, { "input": "2 7\n1 1", "output": "1" }, { "input": "3 3\n18 1 9", "output": "10" }, { "input": "3 3\n15 17 9", "output": "3" }, { "input": "3 3\n10 9 12", "output": "2" }, { "input": "10 3\n2 1 17 10 5 16 8 4 15 17", "output": "31" }, { "input": "10 3\n6 11 4 12 22 15 23 26 24 26", "output": "13" }, { "input": "10 3\n10 24 13 15 18 14 15 26 33 35", "output": "29" }, { "input": "100 3\n529 178 280 403 326 531 671 427 188 866 669 646 421 804 494 609 53 1012 211 243 887 833 900 543 226 42 859 718 454 372 971 692 846 770 511 395 499 479 641 756 115 269 206 45 1039 727 400 779 859 614 146 214 196 919 702 959 380 830 535 878 859 784 316 305 782 924 536 243 236 978 564 150 291 877 808 983 537 839 490 120 168 838 267 650 900 170 211 504 326 771 895 984 994 483 776 100 471 1078 317 580", "output": "15717" }, { "input": "100 3\n329 226 331 909 962 112 837 1005 194 818 506 416 125 648 367 459 400 582 989 547 329 438 234 121 272 226 821 376 834 427 718 164 834 113 654 177 737 212 169 696 744 180 89 944 233 147 667 990 809 1072 1085 1093 814 265 1067 312 833 572 303 901 1032 504 185 817 389 158 613 723 239 269 911 352 769 404 225 822 897 606 947 323 913 804 923 1084 552 901 486 249 209 898 847 610 728 1122 986 669 1116 1076 367 327", "output": "16133" } ]

1,596,829,035

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

0

216

20,172,800

import math class Solution: def __init__(self): return def seq(self, n, d, lis): ret, prev = 0, lis[0] for i in range(1, len(lis)): if lis[i] <= prev: v = math.ceil((prev - lis[i])/d) prev = v*d + lis[i] ret += v else: prev = lis[i] return ret s = Solution() n,d = map(int,input().split()) lis= list(map(int,input().split())) print(s.seq(n,d,lis))

Title: Increasing Sequence Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*. You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing? Input Specification: The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106). Output Specification: Output the minimal number of moves needed to make the sequence increasing. Demo Input: ['4 2\n1 3 3 2\n'] Demo Output: ['3\n'] Note: none

```python import math class Solution: def __init__(self): return def seq(self, n, d, lis): ret, prev = 0, lis[0] for i in range(1, len(lis)): if lis[i] <= prev: v = math.ceil((prev - lis[i])/d) prev = v*d + lis[i] ret += v else: prev = lis[i] return ret s = Solution() n,d = map(int,input().split()) lis= list(map(int,input().split())) print(s.seq(n,d,lis)) ```

0

none

0

[ "none" ]

null

You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings. An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.

The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries. *q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query.

For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.

[ "1\n12\n", "2\n6\n8\n", "3\n1\n2\n3\n" ]

[ "3\n", "1\n2\n", "-1\n-1\n-1\n" ]

12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands. 8 = 4 + 4, 6 can't be split into several composite summands. 1, 2, 3 are less than any composite number, so they do not have valid splittings.

0

[ { "input": "1\n12", "output": "3" }, { "input": "2\n6\n8", "output": "1\n2" }, { "input": "3\n1\n2\n3", "output": "-1\n-1\n-1" }, { "input": "6\n1\n2\n3\n5\n7\n11", "output": "-1\n-1\n-1\n-1\n-1\n-1" }, { "input": "3\n4\n6\n9", "output": "1\n1\n1" }, { "input": "20\n8\n13\n20\n12\n9\n16\n4\n19\n7\n15\n10\n6\n14\n11\n3\n2\n5\n17\n18\n1", "output": "2\n2\n5\n3\n1\n4\n1\n3\n-1\n2\n2\n1\n3\n-1\n-1\n-1\n-1\n3\n4\n-1" }, { "input": "100\n611\n513\n544\n463\n38\n778\n347\n317\n848\n664\n382\n108\n718\n33\n334\n876\n234\n22\n944\n305\n159\n245\n513\n691\n639\n135\n308\n324\n813\n459\n304\n116\n331\n993\n184\n224\n853\n769\n121\n687\n93\n930\n751\n308\n485\n914\n400\n695\n95\n981\n175\n972\n121\n654\n242\n610\n617\n999\n237\n548\n742\n767\n613\n172\n223\n391\n102\n907\n673\n116\n230\n355\n189\n552\n399\n493\n903\n201\n985\n459\n776\n641\n693\n919\n253\n540\n427\n394\n655\n101\n461\n854\n417\n249\n66\n380\n213\n906\n212\n528", "output": "151\n127\n136\n114\n9\n194\n85\n78\n212\n166\n95\n27\n179\n7\n83\n219\n58\n5\n236\n75\n38\n60\n127\n171\n158\n32\n77\n81\n202\n113\n76\n29\n81\n247\n46\n56\n212\n191\n29\n170\n22\n232\n186\n77\n120\n228\n100\n172\n22\n244\n42\n243\n29\n163\n60\n152\n153\n248\n58\n137\n185\n190\n152\n43\n54\n96\n25\n225\n167\n29\n57\n87\n46\n138\n98\n122\n224\n49\n245\n113\n194\n159\n172\n228\n62\n135\n105\n98\n162\n24\n114\n213\n103\n61\n16\n95\n52\n226\n53\n132" }, { "input": "1\n10000001", "output": "2499999" } ]

1,509,002,373

6,273

Python 3

COMPILATION_ERROR

TESTS

0

#include <bits/stdc++.h> using namespace std; const int N = 1e9 + 10; const int inf = 0x3f3f3f3f; bool is_not_prime[N] = {1, 0, 0,}; vector<int> prime; void Prime() { for(int i = 2; i < N; i++) { if(!is_not_prime[i]) prime.push_back(i); for(int j = 0, size = prime.size(); j < size && i * prime[j] < N; j++) { is_not_prime[i * prime[j]] = 1; if(!(i % prime[j])) break; } } } int main() { Prime(); int q; scanf("%d", &q); while(q--) { int n; scanf("%d", &n); if(!is_not_prime[n]) puts("-1"); else { int ans = 1; int m = ceil(sqrt(n)); for(int i = 2; i <= m; i++) { if(n % i == 0) { if(is_not_prime[i]) { ans = max(ans, n / i); break; } } } printf("%d\n", ans); } } return 0; }

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings. An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself. Input Specification: The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries. *q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query. Output Specification: For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings. Demo Input: ['1\n12\n', '2\n6\n8\n', '3\n1\n2\n3\n'] Demo Output: ['3\n', '1\n2\n', '-1\n-1\n-1\n'] Note: 12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands. 8 = 4 + 4, 6 can't be split into several composite summands. 1, 2, 3 are less than any composite number, so they do not have valid splittings.

```python #include <bits/stdc++.h> using namespace std; const int N = 1e9 + 10; const int inf = 0x3f3f3f3f; bool is_not_prime[N] = {1, 0, 0,}; vector<int> prime; void Prime() { for(int i = 2; i < N; i++) { if(!is_not_prime[i]) prime.push_back(i); for(int j = 0, size = prime.size(); j < size && i * prime[j] < N; j++) { is_not_prime[i * prime[j]] = 1; if(!(i % prime[j])) break; } } } int main() { Prime(); int q; scanf("%d", &q); while(q--) { int n; scanf("%d", &n); if(!is_not_prime[n]) puts("-1"); else { int ans = 1; int m = ceil(sqrt(n)); for(int i = 2; i <= m; i++) { if(n % i == 0) { if(is_not_prime[i]) { ans = max(ans, n / i); break; } } } printf("%d\n", ans); } } return 0; } ```

-1

407

A

Triangle

PROGRAMMING

1,600

[ "brute force", "geometry", "implementation", "math" ]

null

There is a right triangle with legs of length *a* and *b*. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices.

The first line contains two integers *a*,<=*b* (1<=≤<=*a*,<=*b*<=≤<=1000), separated by a single space.

In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 in their absolute value.

[ "1 1\n", "5 5\n", "5 10\n" ]

[ "NO\n", "YES\n2 1\n5 5\n-2 4\n", "YES\n-10 4\n-2 -2\n1 2\n" ]

none

500

[ { "input": "1 1", "output": "NO" }, { "input": "5 5", "output": "YES\n2 1\n5 5\n-2 4" }, { "input": "5 10", "output": "YES\n-10 4\n-2 -2\n1 2" }, { "input": "2 2", "output": "NO" }, { "input": "5 6", "output": "NO" }, { "input": "5 11", "output": "NO" }, { "input": "10 15", "output": "YES\n0 0\n6 8\n-12 9" }, { "input": "935 938", "output": "NO" }, { "input": "999 1000", "output": "NO" }, { "input": "1000 1000", "output": "YES\n0 0\n280 960\n-960 280" }, { "input": "15 20", "output": "YES\n0 0\n12 9\n-12 16" }, { "input": "20 15", "output": "YES\n0 0\n12 16\n-12 9" }, { "input": "629 865", "output": "NO" }, { "input": "45 872", "output": "NO" }, { "input": "757 582", "output": "NO" }, { "input": "173 588", "output": "NO" }, { "input": "533 298", "output": "NO" }, { "input": "949 360", "output": "NO" }, { "input": "661 175", "output": "NO" }, { "input": "728 299", "output": "YES\n0 0\n280 672\n-276 115" }, { "input": "575 85", "output": "YES\n0 0\n345 460\n-68 51" }, { "input": "385 505", "output": "YES\n0 0\n231 308\n-404 303" }, { "input": "755 865", "output": "YES\n0 0\n453 604\n-692 519" }, { "input": "395 55", "output": "YES\n0 0\n237 316\n-44 33" }, { "input": "600 175", "output": "YES\n0 0\n168 576\n-168 49" }, { "input": "280 210", "output": "YES\n0 0\n168 224\n-168 126" }, { "input": "180 135", "output": "YES\n0 0\n108 144\n-108 81" }, { "input": "140 105", "output": "YES\n0 0\n84 112\n-84 63" }, { "input": "440 330", "output": "YES\n0 0\n264 352\n-264 198" }, { "input": "130 312", "output": "YES\n0 0\n120 50\n-120 288" }, { "input": "65 156", "output": "YES\n0 0\n60 25\n-60 144" }, { "input": "105 140", "output": "YES\n0 0\n84 63\n-84 112" }, { "input": "408 765", "output": "YES\n0 0\n360 192\n-360 675" }, { "input": "195 468", "output": "YES\n0 0\n180 75\n-180 432" }, { "input": "305 949", "output": "NO" }, { "input": "80 60", "output": "YES\n0 0\n48 64\n-48 36" }, { "input": "15 15", "output": "YES\n0 0\n9 12\n-12 9" }, { "input": "120 90", "output": "YES\n0 0\n72 96\n-72 54" }, { "input": "60 80", "output": "YES\n0 0\n48 36\n-48 64" }, { "input": "5 25", "output": "YES\n0 0\n3 4\n-20 15" }, { "input": "5 1000", "output": "YES\n0 0\n3 4\n-800 600" }, { "input": "25 35", "output": "YES\n0 0\n15 20\n-28 21" }, { "input": "25 5", "output": "YES\n0 0\n15 20\n-4 3" }, { "input": "30 40", "output": "YES\n0 0\n24 18\n-24 32" }, { "input": "65 156", "output": "YES\n0 0\n60 25\n-60 144" }, { "input": "455 470", "output": "YES\n0 0\n273 364\n-376 282" }, { "input": "90 120", "output": "YES\n0 0\n72 54\n-72 96" }, { "input": "5 205", "output": "YES\n0 0\n3 4\n-164 123" }, { "input": "111 111", "output": "YES\n0 0\n36 105\n-105 36" }, { "input": "17 17", "output": "YES\n0 0\n8 15\n-15 8" }, { "input": "1000 5", "output": "YES\n0 0\n600 800\n-4 3" } ]

1,657,012,582

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

1

62

0

from math import sqrt a,b = [int(i) for i in input().split()] def triangle(): ans = [] for i in range(1, a): c = sqrt(a*a - i*i) if c == int(c): ans.append((i, int(c))) break else: return [] for i in range(1, b): c = sqrt(b*b - i*i) if c == int(c): ans.append((i, int(c))) break else: return [] return ans x = triangle() if x: print('YES') print(0,0) print(*x[0]) print(*x[1]) else: print('NO')

Title: Triangle Time Limit: None seconds Memory Limit: None megabytes Problem Description: There is a right triangle with legs of length *a* and *b*. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices. Input Specification: The first line contains two integers *a*,<=*b* (1<=≤<=*a*,<=*b*<=≤<=1000), separated by a single space. Output Specification: In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 in their absolute value. Demo Input: ['1 1\n', '5 5\n', '5 10\n'] Demo Output: ['NO\n', 'YES\n2 1\n5 5\n-2 4\n', 'YES\n-10 4\n-2 -2\n1 2\n'] Note: none

```python from math import sqrt a,b = [int(i) for i in input().split()] def triangle(): ans = [] for i in range(1, a): c = sqrt(a*a - i*i) if c == int(c): ans.append((i, int(c))) break else: return [] for i in range(1, b): c = sqrt(b*b - i*i) if c == int(c): ans.append((i, int(c))) break else: return [] return ans x = triangle() if x: print('YES') print(0,0) print(*x[0]) print(*x[1]) else: print('NO') ```

0

34

A

Reconnaissance 2

PROGRAMMING

800

[ "implementation" ]

A. Reconnaissance 2

2

256

*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.

The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.

Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.

[ "5\n10 12 13 15 10\n", "4\n10 20 30 40\n" ]

[ "5 1\n", "1 2\n" ]

none

500

[ { "input": "5\n10 12 13 15 10", "output": "5 1" }, { "input": "4\n10 20 30 40", "output": "1 2" }, { "input": "6\n744 359 230 586 944 442", "output": "2 3" }, { "input": "5\n826 747 849 687 437", "output": "1 2" }, { "input": "5\n999 999 993 969 999", "output": "1 2" }, { "input": "5\n4 24 6 1 15", "output": "3 4" }, { "input": "2\n511 32", "output": "1 2" }, { "input": "3\n907 452 355", "output": "2 3" }, { "input": "4\n303 872 764 401", "output": "4 1" }, { "input": "10\n684 698 429 694 956 812 594 170 937 764", "output": "1 2" }, { "input": "20\n646 840 437 946 640 564 936 917 487 752 844 734 468 969 674 646 728 642 514 695", "output": "7 8" }, { "input": "30\n996 999 998 984 989 1000 996 993 1000 983 992 999 999 1000 979 992 987 1000 996 1000 1000 989 981 996 995 999 999 989 999 1000", "output": "12 13" }, { "input": "50\n93 27 28 4 5 78 59 24 19 134 31 128 118 36 90 32 32 1 44 32 33 13 31 10 12 25 38 50 25 12 4 22 28 53 48 83 4 25 57 31 71 24 8 7 28 86 23 80 101 58", "output": "16 17" }, { "input": "88\n1000 1000 1000 1000 1000 998 998 1000 1000 1000 1000 999 999 1000 1000 1000 999 1000 997 999 997 1000 999 998 1000 999 1000 1000 1000 999 1000 999 999 1000 1000 999 1000 999 1000 1000 998 1000 1000 1000 998 998 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 999 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 998 1000 1000 998 1000 999 1000 1000 1000 1000", "output": "1 2" }, { "input": "99\n4 4 21 6 5 3 13 2 6 1 3 4 1 3 1 9 11 1 6 17 4 5 20 4 1 9 5 11 3 4 14 1 3 3 1 4 3 5 27 1 1 2 10 7 11 4 19 7 11 6 11 13 3 1 10 7 2 1 16 1 9 4 29 13 2 12 14 2 21 1 9 8 26 12 12 5 2 14 7 8 8 8 9 4 12 2 6 6 7 16 8 14 2 10 20 15 3 7 4", "output": "1 2" }, { "input": "100\n713 572 318 890 577 657 646 146 373 783 392 229 455 871 20 593 573 336 26 381 280 916 907 732 820 713 111 840 570 446 184 711 481 399 788 647 492 15 40 530 549 506 719 782 126 20 778 996 712 761 9 74 812 418 488 175 103 585 900 3 604 521 109 513 145 708 990 361 682 827 791 22 596 780 596 385 450 643 158 496 876 975 319 783 654 895 891 361 397 81 682 899 347 623 809 557 435 279 513 438", "output": "86 87" }, { "input": "100\n31 75 86 68 111 27 22 22 26 30 54 163 107 75 160 122 14 23 17 26 27 20 43 58 59 71 21 148 9 32 43 91 133 286 132 70 90 156 84 14 77 93 23 18 13 72 18 131 33 28 72 175 30 86 249 20 14 208 28 57 63 199 6 10 24 30 62 267 43 479 60 28 138 1 45 3 19 47 7 166 116 117 50 140 28 14 95 85 93 43 61 15 2 70 10 51 7 95 9 25", "output": "7 8" }, { "input": "100\n896 898 967 979 973 709 961 968 806 967 896 967 826 975 936 903 986 856 851 931 852 971 786 837 949 978 686 936 952 909 965 749 908 916 943 973 983 975 939 886 964 928 960 976 907 788 994 773 949 871 947 980 945 985 726 981 887 943 907 990 931 874 840 867 948 951 961 904 888 901 976 967 994 921 828 970 972 722 755 970 860 855 914 869 714 899 969 978 898 862 642 939 904 936 819 934 884 983 955 964", "output": "1 2" }, { "input": "100\n994 927 872 970 815 986 952 996 965 1000 877 986 978 999 950 990 936 997 993 960 921 860 895 869 943 998 983 968 973 953 999 990 995 871 853 979 973 963 953 938 997 989 993 964 960 973 946 975 1000 962 920 746 989 957 904 965 920 979 966 961 1000 993 975 952 846 971 991 979 985 969 984 973 956 1000 952 778 983 974 956 927 995 997 980 997 1000 970 960 970 988 983 947 904 935 972 1000 863 992 996 932 967", "output": "81 82" }, { "input": "100\n48 108 63 21 27 8 49 21 75 8 24 42 149 18 8 28 21 18 25 35 59 70 59 33 40 1 67 34 120 82 4 115 72 87 3 15 15 63 37 12 40 27 83 14 38 20 14 58 93 10 31 3 39 6 197 77 54 16 31 146 9 49 14 8 77 82 5 11 80 116 8 61 50 24 7 103 29 11 3 3 1 12 46 24 21 131 39 29 36 2 107 40 16 99 31 41 29 48 17 17", "output": "36 37" } ]

1,679,762,525

2,147,483,647

Python 3

OK

TESTS

20

92

0

n=int(input()) x=list(map(int,input().split())) minn=abs(x[-1]-x[0]) a=n b=1 for i in range(n-1): if abs(x[i+1]-x[i])<minn: a=i+1 b=i+2 minn=abs(x[i+1]-x[i]) print(a,b)

Title: Reconnaissance 2 Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: *n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit. Input Specification: The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction. Output Specification: Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle. Demo Input: ['5\n10 12 13 15 10\n', '4\n10 20 30 40\n'] Demo Output: ['5 1\n', '1 2\n'] Note: none

```python n=int(input()) x=list(map(int,input().split())) minn=abs(x[-1]-x[0]) a=n b=1 for i in range(n-1): if abs(x[i+1]-x[i])<minn: a=i+1 b=i+2 minn=abs(x[i+1]-x[i]) print(a,b) ```

3.977

719

A

Vitya in the Countryside

PROGRAMMING

1,100

[ "implementation" ]

null

Every summer Vitya comes to visit his grandmother in the countryside. This summer, he got a huge wart. Every grandma knows that one should treat warts when the moon goes down. Thus, Vitya has to catch the moment when the moon is down. Moon cycle lasts 30 days. The size of the visible part of the moon (in Vitya's units) for each day is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and then cycle repeats, thus after the second 1 again goes 0. As there is no internet in the countryside, Vitya has been watching the moon for *n* consecutive days and for each of these days he wrote down the size of the visible part of the moon. Help him find out whether the moon will be up or down next day, or this cannot be determined by the data he has.

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=92) — the number of consecutive days Vitya was watching the size of the visible part of the moon. The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=15) — Vitya's records. It's guaranteed that the input data is consistent.

If Vitya can be sure that the size of visible part of the moon on day *n*<=+<=1 will be less than the size of the visible part on day *n*, then print "DOWN" at the only line of the output. If he might be sure that the size of the visible part will increase, then print "UP". If it's impossible to determine what exactly will happen with the moon, print -1.

[ "5\n3 4 5 6 7\n", "7\n12 13 14 15 14 13 12\n", "1\n8\n" ]

[ "UP\n", "DOWN\n", "-1\n" ]

In the first sample, the size of the moon on the next day will be equal to 8, thus the answer is "UP". In the second sample, the size of the moon on the next day will be 11, thus the answer is "DOWN". In the third sample, there is no way to determine whether the size of the moon on the next day will be 7 or 9, thus the answer is -1.

500

[ { "input": "5\n3 4 5 6 7", "output": "UP" }, { "input": "7\n12 13 14 15 14 13 12", "output": "DOWN" }, { "input": "1\n8", "output": "-1" }, { "input": "44\n7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10", "output": "DOWN" }, { "input": "92\n3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4", "output": "UP" }, { "input": "6\n10 11 12 13 14 15", "output": "DOWN" }, { "input": "27\n11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15", "output": "DOWN" }, { "input": "6\n8 7 6 5 4 3", "output": "DOWN" }, { "input": "27\n14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10", "output": "UP" }, { "input": "79\n7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5", "output": "DOWN" }, { "input": "25\n1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7", "output": "DOWN" }, { "input": "21\n3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7", "output": "DOWN" }, { "input": "56\n1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6", "output": "DOWN" }, { "input": "19\n4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14", "output": "UP" }, { "input": "79\n5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13", "output": "UP" }, { "input": "87\n14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10", "output": "UP" }, { "input": "13\n10 9 8 7 6 5 4 3 2 1 0 1 2", "output": "UP" }, { "input": "2\n8 9", "output": "UP" }, { "input": "3\n10 11 12", "output": "UP" }, { "input": "1\n1", "output": "-1" }, { "input": "1\n2", "output": "-1" }, { "input": "1\n3", "output": "-1" }, { "input": "1\n4", "output": "-1" }, { "input": "1\n5", "output": "-1" }, { "input": "1\n6", "output": "-1" }, { "input": "1\n7", "output": "-1" }, { "input": "1\n9", "output": "-1" }, { "input": "1\n10", "output": "-1" }, { "input": "1\n11", "output": "-1" }, { "input": "1\n12", "output": "-1" }, { "input": "1\n13", "output": "-1" }, { "input": "1\n14", "output": "-1" }, { "input": "1\n15", "output": "DOWN" }, { "input": "1\n0", "output": "UP" }, { "input": "3\n11 12 13", "output": "UP" }, { "input": "2\n10 9", "output": "DOWN" }, { "input": "92\n10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11", "output": "UP" }, { "input": "92\n7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6", "output": "DOWN" }, { "input": "2\n14 15", "output": "DOWN" }, { "input": "2\n1 0", "output": "UP" }, { "input": "2\n15 14", "output": "DOWN" }, { "input": "92\n7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8", "output": "UP" }, { "input": "92\n13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12", "output": "DOWN" }, { "input": "92\n4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3", "output": "DOWN" }, { "input": "92\n14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15", "output": "DOWN" }, { "input": "92\n1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0", "output": "UP" }, { "input": "2\n2 1", "output": "DOWN" }, { "input": "3\n2 1 0", "output": "UP" }, { "input": "5\n4 3 2 1 0", "output": "UP" }, { "input": "2\n5 4", "output": "DOWN" }, { "input": "4\n3 2 1 0", "output": "UP" }, { "input": "3\n13 12 11", "output": "DOWN" }, { "input": "2\n1 2", "output": "UP" }, { "input": "2\n0 1", "output": "UP" }, { "input": "2\n13 14", "output": "UP" }, { "input": "14\n13 12 11 10 9 8 7 6 5 4 3 2 1 0", "output": "UP" } ]

1,474,638,222

2,322

Python 3

CHALLENGED

CHALLENGES

6

61

0

h='' t=[] g=input() for a in input() + ' ': if 58>ord(a)>47: h+=a elif ord(a)==32: t.append(h) h='' if len(t)<1: print(-1) elif len(t)==1: if t[0]=='0' or t[0]== '15': print('UP') else: print(-1) else: s=[t[-2],t[-1]] if (s[0]<s[1] and s[1]!='15') or (s[0]=='1' and s[1]=='0'): print('UP') else: print('DOWN')

Title: Vitya in the Countryside Time Limit: None seconds Memory Limit: None megabytes Problem Description: Every summer Vitya comes to visit his grandmother in the countryside. This summer, he got a huge wart. Every grandma knows that one should treat warts when the moon goes down. Thus, Vitya has to catch the moment when the moon is down. Moon cycle lasts 30 days. The size of the visible part of the moon (in Vitya's units) for each day is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and then cycle repeats, thus after the second 1 again goes 0. As there is no internet in the countryside, Vitya has been watching the moon for *n* consecutive days and for each of these days he wrote down the size of the visible part of the moon. Help him find out whether the moon will be up or down next day, or this cannot be determined by the data he has. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=92) — the number of consecutive days Vitya was watching the size of the visible part of the moon. The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=15) — Vitya's records. It's guaranteed that the input data is consistent. Output Specification: If Vitya can be sure that the size of visible part of the moon on day *n*<=+<=1 will be less than the size of the visible part on day *n*, then print "DOWN" at the only line of the output. If he might be sure that the size of the visible part will increase, then print "UP". If it's impossible to determine what exactly will happen with the moon, print -1. Demo Input: ['5\n3 4 5 6 7\n', '7\n12 13 14 15 14 13 12\n', '1\n8\n'] Demo Output: ['UP\n', 'DOWN\n', '-1\n'] Note: In the first sample, the size of the moon on the next day will be equal to 8, thus the answer is "UP". In the second sample, the size of the moon on the next day will be 11, thus the answer is "DOWN". In the third sample, there is no way to determine whether the size of the moon on the next day will be 7 or 9, thus the answer is -1.

```python h='' t=[] g=input() for a in input() + ' ': if 58>ord(a)>47: h+=a elif ord(a)==32: t.append(h) h='' if len(t)<1: print(-1) elif len(t)==1: if t[0]=='0' or t[0]== '15': print('UP') else: print(-1) else: s=[t[-2],t[-1]] if (s[0]<s[1] and s[1]!='15') or (s[0]=='1' and s[1]=='0'): print('UP') else: print('DOWN') ```

-1

567

A

Lineland Mail

PROGRAMMING

900

[ "greedy", "implementation" ]

null

All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* — a coordinate on the *Ox* axis. No two cities are located at a single point. Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in). Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city. For each city calculate two values *min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city

The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order.

Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city.

[ "4\n-5 -2 2 7\n", "2\n-1 1\n" ]

[ "3 12\n3 9\n4 7\n5 12\n", "2 2\n2 2\n" ]

none

500

[ { "input": "4\n-5 -2 2 7", "output": "3 12\n3 9\n4 7\n5 12" }, { "input": "2\n-1 1", "output": "2 2\n2 2" }, { "input": "3\n-1 0 1", "output": "1 2\n1 1\n1 2" }, { "input": "4\n-1 0 1 3", "output": "1 4\n1 3\n1 2\n2 4" }, { "input": "3\n-1000000000 0 1000000000", "output": "1000000000 2000000000\n1000000000 1000000000\n1000000000 2000000000" }, { "input": "2\n-1000000000 1000000000", "output": "2000000000 2000000000\n2000000000 2000000000" }, { "input": "10\n1 10 12 15 59 68 130 912 1239 9123", "output": "9 9122\n2 9113\n2 9111\n3 9108\n9 9064\n9 9055\n62 8993\n327 8211\n327 7884\n7884 9122" }, { "input": "5\n-2 -1 0 1 2", "output": "1 4\n1 3\n1 2\n1 3\n1 4" }, { "input": "5\n-2 -1 0 1 3", "output": "1 5\n1 4\n1 3\n1 3\n2 5" }, { "input": "3\n-10000 1 10000", "output": "10001 20000\n9999 10001\n9999 20000" }, { "input": "5\n-1000000000 -999999999 -999999998 -999999997 -999999996", "output": "1 4\n1 3\n1 2\n1 3\n1 4" }, { "input": "10\n-857422304 -529223472 82412729 145077145 188538640 265299215 527377039 588634631 592896147 702473706", "output": "328198832 1559896010\n328198832 1231697178\n62664416 939835033\n43461495 1002499449\n43461495 1045960944\n76760575 1122721519\n61257592 1384799343\n4261516 1446056935\n4261516 1450318451\n109577559 1559896010" }, { "input": "10\n-876779400 -829849659 -781819137 -570920213 18428128 25280705 121178189 219147240 528386329 923854124", "output": "46929741 1800633524\n46929741 1753703783\n48030522 1705673261\n210898924 1494774337\n6852577 905425996\n6852577 902060105\n95897484 997957589\n97969051 1095926640\n309239089 1405165729\n395467795 1800633524" }, { "input": "30\n-15 1 21 25 30 40 59 60 77 81 97 100 103 123 139 141 157 158 173 183 200 215 226 231 244 256 267 279 289 292", "output": "16 307\n16 291\n4 271\n4 267\n5 262\n10 252\n1 233\n1 232\n4 215\n4 211\n3 195\n3 192\n3 189\n16 169\n2 154\n2 156\n1 172\n1 173\n10 188\n10 198\n15 215\n11 230\n5 241\n5 246\n12 259\n11 271\n11 282\n10 294\n3 304\n3 307" }, { "input": "10\n-1000000000 -999999999 -999999997 -999999996 -999999995 -999999994 -999999992 -999999990 -999999988 -999999986", "output": "1 14\n1 13\n1 11\n1 10\n1 9\n1 8\n2 8\n2 10\n2 12\n2 14" }, { "input": "50\n-50000 -49459 -48875 -48456 -48411 -48096 -47901 -47500 -47150 -46808 -46687 -46679 -46337 -45747 -45604 -45194 -44752 -44242 -44231 -44122 -43636 -43274 -42916 -42881 -42386 -42095 -41830 -41618 -41145 -40897 -40534 -40007 -39761 -39389 -39104 -38909 -38630 -38561 -38364 -38255 -38214 -38084 -37959 -37607 -37202 -36890 -36681 -36136 -36123 -35886", "output": "541 14114\n541 13573\n419 12989\n45 12570\n45 12525\n195 12210\n195 12015\n350 11614\n342 11264\n121 10922\n8 10801\n8 10793\n342 10451\n143 9861\n143 9718\n410 9308\n442 8866\n11 8356\n11 8345\n109 8236\n362 7750\n358 7388\n35 7084\n35 7119\n291 7614\n265 7905\n212 8170\n212 8382\n248 8855\n248 9103\n363 9466\n246 9993\n246 10239\n285 10611\n195 10896\n195 11091\n69 11370\n69 11439\n109 11636\n41 11745\n41 11786\n125 11916\n125 12041\n352 12393\n312 12798\n209 13110\n209 13319\n13 13864\n13 13877\n237 141..." }, { "input": "3\n-1000000000 999999999 1000000000", "output": "1999999999 2000000000\n1 1999999999\n1 2000000000" } ]

1,664,539,737

2,147,483,647

Python 3

OK

TESTS

54

498

7,987,200

input() pontos = list(map(int, input().split(" "))) for i in range(0, len(pontos)): if i == 0: menor = abs(pontos[i] - pontos[i + 1]) maior = abs(pontos[i] - pontos[-1]) elif i == len(pontos) - 1: menor = abs(pontos[i] - pontos[i - 1]) maior = abs(pontos[i] - pontos[0]) else: menor_a = abs(pontos[i] - pontos[i + 1]) menor_b = abs(pontos[i] - pontos[i - 1]) maior_a = abs(pontos[i] - pontos[-1]) maior_b = abs(pontos[i] - pontos[0]) menor = min(menor_a, menor_b) maior = max(maior_a, maior_b) print(str(menor) + " " + str(maior))

Title: Lineland Mail Time Limit: None seconds Memory Limit: None megabytes Problem Description: All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* — a coordinate on the *Ox* axis. No two cities are located at a single point. Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in). Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city. For each city calculate two values *min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city Input Specification: The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order. Output Specification: Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city. Demo Input: ['4\n-5 -2 2 7\n', '2\n-1 1\n'] Demo Output: ['3 12\n3 9\n4 7\n5 12\n', '2 2\n2 2\n'] Note: none

```python input() pontos = list(map(int, input().split(" "))) for i in range(0, len(pontos)): if i == 0: menor = abs(pontos[i] - pontos[i + 1]) maior = abs(pontos[i] - pontos[-1]) elif i == len(pontos) - 1: menor = abs(pontos[i] - pontos[i - 1]) maior = abs(pontos[i] - pontos[0]) else: menor_a = abs(pontos[i] - pontos[i + 1]) menor_b = abs(pontos[i] - pontos[i - 1]) maior_a = abs(pontos[i] - pontos[-1]) maior_b = abs(pontos[i] - pontos[0]) menor = min(menor_a, menor_b) maior = max(maior_a, maior_b) print(str(menor) + " " + str(maior)) ```

3

672

A

Summer Camp

PROGRAMMING

800

[ "implementation" ]

null

Every year, hundreds of people come to summer camps, they learn new algorithms and solve hard problems. This is your first year at summer camp, and you are asked to solve the following problem. All integers starting with 1 are written in one line. The prefix of these line is "123456789101112131415...". Your task is to print the *n*-th digit of this string (digits are numbered starting with 1.

The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the position of the digit you need to print.

Print the *n*-th digit of the line.

[ "3\n", "11\n" ]

[ "3\n", "0\n" ]

In the first sample the digit at position 3 is '3', as both integers 1 and 2 consist on one digit. In the second sample, the digit at position 11 is '0', it belongs to the integer 10.

500

[ { "input": "3", "output": "3" }, { "input": "11", "output": "0" }, { "input": "12", "output": "1" }, { "input": "13", "output": "1" }, { "input": "29", "output": "9" }, { "input": "30", "output": "2" }, { "input": "1000", "output": "3" }, { "input": "999", "output": "9" }, { "input": "100", "output": "5" }, { "input": "123", "output": "6" }, { "input": "8", "output": "8" }, { "input": "157", "output": "3" }, { "input": "289", "output": "1" }, { "input": "179", "output": "4" }, { "input": "942", "output": "0" }, { "input": "879", "output": "9" }, { "input": "394", "output": "1" }, { "input": "423", "output": "7" }, { "input": "952", "output": "3" }, { "input": "121", "output": "5" }, { "input": "613", "output": "2" }, { "input": "945", "output": "1" }, { "input": "270", "output": "6" }, { "input": "781", "output": "2" }, { "input": "453", "output": "7" }, { "input": "171", "output": "0" }, { "input": "643", "output": "2" }, { "input": "570", "output": "6" }, { "input": "750", "output": "6" }, { "input": "500", "output": "0" }, { "input": "2", "output": "2" }, { "input": "1", "output": "1" }, { "input": "108", "output": "5" }, { "input": "500", "output": "0" }, { "input": "189", "output": "9" }, { "input": "491", "output": "0" }, { "input": "191", "output": "0" } ]

1,575,229,096

2,147,483,647

Python 3

OK

TESTS

37

109

0

def solution672a(): n = int(input()) res = "" i = 1 while len(res) < 1000: res += str(i) i += 1 print(res[n - 1]) if __name__ == '__main__': solution672a()

Title: Summer Camp Time Limit: None seconds Memory Limit: None megabytes Problem Description: Every year, hundreds of people come to summer camps, they learn new algorithms and solve hard problems. This is your first year at summer camp, and you are asked to solve the following problem. All integers starting with 1 are written in one line. The prefix of these line is "123456789101112131415...". Your task is to print the *n*-th digit of this string (digits are numbered starting with 1. Input Specification: The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the position of the digit you need to print. Output Specification: Print the *n*-th digit of the line. Demo Input: ['3\n', '11\n'] Demo Output: ['3\n', '0\n'] Note: In the first sample the digit at position 3 is '3', as both integers 1 and 2 consist on one digit. In the second sample, the digit at position 11 is '0', it belongs to the integer 10.

```python def solution672a(): n = int(input()) res = "" i = 1 while len(res) < 1000: res += str(i) i += 1 print(res[n - 1]) if __name__ == '__main__': solution672a() ```

3

603

A

Alternative Thinking

PROGRAMMING

1,600

[ "dp", "greedy", "math" ]

null

Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length *n*. Each character of Kevin's string represents Kevin's score on one of the *n* questions of the olympiad—'1' for a correctly identified cow and '0' otherwise. However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0,<=1,<=0,<=1}, {1,<=0,<=1}, and {1,<=0,<=1,<=0} are alternating sequences, while {1,<=0,<=0} and {0,<=1,<=0,<=1,<=1} are not. Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.

The first line contains the number of questions on the olympiad *n* (1<=≤<=*n*<=≤<=100<=000). The following line contains a binary string of length *n* representing Kevin's results on the USAICO.

Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.

[ "8\n10000011\n", "2\n01\n" ]

[ "5\n", "2\n" ]

In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'. In the second sample, Kevin can flip the entire string and still have the same score.

500

[ { "input": "8\n10000011", "output": "5" }, { "input": "2\n01", "output": "2" }, { "input": "5\n10101", "output": "5" }, { "input": "75\n010101010101010101010101010101010101010101010101010101010101010101010101010", "output": "75" }, { "input": "11\n00000000000", "output": "3" }, { "input": "56\n10101011010101010101010101010101010101011010101010101010", "output": "56" }, { "input": "50\n01011010110101010101010101010101010101010101010100", "output": "49" }, { "input": "7\n0110100", "output": "7" }, { "input": "8\n11011111", "output": "5" }, { "input": "6\n000000", "output": "3" }, { "input": "5\n01000", "output": "5" }, { "input": "59\n10101010101010101010101010101010101010101010101010101010101", "output": "59" }, { "input": "88\n1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010", "output": "88" }, { "input": "93\n010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010", "output": "93" }, { "input": "70\n0101010101010101010101010101010101010101010101010101010101010101010101", "output": "70" }, { "input": "78\n010101010101010101010101010101101010101010101010101010101010101010101010101010", "output": "78" }, { "input": "83\n10101010101010101010101010101010101010101010101010110101010101010101010101010101010", "output": "83" }, { "input": "87\n101010101010101010101010101010101010101010101010101010101010101010101010101010010101010", "output": "87" }, { "input": "65\n01010101101010101010101010101010101010101010101010101010101010101", "output": "65" }, { "input": "69\n010101010101010101101010101010101010101010101010101010101010101010101", "output": "69" }, { "input": "74\n01010101010101010101010101010101010101010101010101010101010101000101010101", "output": "74" }, { "input": "77\n01010101010101001010101010101010100101010101010101010101010101010101010101010", "output": "77" }, { "input": "60\n101010110101010101010101010110101010101010101010101010101010", "output": "60" }, { "input": "89\n01010101010101010101010101010101010101010101010101010101101010101010101010100101010101010", "output": "89" }, { "input": "68\n01010101010101010101010101010101010100101010100101010101010100101010", "output": "67" }, { "input": "73\n0101010101010101010101010101010101010101010111011010101010101010101010101", "output": "72" }, { "input": "55\n1010101010101010010101010101101010101010101010100101010", "output": "54" }, { "input": "85\n1010101010101010101010101010010101010101010101101010101010101010101011010101010101010", "output": "84" }, { "input": "1\n0", "output": "1" }, { "input": "1\n1", "output": "1" }, { "input": "10\n1111111111", "output": "3" }, { "input": "2\n10", "output": "2" }, { "input": "2\n11", "output": "2" }, { "input": "2\n00", "output": "2" }, { "input": "3\n000", "output": "3" }, { "input": "3\n001", "output": "3" }, { "input": "3\n010", "output": "3" }, { "input": "3\n011", "output": "3" }, { "input": "3\n100", "output": "3" }, { "input": "3\n101", "output": "3" }, { "input": "3\n110", "output": "3" }, { "input": "3\n111", "output": "3" }, { "input": "4\n0000", "output": "3" }, { "input": "4\n0001", "output": "4" }, { "input": "4\n0010", "output": "4" }, { "input": "4\n0011", "output": "4" }, { "input": "4\n0100", "output": "4" }, { "input": "4\n0101", "output": "4" }, { "input": "4\n0110", "output": "4" }, { "input": "4\n0111", "output": "4" }, { "input": "4\n1000", "output": "4" }, { "input": "4\n1001", "output": "4" }, { "input": "4\n1010", "output": "4" }, { "input": "4\n1011", "output": "4" }, { "input": "4\n1100", "output": "4" }, { "input": "4\n1101", "output": "4" }, { "input": "4\n1110", "output": "4" }, { "input": "4\n1111", "output": "3" }, { "input": "5\n00000", "output": "3" }, { "input": "5\n00001", "output": "4" }, { "input": "5\n00010", "output": "5" }, { "input": "5\n00011", "output": "4" }, { "input": "5\n00100", "output": "5" }, { "input": "5\n00101", "output": "5" }, { "input": "5\n00110", "output": "5" }, { "input": "5\n00111", "output": "4" }, { "input": "5\n01000", "output": "5" }, { "input": "5\n01001", "output": "5" }, { "input": "5\n01010", "output": "5" }, { "input": "5\n01011", "output": "5" }, { "input": "5\n01100", "output": "5" }, { "input": "5\n01101", "output": "5" }, { "input": "5\n01110", "output": "5" }, { "input": "5\n01111", "output": "4" }, { "input": "5\n10000", "output": "4" }, { "input": "5\n10001", "output": "5" }, { "input": "5\n10010", "output": "5" }, { "input": "5\n10100", "output": "5" }, { "input": "5\n10101", "output": "5" }, { "input": "5\n10110", "output": "5" }, { "input": "5\n10111", "output": "5" }, { "input": "5\n11000", "output": "4" }, { "input": "5\n11001", "output": "5" }, { "input": "5\n11010", "output": "5" }, { "input": "5\n11011", "output": "5" }, { "input": "5\n11100", "output": "4" }, { "input": "5\n11101", "output": "5" }, { "input": "5\n11110", "output": "4" }, { "input": "5\n11111", "output": "3" } ]

1,587,078,678

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

4

139

0

from sys import stdin inp = stdin.readline t = int(inp().strip()) string = list(inp().strip()) longestSub = 1 prev = string[0] l = 0 d = 0 maxGap = -1 for i in range(1,t): if string[i] != prev: if longestSub % 2: l = i else: d = i maxGap = max(maxGap, abs(d - l)) longestSub += 1 prev = string[i] if maxGap >2: longestSub+=2 elif maxGap == 2: longestSub += 1 print(longestSub)

Title: Alternative Thinking Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length *n*. Each character of Kevin's string represents Kevin's score on one of the *n* questions of the olympiad—'1' for a correctly identified cow and '0' otherwise. However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0,<=1,<=0,<=1}, {1,<=0,<=1}, and {1,<=0,<=1,<=0} are alternating sequences, while {1,<=0,<=0} and {0,<=1,<=0,<=1,<=1} are not. Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have. Input Specification: The first line contains the number of questions on the olympiad *n* (1<=≤<=*n*<=≤<=100<=000). The following line contains a binary string of length *n* representing Kevin's results on the USAICO. Output Specification: Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring. Demo Input: ['8\n10000011\n', '2\n01\n'] Demo Output: ['5\n', '2\n'] Note: In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'. In the second sample, Kevin can flip the entire string and still have the same score.

```python from sys import stdin inp = stdin.readline t = int(inp().strip()) string = list(inp().strip()) longestSub = 1 prev = string[0] l = 0 d = 0 maxGap = -1 for i in range(1,t): if string[i] != prev: if longestSub % 2: l = i else: d = i maxGap = max(maxGap, abs(d - l)) longestSub += 1 prev = string[i] if maxGap >2: longestSub+=2 elif maxGap == 2: longestSub += 1 print(longestSub) ```

0

66

B

Petya and Countryside

PROGRAMMING

1,100

[ "brute force", "implementation" ]

B. Petya and Countryside

2

256

Little Petya often travels to his grandmother in the countryside. The grandmother has a large garden, which can be represented as a rectangle 1<=×<=*n* in size, when viewed from above. This rectangle is divided into *n* equal square sections. The garden is very unusual as each of the square sections possesses its own fixed height and due to the newest irrigation system we can create artificial rain above each section. Creating artificial rain is an expensive operation. That's why we limit ourselves to creating the artificial rain only above one section. At that, the water from each watered section will flow into its neighbouring sections if their height does not exceed the height of the section. That is, for example, the garden can be represented by a 1<=×<=5 rectangle, where the section heights are equal to 4, 2, 3, 3, 2. Then if we create an artificial rain over any of the sections with the height of 3, the water will flow over all the sections, except the ones with the height of 4. See the illustration of this example at the picture: As Petya is keen on programming, he decided to find such a section that if we create artificial rain above it, the number of watered sections will be maximal. Help him.

The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=1000). The second line contains *n* positive integers which are the height of the sections. All the numbers are no less than 1 and not more than 1000.

Print a single number, the maximal number of watered sections if we create artificial rain above exactly one section.

[ "1\n2\n", "5\n1 2 1 2 1\n", "8\n1 2 1 1 1 3 3 4\n" ]

[ "1\n", "3\n", "6\n" ]

none

1,000

[ { "input": "1\n2", "output": "1" }, { "input": "5\n1 2 1 2 1", "output": "3" }, { "input": "8\n1 2 1 1 1 3 3 4", "output": "6" }, { "input": "10\n1 2 3 4 5 6 7 8 9 10", "output": "10" }, { "input": "10\n10 9 8 7 6 5 4 3 2 1", "output": "10" }, { "input": "2\n100 100", "output": "2" }, { "input": "3\n100 100 100", "output": "3" }, { "input": "11\n1 2 3 4 5 6 5 4 3 2 1", "output": "11" }, { "input": "100\n1 2 3 4 5 6 7 8 9 10 11 100 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 1 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": "61" }, { "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 1 82 83 84 85 86 87 88 89 90 91 92 93 94 100 5 4 3 2 1", "output": "81" }, { "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 1 86 87 88 89 90 91 92 93 100 6 5 4 3 2 1", "output": "85" }, { "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 1 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 92 100 7 6 5 4 3 2 1", "output": "61" }, { "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 91 100 8 7 6 1 4 3 2 1", "output": "96" }, { "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 89 100 10 9 8 7 6 5 4 3 2 1", "output": "100" }, { "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 1 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 88 100 11 10 9 8 7 6 5 4 3 2 1", "output": "55" }, { "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 1 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 87 100 12 11 10 9 8 7 6 5 4 3 2 1", "output": "59" }, { "input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 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 86 100 13 12 11 10 9 8 7 6 5 4 3 2 1", "output": "86" }, { "input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 100 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": "83" }, { "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 36 100 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 1 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": "74" }, { "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 90 100 9 8 7 6 5 4 3 2 1", "output": "100" }, { "input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 100 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 1 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": "52" }, { "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 41 100 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 1 2 1", "output": "98" }, { "input": "10\n1 4 4 4 4 4 1 2 4 3", "output": "7" } ]

1,627,733,588

2,147,483,647

Python 3

OK

TESTS

80

748

6,963,200

n = int(input()) s = input().split() a = [0] for i in range(0, n) : a.append(int(s[i])) ans = 0 for i in range(1, n + 1) : l = r = i ln = rn = a[i] while ((l > 1) and (a[l - 1] <= ln)) : l -= 1 ln = a[l] while ((r < n) and (a[r + 1] <= rn)) : r += 1 rn = a[r] ans = max(ans, r - l + 1) print(ans)

Title: Petya and Countryside Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Little Petya often travels to his grandmother in the countryside. The grandmother has a large garden, which can be represented as a rectangle 1<=×<=*n* in size, when viewed from above. This rectangle is divided into *n* equal square sections. The garden is very unusual as each of the square sections possesses its own fixed height and due to the newest irrigation system we can create artificial rain above each section. Creating artificial rain is an expensive operation. That's why we limit ourselves to creating the artificial rain only above one section. At that, the water from each watered section will flow into its neighbouring sections if their height does not exceed the height of the section. That is, for example, the garden can be represented by a 1<=×<=5 rectangle, where the section heights are equal to 4, 2, 3, 3, 2. Then if we create an artificial rain over any of the sections with the height of 3, the water will flow over all the sections, except the ones with the height of 4. See the illustration of this example at the picture: As Petya is keen on programming, he decided to find such a section that if we create artificial rain above it, the number of watered sections will be maximal. Help him. Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=1000). The second line contains *n* positive integers which are the height of the sections. All the numbers are no less than 1 and not more than 1000. Output Specification: Print a single number, the maximal number of watered sections if we create artificial rain above exactly one section. Demo Input: ['1\n2\n', '5\n1 2 1 2 1\n', '8\n1 2 1 1 1 3 3 4\n'] Demo Output: ['1\n', '3\n', '6\n'] Note: none

```python n = int(input()) s = input().split() a = [0] for i in range(0, n) : a.append(int(s[i])) ans = 0 for i in range(1, n + 1) : l = r = i ln = rn = a[i] while ((l > 1) and (a[l - 1] <= ln)) : l -= 1 ln = a[l] while ((r < n) and (a[r + 1] <= rn)) : r += 1 rn = a[r] ans = max(ans, r - l + 1) print(ans) ```

3.80003

14

A

Letter

PROGRAMMING

800

[ "implementation" ]

A. Letter

1

64

A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with *n* rows and *m* columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.

The first line of the input data contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50), *n* — amount of lines, and *m* — amount of columns on Bob's sheet. The following *n* lines contain *m* characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.

Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.

[ "6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..\n", "3 3\n***\n*.*\n***\n" ]

[ "***\n*..\n***\n*..\n***\n", "***\n*.*\n***\n" ]

none

0

[ { "input": "6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..", "output": "***\n*..\n***\n*..\n***" }, { "input": "3 3\n***\n*.*\n***", "output": "***\n*.*\n***" }, { "input": "1 1\n*", "output": "*" }, { "input": "2 1\n*\n*", "output": "*\n*" }, { "input": "5 1\n.\n*\n.\n.\n.", "output": "*" }, { "input": "1 6\n*****.", "output": "*****" }, { "input": "2 2\n..\n*.", "output": "*" }, { "input": "8 2\n..\n..\n..\n..\n..\n*.\n..\n..", "output": "*" }, { "input": "50 1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n*\n.\n*\n.\n.\n.\n.\n.\n.\n.\n.\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": "*\n.\n*\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n*" }, { "input": "2 1\n*\n.", "output": "*" }, { "input": "5 1\n*\n*\n*\n*\n*", "output": "*\n*\n*\n*\n*" }, { "input": "2 2\n*.\n..", "output": "*" }, { "input": "3 4\n...*\n*...\n..*.", "output": "...*\n*...\n..*." }, { "input": "8 2\n**\n**\n**\n**\n**\n**\n**\n**", "output": "**\n**\n**\n**\n**\n**\n**\n**" }, { "input": "50 1\n.\n*\n*\n*\n.\n.\n*\n.\n*\n*\n*\n*\n*\n*\n*\n.\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\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": "*\n*\n*\n.\n.\n*\n.\n*\n*\n*\n*\n*\n*\n*\n.\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n.\n.\n.\n.\n*\n*\n.\n*\n*\n.\n.\n.\n.\n*\n*" } ]

1,674,724,344

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

7

92

0

n, m = map(int, input().split()) a = [] first = 52 second = 0 for _ in range(n): s = input() a.append(s) if s == "." * m: continue fi = -1 for i in range(m): if s[i] == '*': if fi == -1: fi = i else: sc = i try: first = min(first, fi) except NameError: pass try: second = max(second, sc) except NameError: pass while a[0] == "." * m: del a[0] while a[-1] == '.' * m: del a[-1] for s in a: print(s[first:second+1]) """ ////////////////////////////////////////// // // // Implemented by brownfox2k6 // // // ////////////////////////////////////////// """

Title: Letter Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with *n* rows and *m* columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides. Input Specification: The first line of the input data contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50), *n* — amount of lines, and *m* — amount of columns on Bob's sheet. The following *n* lines contain *m* characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square. Output Specification: Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better. Demo Input: ['6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..\n', '3 3\n***\n*.*\n***\n'] Demo Output: ['***\n*..\n***\n*..\n***\n', '***\n*.*\n***\n'] Note: none

```python n, m = map(int, input().split()) a = [] first = 52 second = 0 for _ in range(n): s = input() a.append(s) if s == "." * m: continue fi = -1 for i in range(m): if s[i] == '*': if fi == -1: fi = i else: sc = i try: first = min(first, fi) except NameError: pass try: second = max(second, sc) except NameError: pass while a[0] == "." * m: del a[0] while a[-1] == '.' * m: del a[-1] for s in a: print(s[first:second+1]) """ ////////////////////////////////////////// // // // Implemented by brownfox2k6 // // // ////////////////////////////////////////// """ ```

0

381

A

Sereja and Dima

PROGRAMMING

800

[ "greedy", "implementation", "two pointers" ]

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" } ]

1,684,917,093

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

10

46

0

a = int(input()) b= list(map(int, input().split(' '))) first = 0 second = 0 while b: if b[-1] > b[0]: first += b.pop(-1) else: first += b.pop(0) if b: if b[-1] > b[0]: second += b.pop(-1) else: second += b.pop(0) print(max(first, second), min(first, second))

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 a = int(input()) b= list(map(int, input().split(' '))) first = 0 second = 0 while b: if b[-1] > b[0]: first += b.pop(-1) else: first += b.pop(0) if b: if b[-1] > b[0]: second += b.pop(-1) else: second += b.pop(0) print(max(first, second), min(first, second)) ```

0

443

A

Anton and Letters

PROGRAMMING

800

[ "constructive algorithms", "implementation" ]

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" } ]

1,694,700,141

2,147,483,647

Python 3

OK

TESTS

27

46

0

x = input() x=set(x) if ' ' not in x: print(len(x)-2) else: print(len(x)-4)

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 x = input() x=set(x) if ' ' not in x: print(len(x)-2) else: print(len(x)-4) ```

3

369

C

Valera and Elections

PROGRAMMING

1,600

[ "dfs and similar", "graphs", "trees" ]

null

The city Valera lives in is going to hold elections to the city Parliament. The city has *n* districts and *n*<=-<=1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to *n*, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired. There are *n* candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to *n*, inclusive. If the candidate number *i* will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the *i*-th district to the district 1, where the city Parliament is located. Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.

The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105) — the number of districts in the city. Then *n*<=-<=1 lines follow. Each line contains the description of a city road as three positive integers *x**i*, *y**i*, *t**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, 1<=≤<=*t**i*<=≤<=2) — the districts connected by the *i*-th bidirectional road and the road type. If *t**i* equals to one, then the *i*-th road isn't the problem road; if *t**i* equals to two, then the *i*-th road is the problem road. It's guaranteed that the graph structure of the city is a tree.

In the first line print a single non-negative number *k* — the minimum size of the required subset of candidates. Then on the second line print *k* space-separated integers *a*1,<=*a*2,<=... *a**k* — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.

[ "5\n1 2 2\n2 3 2\n3 4 2\n4 5 2\n", "5\n1 2 1\n2 3 2\n2 4 1\n4 5 1\n", "5\n1 2 2\n1 3 2\n1 4 2\n1 5 2\n" ]

[ "1\n5 \n", "1\n3 \n", "4\n5 4 3 2 \n" ]

none

1,500

[ { "input": "5\n1 2 2\n2 3 2\n3 4 2\n4 5 2", "output": "1\n5 " }, { "input": "5\n1 2 1\n2 3 2\n2 4 1\n4 5 1", "output": "1\n3 " }, { "input": "5\n1 2 2\n1 3 2\n1 4 2\n1 5 2", "output": "4\n5 4 3 2 " }, { "input": "5\n1 5 1\n5 4 2\n4 3 1\n3 2 2", "output": "1\n2 " }, { "input": "2\n1 2 1", "output": "0" }, { "input": "10\n7 5 1\n2 1 2\n8 7 2\n2 4 1\n4 5 2\n9 5 1\n3 2 2\n2 10 1\n6 5 2", "output": "3\n8 6 3 " }, { "input": "2\n2 1 1", "output": "0" }, { "input": "2\n1 2 2", "output": "1\n2 " }, { "input": "5\n3 1 1\n4 5 1\n1 4 1\n1 2 1", "output": "0" }, { "input": "5\n1 3 2\n5 4 2\n2 1 2\n4 3 2", "output": "2\n5 2 " }, { "input": "10\n1 9 1\n3 2 2\n1 2 2\n4 7 2\n3 5 2\n4 3 2\n10 3 2\n7 8 2\n3 6 1", "output": "3\n8 10 5 " }, { "input": "10\n7 9 2\n2 6 2\n7 4 1\n5 4 2\n3 2 1\n8 5 2\n4 3 2\n7 10 1\n1 2 2", "output": "3\n9 8 6 " }, { "input": "10\n3 9 1\n2 10 2\n1 7 1\n3 4 1\n7 8 2\n1 2 1\n5 3 1\n5 6 2\n2 3 2", "output": "3\n6 10 8 " }, { "input": "10\n1 10 2\n10 9 2\n10 8 2\n9 7 2\n8 6 1\n7 5 1\n6 4 1\n5 3 1\n4 2 1", "output": "2\n7 8 " }, { "input": "10\n1 10 2\n10 9 2\n10 8 2\n9 7 2\n8 6 2\n7 5 2\n6 4 2\n5 3 2\n4 2 2", "output": "2\n3 2 " }, { "input": "4\n1 2 2\n2 3 1\n2 4 2", "output": "1\n4 " } ]

1,568,496,871

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

77

0

def dfs(no, veri): vis.add(no) global candidato if listaAdj[no][1] == 2 and not veri: veri = True candidato.append(no) if listaAdj[no][0] not in vis: dfs(listaAdj[no][0], veri) dados = open('Teste.txt') #nDistritos = int(input()) nDistritos = int(dados.readline()) listaAdj = [[] for e in range(nDistritos + 1)] for e in range(nDistritos - 1): #x, y, t = map(int, input().split()) x, y, t = map(int, dados.readline().split()) listaAdj[y].extend([x, t]) vis = {0, 1} candidato = [] for i in range(nDistritos, 1, -1): if i not in vis: veri = False dfs(i, veri) print(len(candidato)) print(*candidato)

Title: Valera and Elections Time Limit: None seconds Memory Limit: None megabytes Problem Description: The city Valera lives in is going to hold elections to the city Parliament. The city has *n* districts and *n*<=-<=1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to *n*, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired. There are *n* candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to *n*, inclusive. If the candidate number *i* will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the *i*-th district to the district 1, where the city Parliament is located. Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105) — the number of districts in the city. Then *n*<=-<=1 lines follow. Each line contains the description of a city road as three positive integers *x**i*, *y**i*, *t**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, 1<=≤<=*t**i*<=≤<=2) — the districts connected by the *i*-th bidirectional road and the road type. If *t**i* equals to one, then the *i*-th road isn't the problem road; if *t**i* equals to two, then the *i*-th road is the problem road. It's guaranteed that the graph structure of the city is a tree. Output Specification: In the first line print a single non-negative number *k* — the minimum size of the required subset of candidates. Then on the second line print *k* space-separated integers *a*1,<=*a*2,<=... *a**k* — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them. Demo Input: ['5\n1 2 2\n2 3 2\n3 4 2\n4 5 2\n', '5\n1 2 1\n2 3 2\n2 4 1\n4 5 1\n', '5\n1 2 2\n1 3 2\n1 4 2\n1 5 2\n'] Demo Output: ['1\n5 \n', '1\n3 \n', '4\n5 4 3 2 \n'] Note: none

```python def dfs(no, veri): vis.add(no) global candidato if listaAdj[no][1] == 2 and not veri: veri = True candidato.append(no) if listaAdj[no][0] not in vis: dfs(listaAdj[no][0], veri) dados = open('Teste.txt') #nDistritos = int(input()) nDistritos = int(dados.readline()) listaAdj = [[] for e in range(nDistritos + 1)] for e in range(nDistritos - 1): #x, y, t = map(int, input().split()) x, y, t = map(int, dados.readline().split()) listaAdj[y].extend([x, t]) vis = {0, 1} candidato = [] for i in range(nDistritos, 1, -1): if i not in vis: veri = False dfs(i, veri) print(len(candidato)) print(*candidato) ```

-1

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" } ]

1,608,558,247

2,147,483,647

PyPy 3

OK

TESTS

40

140

0

s = input() l = ['h', 'e', 'l', 'l', 'o'] i = -1 for c in l: i = s.find(c, i + 1) if i == -1: print('NO') exit() print('YES')

Title: Chat room Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. Input Specification: The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. Output Specification: If Vasya managed to say hello, print "YES", otherwise print "NO". Demo Input: ['ahhellllloou\n', 'hlelo\n'] Demo Output: ['YES\n', 'NO\n'] Note: none

```python s = input() l = ['h', 'e', 'l', 'l', 'o'] i = -1 for c in l: i = s.find(c, i + 1) if i == -1: print('NO') exit() print('YES') ```

3.93

854

A

Fraction

PROGRAMMING

800

[ "brute force", "constructive algorithms", "math" ]

null

Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1). During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation. Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.

In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.

Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.

[ "3\n", "4\n", "12\n" ]

[ "1 2\n", "1 3\n", "5 7\n" ]

none

500

[ { "input": "3", "output": "1 2" }, { "input": "4", "output": "1 3" }, { "input": "12", "output": "5 7" }, { "input": "34", "output": "15 19" }, { "input": "13", "output": "6 7" }, { "input": "11", "output": "5 6" }, { "input": "24", "output": "11 13" }, { "input": "17", "output": "8 9" }, { "input": "10", "output": "3 7" }, { "input": "69", "output": "34 35" }, { "input": "100", "output": "49 51" }, { "input": "57", "output": "28 29" }, { "input": "1000", "output": "499 501" }, { "input": "999", "output": "499 500" }, { "input": "998", "output": "497 501" }, { "input": "997", "output": "498 499" }, { "input": "996", "output": "497 499" }, { "input": "995", "output": "497 498" }, { "input": "994", "output": "495 499" }, { "input": "5", "output": "2 3" }, { "input": "6", "output": "1 5" }, { "input": "8", "output": "3 5" }, { "input": "9", "output": "4 5" }, { "input": "423", "output": "211 212" }, { "input": "876", "output": "437 439" }, { "input": "29", "output": "14 15" } ]

1,594,220,298

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

93

6,656,000

n = int(input()) if n % 2 == 0 and n < 10: print(n//2 - 1, n//2 + 1) elif n % 2 == 0 and n >= 10: print(n//2 - 2, n//2 + 2) else: print(n//2, n//2 + 1)

Title: Fraction Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1). During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation. Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem. Input Specification: In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction. Output Specification: Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum. Demo Input: ['3\n', '4\n', '12\n'] Demo Output: ['1 2\n', '1 3\n', '5 7\n'] Note: none

```python n = int(input()) if n % 2 == 0 and n < 10: print(n//2 - 1, n//2 + 1) elif n % 2 == 0 and n >= 10: print(n//2 - 2, n//2 + 2) else: print(n//2, n//2 + 1) ```

0

748

B

Santa Claus and Keyboard Check

PROGRAMMING

1,500

[ "implementation", "strings" ]

null

Santa Claus decided to disassemble his keyboard to clean it. After he returned all the keys back, he suddenly realized that some pairs of keys took each other's place! That is, Santa suspects that each key is either on its place, or on the place of another key, which is located exactly where the first key should be. In order to make sure that he's right and restore the correct order of keys, Santa typed his favorite patter looking only to his keyboard. You are given the Santa's favorite patter and the string he actually typed. Determine which pairs of keys could be mixed. Each key must occur in pairs at most once.

The input consists of only two strings *s* and *t* denoting the favorite Santa's patter and the resulting string. *s* and *t* are not empty and have the same length, which is at most 1000. Both strings consist only of lowercase English letters.

If Santa is wrong, and there is no way to divide some of keys into pairs and swap keys in each pair so that the keyboard will be fixed, print «-1» (without quotes). Otherwise, the first line of output should contain the only integer *k* (*k*<=≥<=0) — the number of pairs of keys that should be swapped. The following *k* lines should contain two space-separated letters each, denoting the keys which should be swapped. All printed letters must be distinct. If there are several possible answers, print any of them. You are free to choose the order of the pairs and the order of keys in a pair. Each letter must occur at most once. Santa considers the keyboard to be fixed if he can print his favorite patter without mistakes.

[ "helloworld\nehoolwlroz\n", "hastalavistababy\nhastalavistababy\n", "merrychristmas\nchristmasmerry\n" ]

[ "3\nh e\nl o\nd z\n", "0\n", "-1\n" ]

none

1,000

[ { "input": "helloworld\nehoolwlroz", "output": "3\nh e\nl o\nd z" }, { "input": "hastalavistababy\nhastalavistababy", "output": "0" }, { "input": "merrychristmas\nchristmasmerry", "output": "-1" }, { "input": "kusyvdgccw\nkusyvdgccw", "output": "0" }, { "input": "bbbbbabbab\naaaaabaaba", "output": "1\nb a" }, { "input": "zzzzzzzzzzzzzzzzzzzzz\nqwertyuiopasdfghjklzx", "output": "-1" }, { "input": "accdccdcdccacddbcacc\naccbccbcbccacbbdcacc", "output": "1\nd b" }, { "input": "giiibdbebjdaihdghahccdeffjhfgidfbdhjdggajfgaidadjd\ngiiibdbebjdaihdghahccdeffjhfgidfbdhjdggajfgaidadjd", "output": "0" }, { "input": "gndggadlmdefgejidmmcglbjdcmglncfmbjjndjcibnjbabfab\nfihffahlmhogfojnhmmcflkjhcmflicgmkjjihjcnkijkakgak", "output": "5\ng f\nn i\nd h\ne o\nb k" }, { "input": "ijpanyhovzwjjxsvaiyhchfaulcsdgfszjnwtoqbtaqygfmxuwvynvlhqhvmkjbooklxfhmqlqvfoxlnoclfxtbhvnkmhjcmrsdc\nijpanyhovzwjjxsvaiyhchfaulcsdgfszjnwtoqbtaqygfmxuwvynvlhqhvmkjbooklxfhmqlqvfoxlnoclfxtbhvnkmhjcmrsdc", "output": "0" }, { "input": "ab\naa", "output": "-1" }, { "input": "a\nz", "output": "1\na z" }, { "input": "zz\nzy", "output": "-1" }, { "input": "as\ndf", "output": "2\na d\ns f" }, { "input": "abc\nbca", "output": "-1" }, { "input": "rtfg\nrftg", "output": "1\nt f" }, { "input": "y\ny", "output": "0" }, { "input": "qwertyuiopasdfghjklzx\nzzzzzzzzzzzzzzzzzzzzz", "output": "-1" }, { "input": "qazwsxedcrfvtgbyhnujmik\nqwertyuiasdfghjkzxcvbnm", "output": "-1" }, { "input": "aaaaaa\nabcdef", "output": "-1" }, { "input": "qwerty\nffffff", "output": "-1" }, { "input": "dofbgdppdvmwjwtdyphhmqliydxyjfxoopxiscevowleccmhwybsxitvujkfliamvqinlrpytyaqdlbywccprukoisyaseibuqbfqjcabkieimsggsakpnqliwhehnemewhychqrfiuyaecoydnromrh\ndofbgdppdvmwjwtdyphhmqliydxyjfxoopxiscevowleccmhwybsxitvujkfliamvqinlrpytyaqdlbywccprukoisyaseibuqbfqjcabkieimsggsakpnqliwhehnemewhychqrfiuyaecoydnromrh", "output": "0" }, { "input": "acdbccddadbcbabbebbaebdcedbbcebeaccecdabadeabeecbacacdcbccedeadadedeccedecdaabcedccccbbcbcedcaccdede\ndcbaccbbdbacadaaeaadeabcebaaceaedccecbdadbedaeecadcdcbcaccebedbdbebeccebecbddacebccccaacacebcdccbebe", "output": "-1" }, { "input": "bacccbbacabbcaacbbba\nbacccbbacabbcaacbbba", "output": "0" }, { "input": "dbadbddddb\nacbacaaaac", "output": "-1" }, { "input": "dacbdbbbdd\nadbdadddaa", "output": "-1" }, { "input": "bbbbcbcbbc\ndaddbabddb", "output": "-1" }, { "input": "dddddbcdbd\nbcbbbdacdb", "output": "-1" }, { "input": "cbadcbcdaa\nabbbababbb", "output": "-1" }, { "input": "dmkgadidjgdjikgkehhfkhgkeamhdkfemikkjhhkdjfaenmkdgenijinamngjgkmgmmedfdehkhdigdnnkhmdkdindhkhndnakdgdhkdefagkedndnijekdmkdfedkhekgdkhgkimfeakdhhhgkkff\nbdenailbmnbmlcnehjjkcgnehadgickhdlecmggcimkahfdeinhflmlfadfnmncdnddhbkbhgejblnbffcgdbeilfigegfifaebnijeihkanehififlmhcbdcikhieghenbejneldkhaebjggncckk", "output": "-1" }, { "input": "acbbccabaa\nabbbbbabaa", "output": "-1" }, { "input": "ccccaccccc\naaaabaaaac", "output": "-1" }, { "input": "acbacacbbb\nacbacacbbb", "output": "0" }, { "input": "abbababbcc\nccccccccbb", "output": "-1" }, { "input": "jbcbbjiifdcbeajgdeabddbfcecafejddcigfcaedbgicjihifgbahjihcjefgabgbccdiibfjgacehbbdjceacdbdeaiibaicih\nhhihhhddcfihddhjfddhffhcididcdhffidjciddfhjdihdhdcjhdhhdhihdcjdhjhiifddhchjdidhhhfhiddifhfddddhddidh", "output": "-1" }, { "input": "ahaeheedefeehahfefhjhhedheeeedhehhfhdejdhffhhejhhhejadhefhahhadjjhdhheeeehfdaffhhefehhhefhhhhehehjda\neiefbdfgdhffieihfhjajifgjddffgifjbhigfagjhhjicaijbdaegidhiejiegaabgjidcfcjhgehhjjchcbjjdhjbiidjdjage", "output": "-1" }, { "input": "fficficbidbcbfaddifbffdbbiaccbbciiaidbcbbiadcccbccbbaibabcbbdbcibcciibiccfifbiiicadibbiaafadacdficbc\nddjhdghbgcbhadeccjdbddcbfjeiiaaigjejcaiabgechiiahibfejbeahafcfhjbihgjfgihdgdagjjhecjafjeedecehcdjhai", "output": "-1" }, { "input": "z\nz", "output": "0" }, { "input": "a\nz", "output": "1\na z" }, { "input": "z\na", "output": "1\nz a" }, { "input": "aa\nzz", "output": "1\na z" }, { "input": "az\nza", "output": "1\na z" }, { "input": "aa\nza", "output": "-1" }, { "input": "za\nzz", "output": "-1" }, { "input": "aa\nab", "output": "-1" }, { "input": "hehe\nheeh", "output": "-1" }, { "input": "bd\ncc", "output": "-1" }, { "input": "he\nhh", "output": "-1" }, { "input": "hee\nheh", "output": "-1" }, { "input": "aa\nac", "output": "-1" }, { "input": "ab\naa", "output": "-1" }, { "input": "hello\nehlol", "output": "-1" }, { "input": "ac\naa", "output": "-1" }, { "input": "aaabbb\nbbbaab", "output": "-1" }, { "input": "aa\nfa", "output": "-1" }, { "input": "hg\nee", "output": "-1" }, { "input": "helloworld\nehoolwlrow", "output": "-1" }, { "input": "abb\nbab", "output": "-1" }, { "input": "aaa\naae", "output": "-1" }, { "input": "aba\nbaa", "output": "-1" }, { "input": "aa\nba", "output": "-1" }, { "input": "da\naa", "output": "-1" }, { "input": "aaa\naab", "output": "-1" }, { "input": "xy\nzz", "output": "-1" } ]

1,665,779,340

2,147,483,647

PyPy 3-64

OK

TESTS

86

77

0

from collections import defaultdict as dc check=dc(lambda:set()) mp=dict() s1=input() s2=input() flag=0 for i in range(len(s1)): check[s1[i]].add(s2[i]) check[s2[i]].add(s1[i]) if s1[i]==s2[i]:continue if s2[i] in mp:continue else:mp[s1[i]]=s2[i] for i in check: if len(check[i])>1:flag=-1 if flag:print(-1) else: print(len(mp)) for i in mp:print(i,mp[i])

Title: Santa Claus and Keyboard Check Time Limit: None seconds Memory Limit: None megabytes Problem Description: Santa Claus decided to disassemble his keyboard to clean it. After he returned all the keys back, he suddenly realized that some pairs of keys took each other's place! That is, Santa suspects that each key is either on its place, or on the place of another key, which is located exactly where the first key should be. In order to make sure that he's right and restore the correct order of keys, Santa typed his favorite patter looking only to his keyboard. You are given the Santa's favorite patter and the string he actually typed. Determine which pairs of keys could be mixed. Each key must occur in pairs at most once. Input Specification: The input consists of only two strings *s* and *t* denoting the favorite Santa's patter and the resulting string. *s* and *t* are not empty and have the same length, which is at most 1000. Both strings consist only of lowercase English letters. Output Specification: If Santa is wrong, and there is no way to divide some of keys into pairs and swap keys in each pair so that the keyboard will be fixed, print «-1» (without quotes). Otherwise, the first line of output should contain the only integer *k* (*k*<=≥<=0) — the number of pairs of keys that should be swapped. The following *k* lines should contain two space-separated letters each, denoting the keys which should be swapped. All printed letters must be distinct. If there are several possible answers, print any of them. You are free to choose the order of the pairs and the order of keys in a pair. Each letter must occur at most once. Santa considers the keyboard to be fixed if he can print his favorite patter without mistakes. Demo Input: ['helloworld\nehoolwlroz\n', 'hastalavistababy\nhastalavistababy\n', 'merrychristmas\nchristmasmerry\n'] Demo Output: ['3\nh e\nl o\nd z\n', '0\n', '-1\n'] Note: none

```python from collections import defaultdict as dc check=dc(lambda:set()) mp=dict() s1=input() s2=input() flag=0 for i in range(len(s1)): check[s1[i]].add(s2[i]) check[s2[i]].add(s1[i]) if s1[i]==s2[i]:continue if s2[i] in mp:continue else:mp[s1[i]]=s2[i] for i in check: if len(check[i])>1:flag=-1 if flag:print(-1) else: print(len(mp)) for i in mp:print(i,mp[i]) ```

3

593

A

2Char

PROGRAMMING

1,200

[ "brute force", "implementation" ]

null

Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters. Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.

The first line of the input contains number *n* (1<=≤<=*n*<=≤<=100) — the number of words in the article chosen by Andrew. Following are *n* lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.

Print a single integer — the maximum possible total length of words in Andrew's article.

[ "4\nabb\ncacc\naaa\nbbb\n", "5\na\na\nbcbcb\ncdecdecdecdecdecde\naaaa\n" ]

[ "9", "6" ]

In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}. In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.

250

[ { "input": "4\nabb\ncacc\naaa\nbbb", "output": "9" }, { "input": "5\na\na\nbcbcb\ncdecdecdecdecdecde\naaaa", "output": "6" }, { "input": "1\na", "output": "1" }, { "input": "2\nz\nz", "output": "2" }, { "input": "5\nabcde\nfghij\nklmno\npqrst\nuvwxy", "output": "0" }, { "input": "6\ngggggg\ngggggg\ngggggg\ngggggg\ngggggg\ngggggg", "output": "36" }, { "input": "6\naaaaaa\naaaaaa\nbbbbbb\nbbbbbb\naaabbb\nababab", "output": "36" }, { "input": "1\nabc", "output": "0" }, { "input": "2\nabc\nbca", "output": "0" }, { "input": "3\nab\nba\nzzz", "output": "4" }, { "input": "3\nab\nba\nzzzzz", "output": "5" }, { "input": "5\nzzz\nzzzz\nzz\nz\naaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "37" }, { "input": "26\nq\nw\ne\nr\nt\ny\nu\ni\no\np\na\ns\nd\nf\ng\nh\nj\nk\nl\nz\nx\nc\nv\nb\nn\nm", "output": "2" }, { "input": "5\nzzz\nzzzz\nzz\nz\naaaaaaaaaaaaaaaaaaaaaaaaaaaf", "output": "28" }, { "input": "7\npavel\nerika\nalexxxxxxx\ngracio\nzhenya\nsudarev\nchelyaba", "output": "0" }, { "input": "31\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml\nfml", "output": "0" }, { "input": "5\nzloyfreid\ngraciocode\nschooldiary\nkazakov\nevgesha", "output": "0" }, { "input": "4\nurkop\nvisualac\ngnutl\nwtf", "output": "0" }, { "input": "3\naa\nb\nccc", "output": "5" }, { "input": "3\na\nbd\ncaaaaaaa", "output": "9" }, { "input": "4\naa\nax\nay\nxxxx", "output": "8" }, { "input": "5\nc\nbb\ne\ndd\nf", "output": "4" }, { "input": "2\naaaaa\naaaaa", "output": "10" } ]

1,681,909,847

2,147,483,647

PyPy 3-64

OK

TESTS

49

108

0

dc, ma, ones = dict(), 0, [] for _ in range(int(input())): word = input() s = set(list(word)) if len(s) <= 2: L = list(s) L.sort() k = ''.join(L) if k in dc: dc[k] += len(word) else: dc[k] = len(word) for k, v in dc.items(): if len(k) == 2: tmp = v if k[0] in dc: tmp += dc[k[0]] if k[1] in dc: tmp += dc[k[1]] ma = max(ma, tmp) else: ones.append(v) ones.sort(reverse=True) print(max(ma, sum(ones[0:2])))

Title: 2Char Time Limit: None seconds Memory Limit: None megabytes Problem Description: Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters. Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length. Input Specification: The first line of the input contains number *n* (1<=≤<=*n*<=≤<=100) — the number of words in the article chosen by Andrew. Following are *n* lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input. Output Specification: Print a single integer — the maximum possible total length of words in Andrew's article. Demo Input: ['4\nabb\ncacc\naaa\nbbb\n', '5\na\na\nbcbcb\ncdecdecdecdecdecde\naaaa\n'] Demo Output: ['9', '6'] Note: In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}. In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.

```python dc, ma, ones = dict(), 0, [] for _ in range(int(input())): word = input() s = set(list(word)) if len(s) <= 2: L = list(s) L.sort() k = ''.join(L) if k in dc: dc[k] += len(word) else: dc[k] = len(word) for k, v in dc.items(): if len(k) == 2: tmp = v if k[0] in dc: tmp += dc[k[0]] if k[1] in dc: tmp += dc[k[1]] ma = max(ma, tmp) else: ones.append(v) ones.sort(reverse=True) print(max(ma, sum(ones[0:2]))) ```

3

27

D

Ring Road 2

PROGRAMMING

2,200

[ "2-sat", "dfs and similar", "dsu", "graphs" ]

D. Ring Road 2

2

256

It is well known that Berland has *n* cities, which form the Silver ring — cities *i* and *i*<=+<=1 (1<=≤<=*i*<=<<=*n*) are connected by a road, as well as the cities *n* and 1. The goverment have decided to build *m* new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road). Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?

The first line contains two integers *n* and *m* (4<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100). Each of the following *m* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.

If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print *m* characters. *i*-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.

[ "4 2\n1 3\n2 4\n", "6 3\n1 3\n3 5\n5 1\n" ]

[ "io\n", "ooo\n" ]

none

2,000

[ { "input": "4 1\n4 2", "output": "o" }, { "input": "4 2\n1 3\n2 4", "output": "io" }, { "input": "5 1\n3 5", "output": "o" }, { "input": "5 2\n2 4\n4 1", "output": "oo" }, { "input": "5 3\n4 2\n1 3\n5 2", "output": "oio" }, { "input": "5 4\n1 3\n3 5\n1 4\n2 4", "output": "iioo" }, { "input": "6 1\n6 2", "output": "o" }, { "input": "6 2\n3 5\n2 4", "output": "oi" }, { "input": "6 3\n5 1\n4 6\n3 1", "output": "oio" }, { "input": "6 4\n6 3\n1 3\n6 4\n5 3", "output": "oooi" }, { "input": "6 5\n5 3\n4 1\n2 6\n5 1\n5 2", "output": "ioioi" }, { "input": "6 6\n1 4\n4 6\n2 6\n2 4\n5 3\n6 3", "output": "iioioo" }, { "input": "7 1\n4 7", "output": "o" }, { "input": "7 3\n2 7\n1 5\n5 7", "output": "oio" }, { "input": "7 5\n3 7\n5 7\n6 4\n4 1\n3 1", "output": "iiooo" }, { "input": "7 8\n4 7\n3 5\n7 2\n2 4\n4 6\n1 5\n1 3\n7 5", "output": "oioooiii" }, { "input": "5 4\n2 4\n2 5\n3 1\n5 3", "output": "iioo" }, { "input": "10 4\n6 9\n10 2\n4 8\n2 9", "output": "ooio" }, { "input": "17 10\n4 11\n7 16\n8 5\n12 15\n7 4\n1 12\n11 1\n15 1\n7 14\n2 9", "output": "Impossible" }, { "input": "100 50\n79 24\n65 69\n44 6\n49 53\n26 22\n24 76\n10 2\n19 44\n44 10\n87 82\n91 87\n23 92\n69 66\n95 30\n90 24\n95 20\n76 36\n82 24\n63 80\n70 72\n34 76\n42 40\n17 100\n66 64\n85 96\n34 30\n36 69\n45 56\n36 62\n26 38\n31 28\n79 64\n78 69\n41 62\n36 40\n64 67\n21 13\n90 88\n28 30\n59 6\n59 56\n63 1\n24 91\n12 19\n100 80\n98 95\n75 78\n54 44\n81 97\n20 38", "output": "Impossible" }, { "input": "20 14\n18 3\n15 12\n8 15\n17 7\n16 14\n15 7\n18 5\n6 17\n4 18\n4 2\n20 2\n9 19\n20 7\n16 18", "output": "ooooiooooioiii" }, { "input": "10 7\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9", "output": "ooooooo" }, { "input": "6 3\n1 4\n2 5\n3 6", "output": "Impossible" }, { "input": "7 9\n4 7\n3 5\n7 2\n2 4\n4 6\n1 5\n1 3\n7 5\n2 5", "output": "Impossible" } ]

1,602,387,283

2,683

PyPy 3

WRONG_ANSWER

TESTS

9

280

0

import sys from array import array # noqa: F401 def input(): return sys.stdin.buffer.readline().decode('utf-8') n, m = map(int, input().split()) edges = [tuple(map(lambda x: int(x) - 1, input().split())) for _ in range(m)] ans = ['i'] * m for i, (u, v) in enumerate(edges): in_cross = out_cross = 0 for j in range(i): if u in edges[j] or v in edges[j]: continue if (1 if u <= edges[j][0] <= v else 0) + (1 if u <= edges[j][1] <= v else 0) == 1: if ans[j] == 'i': in_cross += 1 else: out_cross += 1 if in_cross == 0: continue elif out_cross == 0: ans[i] = 'o' else: print('Impossible') print(*ans, sep='')

Title: Ring Road 2 Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: It is well known that Berland has *n* cities, which form the Silver ring — cities *i* and *i*<=+<=1 (1<=≤<=*i*<=<<=*n*) are connected by a road, as well as the cities *n* and 1. The goverment have decided to build *m* new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road). Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside? Input Specification: The first line contains two integers *n* and *m* (4<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100). Each of the following *m* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring. Output Specification: If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print *m* characters. *i*-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them. Demo Input: ['4 2\n1 3\n2 4\n', '6 3\n1 3\n3 5\n5 1\n'] Demo Output: ['io\n', 'ooo\n'] Note: none

```python import sys from array import array # noqa: F401 def input(): return sys.stdin.buffer.readline().decode('utf-8') n, m = map(int, input().split()) edges = [tuple(map(lambda x: int(x) - 1, input().split())) for _ in range(m)] ans = ['i'] * m for i, (u, v) in enumerate(edges): in_cross = out_cross = 0 for j in range(i): if u in edges[j] or v in edges[j]: continue if (1 if u <= edges[j][0] <= v else 0) + (1 if u <= edges[j][1] <= v else 0) == 1: if ans[j] == 'i': in_cross += 1 else: out_cross += 1 if in_cross == 0: continue elif out_cross == 0: ans[i] = 'o' else: print('Impossible') print(*ans, sep='') ```

0

603

A

Alternative Thinking

PROGRAMMING

1,600

[ "dp", "greedy", "math" ]

null

Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length *n*. Each character of Kevin's string represents Kevin's score on one of the *n* questions of the olympiad—'1' for a correctly identified cow and '0' otherwise. However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0,<=1,<=0,<=1}, {1,<=0,<=1}, and {1,<=0,<=1,<=0} are alternating sequences, while {1,<=0,<=0} and {0,<=1,<=0,<=1,<=1} are not. Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.

The first line contains the number of questions on the olympiad *n* (1<=≤<=*n*<=≤<=100<=000). The following line contains a binary string of length *n* representing Kevin's results on the USAICO.

Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.

[ "8\n10000011\n", "2\n01\n" ]

[ "5\n", "2\n" ]

In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'. In the second sample, Kevin can flip the entire string and still have the same score.

500

[ { "input": "8\n10000011", "output": "5" }, { "input": "2\n01", "output": "2" }, { "input": "5\n10101", "output": "5" }, { "input": "75\n010101010101010101010101010101010101010101010101010101010101010101010101010", "output": "75" }, { "input": "11\n00000000000", "output": "3" }, { "input": "56\n10101011010101010101010101010101010101011010101010101010", "output": "56" }, { "input": "50\n01011010110101010101010101010101010101010101010100", "output": "49" }, { "input": "7\n0110100", "output": "7" }, { "input": "8\n11011111", "output": "5" }, { "input": "6\n000000", "output": "3" }, { "input": "5\n01000", "output": "5" }, { "input": "59\n10101010101010101010101010101010101010101010101010101010101", "output": "59" }, { "input": "88\n1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010", "output": "88" }, { "input": "93\n010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010", "output": "93" }, { "input": "70\n0101010101010101010101010101010101010101010101010101010101010101010101", "output": "70" }, { "input": "78\n010101010101010101010101010101101010101010101010101010101010101010101010101010", "output": "78" }, { "input": "83\n10101010101010101010101010101010101010101010101010110101010101010101010101010101010", "output": "83" }, { "input": "87\n101010101010101010101010101010101010101010101010101010101010101010101010101010010101010", "output": "87" }, { "input": "65\n01010101101010101010101010101010101010101010101010101010101010101", "output": "65" }, { "input": "69\n010101010101010101101010101010101010101010101010101010101010101010101", "output": "69" }, { "input": "74\n01010101010101010101010101010101010101010101010101010101010101000101010101", "output": "74" }, { "input": "77\n01010101010101001010101010101010100101010101010101010101010101010101010101010", "output": "77" }, { "input": "60\n101010110101010101010101010110101010101010101010101010101010", "output": "60" }, { "input": "89\n01010101010101010101010101010101010101010101010101010101101010101010101010100101010101010", "output": "89" }, { "input": "68\n01010101010101010101010101010101010100101010100101010101010100101010", "output": "67" }, { "input": "73\n0101010101010101010101010101010101010101010111011010101010101010101010101", "output": "72" }, { "input": "55\n1010101010101010010101010101101010101010101010100101010", "output": "54" }, { "input": "85\n1010101010101010101010101010010101010101010101101010101010101010101011010101010101010", "output": "84" }, { "input": "1\n0", "output": "1" }, { "input": "1\n1", "output": "1" }, { "input": "10\n1111111111", "output": "3" }, { "input": "2\n10", "output": "2" }, { "input": "2\n11", "output": "2" }, { "input": "2\n00", "output": "2" }, { "input": "3\n000", "output": "3" }, { "input": "3\n001", "output": "3" }, { "input": "3\n010", "output": "3" }, { "input": "3\n011", "output": "3" }, { "input": "3\n100", "output": "3" }, { "input": "3\n101", "output": "3" }, { "input": "3\n110", "output": "3" }, { "input": "3\n111", "output": "3" }, { "input": "4\n0000", "output": "3" }, { "input": "4\n0001", "output": "4" }, { "input": "4\n0010", "output": "4" }, { "input": "4\n0011", "output": "4" }, { "input": "4\n0100", "output": "4" }, { "input": "4\n0101", "output": "4" }, { "input": "4\n0110", "output": "4" }, { "input": "4\n0111", "output": "4" }, { "input": "4\n1000", "output": "4" }, { "input": "4\n1001", "output": "4" }, { "input": "4\n1010", "output": "4" }, { "input": "4\n1011", "output": "4" }, { "input": "4\n1100", "output": "4" }, { "input": "4\n1101", "output": "4" }, { "input": "4\n1110", "output": "4" }, { "input": "4\n1111", "output": "3" }, { "input": "5\n00000", "output": "3" }, { "input": "5\n00001", "output": "4" }, { "input": "5\n00010", "output": "5" }, { "input": "5\n00011", "output": "4" }, { "input": "5\n00100", "output": "5" }, { "input": "5\n00101", "output": "5" }, { "input": "5\n00110", "output": "5" }, { "input": "5\n00111", "output": "4" }, { "input": "5\n01000", "output": "5" }, { "input": "5\n01001", "output": "5" }, { "input": "5\n01010", "output": "5" }, { "input": "5\n01011", "output": "5" }, { "input": "5\n01100", "output": "5" }, { "input": "5\n01101", "output": "5" }, { "input": "5\n01110", "output": "5" }, { "input": "5\n01111", "output": "4" }, { "input": "5\n10000", "output": "4" }, { "input": "5\n10001", "output": "5" }, { "input": "5\n10010", "output": "5" }, { "input": "5\n10100", "output": "5" }, { "input": "5\n10101", "output": "5" }, { "input": "5\n10110", "output": "5" }, { "input": "5\n10111", "output": "5" }, { "input": "5\n11000", "output": "4" }, { "input": "5\n11001", "output": "5" }, { "input": "5\n11010", "output": "5" }, { "input": "5\n11011", "output": "5" }, { "input": "5\n11100", "output": "4" }, { "input": "5\n11101", "output": "5" }, { "input": "5\n11110", "output": "4" }, { "input": "5\n11111", "output": "3" } ]

1,593,724,114

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

5

124

0

a=int(input()) s=input() ans=[] count=1 for i in range(1,len(s)): if(s[i]==s[i-1]): count+=1 else: ans.append(count) count=1 ans.append(count) if(max(ans)==1): print(len(ans)) elif(max(ans)==2): if(ans[-1]==2 or ans[0]==2): print(len(ans)+1) else: print(len(ans)+2)

Title: Alternative Thinking Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length *n*. Each character of Kevin's string represents Kevin's score on one of the *n* questions of the olympiad—'1' for a correctly identified cow and '0' otherwise. However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0,<=1,<=0,<=1}, {1,<=0,<=1}, and {1,<=0,<=1,<=0} are alternating sequences, while {1,<=0,<=0} and {0,<=1,<=0,<=1,<=1} are not. Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have. Input Specification: The first line contains the number of questions on the olympiad *n* (1<=≤<=*n*<=≤<=100<=000). The following line contains a binary string of length *n* representing Kevin's results on the USAICO. Output Specification: Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring. Demo Input: ['8\n10000011\n', '2\n01\n'] Demo Output: ['5\n', '2\n'] Note: In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'. In the second sample, Kevin can flip the entire string and still have the same score.

```python a=int(input()) s=input() ans=[] count=1 for i in range(1,len(s)): if(s[i]==s[i-1]): count+=1 else: ans.append(count) count=1 ans.append(count) if(max(ans)==1): print(len(ans)) elif(max(ans)==2): if(ans[-1]==2 or ans[0]==2): print(len(ans)+1) else: print(len(ans)+2) ```

0

911

E

Stack Sorting

PROGRAMMING

2,000

[ "constructive algorithms", "data structures", "greedy", "implementation" ]

null

Let's suppose you have an array *a*, a stack *s* (initially empty) and an array *b* (also initially empty). You may perform the following operations until both *a* and *s* are empty: - Take the first element of *a*, push it into *s* and remove it from *a* (if *a* is not empty); - Take the top element from *s*, append it to the end of array *b* and remove it from *s* (if *s* is not empty). You can perform these operations in arbitrary order. If there exists a way to perform the operations such that array *b* is sorted in non-descending order in the end, then array *a* is called stack-sortable. For example, [3,<=1,<=2] is stack-sortable, because *b* will be sorted if we perform the following operations: 1. Remove 3 from *a* and push it into *s*; 1. Remove 1 from *a* and push it into *s*; 1. Remove 1 from *s* and append it to the end of *b*; 1. Remove 2 from *a* and push it into *s*; 1. Remove 2 from *s* and append it to the end of *b*; 1. Remove 3 from *s* and append it to the end of *b*. After all these operations *b*<==<=[1,<=2,<=3], so [3,<=1,<=2] is stack-sortable. [2,<=3,<=1] is not stack-sortable. You are given *k* first elements of some permutation *p* of size *n* (recall that a permutation of size *n* is an array of size *n* where each integer from 1 to *n* occurs exactly once). You have to restore the remaining *n*<=-<=*k* elements of this permutation so it is stack-sortable. If there are multiple answers, choose the answer such that *p* is lexicographically maximal (an array *q* is lexicographically greater than an array *p* iff there exists some integer *k* such that for every *i*<=<<=*k* *q**i*<==<=*p**i*, and *q**k*<=><=*p**k*). You may not swap or change any of first *k* elements of the permutation. Print the lexicographically maximal permutation *p* you can obtain. If there exists no answer then output -1.

The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=200000, 1<=≤<=*k*<=<<=*n*) — the size of a desired permutation, and the number of elements you are given, respectively. The second line contains *k* integers *p*1, *p*2, ..., *p**k* (1<=≤<=*p**i*<=≤<=*n*) — the first *k* elements of *p*. These integers are pairwise distinct.

If it is possible to restore a stack-sortable permutation *p* of size *n* such that the first *k* elements of *p* are equal to elements given in the input, print lexicographically maximal such permutation. Otherwise print -1.

[ "5 3\n3 2 1\n", "5 3\n2 3 1\n", "5 1\n3\n", "5 2\n3 4\n" ]

[ "3 2 1 5 4 ", "-1\n", "3 2 1 5 4 ", "-1\n" ]

none

0

[ { "input": "5 3\n3 2 1", "output": "3 2 1 5 4 " }, { "input": "5 3\n2 3 1", "output": "-1" }, { "input": "5 1\n3", "output": "3 2 1 5 4 " }, { "input": "5 2\n3 4", "output": "-1" }, { "input": "100000 1\n98419", "output": "98419 98418 98417 98416 98415 98414 98413 98412 98411 98410 98409 98408 98407 98406 98405 98404 98403 98402 98401 98400 98399 98398 98397 98396 98395 98394 98393 98392 98391 98390 98389 98388 98387 98386 98385 98384 98383 98382 98381 98380 98379 98378 98377 98376 98375 98374 98373 98372 98371 98370 98369 98368 98367 98366 98365 98364 98363 98362 98361 98360 98359 98358 98357 98356 98355 98354 98353 98352 98351 98350 98349 98348 98347 98346 98345 98344 98343 98342 98341 98340 98339 98338 98337 98336 98335 9..." }, { "input": "20 19\n2 18 19 11 9 20 15 1 8 14 4 6 5 12 17 16 7 13 3", "output": "-1" }, { "input": "10 1\n6", "output": "6 5 4 3 2 1 10 9 8 7 " }, { "input": "100000 2\n64398 63673", "output": "64398 63673 63672 63671 63670 63669 63668 63667 63666 63665 63664 63663 63662 63661 63660 63659 63658 63657 63656 63655 63654 63653 63652 63651 63650 63649 63648 63647 63646 63645 63644 63643 63642 63641 63640 63639 63638 63637 63636 63635 63634 63633 63632 63631 63630 63629 63628 63627 63626 63625 63624 63623 63622 63621 63620 63619 63618 63617 63616 63615 63614 63613 63612 63611 63610 63609 63608 63607 63606 63605 63604 63603 63602 63601 63600 63599 63598 63597 63596 63595 63594 63593 63592 63591 63590 6..." }, { "input": "20 18\n8 14 18 10 1 3 7 15 2 12 17 19 5 4 11 13 20 16", "output": "-1" }, { "input": "10 2\n3 7", "output": "-1" }, { "input": "100000 3\n43791 91790 34124", "output": "-1" }, { "input": "20 17\n9 11 19 4 8 16 13 3 1 6 18 2 20 10 17 7 5", "output": "-1" }, { "input": "10 3\n2 10 3", "output": "-1" }, { "input": "100000 4\n8269 53984 47865 42245", "output": "-1" }, { "input": "20 16\n8 1 5 11 15 14 7 20 16 9 12 13 18 4 6 10", "output": "-1" }, { "input": "10 4\n2 4 1 10", "output": "-1" }, { "input": "100000 5\n82211 48488 99853 11566 42120", "output": "-1" }, { "input": "20 15\n6 7 14 13 8 4 15 2 11 9 12 16 5 1 20", "output": "-1" }, { "input": "10 5\n2 10 5 8 4", "output": "-1" }, { "input": "100000 6\n98217 55264 24242 71840 2627 67839", "output": "-1" }, { "input": "20 14\n10 15 4 3 1 5 11 12 13 14 6 2 19 20", "output": "-1" }, { "input": "10 6\n4 5 2 1 6 3", "output": "-1" }, { "input": "100000 7\n44943 51099 61988 40497 85738 74092 2771", "output": "-1" }, { "input": "20 13\n6 16 5 19 8 1 4 18 2 20 10 11 13", "output": "-1" }, { "input": "10 7\n10 4 3 8 2 5 6", "output": "-1" }, { "input": "100000 8\n88153 88461 80211 24770 13872 57414 32941 63030", "output": "-1" }, { "input": "20 12\n20 11 14 7 16 13 9 1 4 18 6 12", "output": "-1" }, { "input": "10 8\n7 9 3 6 2 4 1 8", "output": "-1" }, { "input": "200000 1\n177300", "output": "177300 177299 177298 177297 177296 177295 177294 177293 177292 177291 177290 177289 177288 177287 177286 177285 177284 177283 177282 177281 177280 177279 177278 177277 177276 177275 177274 177273 177272 177271 177270 177269 177268 177267 177266 177265 177264 177263 177262 177261 177260 177259 177258 177257 177256 177255 177254 177253 177252 177251 177250 177249 177248 177247 177246 177245 177244 177243 177242 177241 177240 177239 177238 177237 177236 177235 177234 177233 177232 177231 177230 177229 177228 ..." }, { "input": "40 39\n25 4 26 34 35 11 22 23 21 2 1 28 20 8 36 5 27 15 39 7 24 14 17 19 33 6 38 16 18 3 32 10 30 13 37 31 29 9 12", "output": "-1" }, { "input": "20 1\n20", "output": "20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 " }, { "input": "200000 2\n102991 8046", "output": "102991 8046 8045 8044 8043 8042 8041 8040 8039 8038 8037 8036 8035 8034 8033 8032 8031 8030 8029 8028 8027 8026 8025 8024 8023 8022 8021 8020 8019 8018 8017 8016 8015 8014 8013 8012 8011 8010 8009 8008 8007 8006 8005 8004 8003 8002 8001 8000 7999 7998 7997 7996 7995 7994 7993 7992 7991 7990 7989 7988 7987 7986 7985 7984 7983 7982 7981 7980 7979 7978 7977 7976 7975 7974 7973 7972 7971 7970 7969 7968 7967 7966 7965 7964 7963 7962 7961 7960 7959 7958 7957 7956 7955 7954 7953 7952 7951 7950 7949 7948 7947 7946..." }, { "input": "40 38\n32 35 36 4 22 6 15 21 40 13 33 17 5 24 28 9 1 23 25 14 26 3 8 11 37 30 18 16 19 20 27 12 39 2 10 38 29 31", "output": "-1" }, { "input": "20 2\n1 13", "output": "1 13 12 11 10 9 8 7 6 5 4 3 2 20 19 18 17 16 15 14 " }, { "input": "200000 3\n60323 163214 48453", "output": "-1" }, { "input": "40 37\n26 16 40 10 9 30 8 33 39 19 4 11 2 3 38 21 22 12 1 27 20 37 24 17 23 14 13 29 7 28 34 31 25 35 6 32 5", "output": "-1" }, { "input": "20 3\n16 6 14", "output": "-1" }, { "input": "200000 4\n194118 175603 110154 129526", "output": "-1" }, { "input": "40 36\n27 33 34 40 16 39 1 10 9 12 8 37 17 7 24 30 2 31 13 23 20 18 29 21 4 28 25 35 6 22 36 15 3 11 5 26", "output": "-1" }, { "input": "20 4\n2 10 4 9", "output": "-1" }, { "input": "200000 5\n53765 19781 63409 69811 120021", "output": "-1" }, { "input": "40 35\n2 1 5 3 11 32 13 16 37 26 6 10 8 35 25 24 7 38 21 17 40 14 9 34 33 20 29 12 22 28 36 31 30 19 27", "output": "-1" }, { "input": "20 5\n11 19 6 2 12", "output": "-1" }, { "input": "200000 6\n33936 11771 42964 153325 684 8678", "output": "-1" }, { "input": "40 34\n35 31 38 25 29 9 32 23 24 16 3 26 39 2 17 28 14 1 30 34 5 36 33 7 22 13 21 12 27 19 40 10 18 15", "output": "-1" }, { "input": "20 6\n3 6 9 13 20 14", "output": "-1" }, { "input": "200000 7\n175932 99083 128533 75304 164663 7578 174396", "output": "-1" }, { "input": "40 33\n11 15 22 26 21 6 8 5 32 39 28 29 30 13 2 40 33 27 17 31 7 36 9 19 3 38 37 12 10 16 1 23 35", "output": "-1" }, { "input": "20 7\n7 5 6 13 16 3 17", "output": "-1" }, { "input": "200000 8\n197281 11492 67218 100058 179300 182264 17781 192818", "output": "-1" }, { "input": "40 32\n22 7 35 31 14 28 9 20 10 3 38 6 15 36 33 16 37 2 11 13 26 23 30 12 40 5 21 1 34 19 27 24", "output": "-1" }, { "input": "20 8\n1 16 14 11 7 9 2 12", "output": "-1" }, { "input": "30 3\n17 5 3", "output": "17 5 3 2 1 4 16 15 14 13 12 11 10 9 8 7 6 30 29 28 27 26 25 24 23 22 21 20 19 18 " }, { "input": "30 3\n29 25 21", "output": "29 25 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 24 23 22 28 27 26 30 " }, { "input": "10 6\n2 1 4 3 6 5", "output": "2 1 4 3 6 5 10 9 8 7 " }, { "input": "4 3\n2 1 3", "output": "2 1 3 4 " }, { "input": "6 4\n5 4 3 1", "output": "5 4 3 1 2 6 " }, { "input": "4 3\n1 2 3", "output": "1 2 3 4 " }, { "input": "6 4\n1 3 2 6", "output": "1 3 2 6 5 4 " }, { "input": "5 4\n3 2 1 5", "output": "3 2 1 5 4 " }, { "input": "10 4\n6 4 1 3", "output": "6 4 1 3 2 5 10 9 8 7 " }, { "input": "4 3\n3 4 2", "output": "-1" }, { "input": "4 3\n3 1 4", "output": "-1" }, { "input": "3 2\n2 3", "output": "-1" }, { "input": "4 3\n1 4 2", "output": "1 4 2 3 " }, { "input": "4 3\n3 1 2", "output": "3 1 2 4 " }, { "input": "2 1\n1", "output": "1 2 " }, { "input": "3 2\n3 2", "output": "3 2 1 " }, { "input": "4 3\n4 1 2", "output": "4 1 2 3 " }, { "input": "3 2\n3 1", "output": "3 1 2 " }, { "input": "4 3\n2 1 4", "output": "2 1 4 3 " }, { "input": "8 5\n3 1 4 2 7", "output": "-1" }, { "input": "6 4\n2 5 1 4", "output": "-1" }, { "input": "10 5\n10 1 8 5 6", "output": "-1" }, { "input": "10 3\n6 4 3", "output": "6 4 3 2 1 5 10 9 8 7 " }, { "input": "10 3\n2 1 6", "output": "2 1 6 5 4 3 10 9 8 7 " }, { "input": "10 3\n8 1 7", "output": "8 1 7 6 5 4 3 2 10 9 " }, { "input": "10 2\n5 4", "output": "5 4 3 2 1 10 9 8 7 6 " }, { "input": "10 3\n1 2 10", "output": "1 2 10 9 8 7 6 5 4 3 " }, { "input": "10 4\n4 1 6 3", "output": "-1" }, { "input": "10 3\n8 1 5", "output": "8 1 5 4 3 2 7 6 10 9 " }, { "input": "10 4\n1 4 9 8", "output": "-1" }, { "input": "10 3\n3 1 6", "output": "-1" }, { "input": "10 6\n1 2 5 4 3 6", "output": "1 2 5 4 3 6 10 9 8 7 " }, { "input": "10 9\n9 8 7 5 4 3 2 1 6", "output": "9 8 7 5 4 3 2 1 6 10 " }, { "input": "10 4\n4 7 5 10", "output": "-1" }, { "input": "10 5\n8 6 2 1 5", "output": "8 6 2 1 5 4 3 7 10 9 " }, { "input": "10 7\n7 5 2 1 4 3 6", "output": "7 5 2 1 4 3 6 10 9 8 " }, { "input": "10 4\n1 2 10 6", "output": "1 2 10 6 5 4 3 9 8 7 " }, { "input": "10 6\n1 10 9 5 4 3", "output": "1 10 9 5 4 3 2 8 7 6 " }, { "input": "10 8\n6 10 4 7 9 8 5 3", "output": "-1" }, { "input": "10 4\n6 1 10 3", "output": "-1" }, { "input": "10 9\n9 6 1 4 2 3 5 10 7", "output": "-1" }, { "input": "10 9\n10 1 9 3 2 4 5 8 6", "output": "10 1 9 3 2 4 5 8 6 7 " }, { "input": "10 4\n10 8 1 7", "output": "10 8 1 7 6 5 4 3 2 9 " }, { "input": "10 4\n2 1 3 6", "output": "2 1 3 6 5 4 10 9 8 7 " }, { "input": "10 3\n2 1 4", "output": "2 1 4 3 10 9 8 7 6 5 " }, { "input": "10 3\n4 1 5", "output": "-1" }, { "input": "10 5\n9 8 1 2 10", "output": "-1" }, { "input": "10 3\n9 8 3", "output": "9 8 3 2 1 7 6 5 4 10 " }, { "input": "10 4\n8 2 1 5", "output": "8 2 1 5 4 3 7 6 10 9 " }, { "input": "10 6\n6 5 3 1 2 4", "output": "6 5 3 1 2 4 10 9 8 7 " }, { "input": "10 2\n1 2", "output": "1 2 10 9 8 7 6 5 4 3 " }, { "input": "10 6\n9 6 5 2 1 4", "output": "9 6 5 2 1 4 3 8 7 10 " }, { "input": "10 4\n2 1 7 3", "output": "2 1 7 3 6 5 4 10 9 8 " }, { "input": "10 2\n6 5", "output": "6 5 4 3 2 1 10 9 8 7 " }, { "input": "10 3\n2 1 5", "output": "2 1 5 4 3 10 9 8 7 6 " }, { "input": "10 4\n3 1 2 4", "output": "3 1 2 4 10 9 8 7 6 5 " }, { "input": "10 3\n8 5 4", "output": "8 5 4 3 2 1 7 6 10 9 " }, { "input": "10 4\n2 1 8 4", "output": "2 1 8 4 3 7 6 5 10 9 " }, { "input": "10 3\n8 3 2", "output": "8 3 2 1 7 6 5 4 10 9 " }, { "input": "10 3\n5 4 2", "output": "5 4 2 1 3 10 9 8 7 6 " }, { "input": "10 9\n10 8 7 5 6 2 1 9 4", "output": "-1" }, { "input": "10 4\n2 1 6 4", "output": "2 1 6 4 3 5 10 9 8 7 " }, { "input": "10 4\n2 1 3 9", "output": "2 1 3 9 8 7 6 5 4 10 " }, { "input": "10 3\n1 4 3", "output": "1 4 3 2 10 9 8 7 6 5 " }, { "input": "10 7\n3 2 1 9 8 6 5", "output": "3 2 1 9 8 6 5 4 7 10 " }, { "input": "10 4\n10 7 1 5", "output": "10 7 1 5 4 3 2 6 9 8 " }, { "input": "10 4\n8 7 1 2", "output": "8 7 1 2 6 5 4 3 10 9 " }, { "input": "10 4\n1 5 4 2", "output": "1 5 4 2 3 10 9 8 7 6 " }, { "input": "10 5\n2 1 9 3 7", "output": "2 1 9 3 7 6 5 4 8 10 " }, { "input": "10 4\n2 1 5 3", "output": "2 1 5 3 4 10 9 8 7 6 " }, { "input": "10 5\n9 6 1 8 2", "output": "-1" }, { "input": "20 13\n3 2 1 7 4 5 6 11 10 9 8 13 12", "output": "3 2 1 7 4 5 6 11 10 9 8 13 12 20 19 18 17 16 15 14 " }, { "input": "20 14\n3 2 1 7 4 5 6 14 11 10 9 8 13 12", "output": "3 2 1 7 4 5 6 14 11 10 9 8 13 12 20 19 18 17 16 15 " }, { "input": "10 5\n9 4 2 1 5", "output": "-1" }, { "input": "10 5\n1 5 2 10 3", "output": "-1" }, { "input": "10 8\n6 5 3 1 2 4 9 8", "output": "6 5 3 1 2 4 9 8 7 10 " }, { "input": "10 4\n10 9 3 7", "output": "-1" }, { "input": "10 7\n10 8 5 1 2 7 3", "output": "-1" }, { "input": "10 3\n3 1 5", "output": "-1" }, { "input": "10 5\n1 9 8 4 3", "output": "1 9 8 4 3 2 7 6 5 10 " }, { "input": "10 3\n1 8 4", "output": "1 8 4 3 2 7 6 5 10 9 " }, { "input": "10 4\n6 2 1 4", "output": "6 2 1 4 3 5 10 9 8 7 " }, { "input": "10 3\n1 6 4", "output": "1 6 4 3 2 5 10 9 8 7 " }, { "input": "10 3\n10 9 3", "output": "10 9 3 2 1 8 7 6 5 4 " }, { "input": "10 9\n8 10 4 1 3 2 9 7 5", "output": "-1" }, { "input": "10 3\n7 10 6", "output": "-1" }, { "input": "10 3\n9 10 8", "output": "-1" }, { "input": "10 6\n10 8 1 6 2 7", "output": "-1" }, { "input": "10 6\n6 5 1 2 9 3", "output": "-1" }, { "input": "10 3\n10 1 8", "output": "10 1 8 7 6 5 4 3 2 9 " }, { "input": "10 9\n1 9 7 10 5 8 4 6 3", "output": "-1" }, { "input": "10 5\n1 9 3 2 5", "output": "1 9 3 2 5 4 8 7 6 10 " }, { "input": "10 4\n10 1 9 7", "output": "10 1 9 7 6 5 4 3 2 8 " }, { "input": "10 8\n1 10 3 2 9 4 8 5", "output": "1 10 3 2 9 4 8 5 7 6 " }, { "input": "10 1\n1", "output": "1 10 9 8 7 6 5 4 3 2 " }, { "input": "10 7\n9 7 1 6 5 4 2", "output": "9 7 1 6 5 4 2 3 8 10 " }, { "input": "10 9\n10 2 1 7 8 3 5 6 9", "output": "-1" }, { "input": "10 4\n2 1 3 10", "output": "2 1 3 10 9 8 7 6 5 4 " }, { "input": "10 9\n5 1 4 6 3 9 8 10 7", "output": "-1" }, { "input": "10 6\n8 2 1 7 6 5", "output": "8 2 1 7 6 5 4 3 10 9 " }, { "input": "10 5\n2 9 8 6 1", "output": "-1" }, { "input": "10 4\n9 2 1 6", "output": "9 2 1 6 5 4 3 8 7 10 " }, { "input": "10 3\n2 1 7", "output": "2 1 7 6 5 4 3 10 9 8 " }, { "input": "10 7\n4 1 2 10 9 6 3", "output": "-1" }, { "input": "10 6\n10 2 1 3 9 4", "output": "10 2 1 3 9 4 8 7 6 5 " }, { "input": "10 4\n9 2 1 4", "output": "9 2 1 4 3 8 7 6 5 10 " }, { "input": "10 3\n5 1 4", "output": "5 1 4 3 2 10 9 8 7 6 " }, { "input": "10 4\n4 1 2 10", "output": "-1" }, { "input": "8 6\n5 4 3 2 1 8", "output": "5 4 3 2 1 8 7 6 " }, { "input": "10 4\n1 6 5 4", "output": "1 6 5 4 3 2 10 9 8 7 " }, { "input": "10 2\n10 2", "output": "10 2 1 9 8 7 6 5 4 3 " }, { "input": "10 5\n1 6 2 10 5", "output": "-1" }, { "input": "10 9\n6 1 2 10 9 5 3 4 8", "output": "-1" }, { "input": "10 5\n4 1 7 2 3", "output": "-1" }, { "input": "10 4\n2 1 3 4", "output": "2 1 3 4 10 9 8 7 6 5 " }, { "input": "11 2\n3 2", "output": "3 2 1 11 10 9 8 7 6 5 4 " }, { "input": "6 5\n3 2 1 4 5", "output": "3 2 1 4 5 6 " }, { "input": "5 4\n2 1 3 5", "output": "2 1 3 5 4 " }, { "input": "10 6\n3 2 1 5 4 6", "output": "3 2 1 5 4 6 10 9 8 7 " }, { "input": "11 5\n1 8 7 6 5", "output": "1 8 7 6 5 4 3 2 11 10 9 " }, { "input": "10 3\n2 1 3", "output": "2 1 3 10 9 8 7 6 5 4 " }, { "input": "10 4\n2 1 7 6", "output": "2 1 7 6 5 4 3 10 9 8 " }, { "input": "10 4\n5 4 1 8", "output": "-1" }, { "input": "10 4\n9 1 5 4", "output": "9 1 5 4 3 2 8 7 6 10 " }, { "input": "10 3\n6 1 4", "output": "6 1 4 3 2 5 10 9 8 7 " }, { "input": "10 6\n1 9 3 2 4 6", "output": "1 9 3 2 4 6 5 8 7 10 " }, { "input": "10 3\n10 1 9", "output": "10 1 9 8 7 6 5 4 3 2 " }, { "input": "10 3\n1 9 7", "output": "1 9 7 6 5 4 3 2 8 10 " }, { "input": "10 2\n2 10", "output": "-1" }, { "input": "10 5\n9 2 1 4 3", "output": "9 2 1 4 3 8 7 6 5 10 " }, { "input": "10 6\n1 2 3 6 5 4", "output": "1 2 3 6 5 4 10 9 8 7 " }, { "input": "10 5\n7 6 5 1 4", "output": "7 6 5 1 4 3 2 10 9 8 " }, { "input": "10 9\n8 1 3 4 10 5 9 7 2", "output": "-1" } ]

1,514,641,746

2,147,483,647

Python 3

COMPILATION_ERROR

TESTS

0

#include <bits/stdc++.h> using namespace std; int main(){ int k,n; cin >>n; cin >> k; bool x=true; bool y=false; int i; int m=0; int p[200000]; for(i=0;i<k;i++){ cin >> p[i]; if (p[i]>m){ m=p[i]; } if(i>=1){ if((p[i-1]-p[i])<1){ x=false; } if(i==1){ if (p[0]==1){ x=true; y=true; } } } } if(x){ for(i=0;i<k;i++){ cout << p[i]<<" "; } if(y){ for(i=p[k-1]-1;i>1;i--){ cout <0;i--){ cout <0;i--){ for(int j=p[i-1]-1;j>p[i];j--){ cout << j <<" " ; } } for(i=n;i>m;i--){ cout << i << " " ; } }else{ cout << -1 << endl; } return 0; }

Title: Stack Sorting Time Limit: None seconds Memory Limit: None megabytes Problem Description: Let's suppose you have an array *a*, a stack *s* (initially empty) and an array *b* (also initially empty). You may perform the following operations until both *a* and *s* are empty: - Take the first element of *a*, push it into *s* and remove it from *a* (if *a* is not empty); - Take the top element from *s*, append it to the end of array *b* and remove it from *s* (if *s* is not empty). You can perform these operations in arbitrary order. If there exists a way to perform the operations such that array *b* is sorted in non-descending order in the end, then array *a* is called stack-sortable. For example, [3,<=1,<=2] is stack-sortable, because *b* will be sorted if we perform the following operations: 1. Remove 3 from *a* and push it into *s*; 1. Remove 1 from *a* and push it into *s*; 1. Remove 1 from *s* and append it to the end of *b*; 1. Remove 2 from *a* and push it into *s*; 1. Remove 2 from *s* and append it to the end of *b*; 1. Remove 3 from *s* and append it to the end of *b*. After all these operations *b*<==<=[1,<=2,<=3], so [3,<=1,<=2] is stack-sortable. [2,<=3,<=1] is not stack-sortable. You are given *k* first elements of some permutation *p* of size *n* (recall that a permutation of size *n* is an array of size *n* where each integer from 1 to *n* occurs exactly once). You have to restore the remaining *n*<=-<=*k* elements of this permutation so it is stack-sortable. If there are multiple answers, choose the answer such that *p* is lexicographically maximal (an array *q* is lexicographically greater than an array *p* iff there exists some integer *k* such that for every *i*<=<<=*k* *q**i*<==<=*p**i*, and *q**k*<=><=*p**k*). You may not swap or change any of first *k* elements of the permutation. Print the lexicographically maximal permutation *p* you can obtain. If there exists no answer then output -1. Input Specification: The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=200000, 1<=≤<=*k*<=<<=*n*) — the size of a desired permutation, and the number of elements you are given, respectively. The second line contains *k* integers *p*1, *p*2, ..., *p**k* (1<=≤<=*p**i*<=≤<=*n*) — the first *k* elements of *p*. These integers are pairwise distinct. Output Specification: If it is possible to restore a stack-sortable permutation *p* of size *n* such that the first *k* elements of *p* are equal to elements given in the input, print lexicographically maximal such permutation. Otherwise print -1. Demo Input: ['5 3\n3 2 1\n', '5 3\n2 3 1\n', '5 1\n3\n', '5 2\n3 4\n'] Demo Output: ['3 2 1 5 4 ', '-1\n', '3 2 1 5 4 ', '-1\n'] Note: none

```python #include <bits/stdc++.h> using namespace std; int main(){ int k,n; cin >>n; cin >> k; bool x=true; bool y=false; int i; int m=0; int p[200000]; for(i=0;i<k;i++){ cin >> p[i]; if (p[i]>m){ m=p[i]; } if(i>=1){ if((p[i-1]-p[i])<1){ x=false; } if(i==1){ if (p[0]==1){ x=true; y=true; } } } } if(x){ for(i=0;i<k;i++){ cout << p[i]<<" "; } if(y){ for(i=p[k-1]-1;i>1;i--){ cout <0;i--){ cout <0;i--){ for(int j=p[i-1]-1;j>p[i];j--){ cout << j <<" " ; } } for(i=n;i>m;i--){ cout << i << " " ; } }else{ cout << -1 << endl; } return 0; } ```

-1

685

A

Robbers' watch

PROGRAMMING

1,700

[ "brute force", "combinatorics", "dp", "math" ]

null

Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches. First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in *n* hours, and each hour in *m* minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to *n*<=-<=1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to *m*<=-<=1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation. Note that to display number 0 section of the watches is required to have at least one place. Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.

The first line of the input contains two integers, given in the decimal notation, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=109) — the number of hours in one day and the number of minutes in one hour, respectively.

Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.

[ "2 3\n", "8 2\n" ]

[ "4\n", "5\n" ]

In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2). In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1).

500

[ { "input": "2 3", "output": "4" }, { "input": "8 2", "output": "5" }, { "input": "1 1", "output": "0" }, { "input": "1 2", "output": "1" }, { "input": "8 8", "output": "0" }, { "input": "50 50", "output": "0" }, { "input": "344 344", "output": "0" }, { "input": "282475250 282475250", "output": "0" }, { "input": "8 282475250", "output": "0" }, { "input": "1000000000 1000000000", "output": "0" }, { "input": "16808 7", "output": "720" }, { "input": "2402 50", "output": "0" }, { "input": "343 2401", "output": "5040" }, { "input": "1582 301", "output": "2874" }, { "input": "421414245 4768815", "output": "0" }, { "input": "2401 343", "output": "5040" }, { "input": "2 1", "output": "1" }, { "input": "282475250 8", "output": "0" }, { "input": "8 7", "output": "35" }, { "input": "50 7", "output": "120" }, { "input": "16808 8", "output": "0" }, { "input": "2402 49", "output": "720" }, { "input": "123 123", "output": "360" }, { "input": "123 456", "output": "150" }, { "input": "1 9", "output": "0" }, { "input": "1 10", "output": "1" }, { "input": "50 67", "output": "6" }, { "input": "7 117649", "output": "5040" }, { "input": "2400 342", "output": "5040" }, { "input": "2400 227", "output": "3360" }, { "input": "117648 5", "output": "3600" }, { "input": "16808 41", "output": "0" }, { "input": "3 16808", "output": "240" }, { "input": "823542 3", "output": "0" }, { "input": "3 823544", "output": "0" }, { "input": "117650 5", "output": "0" }, { "input": "50 50", "output": "0" }, { "input": "50 3", "output": "40" }, { "input": "2402 343", "output": "0" } ]

1,607,878,209

2,147,483,647

PyPy 3

OK

TESTS

39

155

1,433,600

from itertools import permutations def find_power(n): res = 1 n = n // 7 while n > 0: n //= 7 res +=1 return res a, b = list(int(x) for x in input().split()) x = find_power(a-1) y = find_power(b-1) res = 0 if x + y > 7: print(0) exit() for i in permutations('0123456', x + y): s = ''.join(list(i)) p, q = int(s[:x], 7), int(s[x:], 7) if p <= a-1 and q <= b-1: res += 1 print(res)

Title: Robbers' watch Time Limit: None seconds Memory Limit: None megabytes Problem Description: Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches. First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in *n* hours, and each hour in *m* minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to *n*<=-<=1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to *m*<=-<=1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation. Note that to display number 0 section of the watches is required to have at least one place. Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number. Input Specification: The first line of the input contains two integers, given in the decimal notation, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=109) — the number of hours in one day and the number of minutes in one hour, respectively. Output Specification: Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct. Demo Input: ['2 3\n', '8 2\n'] Demo Output: ['4\n', '5\n'] Note: In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2). In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1).

```python from itertools import permutations def find_power(n): res = 1 n = n // 7 while n > 0: n //= 7 res +=1 return res a, b = list(int(x) for x in input().split()) x = find_power(a-1) y = find_power(b-1) res = 0 if x + y > 7: print(0) exit() for i in permutations('0123456', x + y): s = ''.join(list(i)) p, q = int(s[:x], 7), int(s[x:], 7) if p <= a-1 and q <= b-1: res += 1 print(res) ```

3

0

none

0

[ "none" ]

null

You are given two squares, one with sides parallel to the coordinate axes, and another one with sides at 45 degrees to the coordinate axes. Find whether the two squares intersect. The interior of the square is considered to be part of the square, i.e. if one square is completely inside another, they intersect. If the two squares only share one common point, they are also considered to intersect.

The input data consists of two lines, one for each square, both containing 4 pairs of integers. Each pair represents coordinates of one vertex of the square. Coordinates within each line are either in clockwise or counterclockwise order. The first line contains the coordinates of the square with sides parallel to the coordinate axes, the second line contains the coordinates of the square at 45 degrees. All the values are integer and between $-100$ and $100$.

Print "Yes" if squares intersect, otherwise print "No". You can print each letter in any case (upper or lower).

[ "0 0 6 0 6 6 0 6\n1 3 3 5 5 3 3 1\n", "0 0 6 0 6 6 0 6\n7 3 9 5 11 3 9 1\n", "6 0 6 6 0 6 0 0\n7 4 4 7 7 10 10 7\n" ]

[ "YES\n", "NO\n", "YES\n" ]

In the first example the second square lies entirely within the first square, so they do intersect. In the second sample squares do not have any points in common. Here are images corresponding to the samples:

0

[ { "input": "0 0 6 0 6 6 0 6\n1 3 3 5 5 3 3 1", "output": "YES" }, { "input": "0 0 6 0 6 6 0 6\n7 3 9 5 11 3 9 1", "output": "NO" }, { "input": "6 0 6 6 0 6 0 0\n7 4 4 7 7 10 10 7", "output": "YES" }, { "input": "0 0 6 0 6 6 0 6\n8 4 4 8 8 12 12 8", "output": "YES" }, { "input": "2 2 4 2 4 4 2 4\n0 3 3 6 6 3 3 0", "output": "YES" }, { "input": "-5 -5 5 -5 5 5 -5 5\n-5 7 0 2 5 7 0 12", "output": "YES" }, { "input": "-5 -5 5 -5 5 5 -5 5\n-5 12 0 7 5 12 0 17", "output": "NO" }, { "input": "-5 -5 5 -5 5 5 -5 5\n6 0 0 6 -6 0 0 -6", "output": "YES" }, { "input": "-100 -100 100 -100 100 100 -100 100\n-100 0 0 -100 100 0 0 100", "output": "YES" }, { "input": "92 1 92 98 -5 98 -5 1\n44 60 56 48 44 36 32 48", "output": "YES" }, { "input": "-12 -54 -12 33 -99 33 -99 -54\n-77 -40 -86 -31 -77 -22 -68 -31", "output": "YES" }, { "input": "3 45 19 45 19 61 3 61\n-29 45 -13 29 3 45 -13 61", "output": "YES" }, { "input": "79 -19 79 15 45 15 45 -19\n-1 24 -29 52 -1 80 27 52", "output": "NO" }, { "input": "75 -57 75 -21 39 -21 39 -57\n10 -42 -32 0 10 42 52 0", "output": "NO" }, { "input": "-11 53 9 53 9 73 -11 73\n-10 9 -43 42 -10 75 23 42", "output": "YES" }, { "input": "-10 -36 -10 27 -73 27 -73 -36\n44 -28 71 -55 44 -82 17 -55", "output": "NO" }, { "input": "-63 -15 6 -15 6 54 -63 54\n15 -13 -8 10 15 33 38 10", "output": "YES" }, { "input": "47 15 51 15 51 19 47 19\n19 0 -27 46 19 92 65 46", "output": "NO" }, { "input": "87 -5 87 79 3 79 3 -5\n36 36 78 -6 36 -48 -6 -6", "output": "YES" }, { "input": "-4 56 10 56 10 70 -4 70\n-11 47 -35 71 -11 95 13 71", "output": "YES" }, { "input": "-41 6 -41 8 -43 8 -43 6\n-7 27 43 -23 -7 -73 -57 -23", "output": "NO" }, { "input": "44 -58 44 7 -21 7 -21 -58\n22 19 47 -6 22 -31 -3 -6", "output": "YES" }, { "input": "-37 -63 49 -63 49 23 -37 23\n-52 68 -21 37 -52 6 -83 37", "output": "YES" }, { "input": "93 20 93 55 58 55 58 20\n61 -17 39 5 61 27 83 5", "output": "YES" }, { "input": "-7 4 -7 58 -61 58 -61 4\n-28 45 -17 34 -28 23 -39 34", "output": "YES" }, { "input": "24 -79 87 -79 87 -16 24 -16\n-59 21 -85 47 -59 73 -33 47", "output": "NO" }, { "input": "-68 -15 6 -15 6 59 -68 59\n48 -18 57 -27 48 -36 39 -27", "output": "NO" }, { "input": "25 1 25 91 -65 91 -65 1\n24 3 15 12 24 21 33 12", "output": "YES" }, { "input": "55 24 73 24 73 42 55 42\n49 17 10 56 49 95 88 56", "output": "YES" }, { "input": "69 -65 69 -28 32 -28 32 -65\n-1 50 43 6 -1 -38 -45 6", "output": "NO" }, { "input": "86 -26 86 18 42 18 42 -26\n3 -22 -40 21 3 64 46 21", "output": "YES" }, { "input": "52 -47 52 -30 35 -30 35 -47\n49 -22 64 -37 49 -52 34 -37", "output": "YES" }, { "input": "27 -59 27 9 -41 9 -41 -59\n-10 -17 2 -29 -10 -41 -22 -29", "output": "YES" }, { "input": "-90 2 0 2 0 92 -90 92\n-66 31 -86 51 -66 71 -46 51", "output": "YES" }, { "input": "-93 -86 -85 -86 -85 -78 -93 -78\n-13 61 0 48 -13 35 -26 48", "output": "NO" }, { "input": "-3 -45 85 -45 85 43 -3 43\n-22 0 -66 44 -22 88 22 44", "output": "YES" }, { "input": "-27 -73 72 -73 72 26 -27 26\n58 11 100 -31 58 -73 16 -31", "output": "YES" }, { "input": "-40 -31 8 -31 8 17 -40 17\n0 18 -35 53 0 88 35 53", "output": "NO" }, { "input": "-15 -63 -15 7 -85 7 -85 -63\n-35 -40 -33 -42 -35 -44 -37 -42", "output": "YES" }, { "input": "-100 -100 -100 100 100 100 100 -100\n-100 0 0 100 100 0 0 -100", "output": "YES" }, { "input": "67 33 67 67 33 67 33 33\n43 11 9 45 43 79 77 45", "output": "YES" }, { "input": "14 8 9 8 9 3 14 3\n-2 -13 14 3 30 -13 14 -29", "output": "YES" }, { "input": "4 3 7 3 7 6 4 6\n7 29 20 16 7 3 -6 16", "output": "YES" }, { "input": "14 30 3 30 3 19 14 19\n19 -13 11 -5 19 3 27 -5", "output": "NO" }, { "input": "-54 3 -50 3 -50 -1 -54 -1\n3 -50 -6 -41 -15 -50 -6 -59", "output": "NO" }, { "input": "3 8 3 -10 21 -10 21 8\n-9 2 -21 -10 -9 -22 3 -10", "output": "YES" }, { "input": "-35 3 -21 3 -21 -11 -35 -11\n-8 -10 3 -21 -8 -32 -19 -21", "output": "NO" }, { "input": "-5 -23 -5 -31 3 -31 3 -23\n-7 -23 -2 -28 3 -23 -2 -18", "output": "YES" }, { "input": "3 20 10 20 10 13 3 13\n3 20 21 38 39 20 21 2", "output": "YES" }, { "input": "25 3 16 3 16 12 25 12\n21 -2 16 -7 11 -2 16 3", "output": "YES" }, { "input": "-1 18 -1 3 14 3 14 18\n14 3 19 8 14 13 9 8", "output": "YES" }, { "input": "-44 -17 -64 -17 -64 3 -44 3\n-56 15 -44 27 -32 15 -44 3", "output": "YES" }, { "input": "17 3 2 3 2 18 17 18\n22 23 2 3 -18 23 2 43", "output": "YES" }, { "input": "3 -22 3 -36 -11 -36 -11 -22\n11 -44 19 -36 11 -28 3 -36", "output": "YES" }, { "input": "3 45 3 48 0 48 0 45\n13 38 4 47 13 56 22 47", "output": "NO" }, { "input": "3 -10 2 -10 2 -9 3 -9\n38 -10 20 -28 2 -10 20 8", "output": "YES" }, { "input": "-66 3 -47 3 -47 22 -66 22\n-52 -2 -45 5 -52 12 -59 5", "output": "YES" }, { "input": "3 37 -1 37 -1 41 3 41\n6 31 9 34 6 37 3 34", "output": "NO" }, { "input": "13 1 15 1 15 3 13 3\n13 19 21 11 13 3 5 11", "output": "YES" }, { "input": "20 8 3 8 3 -9 20 -9\n2 -11 3 -10 2 -9 1 -10", "output": "NO" }, { "input": "3 41 3 21 -17 21 -17 41\n26 12 10 28 26 44 42 28", "output": "NO" }, { "input": "11 11 11 3 3 3 3 11\n-12 26 -27 11 -12 -4 3 11", "output": "YES" }, { "input": "-29 3 -29 12 -38 12 -38 3\n-35 9 -29 15 -23 9 -29 3", "output": "YES" }, { "input": "3 -32 1 -32 1 -30 3 -30\n4 -32 -16 -52 -36 -32 -16 -12", "output": "YES" }, { "input": "-16 -10 -16 9 3 9 3 -10\n-8 -1 2 9 12 -1 2 -11", "output": "YES" }, { "input": "3 -42 -5 -42 -5 -34 3 -34\n-8 -54 -19 -43 -8 -32 3 -43", "output": "YES" }, { "input": "-47 3 -37 3 -37 -7 -47 -7\n-37 3 -33 -1 -37 -5 -41 -1", "output": "YES" }, { "input": "10 3 12 3 12 5 10 5\n12 4 20 12 12 20 4 12", "output": "YES" }, { "input": "3 -41 -9 -41 -9 -53 3 -53\n18 -16 38 -36 18 -56 -2 -36", "output": "YES" }, { "input": "3 40 2 40 2 41 3 41\n22 39 13 48 4 39 13 30", "output": "NO" }, { "input": "21 26 21 44 3 44 3 26\n-20 38 -32 26 -20 14 -8 26", "output": "NO" }, { "input": "0 7 3 7 3 10 0 10\n3 9 -17 29 -37 9 -17 -11", "output": "YES" }, { "input": "3 21 3 18 6 18 6 21\n-27 18 -11 2 5 18 -11 34", "output": "YES" }, { "input": "-29 13 -39 13 -39 3 -29 3\n-36 -4 -50 -18 -36 -32 -22 -18", "output": "NO" }, { "input": "3 -26 -2 -26 -2 -21 3 -21\n-5 -37 -16 -26 -5 -15 6 -26", "output": "YES" }, { "input": "3 9 -1 9 -1 13 3 13\n-9 17 -1 9 -9 1 -17 9", "output": "YES" }, { "input": "48 8 43 8 43 3 48 3\n31 -4 43 8 55 -4 43 -16", "output": "YES" }, { "input": "-3 1 3 1 3 -5 -3 -5\n20 -22 3 -5 20 12 37 -5", "output": "YES" }, { "input": "14 3 14 -16 -5 -16 -5 3\n14 2 15 1 14 0 13 1", "output": "YES" }, { "input": "-10 12 -10 -1 3 -1 3 12\n1 10 -2 7 -5 10 -2 13", "output": "YES" }, { "input": "39 21 21 21 21 3 39 3\n27 3 47 -17 27 -37 7 -17", "output": "YES" }, { "input": "3 1 3 17 -13 17 -13 1\n17 20 10 27 3 20 10 13", "output": "NO" }, { "input": "15 -18 3 -18 3 -6 15 -6\n29 -1 16 -14 3 -1 16 12", "output": "YES" }, { "input": "41 -6 41 3 32 3 32 -6\n33 3 35 5 33 7 31 5", "output": "YES" }, { "input": "7 35 3 35 3 39 7 39\n23 15 3 35 23 55 43 35", "output": "YES" }, { "input": "19 19 35 19 35 3 19 3\n25 -9 16 -18 7 -9 16 0", "output": "NO" }, { "input": "-20 3 -20 9 -26 9 -26 3\n-19 4 -21 2 -19 0 -17 2", "output": "YES" }, { "input": "13 3 22 3 22 -6 13 -6\n26 3 22 -1 18 3 22 7", "output": "YES" }, { "input": "-4 -8 -4 -15 3 -15 3 -8\n-10 5 -27 -12 -10 -29 7 -12", "output": "YES" }, { "input": "3 15 7 15 7 19 3 19\n-12 30 -23 19 -12 8 -1 19", "output": "NO" }, { "input": "-12 3 5 3 5 -14 -12 -14\n-14 22 5 3 24 22 5 41", "output": "YES" }, { "input": "-37 3 -17 3 -17 -17 -37 -17\n-9 -41 9 -23 -9 -5 -27 -23", "output": "YES" }, { "input": "3 57 3 45 -9 45 -9 57\n8 50 21 37 8 24 -5 37", "output": "YES" }, { "input": "42 3 42 -6 33 -6 33 3\n42 4 41 3 40 4 41 5", "output": "YES" }, { "input": "3 59 3 45 -11 45 -11 59\n-2 50 -8 44 -2 38 4 44", "output": "YES" }, { "input": "-51 3 -39 3 -39 15 -51 15\n-39 14 -53 0 -39 -14 -25 0", "output": "YES" }, { "input": "-7 -15 -7 3 11 3 11 -15\n15 -1 22 -8 15 -15 8 -8", "output": "YES" }, { "input": "3 -39 14 -39 14 -50 3 -50\n17 -39 5 -27 -7 -39 5 -51", "output": "YES" }, { "input": "91 -27 91 29 35 29 35 -27\n59 39 95 3 59 -33 23 3", "output": "YES" }, { "input": "-81 -60 -31 -60 -31 -10 -81 -10\n-58 -68 -95 -31 -58 6 -21 -31", "output": "YES" }, { "input": "78 -59 78 -2 21 -2 21 -59\n48 1 86 -37 48 -75 10 -37", "output": "YES" }, { "input": "-38 -26 32 -26 32 44 -38 44\n2 -27 -44 19 2 65 48 19", "output": "YES" }, { "input": "73 -54 73 -4 23 -4 23 -54\n47 1 77 -29 47 -59 17 -29", "output": "YES" }, { "input": "-6 -25 46 -25 46 27 -6 27\n21 -43 -21 -1 21 41 63 -1", "output": "YES" }, { "input": "-17 -91 -17 -27 -81 -27 -81 -91\n-48 -21 -12 -57 -48 -93 -84 -57", "output": "YES" }, { "input": "-7 16 43 16 43 66 -7 66\n18 -7 -27 38 18 83 63 38", "output": "YES" }, { "input": "-46 11 16 11 16 73 -46 73\n-18 -8 -67 41 -18 90 31 41", "output": "YES" }, { "input": "-33 -64 25 -64 25 -6 -33 -6\n-5 -74 -51 -28 -5 18 41 -28", "output": "YES" }, { "input": "99 -100 100 -100 100 -99 99 -99\n99 -99 100 -98 99 -97 98 -98", "output": "YES" }, { "input": "-100 -100 -100 -99 -99 -99 -99 -100\n-10 -10 -9 -9 -10 -8 -11 -9", "output": "NO" }, { "input": "-4 3 -3 3 -3 4 -4 4\n0 -4 4 0 0 4 -4 0", "output": "NO" }, { "input": "0 0 10 0 10 10 0 10\n11 9 13 7 15 9 13 11", "output": "NO" }, { "input": "1 1 1 6 6 6 6 1\n5 8 8 11 11 8 8 5", "output": "NO" }, { "input": "99 99 99 100 100 100 100 99\n-100 0 0 100 100 0 0 -100", "output": "NO" }, { "input": "0 0 0 2 2 2 2 0\n5 1 9 5 5 9 1 5", "output": "NO" }, { "input": "3 2 3 3 4 3 4 2\n0 4 4 0 0 -4 -4 0", "output": "NO" }, { "input": "0 0 2 0 2 2 0 2\n4 1 7 4 4 7 1 4", "output": "NO" }, { "input": "3 6 3 8 5 8 5 6\n2 9 4 11 6 9 4 7", "output": "YES" }, { "input": "0 0 10 0 10 10 0 10\n-1 5 5 -1 11 5 5 11", "output": "YES" }, { "input": "0 0 1 0 1 1 0 1\n3 0 6 3 3 6 0 3", "output": "NO" }, { "input": "3 7 4 7 4 6 3 6\n0 0 10 10 20 0 10 -10", "output": "NO" }, { "input": "0 0 0 1 1 1 1 0\n0 3 3 6 6 3 3 0", "output": "NO" }, { "input": "0 0 0 4 4 4 4 0\n3 6 7 10 11 6 7 2", "output": "NO" }, { "input": "0 0 0 1 1 1 1 0\n0 10 10 0 20 10 10 20", "output": "NO" } ]

1,529,220,438

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

62

0

class Point: def __init__(self, x, y): self.x = x self.y = y def __add__(self, other): return Point(self.x+other.x, self.y+other.y) def __sub__(self, other): return Point(self.x-other.x, self.y-other.y) def dot(self, other): return self.x*other.x+self.y*other.y def cross(self, other): return self.x*other.y-self.y*other.x def norm_sq(self): return self.x**2+self.y**2 def __mul__(self, other): if isinstance(other, Point): raise TypeError('use .dot(x) or .cross(x) to vectors') return Point(self.x*other, self.y*other) a = list(map(int, input().split())) a1 = a[::2] a2 = a[1::2] x_min = min(a1) x_max = max(a1) y_min = min(a2) y_max = max(a2) a = list(map(int, input().split())) a1 = a[::2] a2 = a[1::2] x_min_2 = min(a1) x_max_2 = max(a1) y_min_2 = min(a2) y_max_2 = max(a2) # x1 = min(a1) # y2 = min(a2) # x3 = max(a1) # y4 = max(a2) # x2 = x4 = (x1+x3)//2 # y1 = y3 = (y2+y4)//2 # p1, p2, p3, p4 = Point(x1, y1), Point(x2, y2), Point(x3, y3), Point(x4, y4) if x_max+y_max > x_min_2+y_min_2 and x_min+y_min < x_max_2+y_max_2 and \ x_max-y_min > x_min_2-y_max_2 and x_min-y_max < x_max_2-y_min_2: print("YES") else: print("NO")

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given two squares, one with sides parallel to the coordinate axes, and another one with sides at 45 degrees to the coordinate axes. Find whether the two squares intersect. The interior of the square is considered to be part of the square, i.e. if one square is completely inside another, they intersect. If the two squares only share one common point, they are also considered to intersect. Input Specification: The input data consists of two lines, one for each square, both containing 4 pairs of integers. Each pair represents coordinates of one vertex of the square. Coordinates within each line are either in clockwise or counterclockwise order. The first line contains the coordinates of the square with sides parallel to the coordinate axes, the second line contains the coordinates of the square at 45 degrees. All the values are integer and between $-100$ and $100$. Output Specification: Print "Yes" if squares intersect, otherwise print "No". You can print each letter in any case (upper or lower). Demo Input: ['0 0 6 0 6 6 0 6\n1 3 3 5 5 3 3 1\n', '0 0 6 0 6 6 0 6\n7 3 9 5 11 3 9 1\n', '6 0 6 6 0 6 0 0\n7 4 4 7 7 10 10 7\n'] Demo Output: ['YES\n', 'NO\n', 'YES\n'] Note: In the first example the second square lies entirely within the first square, so they do intersect. In the second sample squares do not have any points in common. Here are images corresponding to the samples:

```python class Point: def __init__(self, x, y): self.x = x self.y = y def __add__(self, other): return Point(self.x+other.x, self.y+other.y) def __sub__(self, other): return Point(self.x-other.x, self.y-other.y) def dot(self, other): return self.x*other.x+self.y*other.y def cross(self, other): return self.x*other.y-self.y*other.x def norm_sq(self): return self.x**2+self.y**2 def __mul__(self, other): if isinstance(other, Point): raise TypeError('use .dot(x) or .cross(x) to vectors') return Point(self.x*other, self.y*other) a = list(map(int, input().split())) a1 = a[::2] a2 = a[1::2] x_min = min(a1) x_max = max(a1) y_min = min(a2) y_max = max(a2) a = list(map(int, input().split())) a1 = a[::2] a2 = a[1::2] x_min_2 = min(a1) x_max_2 = max(a1) y_min_2 = min(a2) y_max_2 = max(a2) # x1 = min(a1) # y2 = min(a2) # x3 = max(a1) # y4 = max(a2) # x2 = x4 = (x1+x3)//2 # y1 = y3 = (y2+y4)//2 # p1, p2, p3, p4 = Point(x1, y1), Point(x2, y2), Point(x3, y3), Point(x4, y4) if x_max+y_max > x_min_2+y_min_2 and x_min+y_min < x_max_2+y_max_2 and \ x_max-y_min > x_min_2-y_max_2 and x_min-y_max < x_max_2-y_min_2: print("YES") else: print("NO") ```

0

271

A

Beautiful Year

PROGRAMMING

800

[ "brute force" ]

null

It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits. Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.

The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number.

Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists.

[ "1987\n", "2013\n" ]

[ "2013\n", "2014\n" ]

none

500

[ { "input": "1987", "output": "2013" }, { "input": "2013", "output": "2014" }, { "input": "1000", "output": "1023" }, { "input": "1001", "output": "1023" }, { "input": "1234", "output": "1235" }, { "input": "5555", "output": "5601" }, { "input": "9000", "output": "9012" }, { "input": "1111", "output": "1203" }, { "input": "8999", "output": "9012" }, { "input": "4572", "output": "4573" }, { "input": "6666", "output": "6701" }, { "input": "2001", "output": "2013" }, { "input": "3000", "output": "3012" }, { "input": "7712", "output": "7801" }, { "input": "8088", "output": "8091" }, { "input": "1594", "output": "1596" }, { "input": "6016", "output": "6017" }, { "input": "8800", "output": "8901" }, { "input": "3331", "output": "3401" }, { "input": "5090", "output": "5091" }, { "input": "2342", "output": "2345" }, { "input": "2334", "output": "2340" }, { "input": "1123", "output": "1203" }, { "input": "8989", "output": "9012" }, { "input": "8977", "output": "9012" }, { "input": "6869", "output": "6870" }, { "input": "8999", "output": "9012" } ]

1,694,685,877

2,147,483,647

PyPy 3-64

OK

TESTS

27

124

0

def has_distinct_digits(year): return len(set(str(year))) == 4 def next_distinct_year(year): year += 1 while not has_distinct_digits(year): year += 1 return year y = int(input()) print(next_distinct_year(y))

Title: Beautiful Year Time Limit: None seconds Memory Limit: None megabytes Problem Description: It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits. Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits. Input Specification: The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number. Output Specification: Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists. Demo Input: ['1987\n', '2013\n'] Demo Output: ['2013\n', '2014\n'] Note: none

```python def has_distinct_digits(year): return len(set(str(year))) == 4 def next_distinct_year(year): year += 1 while not has_distinct_digits(year): year += 1 return year y = int(input()) print(next_distinct_year(y)) ```

3

767

D

Cartons of milk

PROGRAMMING

2,100

[ "binary search", "data structures", "greedy", "sortings", "two pointers" ]

null

Olya likes milk very much. She drinks *k* cartons of milk each day if she has at least *k* and drinks all of them if she doesn't. But there's an issue — expiration dates. Each carton has a date after which you can't drink it (you still can drink it exactly at the date written on the carton). Due to this, if Olya's fridge contains a carton past its expiry date, she throws it away. Olya hates throwing out cartons, so when she drinks a carton, she chooses the one which expires the fastest. It's easy to understand that this strategy minimizes the amount of cartons thrown out and lets her avoid it if it's even possible. The main issue Olya has is the one of buying new cartons. Currently, there are *n* cartons of milk in Olya's fridge, for each one an expiration date is known (how soon does it expire, measured in days). In the shop that Olya visited there are *m* cartons, and the expiration date is known for each of those cartons as well. Find the maximum number of cartons Olya can buy so that she wouldn't have to throw away any cartons. Assume that Olya drank no cartons today.

In the first line there are three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*<=≤<=106, 1<=≤<=*k*<=≤<=*n*<=+<=*m*) — the amount of cartons in Olya's fridge, the amount of cartons in the shop and the number of cartons Olya drinks each day. In the second line there are *n* integers *f*1,<=*f*2,<=...,<=*f**n* (0<=≤<=*f**i*<=≤<=107) — expiration dates of the cartons in Olya's fridge. The expiration date is expressed by the number of days the drinking of this carton can be delayed. For example, a 0 expiration date means it must be drunk today, 1 — no later than tomorrow, etc. In the third line there are *m* integers *s*1,<=*s*2,<=...,<=*s**m* (0<=≤<=*s**i*<=≤<=107) — expiration dates of the cartons in the shop in a similar format.

If there's no way for Olya to drink the cartons she already has in her fridge, print -1. Otherwise, in the first line print the maximum number *x* of cartons which Olya can buy so that she wouldn't have to throw a carton away. The next line should contain exactly *x* integers — the numbers of the cartons that should be bought (cartons are numbered in an order in which they are written in the input, starting with 1). Numbers should not repeat, but can be in arbitrary order. If there are multiple correct answers, print any of them.

[ "3 6 2\n1 0 1\n2 0 2 0 0 2\n", "3 1 2\n0 0 0\n1\n", "2 1 2\n0 1\n0\n" ]

[ "3\n1 2 3", "-1", "1\n1 " ]

In the first example *k* = 2 and Olya has three cartons with expiry dates 0, 1 and 1 (they expire today, tomorrow and tomorrow), and the shop has 3 cartons with expiry date 0 and 3 cartons with expiry date 2. Olya can buy three cartons, for example, one with the expiry date 0 and two with expiry date 2. In the second example all three cartons Olya owns expire today and it means she would have to throw packets away regardless of whether she buys an extra one or not. In the third example Olya would drink *k* = 2 cartons today (one she alreay has in her fridge and one from the shop) and the remaining one tomorrow.

2,000

[]

1,690,499,018

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

0

46

0

print("_RANDOM_GUESS_1690499018.544926")# 1690499018.5449464

Title: Cartons of milk Time Limit: None seconds Memory Limit: None megabytes Problem Description: Olya likes milk very much. She drinks *k* cartons of milk each day if she has at least *k* and drinks all of them if she doesn't. But there's an issue — expiration dates. Each carton has a date after which you can't drink it (you still can drink it exactly at the date written on the carton). Due to this, if Olya's fridge contains a carton past its expiry date, she throws it away. Olya hates throwing out cartons, so when she drinks a carton, she chooses the one which expires the fastest. It's easy to understand that this strategy minimizes the amount of cartons thrown out and lets her avoid it if it's even possible. The main issue Olya has is the one of buying new cartons. Currently, there are *n* cartons of milk in Olya's fridge, for each one an expiration date is known (how soon does it expire, measured in days). In the shop that Olya visited there are *m* cartons, and the expiration date is known for each of those cartons as well. Find the maximum number of cartons Olya can buy so that she wouldn't have to throw away any cartons. Assume that Olya drank no cartons today. Input Specification: In the first line there are three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*<=≤<=106, 1<=≤<=*k*<=≤<=*n*<=+<=*m*) — the amount of cartons in Olya's fridge, the amount of cartons in the shop and the number of cartons Olya drinks each day. In the second line there are *n* integers *f*1,<=*f*2,<=...,<=*f**n* (0<=≤<=*f**i*<=≤<=107) — expiration dates of the cartons in Olya's fridge. The expiration date is expressed by the number of days the drinking of this carton can be delayed. For example, a 0 expiration date means it must be drunk today, 1 — no later than tomorrow, etc. In the third line there are *m* integers *s*1,<=*s*2,<=...,<=*s**m* (0<=≤<=*s**i*<=≤<=107) — expiration dates of the cartons in the shop in a similar format. Output Specification: If there's no way for Olya to drink the cartons she already has in her fridge, print -1. Otherwise, in the first line print the maximum number *x* of cartons which Olya can buy so that she wouldn't have to throw a carton away. The next line should contain exactly *x* integers — the numbers of the cartons that should be bought (cartons are numbered in an order in which they are written in the input, starting with 1). Numbers should not repeat, but can be in arbitrary order. If there are multiple correct answers, print any of them. Demo Input: ['3 6 2\n1 0 1\n2 0 2 0 0 2\n', '3 1 2\n0 0 0\n1\n', '2 1 2\n0 1\n0\n'] Demo Output: ['3\n1 2 3', '-1', '1\n1 '] Note: In the first example *k* = 2 and Olya has three cartons with expiry dates 0, 1 and 1 (they expire today, tomorrow and tomorrow), and the shop has 3 cartons with expiry date 0 and 3 cartons with expiry date 2. Olya can buy three cartons, for example, one with the expiry date 0 and two with expiry date 2. In the second example all three cartons Olya owns expire today and it means she would have to throw packets away regardless of whether she buys an extra one or not. In the third example Olya would drink *k* = 2 cartons today (one she alreay has in her fridge and one from the shop) and the remaining one tomorrow.

```python print("_RANDOM_GUESS_1690499018.544926")# 1690499018.5449464 ```

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" } ]

1,599,717,612

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

39

218

307,200

n, m = input().split() n = int(n) m = int(m) prime = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47] flag = False for i in range(len(prime)): if prime[i] == n and prime[i+1] == m : flag = True break if flag: print("YES") else: print("NO")

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 n, m = input().split() n = int(n) m = int(m) prime = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47] flag = False for i in range(len(prime)): if prime[i] == n and prime[i+1] == m : flag = True break if flag: print("YES") else: print("NO") ```

-1

501

A

Contest

PROGRAMMING

900

[ "implementation" ]

null

Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points. Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.

The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180). It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round).

Output on a single line: "Misha" (without the quotes), if Misha got more points than Vasya. "Vasya" (without the quotes), if Vasya got more points than Misha. "Tie" (without the quotes), if both of them got the same number of points.

[ "500 1000 20 30\n", "1000 1000 1 1\n", "1500 1000 176 177\n" ]

[ "Vasya\n", "Tie\n", "Misha\n" ]

none

500

[ { "input": "500 1000 20 30", "output": "Vasya" }, { "input": "1000 1000 1 1", "output": "Tie" }, { "input": "1500 1000 176 177", "output": "Misha" }, { "input": "1500 1000 74 177", "output": "Misha" }, { "input": "750 2500 175 178", "output": "Vasya" }, { "input": "750 1000 54 103", "output": "Tie" }, { "input": "2000 1250 176 130", "output": "Tie" }, { "input": "1250 1750 145 179", "output": "Tie" }, { "input": "2000 2000 176 179", "output": "Tie" }, { "input": "1500 1500 148 148", "output": "Tie" }, { "input": "2750 1750 134 147", "output": "Misha" }, { "input": "3250 250 175 173", "output": "Misha" }, { "input": "500 500 170 176", "output": "Misha" }, { "input": "250 1000 179 178", "output": "Vasya" }, { "input": "3250 1000 160 138", "output": "Misha" }, { "input": "3000 2000 162 118", "output": "Tie" }, { "input": "1500 1250 180 160", "output": "Tie" }, { "input": "1250 2500 100 176", "output": "Tie" }, { "input": "3500 3500 177 178", "output": "Tie" }, { "input": "3000 3250 16 34", "output": "Tie" }, { "input": "1750 3000 137 49", "output": "Vasya" }, { "input": "500 1500 179 71", "output": "Vasya" }, { "input": "1250 2000 101 180", "output": "Misha" }, { "input": "250 750 180 176", "output": "Vasya" }, { "input": "2250 2250 163 145", "output": "Vasya" }, { "input": "3000 3000 176 78", "output": "Vasya" }, { "input": "250 3500 8 178", "output": "Vasya" }, { "input": "1750 1250 179 180", "output": "Misha" }, { "input": "2750 1750 13 164", "output": "Misha" }, { "input": "1750 2250 178 53", "output": "Vasya" }, { "input": "2500 2750 73 179", "output": "Misha" }, { "input": "1000 3500 178 175", "output": "Vasya" }, { "input": "1000 500 7 162", "output": "Misha" }, { "input": "1000 250 175 48", "output": "Misha" }, { "input": "1750 500 166 177", "output": "Misha" }, { "input": "250 250 0 0", "output": "Tie" }, { "input": "250 3500 0 0", "output": "Vasya" }, { "input": "250 3500 0 180", "output": "Vasya" }, { "input": "3500 3500 180 180", "output": "Tie" }, { "input": "3500 250 0 180", "output": "Misha" } ]

1,694,823,781

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

0

30

0

a,b,c,d = map(int,input().split()) if max((3*a)//10,(a-(a//250)*b))>max((3*c)//10,(c-(c//250)*d)): print("Misha") elif max((3*a)//10,(a-(a//250)*b))<max((3*c)//10,(c-(c//250)*d)): print("Vasya") else: print("Tie")

Title: Contest Time Limit: None seconds Memory Limit: None megabytes Problem Description: Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points. Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth. Input Specification: The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180). It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round). Output Specification: Output on a single line: "Misha" (without the quotes), if Misha got more points than Vasya. "Vasya" (without the quotes), if Vasya got more points than Misha. "Tie" (without the quotes), if both of them got the same number of points. Demo Input: ['500 1000 20 30\n', '1000 1000 1 1\n', '1500 1000 176 177\n'] Demo Output: ['Vasya\n', 'Tie\n', 'Misha\n'] Note: none

```python a,b,c,d = map(int,input().split()) if max((3*a)//10,(a-(a//250)*b))>max((3*c)//10,(c-(c//250)*d)): print("Misha") elif max((3*a)//10,(a-(a//250)*b))<max((3*c)//10,(c-(c//250)*d)): print("Vasya") else: print("Tie") ```

0

292

B

Network Topology

PROGRAMMING

1,200

[ "graphs", "implementation" ]

null

This problem uses a simplified network topology model, please read the problem statement carefully and use it as a formal document as you develop the solution. Polycarpus continues working as a system administrator in a large corporation. The computer network of this corporation consists of *n* computers, some of them are connected by a cable. The computers are indexed by integers from 1 to *n*. It's known that any two computers connected by cable directly or through other computers Polycarpus decided to find out the network's topology. A network topology is the way of describing the network configuration, the scheme that shows the location and the connections of network devices. Polycarpus knows three main network topologies: bus, ring and star. A bus is the topology that represents a shared cable with all computers connected with it. In the ring topology the cable connects each computer only with two other ones. A star is the topology where all computers of a network are connected to the single central node. Let's represent each of these network topologies as a connected non-directed graph. A bus is a connected graph that is the only path, that is, the graph where all nodes are connected with two other ones except for some two nodes that are the beginning and the end of the path. A ring is a connected graph, where all nodes are connected with two other ones. A star is a connected graph, where a single central node is singled out and connected with all other nodes. For clarifications, see the picture. You've got a connected non-directed graph that characterizes the computer network in Polycarpus' corporation. Help him find out, which topology type the given network is. If that is impossible to do, say that the network's topology is unknown.

The first line contains two space-separated integers *n* and *m* (4<=≤<=*n*<=≤<=105; 3<=≤<=*m*<=≤<=105) — the number of nodes and edges in the graph, correspondingly. Next *m* lines contain the description of the graph's edges. The *i*-th line contains a space-separated pair of integers *x**i*, *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*) — the numbers of nodes that are connected by the *i*-the edge. It is guaranteed that the given graph is connected. There is at most one edge between any two nodes. No edge connects a node with itself.

In a single line print the network topology name of the given graph. If the answer is the bus, print "bus topology" (without the quotes), if the answer is the ring, print "ring topology" (without the quotes), if the answer is the star, print "star topology" (without the quotes). If no answer fits, print "unknown topology" (without the quotes).

[ "4 3\n1 2\n2 3\n3 4\n", "4 4\n1 2\n2 3\n3 4\n4 1\n", "4 3\n1 2\n1 3\n1 4\n", "4 4\n1 2\n2 3\n3 1\n1 4\n" ]

[ "bus topology\n", "ring topology\n", "star topology\n", "unknown topology\n" ]

none

1,000

[ { "input": "4 3\n1 2\n2 3\n3 4", "output": "bus topology" }, { "input": "4 4\n1 2\n2 3\n3 4\n4 1", "output": "ring topology" }, { "input": "4 3\n1 2\n1 3\n1 4", "output": "star topology" }, { "input": "4 4\n1 2\n2 3\n3 1\n1 4", "output": "unknown topology" }, { "input": "5 4\n1 2\n3 5\n1 4\n5 4", "output": "bus topology" }, { "input": "5 5\n3 4\n5 2\n2 1\n5 4\n3 1", "output": "ring topology" }, { "input": "5 4\n4 2\n5 2\n1 2\n2 3", "output": "star topology" }, { "input": "5 9\n5 3\n4 5\n3 1\n3 2\n2 1\n2 5\n1 5\n1 4\n4 2", "output": "unknown topology" }, { "input": "4 3\n2 4\n1 3\n4 1", "output": "bus topology" }, { "input": "4 4\n2 4\n4 1\n1 3\n2 3", "output": "ring topology" }, { "input": "4 3\n1 2\n2 4\n3 2", "output": "star topology" }, { "input": "4 4\n3 2\n2 4\n4 1\n1 2", "output": "unknown topology" }, { "input": "10 9\n10 6\n3 4\n8 9\n8 4\n6 1\n2 9\n5 1\n7 5\n10 3", "output": "bus topology" }, { "input": "10 10\n1 4\n3 6\n10 7\n5 8\n2 10\n3 4\n7 5\n9 6\n8 1\n2 9", "output": "ring topology" }, { "input": "10 9\n1 4\n4 10\n4 9\n8 4\n4 7\n4 5\n4 2\n4 6\n4 3", "output": "star topology" }, { "input": "10 14\n3 2\n7 2\n6 4\n8 1\n3 9\n5 6\n6 3\n4 1\n2 5\n7 10\n9 5\n7 1\n8 10\n3 4", "output": "unknown topology" }, { "input": "4 4\n1 2\n2 3\n2 4\n3 4", "output": "unknown topology" }, { "input": "5 4\n1 2\n1 3\n1 4\n4 5", "output": "unknown topology" }, { "input": "10 9\n10 1\n10 2\n10 3\n10 4\n10 5\n10 6\n10 7\n10 8\n10 9", "output": "star topology" }, { "input": "6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 1", "output": "unknown topology" }, { "input": "6 5\n1 2\n1 3\n1 4\n4 5\n4 6", "output": "unknown topology" }, { "input": "4 4\n1 2\n2 3\n3 4\n4 2", "output": "unknown topology" }, { "input": "4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4", "output": "unknown topology" } ]

1,631,910,961

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

3

184

20,172,800

n, m = map(int, input().split()) grafo = [list() for _ in range (n+1)] for i in range(m): a, b = map(int, input().split()) grafo[a].append(b) grafo[b].append(a) s0 = 0 s1 = 0 s2 = 0 sn = 0 sx = 0 for ponto in grafo: if len(ponto) == 0: s0 += 1 elif len(ponto) == 1: s1 += 1 elif len(ponto) == 2: s2 += 1 elif len(ponto) == n -1: sn += 1 else: sx += 1 if sn == 1 and s1 == n - 1: print("star topology") quit() if s0 > 1 or s2 < n - 2 or sx: print("unknown topology") quit() # ring and bus topology visitados = [0 for _ in range(n+1)] ponto = 1 visitados[1] = 1 while True: prox = grafo[ponto][0] i = 1 while visitados[prox] and i < len(grafo[ponto]): prox = grafo[ponto][i] i += 1 #print(ponto, prox) if i > len(grafo[ponto]) or visitados[prox]: if prox == 1: visitados[1] += 1 break ponto = prox visitados[prox] = 1 if visitados[1] == 2: ring = True for item in visitados[2:]: if not item: ring = False break if ring: print("ring topology") quit() if len(grafo[1]) == 2: ponto = 1 grafo[1].pop(0) while True: prox = grafo[ponto][0] i = 1 while visitados[prox] and i < len(grafo[ponto]): prox = grafo[ponto][i] i += 1 #print(ponto, prox) if i > len(grafo[ponto]) or visitados[prox]: if prox == 1: visitados[1] += 1 break ponto = prox visitados[prox] = 1 bus = True for item in visitados[2:]: if not item: bus = False break if bus: print("bus topology") quit() #print(visitados) print("unkown topology") # print(*grafo, sep='\n')

Title: Network Topology Time Limit: None seconds Memory Limit: None megabytes Problem Description: This problem uses a simplified network topology model, please read the problem statement carefully and use it as a formal document as you develop the solution. Polycarpus continues working as a system administrator in a large corporation. The computer network of this corporation consists of *n* computers, some of them are connected by a cable. The computers are indexed by integers from 1 to *n*. It's known that any two computers connected by cable directly or through other computers Polycarpus decided to find out the network's topology. A network topology is the way of describing the network configuration, the scheme that shows the location and the connections of network devices. Polycarpus knows three main network topologies: bus, ring and star. A bus is the topology that represents a shared cable with all computers connected with it. In the ring topology the cable connects each computer only with two other ones. A star is the topology where all computers of a network are connected to the single central node. Let's represent each of these network topologies as a connected non-directed graph. A bus is a connected graph that is the only path, that is, the graph where all nodes are connected with two other ones except for some two nodes that are the beginning and the end of the path. A ring is a connected graph, where all nodes are connected with two other ones. A star is a connected graph, where a single central node is singled out and connected with all other nodes. For clarifications, see the picture. You've got a connected non-directed graph that characterizes the computer network in Polycarpus' corporation. Help him find out, which topology type the given network is. If that is impossible to do, say that the network's topology is unknown. Input Specification: The first line contains two space-separated integers *n* and *m* (4<=≤<=*n*<=≤<=105; 3<=≤<=*m*<=≤<=105) — the number of nodes and edges in the graph, correspondingly. Next *m* lines contain the description of the graph's edges. The *i*-th line contains a space-separated pair of integers *x**i*, *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*) — the numbers of nodes that are connected by the *i*-the edge. It is guaranteed that the given graph is connected. There is at most one edge between any two nodes. No edge connects a node with itself. Output Specification: In a single line print the network topology name of the given graph. If the answer is the bus, print "bus topology" (without the quotes), if the answer is the ring, print "ring topology" (without the quotes), if the answer is the star, print "star topology" (without the quotes). If no answer fits, print "unknown topology" (without the quotes). Demo Input: ['4 3\n1 2\n2 3\n3 4\n', '4 4\n1 2\n2 3\n3 4\n4 1\n', '4 3\n1 2\n1 3\n1 4\n', '4 4\n1 2\n2 3\n3 1\n1 4\n'] Demo Output: ['bus topology\n', 'ring topology\n', 'star topology\n', 'unknown topology\n'] Note: none

```python n, m = map(int, input().split()) grafo = [list() for _ in range (n+1)] for i in range(m): a, b = map(int, input().split()) grafo[a].append(b) grafo[b].append(a) s0 = 0 s1 = 0 s2 = 0 sn = 0 sx = 0 for ponto in grafo: if len(ponto) == 0: s0 += 1 elif len(ponto) == 1: s1 += 1 elif len(ponto) == 2: s2 += 1 elif len(ponto) == n -1: sn += 1 else: sx += 1 if sn == 1 and s1 == n - 1: print("star topology") quit() if s0 > 1 or s2 < n - 2 or sx: print("unknown topology") quit() # ring and bus topology visitados = [0 for _ in range(n+1)] ponto = 1 visitados[1] = 1 while True: prox = grafo[ponto][0] i = 1 while visitados[prox] and i < len(grafo[ponto]): prox = grafo[ponto][i] i += 1 #print(ponto, prox) if i > len(grafo[ponto]) or visitados[prox]: if prox == 1: visitados[1] += 1 break ponto = prox visitados[prox] = 1 if visitados[1] == 2: ring = True for item in visitados[2:]: if not item: ring = False break if ring: print("ring topology") quit() if len(grafo[1]) == 2: ponto = 1 grafo[1].pop(0) while True: prox = grafo[ponto][0] i = 1 while visitados[prox] and i < len(grafo[ponto]): prox = grafo[ponto][i] i += 1 #print(ponto, prox) if i > len(grafo[ponto]) or visitados[prox]: if prox == 1: visitados[1] += 1 break ponto = prox visitados[prox] = 1 bus = True for item in visitados[2:]: if not item: bus = False break if bus: print("bus topology") quit() #print(visitados) print("unkown topology") # print(*grafo, sep='\n') ```

0

126

B

Password

PROGRAMMING

1,700

[ "binary search", "dp", "hashing", "string suffix structures", "strings" ]

null

Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them. A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*. Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end. Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened. You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend.

You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.

Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes.

[ "fixprefixsuffix\n", "abcdabc\n" ]

[ "fix", "Just a legend" ]

none

1,000

[ { "input": "fixprefixsuffix", "output": "fix" }, { "input": "abcdabc", "output": "Just a legend" }, { "input": "qwertyqwertyqwerty", "output": "qwerty" }, { "input": "papapapap", "output": "papap" }, { "input": "aaaaaaaaaa", "output": "aaaaaaaa" }, { "input": "ghbdtn", "output": "Just a legend" }, { "input": "a", "output": "Just a legend" }, { "input": "aa", "output": "Just a legend" }, { "input": "ab", "output": "Just a legend" }, { "input": "aaa", "output": "a" }, { "input": "aba", "output": "Just a legend" }, { "input": "aab", "output": "Just a legend" }, { "input": "abb", "output": "Just a legend" }, { "input": "abc", "output": "Just a legend" }, { "input": "aaabaabaaaaab", "output": "Just a legend" }, { "input": "aabaaabaaaaab", "output": "aab" }, { "input": "aaabaaaabab", "output": "Just a legend" }, { "input": "abcabcabcabcabc", "output": "abcabcabc" }, { "input": "aaaaabaaaa", "output": "aaaa" }, { "input": "aaaabaaaaaaa", "output": "aaaa" }, { "input": "ghghghgxghghghg", "output": "ghghg" }, { "input": "kincenvizh", "output": "Just a legend" }, { "input": "amcksgurlgqzqizdauqminfzshiweejkevbazyzylrrghumnvqeqqdedyopgtvxakqwpvxntxgrkrcxabhrgoxngrwrxrvcguuyw", "output": "Just a legend" }, { "input": "kwuaizneqxfflhmyruotjlkqksinoanvkyvqptkkntnpjdyzicceelgooajdgpkneuhyvhdtmasiglplajxolxovlhkwuaizneqx", "output": "Just a legend" }, { "input": "nfbdzgdlbjhrlvfryyjbvtsmzacxglcvukmyexdgpuiwvqbnfbdzgdlbjhrlvfryyjbtuomcwbwvlhefnfbdzgdlbjhrlvfryyjb", "output": "nfbdzgdlbjhrlvfryyjb" }, { "input": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpmvbxbyszdugvmoyjeeqsywcvuqskcjotyhvdijyslgfvyujxthngnonasgnbzzygyikheelirdfsvuicbiazfdeqecbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpccdtdteohtjjuicfjfcbzpynlqhecyqqrjpmexgfqywnfmusldgzvgnzdmuuujtkdfnlytzgstquwqpuccxbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp", "output": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp" }, { "input": "abcazabca", "output": "a" } ]

1,698,431,720

2,147,483,647

PyPy 3-64

OK

TESTS

97

1,308

13,926,400

def kmp(): now = -1 sub[0] = now for i in range(1, len(p)): while now != -1 and p[i] != p[now + 1]: now = sub[now] if p[i] == p[now + 1]: now += 1 sub[i] = now else: sub[i] = -1 p = input() sub = [-1] * 1000001 kmp() maxi = sub[len(p) - 1] found = sub[maxi] if maxi != -1 else -1 for i in range(1, len(p) - 1): if sub[i] == maxi: found = maxi if found == -1: print("Just a legend") else: print(p[:found + 1])

Title: Password Time Limit: None seconds Memory Limit: None megabytes Problem Description: Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them. A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*. Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end. Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened. You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend. Input Specification: You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters. Output Specification: Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes. Demo Input: ['fixprefixsuffix\n', 'abcdabc\n'] Demo Output: ['fix', 'Just a legend'] Note: none

```python def kmp(): now = -1 sub[0] = now for i in range(1, len(p)): while now != -1 and p[i] != p[now + 1]: now = sub[now] if p[i] == p[now + 1]: now += 1 sub[i] = now else: sub[i] = -1 p = input() sub = [-1] * 1000001 kmp() maxi = sub[len(p) - 1] found = sub[maxi] if maxi != -1 else -1 for i in range(1, len(p) - 1): if sub[i] == maxi: found = maxi if found == -1: print("Just a legend") else: print(p[:found + 1]) ```

3

465

A

inc ARG

PROGRAMMING

900

[ "implementation" ]

null

Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of *n* bits. These bits are numbered from 1 to *n*. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the *n*-th bit. Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded. Sergey wrote certain values of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation?

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of bits in the cell. The second line contains a string consisting of *n* characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit.

Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell.

[ "4\n1100\n", "4\n1111\n" ]

[ "3\n", "4\n" ]

In the first sample the cell ends up with value 0010, in the second sample — with 0000.

500

[ { "input": "4\n1100", "output": "3" }, { "input": "4\n1111", "output": "4" }, { "input": "1\n0", "output": "1" }, { "input": "1\n1", "output": "1" }, { "input": "2\n00", "output": "1" }, { "input": "2\n01", "output": "1" }, { "input": "2\n10", "output": "2" }, { "input": "2\n11", "output": "2" }, { "input": "10\n0000000000", "output": "1" }, { "input": "20\n11111111110110001100", "output": "11" }, { "input": "50\n01011110100010000001010000100001001101001101101011", "output": "1" }, { "input": "60\n111111111101111111111111111111111111111111111111111111111111", "output": "11" }, { "input": "60\n111111111111111111111111111111111111111111111111111111111111", "output": "60" }, { "input": "66\n111111010010011001110011000111000100011110011001111110011111111101", "output": "7" }, { "input": "90\n000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "1" }, { "input": "91\n1011111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "2" }, { "input": "92\n11111111101011111011111111111111111111111011101111111111011111111111111101111111101011111101", "output": "10" }, { "input": "100\n0001011110100011001100100010111001000001111101101001001001001011110100101101010000000110100101110010", "output": "1" }, { "input": "100\n0111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "1" }, { "input": "100\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "1" }, { "input": "100\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "1" }, { "input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "100" }, { "input": "100\n1011001110001000011111110011000100001110010110111101110110011011011000010100110001111100000010110010", "output": "2" }, { "input": "100\n1101111011001111111111110011110111101110111111111111111111111111111111011111111111110111111111111111", "output": "3" }, { "input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "100" }, { "input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110", "output": "100" }, { "input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111101", "output": "99" }, { "input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111100000", "output": "96" }, { "input": "100\n1111111111111111111111111111111110011000110110001010101100100100001000010010010010000100110000010101", "output": "34" } ]

1,493,746,063

2,147,483,647

Python 3

OK

TESTS

29

61

5,529,600

n=int(input()) s=input()+"0" t=0 while s[t]=='1' and t!=n: t+=1 if t<n: t+=1 print(t)

Title: inc ARG Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of *n* bits. These bits are numbered from 1 to *n*. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the *n*-th bit. Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded. Sergey wrote certain values of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation? Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of bits in the cell. The second line contains a string consisting of *n* characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit. Output Specification: Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell. Demo Input: ['4\n1100\n', '4\n1111\n'] Demo Output: ['3\n', '4\n'] Note: In the first sample the cell ends up with value 0010, in the second sample — with 0000.

```python n=int(input()) s=input()+"0" t=0 while s[t]=='1' and t!=n: t+=1 if t<n: t+=1 print(t) ```

3

669

A

Little Artem and Presents

PROGRAMMING

800

[ "math" ]

null

Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that. How many times can Artem give presents to Masha?

The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday.

Print the maximum possible number of times Artem can give presents to Masha.

[ "1\n", "2\n", "3\n", "4\n" ]

[ "1\n", "1\n", "2\n", "3\n" ]

In the first sample, Artem can only give 1 stone to Masha. In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times. In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone. In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.

500

[ { "input": "1", "output": "1" }, { "input": "2", "output": "1" }, { "input": "3", "output": "2" }, { "input": "4", "output": "3" }, { "input": "100", "output": "67" }, { "input": "101", "output": "67" }, { "input": "102", "output": "68" }, { "input": "1000000000", "output": "666666667" }, { "input": "5", "output": "3" }, { "input": "6", "output": "4" }, { "input": "999999999", "output": "666666666" }, { "input": "999999998", "output": "666666665" }, { "input": "999999997", "output": "666666665" }, { "input": "999999996", "output": "666666664" }, { "input": "999999995", "output": "666666663" }, { "input": "999999994", "output": "666666663" }, { "input": "999999993", "output": "666666662" }, { "input": "999999992", "output": "666666661" }, { "input": "999999991", "output": "666666661" }, { "input": "1000", "output": "667" }, { "input": "10000", "output": "6667" }, { "input": "100000", "output": "66667" }, { "input": "1000000", "output": "666667" }, { "input": "10000000", "output": "6666667" }, { "input": "100000000", "output": "66666667" }, { "input": "7", "output": "5" } ]

1,666,455,247

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

7

2,000

0

last = 0; s = int(input()); g = 0 while (True): if not last == 1 and s >= 1: s -= 1; g += 1; last = 1 elif s >= 2: s -= 2; g += 1; last = 2 else: break; print(g);

Title: Little Artem and Presents Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that. How many times can Artem give presents to Masha? Input Specification: The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday. Output Specification: Print the maximum possible number of times Artem can give presents to Masha. Demo Input: ['1\n', '2\n', '3\n', '4\n'] Demo Output: ['1\n', '1\n', '2\n', '3\n'] Note: In the first sample, Artem can only give 1 stone to Masha. In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times. In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone. In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.

```python last = 0; s = int(input()); g = 0 while (True): if not last == 1 and s >= 1: s -= 1; g += 1; last = 1 elif s >= 2: s -= 2; g += 1; last = 2 else: break; print(g); ```

0

450

A

Jzzhu and Children

PROGRAMMING

1,000

[ "implementation" ]

null

There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies. Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm: 1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty. Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?

The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).

Output a single integer, representing the number of the last child.

[ "5 2\n1 3 1 4 2\n", "6 4\n1 1 2 2 3 3\n" ]

[ "4\n", "6\n" ]

Let's consider the first sample. Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home. Child 4 is the last one who goes home.

500

[ { "input": "5 2\n1 3 1 4 2", "output": "4" }, { "input": "6 4\n1 1 2 2 3 3", "output": "6" }, { "input": "7 3\n6 1 5 4 2 3 1", "output": "4" }, { "input": "10 5\n2 7 3 6 2 5 1 3 4 5", "output": "4" }, { "input": "100 1\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": "100" }, { "input": "9 3\n9 5 2 3 7 1 8 4 6", "output": "7" }, { "input": "20 10\n58 4 32 10 73 7 30 39 47 6 59 21 24 66 79 79 46 13 29 58", "output": "16" }, { "input": "50 5\n89 56 3 2 40 37 56 52 83 59 43 83 43 59 29 74 22 58 53 41 53 67 78 30 57 32 58 29 95 46 45 85 60 49 41 82 8 71 52 40 45 26 6 71 84 91 4 93 40 54", "output": "48" }, { "input": "50 1\n4 3 9 7 6 8 3 7 10 9 8 8 10 2 9 3 2 4 4 10 4 6 8 10 9 9 4 2 8 9 4 4 9 5 1 5 2 4 4 9 10 2 5 10 7 2 8 6 8 1", "output": "44" }, { "input": "50 5\n3 9 10 8 3 3 4 6 8 2 9 9 3 1 2 10 6 8 7 2 7 4 2 7 5 10 2 2 2 5 10 5 6 6 8 7 10 4 3 2 10 8 6 6 8 6 4 4 1 3", "output": "46" }, { "input": "50 2\n56 69 72 15 95 92 51 1 74 87 100 29 46 54 18 81 84 72 84 83 20 63 71 27 45 74 50 89 48 8 21 15 47 3 39 73 80 84 6 99 17 25 56 3 74 64 71 39 89 78", "output": "40" }, { "input": "50 3\n31 39 64 16 86 3 1 9 25 54 98 42 20 3 49 41 73 37 55 62 33 77 64 22 33 82 26 13 10 13 7 40 48 18 46 79 94 72 19 12 11 61 16 37 10 49 14 94 48 69", "output": "11" }, { "input": "50 100\n67 67 61 68 42 29 70 77 12 61 71 27 4 73 87 52 59 38 93 90 31 27 87 47 26 57 76 6 28 72 81 68 50 84 69 79 39 93 52 6 88 12 46 13 90 68 71 38 90 95", "output": "50" }, { "input": "100 3\n4 14 20 11 19 11 14 20 5 7 6 12 11 17 5 11 7 6 2 10 13 5 12 8 5 17 20 18 7 19 11 7 7 20 20 8 10 17 17 19 20 5 15 16 19 7 11 16 4 17 2 10 1 20 20 16 19 9 9 11 5 7 12 9 9 6 20 18 13 19 8 4 8 1 2 4 10 11 15 14 1 7 17 12 13 19 12 2 3 14 15 15 5 17 14 12 17 14 16 9", "output": "86" }, { "input": "100 5\n16 8 14 16 12 11 17 19 19 2 8 9 5 6 19 9 11 18 6 9 14 16 14 18 17 17 17 5 15 20 19 7 7 10 10 5 14 20 5 19 11 16 16 19 17 9 7 12 14 10 2 11 14 5 20 8 10 11 19 2 14 14 19 17 5 10 8 8 4 2 1 10 20 12 14 11 7 6 6 15 1 5 9 15 3 17 16 17 5 14 11 9 16 15 1 11 10 6 15 7", "output": "93" }, { "input": "100 1\n58 94 18 50 17 14 96 62 83 80 75 5 9 22 25 41 3 96 74 45 66 37 2 37 13 85 68 54 77 11 85 19 25 21 52 59 90 61 72 89 82 22 10 16 3 68 61 29 55 76 28 85 65 76 27 3 14 10 56 37 86 18 35 38 56 68 23 88 33 38 52 87 55 83 94 34 100 41 83 56 91 77 32 74 97 13 67 31 57 81 53 39 5 88 46 1 79 4 49 42", "output": "77" }, { "input": "100 2\n1 51 76 62 34 93 90 43 57 59 52 78 3 48 11 60 57 48 5 54 28 81 87 23 44 77 67 61 14 73 29 53 21 89 67 41 47 9 63 37 1 71 40 85 4 14 77 40 78 75 89 74 4 70 32 65 81 95 49 90 72 41 76 55 69 83 73 84 85 93 46 6 74 90 62 37 97 7 7 37 83 30 37 88 34 16 11 59 85 19 57 63 85 20 63 97 97 65 61 48", "output": "97" }, { "input": "100 3\n30 83 14 55 61 66 34 98 90 62 89 74 45 93 33 31 75 35 82 100 63 69 48 18 99 2 36 71 14 30 70 76 96 85 97 90 49 36 6 76 37 94 70 3 63 73 75 48 39 29 13 2 46 26 9 56 1 18 54 53 85 34 2 12 1 93 75 67 77 77 14 26 33 25 55 9 57 70 75 6 87 66 18 3 41 69 73 24 49 2 20 72 39 58 91 54 74 56 66 78", "output": "20" }, { "input": "100 4\n69 92 76 3 32 50 15 38 21 22 14 3 67 41 95 12 10 62 83 52 78 1 18 58 94 35 62 71 58 75 13 73 60 34 50 97 50 70 19 96 53 10 100 26 20 39 62 59 88 26 24 83 70 68 66 8 6 38 16 93 2 91 81 89 78 74 21 8 31 56 28 53 77 5 81 5 94 42 77 75 92 15 59 36 61 18 55 45 69 68 81 51 12 42 85 74 98 31 17 41", "output": "97" }, { "input": "100 5\n2 72 10 60 6 50 72 34 97 77 35 43 80 64 40 53 46 6 90 22 29 70 26 68 52 19 72 88 83 18 55 32 99 81 11 21 39 42 41 63 60 97 30 23 55 78 89 35 24 50 99 52 27 76 24 8 20 27 51 37 17 82 69 18 46 19 26 77 52 83 76 65 43 66 84 84 13 30 66 88 84 23 37 1 17 26 11 50 73 56 54 37 40 29 35 8 1 39 50 82", "output": "51" }, { "input": "100 7\n6 73 7 54 92 33 66 65 80 47 2 53 28 59 61 16 54 89 37 48 77 40 49 59 27 52 17 22 78 80 81 80 8 93 50 7 87 57 29 16 89 55 20 7 51 54 30 98 44 96 27 70 1 1 32 61 22 92 84 98 31 89 91 90 28 56 49 25 86 49 55 16 19 1 18 8 88 47 16 18 73 86 2 96 16 91 74 49 38 98 94 25 34 85 29 27 99 31 31 58", "output": "97" }, { "input": "100 9\n36 4 45 16 19 6 10 87 44 82 71 49 70 35 83 19 40 76 45 94 44 96 10 54 82 77 86 63 11 37 21 3 15 89 80 88 89 16 72 23 25 9 51 25 10 45 96 5 6 18 51 31 42 57 41 51 42 15 89 61 45 82 16 48 61 67 19 40 9 33 90 36 78 36 79 79 16 10 83 87 9 22 84 12 23 76 36 14 2 81 56 33 56 23 57 84 76 55 35 88", "output": "47" }, { "input": "100 10\n75 81 39 64 90 58 92 28 75 9 96 78 92 83 77 68 76 71 14 46 58 60 80 25 78 11 13 63 22 82 65 68 47 6 33 63 90 50 85 43 73 94 80 48 67 11 83 17 22 15 94 80 66 99 66 4 46 35 52 1 62 39 96 57 37 47 97 49 64 12 36 63 90 16 4 75 85 82 85 56 13 4 92 45 44 93 17 35 22 46 18 44 29 7 52 4 100 98 87 51", "output": "98" }, { "input": "100 20\n21 19 61 70 54 97 98 14 61 72 25 94 24 56 55 25 12 80 76 11 35 17 80 26 11 94 52 47 84 61 10 2 74 25 10 21 2 79 55 50 30 75 10 64 44 5 60 96 52 16 74 41 20 77 20 44 8 86 74 36 49 61 99 13 54 64 19 99 50 43 12 73 48 48 83 55 72 73 63 81 30 27 95 9 97 82 24 3 89 90 33 14 47 88 22 78 12 75 58 67", "output": "94" }, { "input": "100 30\n56 79 59 23 11 23 67 82 81 80 99 79 8 58 93 36 98 81 46 39 34 67 3 50 4 68 70 71 2 21 52 30 75 23 33 21 16 100 56 43 8 27 40 8 56 24 17 40 94 10 67 49 61 36 95 87 17 41 7 94 33 19 17 50 26 11 94 54 38 46 77 9 53 35 98 42 50 20 43 6 78 6 38 24 100 45 43 16 1 50 16 46 14 91 95 88 10 1 50 19", "output": "95" }, { "input": "100 40\n86 11 97 17 38 95 11 5 13 83 67 75 50 2 46 39 84 68 22 85 70 23 64 46 59 93 39 80 35 78 93 21 83 19 64 1 49 59 99 83 44 81 70 58 15 82 83 47 55 65 91 10 2 92 4 77 37 32 12 57 78 11 42 8 59 21 96 69 61 30 44 29 12 70 91 14 10 83 11 75 14 10 19 39 8 98 5 81 66 66 79 55 36 29 22 45 19 24 55 49", "output": "88" }, { "input": "100 50\n22 39 95 69 94 53 80 73 33 90 40 60 2 4 84 50 70 38 92 12 36 74 87 70 51 36 57 5 54 6 35 81 52 17 55 100 95 81 32 76 21 1 100 1 95 1 40 91 98 59 84 19 11 51 79 19 47 86 45 15 62 2 59 77 31 68 71 92 17 33 10 33 85 57 5 2 88 97 91 99 63 20 63 54 79 93 24 62 46 27 30 87 3 64 95 88 16 50 79 1", "output": "99" }, { "input": "100 70\n61 48 89 17 97 6 93 13 64 50 66 88 24 52 46 99 6 65 93 64 82 37 57 41 47 1 84 5 97 83 79 46 16 35 40 7 64 15 44 96 37 17 30 92 51 67 26 3 14 56 27 68 66 93 36 39 51 6 40 55 79 26 71 54 8 48 18 2 71 12 55 60 29 37 31 97 26 37 25 68 67 70 3 87 100 41 5 82 65 92 24 66 76 48 89 8 40 93 31 95", "output": "100" }, { "input": "100 90\n87 32 30 15 10 52 93 63 84 1 82 41 27 51 75 32 42 94 39 53 70 13 4 22 99 35 44 38 5 23 18 100 61 80 9 12 42 93 9 77 3 7 60 95 66 78 95 42 69 8 1 88 93 66 96 20 76 63 15 36 92 52 2 72 36 57 48 63 29 20 74 88 49 47 81 61 94 74 70 93 47 3 19 52 59 41 5 40 22 3 76 97 91 37 95 88 91 99 76 15", "output": "98" }, { "input": "100 100\n79 75 7 28 6 96 38 35 57 95 41 74 24 96 32 78 81 13 63 84 24 95 3 23 66 1 60 6 96 49 41 5 14 18 31 97 66 19 49 89 49 70 51 28 20 99 18 1 28 77 24 46 69 21 40 32 31 66 28 6 66 97 9 16 70 90 91 30 34 82 93 41 65 11 39 52 1 88 63 43 80 50 60 49 28 56 18 76 24 57 74 1 28 99 36 35 79 54 18 16", "output": "100" }, { "input": "1 3\n5", "output": "1" }, { "input": "1 1\n100", "output": "1" }, { "input": "2 3\n4 2", "output": "1" }, { "input": "2 5\n99 97", "output": "2" }, { "input": "3 4\n7 5 2", "output": "2" }, { "input": "3 50\n47 86 51", "output": "3" }, { "input": "5 100\n82 100 85 1 37", "output": "5" }, { "input": "5 20\n40 39 21 5 20", "output": "3" }, { "input": "1 27\n81", "output": "1" }, { "input": "20 13\n7 8 29 83 74 28 93 85 7 8 3 9 8 70 49 50 39 41 57 1", "output": "7" }, { "input": "2 1\n100 2", "output": "1" }, { "input": "2 2\n6 4", "output": "1" }, { "input": "5 2\n6 4 4 1 1", "output": "1" }, { "input": "1 4\n3", "output": "1" }, { "input": "3 2\n1 5 3", "output": "2" }, { "input": "3 1\n3 2 2", "output": "1" }, { "input": "3 1\n2 3 2", "output": "2" }, { "input": "5 1\n5 1 1 1 1", "output": "1" }, { "input": "2 3\n7 4", "output": "1" } ]

1,646,919,245

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

15

0

n, m = input().split() m = int(m) n = int(n) q = {i+1: int(x) for i, x in enumerate(input().split())} print(q) while True: for i in range(1, n+1): if len(q) == 1: break if not i in q: continue q[i] -= m if q[i] < 1: q.pop(i) if len(q) == 1: print(list(q.keys())[0]) break

Title: Jzzhu and Children Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies. Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm: 1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty. Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order? Input Specification: The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100). Output Specification: Output a single integer, representing the number of the last child. Demo Input: ['5 2\n1 3 1 4 2\n', '6 4\n1 1 2 2 3 3\n'] Demo Output: ['4\n', '6\n'] Note: Let's consider the first sample. Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home. Child 4 is the last one who goes home.

```python n, m = input().split() m = int(m) n = int(n) q = {i+1: int(x) for i, x in enumerate(input().split())} print(q) while True: for i in range(1, n+1): if len(q) == 1: break if not i in q: continue q[i] -= m if q[i] < 1: q.pop(i) if len(q) == 1: print(list(q.keys())[0]) break ```

0

59

A

Word

PROGRAMMING

800

[ "implementation", "strings" ]

A. Word

2

256

Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.

The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.

Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.

[ "HoUse\n", "ViP\n", "maTRIx\n" ]

[ "house\n", "VIP\n", "matrix\n" ]

none

500

[ { "input": "HoUse", "output": "house" }, { "input": "ViP", "output": "VIP" }, { "input": "maTRIx", "output": "matrix" }, { "input": "BNHWpnpawg", "output": "bnhwpnpawg" }, { "input": "VTYGP", "output": "VTYGP" }, { "input": "CHNenu", "output": "chnenu" }, { "input": "ERPZGrodyu", "output": "erpzgrodyu" }, { "input": "KSXBXWpebh", "output": "KSXBXWPEBH" }, { "input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv", "output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv" }, { "input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd", "output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd" }, { "input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv", "output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv" }, { "input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg", "output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg" }, { "input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak", "output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK" }, { "input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw", "output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW" }, { "input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB", "output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB" }, { "input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge", "output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge" }, { "input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw", "output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw" }, { "input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks", "output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks" }, { "input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz", "output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ" }, { "input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg", "output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG" }, { "input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO", "output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO" }, { "input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd", "output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD" }, { "input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas", "output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas" }, { "input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm", "output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM" }, { "input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn", "output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn" }, { "input": "t", "output": "t" }, { "input": "N", "output": "N" }, { "input": "kv", "output": "kv" }, { "input": "Ur", "output": "ur" }, { "input": "CN", "output": "CN" } ]

1,678,010,051

2,147,483,647

Python 3

OK

TESTS

30

92

0

s = input() l,c = 0,0 for i in s: if i.isupper(): c = c + 1 if i.islower(): l = l + 1 print(s.lower() if l>=c else s.upper())

Title: Word Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. Input Specification: The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. Output Specification: Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. Demo Input: ['HoUse\n', 'ViP\n', 'maTRIx\n'] Demo Output: ['house\n', 'VIP\n', 'matrix\n'] Note: none

```python s = input() l,c = 0,0 for i in s: if i.isupper(): c = c + 1 if i.islower(): l = l + 1 print(s.lower() if l>=c else s.upper()) ```

3.977

204

A

Little Elephant and Interval

PROGRAMMING

1,500

[ "binary search", "combinatorics", "dp" ]

null

The Little Elephant very much loves sums on intervals. This time he has a pair of integers *l* and *r* (*l*<=≤<=*r*). The Little Elephant has to find the number of such integers *x* (*l*<=≤<=*x*<=≤<=*r*), that the first digit of integer *x* equals the last one (in decimal notation). For example, such numbers as 101, 477474 or 9 will be included in the answer and 47, 253 or 1020 will not. Help him and count the number of described numbers *x* for a given pair *l* and *r*.

The single line contains a pair of integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018) — the boundaries of the interval. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

On a single line print a single integer — the answer to the problem.

[ "2 47\n", "47 1024\n" ]

[ "12\n", "98\n" ]

In the first sample the answer includes integers 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44.

500

[ { "input": "2 47", "output": "12" }, { "input": "47 1024", "output": "98" }, { "input": "1 1000", "output": "108" }, { "input": "1 10000", "output": "1008" }, { "input": "47 8545", "output": "849" }, { "input": "1000 1000", "output": "0" }, { "input": "47547 4587554587754542", "output": "458755458770699" }, { "input": "1 1000000", "output": "100008" }, { "input": "47 74", "output": "2" }, { "input": "10001 10000002", "output": "999001" }, { "input": "10000 100000", "output": "9000" }, { "input": "458754 4588754", "output": "413001" }, { "input": "111 111", "output": "1" }, { "input": "110 147", "output": "4" }, { "input": "1 1000000000", "output": "100000008" }, { "input": "12 10000000000", "output": "999999998" }, { "input": "1000000000 1000000000", "output": "0" }, { "input": "1 1000000000000000000", "output": "100000000000000008" }, { "input": "11 111111111111111100", "output": "11111111111111109" }, { "input": "100000000000000000 1000000000000000000", "output": "90000000000000000" }, { "input": "45481484484 848469844684844", "output": "84842436320036" }, { "input": "975400104587000 48754000000000001", "output": "4777859989541300" }, { "input": "11220451511 51511665251233335", "output": "5151165403078183" }, { "input": "77 77", "output": "1" }, { "input": "99 102", "output": "2" }, { "input": "9997 87878000008", "output": "8787799002" }, { "input": "10000000001 111111111111100001", "output": "11111110111110001" }, { "input": "7777 88888", "output": "8112" }, { "input": "999999999 10000000000", "output": "900000001" }, { "input": "235 236", "output": "0" }, { "input": "1 1", "output": "1" }, { "input": "2 2", "output": "1" }, { "input": "1 2", "output": "2" }, { "input": "4 7", "output": "4" }, { "input": "7 10", "output": "3" }, { "input": "1 11", "output": "10" }, { "input": "1 10", "output": "9" }, { "input": "7 8", "output": "2" }, { "input": "88 990", "output": "91" }, { "input": "458985985498001244 985458425544874008", "output": "52647244004687276" }, { "input": "115998725487587451 245744899758754501", "output": "12974617427116705" }, { "input": "595754249475458004 615044544745124547", "output": "1929029526966655" }, { "input": "9754875457700 1000000000000000000", "output": "99999024512454230" }, { "input": "8758754570000 999999999999999999", "output": "99999124124543000" }, { "input": "111111111111111111 333333333444444445", "output": "22222222233333334" }, { "input": "822981258385599125 841978899930248528", "output": "1899764154464941" }, { "input": "779547115376367013 980561039207670775", "output": "20101392383130376" }, { "input": "335408916782916802 416495628489807285", "output": "8108671170689049" }, { "input": "252509053898415172 285803555062529649", "output": "3329450116411448" }, { "input": "919845424847912645 970651082117950285", "output": "5080565727003764" }, { "input": "522842183413115088 853628713003942530", "output": "33078652959082744" }, { "input": "84324827171274023 607953653548585226", "output": "52362882637731121" }, { "input": "1312148742261681 277460340506883334", "output": "27614819176462166" }, { "input": "645762257531682046 885295120956158518", "output": "23953286342447648" }, { "input": "819875140559301752 946247219812473271", "output": "12637207925317152" }, { "input": "4 19", "output": "7" }, { "input": "5 45", "output": "9" }, { "input": "9999999999999987 99999999999999711", "output": "8999999999999973" }, { "input": "2 3", "output": "2" }, { "input": "1827171 232817181719384635", "output": "23281718171755747" }, { "input": "999999999999999999 1000000000000000000", "output": "1" }, { "input": "73 678", "output": "61" } ]

1,589,973,619

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

4

280

0

import math a,b=map(int,input().split()) n=b sum=0 ans=0 l1=[] while(n!=0): l1.append(n%10) sum+=1 n=n//10 if(sum==1): print(b-a+1) elif(sum>1): ans+=9 j=1 while(sum-2!=0): ans=ans+9*pow(10,j-1); j+=1 sum-=1 l1=l1[::-1] pro=1 if(l1[0]<=l1[len(l1)-1]): pro=l1[0] for i in range(1,len(l1)-1): pro=pro*(l1[i]+1) else: l1[len(l1)-2]-=1 pro=l1[0] for i in range(1,len(l1)-1): pro=pro*(l1[i]+1) ans=ans+pro n1=a-1 sum1=0 ans1=0 l2=[] while(n1!=0): l2.append(n1%10) sum1+=1 n1=n1//10 if(sum1==1): ans1=a-1 elif(sum1>1): ans1+=9 j=1 while(sum1-2!=0): ans1=ans1+9*pow(10,j-1); j+=1 sum1-=1 l2=l2[::-1] pro1=1 if(l2[0]<=l2[len(l2)-1]): pro1=l2[0] for i in range(1,len(l2)-1): pro1=pro1*(l2[i]+1) else: l2[len(l2)-2]-=1 pro1=l2[0] for i in range(1,len(l2)-1): pro1=pro1*(l2[i]+1) ans1=ans1+pro1 print(ans-ans1)

Title: Little Elephant and Interval Time Limit: None seconds Memory Limit: None megabytes Problem Description: The Little Elephant very much loves sums on intervals. This time he has a pair of integers *l* and *r* (*l*<=≤<=*r*). The Little Elephant has to find the number of such integers *x* (*l*<=≤<=*x*<=≤<=*r*), that the first digit of integer *x* equals the last one (in decimal notation). For example, such numbers as 101, 477474 or 9 will be included in the answer and 47, 253 or 1020 will not. Help him and count the number of described numbers *x* for a given pair *l* and *r*. Input Specification: The single line contains a pair of integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018) — the boundaries of the interval. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. Output Specification: On a single line print a single integer — the answer to the problem. Demo Input: ['2 47\n', '47 1024\n'] Demo Output: ['12\n', '98\n'] Note: In the first sample the answer includes integers 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44.

```python import math a,b=map(int,input().split()) n=b sum=0 ans=0 l1=[] while(n!=0): l1.append(n%10) sum+=1 n=n//10 if(sum==1): print(b-a+1) elif(sum>1): ans+=9 j=1 while(sum-2!=0): ans=ans+9*pow(10,j-1); j+=1 sum-=1 l1=l1[::-1] pro=1 if(l1[0]<=l1[len(l1)-1]): pro=l1[0] for i in range(1,len(l1)-1): pro=pro*(l1[i]+1) else: l1[len(l1)-2]-=1 pro=l1[0] for i in range(1,len(l1)-1): pro=pro*(l1[i]+1) ans=ans+pro n1=a-1 sum1=0 ans1=0 l2=[] while(n1!=0): l2.append(n1%10) sum1+=1 n1=n1//10 if(sum1==1): ans1=a-1 elif(sum1>1): ans1+=9 j=1 while(sum1-2!=0): ans1=ans1+9*pow(10,j-1); j+=1 sum1-=1 l2=l2[::-1] pro1=1 if(l2[0]<=l2[len(l2)-1]): pro1=l2[0] for i in range(1,len(l2)-1): pro1=pro1*(l2[i]+1) else: l2[len(l2)-2]-=1 pro1=l2[0] for i in range(1,len(l2)-1): pro1=pro1*(l2[i]+1) ans1=ans1+pro1 print(ans-ans1) ```

0

446

A

DZY Loves Sequences

PROGRAMMING

1,600

[ "dp", "implementation", "two pointers" ]

null

DZY has a sequence *a*, consisting of *n* integers. We'll call a sequence *a**i*,<=*a**i*<=+<=1,<=...,<=*a**j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) a subsegment of the sequence *a*. The value (*j*<=-<=*i*<=+<=1) denotes the length of the subsegment. Your task is to find the longest subsegment of *a*, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing. You only need to output the length of the subsegment you find.

The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).

In a single line print the answer to the problem — the maximum length of the required subsegment.

[ "6\n7 2 3 1 5 6\n" ]

[ "5\n" ]

You can choose subsegment *a*2, *a*3, *a*4, *a*5, *a*6 and change its 3rd element (that is *a*4) to 4.

500

[ { "input": "6\n7 2 3 1 5 6", "output": "5" }, { "input": "10\n424238336 649760493 681692778 714636916 719885387 804289384 846930887 957747794 596516650 189641422", "output": "9" }, { "input": "50\n804289384 846930887 681692778 714636916 957747794 424238336 719885387 649760493 596516650 189641422 25202363 350490028 783368691 102520060 44897764 967513927 365180541 540383427 304089173 303455737 35005212 521595369 294702568 726956430 336465783 861021531 59961394 89018457 101513930 125898168 131176230 145174068 233665124 278722863 315634023 369133070 468703136 628175012 635723059 653377374 656478043 801979803 859484422 914544920 608413785 756898538 734575199 973594325 149798316 38664371", "output": "19" }, { "input": "1\n1", "output": "1" }, { "input": "2\n1000000000 1000000000", "output": "2" }, { "input": "5\n1 2 3 4 1", "output": "5" }, { "input": "10\n1 2 3 4 5 5 6 7 8 9", "output": "6" }, { "input": "5\n1 1 1 1 1", "output": "2" }, { "input": "5\n1 1 2 3 4", "output": "5" }, { "input": "5\n1 2 3 1 6", "output": "5" }, { "input": "1\n42", "output": "1" }, { "input": "5\n1 2 42 3 4", "output": "4" }, { "input": "5\n1 5 9 6 10", "output": "4" }, { "input": "5\n5 2 3 4 5", "output": "5" }, { "input": "3\n2 1 3", "output": "3" }, { "input": "5\n1 2 3 3 4", "output": "4" }, { "input": "8\n1 2 3 4 1 5 6 7", "output": "5" }, { "input": "1\n3", "output": "1" }, { "input": "3\n5 1 2", "output": "3" }, { "input": "4\n1 4 3 4", "output": "4" }, { "input": "6\n7 2 12 4 5 6", "output": "5" }, { "input": "6\n7 2 3 1 4 5", "output": "4" }, { "input": "6\n2 3 5 5 6 7", "output": "6" }, { "input": "5\n2 4 7 6 8", "output": "5" }, { "input": "3\n3 1 2", "output": "3" }, { "input": "3\n1 1 2", "output": "3" }, { "input": "2\n1 2", "output": "2" }, { "input": "5\n4 1 2 3 4", "output": "5" }, { "input": "20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6", "output": "7" }, { "input": "4\n1 2 1 3", "output": "3" }, { "input": "4\n4 3 1 2", "output": "3" }, { "input": "6\n1 2 2 3 4 5", "output": "5" }, { "input": "4\n1 1 1 2", "output": "3" }, { "input": "4\n5 1 2 3", "output": "4" }, { "input": "5\n9 1 2 3 4", "output": "5" }, { "input": "2\n1 1", "output": "2" }, { "input": "5\n1 3 2 4 5", "output": "4" }, { "input": "6\n1 2 1 2 4 5", "output": "5" }, { "input": "10\n1 1 5 3 2 9 9 7 7 6", "output": "3" }, { "input": "6\n1 2 3 100000 100 101", "output": "6" }, { "input": "4\n3 3 3 4", "output": "3" }, { "input": "3\n4 3 5", "output": "3" }, { "input": "5\n1 3 2 3 4", "output": "4" }, { "input": "10\n1 2 3 4 5 10 10 11 12 13", "output": "10" }, { "input": "7\n11 2 1 2 13 4 14", "output": "5" }, { "input": "3\n5 1 3", "output": "3" }, { "input": "4\n1 5 3 4", "output": "4" }, { "input": "10\n1 2 3 4 100 6 7 8 9 10", "output": "10" }, { "input": "3\n5 3 5", "output": "3" }, { "input": "5\n100 100 7 8 9", "output": "4" }, { "input": "5\n1 2 3 4 5", "output": "5" }, { "input": "5\n1 2 4 4 5", "output": "5" }, { "input": "6\n7 4 5 6 7 8", "output": "6" }, { "input": "9\n3 4 1 6 3 4 5 6 7", "output": "7" }, { "input": "3\n1000 1 2", "output": "3" }, { "input": "3\n20 1 9", "output": "3" }, { "input": "6\n7 2 3 1 4 6", "output": "4" }, { "input": "3\n100 5 10", "output": "3" }, { "input": "4\n2 2 2 3", "output": "3" }, { "input": "6\n4 2 8 1 2 5", "output": "4" }, { "input": "3\n25 1 6", "output": "3" }, { "input": "10\n17 99 23 72 78 36 5 43 95 9", "output": "5" }, { "input": "7\n21 16 22 21 11 13 19", "output": "4" }, { "input": "5\n1 2 5 3 4", "output": "4" }, { "input": "6\n2 2 2 3 4 5", "output": "5" }, { "input": "5\n1 3 1 2 3", "output": "4" }, { "input": "3\n81 33 64", "output": "3" }, { "input": "7\n14 3 3 19 13 19 15", "output": "4" }, { "input": "9\n1 2 3 4 5 42 7 8 9", "output": "9" }, { "input": "5\n2 3 7 5 6", "output": "5" }, { "input": "5\n1 3 3 4 5", "output": "5" }, { "input": "6\n1 5 4 3 4 5", "output": "4" } ]

1,580,379,078

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

26

265

10,649,600

n = int(input()) a = list(map(int, input().split())) start = [1] * n for i in range(n - 2, -1, -1): if a[i] < a[i + 1]: start[i] += start[i + 1] end = [1] * n for i in range(1, n): if a[i] > a[i - 1]: end[i] += end[i - 1] ret = max(start) for i in range(1, n): if a[i] <= a[i - 1]: ret = max(ret, end[i - 1] + 1) if i + 1 < n and a[i + 1] - a[i - 1] > 1: ret = max(ret, end[i - 1] + 1 + start[i + 1]) print(ret)

Title: DZY Loves Sequences Time Limit: None seconds Memory Limit: None megabytes Problem Description: DZY has a sequence *a*, consisting of *n* integers. We'll call a sequence *a**i*,<=*a**i*<=+<=1,<=...,<=*a**j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) a subsegment of the sequence *a*. The value (*j*<=-<=*i*<=+<=1) denotes the length of the subsegment. Your task is to find the longest subsegment of *a*, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing. You only need to output the length of the subsegment you find. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). Output Specification: In a single line print the answer to the problem — the maximum length of the required subsegment. Demo Input: ['6\n7 2 3 1 5 6\n'] Demo Output: ['5\n'] Note: You can choose subsegment *a*2, *a*3, *a*4, *a*5, *a*6 and change its 3rd element (that is *a*4) to 4.

```python n = int(input()) a = list(map(int, input().split())) start = [1] * n for i in range(n - 2, -1, -1): if a[i] < a[i + 1]: start[i] += start[i + 1] end = [1] * n for i in range(1, n): if a[i] > a[i - 1]: end[i] += end[i - 1] ret = max(start) for i in range(1, n): if a[i] <= a[i - 1]: ret = max(ret, end[i - 1] + 1) if i + 1 < n and a[i + 1] - a[i - 1] > 1: ret = max(ret, end[i - 1] + 1 + start[i + 1]) print(ret) ```

0

385

A

Bear and Raspberry

PROGRAMMING

1,000

[ "brute force", "greedy", "implementation" ]

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" } ]

1,649,881,600

2,147,483,647

PyPy 3-64

OK

TESTS

32

62

0

n, c = map(int, input().split()) arr = list(map(int, input().split())) mx = 0 for i in range(n - 1): mx = max(mx, arr[i] - arr[i + 1] - c) print(mx)

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().split()) arr = list(map(int, input().split())) mx = 0 for i in range(n - 1): mx = max(mx, arr[i] - arr[i + 1] - c) print(mx) ```

3

691

A

Fashion in Berland

PROGRAMMING

1,000

[ "implementation" ]

null

According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open. You are given a jacket with *n* buttons. Determine if it is fastened in a right way.

The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of buttons on the jacket. The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1). The number *a**i*<==<=0 if the *i*-th button is not fastened. Otherwise *a**i*<==<=1.

In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".

[ "3\n1 0 1\n", "3\n1 0 0\n" ]

[ "YES\n", "NO\n" ]

none

0

[ { "input": "3\n1 0 1", "output": "YES" }, { "input": "3\n1 0 0", "output": "NO" }, { "input": "3\n1 1 0", "output": "YES" }, { "input": "3\n1 1 1", "output": "NO" }, { "input": "3\n0 0 1", "output": "NO" }, { "input": "3\n0 0 0", "output": "NO" }, { "input": "3\n0 1 1", "output": "YES" }, { "input": "3\n0 1 0", "output": "NO" }, { "input": "2\n0 1", "output": "YES" }, { "input": "2\n1 0", "output": "YES" }, { "input": "2\n0 0", "output": "NO" }, { "input": "2\n1 1", "output": "NO" }, { "input": "1\n1", "output": "YES" }, { "input": "1\n0", "output": "NO" }, { "input": "10\n1 0 1 1 1 1 1 1 1 1", "output": "YES" }, { "input": "5\n1 1 1 1 0", "output": "YES" }, { "input": "4\n1 0 1 0", "output": "NO" }, { "input": "4\n1 1 0 0", "output": "NO" }, { "input": "10\n0 0 1 1 1 1 1 1 1 1", "output": "NO" }, { "input": "10\n1 1 1 1 1 1 1 1 1 1", "output": "NO" }, { "input": "123\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "NO" }, { "input": "124\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "NO" }, { "input": "94\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "NO" }, { "input": "64\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "NO" }, { "input": "34\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", "output": "NO" }, { "input": "35\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", "output": "NO" }, { "input": "5\n1 1 1 1 1", "output": "NO" }, { "input": "10\n0 0 0 0 0 0 0 0 0 0", "output": "NO" }, { "input": "231\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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "NO" }, { "input": "10\n1 1 0 1 1 1 0 1 1 1", "output": "NO" } ]

1,474,011,922

2,147,483,647

Python 3

COMPILATION_ERROR

TESTS

0

#include <stdio.h> int main(void) { int i,n,a[1000],count=0; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d",&a[i]); } for(i=0; i < n; i++) { if(a[i] == 1) { count++; } } if((n -1) == count ) { printf("YES"); } else { printf("NO"); } }

Title: Fashion in Berland Time Limit: None seconds Memory Limit: None megabytes Problem Description: According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open. You are given a jacket with *n* buttons. Determine if it is fastened in a right way. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of buttons on the jacket. The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1). The number *a**i*<==<=0 if the *i*-th button is not fastened. Otherwise *a**i*<==<=1. Output Specification: In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO". Demo Input: ['3\n1 0 1\n', '3\n1 0 0\n'] Demo Output: ['YES\n', 'NO\n'] Note: none

```python #include <stdio.h> int main(void) { int i,n,a[1000],count=0; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d",&a[i]); } for(i=0; i < n; i++) { if(a[i] == 1) { count++; } } if((n -1) == count ) { printf("YES"); } else { printf("NO"); } } ```

-1

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" } ]

1,679,835,018

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

15

0

import sys n = int(input()) for i in range (n) : def get_string(): return sys.stdin.readline().strip() string = get_string() long = int (len(string)) print(string[0] + str(long-2) + string[long - 1])

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 import sys n = int(input()) for i in range (n) : def get_string(): return sys.stdin.readline().strip() string = get_string() long = int (len(string)) print(string[0] + str(long-2) + string[long - 1]) ```

0

266

A

Stones on the Table

PROGRAMMING

800

[ "implementation" ]

null

There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.

The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table. The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.

Print a single integer — the answer to the problem.

[ "3\nRRG\n", "5\nRRRRR\n", "4\nBRBG\n" ]

[ "1\n", "4\n", "0\n" ]

none

500

[ { "input": "3\nRRG", "output": "1" }, { "input": "5\nRRRRR", "output": "4" }, { "input": "4\nBRBG", "output": "0" }, { "input": "1\nB", "output": "0" }, { "input": "2\nBG", "output": "0" }, { "input": "3\nBGB", "output": "0" }, { "input": "4\nRBBR", "output": "1" }, { "input": "5\nRGGBG", "output": "1" }, { "input": "10\nGGBRBRGGRB", "output": "2" }, { "input": "50\nGRBGGRBRGRBGGBBBBBGGGBBBBRBRGBRRBRGBBBRBBRRGBGGGRB", "output": "18" }, { "input": "15\nBRRBRGGBBRRRRGR", "output": "6" }, { "input": "20\nRRGBBRBRGRGBBGGRGRRR", "output": "6" }, { "input": "25\nBBGBGRBGGBRRBGRRBGGBBRBRB", "output": "6" }, { "input": "30\nGRGGGBGGRGBGGRGRBGBGBRRRRRRGRB", "output": "9" }, { "input": "35\nGBBGBRGBBGGRBBGBRRGGRRRRRRRBRBBRRGB", "output": "14" }, { "input": "40\nGBBRRGBGGGRGGGRRRRBRBGGBBGGGBGBBBBBRGGGG", "output": "20" }, { "input": "45\nGGGBBRBBRRGRBBGGBGRBRGGBRBRGBRRGBGRRBGRGRBRRG", "output": "11" }, { "input": "50\nRBGGBGGRBGRBBBGBBGRBBBGGGRBBBGBBBGRGGBGGBRBGBGRRGG", "output": "17" }, { "input": "50\nGGGBBRGGGGGRRGGRBGGRGBBRBRRBGRGBBBGBRBGRGBBGRGGBRB", "output": "16" }, { "input": "50\nGBGRGRRBRRRRRGGBBGBRRRBBBRBBBRRGRBBRGBRBGGRGRBBGGG", "output": "19" }, { "input": "10\nGRRBRBRBGR", "output": "1" }, { "input": "10\nBRBGBGRRBR", "output": "1" }, { "input": "20\nGBGBGGRRRRGRBBGRGRGR", "output": "5" }, { "input": "20\nRRGGRBBGBBRBGRRBRRBG", "output": "6" }, { "input": "30\nBGBRGBBBGRGBBRGBGRBBBRGGRRGRRB", "output": "8" }, { "input": "30\nBBBBGGBRBGBBGBGBGBGGGRGRRGGBBB", "output": "11" }, { "input": "40\nGBRRGRBGBRRGBRGGGBRGBGBRGBBRRGRGGBBGBGBB", "output": "9" }, { "input": "40\nBRGRGGRGGRBBRRRBRBBGGGRRGBGBBGRBBRGBRRGG", "output": "13" }, { "input": "50\nRBGBGGRRGGRGGBGBGRRBGGBGBRRBBGBBGBBBGBBRBBRBRBRGRG", "output": "13" }, { "input": "50\nRBRRGBGRRRBGRRBGRRGRBBRBBRRBRGGBRBRRBGGRBGGBRBRGRB", "output": "12" }, { "input": "2\nBB", "output": "1" }, { "input": "50\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR", "output": "49" }, { "input": "50\nRRRRRRRRGRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR", "output": "47" }, { "input": "50\nRRRRRRRRRRRRGGRRRRRRRRRBRRRRRRRRRRRRRRBBRRRRRRRRRR", "output": "43" } ]

1,697,286,455

2,147,483,647

PyPy 3-64

OK

TESTS

34

122

0

def main(n: int, s: str) -> int: a = list(s) def check_function(n_sub: int, a_sub: list) -> list: index_list = [] for i in range(int(n_sub) - 1): if a_sub[i] == a_sub[i+1]: index_list.append(i) else: continue return index_list print(len(check_function(n, a))) if __name__ == "__main__": n = input() s = input() main(n, s)

Title: Stones on the Table Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table. The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue. Output Specification: Print a single integer — the answer to the problem. Demo Input: ['3\nRRG\n', '5\nRRRRR\n', '4\nBRBG\n'] Demo Output: ['1\n', '4\n', '0\n'] Note: none

```python def main(n: int, s: str) -> int: a = list(s) def check_function(n_sub: int, a_sub: list) -> list: index_list = [] for i in range(int(n_sub) - 1): if a_sub[i] == a_sub[i+1]: index_list.append(i) else: continue return index_list print(len(check_function(n, a))) if __name__ == "__main__": n = input() s = input() main(n, s) ```

3