contestId int64 0 1.01k | index stringclasses 57 values | name stringlengths 2 58 | type stringclasses 2 values | rating int64 0 3.5k | tags listlengths 0 11 | title stringclasses 522 values | time-limit stringclasses 8 values | memory-limit stringclasses 8 values | problem-description stringlengths 0 7.15k | input-specification stringlengths 0 2.05k | output-specification stringlengths 0 1.5k | demo-input listlengths 0 7 | demo-output listlengths 0 7 | note stringlengths 0 5.24k | points float64 0 425k | test_cases listlengths 0 402 | creationTimeSeconds int64 1.37B 1.7B | relativeTimeSeconds int64 8 2.15B | programmingLanguage stringclasses 3 values | verdict stringclasses 14 values | testset stringclasses 12 values | passedTestCount int64 0 1k | timeConsumedMillis int64 0 15k | memoryConsumedBytes int64 0 805M | code stringlengths 3 65.5k | prompt stringlengths 262 8.2k | response stringlengths 17 65.5k | score float64 -1 3.99 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
200 | B | Drinks | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space. | Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4. | [
"3\n50 50 100\n",
"4\n0 25 50 75\n"
] | [
"66.666666666667\n",
"37.500000000000\n"
] | Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent. | 500 | [
{
"input": "3\n50 50 100",
"output": "66.666666666667"
},
{
"input": "4\n0 25 50 75",
"output": "37.500000000000"
},
{
"input": "3\n0 1 8",
"output": "3.000000000000"
},
{
"input": "5\n96 89 93 95 70",
"output": "88.600000000000"
},
{
"input": "7\n62 41 78 4 38 39 75",
"output": "48.142857142857"
},
{
"input": "13\n2 22 7 0 1 17 3 17 11 2 21 26 22",
"output": "11.615384615385"
},
{
"input": "21\n5 4 11 7 0 5 45 21 0 14 51 6 0 16 10 19 8 9 7 12 18",
"output": "12.761904761905"
},
{
"input": "26\n95 70 93 74 94 70 91 70 39 79 80 57 87 75 37 93 48 67 51 90 85 26 23 64 66 84",
"output": "69.538461538462"
},
{
"input": "29\n84 99 72 96 83 92 95 98 97 93 76 84 99 93 81 76 93 99 99 100 95 100 96 95 97 100 71 98 94",
"output": "91.551724137931"
},
{
"input": "33\n100 99 100 100 99 99 99 100 100 100 99 99 99 100 100 100 100 99 100 99 100 100 97 100 100 100 100 100 100 100 98 98 100",
"output": "99.515151515152"
},
{
"input": "34\n14 9 10 5 4 26 18 23 0 1 0 20 18 15 2 2 3 5 14 1 9 4 2 15 7 1 7 19 10 0 0 11 0 2",
"output": "8.147058823529"
},
{
"input": "38\n99 98 100 100 99 92 99 99 98 84 88 94 86 99 93 100 98 99 65 98 85 84 64 97 96 89 79 96 91 84 99 93 72 96 94 97 96 93",
"output": "91.921052631579"
},
{
"input": "52\n100 94 99 98 99 99 99 95 97 97 98 100 100 98 97 100 98 90 100 99 97 94 90 98 100 100 90 99 100 95 98 95 94 85 97 94 96 94 99 99 99 98 100 100 94 99 99 100 98 87 100 100",
"output": "97.019230769231"
},
{
"input": "58\n10 70 12 89 1 82 100 53 40 100 21 69 92 91 67 66 99 77 25 48 8 63 93 39 46 79 82 14 44 42 1 79 0 69 56 73 67 17 59 4 65 80 20 60 77 52 3 61 16 76 33 18 46 100 28 59 9 6",
"output": "50.965517241379"
},
{
"input": "85\n7 8 1 16 0 15 1 7 0 11 15 6 2 12 2 8 9 8 2 0 3 7 15 7 1 8 5 7 2 26 0 3 11 1 8 10 31 0 7 6 1 8 1 0 9 14 4 8 7 16 9 1 0 16 10 9 6 1 1 4 2 7 4 5 4 1 20 6 16 16 1 1 10 17 8 12 14 19 3 8 1 7 10 23 10",
"output": "7.505882352941"
},
{
"input": "74\n5 3 0 7 13 10 12 10 18 5 0 18 2 13 7 17 2 7 5 2 40 19 0 2 2 3 0 45 4 20 0 4 2 8 1 19 3 9 17 1 15 0 16 1 9 4 0 9 32 2 6 18 11 18 1 15 16 12 7 19 5 3 9 28 26 8 3 10 33 29 4 13 28 6",
"output": "10.418918918919"
},
{
"input": "98\n42 9 21 11 9 11 22 12 52 20 10 6 56 9 26 27 1 29 29 14 38 17 41 21 7 45 15 5 29 4 51 20 6 8 34 17 13 53 30 45 0 10 16 41 4 5 6 4 14 2 31 6 0 11 13 3 3 43 13 36 51 0 7 16 28 23 8 36 30 22 8 54 21 45 39 4 50 15 1 30 17 8 18 10 2 20 16 50 6 68 15 6 38 7 28 8 29 41",
"output": "20.928571428571"
},
{
"input": "99\n60 65 40 63 57 44 30 84 3 10 39 53 40 45 72 20 76 11 61 32 4 26 97 55 14 57 86 96 34 69 52 22 26 79 31 4 21 35 82 47 81 28 72 70 93 84 40 4 69 39 83 58 30 7 32 73 74 12 92 23 61 88 9 58 70 32 75 40 63 71 46 55 39 36 14 97 32 16 95 41 28 20 85 40 5 50 50 50 75 6 10 64 38 19 77 91 50 72 96",
"output": "49.191919191919"
},
{
"input": "99\n100 88 40 30 81 80 91 98 69 73 88 96 79 58 14 100 87 84 52 91 83 88 72 83 99 35 54 80 46 79 52 72 85 32 99 39 79 79 45 83 88 50 75 75 50 59 65 75 97 63 92 58 89 46 93 80 89 33 69 86 99 99 66 85 72 74 79 98 85 95 46 63 77 97 49 81 89 39 70 76 68 91 90 56 31 93 51 87 73 95 74 69 87 95 57 68 49 95 92",
"output": "73.484848484848"
},
{
"input": "100\n18 15 17 0 3 3 0 4 1 8 2 22 7 21 5 0 0 8 3 16 1 0 2 9 9 3 10 8 17 20 5 4 8 12 2 3 1 1 3 2 23 0 1 0 5 7 4 0 1 3 3 4 25 2 2 14 8 4 9 3 0 11 0 3 12 3 14 16 7 7 14 1 17 9 0 35 42 12 3 1 25 9 3 8 5 3 2 8 22 14 11 6 3 9 6 8 7 7 4 6",
"output": "7.640000000000"
},
{
"input": "100\n88 77 65 87 100 63 91 96 92 89 77 95 76 80 84 83 100 71 85 98 26 54 74 78 69 59 96 86 88 91 95 26 52 88 64 70 84 81 76 84 94 82 100 66 97 98 43 94 59 94 100 80 98 73 69 83 94 70 74 79 91 31 62 88 69 55 62 97 40 64 62 83 87 85 50 90 69 72 67 49 100 51 69 96 81 90 83 91 86 34 79 69 100 66 97 98 47 97 74 100",
"output": "77.660000000000"
},
{
"input": "100\n91 92 90 91 98 84 85 96 83 98 99 87 94 70 87 75 86 90 89 88 82 83 91 94 88 86 90 99 100 98 97 75 95 99 95 100 91 92 76 93 95 97 88 93 95 81 96 89 88 100 98 87 90 96 100 99 58 90 96 77 92 82 100 100 93 93 98 99 79 88 97 95 98 66 96 83 96 100 99 92 98 98 92 93 100 97 98 100 98 97 100 100 94 90 99 100 98 79 80 81",
"output": "91.480000000000"
},
{
"input": "1\n0",
"output": "0.000000000000"
},
{
"input": "1\n100",
"output": "100.000000000000"
},
{
"input": "1\n78",
"output": "78.000000000000"
},
{
"input": "2\n0 100",
"output": "50.000000000000"
},
{
"input": "2\n100 100",
"output": "100.000000000000"
},
{
"input": "5\n0 0 0 0 1",
"output": "0.200000000000"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99",
"output": "99.990000000000"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "0.010000000000"
},
{
"input": "5\n100 100 100 100 100",
"output": "100.000000000000"
}
] | 1,697,704,901 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 92 | 0 | n = int(input())
d = list(map(int,input().split()))
s = sum(d)
print(round(s/n,4))
| Title: Drinks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
Output Specification:
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
Demo Input:
['3\n50 50 100\n', '4\n0 25 50 75\n']
Demo Output:
['66.666666666667\n', '37.500000000000\n']
Note:
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent. | ```python
n = int(input())
d = list(map(int,input().split()))
s = sum(d)
print(round(s/n,4))
``` | 3 | |
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"
}
] | 1,582,127,002 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 248 | 0 | n=int(input())
L=list(map(int,input().split()))
L1=[]
for i in L:
if i%2==0:
L1.append(1)
else:
L1.append(0)
if L1.count(1)==1:
print(L1.index(1)+1)
elif L1.count(0)==1:
print(L1.index(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())
L=list(map(int,input().split()))
L1=[]
for i in L:
if i%2==0:
L1.append(1)
else:
L1.append(0)
if L1.count(1)==1:
print(L1.index(1)+1)
elif L1.count(0)==1:
print(L1.index(0)+1)
``` | 3.938 |
251 | A | Points on Line | PROGRAMMING | 1,300 | [
"binary search",
"combinatorics",
"two pointers"
] | null | null | Little Petya likes points a lot. Recently his mom has presented him *n* points lying on the line *OX*. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed *d*.
Note that the order of the points inside the group of three chosen points doesn't matter. | The first line contains two integers: *n* and *d* (1<=≤<=*n*<=≤<=105; 1<=≤<=*d*<=≤<=109). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n*, their absolute value doesn't exceed 109 — the *x*-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase. | Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed *d*.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | [
"4 3\n1 2 3 4\n",
"4 2\n-3 -2 -1 0\n",
"5 19\n1 10 20 30 50\n"
] | [
"4\n",
"2\n",
"1\n"
] | In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}. | 500 | [
{
"input": "4 3\n1 2 3 4",
"output": "4"
},
{
"input": "4 2\n-3 -2 -1 0",
"output": "2"
},
{
"input": "5 19\n1 10 20 30 50",
"output": "1"
},
{
"input": "10 5\n31 36 43 47 48 50 56 69 71 86",
"output": "2"
},
{
"input": "10 50\n1 4 20 27 65 79 82 83 99 100",
"output": "25"
},
{
"input": "10 90\n24 27 40 41 61 69 73 87 95 97",
"output": "120"
},
{
"input": "100 100\n-98 -97 -96 -93 -92 -91 -90 -87 -86 -84 -81 -80 -79 -78 -76 -75 -73 -71 -69 -67 -65 -64 -63 -62 -61 -54 -51 -50 -49 -48 -46 -45 -44 -37 -36 -33 -30 -28 -27 -16 -15 -13 -12 -10 -9 -7 -6 -5 -4 2 3 5 8 9 10 11 13 14 15 16 17 19 22 24 25 26 27 28 30 31 32 36 40 43 45 46 47 50 51 52 53 58 60 63 69 70 73 78 80 81 82 85 88 89 90 91 95 96 97 99",
"output": "79351"
},
{
"input": "1 14751211\n847188590",
"output": "0"
},
{
"input": "2 1000000000\n-907894512 -289906312",
"output": "0"
},
{
"input": "2 1000000000\n-14348867 1760823",
"output": "0"
},
{
"input": "3 1000000000\n-5 -1 1",
"output": "1"
}
] | 1,647,619,730 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 39 | 372 | 18,329,600 | n, k = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
j=0
ans=0
for i in range(2,n):
while abs(a[j]-a[i])>k:
j=j+1
d=i-j
ans=ans+(d)*(d-1)//2
print(ans) | Title: Points on Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya likes points a lot. Recently his mom has presented him *n* points lying on the line *OX*. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed *d*.
Note that the order of the points inside the group of three chosen points doesn't matter.
Input Specification:
The first line contains two integers: *n* and *d* (1<=≤<=*n*<=≤<=105; 1<=≤<=*d*<=≤<=109). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n*, their absolute value doesn't exceed 109 — the *x*-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase.
Output Specification:
Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed *d*.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['4 3\n1 2 3 4\n', '4 2\n-3 -2 -1 0\n', '5 19\n1 10 20 30 50\n']
Demo Output:
['4\n', '2\n', '1\n']
Note:
In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}. | ```python
n, k = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
j=0
ans=0
for i in range(2,n):
while abs(a[j]-a[i])>k:
j=j+1
d=i-j
ans=ans+(d)*(d-1)//2
print(ans)
``` | 3 | |
489 | C | Given Length and Sum of Digits... | PROGRAMMING | 1,400 | [
"dp",
"greedy",
"implementation"
] | null | null | You have a positive integer *m* and a non-negative integer *s*. Your task is to find the smallest and the largest of the numbers that have length *m* and sum of digits *s*. The required numbers should be non-negative integers written in the decimal base without leading zeroes. | The single line of the input contains a pair of integers *m*, *s* (1<=≤<=*m*<=≤<=100,<=0<=≤<=*s*<=≤<=900) — the length and the sum of the digits of the required numbers. | In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes). | [
"2 15\n",
"3 0\n"
] | [
"69 96\n",
"-1 -1\n"
] | none | 1,500 | [
{
"input": "2 15",
"output": "69 96"
},
{
"input": "3 0",
"output": "-1 -1"
},
{
"input": "2 1",
"output": "10 10"
},
{
"input": "3 10",
"output": "109 910"
},
{
"input": "100 100",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000099999999999 9999999999910000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "1 900",
"output": "-1 -1"
},
{
"input": "1 9",
"output": "9 9"
},
{
"input": "1 0",
"output": "0 0"
},
{
"input": "1 1",
"output": "1 1"
},
{
"input": "1 2",
"output": "2 2"
},
{
"input": "1 8",
"output": "8 8"
},
{
"input": "1 10",
"output": "-1 -1"
},
{
"input": "1 11",
"output": "-1 -1"
},
{
"input": "2 0",
"output": "-1 -1"
},
{
"input": "2 1",
"output": "10 10"
},
{
"input": "2 2",
"output": "11 20"
},
{
"input": "2 8",
"output": "17 80"
},
{
"input": "2 10",
"output": "19 91"
},
{
"input": "2 11",
"output": "29 92"
},
{
"input": "2 16",
"output": "79 97"
},
{
"input": "2 17",
"output": "89 98"
},
{
"input": "2 18",
"output": "99 99"
},
{
"input": "2 19",
"output": "-1 -1"
},
{
"input": "2 20",
"output": "-1 -1"
},
{
"input": "2 900",
"output": "-1 -1"
},
{
"input": "3 1",
"output": "100 100"
},
{
"input": "3 2",
"output": "101 200"
},
{
"input": "3 3",
"output": "102 300"
},
{
"input": "3 9",
"output": "108 900"
},
{
"input": "3 10",
"output": "109 910"
},
{
"input": "3 20",
"output": "299 992"
},
{
"input": "3 21",
"output": "399 993"
},
{
"input": "3 26",
"output": "899 998"
},
{
"input": "3 27",
"output": "999 999"
},
{
"input": "3 28",
"output": "-1 -1"
},
{
"input": "3 100",
"output": "-1 -1"
},
{
"input": "100 0",
"output": "-1 -1"
},
{
"input": "100 1",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 2",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 9",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000008 9000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 10",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000009 9100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 11",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000019 9200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 296",
"output": "1000000000000000000000000000000000000000000000000000000000000000000799999999999999999999999999999999 9999999999999999999999999999999980000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 297",
"output": "1000000000000000000000000000000000000000000000000000000000000000000899999999999999999999999999999999 9999999999999999999999999999999990000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 298",
"output": "1000000000000000000000000000000000000000000000000000000000000000000999999999999999999999999999999999 9999999999999999999999999999999991000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 299",
"output": "1000000000000000000000000000000000000000000000000000000000000000001999999999999999999999999999999999 9999999999999999999999999999999992000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 300",
"output": "1000000000000000000000000000000000000000000000000000000000000000002999999999999999999999999999999999 9999999999999999999999999999999993000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 301",
"output": "1000000000000000000000000000000000000000000000000000000000000000003999999999999999999999999999999999 9999999999999999999999999999999994000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 895",
"output": "4999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999994"
},
{
"input": "100 896",
"output": "5999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999995"
},
{
"input": "100 897",
"output": "6999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999996"
},
{
"input": "100 898",
"output": "7999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999997"
},
{
"input": "100 899",
"output": "8999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998"
},
{
"input": "100 900",
"output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"
},
{
"input": "99 900",
"output": "-1 -1"
},
{
"input": "99 891",
"output": "999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"
},
{
"input": "99 892",
"output": "-1 -1"
},
{
"input": "96 364",
"output": "100000000000000000000000000000000000000000000000000000039999999999999999999999999999999999999999 999999999999999999999999999999999999999940000000000000000000000000000000000000000000000000000000"
},
{
"input": "97 206",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000079999999999999999999999 9999999999999999999999800000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "98 250",
"output": "10000000000000000000000000000000000000000000000000000000000000000000006999999999999999999999999999 99999999999999999999999999970000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "99 96",
"output": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000059999999999 999999999960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 215",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000799999999999999999999999 9999999999999999999999980000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "5 18",
"output": "10089 99000"
},
{
"input": "2 19",
"output": "-1 -1"
},
{
"input": "3 3",
"output": "102 300"
}
] | 1,688,488,760 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | """
author: manoj.py
"""
import sys
m, s = map(int, input().split())
min_ans = []
def min_n(n):
print(n, )
if n<=0:
return
min_ans.append(n%9 if n%9 else 9)
min_n(n-(n%9 if n%9 else 9))
min_n(s)
if len(min_ans)>m:
print(-1, -1)
sys.exit(0)
temp = []
if len(min_ans)<m:
min_ans += [0]*(m-len(min_ans))
print(*min_ans, sep='', end=' ')
print(*(min_ans[::-1]), sep='')
| Title: Given Length and Sum of Digits...
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a positive integer *m* and a non-negative integer *s*. Your task is to find the smallest and the largest of the numbers that have length *m* and sum of digits *s*. The required numbers should be non-negative integers written in the decimal base without leading zeroes.
Input Specification:
The single line of the input contains a pair of integers *m*, *s* (1<=≤<=*m*<=≤<=100,<=0<=≤<=*s*<=≤<=900) — the length and the sum of the digits of the required numbers.
Output Specification:
In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes).
Demo Input:
['2 15\n', '3 0\n']
Demo Output:
['69 96\n', '-1 -1\n']
Note:
none | ```python
"""
author: manoj.py
"""
import sys
m, s = map(int, input().split())
min_ans = []
def min_n(n):
print(n, )
if n<=0:
return
min_ans.append(n%9 if n%9 else 9)
min_n(n-(n%9 if n%9 else 9))
min_n(s)
if len(min_ans)>m:
print(-1, -1)
sys.exit(0)
temp = []
if len(min_ans)<m:
min_ans += [0]*(m-len(min_ans))
print(*min_ans, sep='', end=' ')
print(*(min_ans[::-1]), sep='')
``` | 0 | |
633 | D | Fibonacci-ish | PROGRAMMING | 2,000 | [
"brute force",
"dp",
"hashing",
"implementation",
"math"
] | null | null | Yash has recently learnt about the Fibonacci sequence and is very excited about it. He calls a sequence Fibonacci-ish if
1. the sequence consists of at least two elements 1. *f*0 and *f*1 are arbitrary 1. *f**n*<=+<=2<==<=*f**n*<=+<=1<=+<=*f**n* for all *n*<=≥<=0.
You are given some sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. Your task is rearrange elements of this sequence in such a way that its longest possible prefix is Fibonacci-ish sequence. | The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the length of the sequence *a**i*.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=109). | Print the length of the longest possible Fibonacci-ish prefix of the given sequence after rearrangement. | [
"3\n1 2 -1\n",
"5\n28 35 7 14 21\n"
] | [
"3\n",
"4\n"
] | In the first sample, if we rearrange elements of the sequence as - 1, 2, 1, the whole sequence *a*<sub class="lower-index">*i*</sub> would be Fibonacci-ish.
In the second sample, the optimal way to rearrange elements is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/16f1f7e35511b29cb1396890ca2fb7dfa4d428de.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4003973f16750522e492d7d79318d7e2f0ff99cd.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/87b18fd9524b11e12faf154302fb14c1b55556fb.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8950ea952476baea26e03281fa2f7640b6241ef.png" style="max-width: 100.0%;max-height: 100.0%;"/>, 28. | 1,750 | [
{
"input": "3\n1 2 -1",
"output": "3"
},
{
"input": "5\n28 35 7 14 21",
"output": "4"
},
{
"input": "11\n-9 -1 -10 9 7 -4 0 -8 -3 3 5",
"output": "5"
},
{
"input": "10\n-4 -8 -8 8 -9 0 -7 9 1 0",
"output": "4"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "4\n1 -1 0 -2",
"output": "4"
},
{
"input": "2\n1000000000 1000000000",
"output": "2"
},
{
"input": "3\n1 1 2",
"output": "3"
},
{
"input": "5\n0 0 0 0 0",
"output": "5"
},
{
"input": "6\n1 -1 0 -1 -1 -2",
"output": "6"
},
{
"input": "5\n-7 0 -7 -7 -14",
"output": "5"
},
{
"input": "3\n0 -44 -49",
"output": "2"
},
{
"input": "5\n-1 1 0 0 0",
"output": "3"
},
{
"input": "2\n0 0",
"output": "2"
},
{
"input": "3\n0 0 0",
"output": "3"
},
{
"input": "4\n0 0 0 0",
"output": "4"
},
{
"input": "5\n0 0 0 0 0",
"output": "5"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 0",
"output": "10"
}
] | 1,456,511,417 | 4,517 | Python 3 | WRONG_ANSWER | PRETESTS | 0 | 46 | 0 | m=int(input())
if m%5==0:
print(0)
else:
print(5)
startvalue=0
howmany=m
subtract=0
for i in range(1,1000):
subtract=subtract+(m//(5**i))
startvalue=(howmany-subtract)*5
for i in range(5):
print(startvalue+i)
| Title: Fibonacci-ish
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yash has recently learnt about the Fibonacci sequence and is very excited about it. He calls a sequence Fibonacci-ish if
1. the sequence consists of at least two elements 1. *f*0 and *f*1 are arbitrary 1. *f**n*<=+<=2<==<=*f**n*<=+<=1<=+<=*f**n* for all *n*<=≥<=0.
You are given some sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. Your task is rearrange elements of this sequence in such a way that its longest possible prefix is Fibonacci-ish sequence.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the length of the sequence *a**i*.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=109).
Output Specification:
Print the length of the longest possible Fibonacci-ish prefix of the given sequence after rearrangement.
Demo Input:
['3\n1 2 -1\n', '5\n28 35 7 14 21\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample, if we rearrange elements of the sequence as - 1, 2, 1, the whole sequence *a*<sub class="lower-index">*i*</sub> would be Fibonacci-ish.
In the second sample, the optimal way to rearrange elements is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/16f1f7e35511b29cb1396890ca2fb7dfa4d428de.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4003973f16750522e492d7d79318d7e2f0ff99cd.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/87b18fd9524b11e12faf154302fb14c1b55556fb.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8950ea952476baea26e03281fa2f7640b6241ef.png" style="max-width: 100.0%;max-height: 100.0%;"/>, 28. | ```python
m=int(input())
if m%5==0:
print(0)
else:
print(5)
startvalue=0
howmany=m
subtract=0
for i in range(1,1000):
subtract=subtract+(m//(5**i))
startvalue=(howmany-subtract)*5
for i in range(5):
print(startvalue+i)
``` | 0 | |
915 | A | Garden | PROGRAMMING | 900 | [
"implementation"
] | null | null | Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding. | The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket. | Print one integer number — the minimum number of hours required to water the garden. | [
"3 6\n2 3 5\n",
"6 7\n1 2 3 4 5 6\n"
] | [
"2\n",
"7\n"
] | In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1. | 0 | [
{
"input": "3 6\n2 3 5",
"output": "2"
},
{
"input": "6 7\n1 2 3 4 5 6",
"output": "7"
},
{
"input": "5 97\n1 10 50 97 2",
"output": "1"
},
{
"input": "5 97\n1 10 50 100 2",
"output": "97"
},
{
"input": "100 100\n2 46 24 18 86 90 31 38 84 49 58 28 15 80 14 24 87 56 62 87 41 87 55 71 87 32 41 56 91 32 24 75 43 42 35 30 72 53 31 26 54 61 87 85 36 75 44 31 7 38 77 57 61 54 70 77 45 96 39 57 11 8 91 42 52 15 42 30 92 41 27 26 34 27 3 80 32 86 26 97 63 91 30 75 14 7 19 23 45 11 8 43 44 73 11 56 3 55 63 16",
"output": "50"
},
{
"input": "100 91\n13 13 62 96 74 47 81 46 78 21 20 42 4 73 25 30 76 74 58 28 25 52 42 48 74 40 82 9 25 29 17 22 46 64 57 95 81 39 47 86 40 95 97 35 31 98 45 98 47 78 52 63 58 14 89 97 17 95 28 22 20 36 68 38 95 16 2 26 54 47 42 31 31 81 21 21 65 40 82 53 60 71 75 33 96 98 6 22 95 12 5 48 18 27 58 62 5 96 36 75",
"output": "7"
},
{
"input": "8 8\n8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "3 8\n4 3 2",
"output": "2"
},
{
"input": "3 8\n2 4 2",
"output": "2"
},
{
"input": "3 6\n1 3 2",
"output": "2"
},
{
"input": "3 6\n3 2 5",
"output": "2"
},
{
"input": "3 8\n4 2 1",
"output": "2"
},
{
"input": "5 6\n2 3 5 1 2",
"output": "2"
},
{
"input": "2 6\n5 3",
"output": "2"
},
{
"input": "4 12\n6 4 3 1",
"output": "2"
},
{
"input": "3 18\n1 9 6",
"output": "2"
},
{
"input": "3 9\n3 2 1",
"output": "3"
},
{
"input": "3 6\n5 3 2",
"output": "2"
},
{
"input": "2 10\n5 2",
"output": "2"
},
{
"input": "2 18\n6 3",
"output": "3"
},
{
"input": "4 12\n1 2 12 3",
"output": "1"
},
{
"input": "3 7\n3 2 1",
"output": "7"
},
{
"input": "3 6\n3 2 1",
"output": "2"
},
{
"input": "5 10\n5 4 3 2 1",
"output": "2"
},
{
"input": "5 16\n8 4 2 1 7",
"output": "2"
},
{
"input": "6 7\n6 5 4 3 7 1",
"output": "1"
},
{
"input": "2 6\n3 2",
"output": "2"
},
{
"input": "2 4\n4 1",
"output": "1"
},
{
"input": "6 8\n2 4 1 3 5 7",
"output": "2"
},
{
"input": "6 8\n6 5 4 3 2 1",
"output": "2"
},
{
"input": "6 15\n5 2 3 6 4 3",
"output": "3"
},
{
"input": "4 8\n2 4 8 1",
"output": "1"
},
{
"input": "2 5\n5 1",
"output": "1"
},
{
"input": "4 18\n3 1 1 2",
"output": "6"
},
{
"input": "2 1\n2 1",
"output": "1"
},
{
"input": "3 10\n2 10 5",
"output": "1"
},
{
"input": "5 12\n12 4 4 4 3",
"output": "1"
},
{
"input": "3 6\n6 3 2",
"output": "1"
},
{
"input": "2 2\n2 1",
"output": "1"
},
{
"input": "3 18\n1 9 3",
"output": "2"
},
{
"input": "3 8\n7 2 4",
"output": "2"
},
{
"input": "2 100\n99 1",
"output": "100"
},
{
"input": "4 12\n1 3 4 2",
"output": "3"
},
{
"input": "3 6\n2 3 1",
"output": "2"
},
{
"input": "4 6\n3 2 5 12",
"output": "2"
},
{
"input": "4 97\n97 1 50 10",
"output": "1"
},
{
"input": "3 12\n1 12 2",
"output": "1"
},
{
"input": "4 12\n1 4 3 2",
"output": "3"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "3 19\n7 1 1",
"output": "19"
},
{
"input": "5 12\n12 4 3 4 4",
"output": "1"
},
{
"input": "3 8\n8 4 2",
"output": "1"
},
{
"input": "3 3\n3 2 1",
"output": "1"
},
{
"input": "5 6\n3 2 4 2 2",
"output": "2"
},
{
"input": "2 16\n8 4",
"output": "2"
},
{
"input": "3 6\n10 2 3",
"output": "2"
},
{
"input": "5 3\n2 4 5 3 6",
"output": "1"
},
{
"input": "11 99\n1 2 3 6 5 4 7 8 99 33 66",
"output": "1"
},
{
"input": "3 12\n3 12 2",
"output": "1"
},
{
"input": "5 25\n24 5 15 25 23",
"output": "1"
},
{
"input": "2 4\n8 1",
"output": "4"
},
{
"input": "4 100\n2 50 4 1",
"output": "2"
},
{
"input": "3 28\n7 14 1",
"output": "2"
},
{
"input": "4 8\n2 8 4 1",
"output": "1"
},
{
"input": "4 6\n6 1 2 3",
"output": "1"
},
{
"input": "2 12\n4 3",
"output": "3"
},
{
"input": "4 12\n1 2 4 3",
"output": "3"
},
{
"input": "5 12\n2 3 12 6 4",
"output": "1"
},
{
"input": "4 4\n1 2 2 4",
"output": "1"
},
{
"input": "3 6\n2 3 2",
"output": "2"
},
{
"input": "4 21\n21 20 21 2",
"output": "1"
},
{
"input": "3 8\n3 4 2",
"output": "2"
},
{
"input": "1 25\n25",
"output": "1"
},
{
"input": "99 12\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",
"output": "1"
},
{
"input": "98 12\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",
"output": "1"
},
{
"input": "79 12\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",
"output": "1"
},
{
"input": "4 32\n1 1 1 1",
"output": "32"
},
{
"input": "1 100\n1",
"output": "100"
},
{
"input": "2 100\n7 1",
"output": "100"
},
{
"input": "7 24\n1 3 6 4 5 2 7",
"output": "4"
},
{
"input": "6 87\n1 2 8 4 5 7",
"output": "87"
},
{
"input": "1 88\n1",
"output": "88"
},
{
"input": "1 89\n1",
"output": "89"
}
] | 1,553,839,215 | 2,147,483,647 | Python 3 | OK | TESTS | 83 | 109 | 0 |
_=list(map(int,input().split()))
__=sorted(list(map(int,input().split())))
for ___ in range(_[0]-1,-1,-1):
if _[1]%__[___]==0:
print(_[1]//__[___])
break
#ye jaye door bedoone IDE va net E :( | Title: Garden
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
Input Specification:
The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
Output Specification:
Print one integer number — the minimum number of hours required to water the garden.
Demo Input:
['3 6\n2 3 5\n', '6 7\n1 2 3 4 5 6\n']
Demo Output:
['2\n', '7\n']
Note:
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1. | ```python
_=list(map(int,input().split()))
__=sorted(list(map(int,input().split())))
for ___ in range(_[0]-1,-1,-1):
if _[1]%__[___]==0:
print(_[1]//__[___])
break
#ye jaye door bedoone IDE va net E :(
``` | 3 | |
108 | A | Palindromic Times | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | A. Palindromic Times | 2 | 256 | Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him. | The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits. | Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time. | [
"12:21\n",
"23:59\n"
] | [
"13:31\n",
"00:00\n"
] | none | 500 | [
{
"input": "12:21",
"output": "13:31"
},
{
"input": "23:59",
"output": "00:00"
},
{
"input": "15:51",
"output": "20:02"
},
{
"input": "10:44",
"output": "11:11"
},
{
"input": "04:02",
"output": "04:40"
},
{
"input": "02:11",
"output": "02:20"
},
{
"input": "12:15",
"output": "12:21"
},
{
"input": "07:07",
"output": "10:01"
},
{
"input": "00:17",
"output": "01:10"
},
{
"input": "04:55",
"output": "05:50"
},
{
"input": "02:17",
"output": "02:20"
},
{
"input": "07:56",
"output": "10:01"
},
{
"input": "00:29",
"output": "01:10"
},
{
"input": "23:31",
"output": "23:32"
},
{
"input": "19:30",
"output": "20:02"
},
{
"input": "12:14",
"output": "12:21"
},
{
"input": "17:32",
"output": "20:02"
},
{
"input": "03:44",
"output": "04:40"
},
{
"input": "07:15",
"output": "10:01"
},
{
"input": "18:42",
"output": "20:02"
},
{
"input": "08:56",
"output": "10:01"
},
{
"input": "04:50",
"output": "05:50"
},
{
"input": "14:32",
"output": "14:41"
},
{
"input": "23:23",
"output": "23:32"
},
{
"input": "08:35",
"output": "10:01"
},
{
"input": "03:32",
"output": "04:40"
},
{
"input": "07:59",
"output": "10:01"
},
{
"input": "14:12",
"output": "14:41"
},
{
"input": "23:52",
"output": "00:00"
},
{
"input": "16:36",
"output": "20:02"
},
{
"input": "17:50",
"output": "20:02"
},
{
"input": "06:59",
"output": "10:01"
},
{
"input": "16:50",
"output": "20:02"
},
{
"input": "00:00",
"output": "01:10"
},
{
"input": "23:59",
"output": "00:00"
},
{
"input": "23:33",
"output": "00:00"
}
] | 1,597,953,629 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 186 | 204,800 | z=input()
qwe=False
x=z.split(":")
y=list(z)
d=""
a=int(x[0])
for i in range(1,24+1):
if qwe:
break
a+=i
if a>=24:
a=0
for o in range(0,60):
if a<10:
if o<10:
d="0"+str(a)+":"+"0"+str(o)
else:
d="0"+str(a)+":"+str(o)
else:
if o<10:
d=str(a)+":"+"0"+str(o)
else:
d=str(a)+":"+str(o)
ow=list(d)
q=ow.copy()
ow.reverse()
if ow==q:
print("".join(q))
qwe=True
break
else:
d=""
| Title: Palindromic Times
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him.
Input Specification:
The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits.
Output Specification:
Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time.
Demo Input:
['12:21\n', '23:59\n']
Demo Output:
['13:31\n', '00:00\n']
Note:
none | ```python
z=input()
qwe=False
x=z.split(":")
y=list(z)
d=""
a=int(x[0])
for i in range(1,24+1):
if qwe:
break
a+=i
if a>=24:
a=0
for o in range(0,60):
if a<10:
if o<10:
d="0"+str(a)+":"+"0"+str(o)
else:
d="0"+str(a)+":"+str(o)
else:
if o<10:
d=str(a)+":"+"0"+str(o)
else:
d=str(a)+":"+str(o)
ow=list(d)
q=ow.copy()
ow.reverse()
if ow==q:
print("".join(q))
qwe=True
break
else:
d=""
``` | 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"
}
] | 1,559,546,741 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 218 | 0 | cordsC = [0, 0, 0]
for i in range(int(input())):
cords = list(map(int, input().split()))
cordsC[0] += cords[0]; cordsC[1] += cords[1]; cordsC[2] += cords[2]
print("YES" if (cordsC[0] == 0 and cordsC[1] == 0 and cordsC[2] == 0) else "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
cordsC = [0, 0, 0]
for i in range(int(input())):
cords = list(map(int, input().split()))
cordsC[0] += cords[0]; cordsC[1] += cords[1]; cordsC[2] += cords[2]
print("YES" if (cordsC[0] == 0 and cordsC[1] == 0 and cordsC[2] == 0) else "NO")
``` | 3.9455 |
141 | A | Amusing Joke | PROGRAMMING | 800 | [
"implementation",
"sortings",
"strings"
] | null | null | So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door.
The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters.
Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning. | The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100. | Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes. | [
"SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n",
"PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n",
"BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left.
In the second sample letter "P" is missing from the pile and there's an extra letter "L".
In the third sample there's an extra letter "L". | 500 | [
{
"input": "SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS",
"output": "YES"
},
{
"input": "PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI",
"output": "NO"
},
{
"input": "BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER",
"output": "NO"
},
{
"input": "B\nA\nAB",
"output": "YES"
},
{
"input": "ONDOL\nJNPB\nONLNJBODP",
"output": "YES"
},
{
"input": "Y\nW\nYW",
"output": "YES"
},
{
"input": "OI\nM\nIMO",
"output": "YES"
},
{
"input": "VFQRWWWACX\nGHZJPOQUSXRAQDGOGMR\nOPAWDOUSGWWCGQXXQAZJRQRGHRMVF",
"output": "YES"
},
{
"input": "JUTCN\nPIGMZOPMEUFADQBW\nNWQGZMAIPUPOMCDUB",
"output": "NO"
},
{
"input": "Z\nO\nZOCNDOLTBZKQLTBOLDEGXRHZGTTPBJBLSJCVSVXISQZCSFDEBXRCSGBGTHWOVIXYHACAGBRYBKBJAEPIQZHVEGLYH",
"output": "NO"
},
{
"input": "IQ\nOQ\nQOQIGGKFNHJSGCGM",
"output": "NO"
},
{
"input": "ROUWANOPNIGTVMIITVMZ\nOQTUPZMTKUGY\nVTVNGZITGPUNPMQOOATUUIYIWMMKZOTR",
"output": "YES"
},
{
"input": "OVQELLOGFIOLEHXMEMBJDIGBPGEYFG\nJNKFPFFIJOFHRIFHXEWYZOPDJBZTJZKBWQTECNHRFSJPJOAPQT\nYAIPFFFEXJJNEJPLREIGODEGQZVMCOBDFKWTMWJSBEBTOFFQOHIQJLHFNXIGOHEZRZLFOKJBJPTPHPGY",
"output": "YES"
},
{
"input": "NBJGVNGUISUXQTBOBKYHQCOOVQWUXWPXBUDLXPKX\nNSFQDFUMQDQWQ\nWXKKVNTDQQFXCUQBIMQGQHSLVGWSBFYBUPOWPBDUUJUXQNOQDNXOX",
"output": "YES"
},
{
"input": "IJHHGKCXWDBRWJUPRDBZJLNTTNWKXLUGJSBWBOAUKWRAQWGFNL\nNJMWRMBCNPHXTDQQNZ\nWDNJRCLILNQRHWBANLTXWMJBPKUPGKJDJZAQWKTZFBRCTXHHBNXRGUQUNBNMWODGSJWW",
"output": "YES"
},
{
"input": "SRROWANGUGZHCIEFYMQVTWVOMDWPUZJFRDUMVFHYNHNTTGNXCJ\nDJYWGLBFCCECXFHOLORDGDCNRHPWXNHXFCXQCEZUHRRNAEKUIX\nWCUJDNYHNHYOPWMHLDCDYRWBVOGHFFUKOZTXJRXJHRGWICCMRNEVNEGQWTZPNFCSHDRFCFQDCXMHTLUGZAXOFNXNVGUEXIACRERU",
"output": "YES"
},
{
"input": "H\nJKFGHMIAHNDBMFXWYQLZRSVNOTEGCQSVUBYUOZBTNKTXPFQDCMKAGFITEUGOYDFIYQIORMFJEOJDNTFVIQEBICSNGKOSNLNXJWC\nBQSVDOGIHCHXSYNYTQFCHNJGYFIXTSOQINZOKSVQJMTKNTGFNXAVTUYEONMBQMGJLEWJOFGEARIOPKFUFCEMUBRBDNIIDFZDCLWK",
"output": "YES"
},
{
"input": "DSWNZRFVXQ\nPVULCZGOOU\nUOLVZXNUPOQRZGWFVDSCANQTCLEIE",
"output": "NO"
},
{
"input": "EUHTSCENIPXLTSBMLFHD\nIZAVSZPDLXOAGESUSE\nLXAELAZ",
"output": "NO"
},
{
"input": "WYSJFEREGELSKRQRXDXCGBODEFZVSI\nPEJKMGFLBFFDWRCRFSHVEFLEBTJCVCHRJTLDTISHPOGFWPLEWNYJLMXWIAOTYOXMV\nHXERTZWLEXTPIOTFRVMEJVYFFJLRPFMXDEBNSGCEOFFCWTKIDDGCFYSJKGLHBORWEPLDRXRSJYBGASSVCMHEEJFLVI",
"output": "NO"
},
{
"input": "EPBMDIUQAAUGLBIETKOKFLMTCVEPETWJRHHYKCKU\nHGMAETVPCFZYNNKDQXVXUALHYLOTCHM\nECGXACVKEYMCEDOTMKAUFHLHOMT",
"output": "NO"
},
{
"input": "NUBKQEJHALANSHEIFUZHYEZKKDRFHQKAJHLAOWTZIMOCWOVVDW\nEFVOBIGAUAUSQGVSNBKNOBDMINODMFSHDL\nKLAMKNTHBFFOHVKWICHBKNDDQNEISODUSDNLUSIOAVWY",
"output": "NO"
},
{
"input": "VXINHOMEQCATZUGAJEIUIZZLPYFGUTVLNBNWCUVMEENUXKBWBGZTMRJJVJDLVSLBABVCEUDDSQFHOYPYQTWVAGTWOLKYISAGHBMC\nZMRGXPZSHOGCSAECAPGVOIGCWEOWWOJXLGYRDMPXBLOKZVRACPYQLEQGFQCVYXAGBEBELUTDAYEAGPFKXRULZCKFHZCHVCWIRGPK\nRCVUXGQVNWFGRUDLLENNDQEJHYYVWMKTLOVIPELKPWCLSQPTAXAYEMGWCBXEVAIZGGDDRBRT",
"output": "NO"
},
{
"input": "PHBDHHWUUTZAHELGSGGOPOQXSXEZIXHZTOKYFBQLBDYWPVCNQSXHEAXRRPVHFJBVBYCJIFOTQTWSUOWXLKMVJJBNLGTVITWTCZZ\nFUPDLNVIHRWTEEEHOOEC\nLOUSUUSZCHJBPEWIILUOXEXRQNCJEGTOBRVZLTTZAHTKVEJSNGHFTAYGY",
"output": "NO"
},
{
"input": "GDSLNIIKTO\nJF\nPDQYFKDTNOLI",
"output": "NO"
},
{
"input": "AHOKHEKKPJLJIIWJRCGY\nORELJCSIX\nZVWPXVFWFSWOXXLIHJKPXIOKRELYE",
"output": "NO"
},
{
"input": "ZWCOJFORBPHXCOVJIDPKVECMHVHCOC\nTEV\nJVGTBFTLFVIEPCCHODOFOMCVZHWXVCPEH",
"output": "NO"
},
{
"input": "AGFIGYWJLVMYZGNQHEHWKJIAWBPUAQFERMCDROFN\nPMJNHMVNRGCYZAVRWNDSMLSZHFNYIUWFPUSKKIGU\nMCDVPPRXGUAYLSDRHRURZASXUWZSIIEZCPXUVEONKNGNWRYGOSFMCKESMVJZHWWUCHWDQMLASLNNMHAU",
"output": "NO"
},
{
"input": "XLOWVFCZSSXCSYQTIIDKHNTKNKEEDFMDZKXSPVLBIDIREDUAIN\nZKIWNDGBISDB\nSLPKLYFYSRNRMOSWYLJJDGFFENPOXYLPZFTQDANKBDNZDIIEWSUTTKYBKVICLG",
"output": "NO"
},
{
"input": "PMUKBTRKFIAYVGBKHZHUSJYSSEPEOEWPOSPJLWLOCTUYZODLTUAFCMVKGQKRRUSOMPAYOTBTFPXYAZXLOADDEJBDLYOTXJCJYTHA\nTWRRAJLCQJTKOKWCGUH\nEWDPNXVCXWCDQCOYKKSOYTFSZTOOPKPRDKFJDETKSRAJRVCPDOBWUGPYRJPUWJYWCBLKOOTUPBESTOFXZHTYLLMCAXDYAEBUTAHM",
"output": "NO"
},
{
"input": "QMIMGQRQDMJDPNFEFXSXQMCHEJKTWCTCVZPUAYICOIRYOWKUSIWXJLHDYWSBOITHTMINXFKBKAWZTXXBJIVYCRWKXNKIYKLDDXL\nV\nFWACCXBVDOJFIUAVYRALBYJKXXWIIFORRUHKHCXLDBZMXIYJWISFEAWTIQFIZSBXMKNOCQKVKRWDNDAMQSTKYLDNYVTUCGOJXJTW",
"output": "NO"
},
{
"input": "XJXPVOOQODELPPWUISSYVVXRJTYBPDHJNENQEVQNVFIXSESKXVYPVVHPMOSX\nLEXOPFPVPSZK\nZVXVPYEYOYXVOISVLXPOVHEQVXPNQJIOPFDTXEUNMPEPPHELNXKKWSVSOXSBPSJDPVJVSRFQ",
"output": "YES"
},
{
"input": "OSKFHGYNQLSRFSAHPXKGPXUHXTRBJNAQRBSSWJVEENLJCDDHFXVCUNPZAIVVO\nFNUOCXAGRRHNDJAHVVLGGEZQHWARYHENBKHP\nUOEFNWVXCUNERLKVTHAGPSHKHDYFPYWZHJKHQLSNFBJHVJANRXCNSDUGVDABGHVAOVHBJZXGRACHRXEGNRPQEAPORQSILNXFS",
"output": "YES"
},
{
"input": "VYXYVVACMLPDHONBUTQFZTRREERBLKUJYKAHZRCTRLRCLOZYWVPBRGDQPFPQIF\nFE\nRNRPEVDRLYUQFYRZBCQLCYZEABKLRXCJLKVZBVFUEYRATOMDRTHFPGOWQVTIFPPH",
"output": "YES"
},
{
"input": "WYXUZQJQNLASEGLHPMSARWMTTQMQLVAZLGHPIZTRVTCXDXBOLNXZPOFCTEHCXBZ\nBLQZRRWP\nGIQZXPLTTMNHQVWPPEAPLOCDMBSTHRCFLCQRRZXLVAOQEGZBRUZJXXZTMAWLZHSLWNQTYXB",
"output": "YES"
},
{
"input": "MKVJTSSTDGKPVVDPYSRJJYEVGKBMSIOKHLZQAEWLRIBINVRDAJIBCEITKDHUCCVY\nPUJJQFHOGZKTAVNUGKQUHMKTNHCCTI\nQVJKUSIGTSVYUMOMLEGHWYKSKQTGATTKBNTKCJKJPCAIRJIRMHKBIZISEGFHVUVQZBDERJCVAKDLNTHUDCHONDCVVJIYPP",
"output": "YES"
},
{
"input": "OKNJOEYVMZXJMLVJHCSPLUCNYGTDASKSGKKCRVIDGEIBEWRVBVRVZZTLMCJLXHJIA\nDJBFVRTARTFZOWN\nAGHNVUNJVCPLWSVYBJKZSVTFGLELZASLWTIXDDJXCZDICTVIJOTMVEYOVRNMJGRKKHRMEBORAKFCZJBR",
"output": "YES"
},
{
"input": "OQZACLPSAGYDWHFXDFYFRRXWGIEJGSXWUONAFWNFXDTGVNDEWNQPHUXUJNZWWLBPYL\nOHBKWRFDRQUAFRCMT\nWIQRYXRJQWWRUWCYXNXALKFZGXFTLOODWRDPGURFUFUQOHPWBASZNVWXNCAGHWEHFYESJNFBMNFDDAPLDGT",
"output": "YES"
},
{
"input": "OVIRQRFQOOWVDEPLCJETWQSINIOPLTLXHSQWUYUJNFBMKDNOSHNJQQCDHZOJVPRYVSV\nMYYDQKOOYPOOUELCRIT\nNZSOTVLJTTVQLFHDQEJONEOUOFOLYVSOIYUDNOSIQVIRMVOERCLMYSHPCQKIDRDOQPCUPQBWWRYYOXJWJQPNKH",
"output": "YES"
},
{
"input": "WGMBZWNMSJXNGDUQUJTCNXDSJJLYRDOPEGPQXYUGBESDLFTJRZDDCAAFGCOCYCQMDBWK\nYOBMOVYTUATTFGJLYUQD\nDYXVTLQCYFJUNJTUXPUYOPCBCLBWNSDUJRJGWDOJDSQAAMUOJWSYERDYDXYTMTOTMQCGQZDCGNFBALGGDFKZMEBG",
"output": "YES"
},
{
"input": "CWLRBPMEZCXAPUUQFXCUHAQTLPBTXUUKWVXKBHKNSSJFEXLZMXGVFHHVTPYAQYTIKXJJE\nMUFOSEUEXEQTOVLGDSCWM\nJUKEQCXOXWEHCGKFPBIGMWVJLXUONFXBYTUAXERYTXKCESKLXAEHVPZMMUFTHLXTTZSDMBJLQPEUWCVUHSQQVUASPF",
"output": "YES"
},
{
"input": "IDQRX\nWETHO\nODPDGBHVUVSSISROHQJTUKPUCLXABIZQQPPBPKOSEWGEHRSRRNBAVLYEMZISMWWGKHVTXKUGUXEFBSWOIWUHRJGMWBMHQLDZHBWA",
"output": "NO"
},
{
"input": "IXFDY\nJRMOU\nDF",
"output": "NO"
},
{
"input": "JPSPZ\nUGCUB\nJMZZZZZZZZ",
"output": "NO"
},
{
"input": "AC\nA\nBBA",
"output": "NO"
},
{
"input": "UIKWWKXLSHTOOZOVGXKYSOJEHAUEEG\nKZXQDWJJWRXFHKJDQHJK\nXMZHTFOGEXAUJXXJUYVJIFOTKLZHDKELJWERHMGAWGKWAQKEKHIDWGGZVYOHKXRPWSJDPESFJUMKQYWBYUTHQYEFZUGKQOBHYDWB",
"output": "NO"
},
{
"input": "PXWRXRPFLR\nPJRWWXIVHODV\nXW",
"output": "NO"
},
{
"input": "CHTAZVHGSHCVIBK\nEQINEBKXEPYJSAZIMLDF\nZCZZZZDZMCZZEZDZZEZZZZQZZBZZZOZZCZE",
"output": "NO"
},
{
"input": "GXPZFSELJJNDAXYRV\nUYBKPMVBSOVOJWMONLTJOJCNQKMTAHEWLHOWIIBH\nHCWNFWJPEJIWOVPTBMVCRJLSISSVNOHCKLBFMIUAIMASQWPXEYXBOXQGFEMYJLBKDCZIMJNHOJEDGGANIVYKQTUOSOVOPWHVJGXH",
"output": "NO"
},
{
"input": "LFGJCJJDUTUP\nOVSBILTIYCJCRHKCIXCETJQJJ\nGIJJTJCLTJJJ",
"output": "NO"
},
{
"input": "GIO\nPRL\nPRL",
"output": "NO"
},
{
"input": "A\nB\nABC",
"output": "NO"
},
{
"input": "KKK\nKKK\nZZZZZ",
"output": "NO"
},
{
"input": "ZMYGQLDBLAPN\nZFJBKWHROVNPSJQUDFTHOCGREUFLYIWYICD\nZMJZZEDAZANKZZZZZZEZZBZDZZZZZZKHZZFZZZDZNZMDZZA",
"output": "NO"
}
] | 1,690,119,395 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 62 | 0 | def can_form_names(guest_name, host_name, pile_letters):
# Concatenate the guest and host names
combined_names = guest_name + host_name
# Create dictionaries to store the counts of letters in the names and the pile
name_counts = {}
pile_counts = {}
# Count the occurrences of letters in the combined names
for letter in combined_names:
name_counts[letter] = name_counts.get(letter, 0) + 1
# Count the occurrences of letters in the pile
for letter in pile_letters:
pile_counts[letter] = pile_counts.get(letter, 0) + 1
# Check if all the letters needed for the names are present in the pile
for letter, count in name_counts.items():
if letter not in pile_counts or pile_counts[letter] < count:
return False
return True
def main():
guest_name = input().strip()
host_name = input().strip()
pile_letters = input().strip()
if can_form_names(guest_name, host_name, pile_letters):
print("YES")
else:
print("NO")
if __name__ == "__main__":
main()
| Title: Amusing Joke
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door.
The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters.
Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning.
Input Specification:
The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100.
Output Specification:
Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes.
Demo Input:
['SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n', 'PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n', 'BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left.
In the second sample letter "P" is missing from the pile and there's an extra letter "L".
In the third sample there's an extra letter "L". | ```python
def can_form_names(guest_name, host_name, pile_letters):
# Concatenate the guest and host names
combined_names = guest_name + host_name
# Create dictionaries to store the counts of letters in the names and the pile
name_counts = {}
pile_counts = {}
# Count the occurrences of letters in the combined names
for letter in combined_names:
name_counts[letter] = name_counts.get(letter, 0) + 1
# Count the occurrences of letters in the pile
for letter in pile_letters:
pile_counts[letter] = pile_counts.get(letter, 0) + 1
# Check if all the letters needed for the names are present in the pile
for letter, count in name_counts.items():
if letter not in pile_counts or pile_counts[letter] < count:
return False
return True
def main():
guest_name = input().strip()
host_name = input().strip()
pile_letters = input().strip()
if can_form_names(guest_name, host_name, pile_letters):
print("YES")
else:
print("NO")
if __name__ == "__main__":
main()
``` | 0 | |
981 | A | Antipalindrome | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into? | The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only. | If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique. | [
"mew\n",
"wuffuw\n",
"qqqqqqqq\n"
] | [
"3\n",
"5\n",
"0\n"
] | "mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$. | 500 | [
{
"input": "mew",
"output": "3"
},
{
"input": "wuffuw",
"output": "5"
},
{
"input": "qqqqqqqq",
"output": "0"
},
{
"input": "ijvji",
"output": "4"
},
{
"input": "iiiiiii",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvvgiaahqmtjyuchikvvhvrebow",
"output": "49"
},
{
"input": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvahheoqleromusrartldojsjvy",
"output": "50"
},
{
"input": "ijvxljt",
"output": "7"
},
{
"input": "fyhcncnchyf",
"output": "10"
},
{
"input": "ffffffffffff",
"output": "0"
},
{
"input": "fyhcncfsepqj",
"output": "12"
},
{
"input": "ybejrrlbcinttnicblrrjeby",
"output": "23"
},
{
"input": "yyyyyyyyyyyyyyyyyyyyyyyyy",
"output": "0"
},
{
"input": "ybejrrlbcintahovgjddrqatv",
"output": "25"
},
{
"input": "oftmhcmclgyqaojljoaqyglcmchmtfo",
"output": "30"
},
{
"input": "oooooooooooooooooooooooooooooooo",
"output": "0"
},
{
"input": "oftmhcmclgyqaojllbotztajglsmcilv",
"output": "32"
},
{
"input": "gxandbtgpbknxvnkjaajknvxnkbpgtbdnaxg",
"output": "35"
},
{
"input": "gggggggggggggggggggggggggggggggggggg",
"output": "0"
},
{
"input": "gxandbtgpbknxvnkjaygommzqitqzjfalfkk",
"output": "36"
},
{
"input": "fcliblymyqckxvieotjooojtoeivxkcqymylbilcf",
"output": "40"
},
{
"input": "fffffffffffffffffffffffffffffffffffffffffff",
"output": "0"
},
{
"input": "fcliblymyqckxvieotjootiqwtyznhhvuhbaixwqnsy",
"output": "43"
},
{
"input": "rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr",
"output": "0"
},
{
"input": "rajccqwqnqmshmerpvjyfepxwpxyldzpzhctqjnstxyfmlhiy",
"output": "49"
},
{
"input": "a",
"output": "0"
},
{
"input": "abca",
"output": "4"
},
{
"input": "aaaaabaaaaa",
"output": "10"
},
{
"input": "aba",
"output": "2"
},
{
"input": "asaa",
"output": "4"
},
{
"input": "aabaa",
"output": "4"
},
{
"input": "aabbaa",
"output": "5"
},
{
"input": "abcdaaa",
"output": "7"
},
{
"input": "aaholaa",
"output": "7"
},
{
"input": "abcdefghijka",
"output": "12"
},
{
"input": "aaadcba",
"output": "7"
},
{
"input": "aaaabaaaa",
"output": "8"
},
{
"input": "abaa",
"output": "4"
},
{
"input": "abcbaa",
"output": "6"
},
{
"input": "ab",
"output": "2"
},
{
"input": "l",
"output": "0"
},
{
"input": "aaaabcaaaa",
"output": "10"
},
{
"input": "abbaaaaaabba",
"output": "11"
},
{
"input": "abaaa",
"output": "5"
},
{
"input": "baa",
"output": "3"
},
{
"input": "aaaaaaabbba",
"output": "11"
},
{
"input": "ccbcc",
"output": "4"
},
{
"input": "bbbaaab",
"output": "7"
},
{
"input": "abaaaaaaaa",
"output": "10"
},
{
"input": "abaaba",
"output": "5"
},
{
"input": "aabsdfaaaa",
"output": "10"
},
{
"input": "aaaba",
"output": "5"
},
{
"input": "aaabaaa",
"output": "6"
},
{
"input": "baaabbb",
"output": "7"
},
{
"input": "ccbbabbcc",
"output": "8"
},
{
"input": "cabc",
"output": "4"
},
{
"input": "aabcd",
"output": "5"
},
{
"input": "abcdea",
"output": "6"
},
{
"input": "bbabb",
"output": "4"
},
{
"input": "aaaaabababaaaaa",
"output": "14"
},
{
"input": "bbabbb",
"output": "6"
},
{
"input": "aababd",
"output": "6"
},
{
"input": "abaaaa",
"output": "6"
},
{
"input": "aaaaaaaabbba",
"output": "12"
},
{
"input": "aabca",
"output": "5"
},
{
"input": "aaabccbaaa",
"output": "9"
},
{
"input": "aaaaaaaaaaaaaaaaaaaab",
"output": "21"
},
{
"input": "babb",
"output": "4"
},
{
"input": "abcaa",
"output": "5"
},
{
"input": "qwqq",
"output": "4"
},
{
"input": "aaaaaaaaaaabbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaa",
"output": "48"
},
{
"input": "aaab",
"output": "4"
},
{
"input": "aaaaaabaaaaa",
"output": "12"
},
{
"input": "wwuww",
"output": "4"
},
{
"input": "aaaaabcbaaaaa",
"output": "12"
},
{
"input": "aaabbbaaa",
"output": "8"
},
{
"input": "aabcbaa",
"output": "6"
},
{
"input": "abccdefccba",
"output": "11"
},
{
"input": "aabbcbbaa",
"output": "8"
},
{
"input": "aaaabbaaaa",
"output": "9"
},
{
"input": "aabcda",
"output": "6"
},
{
"input": "abbca",
"output": "5"
},
{
"input": "aaaaaabbaaa",
"output": "11"
},
{
"input": "sssssspssssss",
"output": "12"
},
{
"input": "sdnmsdcs",
"output": "8"
},
{
"input": "aaabbbccbbbaaa",
"output": "13"
},
{
"input": "cbdbdc",
"output": "6"
},
{
"input": "abb",
"output": "3"
},
{
"input": "abcdefaaaa",
"output": "10"
},
{
"input": "abbbaaa",
"output": "7"
},
{
"input": "v",
"output": "0"
},
{
"input": "abccbba",
"output": "7"
},
{
"input": "axyza",
"output": "5"
},
{
"input": "abcdefgaaaa",
"output": "11"
},
{
"input": "aaabcdaaa",
"output": "9"
},
{
"input": "aaaacaaaa",
"output": "8"
},
{
"input": "aaaaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaaaaaa",
"output": "42"
},
{
"input": "abbbaa",
"output": "6"
},
{
"input": "abcdee",
"output": "6"
},
{
"input": "oom",
"output": "3"
},
{
"input": "aabcaa",
"output": "6"
},
{
"input": "abba",
"output": "3"
},
{
"input": "aaca",
"output": "4"
},
{
"input": "aacbca",
"output": "6"
},
{
"input": "ababa",
"output": "4"
},
{
"input": "abcda",
"output": "5"
},
{
"input": "cccaaccc",
"output": "7"
},
{
"input": "aaabcda",
"output": "7"
},
{
"input": "aa",
"output": "0"
},
{
"input": "aabaaaa",
"output": "7"
},
{
"input": "abbaaaa",
"output": "7"
},
{
"input": "aaabcbaaa",
"output": "8"
},
{
"input": "aabba",
"output": "5"
},
{
"input": "xyxx",
"output": "4"
},
{
"input": "aaaaaaaaaaaabc",
"output": "14"
},
{
"input": "bbaaaabb",
"output": "7"
},
{
"input": "aaabaa",
"output": "6"
},
{
"input": "sssssabsssss",
"output": "12"
},
{
"input": "bbbaaaabbb",
"output": "9"
},
{
"input": "abbbbaaaa",
"output": "9"
},
{
"input": "wwufuww",
"output": "6"
},
{
"input": "oowoo",
"output": "4"
},
{
"input": "cccaccc",
"output": "6"
},
{
"input": "aaa",
"output": "0"
},
{
"input": "bbbcc",
"output": "5"
},
{
"input": "abcdef",
"output": "6"
},
{
"input": "abbba",
"output": "4"
},
{
"input": "aab",
"output": "3"
},
{
"input": "aaba",
"output": "4"
},
{
"input": "azbyaaa",
"output": "7"
},
{
"input": "oooooiooooo",
"output": "10"
},
{
"input": "aabbbbbaaaaaa",
"output": "13"
}
] | 1,588,089,702 | 2,147,483,647 | PyPy 3 | OK | TESTS | 133 | 140 | 0 | s=str(input())
if(s!=s[::-1]):
print(len(s))
else:
flag=0
for i in range(1,len(s)):
d=s[i:]
if(d!=d[::-1]):
flag=1
break
else:
pass
if(flag>0):
print(len(s)-i)
else:
print(0)
| Title: Antipalindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
Input Specification:
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
Output Specification:
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
Demo Input:
['mew\n', 'wuffuw\n', 'qqqqqqqq\n']
Demo Output:
['3\n', '5\n', '0\n']
Note:
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$. | ```python
s=str(input())
if(s!=s[::-1]):
print(len(s))
else:
flag=0
for i in range(1,len(s)):
d=s[i:]
if(d!=d[::-1]):
flag=1
break
else:
pass
if(flag>0):
print(len(s)-i)
else:
print(0)
``` | 3 | |
150 | A | Win or Freeze | PROGRAMMING | 1,400 | [
"games",
"math",
"number theory"
] | null | null | You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer *q*. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move. | The first line contains the only integer *q* (1<=≤<=*q*<=≤<=1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator. | In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them. | [
"6\n",
"30\n",
"1\n"
] | [
"2\n",
"1\n6\n",
"1\n0\n"
] | Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | 500 | [
{
"input": "6",
"output": "2"
},
{
"input": "30",
"output": "1\n6"
},
{
"input": "1",
"output": "1\n0"
},
{
"input": "2",
"output": "1\n0"
},
{
"input": "3",
"output": "1\n0"
},
{
"input": "5",
"output": "1\n0"
},
{
"input": "445538663413",
"output": "1\n0"
},
{
"input": "5138168457911",
"output": "2"
},
{
"input": "472670214391",
"output": "1\n23020027"
},
{
"input": "1468526771489",
"output": "1\n613783"
},
{
"input": "1307514188557",
"output": "1\n39283"
},
{
"input": "8110708459517",
"output": "2"
},
{
"input": "7938986881993",
"output": "1\n378028993"
},
{
"input": "34280152201",
"output": "2"
},
{
"input": "8587340257",
"output": "1\n9409"
},
{
"input": "1245373417369",
"output": "1\n908209"
},
{
"input": "48855707",
"output": "1\n18769"
},
{
"input": "614125",
"output": "1\n25"
},
{
"input": "1716443237161",
"output": "1\n5329"
},
{
"input": "49380563",
"output": "1\n289"
},
{
"input": "2975",
"output": "1\n25"
},
{
"input": "5839252225",
"output": "1\n25"
},
{
"input": "266418",
"output": "1\n9"
},
{
"input": "319757451841",
"output": "1\n289"
},
{
"input": "3047527844089",
"output": "2"
},
{
"input": "1408514752349",
"output": "1\n72361"
},
{
"input": "9999925100701",
"output": "1\n0"
},
{
"input": "9999926826034",
"output": "2"
},
{
"input": "6599669076000",
"output": "1\n4"
},
{
"input": "324",
"output": "1\n4"
},
{
"input": "99",
"output": "1\n9"
},
{
"input": "4",
"output": "2"
},
{
"input": "8",
"output": "1\n4"
},
{
"input": "1000000000000",
"output": "1\n4"
},
{
"input": "9",
"output": "2"
},
{
"input": "50",
"output": "1\n25"
},
{
"input": "16",
"output": "1\n4"
},
{
"input": "25",
"output": "2"
},
{
"input": "44",
"output": "1\n4"
},
{
"input": "388",
"output": "1\n4"
},
{
"input": "2000000014",
"output": "2"
},
{
"input": "128",
"output": "1\n4"
},
{
"input": "27",
"output": "1\n9"
},
{
"input": "274875809788",
"output": "1\n4"
},
{
"input": "64",
"output": "1\n4"
},
{
"input": "802241960524",
"output": "1\n4"
},
{
"input": "401120980262",
"output": "2"
},
{
"input": "236",
"output": "1\n4"
},
{
"input": "7420738134810",
"output": "1\n6"
},
{
"input": "12",
"output": "1\n4"
},
{
"input": "57461344602",
"output": "1\n6"
},
{
"input": "81",
"output": "1\n9"
},
{
"input": "30971726",
"output": "2"
}
] | 1,586,884,547 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 13 | 592 | 0 | import math
def checkprime(x):
for j in range(2,math.ceil(math.sqrt(x))):
if(x%j==0):
return (j)
return x
x=int(input())
y=checkprime(x)
if y==x:
print(1,0,sep='\n')
else:
y1=x/y
if checkprime(y1)==y1:
print(2)
else:
print(1,y*checkprime(y1),sep="\n")
| Title: Win or Freeze
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer *q*. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input Specification:
The first line contains the only integer *q* (1<=≤<=*q*<=≤<=1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output Specification:
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Demo Input:
['6\n', '30\n', '1\n']
Demo Output:
['2\n', '1\n6\n', '1\n0\n']
Note:
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | ```python
import math
def checkprime(x):
for j in range(2,math.ceil(math.sqrt(x))):
if(x%j==0):
return (j)
return x
x=int(input())
y=checkprime(x)
if y==x:
print(1,0,sep='\n')
else:
y1=x/y
if checkprime(y1)==y1:
print(2)
else:
print(1,y*checkprime(y1),sep="\n")
``` | 0 | |
573 | B | Bear and Blocks | PROGRAMMING | 1,600 | [
"binary search",
"data structures",
"dp",
"math"
] | null | null | Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built *n* towers in a row. The *i*-th tower is made of *h**i* identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=109) — sizes of towers. | Print the number of operations needed to destroy all towers. | [
"6\n2 1 4 6 2 2\n",
"7\n3 3 3 1 3 3 3\n"
] | [
"3\n",
"2\n"
] | The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. | 1,000 | [
{
"input": "6\n2 1 4 6 2 2",
"output": "3"
},
{
"input": "7\n3 3 3 1 3 3 3",
"output": "2"
},
{
"input": "7\n5128 5672 5805 5452 5882 5567 5032",
"output": "4"
},
{
"input": "10\n1 2 2 3 5 5 5 4 2 1",
"output": "5"
},
{
"input": "14\n20 20 20 20 20 20 3 20 20 20 20 20 20 20",
"output": "5"
},
{
"input": "50\n3 2 4 3 5 3 4 5 3 2 3 3 3 4 5 4 2 2 3 3 4 4 3 2 3 3 2 3 4 4 5 2 5 2 3 5 4 4 2 2 3 5 2 5 2 2 5 4 5 4",
"output": "4"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n1049 1098",
"output": "1"
},
{
"input": "2\n100 100",
"output": "1"
},
{
"input": "5\n1 2 3 2 1",
"output": "3"
},
{
"input": "15\n2 2 1 1 2 2 2 2 2 2 2 2 2 1 2",
"output": "2"
},
{
"input": "28\n415546599 415546599 415546599 415546599 415546599 415546599 415546599 415546599 415546599 2 802811737 802811737 802811737 802811737 802811737 802811737 802811737 802811737 1 550595901 550595901 550595901 550595901 550595901 550595901 550595901 550595901 550595901",
"output": "6"
},
{
"input": "45\n3 12 13 11 13 13 10 11 14 15 15 13 14 12 13 11 14 10 10 14 14 11 10 12 11 11 13 14 10 11 14 13 14 11 11 11 12 15 1 10 15 12 14 14 14",
"output": "13"
},
{
"input": "84\n1 3 4 5 6 5 6 7 8 9 7 4 5 4 2 5 1 1 1 3 2 7 7 8 10 9 5 6 5 2 3 3 3 3 3 2 4 8 6 5 8 9 8 7 9 3 4 4 4 2 2 1 6 4 9 5 9 9 10 7 10 4 5 4 2 4 3 3 4 4 6 6 6 9 10 12 7 5 9 8 5 3 3 2",
"output": "8"
},
{
"input": "170\n1 2 1 2 1 1 1 1 2 3 2 1 1 2 2 1 2 1 2 1 1 2 3 3 2 1 1 1 1 1 1 1 1 2 1 2 3 3 2 1 2 2 1 2 3 2 1 1 2 3 2 1 2 1 1 1 2 3 3 2 1 2 1 2 1 1 1 2 1 2 1 1 2 2 1 1 2 1 2 2 1 2 1 2 2 1 2 1 2 3 2 1 1 2 3 4 4 3 2 1 2 1 2 1 2 3 3 2 1 2 1 1 1 1 1 1 1 2 2 1 1 2 1 1 1 1 2 1 1 2 3 2 1 2 2 1 2 1 1 1 2 2 1 2 1 2 3 2 1 2 1 1 1 2 3 4 5 4 3 2 1 1 2 1 2 3 4 3 2 1",
"output": "5"
},
{
"input": "1\n5",
"output": "1"
}
] | 1,443,797,815 | 2,095 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 1,000 | 9,216,000 | def solve(n,a):
a.insert(0,0)
a.append(0)
count = 0
b = a[:]
while True:
found = False
for i in range(1,len(a)-1):
if a[i] != 0 and not found:
found = True
count += 1
cur = max(a[i]-1, 0)
b[i] = min([cur, a[i-1], a[i+1]])
a = b[:]
if not found:
break
return count
def main(infile, outfile):
n = int(infile.readline())
a = list(map(int,infile.readline().split()))
outfile.write(str(solve(n,a)) + '\n')
if __name__ == '__main__':
from sys import stdin, stdout
main(stdin, stdout)
| Title: Bear and Blocks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built *n* towers in a row. The *i*-th tower is made of *h**i* identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=109) — sizes of towers.
Output Specification:
Print the number of operations needed to destroy all towers.
Demo Input:
['6\n2 1 4 6 2 2\n', '7\n3 3 3 1 3 3 3\n']
Demo Output:
['3\n', '2\n']
Note:
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. | ```python
def solve(n,a):
a.insert(0,0)
a.append(0)
count = 0
b = a[:]
while True:
found = False
for i in range(1,len(a)-1):
if a[i] != 0 and not found:
found = True
count += 1
cur = max(a[i]-1, 0)
b[i] = min([cur, a[i-1], a[i+1]])
a = b[:]
if not found:
break
return count
def main(infile, outfile):
n = int(infile.readline())
a = list(map(int,infile.readline().split()))
outfile.write(str(solve(n,a)) + '\n')
if __name__ == '__main__':
from sys import stdin, stdout
main(stdin, stdout)
``` | 0 | |
263 | A | Beautiful Matrix | PROGRAMMING | 800 | [
"implementation"
] | null | null | You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful. | The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one. | Print a single integer — the minimum number of moves needed to make the matrix beautiful. | [
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] | [
"3\n",
"1\n"
] | none | 500 | [
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,694,883,529 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | # Función para calcular la distancia entre dos puntos en una matriz
def calculate_distance(x1, y1, x2, y2):
return abs(x1 - x2) + abs(y1 - y2)
# Función para calcular el mínimo número de movimientos para colocar el 1 en el centro
def min_moves_to_beautiful(matrix):
# Encontrar la posición actual del número 1
one_row, one_col = None, None
for i in range(5):
for j in range(5):
if matrix[i][j] == 1:
one_row, one_col = i, j
break
# Calcular la distancia desde la posición actual a la posición deseada (3, 3)
return calculate_distance(one_row, one_col, 2, 2)
| Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none | ```python
# Función para calcular la distancia entre dos puntos en una matriz
def calculate_distance(x1, y1, x2, y2):
return abs(x1 - x2) + abs(y1 - y2)
# Función para calcular el mínimo número de movimientos para colocar el 1 en el centro
def min_moves_to_beautiful(matrix):
# Encontrar la posición actual del número 1
one_row, one_col = None, None
for i in range(5):
for j in range(5):
if matrix[i][j] == 1:
one_row, one_col = i, j
break
# Calcular la distancia desde la posición actual a la posición deseada (3, 3)
return calculate_distance(one_row, one_col, 2, 2)
``` | 0 | |
907 | A | Masha and Bears | PROGRAMMING | 1,300 | [
"brute force",
"implementation"
] | null | null | A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars. | You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3. | Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes). | [
"50 30 10 10\n",
"100 50 10 21\n"
] | [
"50\n30\n10\n",
"-1\n"
] | In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20. | 500 | [
{
"input": "50 30 10 10",
"output": "50\n30\n10"
},
{
"input": "100 50 10 21",
"output": "-1"
},
{
"input": "100 50 19 10",
"output": "100\n50\n19"
},
{
"input": "99 50 25 49",
"output": "100\n99\n49"
},
{
"input": "3 2 1 1",
"output": "4\n3\n1"
},
{
"input": "100 99 98 100",
"output": "-1"
},
{
"input": "100 40 30 40",
"output": "-1"
},
{
"input": "100 50 19 25",
"output": "100\n51\n25"
},
{
"input": "100 50 19 30",
"output": "100\n61\n30"
},
{
"input": "49 48 25 49",
"output": "-1"
},
{
"input": "48 47 23 46",
"output": "94\n93\n46"
},
{
"input": "37 23 16 20",
"output": "42\n41\n20"
},
{
"input": "98 2 1 1",
"output": "98\n3\n1"
},
{
"input": "99 22 13 14",
"output": "99\n29\n14"
},
{
"input": "97 95 3 2",
"output": "97\n95\n3"
},
{
"input": "27 3 2 3",
"output": "-1"
},
{
"input": "13 7 6 2",
"output": "-1"
},
{
"input": "19 17 11 6",
"output": "19\n17\n11"
},
{
"input": "15 6 4 5",
"output": "15\n11\n5"
},
{
"input": "21 3 1 3",
"output": "-1"
},
{
"input": "100 99 98 97",
"output": "196\n195\n98"
},
{
"input": "99 32 13 5",
"output": "-1"
},
{
"input": "5 4 3 1",
"output": "-1"
},
{
"input": "50 20 4 8",
"output": "50\n20\n8"
},
{
"input": "60 30 10 1",
"output": "-1"
},
{
"input": "100 50 10 1",
"output": "-1"
},
{
"input": "40 30 20 10",
"output": "40\n30\n20"
},
{
"input": "50 49 5 10",
"output": "50\n49\n10"
},
{
"input": "99 98 50 97",
"output": "196\n195\n97"
},
{
"input": "50 30 10 5",
"output": "50\n30\n10"
},
{
"input": "4 3 2 1",
"output": "4\n3\n2"
},
{
"input": "100 50 3 1",
"output": "-1"
}
] | 1,514,038,244 | 344 | Python 3 | WRONG_ANSWER | PRETESTS | 0 | 62 | 5,529,600 | v1, v2, v3, vm = map(int, input().split())
if vm > 2*v3:
print(-1)
else:
print(vm)
print(2*v2)
print(2*v1) | Title: Masha and Bears
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input Specification:
You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3.
Output Specification:
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Demo Input:
['50 30 10 10\n', '100 50 10 21\n']
Demo Output:
['50\n30\n10\n', '-1\n']
Note:
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20. | ```python
v1, v2, v3, vm = map(int, input().split())
if vm > 2*v3:
print(-1)
else:
print(vm)
print(2*v2)
print(2*v1)
``` | 0 | |
745 | A | Hongcow Learns the Cyclic Shift | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | Hongcow is learning to spell! One day, his teacher gives him a word that he needs to learn to spell. Being a dutiful student, he immediately learns how to spell the word.
Hongcow has decided to try to make new words from this one. He starts by taking the word he just learned how to spell, and moves the last character of the word to the beginning of the word. He calls this a cyclic shift. He can apply cyclic shift many times. For example, consecutively applying cyclic shift operation to the word "abracadabra" Hongcow will get words "aabracadabr", "raabracadab" and so on.
Hongcow is now wondering how many distinct words he can generate by doing the cyclic shift arbitrarily many times. The initial string is also counted. | The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=50), the word Hongcow initially learns how to spell. The string *s* consists only of lowercase English letters ('a'–'z'). | Output a single integer equal to the number of distinct strings that Hongcow can obtain by applying the cyclic shift arbitrarily many times to the given string. | [
"abcd\n",
"bbb\n",
"yzyz\n"
] | [
"4\n",
"1\n",
"2\n"
] | For the first sample, the strings Hongcow can generate are "abcd", "dabc", "cdab", and "bcda".
For the second sample, no matter how many times Hongcow does the cyclic shift, Hongcow can only generate "bbb".
For the third sample, the two strings Hongcow can generate are "yzyz" and "zyzy". | 500 | [
{
"input": "abcd",
"output": "4"
},
{
"input": "bbb",
"output": "1"
},
{
"input": "yzyz",
"output": "2"
},
{
"input": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxy",
"output": "25"
},
{
"input": "zclkjadoprqronzclkjadoprqronzclkjadoprqron",
"output": "14"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "1"
},
{
"input": "xyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxy",
"output": "2"
},
{
"input": "y",
"output": "1"
},
{
"input": "ervbfotfedpozygoumbmxeaqegouaqqzqerlykhmvxvvlcaos",
"output": "49"
},
{
"input": "zyzzzyyzyyyzyyzyzyzyzyzzzyyyzzyzyyzzzzzyyyzzzzyzyy",
"output": "50"
},
{
"input": "zzfyftdezzfyftdezzfyftdezzfyftdezzfyftdezzfyftde",
"output": "8"
},
{
"input": "yehcqdlllqpuxdsaicyjjxiylahgxbygmsopjbxhtimzkashs",
"output": "49"
},
{
"input": "yyyyzzzyzzzyzyzyzyyyyyzzyzyzyyyyyzyzyyyzyzzyyzzzz",
"output": "49"
},
{
"input": "zkqcrhzlzsnwzkqcrhzlzsnwzkqcrhzlzsnwzkqcrhzlzsnw",
"output": "12"
},
{
"input": "xxyxxyxxyxxyxxyxxyxxyxxyxxyxxyxxyxxyxxyxxyxxyxxy",
"output": "3"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaaaaaaaaab",
"output": "25"
},
{
"input": "aabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaaba",
"output": "4"
},
{
"input": "pqqpqqpqqpqqpqqpqqpqqpqqpqqpqqpqqppqppqppqppqppq",
"output": "48"
},
{
"input": "zxkljaqzxkljaqzxkljaqzxkljaqzxrljaqzxkljaqzxkljaq",
"output": "49"
},
{
"input": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwx",
"output": "50"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaz",
"output": "50"
},
{
"input": "abcddcba",
"output": "8"
},
{
"input": "aabaabaabaacaabaabaabaacaabaabaabaacaabaabaabaac",
"output": "12"
},
{
"input": "aabaabcaabaabcdaabaabcaabaabcd",
"output": "15"
},
{
"input": "ababaababaaababaababaaaababaababaaababaababaaaa",
"output": "47"
},
{
"input": "ababaababaaababaababaaaababaababaaababaababaaa",
"output": "23"
},
{
"input": "aaababaab",
"output": "9"
},
{
"input": "aba",
"output": "3"
}
] | 1,568,146,324 | 2,147,483,647 | Python 3 | OK | TESTS | 28 | 109 | 0 | l=list(input())
s="".join(l)
d=[]
d.append(s)
for i in range(len(l)):
l.append(l[0])
l.pop(0)
s="".join(l)
if s not in d:
d.append(s)
print(len(d)) | Title: Hongcow Learns the Cyclic Shift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hongcow is learning to spell! One day, his teacher gives him a word that he needs to learn to spell. Being a dutiful student, he immediately learns how to spell the word.
Hongcow has decided to try to make new words from this one. He starts by taking the word he just learned how to spell, and moves the last character of the word to the beginning of the word. He calls this a cyclic shift. He can apply cyclic shift many times. For example, consecutively applying cyclic shift operation to the word "abracadabra" Hongcow will get words "aabracadabr", "raabracadab" and so on.
Hongcow is now wondering how many distinct words he can generate by doing the cyclic shift arbitrarily many times. The initial string is also counted.
Input Specification:
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=50), the word Hongcow initially learns how to spell. The string *s* consists only of lowercase English letters ('a'–'z').
Output Specification:
Output a single integer equal to the number of distinct strings that Hongcow can obtain by applying the cyclic shift arbitrarily many times to the given string.
Demo Input:
['abcd\n', 'bbb\n', 'yzyz\n']
Demo Output:
['4\n', '1\n', '2\n']
Note:
For the first sample, the strings Hongcow can generate are "abcd", "dabc", "cdab", and "bcda".
For the second sample, no matter how many times Hongcow does the cyclic shift, Hongcow can only generate "bbb".
For the third sample, the two strings Hongcow can generate are "yzyz" and "zyzy". | ```python
l=list(input())
s="".join(l)
d=[]
d.append(s)
for i in range(len(l)):
l.append(l[0])
l.pop(0)
s="".join(l)
if s not in d:
d.append(s)
print(len(d))
``` | 3 | |
877 | B | Nikita and string | PROGRAMMING | 1,500 | [
"brute force",
"dp"
] | null | null | One day Nikita found the string containing letters "a" and "b" only.
Nikita thinks that string is beautiful if it can be cut into 3 strings (possibly empty) without changing the order of the letters, where the 1-st and the 3-rd one contain only letters "a" and the 2-nd contains only letters "b".
Nikita wants to make the string beautiful by removing some (possibly none) of its characters, but without changing their order. What is the maximum length of the string he can get? | The first line contains a non-empty string of length not greater than 5<=000 containing only lowercase English letters "a" and "b". | Print a single integer — the maximum possible size of beautiful string Nikita can get. | [
"abba\n",
"bab\n"
] | [
"4",
"2"
] | It the first sample the string is already beautiful.
In the second sample he needs to delete one of "b" to make it beautiful. | 1,000 | [
{
"input": "abba",
"output": "4"
},
{
"input": "bab",
"output": "2"
},
{
"input": "bbabbbaabbbb",
"output": "9"
},
{
"input": "bbabbbbbaaba",
"output": "10"
},
{
"input": "bbabbbababaa",
"output": "9"
},
{
"input": "aabbaababbab",
"output": "8"
},
{
"input": "a",
"output": "1"
},
{
"input": "b",
"output": "1"
},
{
"input": "ab",
"output": "2"
},
{
"input": "ba",
"output": "2"
},
{
"input": "bb",
"output": "2"
},
{
"input": "aa",
"output": "2"
},
{
"input": "babbbaab",
"output": "6"
},
{
"input": "abaaaa",
"output": "6"
},
{
"input": "aaa",
"output": "3"
}
] | 1,586,092,192 | 2,147,483,647 | Python 3 | OK | TESTS | 46 | 109 | 307,200 |
max1=0
string=input()
first=0
second=0
third=0
i=0
length=len(string)
while(i<length):
if string[i]=='a':
if second>third:
third=second+1
else:
third+=1
first+=1
if string[i]=='b':
if first>second:
second=first+1
else:
second+=1
i+=1
print(max(third,first,second)) | Title: Nikita and string
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Nikita found the string containing letters "a" and "b" only.
Nikita thinks that string is beautiful if it can be cut into 3 strings (possibly empty) without changing the order of the letters, where the 1-st and the 3-rd one contain only letters "a" and the 2-nd contains only letters "b".
Nikita wants to make the string beautiful by removing some (possibly none) of its characters, but without changing their order. What is the maximum length of the string he can get?
Input Specification:
The first line contains a non-empty string of length not greater than 5<=000 containing only lowercase English letters "a" and "b".
Output Specification:
Print a single integer — the maximum possible size of beautiful string Nikita can get.
Demo Input:
['abba\n', 'bab\n']
Demo Output:
['4', '2']
Note:
It the first sample the string is already beautiful.
In the second sample he needs to delete one of "b" to make it beautiful. | ```python
max1=0
string=input()
first=0
second=0
third=0
i=0
length=len(string)
while(i<length):
if string[i]=='a':
if second>third:
third=second+1
else:
third+=1
first+=1
if string[i]=='b':
if first>second:
second=first+1
else:
second+=1
i+=1
print(max(third,first,second))
``` | 3 | |
630 | F | Selection of Personnel | PROGRAMMING | 1,300 | [
"combinatorics",
"math"
] | null | null | One company of IT City decided to create a group of innovative developments consisting from 5 to 7 people and hire new employees for it. After placing an advertisment the company received *n* resumes. Now the HR department has to evaluate each possible group composition and select one of them. Your task is to count the number of variants of group composition to evaluate. | The only line of the input contains one integer *n* (7<=≤<=*n*<=≤<=777) — the number of potential employees that sent resumes. | Output one integer — the number of different variants of group composition. | [
"7\n"
] | [
"29"
] | none | 0 | [
{
"input": "7",
"output": "29"
},
{
"input": "8",
"output": "92"
},
{
"input": "9",
"output": "246"
},
{
"input": "10",
"output": "582"
},
{
"input": "321",
"output": "66715035255088"
},
{
"input": "624",
"output": "7147161340917624"
},
{
"input": "666",
"output": "11292070960994226"
},
{
"input": "700",
"output": "16017044425409540"
},
{
"input": "776",
"output": "33019955679376860"
},
{
"input": "777",
"output": "33319741730082870"
}
] | 1,695,471,646 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 10 | 46 | 0 | from math import comb
n = int(input())
print(comb(n, 5) + comb(n, 6) + comb(n, 7))
| Title: Selection of Personnel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One company of IT City decided to create a group of innovative developments consisting from 5 to 7 people and hire new employees for it. After placing an advertisment the company received *n* resumes. Now the HR department has to evaluate each possible group composition and select one of them. Your task is to count the number of variants of group composition to evaluate.
Input Specification:
The only line of the input contains one integer *n* (7<=≤<=*n*<=≤<=777) — the number of potential employees that sent resumes.
Output Specification:
Output one integer — the number of different variants of group composition.
Demo Input:
['7\n']
Demo Output:
['29']
Note:
none | ```python
from math import comb
n = int(input())
print(comb(n, 5) + comb(n, 6) + comb(n, 7))
``` | 3 | |
331 | C1 | The Great Julya Calendar | PROGRAMMING | 1,100 | [
"dp"
] | null | null | Yet another Armageddon is coming! This time the culprit is the Julya tribe calendar.
The beavers in this tribe knew math very well. Smart Beaver, an archaeologist, got a sacred plate with a magic integer on it. The translation from Old Beaverish is as follows:
"May the Great Beaver bless you! May your chacres open and may your third eye never turn blind from beholding the Truth! Take the magic number, subtract a digit from it (the digit must occur in the number) and get a new magic number. Repeat this operation until a magic number equals zero. The Earth will stand on Three Beavers for the time, equal to the number of subtractions you perform!"
Distinct subtraction sequences can obviously get you different number of operations. But the Smart Beaver is ready to face the worst and is asking you to count the minimum number of operations he needs to reduce the magic number to zero. | The single line contains the magic integer *n*, 0<=≤<=*n*.
- to get 20 points, you need to solve the problem with constraints: *n*<=≤<=106 (subproblem C1); - to get 40 points, you need to solve the problem with constraints: *n*<=≤<=1012 (subproblems C1+C2); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=1018 (subproblems C1+C2+C3). | Print a single integer — the minimum number of subtractions that turns the magic number to a zero. | [
"24\n"
] | [
"5"
] | In the first test sample the minimum number of operations can be reached by the following sequence of subtractions: | 20 | [
{
"input": "24",
"output": "5"
},
{
"input": "0",
"output": "0"
},
{
"input": "3",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "10",
"output": "2"
},
{
"input": "31",
"output": "6"
},
{
"input": "701",
"output": "116"
},
{
"input": "222",
"output": "39"
},
{
"input": "156",
"output": "28"
},
{
"input": "12343",
"output": "1778"
},
{
"input": "7984",
"output": "1178"
},
{
"input": "1000000",
"output": "128207"
},
{
"input": "1",
"output": "1"
},
{
"input": "3",
"output": "1"
},
{
"input": "5",
"output": "1"
},
{
"input": "11",
"output": "3"
},
{
"input": "77",
"output": "14"
},
{
"input": "216",
"output": "37"
},
{
"input": "1468",
"output": "233"
},
{
"input": "1995",
"output": "305"
},
{
"input": "11010",
"output": "1568"
},
{
"input": "47320",
"output": "6746"
},
{
"input": "258634",
"output": "35024"
}
] | 1,551,463,575 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS1 | 0 | 0 | 0 | n, k = int(input()), 0
while n:
n- = int(max(str(n)))
k + = 1
print( k )
#python ki jai ho :p | Title: The Great Julya Calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yet another Armageddon is coming! This time the culprit is the Julya tribe calendar.
The beavers in this tribe knew math very well. Smart Beaver, an archaeologist, got a sacred plate with a magic integer on it. The translation from Old Beaverish is as follows:
"May the Great Beaver bless you! May your chacres open and may your third eye never turn blind from beholding the Truth! Take the magic number, subtract a digit from it (the digit must occur in the number) and get a new magic number. Repeat this operation until a magic number equals zero. The Earth will stand on Three Beavers for the time, equal to the number of subtractions you perform!"
Distinct subtraction sequences can obviously get you different number of operations. But the Smart Beaver is ready to face the worst and is asking you to count the minimum number of operations he needs to reduce the magic number to zero.
Input Specification:
The single line contains the magic integer *n*, 0<=≤<=*n*.
- to get 20 points, you need to solve the problem with constraints: *n*<=≤<=106 (subproblem C1); - to get 40 points, you need to solve the problem with constraints: *n*<=≤<=1012 (subproblems C1+C2); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=1018 (subproblems C1+C2+C3).
Output Specification:
Print a single integer — the minimum number of subtractions that turns the magic number to a zero.
Demo Input:
['24\n']
Demo Output:
['5']
Note:
In the first test sample the minimum number of operations can be reached by the following sequence of subtractions: | ```python
n, k = int(input()), 0
while n:
n- = int(max(str(n)))
k + = 1
print( k )
#python ki jai ho :p
``` | -1 | |
166 | E | Tetrahedron | PROGRAMMING | 1,500 | [
"dp",
"math",
"matrices"
] | null | null | You are given a tetrahedron. Let's mark its vertices with letters *A*, *B*, *C* and *D* correspondingly.
An ant is standing in the vertex *D* of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.
You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex *D* to itself in exactly *n* steps. In other words, you are asked to find out the number of different cyclic paths with the length of *n* from vertex *D* to itself. As the number can be quite large, you should print it modulo 1000000007 (109<=+<=7). | The first line contains the only integer *n* (1<=≤<=*n*<=≤<=107) — the required length of the cyclic path. | Print the only integer — the required number of ways modulo 1000000007 (109<=+<=7). | [
"2\n",
"4\n"
] | [
"3\n",
"21\n"
] | The required paths in the first sample are:
- *D* - *A* - *D* - *D* - *B* - *D* - *D* - *C* - *D* | 1,000 | [
{
"input": "2",
"output": "3"
},
{
"input": "4",
"output": "21"
},
{
"input": "1",
"output": "0"
},
{
"input": "3",
"output": "6"
},
{
"input": "5",
"output": "60"
},
{
"input": "6",
"output": "183"
},
{
"input": "7",
"output": "546"
},
{
"input": "8",
"output": "1641"
},
{
"input": "9",
"output": "4920"
},
{
"input": "10",
"output": "14763"
},
{
"input": "15",
"output": "3587226"
},
{
"input": "30",
"output": "782663359"
},
{
"input": "10000000",
"output": "192336614"
},
{
"input": "100",
"output": "721510432"
},
{
"input": "300",
"output": "327873818"
},
{
"input": "900",
"output": "295068084"
},
{
"input": "1500",
"output": "451187545"
},
{
"input": "3000",
"output": "645417275"
},
{
"input": "5000",
"output": "755610910"
},
{
"input": "10000",
"output": "723907367"
},
{
"input": "50000",
"output": "969527595"
},
{
"input": "90000",
"output": "548978368"
},
{
"input": "99999",
"output": "909741855"
},
{
"input": "100000",
"output": "729225554"
},
{
"input": "300000",
"output": "93822635"
},
{
"input": "800000",
"output": "178940616"
},
{
"input": "1000000",
"output": "266233856"
},
{
"input": "4000000",
"output": "882155933"
},
{
"input": "9000000",
"output": "295060537"
},
{
"input": "9999999",
"output": "730778875"
},
{
"input": "1000000",
"output": "266233856"
},
{
"input": "9999999",
"output": "730778875"
},
{
"input": "10000000",
"output": "192336614"
},
{
"input": "9999998",
"output": "576926295"
},
{
"input": "30",
"output": "782663359"
},
{
"input": "31",
"output": "347990060"
}
] | 1,689,849,911 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 0 | n = int(input())
MOD = 10 ** 9 + 7
if n == 1:
print(0)
exit(0)
n -= 2
ans = 3 * (2 ** n)
ans %= MOD
print(ans) | Title: Tetrahedron
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a tetrahedron. Let's mark its vertices with letters *A*, *B*, *C* and *D* correspondingly.
An ant is standing in the vertex *D* of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.
You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex *D* to itself in exactly *n* steps. In other words, you are asked to find out the number of different cyclic paths with the length of *n* from vertex *D* to itself. As the number can be quite large, you should print it modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=107) — the required length of the cyclic path.
Output Specification:
Print the only integer — the required number of ways modulo 1000000007 (109<=+<=7).
Demo Input:
['2\n', '4\n']
Demo Output:
['3\n', '21\n']
Note:
The required paths in the first sample are:
- *D* - *A* - *D* - *D* - *B* - *D* - *D* - *C* - *D* | ```python
n = int(input())
MOD = 10 ** 9 + 7
if n == 1:
print(0)
exit(0)
n -= 2
ans = 3 * (2 ** n)
ans %= MOD
print(ans)
``` | 0 | |
985 | A | Chess Placing | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | You are given a chessboard of size 1<=×<=*n*. It is guaranteed that *n* is even. The chessboard is painted like this: "BWBW...BW".
Some cells of the board are occupied by the chess pieces. Each cell contains no more than one chess piece. It is known that the total number of pieces equals to .
In one step you can move one of the pieces one cell to the left or to the right. You cannot move pieces beyond the borders of the board. You also cannot move pieces to the cells that are already occupied.
Your task is to place all the pieces in the cells of the same color using the minimum number of moves (all the pieces must occupy only the black cells or only the white cells after all the moves are made). | The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100, *n* is even) — the size of the chessboard.
The second line of the input contains integer numbers (1<=≤<=*p**i*<=≤<=*n*) — initial positions of the pieces. It is guaranteed that all the positions are distinct. | Print one integer — the minimum number of moves you have to make to place all the pieces in the cells of the same color. | [
"6\n1 2 6\n",
"10\n1 2 3 4 5\n"
] | [
"2\n",
"10\n"
] | In the first example the only possible strategy is to move the piece at the position 6 to the position 5 and move the piece at the position 2 to the position 3. Notice that if you decide to place the pieces in the white cells the minimum number of moves will be 3.
In the second example the possible strategy is to move <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e1e06f6a15cce30628c7a2360c4ffa57a8ba0ebd.png" style="max-width: 100.0%;max-height: 100.0%;"/> in 4 moves, then <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c84dfbe0c6a917b45fc3f69467c256c4ac460eeb.png" style="max-width: 100.0%;max-height: 100.0%;"/> in 3 moves, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/598731d81393332209d914cb0bbe97d8566c887d.png" style="max-width: 100.0%;max-height: 100.0%;"/> in 2 moves and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/29f71c065c3536e88b54429c734103ad3604f68b.png" style="max-width: 100.0%;max-height: 100.0%;"/> in 1 move. | 0 | [
{
"input": "6\n1 2 6",
"output": "2"
},
{
"input": "10\n1 2 3 4 5",
"output": "10"
},
{
"input": "2\n2",
"output": "0"
},
{
"input": "100\n2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100",
"output": "0"
},
{
"input": "100\n93 54 57 61 68 66 70 96 64 82 80 75 69 77 76 94 67 86 90 73 74 58 100 83 92 89 56 99 88 59 95 72 81 51 85 71 97 60 91 63 65 98 79 84 53 62 87 55 52 78",
"output": "1225"
},
{
"input": "100\n41 13 29 11 25 15 6 23 28 50 48 17 3 9 44 24 5 19 34 22 33 32 20 16 35 37 4 10 46 2 39 40 47 49 36 42 1 30 43 21 14 7 18 45 31 8 12 26 27 38",
"output": "1225"
},
{
"input": "96\n12 58 70 19 65 61 41 46 15 92 64 72 9 26 53 37 2 3 1 40 10 8 94 66 50 34 36 96 47 78 7 57 5 6 17 69 28 88 89 49 55 81 35 22 25 79 86 59",
"output": "152"
},
{
"input": "10\n5 6 7 8 9",
"output": "7"
},
{
"input": "20\n1 2 3 4 5 6 7 8 9 10",
"output": "45"
},
{
"input": "10\n6 7 8 9 10",
"output": "10"
},
{
"input": "10\n9 8 7 6 5",
"output": "7"
},
{
"input": "6\n1 5 6",
"output": "2"
},
{
"input": "12\n1 7 8 9 10 12",
"output": "7"
},
{
"input": "6\n1 4 5",
"output": "1"
},
{
"input": "24\n10 21 15 3 11 4 18 24 16 22 14 9",
"output": "11"
},
{
"input": "20\n3 4 6 7 8 10 11 13 14 17",
"output": "15"
},
{
"input": "10\n10 9 8 1 5",
"output": "5"
},
{
"input": "100\n84 10 26 79 58 93 67 85 7 2 99 4 47 45 75 22 32 82 65 53 63 49 42 52 12 69 86 46 25 76 40 15 13 78 8 81 62 28 60 21 27 80 98 56 3 36 54 16 50 43",
"output": "104"
},
{
"input": "10\n1 7 8 9 10",
"output": "7"
},
{
"input": "10\n1 4 6 8 10",
"output": "1"
},
{
"input": "80\n41 70 18 53 32 79 51 49 21 27 47 65 50 15 62 60 5 40 14 25 64 9 19 58 38 76 66 52 17 34 13 2 80 43 3 42 33 36 6 72",
"output": "47"
},
{
"input": "50\n27 42 41 4 10 45 44 26 49 50 17 28 2 36 18 39 23 12 21 24 19 29 22 40 37",
"output": "59"
},
{
"input": "10\n2 3 4 5 6",
"output": "7"
},
{
"input": "6\n3 5 6",
"output": "2"
},
{
"input": "100\n9 63 62 88 3 67 54 33 79 51 71 80 37 46 43 57 69 17 34 6 18 40 59 83 76 86 8 55 90 89 45 42 28 98 30 38 77 91 73 58 23 61 41 65 64 93 14 44 16 24",
"output": "160"
},
{
"input": "10\n1 6 7 8 9",
"output": "5"
},
{
"input": "6\n3 4 5",
"output": "2"
}
] | 1,603,170,371 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 109 | 307,200 | import math
#n,k = map(int, input().strip().split(' '))
n=int(input())
lst = list(map(int, input().strip().split(' ')))
sum_odd=0
sum_even=0
lst.sort()
for i in range(n//2):
sum_odd+=abs(lst[i]-(2*i+1))
for i in range(n//2):
sum_even+=abs(lst[i]-(2*(i+1)))
print(min(sum_odd,sum_even))
| Title: Chess Placing
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a chessboard of size 1<=×<=*n*. It is guaranteed that *n* is even. The chessboard is painted like this: "BWBW...BW".
Some cells of the board are occupied by the chess pieces. Each cell contains no more than one chess piece. It is known that the total number of pieces equals to .
In one step you can move one of the pieces one cell to the left or to the right. You cannot move pieces beyond the borders of the board. You also cannot move pieces to the cells that are already occupied.
Your task is to place all the pieces in the cells of the same color using the minimum number of moves (all the pieces must occupy only the black cells or only the white cells after all the moves are made).
Input Specification:
The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100, *n* is even) — the size of the chessboard.
The second line of the input contains integer numbers (1<=≤<=*p**i*<=≤<=*n*) — initial positions of the pieces. It is guaranteed that all the positions are distinct.
Output Specification:
Print one integer — the minimum number of moves you have to make to place all the pieces in the cells of the same color.
Demo Input:
['6\n1 2 6\n', '10\n1 2 3 4 5\n']
Demo Output:
['2\n', '10\n']
Note:
In the first example the only possible strategy is to move the piece at the position 6 to the position 5 and move the piece at the position 2 to the position 3. Notice that if you decide to place the pieces in the white cells the minimum number of moves will be 3.
In the second example the possible strategy is to move <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e1e06f6a15cce30628c7a2360c4ffa57a8ba0ebd.png" style="max-width: 100.0%;max-height: 100.0%;"/> in 4 moves, then <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c84dfbe0c6a917b45fc3f69467c256c4ac460eeb.png" style="max-width: 100.0%;max-height: 100.0%;"/> in 3 moves, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/598731d81393332209d914cb0bbe97d8566c887d.png" style="max-width: 100.0%;max-height: 100.0%;"/> in 2 moves and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/29f71c065c3536e88b54429c734103ad3604f68b.png" style="max-width: 100.0%;max-height: 100.0%;"/> in 1 move. | ```python
import math
#n,k = map(int, input().strip().split(' '))
n=int(input())
lst = list(map(int, input().strip().split(' ')))
sum_odd=0
sum_even=0
lst.sort()
for i in range(n//2):
sum_odd+=abs(lst[i]-(2*i+1))
for i in range(n//2):
sum_even+=abs(lst[i]-(2*(i+1)))
print(min(sum_odd,sum_even))
``` | 3 | |
279 | B | Books | PROGRAMMING | 1,400 | [
"binary search",
"brute force",
"implementation",
"two pointers"
] | null | null | When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read. | The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book. | Print a single integer — the maximum number of books Valera can read. | [
"4 5\n3 1 2 1\n",
"3 3\n2 2 3\n"
] | [
"3\n",
"1\n"
] | none | 1,000 | [
{
"input": "4 5\n3 1 2 1",
"output": "3"
},
{
"input": "3 3\n2 2 3",
"output": "1"
},
{
"input": "1 3\n5",
"output": "0"
},
{
"input": "1 10\n4",
"output": "1"
},
{
"input": "2 10\n6 4",
"output": "2"
},
{
"input": "6 10\n2 3 4 2 1 1",
"output": "4"
},
{
"input": "7 13\n6 8 14 9 4 11 10",
"output": "2"
},
{
"input": "10 15\n10 9 1 1 5 10 5 3 7 2",
"output": "3"
},
{
"input": "20 30\n8 1 2 6 9 4 1 9 9 10 4 7 8 9 5 7 1 8 7 4",
"output": "6"
},
{
"input": "30 60\n16 13 22 38 13 35 17 17 20 38 12 19 9 22 20 3 35 34 34 21 35 40 22 3 27 19 12 4 8 19",
"output": "4"
},
{
"input": "100 100\n75 92 18 6 81 67 7 92 100 65 82 32 50 67 85 31 80 91 84 63 39 52 92 81 1 98 24 12 43 48 17 86 51 72 48 95 45 50 12 66 19 79 49 89 34 1 97 75 20 33 96 27 42 23 73 71 93 1 85 19 66 14 17 61 20 39 36 33 42 61 56 64 23 91 80 99 40 74 13 18 98 85 74 39 62 84 46 74 50 23 38 11 79 14 9 25 66 100 25 52",
"output": "3"
},
{
"input": "10 1\n4418 7528 8170 1736 1317 3205 8183 4995 8039 4708",
"output": "0"
},
{
"input": "50 2\n124 214 63 73 996 760 38 571 451 300 970 1 706 937 837 494 619 88 851 411 957 990 842 613 821 649 627 34 693 678 734 116 816 985 705 940 499 493 922 967 854 439 112 644 961 438 189 572 655 550",
"output": "1"
}
] | 1,694,616,724 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 1 | 2,000 | 0 | a,b=[int(x) for x in input().split()]
A=list(map(int, input().split()))
A.append(10**10)
x=0
c=b
while c>=0:
c-=A[x]
x+=1
i=0
while i+x<a:
c+=A[i]-A[i+x]
if c>=0:
x+=1
c-=A[i+x]
print(x-1) | Title: Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
Output Specification:
Print a single integer — the maximum number of books Valera can read.
Demo Input:
['4 5\n3 1 2 1\n', '3 3\n2 2 3\n']
Demo Output:
['3\n', '1\n']
Note:
none | ```python
a,b=[int(x) for x in input().split()]
A=list(map(int, input().split()))
A.append(10**10)
x=0
c=b
while c>=0:
c-=A[x]
x+=1
i=0
while i+x<a:
c+=A[i]-A[i+x]
if c>=0:
x+=1
c-=A[i+x]
print(x-1)
``` | 0 | |
53 | C | Little Frog | PROGRAMMING | 1,200 | [
"constructive algorithms"
] | C. Little Frog | 2 | 256 | Once upon a time a little frog whose name was Vasya decided to travel around his home swamp. Overall there are *n* mounds on the swamp, located on one line. The distance between the neighboring mounds is one meter. Vasya wants to visit all the mounds in one day; besides, he wants to visit each one exactly once. For that he makes a route plan, to decide the order in which to jump on the mounds. Vasya can pick any mound as the first one. He thinks it boring to jump two times at the same distance. That's why he wants any two jumps on his route to have different lengths. Help Vasya the Frog and make the plan for him. | The single line contains a number *n* (1<=≤<=*n*<=≤<=104) which is the number of mounds. | Print *n* integers *p**i* (1<=≤<=*p**i*<=≤<=*n*) which are the frog's route plan.
- All the *p**i*'s should be mutually different. - All the |*p**i*–*p**i*<=+<=1|'s should be mutually different (1<=≤<=*i*<=≤<=*n*<=-<=1).
If there are several solutions, output any. | [
"2\n",
"3\n"
] | [
"1 2 ",
"1 3 2 "
] | none | 1,500 | [
{
"input": "2",
"output": "1 2 "
},
{
"input": "3",
"output": "1 3 2 "
},
{
"input": "4",
"output": "1 4 2 3 "
},
{
"input": "5",
"output": "1 5 2 4 3 "
},
{
"input": "6",
"output": "1 6 2 5 3 4 "
},
{
"input": "1",
"output": "1 "
},
{
"input": "9149",
"output": "1 9149 2 9148 3 9147 4 9146 5 9145 6 9144 7 9143 8 9142 9 9141 10 9140 11 9139 12 9138 13 9137 14 9136 15 9135 16 9134 17 9133 18 9132 19 9131 20 9130 21 9129 22 9128 23 9127 24 9126 25 9125 26 9124 27 9123 28 9122 29 9121 30 9120 31 9119 32 9118 33 9117 34 9116 35 9115 36 9114 37 9113 38 9112 39 9111 40 9110 41 9109 42 9108 43 9107 44 9106 45 9105 46 9104 47 9103 48 9102 49 9101 50 9100 51 9099 52 9098 53 9097 54 9096 55 9095 56 9094 57 9093 58 9092 59 9091 60 9090 61 9089 62 9088 63 9087 64 9086 65 9085 ..."
},
{
"input": "2877",
"output": "1 2877 2 2876 3 2875 4 2874 5 2873 6 2872 7 2871 8 2870 9 2869 10 2868 11 2867 12 2866 13 2865 14 2864 15 2863 16 2862 17 2861 18 2860 19 2859 20 2858 21 2857 22 2856 23 2855 24 2854 25 2853 26 2852 27 2851 28 2850 29 2849 30 2848 31 2847 32 2846 33 2845 34 2844 35 2843 36 2842 37 2841 38 2840 39 2839 40 2838 41 2837 42 2836 43 2835 44 2834 45 2833 46 2832 47 2831 48 2830 49 2829 50 2828 51 2827 52 2826 53 2825 54 2824 55 2823 56 2822 57 2821 58 2820 59 2819 60 2818 61 2817 62 2816 63 2815 64 2814 65 2813 ..."
},
{
"input": "2956",
"output": "1 2956 2 2955 3 2954 4 2953 5 2952 6 2951 7 2950 8 2949 9 2948 10 2947 11 2946 12 2945 13 2944 14 2943 15 2942 16 2941 17 2940 18 2939 19 2938 20 2937 21 2936 22 2935 23 2934 24 2933 25 2932 26 2931 27 2930 28 2929 29 2928 30 2927 31 2926 32 2925 33 2924 34 2923 35 2922 36 2921 37 2920 38 2919 39 2918 40 2917 41 2916 42 2915 43 2914 44 2913 45 2912 46 2911 47 2910 48 2909 49 2908 50 2907 51 2906 52 2905 53 2904 54 2903 55 2902 56 2901 57 2900 58 2899 59 2898 60 2897 61 2896 62 2895 63 2894 64 2893 65 2892 ..."
},
{
"input": "3035",
"output": "1 3035 2 3034 3 3033 4 3032 5 3031 6 3030 7 3029 8 3028 9 3027 10 3026 11 3025 12 3024 13 3023 14 3022 15 3021 16 3020 17 3019 18 3018 19 3017 20 3016 21 3015 22 3014 23 3013 24 3012 25 3011 26 3010 27 3009 28 3008 29 3007 30 3006 31 3005 32 3004 33 3003 34 3002 35 3001 36 3000 37 2999 38 2998 39 2997 40 2996 41 2995 42 2994 43 2993 44 2992 45 2991 46 2990 47 2989 48 2988 49 2987 50 2986 51 2985 52 2984 53 2983 54 2982 55 2981 56 2980 57 2979 58 2978 59 2977 60 2976 61 2975 62 2974 63 2973 64 2972 65 2971 ..."
},
{
"input": "3114",
"output": "1 3114 2 3113 3 3112 4 3111 5 3110 6 3109 7 3108 8 3107 9 3106 10 3105 11 3104 12 3103 13 3102 14 3101 15 3100 16 3099 17 3098 18 3097 19 3096 20 3095 21 3094 22 3093 23 3092 24 3091 25 3090 26 3089 27 3088 28 3087 29 3086 30 3085 31 3084 32 3083 33 3082 34 3081 35 3080 36 3079 37 3078 38 3077 39 3076 40 3075 41 3074 42 3073 43 3072 44 3071 45 3070 46 3069 47 3068 48 3067 49 3066 50 3065 51 3064 52 3063 53 3062 54 3061 55 3060 56 3059 57 3058 58 3057 59 3056 60 3055 61 3054 62 3053 63 3052 64 3051 65 3050 ..."
},
{
"input": "3193",
"output": "1 3193 2 3192 3 3191 4 3190 5 3189 6 3188 7 3187 8 3186 9 3185 10 3184 11 3183 12 3182 13 3181 14 3180 15 3179 16 3178 17 3177 18 3176 19 3175 20 3174 21 3173 22 3172 23 3171 24 3170 25 3169 26 3168 27 3167 28 3166 29 3165 30 3164 31 3163 32 3162 33 3161 34 3160 35 3159 36 3158 37 3157 38 3156 39 3155 40 3154 41 3153 42 3152 43 3151 44 3150 45 3149 46 3148 47 3147 48 3146 49 3145 50 3144 51 3143 52 3142 53 3141 54 3140 55 3139 56 3138 57 3137 58 3136 59 3135 60 3134 61 3133 62 3132 63 3131 64 3130 65 3129 ..."
},
{
"input": "3273",
"output": "1 3273 2 3272 3 3271 4 3270 5 3269 6 3268 7 3267 8 3266 9 3265 10 3264 11 3263 12 3262 13 3261 14 3260 15 3259 16 3258 17 3257 18 3256 19 3255 20 3254 21 3253 22 3252 23 3251 24 3250 25 3249 26 3248 27 3247 28 3246 29 3245 30 3244 31 3243 32 3242 33 3241 34 3240 35 3239 36 3238 37 3237 38 3236 39 3235 40 3234 41 3233 42 3232 43 3231 44 3230 45 3229 46 3228 47 3227 48 3226 49 3225 50 3224 51 3223 52 3222 53 3221 54 3220 55 3219 56 3218 57 3217 58 3216 59 3215 60 3214 61 3213 62 3212 63 3211 64 3210 65 3209 ..."
},
{
"input": "7000",
"output": "1 7000 2 6999 3 6998 4 6997 5 6996 6 6995 7 6994 8 6993 9 6992 10 6991 11 6990 12 6989 13 6988 14 6987 15 6986 16 6985 17 6984 18 6983 19 6982 20 6981 21 6980 22 6979 23 6978 24 6977 25 6976 26 6975 27 6974 28 6973 29 6972 30 6971 31 6970 32 6969 33 6968 34 6967 35 6966 36 6965 37 6964 38 6963 39 6962 40 6961 41 6960 42 6959 43 6958 44 6957 45 6956 46 6955 47 6954 48 6953 49 6952 50 6951 51 6950 52 6949 53 6948 54 6947 55 6946 56 6945 57 6944 58 6943 59 6942 60 6941 61 6940 62 6939 63 6938 64 6937 65 6936 ..."
},
{
"input": "7079",
"output": "1 7079 2 7078 3 7077 4 7076 5 7075 6 7074 7 7073 8 7072 9 7071 10 7070 11 7069 12 7068 13 7067 14 7066 15 7065 16 7064 17 7063 18 7062 19 7061 20 7060 21 7059 22 7058 23 7057 24 7056 25 7055 26 7054 27 7053 28 7052 29 7051 30 7050 31 7049 32 7048 33 7047 34 7046 35 7045 36 7044 37 7043 38 7042 39 7041 40 7040 41 7039 42 7038 43 7037 44 7036 45 7035 46 7034 47 7033 48 7032 49 7031 50 7030 51 7029 52 7028 53 7027 54 7026 55 7025 56 7024 57 7023 58 7022 59 7021 60 7020 61 7019 62 7018 63 7017 64 7016 65 7015 ..."
},
{
"input": "4653",
"output": "1 4653 2 4652 3 4651 4 4650 5 4649 6 4648 7 4647 8 4646 9 4645 10 4644 11 4643 12 4642 13 4641 14 4640 15 4639 16 4638 17 4637 18 4636 19 4635 20 4634 21 4633 22 4632 23 4631 24 4630 25 4629 26 4628 27 4627 28 4626 29 4625 30 4624 31 4623 32 4622 33 4621 34 4620 35 4619 36 4618 37 4617 38 4616 39 4615 40 4614 41 4613 42 4612 43 4611 44 4610 45 4609 46 4608 47 4607 48 4606 49 4605 50 4604 51 4603 52 4602 53 4601 54 4600 55 4599 56 4598 57 4597 58 4596 59 4595 60 4594 61 4593 62 4592 63 4591 64 4590 65 4589 ..."
},
{
"input": "9995",
"output": "1 9995 2 9994 3 9993 4 9992 5 9991 6 9990 7 9989 8 9988 9 9987 10 9986 11 9985 12 9984 13 9983 14 9982 15 9981 16 9980 17 9979 18 9978 19 9977 20 9976 21 9975 22 9974 23 9973 24 9972 25 9971 26 9970 27 9969 28 9968 29 9967 30 9966 31 9965 32 9964 33 9963 34 9962 35 9961 36 9960 37 9959 38 9958 39 9957 40 9956 41 9955 42 9954 43 9953 44 9952 45 9951 46 9950 47 9949 48 9948 49 9947 50 9946 51 9945 52 9944 53 9943 54 9942 55 9941 56 9940 57 9939 58 9938 59 9937 60 9936 61 9935 62 9934 63 9933 64 9932 65 9931 ..."
},
{
"input": "9996",
"output": "1 9996 2 9995 3 9994 4 9993 5 9992 6 9991 7 9990 8 9989 9 9988 10 9987 11 9986 12 9985 13 9984 14 9983 15 9982 16 9981 17 9980 18 9979 19 9978 20 9977 21 9976 22 9975 23 9974 24 9973 25 9972 26 9971 27 9970 28 9969 29 9968 30 9967 31 9966 32 9965 33 9964 34 9963 35 9962 36 9961 37 9960 38 9959 39 9958 40 9957 41 9956 42 9955 43 9954 44 9953 45 9952 46 9951 47 9950 48 9949 49 9948 50 9947 51 9946 52 9945 53 9944 54 9943 55 9942 56 9941 57 9940 58 9939 59 9938 60 9937 61 9936 62 9935 63 9934 64 9933 65 9932 ..."
},
{
"input": "9997",
"output": "1 9997 2 9996 3 9995 4 9994 5 9993 6 9992 7 9991 8 9990 9 9989 10 9988 11 9987 12 9986 13 9985 14 9984 15 9983 16 9982 17 9981 18 9980 19 9979 20 9978 21 9977 22 9976 23 9975 24 9974 25 9973 26 9972 27 9971 28 9970 29 9969 30 9968 31 9967 32 9966 33 9965 34 9964 35 9963 36 9962 37 9961 38 9960 39 9959 40 9958 41 9957 42 9956 43 9955 44 9954 45 9953 46 9952 47 9951 48 9950 49 9949 50 9948 51 9947 52 9946 53 9945 54 9944 55 9943 56 9942 57 9941 58 9940 59 9939 60 9938 61 9937 62 9936 63 9935 64 9934 65 9933 ..."
},
{
"input": "9998",
"output": "1 9998 2 9997 3 9996 4 9995 5 9994 6 9993 7 9992 8 9991 9 9990 10 9989 11 9988 12 9987 13 9986 14 9985 15 9984 16 9983 17 9982 18 9981 19 9980 20 9979 21 9978 22 9977 23 9976 24 9975 25 9974 26 9973 27 9972 28 9971 29 9970 30 9969 31 9968 32 9967 33 9966 34 9965 35 9964 36 9963 37 9962 38 9961 39 9960 40 9959 41 9958 42 9957 43 9956 44 9955 45 9954 46 9953 47 9952 48 9951 49 9950 50 9949 51 9948 52 9947 53 9946 54 9945 55 9944 56 9943 57 9942 58 9941 59 9940 60 9939 61 9938 62 9937 63 9936 64 9935 65 9934 ..."
},
{
"input": "9999",
"output": "1 9999 2 9998 3 9997 4 9996 5 9995 6 9994 7 9993 8 9992 9 9991 10 9990 11 9989 12 9988 13 9987 14 9986 15 9985 16 9984 17 9983 18 9982 19 9981 20 9980 21 9979 22 9978 23 9977 24 9976 25 9975 26 9974 27 9973 28 9972 29 9971 30 9970 31 9969 32 9968 33 9967 34 9966 35 9965 36 9964 37 9963 38 9962 39 9961 40 9960 41 9959 42 9958 43 9957 44 9956 45 9955 46 9954 47 9953 48 9952 49 9951 50 9950 51 9949 52 9948 53 9947 54 9946 55 9945 56 9944 57 9943 58 9942 59 9941 60 9940 61 9939 62 9938 63 9937 64 9936 65 9935 ..."
},
{
"input": "10000",
"output": "1 10000 2 9999 3 9998 4 9997 5 9996 6 9995 7 9994 8 9993 9 9992 10 9991 11 9990 12 9989 13 9988 14 9987 15 9986 16 9985 17 9984 18 9983 19 9982 20 9981 21 9980 22 9979 23 9978 24 9977 25 9976 26 9975 27 9974 28 9973 29 9972 30 9971 31 9970 32 9969 33 9968 34 9967 35 9966 36 9965 37 9964 38 9963 39 9962 40 9961 41 9960 42 9959 43 9958 44 9957 45 9956 46 9955 47 9954 48 9953 49 9952 50 9951 51 9950 52 9949 53 9948 54 9947 55 9946 56 9945 57 9944 58 9943 59 9942 60 9941 61 9940 62 9939 63 9938 64 9937 65 9936..."
}
] | 1,594,450,297 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 186 | 6,656,000 | n=int(input())
odd=[]
even=[]
for i in range(1,n+1):
if i%2:
odd.append(i)
else:
even.append(i)
print(*(odd+even)) | Title: Little Frog
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once upon a time a little frog whose name was Vasya decided to travel around his home swamp. Overall there are *n* mounds on the swamp, located on one line. The distance between the neighboring mounds is one meter. Vasya wants to visit all the mounds in one day; besides, he wants to visit each one exactly once. For that he makes a route plan, to decide the order in which to jump on the mounds. Vasya can pick any mound as the first one. He thinks it boring to jump two times at the same distance. That's why he wants any two jumps on his route to have different lengths. Help Vasya the Frog and make the plan for him.
Input Specification:
The single line contains a number *n* (1<=≤<=*n*<=≤<=104) which is the number of mounds.
Output Specification:
Print *n* integers *p**i* (1<=≤<=*p**i*<=≤<=*n*) which are the frog's route plan.
- All the *p**i*'s should be mutually different. - All the |*p**i*–*p**i*<=+<=1|'s should be mutually different (1<=≤<=*i*<=≤<=*n*<=-<=1).
If there are several solutions, output any.
Demo Input:
['2\n', '3\n']
Demo Output:
['1 2 ', '1 3 2 ']
Note:
none | ```python
n=int(input())
odd=[]
even=[]
for i in range(1,n+1):
if i%2:
odd.append(i)
else:
even.append(i)
print(*(odd+even))
``` | 0 |
672 | B | Different is Good | PROGRAMMING | 1,000 | [
"constructive algorithms",
"implementation",
"strings"
] | null | null | A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.
Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".
If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.
Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*.
The second line contains the string *s* of length *n* consisting of only lowercase English letters. | If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes. | [
"2\naa\n",
"4\nkoko\n",
"5\nmurat\n"
] | [
"1\n",
"2\n",
"0\n"
] | In the first sample one of the possible solutions is to change the first character to 'b'.
In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko". | 1,000 | [
{
"input": "2\naa",
"output": "1"
},
{
"input": "4\nkoko",
"output": "2"
},
{
"input": "5\nmurat",
"output": "0"
},
{
"input": "6\nacbead",
"output": "1"
},
{
"input": "7\ncdaadad",
"output": "4"
},
{
"input": "25\npeoaicnbisdocqofsqdpgobpn",
"output": "12"
},
{
"input": "25\ntcqpchnqskqjacruoaqilgebu",
"output": "7"
},
{
"input": "13\naebaecedabbee",
"output": "8"
},
{
"input": "27\naaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "10\nbababbdaee",
"output": "6"
},
{
"input": "11\ndbadcdbdbca",
"output": "7"
},
{
"input": "12\nacceaabddaaa",
"output": "7"
},
{
"input": "13\nabddfbfaeecfa",
"output": "7"
},
{
"input": "14\neeceecacdbcbbb",
"output": "9"
},
{
"input": "15\ndcbceaaggabaheb",
"output": "8"
},
{
"input": "16\nhgiegfbadgcicbhd",
"output": "7"
},
{
"input": "17\nabhfibbdddfghgfdi",
"output": "10"
},
{
"input": "26\nbbbbbabbaababaaabaaababbaa",
"output": "24"
},
{
"input": "26\nahnxdnbfbcrirerssyzydihuee",
"output": "11"
},
{
"input": "26\nhwqeqhkpxwulbsiwmnlfyhgknc",
"output": "8"
},
{
"input": "26\nrvxmulriorilidecqwmfaemifj",
"output": "10"
},
{
"input": "26\naowpmreooavnmamogdoopuisge",
"output": "12"
},
{
"input": "26\ninimevtuefhvuefirdehmmfudh",
"output": "15"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "25"
},
{
"input": "27\nqdcfjtblgglnilgassirrjekcjt",
"output": "-1"
},
{
"input": "27\nabcdefghijklmnopqrstuvwxyza",
"output": "-1"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "0"
},
{
"input": "5\nzzzzz",
"output": "4"
},
{
"input": "27\naaaaaaaaaaaaaaaaabaaaaaaaaa",
"output": "-1"
},
{
"input": "1\nq",
"output": "0"
},
{
"input": "27\nqwertyuioplkjhgfdsazxcvbnmm",
"output": "-1"
},
{
"input": "9\nxxxyyyzzz",
"output": "6"
},
{
"input": "45\naaabbbcccdddeeefffgghhiijjkkkkkkkkkkkkkkkkkkk",
"output": "-1"
},
{
"input": "27\nqwertyuiopasdfghjklzxcvbnmm",
"output": "-1"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxyz",
"output": "0"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxya",
"output": "1"
},
{
"input": "27\nabcdefghijklmnopqrstuvwxyzz",
"output": "-1"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "25"
},
{
"input": "26\nqwertyuioplkjhgfdsazxcvbnm",
"output": "0"
},
{
"input": "10\nzzzzzzzzzz",
"output": "9"
},
{
"input": "1\na",
"output": "0"
},
{
"input": "30\nabcdefghtyabcdefghtyabcdefghty",
"output": "-1"
}
] | 1,587,543,122 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 108 | 0 | input()
s = input()
a = len(set(s))
if a == 0:
a = len(s)
print(a)
else:
print(a) | Title: Different is Good
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.
Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".
If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.
Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*.
The second line contains the string *s* of length *n* consisting of only lowercase English letters.
Output Specification:
If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.
Demo Input:
['2\naa\n', '4\nkoko\n', '5\nmurat\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
In the first sample one of the possible solutions is to change the first character to 'b'.
In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko". | ```python
input()
s = input()
a = len(set(s))
if a == 0:
a = len(s)
print(a)
else:
print(a)
``` | 0 | |
1,005 | B | Delete from the Left | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | You are given two strings $s$ and $t$. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by $1$. You can't choose a string if it is empty.
For example:
- by applying a move to the string "where", the result is the string "here", - by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings $s$ and $t$ equal. | The first line of the input contains $s$. In the second line of the input contains $t$. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and $2\cdot10^5$, inclusive. | Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings. | [
"test\nwest\n",
"codeforces\nyes\n",
"test\nyes\n",
"b\nab\n"
] | [
"2\n",
"9\n",
"7\n",
"1\n"
] | In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" $8$ times. As a result, the string becomes "codeforces" $\to$ "es". The move should be applied to the string "yes" once. The result is the same string "yes" $\to$ "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted. | 0 | [
{
"input": "test\nwest",
"output": "2"
},
{
"input": "codeforces\nyes",
"output": "9"
},
{
"input": "test\nyes",
"output": "7"
},
{
"input": "b\nab",
"output": "1"
},
{
"input": "z\nz",
"output": "0"
},
{
"input": "abacabadabacaba\nabacabadacaba",
"output": "18"
},
{
"input": "z\na",
"output": "2"
},
{
"input": "ama\nsama",
"output": "1"
},
{
"input": "aaabc\nbc",
"output": "3"
},
{
"input": "lxuyspmieeuyxluv\ndfwxqvavzxctblaa",
"output": "32"
},
{
"input": "ahmedaly\naly",
"output": "5"
},
{
"input": "aa\na",
"output": "1"
},
{
"input": "aa\nba",
"output": "2"
},
{
"input": "cccba\nba",
"output": "3"
},
{
"input": "a\nab",
"output": "3"
},
{
"input": "dadc\ndddc",
"output": "4"
},
{
"input": "aacaaab\nb",
"output": "6"
},
{
"input": "dddc\ncdcde",
"output": "9"
},
{
"input": "bbdab\ndab",
"output": "2"
},
{
"input": "a\naa",
"output": "1"
},
{
"input": "aa\nab",
"output": "4"
},
{
"input": "codeforces\nces",
"output": "7"
}
] | 1,641,827,957 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 93 | 1,126,400 | def max_string(s,t):
if len(s)>len(t):
return(s)
else:
return (t)
def min_string(s,t):
if len(s)<len(t):
return(s)
else:
return (t)
s=input()
t=input()
x=''
y=''
if len(s)!=len(t):
x=max_string(s,t)
y=min_string(s,t)
else:
x=s
y=t
l=len(x)
ll=len(x)-1
ss=len(y)
z=ss
c=0
d=0
if len(y)==1:
d=-1
for i in range(len(y)-1,d,-1):
if x[ll]==y[i]:
c=c+1
else:
break
ll-=1
print(ll+(len(y)-c)+1) | Title: Delete from the Left
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two strings $s$ and $t$. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by $1$. You can't choose a string if it is empty.
For example:
- by applying a move to the string "where", the result is the string "here", - by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings $s$ and $t$ equal.
Input Specification:
The first line of the input contains $s$. In the second line of the input contains $t$. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and $2\cdot10^5$, inclusive.
Output Specification:
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Demo Input:
['test\nwest\n', 'codeforces\nyes\n', 'test\nyes\n', 'b\nab\n']
Demo Output:
['2\n', '9\n', '7\n', '1\n']
Note:
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" $8$ times. As a result, the string becomes "codeforces" $\to$ "es". The move should be applied to the string "yes" once. The result is the same string "yes" $\to$ "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted. | ```python
def max_string(s,t):
if len(s)>len(t):
return(s)
else:
return (t)
def min_string(s,t):
if len(s)<len(t):
return(s)
else:
return (t)
s=input()
t=input()
x=''
y=''
if len(s)!=len(t):
x=max_string(s,t)
y=min_string(s,t)
else:
x=s
y=t
l=len(x)
ll=len(x)-1
ss=len(y)
z=ss
c=0
d=0
if len(y)==1:
d=-1
for i in range(len(y)-1,d,-1):
if x[ll]==y[i]:
c=c+1
else:
break
ll-=1
print(ll+(len(y)-c)+1)
``` | 0 | |
112 | A | Petya and Strings | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Petya and Strings | 2 | 256 | Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison. | Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters. | If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared. | [
"aaaa\naaaA\n",
"abs\nAbz\n",
"abcdefg\nAbCdEfF\n"
] | [
"0\n",
"-1\n",
"1\n"
] | If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order | 500 | [
{
"input": "aaaa\naaaA",
"output": "0"
},
{
"input": "abs\nAbz",
"output": "-1"
},
{
"input": "abcdefg\nAbCdEfF",
"output": "1"
},
{
"input": "asadasdasd\nasdwasdawd",
"output": "-1"
},
{
"input": "aslkjlkasdd\nasdlkjdajwi",
"output": "1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "aAaaaAAaAaaAzZsssSsdDfeEaeqZlpP\nAaaaAaaAaaAaZzSSSSsDdFeeAeQZLpp",
"output": "0"
},
{
"input": "bwuEhEveouaTECagLZiqmUdxEmhRSOzMauJRWLQMppZOumxhAmwuGeDIkvkBLvMXwUoFmpAfDprBcFtEwOULcZWRQhcTbTbX\nHhoDWbcxwiMnCNexOsKsujLiSGcLllXOkRSbnOzThAjnnliLYFFmsYkOfpTxRNEfBsoUHfoLTiqAINRPxWRqrTJhgfkKcDOH",
"output": "-1"
},
{
"input": "kGWUuguKzcvxqKTNpxeDWXpXkrXDvGMFGoXKDfPBZvWSDUyIYBynbKOUonHvmZaKeirUhfmVRKtGhAdBfKMWXDUoqvbfpfHYcg\ncvOULleuIIiYVVxcLZmHVpNGXuEpzcWZZWyMOwIwbpkKPwCfkVbKkUuosvxYCKjqfVmHfJKbdrsAcatPYgrCABaFcoBuOmMfFt",
"output": "1"
},
{
"input": "nCeNVIzHqPceNhjHeHvJvgBsNFiXBATRrjSTXJzhLMDMxiJztphxBRlDlqwDFImWeEPkggZCXSRwelOdpNrYnTepiOqpvkr\nHJbjJFtlvNxIbkKlxQUwmZHJFVNMwPAPDRslIoXISBYHHfymyIaQHLgECPxAmqnOCizwXnIUBRmpYUBVPenoUKhCobKdOjL",
"output": "1"
},
{
"input": "ttXjenUAlfixytHEOrPkgXmkKTSGYuyVXGIHYmWWYGlBYpHkujueqBSgjLguSgiMGJWATIGEUjjAjKXdMiVbHozZUmqQtFrT\nJziDBFBDmDJCcGqFsQwDFBYdOidLxxhBCtScznnDgnsiStlWFnEXQrJxqTXKPxZyIGfLIToETKWZBPUIBmLeImrlSBWCkTNo",
"output": "1"
},
{
"input": "AjQhPqSVhwQQjcgCycjKorWBgFCRuQBwgdVuAPSMJAvTyxGVuFHjfJzkKfsmfhFbKqFrFIohSZBbpjgEHebezmVlGLTPSCTMf\nXhxWuSnMmKFrCUOwkTUmvKAfbTbHWzzOTzxJatLLCdlGnHVaBUnxDlsqpvjLHMThOPAFBggVKDyKBrZAmjnjrhHlrnSkyzBja",
"output": "-1"
},
{
"input": "HCIgYtnqcMyjVngziNflxKHtdTmcRJhzMAjFAsNdWXFJYEhiTzsQUtFNkAbdrFBRmvLirkuirqTDvIpEfyiIqkrwsjvpPWTEdI\nErqiiWKsmIjyZuzgTlTqxYZwlrpvRyaVhRTOYUqtPMVGGtWOkDCOOQRKrkkRzPftyQCkYkzKkzTPqqXmeZhvvEEiEhkdOmoMvy",
"output": "1"
},
{
"input": "mtBeJYILXcECGyEVSyzLFdQJbiVnnfkbsYYsdUJSIRmyzLfTTtFwIBmRLVnwcewIqcuydkcLpflHAFyDaToLiFMgeHvQorTVbI\nClLvyejznjbRfCDcrCzkLvqQaGzTjwmWONBdCctJAPJBcQrcYvHaSLQgPIJbmkFBhFzuQLBiRzAdNHulCjIAkBvZxxlkdzUWLR",
"output": "1"
},
{
"input": "tjucSbGESVmVridTBjTmpVBCwwdWKBPeBvmgdxgIVLwQxveETnSdxkTVJpXoperWSgdpPMKNmwDiGeHfxnuqaDissgXPlMuNZIr\nHfjOOJhomqNIKHvqSgfySjlsWJQBuWYwhLQhlZYlpZwboMpoLoluGsBmhhlYgeIouwdkPfiaAIrkYRlxtiFazOPOllPsNZHcIZd",
"output": "1"
},
{
"input": "AanbDfbZNlUodtBQlvPMyomStKNhgvSGhSbTdabxGFGGXCdpsJDimsAykKjfBDPMulkhBMsqLmVKLDoesHZsRAEEdEzqigueXInY\ncwfyjoppiJNrjrOLNZkqcGimrpTsiyFBVgMWEPXsMrxLJDDbtYzerXiFGuLBcQYitLdqhGHBpdjRnkUegmnwhGHAKXGyFtscWDSI",
"output": "-1"
},
{
"input": "HRfxniwuJCaHOcaOVgjOGHXKrwxrDQxJpppeGDXnTAowyKbCsCQPbchCKeTWOcKbySSYnoaTJDnmRcyGPbfXJyZoPcARHBu\nxkLXvwkvGIWSQaFTznLOctUXNuzzBBOlqvzmVfTSejekTAlwidRrsxkbZTsGGeEWxCXHzqWVuLGoCyrGjKkQoHqduXwYQKC",
"output": "-1"
},
{
"input": "OjYwwNuPESIazoyLFREpObIaMKhCaKAMWMfRGgucEuyNYRantwdwQkmflzfqbcFRaXBnZoIUGsFqXZHGKwlaBUXABBcQEWWPvkjW\nRxLqGcTTpBwHrHltCOllnTpRKLDofBUqqHxnOtVWPgvGaeHIevgUSOeeDOJubfqonFpVNGVbHFcAhjnyFvrrqnRgKhkYqQZmRfUl",
"output": "-1"
},
{
"input": "tatuhQPIzjptlzzJpCAPXSRTKZRlwgfoCIsFjJquRoIDyZZYRSPdFUTjjUPhLBBfeEIfLQpygKXRcyQFiQsEtRtLnZErBqW\ntkHUjllbafLUWhVCnvblKjgYIEoHhsjVmrDBmAWbvtkHxDbRFvsXAjHIrujaDbYwOZmacknhZPeCcorbRgHjjgAgoJdjvLo",
"output": "-1"
},
{
"input": "cymCPGqdXKUdADEWDdUaLEEMHiXHsdAZuDnJDMUvxvrLRBrPSDpXPAgMRoGplLtniFRTomDTAHXWAdgUveTxaqKVSvnOyhOwiRN\nuhmyEWzapiRNPFDisvHTbenXMfeZaHqOFlKjrfQjUBwdFktNpeiRoDWuBftZLcCZZAVfioOihZVNqiNCNDIsUdIhvbcaxpTRWoV",
"output": "-1"
},
{
"input": "sSvpcITJAwghVfJaLKBmyjOkhltTGjYJVLWCYMFUomiJaKQYhXTajvZVHIMHbyckYROGQZzjWyWCcnmDmrkvTKfHSSzCIhsXgEZa\nvhCXkCwAmErGVBPBAnkSYEYvseFKbWSktoqaHYXUmYkHfOkRwuEyBRoGoBrOXBKVxXycjZGStuvDarnXMbZLWrbjrisDoJBdSvWJ",
"output": "-1"
},
{
"input": "hJDANKUNBisOOINDsTixJmYgHNogtpwswwcvVMptfGwIjvqgwTYFcqTdyAqaqlnhOCMtsnWXQqtjFwQlEcBtMFAtSqnqthVb\nrNquIcjNWESjpPVWmzUJFrelpUZeGDmSvCurCqVmKHKVAAPkaHksniOlzjiKYIJtvbuQWZRufMebpTFPqyxIWWjfPaWYiNlK",
"output": "-1"
},
{
"input": "ycLoapxsfsDTHMSfAAPIUpiEhQKUIXUcXEiopMBuuZLHtfPpLmCHwNMNQUwsEXxCEmKHTBSnKhtQhGWUvppUFZUgSpbeChX\ndCZhgVXofkGousCzObxZSJwXcHIaqUDSCPKzXntcVmPxtNcXmVcjsetZYxedmgQzXTZHMvzjoaXCMKsncGciSDqQWIIRlys",
"output": "1"
},
{
"input": "nvUbnrywIePXcoukIhwTfUVcHUEgXcsMyNQhmMlTltZiCooyZiIKRIGVHMCnTKgzXXIuvoNDEZswKoACOBGSyVNqTNQqMhAG\nplxuGSsyyJjdvpddrSebOARSAYcZKEaKjqbCwvjhNykuaECoQVHTVFMKXwvrQXRaqXsHsBaGVhCxGRxNyGUbMlxOarMZNXxy",
"output": "-1"
},
{
"input": "EncmXtAblQzcVRzMQqdDqXfAhXbtJKQwZVWyHoWUckohnZqfoCmNJDzexFgFJYrwNHGgzCJTzQQFnxGlhmvQTpicTkEeVICKac\nNIUNZoMLFMyAjVgQLITELJSodIXcGSDWfhFypRoGYuogJpnqGTotWxVqpvBHjFOWcDRDtARsaHarHaOkeNWEHGTaGOFCOFEwvK",
"output": "-1"
},
{
"input": "UG\nak",
"output": "1"
},
{
"input": "JZR\nVae",
"output": "-1"
},
{
"input": "a\nZ",
"output": "-1"
},
{
"input": "rk\nkv",
"output": "1"
},
{
"input": "RvuT\nbJzE",
"output": "1"
},
{
"input": "PPS\nydq",
"output": "-1"
},
{
"input": "q\nq",
"output": "0"
},
{
"input": "peOw\nIgSJ",
"output": "1"
},
{
"input": "PyK\noKN",
"output": "1"
},
{
"input": "O\ni",
"output": "1"
},
{
"input": "NmGY\npDlP",
"output": "-1"
},
{
"input": "nG\nZf",
"output": "-1"
},
{
"input": "m\na",
"output": "1"
},
{
"input": "MWyB\nWZEV",
"output": "-1"
},
{
"input": "Gre\nfxc",
"output": "1"
},
{
"input": "Ooq\nwap",
"output": "-1"
},
{
"input": "XId\nlbB",
"output": "1"
},
{
"input": "lfFpECEqUMEOJhipvkZjDPcpDNJedOVXiSMgBvBZbtfzIKekcvpWPCazKAhJyHircRtgcBIJwwstpHaLAgxFOngAWUZRgCef\nLfFPEcequmeojHIpVkzjDPcpdNJEDOVXiSmGBVBZBtfZikEKcvPwpCAzKAHJyHIrCRTgCbIJWwSTphALagXfOnGAwUzRGcEF",
"output": "0"
},
{
"input": "DQBdtSEDtFGiNRUeJNbOIfDZnsryUlzJHGTXGFXnwsVyxNtLgmklmFvRCzYETBVdmkpJJIvIOkMDgCFHZOTODiYrkwXd\nDQbDtsEdTFginRUEJNBOIfdZnsryulZJHGtxGFxnwSvYxnTLgmKlmFVRCzyEtBVdmKpJjiVioKMDgCFhzoTODiYrKwXD",
"output": "0"
},
{
"input": "tYWRijFQSzHBpCjUzqBtNvBKyzZRnIdWEuyqnORBQTLyOQglIGfYJIRjuxnbLvkqZakNqPiGDvgpWYkfxYNXsdoKXZtRkSasfa\nTYwRiJfqsZHBPcJuZQBTnVbkyZZRnidwEuYQnorbQTLYOqGligFyjirJUxnblVKqZaknQpigDVGPwyKfxyNXSDoKxztRKSaSFA",
"output": "0"
},
{
"input": "KhScXYiErQIUtmVhNTCXSLAviefIeHIIdiGhsYnPkSBaDTvMkyanfMLBOvDWgRybLtDqvXVdVjccNunDyijhhZEAKBrdz\nkHsCXyiErqIuTMVHNTCxSLaViEFIEhIIDiGHsYNpKsBAdTvMKyANFMLBovdwGRYbLtdQVxvDVJCcNUndYiJHhzeakBrdZ",
"output": "0"
},
{
"input": "cpPQMpjRQJKQVXjWDYECXbagSmNcVfOuBWNZxihdERraVuiOpSVDCPgTGuSQALNoVjySceHcKXwOEpSzXrEqWwwrYeppNiWhDVg\nCPPqmPjRqJkQvxJwdyECXBAGsMNcVfOuBWNzxIhderRavUiOpSvDCpGTgusqAlNovjyScEhCKXwoePSZxrEQwWwryEPPniWHDvG",
"output": "0"
},
{
"input": "SajcCGMepaLjZIWLRBGFcrZRCRvvoCsIyKsQerbrwsIamxxpRmQSZSalasJLVFbCHCuXJlubciQAvLxXYBazLsMKLHLdDQ\nsaJcCgmEpaLJziWlrBgFcRzrCrVVOcSIykSQerBrwSIamxxPrMqSzSalASjLVFbChCUxjLUbCIQAVlxxybAZLsmkLhLDdQ",
"output": "0"
},
{
"input": "kigPrWNTOUNDBskAfefjhHYZNYdnfZWuXWzHiBxFQryBbAkPtenFwWvCSTYGpzOntUNzNUhxRWjKmicTwLwJAnbAxj\nkigpRWntOUNdBsKaFEFjhhYZnYDNfzWuXwZhibxFQRybbakPteNfwwvcStyGPzoNTunznuHXrWjKMIctWLWJANBAxJ",
"output": "0"
},
{
"input": "nTomZZuTTRTAAPoUsySVFGElrpQRNLjqvFmcYytiheQnjUhPLnqNBiYtQkljbcvmjuNAVKbvQOWpqqFlQhAhULIhquoCnjUI\nntOmzZuttrtAAPOUSySVFgeLRPQrNLjQvfmCyYTiHEQnjuHPlNQNbIYtqKLJBCVmjunavkbvQOWPQQFlqHaHULIHQuOcnJUi",
"output": "0"
},
{
"input": "abac\nadaa",
"output": "-1"
},
{
"input": "Bbc\nabc",
"output": "1"
},
{
"input": "aaaba\naaaab",
"output": "1"
}
] | 1,698,654,790 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 122 | 0 | a, b = str(input()), str(input())
a, b = sorted(a.lower()), sorted(b.lower())
for x in range(len(a)):
if a[x] != b[x]:
oA, oB = ord(a[x]), ord(b[x])
if oA > oB: print(1)
elif oB > oA: print(-1)
else: print(0) | Title: Petya and Strings
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input Specification:
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output Specification:
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Demo Input:
['aaaa\naaaA\n', 'abs\nAbz\n', 'abcdefg\nAbCdEfF\n']
Demo Output:
['0\n', '-1\n', '1\n']
Note:
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order | ```python
a, b = str(input()), str(input())
a, b = sorted(a.lower()), sorted(b.lower())
for x in range(len(a)):
if a[x] != b[x]:
oA, oB = ord(a[x]), ord(b[x])
if oA > oB: print(1)
elif oB > oA: print(-1)
else: print(0)
``` | 0 |
621 | A | Wet Shark and Odd and Even | PROGRAMMING | 900 | [
"implementation"
] | null | null | Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0. | The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive. | Print the maximum possible even sum that can be obtained if we use some of the given integers. | [
"3\n1 2 3\n",
"5\n999999999 999999999 999999999 999999999 999999999\n"
] | [
"6",
"3999999996"
] | In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999. | 500 | [
{
"input": "3\n1 2 3",
"output": "6"
},
{
"input": "5\n999999999 999999999 999999999 999999999 999999999",
"output": "3999999996"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "15\n39 52 88 78 46 95 84 98 55 3 68 42 6 18 98",
"output": "870"
},
{
"input": "15\n59 96 34 48 8 72 67 90 15 85 7 90 97 47 25",
"output": "840"
},
{
"input": "15\n87 37 91 29 58 45 51 74 70 71 47 38 91 89 44",
"output": "922"
},
{
"input": "15\n11 81 49 7 11 14 30 67 29 50 90 81 77 18 59",
"output": "674"
},
{
"input": "15\n39 21 95 89 73 90 9 55 85 32 30 21 68 59 82",
"output": "848"
},
{
"input": "15\n59 70 48 54 26 67 84 39 40 18 77 69 70 88 93",
"output": "902"
},
{
"input": "15\n87 22 98 32 88 36 72 31 100 97 17 16 60 22 20",
"output": "798"
},
{
"input": "15\n15 63 51 13 37 9 43 19 55 79 57 60 50 59 31",
"output": "632"
},
{
"input": "1\n4",
"output": "4"
},
{
"input": "2\n1 4",
"output": "4"
},
{
"input": "3\n1 2 4",
"output": "6"
},
{
"input": "2\n9 3",
"output": "12"
},
{
"input": "2\n1000000000 1001",
"output": "1000000000"
},
{
"input": "3\n1 8 4",
"output": "12"
},
{
"input": "3\n7 4 4",
"output": "8"
},
{
"input": "5\n2 3 4 5 3",
"output": "14"
},
{
"input": "2\n4 5",
"output": "4"
},
{
"input": "3\n2 4 5",
"output": "6"
},
{
"input": "3\n2 2 3",
"output": "4"
},
{
"input": "2\n2 3",
"output": "2"
},
{
"input": "4\n2 3 7 7",
"output": "16"
},
{
"input": "2\n999999999 2",
"output": "2"
},
{
"input": "2\n2 5",
"output": "2"
},
{
"input": "3\n5 3 1",
"output": "8"
},
{
"input": "4\n3 2 5 7",
"output": "14"
}
] | 1,667,932,217 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 14 | 93 | 7,884,800 | n = int(input())
listN = list(map(int, input().split()))
listN.sort()
listDiv2 = [i%2 for i in listN]
if(listDiv2.count(1)%2 == 0): print(sum(listN))
else: print(sum(listN) - listN[0])
| Title: Wet Shark and Odd and Even
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.
Input Specification:
The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
Output Specification:
Print the maximum possible even sum that can be obtained if we use some of the given integers.
Demo Input:
['3\n1 2 3\n', '5\n999999999 999999999 999999999 999999999 999999999\n']
Demo Output:
['6', '3999999996']
Note:
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999. | ```python
n = int(input())
listN = list(map(int, input().split()))
listN.sort()
listDiv2 = [i%2 for i in listN]
if(listDiv2.count(1)%2 == 0): print(sum(listN))
else: print(sum(listN) - listN[0])
``` | 0 | |
171 | B | Star | PROGRAMMING | 1,300 | [
"*special",
"combinatorics"
] | null | null | The input contains a single integer *a* (1<=≤<=*a*<=≤<=18257). | Print a single integer *output* (1<=≤<=*output*<=≤<=2·109). | [
"2\n"
] | [
"13"
] | none | 0 | [
{
"input": "2",
"output": "13"
},
{
"input": "1",
"output": "1"
},
{
"input": "3",
"output": "37"
},
{
"input": "4",
"output": "73"
},
{
"input": "5",
"output": "121"
},
{
"input": "6",
"output": "181"
},
{
"input": "7",
"output": "253"
},
{
"input": "8",
"output": "337"
},
{
"input": "9",
"output": "433"
},
{
"input": "15000",
"output": "1349910001"
},
{
"input": "4845",
"output": "140815081"
},
{
"input": "6914",
"output": "286778893"
},
{
"input": "3994",
"output": "95688253"
},
{
"input": "12504",
"output": "938025073"
},
{
"input": "13170",
"output": "1040614381"
},
{
"input": "427",
"output": "1091413"
},
{
"input": "11877",
"output": "846307513"
},
{
"input": "3202",
"output": "61497613"
},
{
"input": "5689",
"output": "194154193"
},
{
"input": "15302",
"output": "1404815413"
},
{
"input": "17042",
"output": "1742476333"
},
{
"input": "1481",
"output": "13151281"
},
{
"input": "15592",
"output": "1458569233"
},
{
"input": "16344",
"output": "1602659953"
},
{
"input": "4222",
"output": "106926373"
},
{
"input": "11808",
"output": "836502337"
},
{
"input": "13366",
"output": "1071819541"
},
{
"input": "3823",
"output": "87669037"
},
{
"input": "581",
"output": "2021881"
},
{
"input": "15479",
"output": "1437503773"
},
{
"input": "6543",
"output": "256825837"
},
{
"input": "11136",
"output": "743996161"
},
{
"input": "16331",
"output": "1600111381"
},
{
"input": "8543",
"output": "437845837"
},
{
"input": "7530",
"output": "340160221"
},
{
"input": "3154",
"output": "59667373"
},
{
"input": "11501",
"output": "793569001"
},
{
"input": "12038",
"output": "869408437"
},
{
"input": "13082",
"output": "1026753853"
},
{
"input": "18257",
"output": "1999798753"
}
] | 1,694,847,487 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 62 | 0 | # LUOGU_RID: 124877803
a=int(input())
output=(2*n-1)*(2*n-1)+4*(n-1)*n/2
print(output) | Title: Star
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Input Specification:
The input contains a single integer *a* (1<=≤<=*a*<=≤<=18257).
Output Specification:
Print a single integer *output* (1<=≤<=*output*<=≤<=2·109).
Demo Input:
['2\n']
Demo Output:
['13']
Note:
none | ```python
# LUOGU_RID: 124877803
a=int(input())
output=(2*n-1)*(2*n-1)+4*(n-1)*n/2
print(output)
``` | -1 | ||
124 | A | The number of positions | PROGRAMMING | 1,000 | [
"math"
] | null | null | Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy. | The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100). | Print the single number — the number of the sought positions. | [
"3 1 1\n",
"5 2 3\n"
] | [
"2\n",
"3\n"
] | The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5. | 500 | [
{
"input": "3 1 1",
"output": "2"
},
{
"input": "5 2 3",
"output": "3"
},
{
"input": "5 4 0",
"output": "1"
},
{
"input": "6 5 5",
"output": "1"
},
{
"input": "9 4 3",
"output": "4"
},
{
"input": "11 4 6",
"output": "7"
},
{
"input": "13 8 7",
"output": "5"
},
{
"input": "14 5 5",
"output": "6"
},
{
"input": "16 6 9",
"output": "10"
},
{
"input": "20 13 17",
"output": "7"
},
{
"input": "22 4 8",
"output": "9"
},
{
"input": "23 8 14",
"output": "15"
},
{
"input": "26 18 22",
"output": "8"
},
{
"input": "28 6 1",
"output": "2"
},
{
"input": "29 5 23",
"output": "24"
},
{
"input": "32 27 15",
"output": "5"
},
{
"input": "33 11 5",
"output": "6"
},
{
"input": "37 21 15",
"output": "16"
},
{
"input": "39 34 33",
"output": "5"
},
{
"input": "41 27 11",
"output": "12"
},
{
"input": "42 25 16",
"output": "17"
},
{
"input": "45 7 43",
"output": "38"
},
{
"input": "47 16 17",
"output": "18"
},
{
"input": "49 11 37",
"output": "38"
},
{
"input": "51 38 39",
"output": "13"
},
{
"input": "52 29 7",
"output": "8"
},
{
"input": "56 43 12",
"output": "13"
},
{
"input": "58 57 28",
"output": "1"
},
{
"input": "59 12 39",
"output": "40"
},
{
"input": "62 9 52",
"output": "53"
},
{
"input": "63 29 44",
"output": "34"
},
{
"input": "65 30 22",
"output": "23"
},
{
"input": "66 27 38",
"output": "39"
},
{
"input": "71 33 53",
"output": "38"
},
{
"input": "73 14 12",
"output": "13"
},
{
"input": "73 37 35",
"output": "36"
},
{
"input": "76 69 44",
"output": "7"
},
{
"input": "79 25 20",
"output": "21"
},
{
"input": "81 60 20",
"output": "21"
},
{
"input": "81 79 14",
"output": "2"
},
{
"input": "84 0 42",
"output": "43"
},
{
"input": "88 79 8",
"output": "9"
},
{
"input": "90 76 59",
"output": "14"
},
{
"input": "92 2 22",
"output": "23"
},
{
"input": "94 5 88",
"output": "89"
},
{
"input": "94 62 48",
"output": "32"
},
{
"input": "96 22 72",
"output": "73"
},
{
"input": "100 11 88",
"output": "89"
},
{
"input": "100 81 91",
"output": "19"
},
{
"input": "1 0 0",
"output": "1"
}
] | 1,661,310,028 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 154 | 0 | # Online Python compiler (interpreter) to run Python online.
# Write Python 3 code in this online editor and run it.
a, b, c = [int(x) for x in input().split()]
print(a-b) | Title: The number of positions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
Input Specification:
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
Output Specification:
Print the single number — the number of the sought positions.
Demo Input:
['3 1 1\n', '5 2 3\n']
Demo Output:
['2\n', '3\n']
Note:
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5. | ```python
# Online Python compiler (interpreter) to run Python online.
# Write Python 3 code in this online editor and run it.
a, b, c = [int(x) for x in input().split()]
print(a-b)
``` | 0 | |
76 | B | Mice | PROGRAMMING | 2,100 | [
"greedy",
"two pointers"
] | B. Mice | 0 | 256 | Modern researches has shown that a flock of hungry mice searching for a piece of cheese acts as follows: if there are several pieces of cheese then each mouse chooses the closest one. After that all mice start moving towards the chosen piece of cheese. When a mouse or several mice achieve the destination point and there is still a piece of cheese in it, they eat it and become well-fed. Each mice that reaches this point after that remains hungry. Moving speeds of all mice are equal.
If there are several ways to choose closest pieces then mice will choose it in a way that would minimize the number of hungry mice. To check this theory scientists decided to conduct an experiment. They located *N* mice and *M* pieces of cheese on a cartesian plane where all mice are located on the line *y*<==<=*Y*0 and all pieces of cheese — on another line *y*<==<=*Y*1. To check the results of the experiment the scientists need a program which simulates the behavior of a flock of hungry mice.
Write a program that computes the minimal number of mice which will remain hungry, i.e. without cheese. | The first line of the input contains four integer numbers *N* (1<=≤<=*N*<=≤<=105), *M* (0<=≤<=*M*<=≤<=105), *Y*0 (0<=≤<=*Y*0<=≤<=107), *Y*1 (0<=≤<=*Y*1<=≤<=107, *Y*0<=≠<=*Y*1). The second line contains a strictly increasing sequence of *N* numbers — *x* coordinates of mice. Third line contains a strictly increasing sequence of *M* numbers — *x* coordinates of cheese. All coordinates are integers and do not exceed 107 by absolute value. | The only line of output should contain one number — the minimal number of mice which will remain without cheese. | [
"3 2 0 2\n0 1 3\n2 5\n"
] | [
"1\n"
] | All the three mice will choose the first piece of cheese. Second and third mice will eat this piece. The first one will remain hungry, because it was running towards the same piece, but it was late. The second piece of cheese will remain uneaten. | 0 | [
{
"input": "3 2 0 2\n0 1 3\n2 5",
"output": "1"
},
{
"input": "7 11 10 20\n6 18 32 63 66 68 87\n6 8 15 23 25 41 53 59 60 75 90",
"output": "1"
},
{
"input": "13 17 14 1\n6 9 10 12 17 25 91 100 118 136 145 163 172\n0 1 2 3 4 10 12 13 16 17 19 22 26 27 28 109 154",
"output": "4"
},
{
"input": "19 23 13 11\n3 6 7 15 21 22 23 33 35 37 40 44 79 86 100 114 121 135 142\n2 3 5 6 7 14 15 17 18 19 20 22 25 27 28 34 36 38 39 41 42 93 128",
"output": "4"
},
{
"input": "20 18 1 2\n-9999944 -9999861 -9999850 -9999763 -9999656 -9999517 -9999375 -9999275 -9999203 -9999080 -9998988 -9998887 -9998714 -9998534 -9998475 -9998352 -9998164 -9998016 -9998002 -9997882\n-9999976 -9999912 -9999788 -9999738 -9999574 -9999460 -9999290 -9999260 -9999146 -9999014 -9998962 -9998812 -9998616 -9998452 -9998252 -9998076 -9997928 -9997836",
"output": "2"
}
] | 1,669,198,216 | 2,147,483,647 | PyPy 3 | OK | TESTS | 20 | 311 | 14,745,600 | def distance(x, y):
return abs(x - y)
n, m, y0, y1 = map(int, input().split())
mouse = list(map(int, input().split()))
cheeses = list(map(int, input().split()))
i = 0
j = 0
res = 0
mouse.append(float("inf"))
cheeses.append(float("inf"))
prevJ = -1
for i in range(n):
while distance(mouse[i], cheeses[j]) > distance(mouse[i], cheeses[j + 1]):
j += 1
if prevJ < j or (0 < i and distance(mouse[i - 1], cheeses[j]) == distance(mouse[i], cheeses[j])):
res += 1
elif j + 1 < m and distance(mouse[i], cheeses[j + 1]) == distance(mouse[i], cheeses[j]):
res += 1
j += 1
prevJ = j
print(n - res)
| Title: Mice
Time Limit: 0 seconds
Memory Limit: 256 megabytes
Problem Description:
Modern researches has shown that a flock of hungry mice searching for a piece of cheese acts as follows: if there are several pieces of cheese then each mouse chooses the closest one. After that all mice start moving towards the chosen piece of cheese. When a mouse or several mice achieve the destination point and there is still a piece of cheese in it, they eat it and become well-fed. Each mice that reaches this point after that remains hungry. Moving speeds of all mice are equal.
If there are several ways to choose closest pieces then mice will choose it in a way that would minimize the number of hungry mice. To check this theory scientists decided to conduct an experiment. They located *N* mice and *M* pieces of cheese on a cartesian plane where all mice are located on the line *y*<==<=*Y*0 and all pieces of cheese — on another line *y*<==<=*Y*1. To check the results of the experiment the scientists need a program which simulates the behavior of a flock of hungry mice.
Write a program that computes the minimal number of mice which will remain hungry, i.e. without cheese.
Input Specification:
The first line of the input contains four integer numbers *N* (1<=≤<=*N*<=≤<=105), *M* (0<=≤<=*M*<=≤<=105), *Y*0 (0<=≤<=*Y*0<=≤<=107), *Y*1 (0<=≤<=*Y*1<=≤<=107, *Y*0<=≠<=*Y*1). The second line contains a strictly increasing sequence of *N* numbers — *x* coordinates of mice. Third line contains a strictly increasing sequence of *M* numbers — *x* coordinates of cheese. All coordinates are integers and do not exceed 107 by absolute value.
Output Specification:
The only line of output should contain one number — the minimal number of mice which will remain without cheese.
Demo Input:
['3 2 0 2\n0 1 3\n2 5\n']
Demo Output:
['1\n']
Note:
All the three mice will choose the first piece of cheese. Second and third mice will eat this piece. The first one will remain hungry, because it was running towards the same piece, but it was late. The second piece of cheese will remain uneaten. | ```python
def distance(x, y):
return abs(x - y)
n, m, y0, y1 = map(int, input().split())
mouse = list(map(int, input().split()))
cheeses = list(map(int, input().split()))
i = 0
j = 0
res = 0
mouse.append(float("inf"))
cheeses.append(float("inf"))
prevJ = -1
for i in range(n):
while distance(mouse[i], cheeses[j]) > distance(mouse[i], cheeses[j + 1]):
j += 1
if prevJ < j or (0 < i and distance(mouse[i - 1], cheeses[j]) == distance(mouse[i], cheeses[j])):
res += 1
elif j + 1 < m and distance(mouse[i], cheeses[j + 1]) == distance(mouse[i], cheeses[j]):
res += 1
j += 1
prevJ = j
print(n - res)
``` | 3 |
272 | A | Dima and Friends | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space. | In a single line print the answer to the problem. | [
"1\n1\n",
"1\n2\n",
"2\n3 5\n"
] | [
"3\n",
"2\n",
"3\n"
] | In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers. | 500 | [
{
"input": "1\n1",
"output": "3"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "1\n5",
"output": "3"
},
{
"input": "5\n4 4 3 5 1",
"output": "4"
},
{
"input": "6\n2 3 2 2 1 3",
"output": "4"
},
{
"input": "8\n2 2 5 3 4 3 3 2",
"output": "4"
},
{
"input": "7\n4 1 3 2 2 4 5",
"output": "4"
},
{
"input": "3\n3 5 1",
"output": "4"
},
{
"input": "95\n4 2 3 4 4 5 2 2 4 4 3 5 3 3 3 5 4 2 5 4 2 1 1 3 4 2 1 3 5 4 2 1 1 5 1 1 2 2 4 4 5 4 5 5 2 1 2 2 2 4 5 5 2 4 3 4 4 3 5 2 4 1 5 4 5 1 3 2 4 2 2 1 5 3 1 5 3 4 3 3 2 1 2 2 1 3 1 5 2 3 1 1 2 5 2",
"output": "5"
},
{
"input": "31\n3 2 3 3 3 3 4 4 1 5 5 4 2 4 3 2 2 1 4 4 1 2 3 1 1 5 5 3 4 4 1",
"output": "4"
},
{
"input": "42\n3 1 2 2 5 1 2 2 4 5 4 5 2 5 4 5 4 4 1 4 3 3 4 4 4 4 3 2 1 3 4 5 5 2 1 2 1 5 5 2 4 4",
"output": "5"
},
{
"input": "25\n4 5 5 5 3 1 1 4 4 4 3 5 4 4 1 4 4 1 2 4 2 5 4 5 3",
"output": "5"
},
{
"input": "73\n3 4 3 4 5 1 3 4 2 1 4 2 2 3 5 3 1 4 2 3 2 1 4 5 3 5 2 2 4 3 2 2 5 3 2 3 5 1 3 1 1 4 5 2 4 2 5 1 4 3 1 3 1 4 2 3 3 3 3 5 5 2 5 2 5 4 3 1 1 5 5 2 3",
"output": "4"
},
{
"input": "46\n1 4 4 5 4 5 2 3 5 5 3 2 5 4 1 3 2 2 1 4 3 1 5 5 2 2 2 2 4 4 1 1 4 3 4 3 1 4 2 2 4 2 3 2 5 2",
"output": "4"
},
{
"input": "23\n5 2 1 1 4 2 5 5 3 5 4 5 5 1 1 5 2 4 5 3 4 4 3",
"output": "5"
},
{
"input": "6\n4 2 3 1 3 5",
"output": "4"
},
{
"input": "15\n5 5 5 3 5 4 1 3 3 4 3 4 1 4 4",
"output": "5"
},
{
"input": "93\n1 3 1 4 3 3 5 3 1 4 5 4 3 2 2 4 3 1 4 1 2 3 3 3 2 5 1 3 1 4 5 1 1 1 4 2 1 2 3 1 1 1 5 1 5 5 1 2 5 4 3 2 2 4 4 2 5 4 5 5 3 1 3 1 2 1 3 1 1 2 3 4 4 5 5 3 2 1 3 3 5 1 3 5 4 4 1 3 3 4 2 3 2",
"output": "5"
},
{
"input": "96\n1 5 1 3 2 1 2 2 2 2 3 4 1 1 5 4 4 1 2 3 5 1 4 4 4 1 3 3 1 4 5 4 1 3 5 3 4 4 3 2 1 1 4 4 5 1 1 2 5 1 2 3 1 4 1 2 2 2 3 2 3 3 2 5 2 2 3 3 3 3 2 1 2 4 5 5 1 5 3 2 1 4 3 5 5 5 3 3 5 3 4 3 4 2 1 3",
"output": "5"
},
{
"input": "49\n1 4 4 3 5 2 2 1 5 1 2 1 2 5 1 4 1 4 5 2 4 5 3 5 2 4 2 1 3 4 2 1 4 2 1 1 3 3 2 3 5 4 3 4 2 4 1 4 1",
"output": "5"
},
{
"input": "73\n4 1 3 3 3 1 5 2 1 4 1 1 3 5 1 1 4 5 2 1 5 4 1 5 3 1 5 2 4 5 1 4 3 3 5 2 2 3 3 2 5 1 4 5 2 3 1 4 4 3 5 2 3 5 1 4 3 5 1 2 4 1 3 3 5 4 2 4 2 4 1 2 5",
"output": "5"
},
{
"input": "41\n5 3 5 4 2 5 4 3 1 1 1 5 4 3 4 3 5 4 2 5 4 1 1 3 2 4 5 3 5 1 5 5 1 1 1 4 4 1 2 4 3",
"output": "5"
},
{
"input": "100\n3 3 1 4 2 4 4 3 1 5 1 1 4 4 3 4 4 3 5 4 5 2 4 3 4 1 2 4 5 4 2 1 5 4 1 1 4 3 2 4 1 2 1 4 4 5 5 4 4 5 3 2 5 1 4 2 2 1 1 2 5 2 5 1 5 3 1 4 3 2 4 3 2 2 4 5 5 1 2 3 1 4 1 2 2 2 5 5 2 3 2 4 3 1 1 2 1 2 1 2",
"output": "5"
},
{
"input": "100\n2 1 1 3 5 4 4 2 3 4 3 4 5 4 5 4 2 4 5 3 4 5 4 1 1 4 4 1 1 2 5 4 2 4 5 3 2 5 4 3 4 5 1 3 4 2 5 4 5 4 5 2 4 1 2 5 3 1 4 4 5 3 4 3 1 2 5 4 2 5 4 1 5 3 5 4 1 2 5 3 1 1 1 1 5 3 4 3 5 1 1 5 5 1 1 2 2 1 5 1",
"output": "5"
},
{
"input": "100\n4 4 3 3 2 5 4 4 2 1 4 4 4 5 4 1 2 1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3 4 5 5 2 1 2 2 2 2 2 3 2 5 1 2 2 3 2 5 5 1 3 4 5 2 1 3 4 2 2 4 4 3 3 3 2 3 2 1 5 5 5 2 1 4 2 3 5 1 4 4 2 3 2 5 5 4 3 5 1 3 5 5",
"output": "5"
},
{
"input": "100\n4 4 2 5 4 2 2 3 4 4 3 2 3 3 1 3 4 3 3 4 1 3 1 4 5 3 4 3 1 1 1 3 3 2 3 4 3 4 2 2 1 5 1 4 5 1 1 1 3 3 1 1 3 2 5 4 2 5 2 4 5 4 4 1 1 2 1 1 4 5 1 1 5 3 3 2 5 5 5 1 4 1 4 1 1 3 2 3 4 4 2 5 5 2 5 1 1 3 5 3",
"output": "5"
},
{
"input": "100\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "4"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "5"
}
] | 1,609,222,648 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 154 | 0 | n=int(input())
l=list(map(int,input().split()))
s=sum(l)
if(s%(n+1)==0):
print(n+1)
else:
print(s%(n+1)) | Title: Dima and Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
Output Specification:
In a single line print the answer to the problem.
Demo Input:
['1\n1\n', '1\n2\n', '2\n3 5\n']
Demo Output:
['3\n', '2\n', '3\n']
Note:
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers. | ```python
n=int(input())
l=list(map(int,input().split()))
s=sum(l)
if(s%(n+1)==0):
print(n+1)
else:
print(s%(n+1))
``` | 0 | |
660 | A | Co-prime Array | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"math",
"number theory"
] | null | null | You are given an array of *n* elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers *a* and *b* are said to be co-prime if the only positive integer that divides both of them is 1. | The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the given array.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*. | Print integer *k* on the first line — the least number of elements needed to add to the array *a* to make it co-prime.
The second line should contain *n*<=+<=*k* integers *a**j* — the elements of the array *a* after adding *k* elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array *a* by adding *k* elements to it.
If there are multiple answers you can print any one of them. | [
"3\n2 7 28\n"
] | [
"1\n2 7 9 28\n"
] | none | 0 | [
{
"input": "3\n2 7 28",
"output": "1\n2 7 1 28"
},
{
"input": "1\n1",
"output": "0\n1"
},
{
"input": "1\n548",
"output": "0\n548"
},
{
"input": "1\n963837006",
"output": "0\n963837006"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "0\n1 1 1 1 1 1 1 1 1 1"
},
{
"input": "10\n26 723 970 13 422 968 875 329 234 983",
"output": "2\n26 723 970 13 422 1 968 875 1 329 234 983"
},
{
"input": "10\n319645572 758298525 812547177 459359946 355467212 304450522 807957797 916787906 239781206 242840396",
"output": "7\n319645572 1 758298525 1 812547177 1 459359946 1 355467212 1 304450522 807957797 916787906 1 239781206 1 242840396"
},
{
"input": "100\n1 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1",
"output": "19\n1 1 1 1 2 1 1 1 1 1 2 1 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 1 1 2 1 2 1 2 1 1 1 2 1 2 1 2 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 2 1"
},
{
"input": "100\n591 417 888 251 792 847 685 3 182 461 102 348 555 956 771 901 712 878 580 631 342 333 285 899 525 725 537 718 929 653 84 788 104 355 624 803 253 853 201 995 536 184 65 205 540 652 549 777 248 405 677 950 431 580 600 846 328 429 134 983 526 103 500 963 400 23 276 704 570 757 410 658 507 620 984 244 486 454 802 411 985 303 635 283 96 597 855 775 139 839 839 61 219 986 776 72 729 69 20 917",
"output": "38\n591 1 417 1 888 251 792 1 847 685 3 182 461 102 1 348 1 555 956 771 901 712 1 878 1 580 631 342 1 333 1 285 899 525 1 725 537 718 929 653 84 1 788 1 104 355 624 803 1 253 853 201 995 536 1 184 65 1 205 1 540 1 652 549 1 777 248 405 677 950 431 580 1 600 1 846 1 328 429 134 983 526 103 500 963 400 23 1 276 1 704 1 570 757 410 1 658 507 620 1 984 1 244 1 486 1 454 1 802 411 985 303 635 283 96 1 597 1 855 1 775 139 839 1 839 61 219 986 1 776 1 72 1 729 1 69 20 917"
},
{
"input": "5\n472882027 472882027 472882027 472882027 472882027",
"output": "4\n472882027 1 472882027 1 472882027 1 472882027 1 472882027"
},
{
"input": "2\n1000000000 1000000000",
"output": "1\n1000000000 1 1000000000"
},
{
"input": "2\n8 6",
"output": "1\n8 1 6"
},
{
"input": "3\n100000000 1000000000 1000000000",
"output": "2\n100000000 1 1000000000 1 1000000000"
},
{
"input": "5\n1 2 3 4 5",
"output": "0\n1 2 3 4 5"
},
{
"input": "20\n2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000",
"output": "19\n2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000"
},
{
"input": "2\n223092870 23",
"output": "1\n223092870 1 23"
},
{
"input": "2\n100000003 100000003",
"output": "1\n100000003 1 100000003"
},
{
"input": "2\n999999937 999999937",
"output": "1\n999999937 1 999999937"
},
{
"input": "4\n999 999999937 999999937 999",
"output": "1\n999 999999937 1 999999937 999"
},
{
"input": "2\n999999929 999999929",
"output": "1\n999999929 1 999999929"
},
{
"input": "2\n1049459 2098918",
"output": "1\n1049459 1 2098918"
},
{
"input": "2\n352229 704458",
"output": "1\n352229 1 704458"
},
{
"input": "2\n7293 4011",
"output": "1\n7293 1 4011"
},
{
"input": "2\n5565651 3999930",
"output": "1\n5565651 1 3999930"
},
{
"input": "2\n997 997",
"output": "1\n997 1 997"
},
{
"input": "3\n9994223 9994223 9994223",
"output": "2\n9994223 1 9994223 1 9994223"
},
{
"input": "2\n99999998 1000000000",
"output": "1\n99999998 1 1000000000"
},
{
"input": "3\n1000000000 1000000000 1000000000",
"output": "2\n1000000000 1 1000000000 1 1000000000"
},
{
"input": "2\n130471 130471",
"output": "1\n130471 1 130471"
},
{
"input": "3\n1000000000 2 2",
"output": "2\n1000000000 1 2 1 2"
},
{
"input": "2\n223092870 66526",
"output": "1\n223092870 1 66526"
},
{
"input": "14\n1000000000 1000000000 223092870 223092870 6 105 2 2 510510 510510 999999491 999999491 436077930 570018449",
"output": "10\n1000000000 1 1000000000 1 223092870 1 223092870 1 6 1 105 2 1 2 1 510510 1 510510 999999491 1 999999491 436077930 1 570018449"
},
{
"input": "2\n3996017 3996017",
"output": "1\n3996017 1 3996017"
},
{
"input": "2\n999983 999983",
"output": "1\n999983 1 999983"
},
{
"input": "2\n618575685 773990454",
"output": "1\n618575685 1 773990454"
},
{
"input": "3\n9699690 3 7",
"output": "1\n9699690 1 3 7"
},
{
"input": "2\n999999999 999999996",
"output": "1\n999999999 1 999999996"
},
{
"input": "2\n99999910 99999910",
"output": "1\n99999910 1 99999910"
},
{
"input": "12\n1000000000 1000000000 223092870 223092870 6 105 2 2 510510 510510 999999491 999999491",
"output": "9\n1000000000 1 1000000000 1 223092870 1 223092870 1 6 1 105 2 1 2 1 510510 1 510510 999999491 1 999999491"
},
{
"input": "3\n999999937 999999937 999999937",
"output": "2\n999999937 1 999999937 1 999999937"
},
{
"input": "2\n99839 99839",
"output": "1\n99839 1 99839"
},
{
"input": "3\n19999909 19999909 19999909",
"output": "2\n19999909 1 19999909 1 19999909"
},
{
"input": "4\n1 1000000000 1 1000000000",
"output": "0\n1 1000000000 1 1000000000"
},
{
"input": "2\n64006 64006",
"output": "1\n64006 1 64006"
},
{
"input": "2\n1956955 1956955",
"output": "1\n1956955 1 1956955"
},
{
"input": "3\n1 1000000000 1000000000",
"output": "1\n1 1000000000 1 1000000000"
},
{
"input": "2\n982451707 982451707",
"output": "1\n982451707 1 982451707"
},
{
"input": "2\n999999733 999999733",
"output": "1\n999999733 1 999999733"
},
{
"input": "3\n999999733 999999733 999999733",
"output": "2\n999999733 1 999999733 1 999999733"
},
{
"input": "2\n3257 3257",
"output": "1\n3257 1 3257"
},
{
"input": "2\n223092870 181598",
"output": "1\n223092870 1 181598"
},
{
"input": "3\n959919409 105935 105935",
"output": "2\n959919409 1 105935 1 105935"
},
{
"input": "2\n510510 510510",
"output": "1\n510510 1 510510"
},
{
"input": "3\n223092870 1000000000 1000000000",
"output": "2\n223092870 1 1000000000 1 1000000000"
},
{
"input": "14\n1000000000 2 1000000000 3 1000000000 6 1000000000 1000000000 15 1000000000 1000000000 1000000000 100000000 1000",
"output": "11\n1000000000 1 2 1 1000000000 3 1000000000 1 6 1 1000000000 1 1000000000 1 15 1 1000000000 1 1000000000 1 1000000000 1 100000000 1 1000"
},
{
"input": "7\n1 982451653 982451653 1 982451653 982451653 982451653",
"output": "3\n1 982451653 1 982451653 1 982451653 1 982451653 1 982451653"
},
{
"input": "2\n100000007 100000007",
"output": "1\n100000007 1 100000007"
},
{
"input": "3\n999999757 999999757 999999757",
"output": "2\n999999757 1 999999757 1 999999757"
},
{
"input": "3\n99999989 99999989 99999989",
"output": "2\n99999989 1 99999989 1 99999989"
},
{
"input": "5\n2 4 982451707 982451707 3",
"output": "2\n2 1 4 982451707 1 982451707 3"
},
{
"input": "2\n20000014 20000014",
"output": "1\n20000014 1 20000014"
},
{
"input": "2\n99999989 99999989",
"output": "1\n99999989 1 99999989"
},
{
"input": "2\n111546435 111546435",
"output": "1\n111546435 1 111546435"
},
{
"input": "2\n55288874 33538046",
"output": "1\n55288874 1 33538046"
},
{
"input": "5\n179424673 179424673 179424673 179424673 179424673",
"output": "4\n179424673 1 179424673 1 179424673 1 179424673 1 179424673"
},
{
"input": "2\n199999978 199999978",
"output": "1\n199999978 1 199999978"
},
{
"input": "2\n1000000000 2",
"output": "1\n1000000000 1 2"
},
{
"input": "3\n19999897 19999897 19999897",
"output": "2\n19999897 1 19999897 1 19999897"
},
{
"input": "2\n19999982 19999982",
"output": "1\n19999982 1 19999982"
},
{
"input": "2\n10000007 10000007",
"output": "1\n10000007 1 10000007"
},
{
"input": "3\n999999937 999999937 2",
"output": "1\n999999937 1 999999937 2"
},
{
"input": "5\n2017 2017 2017 2017 2017",
"output": "4\n2017 1 2017 1 2017 1 2017 1 2017"
},
{
"input": "2\n19999909 39999818",
"output": "1\n19999909 1 39999818"
},
{
"input": "2\n62615533 7919",
"output": "1\n62615533 1 7919"
},
{
"input": "5\n39989 39989 33 31 29",
"output": "1\n39989 1 39989 33 31 29"
},
{
"input": "2\n1000000000 100000",
"output": "1\n1000000000 1 100000"
},
{
"input": "2\n1938 10010",
"output": "1\n1938 1 10010"
},
{
"input": "2\n199999 199999",
"output": "1\n199999 1 199999"
},
{
"input": "2\n107273 107273",
"output": "1\n107273 1 107273"
},
{
"input": "3\n49999 49999 49999",
"output": "2\n49999 1 49999 1 49999"
},
{
"input": "2\n1999966 1999958",
"output": "1\n1999966 1 1999958"
},
{
"input": "2\n86020 300846",
"output": "1\n86020 1 300846"
},
{
"input": "2\n999999997 213",
"output": "1\n999999997 1 213"
},
{
"input": "2\n200000014 200000434",
"output": "1\n200000014 1 200000434"
}
] | 1,656,467,021 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 93 | 124 | 4,812,800 | def gcd(a, b):
if b == 0:
return a
return gcd(b, a%b)
n = int(input())
arr = list(map(int, input().split()))
length = n
count = 0
i = 0
while i < length - 1:
if gcd(arr[i], arr[i+1]) > 1:
arr.insert(i+1, 1)
count += 1
length += 1
i += 1
print(count)
print(*arr) | Title: Co-prime Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array of *n* elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers *a* and *b* are said to be co-prime if the only positive integer that divides both of them is 1.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the given array.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.
Output Specification:
Print integer *k* on the first line — the least number of elements needed to add to the array *a* to make it co-prime.
The second line should contain *n*<=+<=*k* integers *a**j* — the elements of the array *a* after adding *k* elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array *a* by adding *k* elements to it.
If there are multiple answers you can print any one of them.
Demo Input:
['3\n2 7 28\n']
Demo Output:
['1\n2 7 9 28\n']
Note:
none | ```python
def gcd(a, b):
if b == 0:
return a
return gcd(b, a%b)
n = int(input())
arr = list(map(int, input().split()))
length = n
count = 0
i = 0
while i < length - 1:
if gcd(arr[i], arr[i+1]) > 1:
arr.insert(i+1, 1)
count += 1
length += 1
i += 1
print(count)
print(*arr)
``` | 3 | |
898 | A | Rounding | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Vasya has a non-negative integer *n*. He wants to round it to nearest integer, which ends up with 0. If *n* already ends up with 0, Vasya considers it already rounded.
For example, if *n*<==<=4722 answer is 4720. If *n*<==<=5 Vasya can round it to 0 or to 10. Both ways are correct.
For given *n* find out to which integer will Vasya round it. | The first line contains single integer *n* (0<=≤<=*n*<=≤<=109) — number that Vasya has. | Print result of rounding *n*. Pay attention that in some cases answer isn't unique. In that case print any correct answer. | [
"5\n",
"113\n",
"1000000000\n",
"5432359\n"
] | [
"0\n",
"110\n",
"1000000000\n",
"5432360\n"
] | In the first example *n* = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10. | 500 | [
{
"input": "5",
"output": "0"
},
{
"input": "113",
"output": "110"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "5432359",
"output": "5432360"
},
{
"input": "999999994",
"output": "999999990"
},
{
"input": "10",
"output": "10"
},
{
"input": "9",
"output": "10"
},
{
"input": "1",
"output": "0"
},
{
"input": "0",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "6",
"output": "10"
},
{
"input": "7",
"output": "10"
},
{
"input": "8",
"output": "10"
},
{
"input": "19",
"output": "20"
},
{
"input": "100",
"output": "100"
},
{
"input": "997",
"output": "1000"
},
{
"input": "9994",
"output": "9990"
},
{
"input": "10002",
"output": "10000"
},
{
"input": "100000",
"output": "100000"
},
{
"input": "99999",
"output": "100000"
},
{
"input": "999999999",
"output": "1000000000"
},
{
"input": "999999998",
"output": "1000000000"
},
{
"input": "999999995",
"output": "999999990"
},
{
"input": "999999990",
"output": "999999990"
},
{
"input": "1000000",
"output": "1000000"
},
{
"input": "1000010",
"output": "1000010"
},
{
"input": "10000010",
"output": "10000010"
},
{
"input": "100000011",
"output": "100000010"
},
{
"input": "400000003",
"output": "400000000"
},
{
"input": "234234",
"output": "234230"
},
{
"input": "675621",
"output": "675620"
},
{
"input": "43532",
"output": "43530"
},
{
"input": "4576453",
"output": "4576450"
},
{
"input": "65754674",
"output": "65754670"
},
{
"input": "3245526",
"output": "3245530"
},
{
"input": "123445",
"output": "123440"
},
{
"input": "234217",
"output": "234220"
},
{
"input": "23451218",
"output": "23451220"
},
{
"input": "1231239",
"output": "1231240"
},
{
"input": "1923140",
"output": "1923140"
},
{
"input": "307910310",
"output": "307910310"
},
{
"input": "780961030",
"output": "780961030"
},
{
"input": "103509421",
"output": "103509420"
},
{
"input": "576560141",
"output": "576560140"
},
{
"input": "48851642",
"output": "48851640"
},
{
"input": "226935072",
"output": "226935070"
},
{
"input": "844450763",
"output": "844450760"
},
{
"input": "22534183",
"output": "22534180"
},
{
"input": "640049874",
"output": "640049870"
},
{
"input": "818133304",
"output": "818133300"
},
{
"input": "730616285",
"output": "730616280"
},
{
"input": "613732415",
"output": "613732410"
},
{
"input": "380991216",
"output": "380991220"
},
{
"input": "559074636",
"output": "559074640"
},
{
"input": "176590327",
"output": "176590330"
},
{
"input": "354673757",
"output": "354673760"
},
{
"input": "267156738",
"output": "267156740"
},
{
"input": "150272868",
"output": "150272870"
},
{
"input": "62755859",
"output": "62755860"
},
{
"input": "945871979",
"output": "945871980"
},
{
"input": "46",
"output": "50"
},
{
"input": "999",
"output": "1000"
},
{
"input": "1397",
"output": "1400"
}
] | 1,637,750,090 | 2,147,483,647 | Python 3 | OK | TESTS | 64 | 46 | 0 | n = int(input())
m = n % 10
if m <= 5:
num = n - m
else:
num = (n // 10 + 1) * 10
print(num)
| Title: Rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has a non-negative integer *n*. He wants to round it to nearest integer, which ends up with 0. If *n* already ends up with 0, Vasya considers it already rounded.
For example, if *n*<==<=4722 answer is 4720. If *n*<==<=5 Vasya can round it to 0 or to 10. Both ways are correct.
For given *n* find out to which integer will Vasya round it.
Input Specification:
The first line contains single integer *n* (0<=≤<=*n*<=≤<=109) — number that Vasya has.
Output Specification:
Print result of rounding *n*. Pay attention that in some cases answer isn't unique. In that case print any correct answer.
Demo Input:
['5\n', '113\n', '1000000000\n', '5432359\n']
Demo Output:
['0\n', '110\n', '1000000000\n', '5432360\n']
Note:
In the first example *n* = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10. | ```python
n = int(input())
m = n % 10
if m <= 5:
num = n - m
else:
num = (n // 10 + 1) * 10
print(num)
``` | 3 | |
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,681,906,768 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | n = int(input())
for i in range(n):
w = str(input())
if len(w) > 10:
print(w[0],end="")
print(len(w)-2,end="")
print(w[-1],end="")
else:
print(w) | Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
n = int(input())
for i in range(n):
w = str(input())
if len(w) > 10:
print(w[0],end="")
print(len(w)-2,end="")
print(w[-1],end="")
else:
print(w)
``` | 0 |
703 | A | Mishka and Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her! | The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively. | If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line. | [
"3\n3 5\n2 1\n4 2\n",
"2\n6 1\n1 6\n",
"3\n1 5\n3 3\n2 2\n"
] | [
"Mishka",
"Friendship is magic!^^",
"Chris"
] | In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris. | 500 | [
{
"input": "3\n3 5\n2 1\n4 2",
"output": "Mishka"
},
{
"input": "2\n6 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "3\n1 5\n3 3\n2 2",
"output": "Chris"
},
{
"input": "6\n4 1\n4 2\n5 3\n5 1\n5 3\n4 1",
"output": "Mishka"
},
{
"input": "8\n2 4\n1 4\n1 5\n2 6\n2 5\n2 5\n2 4\n2 5",
"output": "Chris"
},
{
"input": "8\n4 1\n2 6\n4 2\n2 5\n5 2\n3 5\n5 2\n1 5",
"output": "Friendship is magic!^^"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3",
"output": "Mishka"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "9\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "10\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n2 4\n6 6\n3 2\n1 5\n5 2\n1 5\n1 5\n3 1\n6 5\n4 3\n1 1\n5 1\n3 3\n2 4\n1 5\n3 4\n5 1\n5 5\n2 5\n2 1\n4 3\n6 5\n1 1\n2 1\n1 3\n1 1\n6 4\n4 6\n6 4\n2 1\n2 5\n6 2\n3 4\n5 5\n1 4\n4 6\n3 4\n1 6\n5 1\n4 3\n3 4\n2 2\n1 2\n2 3\n1 3\n4 4\n5 5\n4 5\n4 4\n3 1\n4 5\n2 3\n2 6\n6 5\n6 1\n6 6\n2 3\n6 4\n3 3\n2 5\n4 4\n3 1\n2 4\n6 1\n3 2\n1 3\n5 4\n6 6\n2 5\n5 1\n1 1\n2 5\n6 5\n3 6\n5 6\n4 3\n3 4\n3 4\n6 5\n5 2\n4 2\n1 1\n3 1\n2 6\n1 6\n1 2\n6 1\n3 4\n1 6\n3 1\n5 3\n1 3\n5 6\n2 1\n6 4\n3 1\n1 6\n6 3\n3 3\n4 3",
"output": "Chris"
},
{
"input": "100\n4 1\n3 4\n4 6\n4 5\n6 5\n5 3\n6 2\n6 3\n5 2\n4 5\n1 5\n5 4\n1 4\n4 5\n4 6\n1 6\n4 4\n5 1\n6 4\n6 4\n4 6\n2 3\n6 2\n4 6\n1 4\n2 3\n4 3\n1 3\n6 2\n3 1\n3 4\n2 6\n4 5\n5 4\n2 2\n2 5\n4 1\n2 2\n3 3\n1 4\n5 6\n6 4\n4 2\n6 1\n5 5\n4 1\n2 1\n6 4\n4 4\n4 3\n5 3\n4 5\n5 3\n3 5\n6 3\n1 1\n3 4\n6 3\n6 1\n5 1\n2 4\n4 3\n2 2\n5 5\n1 5\n5 3\n4 6\n1 4\n6 3\n4 3\n2 4\n3 2\n2 4\n3 4\n6 2\n5 6\n1 2\n1 5\n5 5\n2 6\n5 1\n1 6\n5 3\n3 5\n2 6\n4 6\n6 2\n3 1\n5 5\n6 1\n3 6\n4 4\n1 1\n4 6\n5 3\n4 2\n5 1\n3 3\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "100\n6 3\n4 5\n4 3\n5 4\n5 1\n6 3\n4 2\n4 6\n3 1\n2 4\n2 2\n4 6\n5 3\n5 5\n4 2\n6 2\n2 3\n4 4\n6 4\n3 5\n2 4\n2 2\n5 2\n3 5\n2 4\n4 4\n3 5\n6 5\n1 3\n1 6\n2 2\n2 4\n3 2\n5 4\n1 6\n3 4\n4 1\n1 5\n1 4\n5 3\n2 2\n4 5\n6 3\n4 4\n1 1\n4 1\n2 4\n4 1\n4 5\n5 3\n1 1\n1 6\n5 6\n6 6\n4 2\n4 3\n3 4\n3 6\n3 4\n6 5\n3 4\n5 4\n5 1\n5 3\n5 1\n1 2\n2 6\n3 4\n6 5\n4 3\n1 1\n5 5\n5 1\n3 3\n5 2\n1 3\n6 6\n5 6\n1 4\n4 4\n1 4\n3 6\n6 5\n3 3\n3 6\n1 5\n1 2\n3 6\n3 6\n4 1\n5 2\n1 2\n5 2\n3 3\n4 4\n4 2\n6 2\n5 4\n6 1\n6 3",
"output": "Mishka"
},
{
"input": "8\n4 1\n6 2\n4 1\n5 3\n4 1\n5 3\n6 2\n5 3",
"output": "Mishka"
},
{
"input": "5\n3 6\n3 5\n3 5\n1 6\n3 5",
"output": "Chris"
},
{
"input": "4\n4 1\n2 4\n5 3\n3 6",
"output": "Friendship is magic!^^"
},
{
"input": "6\n6 3\n5 1\n6 3\n4 3\n4 3\n5 2",
"output": "Mishka"
},
{
"input": "7\n3 4\n1 4\n2 5\n1 6\n1 6\n1 5\n3 4",
"output": "Chris"
},
{
"input": "6\n6 2\n2 5\n5 2\n3 6\n4 3\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "8\n6 1\n5 3\n4 3\n4 1\n5 1\n4 2\n4 2\n4 1",
"output": "Mishka"
},
{
"input": "9\n2 5\n2 5\n1 4\n2 6\n2 4\n2 5\n2 6\n1 5\n2 5",
"output": "Chris"
},
{
"input": "4\n6 2\n2 4\n4 2\n3 6",
"output": "Friendship is magic!^^"
},
{
"input": "9\n5 2\n4 1\n4 1\n5 1\n6 2\n6 1\n5 3\n6 1\n6 2",
"output": "Mishka"
},
{
"input": "8\n2 4\n3 6\n1 6\n1 6\n2 4\n3 4\n3 6\n3 4",
"output": "Chris"
},
{
"input": "6\n5 3\n3 6\n6 2\n1 6\n5 1\n3 5",
"output": "Friendship is magic!^^"
},
{
"input": "6\n5 2\n5 1\n6 1\n5 2\n4 2\n5 1",
"output": "Mishka"
},
{
"input": "5\n1 4\n2 5\n3 4\n2 6\n3 4",
"output": "Chris"
},
{
"input": "4\n6 2\n3 4\n5 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "93\n4 3\n4 1\n4 2\n5 2\n5 3\n6 3\n4 3\n6 2\n6 3\n5 1\n4 2\n4 2\n5 1\n6 2\n6 3\n6 1\n4 1\n6 2\n5 3\n4 3\n4 1\n4 2\n5 2\n6 3\n5 2\n5 2\n6 3\n5 1\n6 2\n5 2\n4 1\n5 2\n5 1\n4 1\n6 1\n5 2\n4 3\n5 3\n5 3\n5 1\n4 3\n4 3\n4 2\n4 1\n6 2\n6 1\n4 1\n5 2\n5 2\n6 2\n5 3\n5 1\n6 2\n5 1\n6 3\n5 2\n6 2\n6 2\n4 2\n5 2\n6 1\n6 3\n6 3\n5 1\n5 1\n4 1\n5 1\n4 3\n5 3\n6 3\n4 1\n4 3\n6 1\n6 1\n4 2\n6 2\n4 2\n5 2\n4 1\n5 2\n4 1\n5 1\n5 2\n5 1\n4 1\n6 3\n6 2\n4 3\n4 1\n5 2\n4 3\n5 2\n5 1",
"output": "Mishka"
},
{
"input": "11\n1 6\n1 6\n2 4\n2 5\n3 4\n1 5\n1 6\n1 5\n1 6\n2 6\n3 4",
"output": "Chris"
},
{
"input": "70\n6 1\n3 6\n4 3\n2 5\n5 2\n1 4\n6 2\n1 6\n4 3\n1 4\n5 3\n2 4\n5 3\n1 6\n5 1\n3 5\n4 2\n2 4\n5 1\n3 5\n6 2\n1 5\n4 2\n2 5\n5 3\n1 5\n4 2\n1 4\n5 2\n2 6\n4 3\n1 5\n6 2\n3 4\n4 2\n3 5\n6 3\n3 4\n5 1\n1 4\n4 2\n1 4\n6 3\n2 6\n5 2\n1 6\n6 1\n2 6\n5 3\n1 5\n5 1\n1 6\n4 1\n1 5\n4 2\n2 4\n5 1\n2 5\n6 3\n1 4\n6 3\n3 6\n5 1\n1 4\n5 3\n3 5\n4 2\n3 4\n6 2\n1 4",
"output": "Friendship is magic!^^"
},
{
"input": "59\n4 1\n5 3\n6 1\n4 2\n5 1\n4 3\n6 1\n5 1\n4 3\n4 3\n5 2\n5 3\n4 1\n6 2\n5 1\n6 3\n6 3\n5 2\n5 2\n6 1\n4 1\n6 1\n4 3\n5 3\n5 3\n4 3\n4 2\n4 2\n6 3\n6 3\n6 1\n4 3\n5 1\n6 2\n6 1\n4 1\n6 1\n5 3\n4 2\n5 1\n6 2\n6 2\n4 3\n5 3\n4 3\n6 3\n5 2\n5 2\n4 3\n5 1\n5 3\n6 1\n6 3\n6 3\n4 3\n5 2\n5 2\n5 2\n4 3",
"output": "Mishka"
},
{
"input": "42\n1 5\n1 6\n1 6\n1 4\n2 5\n3 6\n1 6\n3 4\n2 5\n2 5\n2 4\n1 4\n3 4\n2 4\n2 6\n1 5\n3 6\n2 6\n2 6\n3 5\n1 4\n1 5\n2 6\n3 6\n1 4\n3 4\n2 4\n1 6\n3 4\n2 4\n2 6\n1 6\n1 4\n1 6\n1 6\n2 4\n1 5\n1 6\n2 5\n3 6\n3 5\n3 4",
"output": "Chris"
},
{
"input": "78\n4 3\n3 5\n4 3\n1 5\n5 1\n1 5\n4 3\n1 4\n6 3\n1 5\n4 1\n2 4\n4 3\n2 4\n5 1\n3 6\n4 2\n3 6\n6 3\n3 4\n4 3\n3 6\n5 3\n1 5\n4 1\n2 6\n4 2\n2 4\n4 1\n3 5\n5 2\n3 6\n4 3\n2 4\n6 3\n1 6\n4 3\n3 5\n6 3\n2 6\n4 1\n2 4\n6 2\n1 6\n4 2\n1 4\n4 3\n1 4\n4 3\n2 4\n6 2\n3 5\n6 1\n3 6\n5 3\n1 6\n6 1\n2 6\n4 2\n1 5\n6 2\n2 6\n6 3\n2 4\n4 2\n3 5\n6 1\n2 5\n5 3\n2 6\n5 1\n3 6\n4 3\n3 6\n6 3\n2 5\n6 1\n2 6",
"output": "Friendship is magic!^^"
},
{
"input": "76\n4 1\n5 2\n4 3\n5 2\n5 3\n5 2\n6 1\n4 2\n6 2\n5 3\n4 2\n6 2\n4 1\n4 2\n5 1\n5 1\n6 2\n5 2\n5 3\n6 3\n5 2\n4 3\n6 3\n6 1\n4 3\n6 2\n6 1\n4 1\n6 1\n5 3\n4 1\n5 3\n4 2\n5 2\n4 3\n6 1\n6 2\n5 2\n6 1\n5 3\n4 3\n5 1\n5 3\n4 3\n5 1\n5 1\n4 1\n4 1\n4 1\n4 3\n5 3\n6 3\n6 3\n5 2\n6 2\n6 3\n5 1\n6 3\n5 3\n6 1\n5 3\n4 1\n5 3\n6 1\n4 2\n6 2\n4 3\n4 1\n6 2\n4 3\n5 3\n5 2\n5 3\n5 1\n6 3\n5 2",
"output": "Mishka"
},
{
"input": "84\n3 6\n3 4\n2 5\n2 4\n1 6\n3 4\n1 5\n1 6\n3 5\n1 6\n2 4\n2 6\n2 6\n2 4\n3 5\n1 5\n3 6\n3 6\n3 4\n3 4\n2 6\n1 6\n1 6\n3 5\n3 4\n1 6\n3 4\n3 5\n2 4\n2 5\n2 5\n3 5\n1 6\n3 4\n2 6\n2 6\n3 4\n3 4\n2 5\n2 5\n2 4\n3 4\n2 5\n3 4\n3 4\n2 6\n2 6\n1 6\n2 4\n1 5\n3 4\n2 5\n2 5\n3 4\n2 4\n2 6\n2 6\n1 4\n3 5\n3 5\n2 4\n2 5\n3 4\n1 5\n1 5\n2 6\n1 5\n3 5\n2 4\n2 5\n3 4\n2 6\n1 6\n2 5\n3 5\n3 5\n3 4\n2 5\n2 6\n3 4\n1 6\n2 5\n2 6\n1 4",
"output": "Chris"
},
{
"input": "44\n6 1\n1 6\n5 2\n1 4\n6 2\n2 5\n5 3\n3 6\n5 2\n1 6\n4 1\n2 4\n6 1\n3 4\n6 3\n3 6\n4 3\n2 4\n6 1\n3 4\n6 1\n1 6\n4 1\n3 5\n6 1\n3 6\n4 1\n1 4\n4 2\n2 6\n6 1\n2 4\n6 2\n1 4\n6 2\n2 4\n5 2\n3 6\n6 3\n2 6\n5 3\n3 4\n5 3\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "42\n5 3\n5 1\n5 2\n4 1\n6 3\n6 1\n6 2\n4 1\n4 3\n4 1\n5 1\n5 3\n5 1\n4 1\n4 2\n6 1\n6 3\n5 1\n4 1\n4 1\n6 3\n4 3\n6 3\n5 2\n6 1\n4 1\n5 3\n4 3\n5 2\n6 3\n6 1\n5 1\n4 2\n4 3\n5 2\n5 3\n6 3\n5 2\n5 1\n5 3\n6 2\n6 1",
"output": "Mishka"
},
{
"input": "50\n3 6\n2 6\n1 4\n1 4\n1 4\n2 5\n3 4\n3 5\n2 6\n1 6\n3 5\n1 5\n2 6\n2 4\n2 4\n3 5\n1 6\n1 5\n1 5\n1 4\n3 5\n1 6\n3 5\n1 4\n1 5\n1 4\n3 6\n1 6\n1 4\n1 4\n1 4\n1 5\n3 6\n1 6\n1 6\n2 4\n1 5\n2 6\n2 5\n3 5\n3 6\n3 4\n2 4\n2 6\n3 4\n2 5\n3 6\n3 5\n2 4\n2 4",
"output": "Chris"
},
{
"input": "86\n6 3\n2 4\n6 3\n3 5\n6 3\n1 5\n5 2\n2 4\n4 3\n2 6\n4 1\n2 6\n5 2\n1 4\n5 1\n2 4\n4 1\n1 4\n6 2\n3 5\n4 2\n2 4\n6 2\n1 5\n5 3\n2 5\n5 1\n1 6\n6 1\n1 4\n4 3\n3 4\n5 2\n2 4\n5 3\n2 5\n4 3\n3 4\n4 1\n1 5\n6 3\n3 4\n4 3\n3 4\n4 1\n3 4\n5 1\n1 6\n4 2\n1 6\n5 1\n2 4\n5 1\n3 6\n4 1\n1 5\n5 2\n1 4\n4 3\n2 5\n5 1\n1 5\n6 2\n2 6\n4 2\n2 4\n4 1\n2 5\n5 3\n3 4\n5 1\n3 4\n6 3\n3 4\n4 3\n2 6\n6 2\n2 5\n5 2\n3 5\n4 2\n3 6\n6 2\n3 4\n4 2\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "84\n6 1\n6 3\n6 3\n4 1\n4 3\n4 2\n6 3\n5 3\n6 1\n6 3\n4 3\n5 2\n5 3\n5 1\n6 2\n6 2\n6 1\n4 1\n6 3\n5 2\n4 1\n5 3\n6 3\n4 2\n6 2\n6 3\n4 3\n4 1\n4 3\n5 1\n5 1\n5 1\n4 1\n6 1\n4 3\n6 2\n5 1\n5 1\n6 2\n5 2\n4 1\n6 1\n6 1\n6 3\n6 2\n4 3\n6 3\n6 2\n5 2\n5 1\n4 3\n6 2\n4 1\n6 2\n6 1\n5 2\n5 1\n6 2\n6 1\n5 3\n5 2\n6 1\n6 3\n5 2\n6 1\n6 3\n4 3\n5 1\n6 3\n6 1\n5 3\n4 3\n5 2\n5 1\n6 2\n5 3\n6 1\n5 1\n4 1\n5 1\n5 1\n5 2\n5 2\n5 1",
"output": "Mishka"
},
{
"input": "92\n1 5\n2 4\n3 5\n1 6\n2 5\n1 6\n3 6\n1 6\n2 4\n3 4\n3 4\n3 6\n1 5\n2 5\n1 5\n1 5\n2 6\n2 4\n3 6\n1 4\n1 6\n2 6\n3 4\n2 6\n2 6\n1 4\n3 5\n2 5\n2 6\n1 5\n1 4\n1 5\n3 6\n3 5\n2 5\n1 5\n3 5\n3 6\n2 6\n2 6\n1 5\n3 4\n2 4\n3 6\n2 5\n1 5\n2 4\n1 4\n2 6\n2 6\n2 6\n1 5\n3 6\n3 6\n2 5\n1 4\n2 4\n3 4\n1 5\n2 5\n2 4\n2 5\n3 5\n3 4\n3 6\n2 6\n3 5\n1 4\n3 4\n1 6\n3 6\n2 6\n1 4\n3 6\n3 6\n2 5\n2 6\n1 6\n2 6\n3 5\n2 5\n3 6\n2 5\n2 6\n1 5\n2 4\n1 4\n2 4\n1 5\n2 5\n2 5\n2 6",
"output": "Chris"
},
{
"input": "20\n5 1\n1 4\n4 3\n1 5\n4 2\n3 6\n6 2\n1 6\n4 1\n1 4\n5 2\n3 4\n5 1\n1 6\n5 1\n2 6\n6 3\n2 5\n6 2\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "100\n4 3\n4 3\n4 2\n4 3\n4 1\n4 3\n5 2\n5 2\n6 2\n4 2\n5 1\n4 2\n5 2\n6 1\n4 1\n6 3\n5 3\n5 1\n5 1\n5 1\n5 3\n6 1\n6 1\n4 1\n5 2\n5 2\n6 1\n6 3\n4 2\n4 1\n5 3\n4 1\n5 3\n5 1\n6 3\n6 3\n6 1\n5 2\n5 3\n5 3\n6 1\n4 1\n6 2\n6 1\n6 2\n6 3\n4 3\n4 3\n6 3\n4 2\n4 2\n5 3\n5 2\n5 2\n4 3\n5 3\n5 2\n4 2\n5 1\n4 2\n5 1\n5 3\n6 3\n5 3\n5 3\n4 2\n4 1\n4 2\n4 3\n6 3\n4 3\n6 2\n6 1\n5 3\n5 2\n4 1\n6 1\n5 2\n6 2\n4 2\n6 3\n4 3\n5 1\n6 3\n5 2\n4 3\n5 3\n5 3\n4 3\n6 3\n4 3\n4 1\n5 1\n6 2\n6 3\n5 3\n6 1\n6 3\n5 3\n6 1",
"output": "Mishka"
},
{
"input": "100\n1 5\n1 4\n1 5\n2 4\n2 6\n3 6\n3 5\n1 5\n2 5\n3 6\n3 5\n1 6\n1 4\n1 5\n1 6\n2 6\n1 5\n3 5\n3 4\n2 6\n2 6\n2 5\n3 4\n1 6\n1 4\n2 4\n1 5\n1 6\n3 5\n1 6\n2 6\n3 5\n1 6\n3 4\n3 5\n1 6\n3 6\n2 4\n2 4\n3 5\n2 6\n1 5\n3 5\n3 6\n2 4\n2 4\n2 6\n3 4\n3 4\n1 5\n1 4\n2 5\n3 4\n1 4\n2 6\n2 5\n2 4\n2 4\n2 5\n1 5\n1 6\n1 5\n1 5\n1 5\n1 6\n3 4\n2 4\n3 5\n3 5\n1 6\n3 5\n1 5\n1 6\n3 6\n3 4\n1 5\n3 5\n3 6\n1 4\n3 6\n1 5\n3 5\n3 6\n3 5\n1 4\n3 4\n2 4\n2 4\n2 5\n3 6\n3 5\n1 5\n2 4\n1 4\n3 4\n1 5\n3 4\n3 6\n3 5\n3 4",
"output": "Chris"
},
{
"input": "100\n4 3\n3 4\n5 1\n2 5\n5 3\n1 5\n6 3\n2 4\n5 2\n2 6\n5 2\n1 5\n6 3\n1 5\n6 3\n3 4\n5 2\n1 5\n6 1\n1 5\n4 2\n3 5\n6 3\n2 6\n6 3\n1 4\n6 2\n3 4\n4 1\n3 6\n5 1\n2 4\n5 1\n3 4\n6 2\n3 5\n4 1\n2 6\n4 3\n2 6\n5 2\n3 6\n6 2\n3 5\n4 3\n1 5\n5 3\n3 6\n4 2\n3 4\n6 1\n3 4\n5 2\n2 6\n5 2\n2 4\n6 2\n3 6\n4 3\n2 4\n4 3\n2 6\n4 2\n3 4\n6 3\n2 4\n6 3\n3 5\n5 2\n1 5\n6 3\n3 6\n4 3\n1 4\n5 2\n1 6\n4 1\n2 5\n4 1\n2 4\n4 2\n2 5\n6 1\n2 4\n6 3\n1 5\n4 3\n2 6\n6 3\n2 6\n5 3\n1 5\n4 1\n1 5\n6 2\n2 5\n5 1\n3 6\n4 3\n3 4",
"output": "Friendship is magic!^^"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3",
"output": "Mishka"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "99\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "84\n6 2\n1 5\n6 2\n2 3\n5 5\n1 2\n3 4\n3 4\n6 5\n6 4\n2 5\n4 1\n1 2\n1 1\n1 4\n2 5\n5 6\n6 3\n2 4\n5 5\n2 6\n3 4\n5 1\n3 3\n5 5\n4 6\n4 6\n2 4\n4 1\n5 2\n2 2\n3 6\n3 3\n4 6\n1 1\n2 4\n6 5\n5 2\n6 5\n5 5\n2 5\n6 4\n1 1\n6 2\n3 6\n6 5\n4 4\n1 5\n5 6\n4 4\n3 5\n6 1\n3 4\n1 5\n4 6\n4 6\n4 1\n3 6\n6 2\n1 1\n4 5\n5 4\n5 3\n3 4\n6 4\n1 1\n5 2\n6 5\n6 1\n2 2\n2 4\n3 3\n4 6\n1 3\n6 6\n5 2\n1 6\n6 2\n6 6\n4 1\n3 6\n6 4\n2 3\n3 4",
"output": "Chris"
},
{
"input": "70\n3 4\n2 3\n2 3\n6 5\n6 6\n4 3\n2 3\n3 1\n3 5\n5 6\n1 6\n2 5\n5 3\n2 5\n4 6\n5 1\n6 1\n3 1\n3 3\n5 3\n2 1\n3 3\n6 4\n6 3\n4 3\n4 5\n3 5\n5 5\n5 2\n1 6\n3 4\n5 2\n2 4\n1 6\n4 3\n4 3\n6 2\n1 3\n1 5\n6 1\n3 1\n1 1\n1 3\n2 2\n3 2\n6 4\n1 1\n4 4\n3 1\n4 5\n4 2\n6 3\n4 4\n3 2\n1 2\n2 6\n3 3\n1 5\n1 1\n6 5\n2 2\n3 1\n5 4\n5 2\n6 4\n6 3\n6 6\n6 3\n3 3\n5 4",
"output": "Mishka"
},
{
"input": "56\n6 4\n3 4\n6 1\n3 3\n1 4\n2 3\n1 5\n2 5\n1 5\n5 5\n2 3\n1 1\n3 2\n3 5\n4 6\n4 4\n5 2\n4 3\n3 1\n3 6\n2 3\n3 4\n5 6\n5 2\n5 6\n1 5\n1 5\n4 1\n6 3\n2 2\n2 1\n5 5\n2 1\n4 1\n5 4\n2 5\n4 1\n6 2\n3 4\n4 2\n6 4\n5 4\n4 2\n4 3\n6 2\n6 2\n3 1\n1 4\n3 6\n5 1\n5 5\n3 6\n6 4\n2 3\n6 5\n3 3",
"output": "Mishka"
},
{
"input": "94\n2 4\n6 4\n1 6\n1 4\n5 1\n3 3\n4 3\n6 1\n6 5\n3 2\n2 3\n5 1\n5 3\n1 2\n4 3\n3 2\n2 3\n4 6\n1 3\n6 3\n1 1\n3 2\n4 3\n1 5\n4 6\n3 2\n6 3\n1 6\n1 1\n1 2\n3 5\n1 3\n3 5\n4 4\n4 2\n1 4\n4 5\n1 3\n1 2\n1 1\n5 4\n5 5\n6 1\n2 1\n2 6\n6 6\n4 2\n3 6\n1 6\n6 6\n1 5\n3 2\n1 2\n4 4\n6 4\n4 1\n1 5\n3 3\n1 3\n3 4\n4 4\n1 1\n2 5\n4 5\n3 1\n3 1\n3 6\n3 2\n1 4\n1 6\n6 3\n2 4\n1 1\n2 2\n2 2\n2 1\n5 4\n1 2\n6 6\n2 2\n3 3\n6 3\n6 3\n1 6\n2 3\n2 4\n2 3\n6 6\n2 6\n6 3\n3 5\n1 4\n1 1\n3 5",
"output": "Chris"
},
{
"input": "81\n4 2\n1 2\n2 3\n4 5\n6 2\n1 6\n3 6\n3 4\n4 6\n4 4\n3 5\n4 6\n3 6\n3 5\n3 1\n1 3\n5 3\n3 4\n1 1\n4 1\n1 2\n6 1\n1 3\n6 5\n4 5\n4 2\n4 5\n6 2\n1 2\n2 6\n5 2\n1 5\n2 4\n4 3\n5 4\n1 2\n5 3\n2 6\n6 4\n1 1\n1 3\n3 1\n3 1\n6 5\n5 5\n6 1\n6 6\n5 2\n1 3\n1 4\n2 3\n5 5\n3 1\n3 1\n4 4\n1 6\n6 4\n2 2\n4 6\n4 4\n2 6\n2 4\n2 4\n4 1\n1 6\n1 4\n1 3\n6 5\n5 1\n1 3\n5 1\n1 4\n3 5\n2 6\n1 3\n5 6\n3 5\n4 4\n5 5\n5 6\n4 3",
"output": "Chris"
},
{
"input": "67\n6 5\n3 6\n1 6\n5 3\n5 4\n5 1\n1 6\n1 1\n3 2\n4 4\n3 1\n4 1\n1 5\n5 3\n3 3\n6 4\n2 4\n2 2\n4 3\n1 4\n1 4\n6 1\n1 2\n2 2\n5 1\n6 2\n3 5\n5 5\n2 2\n6 5\n6 2\n4 4\n3 1\n4 2\n6 6\n6 4\n5 1\n2 2\n4 5\n5 5\n4 6\n1 5\n6 3\n4 4\n1 5\n6 4\n3 6\n3 4\n1 6\n2 4\n2 1\n2 5\n6 5\n6 4\n4 1\n3 2\n1 2\n5 1\n5 6\n1 5\n3 5\n3 1\n5 3\n3 2\n5 1\n4 6\n6 6",
"output": "Mishka"
},
{
"input": "55\n6 6\n6 5\n2 2\n2 2\n6 4\n5 5\n6 5\n5 3\n1 3\n2 2\n5 6\n3 3\n3 3\n6 5\n3 5\n5 5\n1 2\n1 1\n4 6\n1 2\n5 5\n6 2\n6 3\n1 2\n5 1\n1 3\n3 3\n4 4\n2 5\n1 1\n5 3\n4 3\n2 2\n4 5\n5 6\n4 5\n6 3\n1 6\n6 4\n3 6\n1 6\n5 2\n6 3\n2 3\n5 5\n4 3\n3 1\n4 2\n1 1\n2 5\n5 3\n2 2\n6 3\n4 5\n2 2",
"output": "Mishka"
},
{
"input": "92\n2 3\n1 3\n2 6\n5 1\n5 5\n3 2\n5 6\n2 5\n3 1\n3 6\n4 5\n2 5\n1 2\n2 3\n6 5\n3 6\n4 4\n6 2\n4 5\n4 4\n5 1\n6 1\n3 4\n3 5\n6 6\n3 2\n6 4\n2 2\n3 5\n6 4\n6 3\n6 6\n3 4\n3 3\n6 1\n5 4\n6 2\n2 6\n5 6\n1 4\n4 6\n6 3\n3 1\n4 1\n6 6\n3 5\n6 3\n6 1\n1 6\n3 2\n6 6\n4 3\n3 4\n1 3\n3 5\n5 3\n6 5\n4 3\n5 5\n4 1\n1 5\n6 4\n2 3\n2 3\n1 5\n1 2\n5 2\n4 3\n3 6\n5 5\n5 4\n1 4\n3 3\n1 6\n5 6\n5 4\n5 3\n1 1\n6 2\n5 5\n2 5\n4 3\n6 6\n5 1\n1 1\n4 6\n4 6\n3 1\n6 4\n2 4\n2 2\n2 1",
"output": "Chris"
},
{
"input": "79\n5 3\n4 6\n3 6\n2 1\n5 2\n2 3\n4 4\n6 2\n2 5\n1 6\n6 6\n2 6\n3 3\n4 5\n6 2\n2 1\n1 5\n5 1\n2 1\n2 6\n5 3\n6 2\n2 6\n2 3\n1 5\n4 4\n6 3\n5 2\n3 2\n1 3\n1 3\n6 3\n2 6\n3 6\n5 3\n4 5\n6 1\n3 5\n3 5\n6 5\n1 5\n4 2\n6 2\n2 3\n4 6\n3 6\n2 5\n4 4\n1 1\n4 6\n2 6\n6 4\n3 2\n4 1\n1 2\n6 4\n5 6\n1 4\n2 2\n5 4\n3 2\n1 2\n2 4\n2 5\n2 1\n3 6\n3 3\n1 1\n2 2\n4 4\n4 5\n3 3\n5 3\n6 2\n4 5\n6 5\n2 5\n5 6\n2 2",
"output": "Chris"
},
{
"input": "65\n1 1\n5 1\n2 2\n5 4\n4 5\n2 5\n3 2\n5 6\n6 3\n1 1\n6 1\n1 5\n1 1\n5 2\n6 4\n1 6\n1 1\n4 3\n2 3\n5 6\n4 4\n6 2\n1 3\n4 3\n1 3\n6 3\n3 5\n4 2\n4 1\n6 1\n3 2\n2 6\n3 2\n3 5\n6 3\n4 3\n1 5\n2 6\n1 3\n4 1\n4 1\n2 5\n2 5\n6 2\n5 3\n3 1\n3 3\n5 1\n2 4\n5 3\n3 3\n1 1\n6 3\n3 3\n5 1\n1 6\n4 5\n6 6\n5 5\n2 5\n4 1\n2 2\n1 4\n1 6\n6 5",
"output": "Mishka"
},
{
"input": "1\n1 1",
"output": "Friendship is magic!^^"
}
] | 1,671,529,672 | 2,147,483,647 | Python 3 | OK | TESTS | 69 | 46 | 0 | N = int(input())
m = 0
c = 0
for i in range (N):
m1,c1 = map(int, input().split())
if (m1>c1):
m += 1
if (m1<c1):
c += 1
if m<c:
print("Chris")
if c<m:
print("Mishka")
if c==m:
print("Friendship is magic!^^") | Title: Mishka and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
Input Specification:
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
Output Specification:
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
Demo Input:
['3\n3 5\n2 1\n4 2\n', '2\n6 1\n1 6\n', '3\n1 5\n3 3\n2 2\n']
Demo Output:
['Mishka', 'Friendship is magic!^^', 'Chris']
Note:
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris. | ```python
N = int(input())
m = 0
c = 0
for i in range (N):
m1,c1 = map(int, input().split())
if (m1>c1):
m += 1
if (m1<c1):
c += 1
if m<c:
print("Chris")
if c<m:
print("Mishka")
if c==m:
print("Friendship is magic!^^")
``` | 3 | |
302 | B | Eugeny and Play List | PROGRAMMING | 1,200 | [
"binary search",
"implementation",
"two pointers"
] | null | null | Eugeny loves listening to music. He has *n* songs in his play list. We know that song number *i* has the duration of *t**i* minutes. Eugeny listens to each song, perhaps more than once. He listens to song number *i* *c**i* times. Eugeny's play list is organized as follows: first song number 1 plays *c*1 times, then song number 2 plays *c*2 times, ..., in the end the song number *n* plays *c**n* times.
Eugeny took a piece of paper and wrote out *m* moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment *x* means that Eugeny wants to know which song was playing during the *x*-th minute of his listening to the play list.
Help Eugeny and calculate the required numbers of songs. | The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105). The next *n* lines contain pairs of integers. The *i*-th line contains integers *c**i*,<=*t**i* (1<=≤<=*c**i*,<=*t**i*<=≤<=109) — the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 109 .
The next line contains *m* positive integers *v*1,<=*v*2,<=...,<=*v**m*, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time *v**i*, when the music doesn't play any longer. It is guaranteed that *v**i*<=<<=*v**i*<=+<=1 (*i*<=<<=*m*).
The moment of time *v**i* means that Eugeny wants to know which song was playing during the *v**i*-th munite from the start of listening to the playlist. | Print *m* integers — the *i*-th number must equal the number of the song that was playing during the *v**i*-th minute after Eugeny started listening to the play list. | [
"1 2\n2 8\n1 16\n",
"4 9\n1 2\n2 1\n1 1\n2 2\n1 2 3 4 5 6 7 8 9\n"
] | [
"1\n1\n",
"1\n1\n2\n2\n3\n4\n4\n4\n4\n"
] | none | 1,000 | [
{
"input": "1 2\n2 8\n1 16",
"output": "1\n1"
},
{
"input": "4 9\n1 2\n2 1\n1 1\n2 2\n1 2 3 4 5 6 7 8 9",
"output": "1\n1\n2\n2\n3\n4\n4\n4\n4"
},
{
"input": "3 3\n2 8\n5 1\n10 5\n13 16 62",
"output": "1\n1\n3"
},
{
"input": "4 4\n2 8\n2 2\n6 3\n8 7\n13 23 29 85",
"output": "1\n3\n3\n4"
},
{
"input": "5 5\n9 6\n8 7\n2 9\n10 3\n8 10\n69 95 146 162 177",
"output": "2\n2\n4\n5\n5"
},
{
"input": "6 6\n4 9\n8 5\n3 8\n8 10\n4 2\n10 9\n15 45 97 197 231 265",
"output": "1\n2\n3\n6\n6\n6"
},
{
"input": "7 7\n1 10\n1 1\n7 2\n4 9\n10 4\n5 5\n7 1\n48 71 86 87 110 113 127",
"output": "4\n5\n5\n5\n6\n6\n7"
},
{
"input": "8 8\n4 6\n10 9\n5 1\n8 7\n4 7\n2 6\n5 3\n1 10\n21 91 93 142 145 157 181 206",
"output": "1\n2\n2\n4\n4\n4\n5\n6"
},
{
"input": "9 9\n2 5\n7 1\n8 2\n8 8\n8 8\n4 4\n6 10\n10 9\n2 9\n1 10 36 48 76 151 229 276 310",
"output": "1\n1\n4\n4\n4\n5\n7\n8\n8"
},
{
"input": "10 10\n3 1\n2 7\n5 1\n7 2\n9 10\n9 5\n2 5\n4 10\n9 9\n1 9\n26 34 37 141 146 201 239 245 296 299",
"output": "4\n4\n5\n6\n6\n8\n9\n9\n9\n9"
},
{
"input": "2 3\n1 500000000\n1 500000000\n499999999 500000000 500000001",
"output": "1\n1\n2"
},
{
"input": "2 3\n500000000 1\n1 500000000\n499999999 500000000 500000001",
"output": "1\n1\n2"
},
{
"input": "2 3\n500000000 1\n500000000 1\n499999999 500000000 500000001",
"output": "1\n1\n2"
},
{
"input": "3 1\n2 50\n1 50\n1 50\n160",
"output": "3"
}
] | 1,616,802,587 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 11 | 2,000 | 11,366,400 | qtd_musicas, qtd_momentos = [int(i) for i in input().split()]
musicas = [0] # lista de tuplas (vezes, tempo)
aux = 0
for i in range(qtd_musicas):
vezes, tempo = [int(i) for i in input().split()]
aux += vezes * tempo
musicas.append(aux)
momentos = [int(i) for i in input().split()]
for i in momentos:
for j in range(1, len(musicas)):
if i > musicas[j - 1] and i <= musicas[j]:
print(j)
break
| Title: Eugeny and Play List
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eugeny loves listening to music. He has *n* songs in his play list. We know that song number *i* has the duration of *t**i* minutes. Eugeny listens to each song, perhaps more than once. He listens to song number *i* *c**i* times. Eugeny's play list is organized as follows: first song number 1 plays *c*1 times, then song number 2 plays *c*2 times, ..., in the end the song number *n* plays *c**n* times.
Eugeny took a piece of paper and wrote out *m* moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment *x* means that Eugeny wants to know which song was playing during the *x*-th minute of his listening to the play list.
Help Eugeny and calculate the required numbers of songs.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105). The next *n* lines contain pairs of integers. The *i*-th line contains integers *c**i*,<=*t**i* (1<=≤<=*c**i*,<=*t**i*<=≤<=109) — the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 109 .
The next line contains *m* positive integers *v*1,<=*v*2,<=...,<=*v**m*, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time *v**i*, when the music doesn't play any longer. It is guaranteed that *v**i*<=<<=*v**i*<=+<=1 (*i*<=<<=*m*).
The moment of time *v**i* means that Eugeny wants to know which song was playing during the *v**i*-th munite from the start of listening to the playlist.
Output Specification:
Print *m* integers — the *i*-th number must equal the number of the song that was playing during the *v**i*-th minute after Eugeny started listening to the play list.
Demo Input:
['1 2\n2 8\n1 16\n', '4 9\n1 2\n2 1\n1 1\n2 2\n1 2 3 4 5 6 7 8 9\n']
Demo Output:
['1\n1\n', '1\n1\n2\n2\n3\n4\n4\n4\n4\n']
Note:
none | ```python
qtd_musicas, qtd_momentos = [int(i) for i in input().split()]
musicas = [0] # lista de tuplas (vezes, tempo)
aux = 0
for i in range(qtd_musicas):
vezes, tempo = [int(i) for i in input().split()]
aux += vezes * tempo
musicas.append(aux)
momentos = [int(i) for i in input().split()]
for i in momentos:
for j in range(1, len(musicas)):
if i > musicas[j - 1] and i <= musicas[j]:
print(j)
break
``` | 0 | |
26 | A | Almost Prime | PROGRAMMING | 900 | [
"number theory"
] | A. Almost Prime | 2 | 256 | A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive. | Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000). | Output the amount of almost prime numbers between 1 and *n*, inclusive. | [
"10\n",
"21\n"
] | [
"2\n",
"8\n"
] | none | 500 | [
{
"input": "10",
"output": "2"
},
{
"input": "21",
"output": "8"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "8",
"output": "1"
},
{
"input": "19",
"output": "6"
},
{
"input": "40",
"output": "19"
},
{
"input": "77",
"output": "41"
},
{
"input": "222",
"output": "125"
},
{
"input": "987",
"output": "501"
},
{
"input": "1000",
"output": "508"
},
{
"input": "2000",
"output": "958"
},
{
"input": "3000",
"output": "1375"
},
{
"input": "2999",
"output": "1375"
},
{
"input": "2998",
"output": "1375"
},
{
"input": "2997",
"output": "1374"
},
{
"input": "1429",
"output": "706"
},
{
"input": "1673",
"output": "808"
},
{
"input": "1500",
"output": "732"
},
{
"input": "500",
"output": "266"
},
{
"input": "856",
"output": "439"
}
] | 1,538,241,174 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 11 | 2,000 | 0 | def number():
num = int(input())
if num < 6:
return 0
res=0
for i in range(6, num+1):
almost = 0
for j in range(2, i+1):
primo = True
for k in range(2, j):
if j % k == 0:
primo = False
break
if primo:
if i % j == 0:
almost = almost+1
if almost == 3:
break
if almost == 2:
res = res+1
return res
print(str(number())) | Title: Almost Prime
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive.
Input Specification:
Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000).
Output Specification:
Output the amount of almost prime numbers between 1 and *n*, inclusive.
Demo Input:
['10\n', '21\n']
Demo Output:
['2\n', '8\n']
Note:
none | ```python
def number():
num = int(input())
if num < 6:
return 0
res=0
for i in range(6, num+1):
almost = 0
for j in range(2, i+1):
primo = True
for k in range(2, j):
if j % k == 0:
primo = False
break
if primo:
if i % j == 0:
almost = almost+1
if almost == 3:
break
if almost == 2:
res = res+1
return res
print(str(number()))
``` | 0 |
171 | A | Mysterious numbers - 1 | PROGRAMMING | 1,200 | [
"*special",
"constructive algorithms"
] | null | null | The input contains two integers *a*1,<=*a*2 (0<=≤<=*a**i*<=≤<=109), separated by a single space.
Output a single integer. | The input contains two integers *a*1,<=*a*2 (0<=≤<=*a**i*<=≤<=109), separated by a single space. | Output a single integer. | [
"3 14\n",
"27 12\n",
"100 200\n"
] | [
"44\n",
"48\n",
"102\n"
] | none | 0 | [
{
"input": "3 14",
"output": "44"
},
{
"input": "27 12",
"output": "48"
},
{
"input": "100 200",
"output": "102"
},
{
"input": "0 0",
"output": "0"
},
{
"input": "31415 92653",
"output": "67044"
},
{
"input": "1000000000 0",
"output": "1000000000"
},
{
"input": "0 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000",
"output": "1000000001"
},
{
"input": "804289383 846930886",
"output": "1492329031"
},
{
"input": "681692777 714636915",
"output": "1201329194"
},
{
"input": "957747793 424238335",
"output": "1491580217"
},
{
"input": "719885386 649760492",
"output": "1013953332"
},
{
"input": "596516649 189641421",
"output": "720663630"
},
{
"input": "25202362 350490027",
"output": "745296415"
},
{
"input": "783368690 102520059",
"output": "1733393891"
},
{
"input": "44897763 967513926",
"output": "674213532"
},
{
"input": "365180540 540383426",
"output": "989563585"
},
{
"input": "304089172 303455736",
"output": "941643475"
},
{
"input": "35005211 521595368",
"output": "898600336"
},
{
"input": "294702567 726956429",
"output": "1219362194"
},
{
"input": "336465782 861021530",
"output": "371585950"
},
{
"input": "278722862 233665123",
"output": "600289194"
},
{
"input": "145174067 468703135",
"output": "676481931"
},
{
"input": "101513929 801979802",
"output": "310493037"
},
{
"input": "315634022 635723058",
"output": "1165961558"
},
{
"input": "369133069 125898167",
"output": "1131031590"
},
{
"input": "59961393 89018456",
"output": "125442491"
},
{
"input": "628175011 656478042",
"output": "869049667"
},
{
"input": "131176229 653377373",
"output": "504949585"
},
{
"input": "859484421 914544919",
"output": "1778929840"
},
{
"input": "608413784 756898537",
"output": "1344312441"
},
{
"input": "734575198 973594324",
"output": "1158070577"
},
{
"input": "149798315 38664370",
"output": "157144998"
},
{
"input": "129566413 184803526",
"output": "754874894"
},
{
"input": "412776091 424268980",
"output": "502638515"
},
{
"input": "911759956 749241873",
"output": "1289902903"
},
{
"input": "137806862 42999170",
"output": "145006786"
},
{
"input": "982906996 135497281",
"output": "1165701527"
},
{
"input": "511702305 84420925",
"output": "564604753"
},
{
"input": "937477084 827336327",
"output": "1661110812"
},
{
"input": "1000000000 999999999",
"output": "1999999999"
}
] | 1,490,865,387 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 92 | 4,608,000 | a=input()
b=input()[::-1]
print(int(a)+int(b)) | Title: Mysterious numbers - 1
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The input contains two integers *a*1,<=*a*2 (0<=≤<=*a**i*<=≤<=109), separated by a single space.
Output a single integer.
Input Specification:
The input contains two integers *a*1,<=*a*2 (0<=≤<=*a**i*<=≤<=109), separated by a single space.
Output Specification:
Output a single integer.
Demo Input:
['3 14\n', '27 12\n', '100 200\n']
Demo Output:
['44\n', '48\n', '102\n']
Note:
none | ```python
a=input()
b=input()[::-1]
print(int(a)+int(b))
``` | -1 | |
393 | A | Nineteen | PROGRAMMING | 0 | [] | null | null | Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.
For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.
Help her to find the maximum number of "nineteen"s that she can get in her string. | The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100. | Print a single integer — the maximum number of "nineteen"s that she can get in her string. | [
"nniinneetteeeenn\n",
"nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n",
"nineteenineteen\n"
] | [
"2",
"2",
"2"
] | none | 500 | [
{
"input": "nniinneetteeeenn",
"output": "2"
},
{
"input": "nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii",
"output": "2"
},
{
"input": "nineteenineteen",
"output": "2"
},
{
"input": "nssemsnnsitjtihtthij",
"output": "0"
},
{
"input": "eehihnttehtherjsihihnrhimihrjinjiehmtjimnrss",
"output": "1"
},
{
"input": "rrrteiehtesisntnjirtitijnjjjthrsmhtneirjimniemmnrhirssjnhetmnmjejjnjjritjttnnrhnjs",
"output": "2"
},
{
"input": "mmrehtretseihsrjmtsenemniehssnisijmsnntesismmtmthnsieijjjnsnhisi",
"output": "2"
},
{
"input": "hshretttnntmmiertrrnjihnrmshnthirnnirrheinnnrjiirshthsrsijtrrtrmnjrrjnresnintnmtrhsnjrinsseimn",
"output": "1"
},
{
"input": "snmmensntritetnmmmerhhrmhnehehtesmhthseemjhmnrti",
"output": "2"
},
{
"input": "rmeetriiitijmrenmeiijt",
"output": "0"
},
{
"input": "ihimeitimrmhriemsjhrtjtijtesmhemnmmrsetmjttthtjhnnmirtimne",
"output": "1"
},
{
"input": "rhtsnmnesieernhstjnmmirthhieejsjttsiierhihhrrijhrrnejsjer",
"output": "2"
},
{
"input": "emmtjsjhretehmiiiestmtmnmissjrstnsnjmhimjmststsitemtttjrnhsrmsenjtjim",
"output": "2"
},
{
"input": "nmehhjrhirniitshjtrrtitsjsntjhrstjehhhrrerhemehjeermhmhjejjesnhsiirheijjrnrjmminneeehtm",
"output": "3"
},
{
"input": "hsntijjetmehejtsitnthietssmeenjrhhetsnjrsethisjrtrhrierjtmimeenjnhnijeesjttrmn",
"output": "3"
},
{
"input": "jnirirhmirmhisemittnnsmsttesjhmjnsjsmntisheneiinsrjsjirnrmnjmjhmistntersimrjni",
"output": "1"
},
{
"input": "neithjhhhtmejjnmieishethmtetthrienrhjmjenrmtejerernmthmsnrthhtrimmtmshm",
"output": "2"
},
{
"input": "sithnrsnemhijsnjitmijjhejjrinejhjinhtisttteermrjjrtsirmessejireihjnnhhemiirmhhjeet",
"output": "3"
},
{
"input": "jrjshtjstteh",
"output": "0"
},
{
"input": "jsihrimrjnnmhttmrtrenetimemjnshnimeiitmnmjishjjneisesrjemeshjsijithtn",
"output": "2"
},
{
"input": "hhtjnnmsemermhhtsstejehsssmnesereehnnsnnremjmmieethmirjjhn",
"output": "2"
},
{
"input": "tmnersmrtsehhntsietttrehrhneiireijnijjejmjhei",
"output": "1"
},
{
"input": "mtstiresrtmesritnjriirehtermtrtseirtjrhsejhhmnsineinsjsin",
"output": "2"
},
{
"input": "ssitrhtmmhtnmtreijteinimjemsiiirhrttinsnneshintjnin",
"output": "1"
},
{
"input": "rnsrsmretjiitrjthhritniijhjmm",
"output": "0"
},
{
"input": "hntrteieimrimteemenserntrejhhmijmtjjhnsrsrmrnsjseihnjmehtthnnithirnhj",
"output": "3"
},
{
"input": "nmmtsmjrntrhhtmimeresnrinstjnhiinjtnjjjnthsintmtrhijnrnmtjihtinmni",
"output": "0"
},
{
"input": "eihstiirnmteejeehimttrijittjsntjejmessstsemmtristjrhenithrrsssihnthheehhrnmimssjmejjreimjiemrmiis",
"output": "2"
},
{
"input": "srthnimimnemtnmhsjmmmjmmrsrisehjseinemienntetmitjtnnneseimhnrmiinsismhinjjnreehseh",
"output": "3"
},
{
"input": "etrsmrjehntjjimjnmsresjnrthjhehhtreiijjminnheeiinseenmmethiemmistsei",
"output": "3"
},
{
"input": "msjeshtthsieshejsjhsnhejsihisijsertenrshhrthjhiirijjneinjrtrmrs",
"output": "1"
},
{
"input": "mehsmstmeejrhhsjihntjmrjrihssmtnensttmirtieehimj",
"output": "1"
},
{
"input": "mmmsermimjmrhrhejhrrejermsneheihhjemnehrhihesnjsehthjsmmjeiejmmnhinsemjrntrhrhsmjtttsrhjjmejj",
"output": "2"
},
{
"input": "rhsmrmesijmmsnsmmhertnrhsetmisshriirhetmjihsmiinimtrnitrseii",
"output": "1"
},
{
"input": "iihienhirmnihh",
"output": "0"
},
{
"input": "ismtthhshjmhisssnmnhe",
"output": "0"
},
{
"input": "rhsmnrmhejshinnjrtmtsssijimimethnm",
"output": "0"
},
{
"input": "eehnshtiriejhiirntminrirnjihmrnittnmmnjejjhjtennremrnssnejtntrtsiejjijisermj",
"output": "3"
},
{
"input": "rnhmeesnhttrjintnhnrhristjrthhrmehrhjmjhjehmstrijemjmmistes",
"output": "2"
},
{
"input": "ssrmjmjeeetrnimemrhimes",
"output": "0"
},
{
"input": "n",
"output": "0"
},
{
"input": "ni",
"output": "0"
},
{
"input": "nine",
"output": "0"
},
{
"input": "nineteenineteenineteenineteenineteenineteenineteenineteenineteenineteenineteenineteenineteen",
"output": "13"
},
{
"input": "ninetee",
"output": "0"
},
{
"input": "mzbmweyydiadtlcouegmdbyfwurpwbpuvhifnuapwynd",
"output": "0"
},
{
"input": "zenudggmyopddhszhrbmftgzmjorabhgojdtfnzxjkayjlkgczsyshczutkdch",
"output": "0"
},
{
"input": "rtzxovxqfapkdmelxiyjroohufhbakpmmvaxq",
"output": "0"
},
{
"input": "zninetneeineteeniwnteeennieteenineteenineteenineteenineteenineteenineteenineteenineteeninetzeenz",
"output": "13"
},
{
"input": "nnnnnnniiiiiiiiiiiitttttttttteeeeeeeeeeeeeeeeee",
"output": "3"
},
{
"input": "ttttiiiieeeeeeeeeeeennnnnnnnn",
"output": "4"
},
{
"input": "ttttttttteeeeeeeeeeeeeeeeeeeeeiiiiiiiiiiiinnnnnnn",
"output": "3"
},
{
"input": "nnnnnnnnnneeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeiiiiiiiiiiiiiiiiiiiitttttttttttttttttttt",
"output": "4"
},
{
"input": "eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeiiiiiiiiiiiiiiiiittttttttttttttttnnnnnnn",
"output": "3"
},
{
"input": "nineeen",
"output": "0"
},
{
"input": "nineteeeeeeeeeeeeeeeeettttttttttttttttttiiiiiiiiiiiiiiiiii",
"output": "0"
},
{
"input": "nineteenieteenieteenieteenieteenieteenieteen",
"output": "4"
},
{
"input": "nineteenineteenineteenineteenineteen",
"output": "5"
}
] | 1,631,025,243 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 46 | 6,963,200 | s=input()
c1=0
c2=0
c3=0
c4=0
for i in s:
if i=='n':
c1+=1
elif i=='i':
c2+=1
elif i=='e':
c3+=1
elif i=='t':
c4+=1
if c1>=3 and c2>=1 and c3>=3 and c4>=1 and c1%3==0:
print(int(min((c1/3), (c2), (c3/3), (c4))))
elif c1>=3 and c2>=1 and c3>=3 and c4>=1 and (c1+1)%3==0:
print(int(min(((c1+1)/3), (c2), (c3/3), (c4))))
| Title: Nineteen
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.
For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.
Help her to find the maximum number of "nineteen"s that she can get in her string.
Input Specification:
The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100.
Output Specification:
Print a single integer — the maximum number of "nineteen"s that she can get in her string.
Demo Input:
['nniinneetteeeenn\n', 'nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n', 'nineteenineteen\n']
Demo Output:
['2', '2', '2']
Note:
none | ```python
s=input()
c1=0
c2=0
c3=0
c4=0
for i in s:
if i=='n':
c1+=1
elif i=='i':
c2+=1
elif i=='e':
c3+=1
elif i=='t':
c4+=1
if c1>=3 and c2>=1 and c3>=3 and c4>=1 and c1%3==0:
print(int(min((c1/3), (c2), (c3/3), (c4))))
elif c1>=3 and c2>=1 and c3>=3 and c4>=1 and (c1+1)%3==0:
print(int(min(((c1+1)/3), (c2), (c3/3), (c4))))
``` | 0 | |
935 | A | Fafa and his Company | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly. | The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company. | Print a single integer representing the answer to the problem. | [
"2\n",
"10\n"
] | [
"1\n",
"3\n"
] | In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them. | 500 | [
{
"input": "2",
"output": "1"
},
{
"input": "10",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "6",
"output": "3"
},
{
"input": "13",
"output": "1"
},
{
"input": "100000",
"output": "35"
},
{
"input": "1024",
"output": "10"
},
{
"input": "99999",
"output": "11"
},
{
"input": "10007",
"output": "1"
},
{
"input": "4096",
"output": "12"
},
{
"input": "65536",
"output": "16"
},
{
"input": "40320",
"output": "95"
},
{
"input": "30030",
"output": "63"
},
{
"input": "161",
"output": "3"
},
{
"input": "1000",
"output": "15"
},
{
"input": "10000",
"output": "24"
},
{
"input": "777",
"output": "7"
},
{
"input": "121",
"output": "2"
},
{
"input": "25",
"output": "2"
},
{
"input": "40000",
"output": "34"
},
{
"input": "99990",
"output": "47"
},
{
"input": "98765",
"output": "3"
},
{
"input": "56789",
"output": "3"
},
{
"input": "13579",
"output": "3"
},
{
"input": "97531",
"output": "3"
},
{
"input": "12345",
"output": "7"
},
{
"input": "54321",
"output": "7"
},
{
"input": "83160",
"output": "127"
},
{
"input": "9",
"output": "2"
},
{
"input": "21",
"output": "3"
},
{
"input": "11",
"output": "1"
},
{
"input": "15",
"output": "3"
}
] | 1,636,804,742 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 61 | 4,300,800 | n=int(input())
p=0
for i in range(1,n):
if n%i==0:
p+=1
print(p)
| Title: Fafa and his Company
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
Input Specification:
The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company.
Output Specification:
Print a single integer representing the answer to the problem.
Demo Input:
['2\n', '10\n']
Demo Output:
['1\n', '3\n']
Note:
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them. | ```python
n=int(input())
p=0
for i in range(1,n):
if n%i==0:
p+=1
print(p)
``` | 3 | |
676 | A | Nicholas and Permutation | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | Nicholas has an array *a* that contains *n* distinct integers from 1 to *n*. In other words, Nicholas has a permutation of size *n*.
Nicholas want the minimum element (integer 1) and the maximum element (integer *n*) to be as far as possible from each other. He wants to perform exactly one swap in order to maximize the distance between the minimum and the maximum elements. The distance between two elements is considered to be equal to the absolute difference between their positions. | The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100) — the size of the permutation.
The second line of the input contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*), where *a**i* is equal to the element at the *i*-th position. | Print a single integer — the maximum possible distance between the minimum and the maximum elements Nicholas can achieve by performing exactly one swap. | [
"5\n4 5 1 3 2\n",
"7\n1 6 5 3 4 7 2\n",
"6\n6 5 4 3 2 1\n"
] | [
"3\n",
"6\n",
"5\n"
] | In the first sample, one may obtain the optimal answer by swapping elements 1 and 2.
In the second sample, the minimum and the maximum elements will be located in the opposite ends of the array if we swap 7 and 2.
In the third sample, the distance between the minimum and the maximum elements is already maximum possible, so we just perform some unnecessary swap, for example, one can swap 5 and 2. | 500 | [
{
"input": "5\n4 5 1 3 2",
"output": "3"
},
{
"input": "7\n1 6 5 3 4 7 2",
"output": "6"
},
{
"input": "6\n6 5 4 3 2 1",
"output": "5"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n2 3 1",
"output": "2"
},
{
"input": "4\n4 1 3 2",
"output": "3"
},
{
"input": "5\n1 4 5 2 3",
"output": "4"
},
{
"input": "6\n4 6 3 5 2 1",
"output": "5"
},
{
"input": "7\n1 5 3 6 2 4 7",
"output": "6"
},
{
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"input": "100\n41 67 94 18 14 83 59 12 19 54 13 68 75 26 15 65 80 40 23 30 34 78 47 21 63 79 4 70 3 31 86 69 92 10 61 74 97 100 9 99 32 27 91 55 85 52 16 17 28 1 64 29 58 76 98 25 84 7 2 96 20 72 36 46 49 82 93 44 45 6 38 87 57 50 53 35 60 33 8 89 39 42 37 48 62 81 73 43 95 11 66 88 90 22 24 77 71 51 5 56",
"output": "62"
},
{
"input": "100\n1 88 38 56 62 99 39 80 12 33 57 24 28 84 37 42 10 95 83 58 8 40 20 2 30 78 60 79 36 71 51 31 27 65 22 47 6 19 61 94 75 4 74 35 15 23 92 9 70 13 11 59 90 18 66 81 64 72 16 32 34 67 46 91 21 87 77 97 82 41 7 86 26 43 45 3 93 17 52 96 50 63 48 5 53 44 29 25 98 54 49 14 73 69 89 55 76 85 68 100",
"output": "99"
},
{
"input": "100\n22 59 25 77 68 79 32 45 20 28 61 60 38 86 33 10 100 15 53 75 78 39 67 13 66 34 96 4 63 23 73 29 31 35 71 55 16 14 72 56 94 97 17 93 47 84 57 8 21 51 54 85 26 76 49 81 2 92 62 44 91 87 11 24 95 69 5 7 99 6 65 48 70 12 41 18 74 27 42 3 80 30 50 98 58 37 82 89 83 36 40 52 19 9 88 46 43 1 90 64",
"output": "97"
},
{
"input": "100\n12 1 76 78 97 82 59 80 48 8 91 51 54 74 16 10 89 99 83 63 93 90 55 25 30 33 29 6 9 65 92 79 44 39 15 58 37 46 32 19 27 3 75 49 62 71 98 42 69 50 26 81 96 5 7 61 60 21 20 36 18 34 40 4 47 85 64 38 22 84 2 68 11 56 31 66 17 14 95 43 53 35 23 52 70 13 72 45 41 77 73 87 88 94 28 86 24 67 100 57",
"output": "98"
},
{
"input": "100\n66 100 53 88 7 73 54 41 31 42 8 46 65 90 78 14 94 30 79 39 89 5 83 50 38 61 37 86 22 95 60 98 34 57 91 10 75 25 15 43 23 17 96 35 93 48 87 47 56 13 19 9 82 62 67 80 11 55 99 70 18 26 58 85 12 44 16 45 4 49 20 71 92 24 81 2 76 32 6 21 84 36 52 97 59 63 40 51 27 64 68 3 77 72 28 33 29 1 74 69",
"output": "98"
},
{
"input": "100\n56 64 1 95 72 39 9 49 87 29 94 7 32 6 30 48 50 25 31 78 90 45 60 44 80 68 17 20 73 15 75 98 83 13 71 22 36 26 96 88 35 3 85 54 16 41 92 99 69 86 93 33 43 62 77 46 47 37 12 10 18 40 27 4 63 55 28 59 23 34 61 53 76 42 51 91 21 70 8 58 38 19 5 66 84 11 52 24 81 82 79 67 97 65 57 74 2 89 100 14",
"output": "98"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "3\n1 3 2",
"output": "2"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 3 1",
"output": "2"
},
{
"input": "3\n3 1 2",
"output": "2"
},
{
"input": "3\n3 2 1",
"output": "2"
},
{
"input": "4\n1 2 3 4",
"output": "3"
},
{
"input": "4\n1 2 4 3",
"output": "3"
},
{
"input": "4\n1 3 2 4",
"output": "3"
},
{
"input": "4\n1 3 4 2",
"output": "3"
},
{
"input": "4\n1 4 2 3",
"output": "3"
},
{
"input": "4\n1 4 3 2",
"output": "3"
},
{
"input": "4\n2 1 3 4",
"output": "3"
},
{
"input": "4\n2 1 4 3",
"output": "2"
},
{
"input": "4\n2 4 1 3",
"output": "2"
},
{
"input": "4\n2 4 3 1",
"output": "3"
},
{
"input": "4\n3 1 2 4",
"output": "3"
},
{
"input": "4\n3 1 4 2",
"output": "2"
},
{
"input": "4\n3 2 1 4",
"output": "3"
},
{
"input": "4\n3 2 4 1",
"output": "3"
},
{
"input": "4\n3 4 1 2",
"output": "2"
},
{
"input": "4\n3 4 2 1",
"output": "3"
},
{
"input": "4\n4 1 2 3",
"output": "3"
},
{
"input": "4\n4 1 3 2",
"output": "3"
},
{
"input": "4\n4 2 1 3",
"output": "3"
},
{
"input": "4\n4 2 3 1",
"output": "3"
},
{
"input": "4\n4 3 1 2",
"output": "3"
},
{
"input": "4\n4 3 2 1",
"output": "3"
},
{
"input": "8\n2 5 6 4 8 3 1 7",
"output": "6"
},
{
"input": "5\n2 3 1 5 4",
"output": "3"
},
{
"input": "6\n2 5 3 6 4 1",
"output": "5"
},
{
"input": "6\n5 4 2 6 1 3",
"output": "4"
},
{
"input": "6\n4 2 3 1 6 5",
"output": "4"
},
{
"input": "6\n5 4 2 1 6 3",
"output": "4"
},
{
"input": "9\n7 2 3 4 5 6 1 9 8",
"output": "7"
},
{
"input": "6\n3 2 1 4 6 5",
"output": "4"
},
{
"input": "6\n2 3 4 1 6 5",
"output": "4"
},
{
"input": "10\n5 2 3 4 1 6 7 8 10 9",
"output": "8"
},
{
"input": "6\n5 2 3 1 6 4",
"output": "4"
},
{
"input": "10\n2 9 3 4 1 10 5 6 7 8",
"output": "5"
},
{
"input": "10\n2 3 4 5 6 7 1 8 10 9",
"output": "8"
},
{
"input": "8\n2 3 4 5 1 6 8 7",
"output": "6"
},
{
"input": "6\n2 1 3 4 5 6",
"output": "5"
}
] | 1,620,993,214 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 46 | 6,758,400 | n=int(input())
l=list(map(int,input().split()))
a=min(l.index(max(l)),l.index(min(l)))
print(n-a-1) | Title: Nicholas and Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nicholas has an array *a* that contains *n* distinct integers from 1 to *n*. In other words, Nicholas has a permutation of size *n*.
Nicholas want the minimum element (integer 1) and the maximum element (integer *n*) to be as far as possible from each other. He wants to perform exactly one swap in order to maximize the distance between the minimum and the maximum elements. The distance between two elements is considered to be equal to the absolute difference between their positions.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100) — the size of the permutation.
The second line of the input contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*), where *a**i* is equal to the element at the *i*-th position.
Output Specification:
Print a single integer — the maximum possible distance between the minimum and the maximum elements Nicholas can achieve by performing exactly one swap.
Demo Input:
['5\n4 5 1 3 2\n', '7\n1 6 5 3 4 7 2\n', '6\n6 5 4 3 2 1\n']
Demo Output:
['3\n', '6\n', '5\n']
Note:
In the first sample, one may obtain the optimal answer by swapping elements 1 and 2.
In the second sample, the minimum and the maximum elements will be located in the opposite ends of the array if we swap 7 and 2.
In the third sample, the distance between the minimum and the maximum elements is already maximum possible, so we just perform some unnecessary swap, for example, one can swap 5 and 2. | ```python
n=int(input())
l=list(map(int,input().split()))
a=min(l.index(max(l)),l.index(min(l)))
print(n-a-1)
``` | 0 | |
109 | A | Lucky Sum of Digits | PROGRAMMING | 1,000 | [
"brute force",
"implementation"
] | A. Lucky Sum of Digits | 2 | 256 | Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task. | The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number. | Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1. | [
"11\n",
"10\n"
] | [
"47\n",
"-1\n"
] | none | 500 | [
{
"input": "11",
"output": "47"
},
{
"input": "10",
"output": "-1"
},
{
"input": "64",
"output": "4477777777"
},
{
"input": "1",
"output": "-1"
},
{
"input": "4",
"output": "4"
},
{
"input": "7",
"output": "7"
},
{
"input": "12",
"output": "444"
},
{
"input": "1000000",
"output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "47",
"output": "44477777"
},
{
"input": "100",
"output": "4444777777777777"
},
{
"input": "700",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "485",
"output": "44447777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "111",
"output": "444447777777777777"
},
{
"input": "85",
"output": "4477777777777"
},
{
"input": "114",
"output": "444477777777777777"
},
{
"input": "474",
"output": "444777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "74",
"output": "47777777777"
},
{
"input": "1000",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "1024",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "4444",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "45784",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "10000",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
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"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
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},
{
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},
{
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},
{
"input": "3840",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "100000",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "9876",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "99999",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
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},
{
"input": "854759",
"output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "11000",
"output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
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},
{
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},
{
"input": "888887",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999998",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "40008",
"output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "10691",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "-1"
},
{
"input": "9",
"output": "-1"
},
{
"input": "8",
"output": "44"
},
{
"input": "2",
"output": "-1"
},
{
"input": "3",
"output": "-1"
},
{
"input": "999997",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999996",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999990",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999980",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "800000",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "980000",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
}
] | 1,679,354,597 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 6 | 92 | 0 | from math import ceil
x = int(input())
i = 0
limite = 0
if (x % 7 == 0):
print("7" * (x//7))
else:
limite = ceil(x/7)
valor = 4*limite
while(valor != x and i < limite):
valor += 3
i += 1
if (valor != x):
print(-1)
else:
print("4" * (limite - i), "7" * i, sep="") | Title: Lucky Sum of Digits
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.
Output Specification:
Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.
Demo Input:
['11\n', '10\n']
Demo Output:
['47\n', '-1\n']
Note:
none | ```python
from math import ceil
x = int(input())
i = 0
limite = 0
if (x % 7 == 0):
print("7" * (x//7))
else:
limite = ceil(x/7)
valor = 4*limite
while(valor != x and i < limite):
valor += 3
i += 1
if (valor != x):
print(-1)
else:
print("4" * (limite - i), "7" * i, sep="")
``` | 0 |
340 | A | The Wall | PROGRAMMING | 1,200 | [
"math"
] | null | null | Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question. | The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*). | Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink. | [
"2 3 6 18\n"
] | [
"3"
] | Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | 500 | [
{
"input": "2 3 6 18",
"output": "3"
},
{
"input": "4 6 20 201",
"output": "15"
},
{
"input": "15 27 100 10000",
"output": "74"
},
{
"input": "105 60 3456 78910",
"output": "179"
},
{
"input": "1 1 1000 100000",
"output": "99001"
},
{
"input": "3 2 5 5",
"output": "0"
},
{
"input": "555 777 1 1000000",
"output": "257"
},
{
"input": "1000 1000 1 32323",
"output": "32"
},
{
"input": "45 125 93451125 100000000",
"output": "5821"
},
{
"input": "101 171 1 1000000000",
"output": "57900"
},
{
"input": "165 255 69696 1000000000",
"output": "356482"
},
{
"input": "555 777 666013 1000000000",
"output": "257229"
},
{
"input": "23 46 123321 900000000",
"output": "19562537"
},
{
"input": "321 123 15 1000000",
"output": "75"
},
{
"input": "819 1000 9532 152901000",
"output": "186"
},
{
"input": "819 1000 10000 1000000",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "1 2 2 1000003",
"output": "500001"
},
{
"input": "1 1 1 1000000000",
"output": "1000000000"
},
{
"input": "10 15 69 195610342",
"output": "6520342"
},
{
"input": "2 1 1 1000000000",
"output": "500000000"
},
{
"input": "1000 1000 1 20",
"output": "0"
},
{
"input": "1 1 1 2000000000",
"output": "2000000000"
},
{
"input": "1 2 1 2000000000",
"output": "1000000000"
},
{
"input": "2 1 1 2000000000",
"output": "1000000000"
},
{
"input": "2 3 1 1000000000",
"output": "166666666"
},
{
"input": "2 3 1 2000000000",
"output": "333333333"
},
{
"input": "3 7 1 1000000000",
"output": "47619047"
},
{
"input": "1 1 1000000000 2000000000",
"output": "1000000001"
},
{
"input": "2 2 1 2000000000",
"output": "1000000000"
},
{
"input": "1 1 2 2000000000",
"output": "1999999999"
},
{
"input": "3 2 1 2000000000",
"output": "333333333"
},
{
"input": "1 1 2000000000 2000000000",
"output": "1"
},
{
"input": "2 3 7 7",
"output": "0"
},
{
"input": "3 3 3 7",
"output": "2"
}
] | 1,590,978,544 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 186 | 0 | x,y,a,b=map(int,input().split())
i=a//x
lst=[]
r1=0
while r1 < b:
r1=i * x
lst.append(r1)
i+=1
ii=a//y
lst2=[]
r2=0
while r2 < b:
r2=ii * y
lst2.append(r2)
ii+=1
print(len(set(lst).intersection(lst2))) | Title: The Wall
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input Specification:
The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*).
Output Specification:
Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink.
Demo Input:
['2 3 6 18\n']
Demo Output:
['3']
Note:
Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | ```python
x,y,a,b=map(int,input().split())
i=a//x
lst=[]
r1=0
while r1 < b:
r1=i * x
lst.append(r1)
i+=1
ii=a//y
lst2=[]
r2=0
while r2 < b:
r2=ii * y
lst2.append(r2)
ii+=1
print(len(set(lst).intersection(lst2)))
``` | 0 | |
620 | A | Professor GukiZ's Robot | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Professor GukiZ makes a new robot. The robot are in the point with coordinates (*x*1,<=*y*1) and should go to the point (*x*2,<=*y*2). In a single step the robot can change any of its coordinates (maybe both of them) by one (decrease or increase). So the robot can move in one of the 8 directions. Find the minimal number of steps the robot should make to get the finish position. | The first line contains two integers *x*1,<=*y*1 (<=-<=109<=≤<=*x*1,<=*y*1<=≤<=109) — the start position of the robot.
The second line contains two integers *x*2,<=*y*2 (<=-<=109<=≤<=*x*2,<=*y*2<=≤<=109) — the finish position of the robot. | Print the only integer *d* — the minimal number of steps to get the finish position. | [
"0 0\n4 5\n",
"3 4\n6 1\n"
] | [
"5\n",
"3\n"
] | In the first example robot should increase both of its coordinates by one four times, so it will be in position (4, 4). After that robot should simply increase its *y* coordinate and get the finish position.
In the second example robot should simultaneously increase *x* coordinate and decrease *y* coordinate by one three times. | 0 | [
{
"input": "0 0\n4 5",
"output": "5"
},
{
"input": "3 4\n6 1",
"output": "3"
},
{
"input": "0 0\n4 6",
"output": "6"
},
{
"input": "1 1\n-3 -5",
"output": "6"
},
{
"input": "-1 -1\n-10 100",
"output": "101"
},
{
"input": "1 -1\n100 -100",
"output": "99"
},
{
"input": "-1000000000 -1000000000\n1000000000 1000000000",
"output": "2000000000"
},
{
"input": "-1000000000 -1000000000\n0 999999999",
"output": "1999999999"
},
{
"input": "0 0\n2 1",
"output": "2"
},
{
"input": "10 0\n100 0",
"output": "90"
},
{
"input": "1 5\n6 4",
"output": "5"
},
{
"input": "0 0\n5 4",
"output": "5"
},
{
"input": "10 1\n20 1",
"output": "10"
},
{
"input": "1 1\n-3 4",
"output": "4"
},
{
"input": "-863407280 504312726\n786535210 -661703810",
"output": "1649942490"
},
{
"input": "-588306085 -741137832\n341385643 152943311",
"output": "929691728"
},
{
"input": "0 0\n4 0",
"output": "4"
},
{
"input": "93097194 -48405232\n-716984003 -428596062",
"output": "810081197"
},
{
"input": "9 1\n1 1",
"output": "8"
},
{
"input": "4 6\n0 4",
"output": "4"
},
{
"input": "2 4\n5 2",
"output": "3"
},
{
"input": "-100000000 -100000000\n100000000 100000123",
"output": "200000123"
},
{
"input": "5 6\n5 7",
"output": "1"
},
{
"input": "12 16\n12 1",
"output": "15"
},
{
"input": "0 0\n5 1",
"output": "5"
},
{
"input": "0 1\n1 1",
"output": "1"
},
{
"input": "-44602634 913365223\n-572368780 933284951",
"output": "527766146"
},
{
"input": "-2 0\n2 -2",
"output": "4"
},
{
"input": "0 0\n3 1",
"output": "3"
},
{
"input": "-458 2\n1255 4548",
"output": "4546"
},
{
"input": "-5 -4\n-3 -3",
"output": "2"
},
{
"input": "4 5\n7 3",
"output": "3"
},
{
"input": "-1000000000 -999999999\n1000000000 999999998",
"output": "2000000000"
},
{
"input": "-1000000000 -1000000000\n1000000000 -1000000000",
"output": "2000000000"
},
{
"input": "-464122675 -898521847\n656107323 -625340409",
"output": "1120229998"
},
{
"input": "-463154699 -654742385\n-699179052 -789004997",
"output": "236024353"
},
{
"input": "982747270 -593488945\n342286841 -593604186",
"output": "640460429"
},
{
"input": "-80625246 708958515\n468950878 574646184",
"output": "549576124"
},
{
"input": "0 0\n1 0",
"output": "1"
},
{
"input": "109810 1\n2 3",
"output": "109808"
},
{
"input": "-9 0\n9 9",
"output": "18"
},
{
"input": "9 9\n9 9",
"output": "0"
},
{
"input": "1 1\n4 3",
"output": "3"
},
{
"input": "1 2\n45 1",
"output": "44"
},
{
"input": "207558188 -313753260\n-211535387 -721675423",
"output": "419093575"
},
{
"input": "-11 0\n0 0",
"output": "11"
},
{
"input": "-1000000000 1000000000\n1000000000 -1000000000",
"output": "2000000000"
},
{
"input": "0 0\n1 1",
"output": "1"
},
{
"input": "0 0\n0 1",
"output": "1"
},
{
"input": "0 0\n-1 1",
"output": "1"
},
{
"input": "0 0\n-1 0",
"output": "1"
},
{
"input": "0 0\n-1 -1",
"output": "1"
},
{
"input": "0 0\n0 -1",
"output": "1"
},
{
"input": "0 0\n1 -1",
"output": "1"
},
{
"input": "10 90\n90 10",
"output": "80"
},
{
"input": "851016864 573579544\n-761410925 -380746263",
"output": "1612427789"
},
{
"input": "1 9\n9 9",
"output": "8"
},
{
"input": "1000 1000\n1000 1000",
"output": "0"
},
{
"input": "1 9\n9 1",
"output": "8"
},
{
"input": "1 90\n90 90",
"output": "89"
},
{
"input": "100 100\n1000 1000",
"output": "900"
},
{
"input": "-1 0\n0 0",
"output": "1"
},
{
"input": "-750595959 -2984043\n649569876 -749608783",
"output": "1400165835"
},
{
"input": "958048496 712083589\n423286949 810566863",
"output": "534761547"
},
{
"input": "146316710 53945094\n-523054748 147499505",
"output": "669371458"
},
{
"input": "50383856 -596516251\n-802950224 -557916272",
"output": "853334080"
},
{
"input": "-637204864 -280290367\n-119020929 153679771",
"output": "518183935"
},
{
"input": "-100 -100\n-60 -91",
"output": "40"
},
{
"input": "337537326 74909428\n-765558776 167951547",
"output": "1103096102"
},
{
"input": "0 81\n18 90",
"output": "18"
},
{
"input": "283722202 -902633305\n-831696497 -160868946",
"output": "1115418699"
},
{
"input": "1000 1000\n-1000 1000",
"output": "2000"
},
{
"input": "5 6\n4 8",
"output": "2"
},
{
"input": "40572000 597493595\n-935051731 368493185",
"output": "975623731"
},
{
"input": "-5 5\n5 5",
"output": "10"
}
] | 1,606,128,306 | 2,147,483,647 | PyPy 3 | OK | TESTS | 75 | 140 | 0 | x, y=map(int, input().split())
z, t=map(int, input().split())
print(max(abs(x-z), abs(y-t))) | Title: Professor GukiZ's Robot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ makes a new robot. The robot are in the point with coordinates (*x*1,<=*y*1) and should go to the point (*x*2,<=*y*2). In a single step the robot can change any of its coordinates (maybe both of them) by one (decrease or increase). So the robot can move in one of the 8 directions. Find the minimal number of steps the robot should make to get the finish position.
Input Specification:
The first line contains two integers *x*1,<=*y*1 (<=-<=109<=≤<=*x*1,<=*y*1<=≤<=109) — the start position of the robot.
The second line contains two integers *x*2,<=*y*2 (<=-<=109<=≤<=*x*2,<=*y*2<=≤<=109) — the finish position of the robot.
Output Specification:
Print the only integer *d* — the minimal number of steps to get the finish position.
Demo Input:
['0 0\n4 5\n', '3 4\n6 1\n']
Demo Output:
['5\n', '3\n']
Note:
In the first example robot should increase both of its coordinates by one four times, so it will be in position (4, 4). After that robot should simply increase its *y* coordinate and get the finish position.
In the second example robot should simultaneously increase *x* coordinate and decrease *y* coordinate by one three times. | ```python
x, y=map(int, input().split())
z, t=map(int, input().split())
print(max(abs(x-z), abs(y-t)))
``` | 3 | |
884 | B | Japanese Crosswords Strike Back | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it! | The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding. | Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO. | [
"2 4\n1 3\n",
"3 10\n3 3 2\n",
"2 10\n1 3\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "2 4\n1 3",
"output": "NO"
},
{
"input": "3 10\n3 3 2",
"output": "YES"
},
{
"input": "2 10\n1 3",
"output": "NO"
},
{
"input": "1 1\n1",
"output": "YES"
},
{
"input": "1 10\n10",
"output": "YES"
},
{
"input": "1 10000\n10000",
"output": "YES"
},
{
"input": "10 1\n5 78 3 87 4 9 5 8 9 1235",
"output": "NO"
},
{
"input": "3 12\n3 3 3",
"output": "NO"
},
{
"input": "3 9\n2 2 2",
"output": "NO"
},
{
"input": "2 5\n1 1",
"output": "NO"
},
{
"input": "1 2\n1",
"output": "NO"
},
{
"input": "3 13\n3 3 3",
"output": "NO"
},
{
"input": "3 6\n1 1 1",
"output": "NO"
},
{
"input": "1 6\n5",
"output": "NO"
},
{
"input": "3 11\n3 3 2",
"output": "NO"
},
{
"input": "2 6\n1 3",
"output": "NO"
},
{
"input": "3 10\n2 2 2",
"output": "NO"
},
{
"input": "3 8\n2 1 1",
"output": "NO"
},
{
"input": "1 5\n2",
"output": "NO"
},
{
"input": "1 3\n1",
"output": "NO"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "NO"
},
{
"input": "2 10\n4 4",
"output": "NO"
},
{
"input": "2 8\n2 3",
"output": "NO"
},
{
"input": "2 4\n1 1",
"output": "NO"
},
{
"input": "3 10\n1 2 4",
"output": "NO"
},
{
"input": "3 10\n2 1 3",
"output": "NO"
},
{
"input": "2 6\n1 2",
"output": "NO"
},
{
"input": "3 4\n1 1 1",
"output": "NO"
},
{
"input": "3 11\n1 2 4",
"output": "NO"
},
{
"input": "3 12\n3 3 2",
"output": "NO"
},
{
"input": "4 9\n1 1 1 1",
"output": "NO"
},
{
"input": "1 10\n9",
"output": "NO"
},
{
"input": "1 7\n5",
"output": "NO"
},
{
"input": "2 5\n1 2",
"output": "NO"
},
{
"input": "3 8\n1 1 2",
"output": "NO"
},
{
"input": "3 7\n1 1 1",
"output": "NO"
},
{
"input": "1 10\n1",
"output": "NO"
},
{
"input": "3 10\n2 3 4",
"output": "NO"
},
{
"input": "3 9\n1 2 3",
"output": "NO"
},
{
"input": "3 9\n3 3 2",
"output": "NO"
},
{
"input": "3 6\n3 3 2",
"output": "NO"
},
{
"input": "1 1\n3",
"output": "NO"
},
{
"input": "1 3\n2",
"output": "NO"
},
{
"input": "3 10\n3 3 3",
"output": "NO"
},
{
"input": "3 5\n1 1 1",
"output": "YES"
},
{
"input": "2 1\n100 100",
"output": "NO"
},
{
"input": "3 3\n3 3 3",
"output": "NO"
},
{
"input": "4 17\n3 3 9 1",
"output": "NO"
},
{
"input": "4 1660\n1505 13 37 100",
"output": "NO"
},
{
"input": "3 5\n3 3 2",
"output": "NO"
},
{
"input": "4 10\n3 3 2 5",
"output": "NO"
},
{
"input": "5 5\n5 5 5 5 5",
"output": "NO"
},
{
"input": "1 1\n2",
"output": "NO"
},
{
"input": "5 10\n1 2 2 4 5",
"output": "NO"
},
{
"input": "2 1\n1 1",
"output": "NO"
}
] | 1,510,270,637 | 2,147,483,647 | Python 3 | OK | TESTS | 66 | 92 | 7,372,800 | n,x = list(map(int,input().split()))
a = list(map(int,input().split()))
if x - sum(a) == n-1:
print("YES")
else:
print("NO")
| Title: Japanese Crosswords Strike Back
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input Specification:
The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding.
Output Specification:
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Demo Input:
['2 4\n1 3\n', '3 10\n3 3 2\n', '2 10\n1 3\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none | ```python
n,x = list(map(int,input().split()))
a = list(map(int,input().split()))
if x - sum(a) == n-1:
print("YES")
else:
print("NO")
``` | 3 | |
52 | A | 123-sequence | PROGRAMMING | 900 | [
"implementation"
] | A. 123-sequence | 2 | 256 | There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3). | Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal. | [
"9\n1 3 2 2 2 1 1 2 3\n"
] | [
"5\n"
] | In the example all the numbers equal to 1 and 3 should be replaced by 2. | 500 | [
{
"input": "9\n1 3 2 2 2 1 1 2 3",
"output": "5"
},
{
"input": "6\n3 3 2 2 1 3",
"output": "3"
},
{
"input": "12\n3 1 3 1 2 1 3 2 2 1 2 1",
"output": "7"
},
{
"input": "15\n3 2 1 1 1 1 3 2 2 3 3 1 2 3 2",
"output": "10"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "2\n3 2",
"output": "1"
},
{
"input": "2\n3 1",
"output": "1"
},
{
"input": "18\n2 3 2 1 2 3 2 1 2 3 3 3 1 2 3 3 3 2",
"output": "10"
},
{
"input": "30\n2 1 3 2 3 2 2 2 2 3 2 2 3 2 1 1 3 1 3 2 1 2 3 1 1 3 3 1 3 1",
"output": "19"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n3",
"output": "0"
}
] | 1,674,661,826 | 2,147,483,647 | Python 3 | OK | TESTS | 23 | 278 | 11,059,200 | length = int(input())
numbers = tuple(input().split())
a = numbers.count('1')
b = numbers.count('2')
c = numbers.count('3')
print(min((a + b), (a + c), (b + c)))
| Title: 123-sequence
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3).
Output Specification:
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Demo Input:
['9\n1 3 2 2 2 1 1 2 3\n']
Demo Output:
['5\n']
Note:
In the example all the numbers equal to 1 and 3 should be replaced by 2. | ```python
length = int(input())
numbers = tuple(input().split())
a = numbers.count('1')
b = numbers.count('2')
c = numbers.count('3')
print(min((a + b), (a + c), (b + c)))
``` | 3.909901 |
272 | A | Dima and Friends | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space. | In a single line print the answer to the problem. | [
"1\n1\n",
"1\n2\n",
"2\n3 5\n"
] | [
"3\n",
"2\n",
"3\n"
] | In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers. | 500 | [
{
"input": "1\n1",
"output": "3"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "1\n5",
"output": "3"
},
{
"input": "5\n4 4 3 5 1",
"output": "4"
},
{
"input": "6\n2 3 2 2 1 3",
"output": "4"
},
{
"input": "8\n2 2 5 3 4 3 3 2",
"output": "4"
},
{
"input": "7\n4 1 3 2 2 4 5",
"output": "4"
},
{
"input": "3\n3 5 1",
"output": "4"
},
{
"input": "95\n4 2 3 4 4 5 2 2 4 4 3 5 3 3 3 5 4 2 5 4 2 1 1 3 4 2 1 3 5 4 2 1 1 5 1 1 2 2 4 4 5 4 5 5 2 1 2 2 2 4 5 5 2 4 3 4 4 3 5 2 4 1 5 4 5 1 3 2 4 2 2 1 5 3 1 5 3 4 3 3 2 1 2 2 1 3 1 5 2 3 1 1 2 5 2",
"output": "5"
},
{
"input": "31\n3 2 3 3 3 3 4 4 1 5 5 4 2 4 3 2 2 1 4 4 1 2 3 1 1 5 5 3 4 4 1",
"output": "4"
},
{
"input": "42\n3 1 2 2 5 1 2 2 4 5 4 5 2 5 4 5 4 4 1 4 3 3 4 4 4 4 3 2 1 3 4 5 5 2 1 2 1 5 5 2 4 4",
"output": "5"
},
{
"input": "25\n4 5 5 5 3 1 1 4 4 4 3 5 4 4 1 4 4 1 2 4 2 5 4 5 3",
"output": "5"
},
{
"input": "73\n3 4 3 4 5 1 3 4 2 1 4 2 2 3 5 3 1 4 2 3 2 1 4 5 3 5 2 2 4 3 2 2 5 3 2 3 5 1 3 1 1 4 5 2 4 2 5 1 4 3 1 3 1 4 2 3 3 3 3 5 5 2 5 2 5 4 3 1 1 5 5 2 3",
"output": "4"
},
{
"input": "46\n1 4 4 5 4 5 2 3 5 5 3 2 5 4 1 3 2 2 1 4 3 1 5 5 2 2 2 2 4 4 1 1 4 3 4 3 1 4 2 2 4 2 3 2 5 2",
"output": "4"
},
{
"input": "23\n5 2 1 1 4 2 5 5 3 5 4 5 5 1 1 5 2 4 5 3 4 4 3",
"output": "5"
},
{
"input": "6\n4 2 3 1 3 5",
"output": "4"
},
{
"input": "15\n5 5 5 3 5 4 1 3 3 4 3 4 1 4 4",
"output": "5"
},
{
"input": "93\n1 3 1 4 3 3 5 3 1 4 5 4 3 2 2 4 3 1 4 1 2 3 3 3 2 5 1 3 1 4 5 1 1 1 4 2 1 2 3 1 1 1 5 1 5 5 1 2 5 4 3 2 2 4 4 2 5 4 5 5 3 1 3 1 2 1 3 1 1 2 3 4 4 5 5 3 2 1 3 3 5 1 3 5 4 4 1 3 3 4 2 3 2",
"output": "5"
},
{
"input": "96\n1 5 1 3 2 1 2 2 2 2 3 4 1 1 5 4 4 1 2 3 5 1 4 4 4 1 3 3 1 4 5 4 1 3 5 3 4 4 3 2 1 1 4 4 5 1 1 2 5 1 2 3 1 4 1 2 2 2 3 2 3 3 2 5 2 2 3 3 3 3 2 1 2 4 5 5 1 5 3 2 1 4 3 5 5 5 3 3 5 3 4 3 4 2 1 3",
"output": "5"
},
{
"input": "49\n1 4 4 3 5 2 2 1 5 1 2 1 2 5 1 4 1 4 5 2 4 5 3 5 2 4 2 1 3 4 2 1 4 2 1 1 3 3 2 3 5 4 3 4 2 4 1 4 1",
"output": "5"
},
{
"input": "73\n4 1 3 3 3 1 5 2 1 4 1 1 3 5 1 1 4 5 2 1 5 4 1 5 3 1 5 2 4 5 1 4 3 3 5 2 2 3 3 2 5 1 4 5 2 3 1 4 4 3 5 2 3 5 1 4 3 5 1 2 4 1 3 3 5 4 2 4 2 4 1 2 5",
"output": "5"
},
{
"input": "41\n5 3 5 4 2 5 4 3 1 1 1 5 4 3 4 3 5 4 2 5 4 1 1 3 2 4 5 3 5 1 5 5 1 1 1 4 4 1 2 4 3",
"output": "5"
},
{
"input": "100\n3 3 1 4 2 4 4 3 1 5 1 1 4 4 3 4 4 3 5 4 5 2 4 3 4 1 2 4 5 4 2 1 5 4 1 1 4 3 2 4 1 2 1 4 4 5 5 4 4 5 3 2 5 1 4 2 2 1 1 2 5 2 5 1 5 3 1 4 3 2 4 3 2 2 4 5 5 1 2 3 1 4 1 2 2 2 5 5 2 3 2 4 3 1 1 2 1 2 1 2",
"output": "5"
},
{
"input": "100\n2 1 1 3 5 4 4 2 3 4 3 4 5 4 5 4 2 4 5 3 4 5 4 1 1 4 4 1 1 2 5 4 2 4 5 3 2 5 4 3 4 5 1 3 4 2 5 4 5 4 5 2 4 1 2 5 3 1 4 4 5 3 4 3 1 2 5 4 2 5 4 1 5 3 5 4 1 2 5 3 1 1 1 1 5 3 4 3 5 1 1 5 5 1 1 2 2 1 5 1",
"output": "5"
},
{
"input": "100\n4 4 3 3 2 5 4 4 2 1 4 4 4 5 4 1 2 1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3 4 5 5 2 1 2 2 2 2 2 3 2 5 1 2 2 3 2 5 5 1 3 4 5 2 1 3 4 2 2 4 4 3 3 3 2 3 2 1 5 5 5 2 1 4 2 3 5 1 4 4 2 3 2 5 5 4 3 5 1 3 5 5",
"output": "5"
},
{
"input": "100\n4 4 2 5 4 2 2 3 4 4 3 2 3 3 1 3 4 3 3 4 1 3 1 4 5 3 4 3 1 1 1 3 3 2 3 4 3 4 2 2 1 5 1 4 5 1 1 1 3 3 1 1 3 2 5 4 2 5 2 4 5 4 4 1 1 2 1 1 4 5 1 1 5 3 3 2 5 5 5 1 4 1 4 1 1 3 2 3 4 4 2 5 5 2 5 1 1 3 5 3",
"output": "5"
},
{
"input": "100\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "4"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "5"
}
] | 1,608,734,561 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 186 | 204,800 | n=int(input()) #Number of divisors of a and b
divisors = input().split(" ")
divisors = [int(i) for i in divisors]
divisors.sort()
possibleValue = divisors[0]*divisors[n-1]
temp=[]
i=1
while (i*i)<=possibleValue:
if possibleValue%i==0:
temp.append(i)
if possibleValue//i !=i:
temp.append(possibleValue//i)
i+=1
temp.sort()
for i in temp:
divisors.remove(i)
print(temp[0]*temp[len(temp)-1],divisors[0]*divisors[len(divisors)-1])
| Title: Dima and Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
Output Specification:
In a single line print the answer to the problem.
Demo Input:
['1\n1\n', '1\n2\n', '2\n3 5\n']
Demo Output:
['3\n', '2\n', '3\n']
Note:
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers. | ```python
n=int(input()) #Number of divisors of a and b
divisors = input().split(" ")
divisors = [int(i) for i in divisors]
divisors.sort()
possibleValue = divisors[0]*divisors[n-1]
temp=[]
i=1
while (i*i)<=possibleValue:
if possibleValue%i==0:
temp.append(i)
if possibleValue//i !=i:
temp.append(possibleValue//i)
i+=1
temp.sort()
for i in temp:
divisors.remove(i)
print(temp[0]*temp[len(temp)-1],divisors[0]*divisors[len(divisors)-1])
``` | -1 | |
202 | A | LLPS | PROGRAMMING | 800 | [
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null | This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z". | The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10. | Print the lexicographically largest palindromic subsequence of string *s*. | [
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] | [
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] | Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | 500 | [
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
{
"input": "helloworld",
"output": "w"
},
{
"input": "zzzzzzzazz",
"output": "zzzzzzzzz"
},
{
"input": "testcase",
"output": "tt"
},
{
"input": "hahahahaha",
"output": "hhhhh"
},
{
"input": "abbbbbbbbb",
"output": "bbbbbbbbb"
},
{
"input": "zaz",
"output": "zz"
},
{
"input": "aza",
"output": "z"
},
{
"input": "dcbaedcba",
"output": "e"
},
{
"input": "abcdeabcd",
"output": "e"
},
{
"input": "edcbabcde",
"output": "ee"
},
{
"input": "aaaaaaaaab",
"output": "b"
},
{
"input": "testzzzzzz",
"output": "zzzzzz"
},
{
"input": "zzzzzzwait",
"output": "zzzzzz"
},
{
"input": "rrrrrqponm",
"output": "rrrrr"
},
{
"input": "zzyzyy",
"output": "zzz"
},
{
"input": "aababb",
"output": "bbb"
},
{
"input": "zanzibar",
"output": "zz"
},
{
"input": "hhgfedcbaa",
"output": "hh"
},
{
"input": "aabcdefghh",
"output": "hh"
},
{
"input": "aruaru",
"output": "uu"
},
{
"input": "uraura",
"output": "uu"
},
{
"input": "aru",
"output": "u"
},
{
"input": "aburvabur",
"output": "v"
},
{
"input": "ura",
"output": "u"
},
{
"input": "eurottat",
"output": "u"
},
{
"input": "referee",
"output": "rr"
},
{
"input": "joking",
"output": "o"
},
{
"input": "seriously",
"output": "y"
},
{
"input": "sets",
"output": "t"
},
{
"input": "test",
"output": "tt"
},
{
"input": "klmgameklm",
"output": "mmm"
},
{
"input": "dfkjafdkdd",
"output": "kk"
},
{
"input": "zzzzzzzzzz",
"output": "zzzzzzzzzz"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaaaa"
},
{
"input": "buzz",
"output": "zz"
},
{
"input": "b",
"output": "b"
},
{
"input": "y",
"output": "y"
},
{
"input": "yy",
"output": "yy"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "bb",
"output": "bb"
},
{
"input": "aa",
"output": "aa"
},
{
"input": "a",
"output": "a"
},
{
"input": "z",
"output": "z"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "zzazazzzaz",
"output": "zzzzzzz"
},
{
"input": "hellhellhe",
"output": "llll"
},
{
"input": "hellohello",
"output": "oo"
},
{
"input": "refer",
"output": "rr"
}
] | 1,616,939,969 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 154 | 0 | def palindromic_subseq(word):
largest_char = sorted(word)[-1]
return largest_char * (word.count(largest_char))
word = input()
print(palindromic_subseq(word)) | Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | ```python
def palindromic_subseq(word):
largest_char = sorted(word)[-1]
return largest_char * (word.count(largest_char))
word = input()
print(palindromic_subseq(word))
``` | 3 | |
981 | A | Antipalindrome | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into? | The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only. | If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique. | [
"mew\n",
"wuffuw\n",
"qqqqqqqq\n"
] | [
"3\n",
"5\n",
"0\n"
] | "mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$. | 500 | [
{
"input": "mew",
"output": "3"
},
{
"input": "wuffuw",
"output": "5"
},
{
"input": "qqqqqqqq",
"output": "0"
},
{
"input": "ijvji",
"output": "4"
},
{
"input": "iiiiiii",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvvgiaahqmtjyuchikvvhvrebow",
"output": "49"
},
{
"input": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvahheoqleromusrartldojsjvy",
"output": "50"
},
{
"input": "ijvxljt",
"output": "7"
},
{
"input": "fyhcncnchyf",
"output": "10"
},
{
"input": "ffffffffffff",
"output": "0"
},
{
"input": "fyhcncfsepqj",
"output": "12"
},
{
"input": "ybejrrlbcinttnicblrrjeby",
"output": "23"
},
{
"input": "yyyyyyyyyyyyyyyyyyyyyyyyy",
"output": "0"
},
{
"input": "ybejrrlbcintahovgjddrqatv",
"output": "25"
},
{
"input": "oftmhcmclgyqaojljoaqyglcmchmtfo",
"output": "30"
},
{
"input": "oooooooooooooooooooooooooooooooo",
"output": "0"
},
{
"input": "oftmhcmclgyqaojllbotztajglsmcilv",
"output": "32"
},
{
"input": "gxandbtgpbknxvnkjaajknvxnkbpgtbdnaxg",
"output": "35"
},
{
"input": "gggggggggggggggggggggggggggggggggggg",
"output": "0"
},
{
"input": "gxandbtgpbknxvnkjaygommzqitqzjfalfkk",
"output": "36"
},
{
"input": "fcliblymyqckxvieotjooojtoeivxkcqymylbilcf",
"output": "40"
},
{
"input": "fffffffffffffffffffffffffffffffffffffffffff",
"output": "0"
},
{
"input": "fcliblymyqckxvieotjootiqwtyznhhvuhbaixwqnsy",
"output": "43"
},
{
"input": "rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr",
"output": "0"
},
{
"input": "rajccqwqnqmshmerpvjyfepxwpxyldzpzhctqjnstxyfmlhiy",
"output": "49"
},
{
"input": "a",
"output": "0"
},
{
"input": "abca",
"output": "4"
},
{
"input": "aaaaabaaaaa",
"output": "10"
},
{
"input": "aba",
"output": "2"
},
{
"input": "asaa",
"output": "4"
},
{
"input": "aabaa",
"output": "4"
},
{
"input": "aabbaa",
"output": "5"
},
{
"input": "abcdaaa",
"output": "7"
},
{
"input": "aaholaa",
"output": "7"
},
{
"input": "abcdefghijka",
"output": "12"
},
{
"input": "aaadcba",
"output": "7"
},
{
"input": "aaaabaaaa",
"output": "8"
},
{
"input": "abaa",
"output": "4"
},
{
"input": "abcbaa",
"output": "6"
},
{
"input": "ab",
"output": "2"
},
{
"input": "l",
"output": "0"
},
{
"input": "aaaabcaaaa",
"output": "10"
},
{
"input": "abbaaaaaabba",
"output": "11"
},
{
"input": "abaaa",
"output": "5"
},
{
"input": "baa",
"output": "3"
},
{
"input": "aaaaaaabbba",
"output": "11"
},
{
"input": "ccbcc",
"output": "4"
},
{
"input": "bbbaaab",
"output": "7"
},
{
"input": "abaaaaaaaa",
"output": "10"
},
{
"input": "abaaba",
"output": "5"
},
{
"input": "aabsdfaaaa",
"output": "10"
},
{
"input": "aaaba",
"output": "5"
},
{
"input": "aaabaaa",
"output": "6"
},
{
"input": "baaabbb",
"output": "7"
},
{
"input": "ccbbabbcc",
"output": "8"
},
{
"input": "cabc",
"output": "4"
},
{
"input": "aabcd",
"output": "5"
},
{
"input": "abcdea",
"output": "6"
},
{
"input": "bbabb",
"output": "4"
},
{
"input": "aaaaabababaaaaa",
"output": "14"
},
{
"input": "bbabbb",
"output": "6"
},
{
"input": "aababd",
"output": "6"
},
{
"input": "abaaaa",
"output": "6"
},
{
"input": "aaaaaaaabbba",
"output": "12"
},
{
"input": "aabca",
"output": "5"
},
{
"input": "aaabccbaaa",
"output": "9"
},
{
"input": "aaaaaaaaaaaaaaaaaaaab",
"output": "21"
},
{
"input": "babb",
"output": "4"
},
{
"input": "abcaa",
"output": "5"
},
{
"input": "qwqq",
"output": "4"
},
{
"input": "aaaaaaaaaaabbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaa",
"output": "48"
},
{
"input": "aaab",
"output": "4"
},
{
"input": "aaaaaabaaaaa",
"output": "12"
},
{
"input": "wwuww",
"output": "4"
},
{
"input": "aaaaabcbaaaaa",
"output": "12"
},
{
"input": "aaabbbaaa",
"output": "8"
},
{
"input": "aabcbaa",
"output": "6"
},
{
"input": "abccdefccba",
"output": "11"
},
{
"input": "aabbcbbaa",
"output": "8"
},
{
"input": "aaaabbaaaa",
"output": "9"
},
{
"input": "aabcda",
"output": "6"
},
{
"input": "abbca",
"output": "5"
},
{
"input": "aaaaaabbaaa",
"output": "11"
},
{
"input": "sssssspssssss",
"output": "12"
},
{
"input": "sdnmsdcs",
"output": "8"
},
{
"input": "aaabbbccbbbaaa",
"output": "13"
},
{
"input": "cbdbdc",
"output": "6"
},
{
"input": "abb",
"output": "3"
},
{
"input": "abcdefaaaa",
"output": "10"
},
{
"input": "abbbaaa",
"output": "7"
},
{
"input": "v",
"output": "0"
},
{
"input": "abccbba",
"output": "7"
},
{
"input": "axyza",
"output": "5"
},
{
"input": "abcdefgaaaa",
"output": "11"
},
{
"input": "aaabcdaaa",
"output": "9"
},
{
"input": "aaaacaaaa",
"output": "8"
},
{
"input": "aaaaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaaaaaa",
"output": "42"
},
{
"input": "abbbaa",
"output": "6"
},
{
"input": "abcdee",
"output": "6"
},
{
"input": "oom",
"output": "3"
},
{
"input": "aabcaa",
"output": "6"
},
{
"input": "abba",
"output": "3"
},
{
"input": "aaca",
"output": "4"
},
{
"input": "aacbca",
"output": "6"
},
{
"input": "ababa",
"output": "4"
},
{
"input": "abcda",
"output": "5"
},
{
"input": "cccaaccc",
"output": "7"
},
{
"input": "aaabcda",
"output": "7"
},
{
"input": "aa",
"output": "0"
},
{
"input": "aabaaaa",
"output": "7"
},
{
"input": "abbaaaa",
"output": "7"
},
{
"input": "aaabcbaaa",
"output": "8"
},
{
"input": "aabba",
"output": "5"
},
{
"input": "xyxx",
"output": "4"
},
{
"input": "aaaaaaaaaaaabc",
"output": "14"
},
{
"input": "bbaaaabb",
"output": "7"
},
{
"input": "aaabaa",
"output": "6"
},
{
"input": "sssssabsssss",
"output": "12"
},
{
"input": "bbbaaaabbb",
"output": "9"
},
{
"input": "abbbbaaaa",
"output": "9"
},
{
"input": "wwufuww",
"output": "6"
},
{
"input": "oowoo",
"output": "4"
},
{
"input": "cccaccc",
"output": "6"
},
{
"input": "aaa",
"output": "0"
},
{
"input": "bbbcc",
"output": "5"
},
{
"input": "abcdef",
"output": "6"
},
{
"input": "abbba",
"output": "4"
},
{
"input": "aab",
"output": "3"
},
{
"input": "aaba",
"output": "4"
},
{
"input": "azbyaaa",
"output": "7"
},
{
"input": "oooooiooooo",
"output": "10"
},
{
"input": "aabbbbbaaaaaa",
"output": "13"
}
] | 1,565,480,665 | 2,147,483,647 | Python 3 | OK | TESTS | 133 | 124 | 0 | # https://vjudge.net/contest/319028#problem/D
def is_palindrome(word):
palindrome = True
size = len(word)
middle = int(size / 2)
for i in range(0, middle):
palindrome = palindrome and word[i] == word[size - i - 1]
return palindrome
word = str(input())
result = len(word)
if is_palindrome(word):
i = 0
while is_palindrome(word[i::]) and i < result:
i += 1
result -= i
print(result) | Title: Antipalindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
Input Specification:
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
Output Specification:
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
Demo Input:
['mew\n', 'wuffuw\n', 'qqqqqqqq\n']
Demo Output:
['3\n', '5\n', '0\n']
Note:
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$. | ```python
# https://vjudge.net/contest/319028#problem/D
def is_palindrome(word):
palindrome = True
size = len(word)
middle = int(size / 2)
for i in range(0, middle):
palindrome = palindrome and word[i] == word[size - i - 1]
return palindrome
word = str(input())
result = len(word)
if is_palindrome(word):
i = 0
while is_palindrome(word[i::]) and i < result:
i += 1
result -= i
print(result)
``` | 3 | |
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"
}
] | 1,608,909,022 | 2,147,483,647 | PyPy 3 | OK | TESTS | 32 | 340 | 0 | x, numbers = input(), list(map(int, input().split()))
if numbers[0] % 2 == 0 and (numbers[1] % 2 == 0 or numbers[-1] % 2 == 0):
for i in numbers:
if i % 2 != 0:
print(numbers.index(i)+1)
break
elif numbers[0] % 2 != 0 and (numbers[1] % 2 == 0 and numbers[-1] % 2 == 0):
for i in numbers:
if i % 2 != 0:
print(numbers.index(i)+1)
break
else:
for i in numbers:
if i % 2 == 0:
print(numbers.index(i)+1)
break | 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
x, numbers = input(), list(map(int, input().split()))
if numbers[0] % 2 == 0 and (numbers[1] % 2 == 0 or numbers[-1] % 2 == 0):
for i in numbers:
if i % 2 != 0:
print(numbers.index(i)+1)
break
elif numbers[0] % 2 != 0 and (numbers[1] % 2 == 0 and numbers[-1] % 2 == 0):
for i in numbers:
if i % 2 != 0:
print(numbers.index(i)+1)
break
else:
for i in numbers:
if i % 2 == 0:
print(numbers.index(i)+1)
break
``` | 3.915 |
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"
}
] | 1,650,110,741 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 46 | 0 | mes = input()
word = ''
for i in range(len(mes)):
if mes[i] == 'h':
if 'h' not in word:
word += 'h'
else:
continue
elif mes[i] == 'e' and 'h' in word:
if 'e' not in word:
word += 'e'
else:
continue
elif mes[i] == 'l' and 'e' in word:
if 'll' not in word:
word += 'l'
else:
continue
elif mes[i] == 'o' and 'll' in word:
if 'o' not in word:
word += 'o'
else:
continue
else:
continue
if word == 'hello':
print('YES')
else:
print('NO')
| Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
mes = input()
word = ''
for i in range(len(mes)):
if mes[i] == 'h':
if 'h' not in word:
word += 'h'
else:
continue
elif mes[i] == 'e' and 'h' in word:
if 'e' not in word:
word += 'e'
else:
continue
elif mes[i] == 'l' and 'e' in word:
if 'll' not in word:
word += 'l'
else:
continue
elif mes[i] == 'o' and 'll' in word:
if 'o' not in word:
word += 'o'
else:
continue
else:
continue
if word == 'hello':
print('YES')
else:
print('NO')
``` | 3.977 |
466 | C | Number of Ways | PROGRAMMING | 1,700 | [
"binary search",
"brute force",
"data structures",
"dp",
"two pointers"
] | null | null | You've got array *a*[1],<=*a*[2],<=...,<=*a*[*n*], consisting of *n* integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices *i*,<=*j* (2<=≤<=*i*<=≤<=*j*<=≤<=*n*<=-<=1), that . | The first line contains integer *n* (1<=≤<=*n*<=≤<=5·105), showing how many numbers are in the array. The second line contains *n* integers *a*[1], *a*[2], ..., *a*[*n*] (|*a*[*i*]|<=≤<=<=109) — the elements of array *a*. | Print a single integer — the number of ways to split the array into three parts with the same sum. | [
"5\n1 2 3 0 3\n",
"4\n0 1 -1 0\n",
"2\n4 1\n"
] | [
"2\n",
"1\n",
"0\n"
] | none | 1,500 | [
{
"input": "5\n1 2 3 0 3",
"output": "2"
},
{
"input": "4\n0 1 -1 0",
"output": "1"
},
{
"input": "2\n4 1",
"output": "0"
},
{
"input": "9\n0 0 0 0 0 0 0 0 0",
"output": "28"
},
{
"input": "10\n2 5 -2 2 -3 -2 3 5 -5 -2",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "0"
},
{
"input": "100\n1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "2030"
},
{
"input": "6\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "10\n1 0 0 0 1 1 1 0 1 1",
"output": "2"
},
{
"input": "10\n-2 2 -2 0 -2 -1 1 -1 2 0",
"output": "0"
},
{
"input": "4\n0 2 -1 2",
"output": "0"
},
{
"input": "5\n3 3 -3 3 3",
"output": "3"
},
{
"input": "5\n1 1 1 1 1",
"output": "0"
},
{
"input": "8\n-1 -1 -1 -1 -1 -1 -1 -1",
"output": "0"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "4\n1 -1 0 0",
"output": "1"
},
{
"input": "3\n6 -3 6",
"output": "0"
}
] | 1,699,802,862 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 420 | 62,976,000 | l=int(input())
a=[int(v) for v in input().split(' ')]
c=0;m=0;C=0;s=sum(a)
for i in range(l-1):
c+=a[i]
if c==2*s/3:
C+=m
if c==s/3:
m+=1
print(C) | Title: Number of Ways
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got array *a*[1],<=*a*[2],<=...,<=*a*[*n*], consisting of *n* integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices *i*,<=*j* (2<=≤<=*i*<=≤<=*j*<=≤<=*n*<=-<=1), that .
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5·105), showing how many numbers are in the array. The second line contains *n* integers *a*[1], *a*[2], ..., *a*[*n*] (|*a*[*i*]|<=≤<=<=109) — the elements of array *a*.
Output Specification:
Print a single integer — the number of ways to split the array into three parts with the same sum.
Demo Input:
['5\n1 2 3 0 3\n', '4\n0 1 -1 0\n', '2\n4 1\n']
Demo Output:
['2\n', '1\n', '0\n']
Note:
none | ```python
l=int(input())
a=[int(v) for v in input().split(' ')]
c=0;m=0;C=0;s=sum(a)
for i in range(l-1):
c+=a[i]
if c==2*s/3:
C+=m
if c==s/3:
m+=1
print(C)
``` | 3 | |
181 | A | Series of Crimes | PROGRAMMING | 800 | [
"brute force",
"geometry",
"implementation"
] | null | null | The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed. | The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements. | Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right. | [
"3 2\n.*\n..\n**\n",
"3 3\n*.*\n*..\n...\n"
] | [
"1 1\n",
"2 3\n"
] | none | 500 | [
{
"input": "3 2\n.*\n..\n**",
"output": "1 1"
},
{
"input": "2 5\n*....\n*...*",
"output": "1 5"
},
{
"input": "7 2\n..\n**\n..\n..\n..\n..\n.*",
"output": "7 1"
},
{
"input": "7 2\n*.\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "2 10\n*......*..\n.......*..",
"output": "2 1"
},
{
"input": "10 3\n*..\n...\n...\n...\n...\n...\n...\n...\n...\n**.",
"output": "1 2"
},
{
"input": "100 2\n*.\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "99 3\n**.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\n...\n.*.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "99 3\n**.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\n...\n.*.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "2 100\n...................................................................................................*\n*..................................................................................................*",
"output": "1 1"
},
{
"input": "2 2\n*.\n**",
"output": "1 2"
},
{
"input": "2 2\n**\n.*",
"output": "2 1"
},
{
"input": "2 3\n*.*\n..*",
"output": "2 1"
}
] | 1,636,533,925 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 92 | 4,505,600 | n, m = list(map(int, input().split()))
row, column = [0, 0]
index_value = 0
diff = 0
if m >= n:
g = 0
for i in range(n):
data = list(input())
if data.count("*") == 2:
temp = [i for i, i1 in enumerate(data) if i1 == "*"]
row = temp[0]
column += temp[1]
elif data.count("*") == 1:
g = i + 1
column += data.index("*")
row = g
else:
for i in range(n):
data = list(input())
if data.count("*") == 1:
row = 1 + i
index_value = data.index("*")
if data.count("*") == 2:
temp = [i for i, i1 in enumerate(data) if i1 == "*"]
diff = temp[1] - temp[0] - 1
if index_value in range(0, m - 1):
column = index_value + diff
else:
column = index_value - diff
print(row, column)
| Title: Series of Crimes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
Output Specification:
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
Demo Input:
['3 2\n.*\n..\n**\n', '3 3\n*.*\n*..\n...\n']
Demo Output:
['1 1\n', '2 3\n']
Note:
none | ```python
n, m = list(map(int, input().split()))
row, column = [0, 0]
index_value = 0
diff = 0
if m >= n:
g = 0
for i in range(n):
data = list(input())
if data.count("*") == 2:
temp = [i for i, i1 in enumerate(data) if i1 == "*"]
row = temp[0]
column += temp[1]
elif data.count("*") == 1:
g = i + 1
column += data.index("*")
row = g
else:
for i in range(n):
data = list(input())
if data.count("*") == 1:
row = 1 + i
index_value = data.index("*")
if data.count("*") == 2:
temp = [i for i, i1 in enumerate(data) if i1 == "*"]
diff = temp[1] - temp[0] - 1
if index_value in range(0, m - 1):
column = index_value + diff
else:
column = index_value - diff
print(row, column)
``` | 0 | |
588 | A | Duff and Meat | PROGRAMMING | 900 | [
"greedy"
] | null | null | Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat.
There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.
Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days. | The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days.
In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day. | Print the minimum money needed to keep Duff happy for *n* days, in one line. | [
"3\n1 3\n2 2\n3 1\n",
"3\n1 3\n2 1\n3 2\n"
] | [
"10\n",
"8\n"
] | In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.
In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day. | 750 | [
{
"input": "3\n1 3\n2 2\n3 1",
"output": "10"
},
{
"input": "3\n1 3\n2 1\n3 2",
"output": "8"
},
{
"input": "1\n39 52",
"output": "2028"
},
{
"input": "2\n25 56\n94 17",
"output": "2998"
},
{
"input": "5\n39 21\n95 89\n73 90\n9 55\n85 32",
"output": "6321"
},
{
"input": "12\n70 11\n74 27\n32 11\n26 83\n57 18\n97 28\n75 43\n75 21\n84 29\n16 2\n89 63\n21 88",
"output": "6742"
},
{
"input": "2\n100 68\n38 80",
"output": "9384"
}
] | 1,577,700,248 | 2,147,483,647 | PyPy 3 | OK | TESTS | 30 | 842 | 12,697,600 | n=int(input())
a=[0]*n
p=[0]*n
for i in range(n):
x, b = map(int, input().split())
a[i]=x
p[i]=b
k1=n
k=n
summ=0
while k!=0:
k = p[:k1].index(min(p[:k1]))
summ+=sum(a[k:k1])*p[k]
k1=k
print(summ)
| Title: Duff and Meat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat.
There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.
Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days.
Input Specification:
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days.
In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day.
Output Specification:
Print the minimum money needed to keep Duff happy for *n* days, in one line.
Demo Input:
['3\n1 3\n2 2\n3 1\n', '3\n1 3\n2 1\n3 2\n']
Demo Output:
['10\n', '8\n']
Note:
In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.
In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day. | ```python
n=int(input())
a=[0]*n
p=[0]*n
for i in range(n):
x, b = map(int, input().split())
a[i]=x
p[i]=b
k1=n
k=n
summ=0
while k!=0:
k = p[:k1].index(min(p[:k1]))
summ+=sum(a[k:k1])*p[k]
k1=k
print(summ)
``` | 3 | |
946 | C | String Transformation | PROGRAMMING | 1,300 | [
"greedy",
"strings"
] | null | null | You are given a string *s* consisting of |*s*| small english letters.
In one move you can replace any character of this string to the next character in alphabetical order (a will be replaced with b, s will be replaced with t, etc.). You cannot replace letter z with any other letter.
Your target is to make some number of moves (not necessary minimal) to get string abcdefghijklmnopqrstuvwxyz (english alphabet) as a subsequence. Subsequence of the string is the string that is obtained by deleting characters at some positions. You need to print the string that will be obtained from the given string and will be contain english alphabet as a subsequence or say that it is impossible. | The only one line of the input consisting of the string *s* consisting of |*s*| (1<=≤<=|*s*|<=≤<=105) small english letters. | If you can get a string that can be obtained from the given string and will contain english alphabet as a subsequence, print it. Otherwise print «-1» (without quotes). | [
"aacceeggiikkmmooqqssuuwwyy\n",
"thereisnoanswer\n"
] | [
"abcdefghijklmnopqrstuvwxyz\n",
"-1\n"
] | none | 0 | [
{
"input": "aacceeggiikkmmooqqssuuwwyy",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "thereisnoanswer",
"output": "-1"
},
{
"input": "jqcfvsaveaixhioaaeephbmsmfcgdyawscpyioybkgxlcrhaxs",
"output": "-1"
},
{
"input": "rtdacjpsjjmjdhcoprjhaenlwuvpfqzurnrswngmpnkdnunaendlpbfuylqgxtndhmhqgbsknsy",
"output": "-1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaa"
},
{
"input": "abcdefghijklmnopqrstuvwxxx",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "abcdefghijklmnopqrstuvwxya",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "cdaaaaaaaaabcdjklmnopqrstuvwxyzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "cdabcdefghijklmnopqrstuvwxyzxyzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"
},
{
"input": "zazaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "zazbcdefghijklmnopqrstuvwxyz"
},
{
"input": "abcdefghijklmnopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "abbbefghijklmnopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaa"
},
{
"input": "abcdefghijklmaopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "abcdefghijklmnopqrstuvwxyx",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaz",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaz"
},
{
"input": "zaaaazaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "zabcdzefghijklmnopqrstuvwxyzaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaa"
},
{
"input": "aaaaaafghijklmnopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyzz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaz",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaz"
},
{
"input": "abcdefghijklmnopqrstuvwaxy",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaa"
},
{
"input": "abcdefghijklmnapqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "abcdefghijklmnopqrstuvnxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaa"
},
{
"input": "abcdefghijklmnopqrstuvwxyzzzz",
"output": "abcdefghijklmnopqrstuvwxyzzzz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aacceeggiikkmmooqqssuuwwya",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aacdefghijklmnopqrstuvwxyyy",
"output": "abcdefghijklmnopqrstuvwxyzy"
},
{
"input": "abcaefghijklmnopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "zaaacaaaaaaaaaaaaaaaaaaaayy",
"output": "zabcdefghijklmnopqrstuvwxyz"
},
{
"input": "abcdedccdcdccdcdcdcdcdcddccdcdcdc",
"output": "abcdefghijklmnopqrstuvwxyzcdcdcdc"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "abcdecdcdcddcdcdcdcdcdcdcd",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "abaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "a",
"output": "-1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaa"
},
{
"input": "aaadefghijklmnopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaa"
},
{
"input": "abbbbbbbbbbbbbbbbbbbbbbbbz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aacceeggiikkmmaacceeggiikkmmooaacceeggiikkmmaacceeggiikkmmooqqssuuwwzy",
"output": "abcdefghijklmnopqrstuvwxyzmmooaacceeggiikkmmaacceeggiikkmmooqqssuuwwzy"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "phqghumeaylnlfdxfircvscxggbwkfnqduxwfnfozvsrtkjprepggxrpnrvystmwcysyycqpevikeffmznimkkasvwsrenzkycxf",
"output": "-1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaap",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "zabcdefghijklmnopqrstuvwxyz",
"output": "zabcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyza"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzabcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "rveviaomdienfygifatviahordebxazoxflfgzslhyzowhxbhqzpsgellkoimnwkvhpbijorhpggwfjexivpqbcbmqjyghkbq",
"output": "rveviaomdienfygifbtvichordefxgzoxhlijzslkyzowlxmnqzpsopqrstuvwxyzhpbijorhpggwfjexivpqbcbmqjyghkbq"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "xtlsgypsfadpooefxzbcoejuvpvaboygpoeylfpbnpljvrvipyamyehwqnqrqpmxujjloovaowuxwhmsncbxcoksfzkvatxdknly",
"output": "xtlsgypsfadpooefxzbcoejuvpvdeoygpofylgphnpljvrvipyjmyklwqnqrqpmxunopqrvstwuxwvwxyzbxcoksfzkvatxdknly"
},
{
"input": "jqcfvsaveaixhioaaeephbmsmfcgdyawscpyioybkgxlcrhaxsa",
"output": "jqcfvsavebixhiocdefphgmsmhijkylwsmpynoypqrxstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "wlrbbmqbhcdarzowkkyhiddqscdxrjmowfrxsjybldbefsarcbynecdyggxxpklorellnmpapqfwkhopkmcoqh",
"output": "wlrbbmqbhcdarzowkkyhiddqscdxrjmowfrxsjybldcefsdrefynghiyjkxxplmornopqrstuvwxyzopkmcoqh"
},
{
"input": "abadefghijklmnopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "zazsazcbbbbbbbbbbbbbbbbbbbbbbb",
"output": "zazsbzcdefghijklmnopqrstuvwxyz"
},
{
"input": "zazsazcbbbbbbbbbbbbbbbbbbbbbyb",
"output": "zazsbzcdefghijklmnopqrstuvwxyz"
},
{
"input": "bbcdefghijklmnopqrstuvwxyzzz",
"output": "-1"
},
{
"input": "zaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "zabcdefghijklmnopqrstuvwxyz"
},
{
"input": "zzzzzaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "zzzzzabcdefghijklmnopqrstuvwxyza"
},
{
"input": "kkimnfjbbgggicykcciwtoazomcvisigagkjwhyrmojmoebnqoadpmockfjxibdtvrbedrsdoundbcpkfdqdidqdmxdltink",
"output": "kkimnfjbbgggicykcciwtoazomcvisigbgkjwhyrmojmoecnqodepmofkgjxihitvrjklrsmounopqrstuvwxyzdmxdltink"
},
{
"input": "cawgathqceccscakbazmhwbefvygjbcfyihcbgga",
"output": "-1"
},
{
"input": "acrsbyszsbfslzbqzzamcmrypictkcheddehvxdipaxaannjodzyfxgtfnwababzjraapqbqbfzhbiewlzz",
"output": "acrsbyszscfslzdqzzemfmrypigtkhijklmnvxopqrxstuvwxyzyfxgtfnwababzjraapqbqbfzhbiewlzz"
},
{
"input": "ggcebbheeblbioxdvtlrtkxeuilonazpebcbqpzz",
"output": "-1"
},
{
"input": "zzzzabcdefghijklmnopqrstuvwxy",
"output": "-1"
},
{
"input": "zabcdefghijklmnopqrstuvwxy",
"output": "-1"
},
{
"input": "babcdefghijklmnopqrstuvwxyz",
"output": "babcdefghijklmnopqrstuvwxyz"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "-1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
}
] | 1,632,394,196 | 896 | PyPy 3 | RUNTIME_ERROR | TESTS | 2 | 108 | 22,220,800 | s = str(input())
n = len(s)
if len(s) < 26:
print(-1)
exit()
cur = 0
t = []
for i in range(n):
if s[i] != 'z':
t.append(chr(cur+ord('a')))
cur += 1
else:
t.append('z')
cnt = 0
for i in range(n):
if t[i] == chr(cnt+ord('a')):
cnt += 1
if cnt == 26:
print(''.join(t))
exit()
else:
print(-1)
| Title: String Transformation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *s* consisting of |*s*| small english letters.
In one move you can replace any character of this string to the next character in alphabetical order (a will be replaced with b, s will be replaced with t, etc.). You cannot replace letter z with any other letter.
Your target is to make some number of moves (not necessary minimal) to get string abcdefghijklmnopqrstuvwxyz (english alphabet) as a subsequence. Subsequence of the string is the string that is obtained by deleting characters at some positions. You need to print the string that will be obtained from the given string and will be contain english alphabet as a subsequence or say that it is impossible.
Input Specification:
The only one line of the input consisting of the string *s* consisting of |*s*| (1<=≤<=|*s*|<=≤<=105) small english letters.
Output Specification:
If you can get a string that can be obtained from the given string and will contain english alphabet as a subsequence, print it. Otherwise print «-1» (without quotes).
Demo Input:
['aacceeggiikkmmooqqssuuwwyy\n', 'thereisnoanswer\n']
Demo Output:
['abcdefghijklmnopqrstuvwxyz\n', '-1\n']
Note:
none | ```python
s = str(input())
n = len(s)
if len(s) < 26:
print(-1)
exit()
cur = 0
t = []
for i in range(n):
if s[i] != 'z':
t.append(chr(cur+ord('a')))
cur += 1
else:
t.append('z')
cnt = 0
for i in range(n):
if t[i] == chr(cnt+ord('a')):
cnt += 1
if cnt == 26:
print(''.join(t))
exit()
else:
print(-1)
``` | -1 | |
461 | A | Appleman and Toastman | PROGRAMMING | 1,200 | [
"greedy",
"sortings"
] | null | null | Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman. | Print a single integer — the largest possible score. | [
"3\n3 1 5\n",
"1\n10\n"
] | [
"26\n",
"10\n"
] | Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions. | 500 | [
{
"input": "3\n3 1 5",
"output": "26"
},
{
"input": "1\n10",
"output": "10"
},
{
"input": "10\n8 10 2 5 6 2 4 7 2 1",
"output": "376"
},
{
"input": "10\n171308 397870 724672 431255 228496 892002 542924 718337 888642 161821",
"output": "40204082"
},
{
"input": "10\n1 2 2 2 4 5 6 7 8 10",
"output": "376"
},
{
"input": "10\n161821 171308 228496 397870 431255 542924 718337 724672 888642 892002",
"output": "40204082"
},
{
"input": "1\n397870",
"output": "397870"
},
{
"input": "1\n1000000",
"output": "1000000"
},
{
"input": "10\n10 8 7 6 5 4 2 2 2 1",
"output": "376"
},
{
"input": "10\n892002 888642 724672 718337 542924 431255 397870 228496 171308 161821",
"output": "40204082"
},
{
"input": "10\n5 2 6 10 10 10 10 2 2 5",
"output": "485"
},
{
"input": "10\n431255 724672 228496 397870 397870 397870 397870 724672 888642 431255",
"output": "36742665"
},
{
"input": "10\n2 2 2 5 5 6 10 10 10 10",
"output": "485"
},
{
"input": "10\n228496 397870 397870 397870 397870 431255 431255 724672 724672 888642",
"output": "36742665"
},
{
"input": "10\n10 10 10 10 6 5 5 2 2 2",
"output": "485"
},
{
"input": "10\n888642 724672 724672 431255 431255 397870 397870 397870 397870 228496",
"output": "36742665"
},
{
"input": "10\n10 10 10 10 10 10 10 10 10 10",
"output": "640"
},
{
"input": "10\n1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000",
"output": "64000000"
},
{
"input": "1\n397870",
"output": "397870"
},
{
"input": "2\n1 2",
"output": "6"
},
{
"input": "2\n2 3",
"output": "10"
},
{
"input": "2\n1 1",
"output": "4"
}
] | 1,560,598,298 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 5,836,800 | n = int(input())
a = list(sorted(map(int, input().split())))
s = sum(a)
p = 0
x = max(a)
while len(a) > 0:
p = sum(a)
len(a) - 1
s = s + p
a.remove(min(a))
print(s - x) | Title: Appleman and Toastman
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman.
Output Specification:
Print a single integer — the largest possible score.
Demo Input:
['3\n3 1 5\n', '1\n10\n']
Demo Output:
['26\n', '10\n']
Note:
Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions. | ```python
n = int(input())
a = list(sorted(map(int, input().split())))
s = sum(a)
p = 0
x = max(a)
while len(a) > 0:
p = sum(a)
len(a) - 1
s = s + p
a.remove(min(a))
print(s - x)
``` | 0 | |
525 | A | Vitaliy and Pie | PROGRAMMING | 1,100 | [
"greedy",
"hashing",
"strings"
] | null | null | After a hard day Vitaly got very hungry and he wants to eat his favorite potato pie. But it's not that simple. Vitaly is in the first room of the house with *n* room located in a line and numbered starting from one from left to right. You can go from the first room to the second room, from the second room to the third room and so on — you can go from the (*n*<=-<=1)-th room to the *n*-th room. Thus, you can go to room *x* only from room *x*<=-<=1.
The potato pie is located in the *n*-th room and Vitaly needs to go there.
Each pair of consecutive rooms has a door between them. In order to go to room *x* from room *x*<=-<=1, you need to open the door between the rooms with the corresponding key.
In total the house has several types of doors (represented by uppercase Latin letters) and several types of keys (represented by lowercase Latin letters). The key of type *t* can open the door of type *T* if and only if *t* and *T* are the same letter, written in different cases. For example, key f can open door F.
Each of the first *n*<=-<=1 rooms contains exactly one key of some type that Vitaly can use to get to next rooms. Once the door is open with some key, Vitaly won't get the key from the keyhole but he will immediately run into the next room. In other words, each key can open no more than one door.
Vitaly realizes that he may end up in some room without the key that opens the door to the next room. Before the start his run for the potato pie Vitaly can buy any number of keys of any type that is guaranteed to get to room *n*.
Given the plan of the house, Vitaly wants to know what is the minimum number of keys he needs to buy to surely get to the room *n*, which has a delicious potato pie. Write a program that will help Vitaly find out this number. | The first line of the input contains a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of rooms in the house.
The second line of the input contains string *s* of length 2·*n*<=-<=2. Let's number the elements of the string from left to right, starting from one.
The odd positions in the given string *s* contain lowercase Latin letters — the types of the keys that lie in the corresponding rooms. Thus, each odd position *i* of the given string *s* contains a lowercase Latin letter — the type of the key that lies in room number (*i*<=+<=1)<=/<=2.
The even positions in the given string contain uppercase Latin letters — the types of doors between the rooms. Thus, each even position *i* of the given string *s* contains an uppercase letter — the type of the door that leads from room *i*<=/<=2 to room *i*<=/<=2<=+<=1. | Print the only integer — the minimum number of keys that Vitaly needs to buy to surely get from room one to room *n*. | [
"3\naAbB\n",
"4\naBaCaB\n",
"5\nxYyXzZaZ\n"
] | [
"0\n",
"3\n",
"2\n"
] | none | 250 | [
{
"input": "3\naAbB",
"output": "0"
},
{
"input": "4\naBaCaB",
"output": "3"
},
{
"input": "5\nxYyXzZaZ",
"output": "2"
},
{
"input": "26\naAbBcCdDeEfFgGhHiIjJkKlLmMnNoOpPqQrRsStTuUvVwWxXyY",
"output": "0"
},
{
"input": "26\nzAyBxCwDvEuFtGsHrIqJpKoLnMmNlOkPjQiRhSgTfUeVdWcXbY",
"output": "13"
},
{
"input": "5\naArRaRaR",
"output": "2"
},
{
"input": "2\ndA",
"output": "1"
},
{
"input": "2\ncB",
"output": "1"
},
{
"input": "10\nhNcMeXsSlHsUwYeMcA",
"output": "7"
},
{
"input": "100\nqDpInBmCrFwXpDbFgOzVvOcEmJrUcToAdEwEgTvBvBfWwRpGyEaXgDdRwVlQnYgWmWhMrHaIzPyXvGaFlRsVzHhZrOuVpXrKxFzAmWwPlFtNfPtJxVmLuHjKfYyArHrEnSwSzOvDpQhCgCqLlAcNpGhXrEeFuCmAqIkXyYtSsQwIxJzNiIuTgEbVuWrMwPrAlLyKaZ",
"output": "42"
},
{
"input": "2\ndD",
"output": "0"
},
{
"input": "2\ndE",
"output": "1"
},
{
"input": "3\ndDdD",
"output": "0"
},
{
"input": "3\ndEdD",
"output": "1"
},
{
"input": "3\ndEeD",
"output": "1"
},
{
"input": "3\ndEeF",
"output": "2"
}
] | 1,427,388,081 | 681 | Python 3 | TIME_LIMIT_EXCEEDED | PRETESTS | 7 | 2,000 | 409,600 | n = int(input())
s = input()
list1 = []
count = 0
counter = 0
for i in s:
if counter % 2 == 0:
list1.append(i.upper())
counter += 1
else:
if i not in list1:
count += 1
counter += 1
else:
list1.remove(i.upper())
counter += 1
print(count) | Title: Vitaliy and Pie
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After a hard day Vitaly got very hungry and he wants to eat his favorite potato pie. But it's not that simple. Vitaly is in the first room of the house with *n* room located in a line and numbered starting from one from left to right. You can go from the first room to the second room, from the second room to the third room and so on — you can go from the (*n*<=-<=1)-th room to the *n*-th room. Thus, you can go to room *x* only from room *x*<=-<=1.
The potato pie is located in the *n*-th room and Vitaly needs to go there.
Each pair of consecutive rooms has a door between them. In order to go to room *x* from room *x*<=-<=1, you need to open the door between the rooms with the corresponding key.
In total the house has several types of doors (represented by uppercase Latin letters) and several types of keys (represented by lowercase Latin letters). The key of type *t* can open the door of type *T* if and only if *t* and *T* are the same letter, written in different cases. For example, key f can open door F.
Each of the first *n*<=-<=1 rooms contains exactly one key of some type that Vitaly can use to get to next rooms. Once the door is open with some key, Vitaly won't get the key from the keyhole but he will immediately run into the next room. In other words, each key can open no more than one door.
Vitaly realizes that he may end up in some room without the key that opens the door to the next room. Before the start his run for the potato pie Vitaly can buy any number of keys of any type that is guaranteed to get to room *n*.
Given the plan of the house, Vitaly wants to know what is the minimum number of keys he needs to buy to surely get to the room *n*, which has a delicious potato pie. Write a program that will help Vitaly find out this number.
Input Specification:
The first line of the input contains a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of rooms in the house.
The second line of the input contains string *s* of length 2·*n*<=-<=2. Let's number the elements of the string from left to right, starting from one.
The odd positions in the given string *s* contain lowercase Latin letters — the types of the keys that lie in the corresponding rooms. Thus, each odd position *i* of the given string *s* contains a lowercase Latin letter — the type of the key that lies in room number (*i*<=+<=1)<=/<=2.
The even positions in the given string contain uppercase Latin letters — the types of doors between the rooms. Thus, each even position *i* of the given string *s* contains an uppercase letter — the type of the door that leads from room *i*<=/<=2 to room *i*<=/<=2<=+<=1.
Output Specification:
Print the only integer — the minimum number of keys that Vitaly needs to buy to surely get from room one to room *n*.
Demo Input:
['3\naAbB\n', '4\naBaCaB\n', '5\nxYyXzZaZ\n']
Demo Output:
['0\n', '3\n', '2\n']
Note:
none | ```python
n = int(input())
s = input()
list1 = []
count = 0
counter = 0
for i in s:
if counter % 2 == 0:
list1.append(i.upper())
counter += 1
else:
if i not in list1:
count += 1
counter += 1
else:
list1.remove(i.upper())
counter += 1
print(count)
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Dreamoon likes to play with sets, integers and . is defined as the largest positive integer that divides both *a* and *b*.
Let *S* be a set of exactly four distinct integers greater than 0. Define *S* to be of rank *k* if and only if for all pairs of distinct elements *s**i*, *s**j* from *S*, .
Given *k* and *n*, Dreamoon wants to make up *n* sets of rank *k* using integers from 1 to *m* such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum *m* that makes it possible and print one possible solution. | The single line of the input contains two space separated integers *n*, *k* (1<=≤<=*n*<=≤<=10<=000,<=1<=≤<=*k*<=≤<=100). | On the first line print a single integer — the minimal possible *m*.
On each of the next *n* lines print four space separated integers representing the *i*-th set.
Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal *m*, print any one of them. | [
"1 1\n",
"2 2\n"
] | [
"5\n1 2 3 5\n",
"22\n2 4 6 22\n14 18 10 16\n"
] | For the first example it's easy to see that set {1, 2, 3, 4} isn't a valid set of rank 1 since <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e2af04e5e60e1fe79a4d74bf22dfa575f0b0f7bb.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 0 | [
{
"input": "1 1",
"output": "5\n1 3 4 5"
},
{
"input": "2 2",
"output": "22\n2 6 8 10\n14 18 20 22"
},
{
"input": "7 7",
"output": "287\n7 21 28 35\n49 63 70 77\n91 105 112 119\n133 147 154 161\n175 189 196 203\n217 231 238 245\n259 273 280 287"
},
{
"input": "13 7",
"output": "539\n7 21 28 35\n49 63 70 77\n91 105 112 119\n133 147 154 161\n175 189 196 203\n217 231 238 245\n259 273 280 287\n301 315 322 329\n343 357 364 371\n385 399 406 413\n427 441 448 455\n469 483 490 497\n511 525 532 539"
},
{
"input": "15 27",
"output": "2403\n27 81 108 135\n189 243 270 297\n351 405 432 459\n513 567 594 621\n675 729 756 783\n837 891 918 945\n999 1053 1080 1107\n1161 1215 1242 1269\n1323 1377 1404 1431\n1485 1539 1566 1593\n1647 1701 1728 1755\n1809 1863 1890 1917\n1971 2025 2052 2079\n2133 2187 2214 2241\n2295 2349 2376 2403"
},
{
"input": "19 21",
"output": "2373\n21 63 84 105\n147 189 210 231\n273 315 336 357\n399 441 462 483\n525 567 588 609\n651 693 714 735\n777 819 840 861\n903 945 966 987\n1029 1071 1092 1113\n1155 1197 1218 1239\n1281 1323 1344 1365\n1407 1449 1470 1491\n1533 1575 1596 1617\n1659 1701 1722 1743\n1785 1827 1848 1869\n1911 1953 1974 1995\n2037 2079 2100 2121\n2163 2205 2226 2247\n2289 2331 2352 2373"
},
{
"input": "113 97",
"output": "65669\n97 291 388 485\n679 873 970 1067\n1261 1455 1552 1649\n1843 2037 2134 2231\n2425 2619 2716 2813\n3007 3201 3298 3395\n3589 3783 3880 3977\n4171 4365 4462 4559\n4753 4947 5044 5141\n5335 5529 5626 5723\n5917 6111 6208 6305\n6499 6693 6790 6887\n7081 7275 7372 7469\n7663 7857 7954 8051\n8245 8439 8536 8633\n8827 9021 9118 9215\n9409 9603 9700 9797\n9991 10185 10282 10379\n10573 10767 10864 10961\n11155 11349 11446 11543\n11737 11931 12028 12125\n12319 12513 12610 12707\n12901 13095 13192 13289\n13483 ..."
},
{
"input": "10000 100",
"output": "5999900\n100 300 400 500\n700 900 1000 1100\n1300 1500 1600 1700\n1900 2100 2200 2300\n2500 2700 2800 2900\n3100 3300 3400 3500\n3700 3900 4000 4100\n4300 4500 4600 4700\n4900 5100 5200 5300\n5500 5700 5800 5900\n6100 6300 6400 6500\n6700 6900 7000 7100\n7300 7500 7600 7700\n7900 8100 8200 8300\n8500 8700 8800 8900\n9100 9300 9400 9500\n9700 9900 10000 10100\n10300 10500 10600 10700\n10900 11100 11200 11300\n11500 11700 11800 11900\n12100 12300 12400 12500\n12700 12900 13000 13100\n13300 13500 13600 13700\n..."
},
{
"input": "10000 1",
"output": "59999\n1 3 4 5\n7 9 10 11\n13 15 16 17\n19 21 22 23\n25 27 28 29\n31 33 34 35\n37 39 40 41\n43 45 46 47\n49 51 52 53\n55 57 58 59\n61 63 64 65\n67 69 70 71\n73 75 76 77\n79 81 82 83\n85 87 88 89\n91 93 94 95\n97 99 100 101\n103 105 106 107\n109 111 112 113\n115 117 118 119\n121 123 124 125\n127 129 130 131\n133 135 136 137\n139 141 142 143\n145 147 148 149\n151 153 154 155\n157 159 160 161\n163 165 166 167\n169 171 172 173\n175 177 178 179\n181 183 184 185\n187 189 190 191\n193 195 196 197\n199 201 202 203..."
},
{
"input": "1 100",
"output": "500\n100 300 400 500"
},
{
"input": "9252 39",
"output": "2164929\n39 117 156 195\n273 351 390 429\n507 585 624 663\n741 819 858 897\n975 1053 1092 1131\n1209 1287 1326 1365\n1443 1521 1560 1599\n1677 1755 1794 1833\n1911 1989 2028 2067\n2145 2223 2262 2301\n2379 2457 2496 2535\n2613 2691 2730 2769\n2847 2925 2964 3003\n3081 3159 3198 3237\n3315 3393 3432 3471\n3549 3627 3666 3705\n3783 3861 3900 3939\n4017 4095 4134 4173\n4251 4329 4368 4407\n4485 4563 4602 4641\n4719 4797 4836 4875\n4953 5031 5070 5109\n5187 5265 5304 5343\n5421 5499 5538 5577\n5655 5733 5772 5..."
},
{
"input": "8096 59",
"output": "2865925\n59 177 236 295\n413 531 590 649\n767 885 944 1003\n1121 1239 1298 1357\n1475 1593 1652 1711\n1829 1947 2006 2065\n2183 2301 2360 2419\n2537 2655 2714 2773\n2891 3009 3068 3127\n3245 3363 3422 3481\n3599 3717 3776 3835\n3953 4071 4130 4189\n4307 4425 4484 4543\n4661 4779 4838 4897\n5015 5133 5192 5251\n5369 5487 5546 5605\n5723 5841 5900 5959\n6077 6195 6254 6313\n6431 6549 6608 6667\n6785 6903 6962 7021\n7139 7257 7316 7375\n7493 7611 7670 7729\n7847 7965 8024 8083\n8201 8319 8378 8437\n8555 8673 ..."
},
{
"input": "4237 87",
"output": "2211627\n87 261 348 435\n609 783 870 957\n1131 1305 1392 1479\n1653 1827 1914 2001\n2175 2349 2436 2523\n2697 2871 2958 3045\n3219 3393 3480 3567\n3741 3915 4002 4089\n4263 4437 4524 4611\n4785 4959 5046 5133\n5307 5481 5568 5655\n5829 6003 6090 6177\n6351 6525 6612 6699\n6873 7047 7134 7221\n7395 7569 7656 7743\n7917 8091 8178 8265\n8439 8613 8700 8787\n8961 9135 9222 9309\n9483 9657 9744 9831\n10005 10179 10266 10353\n10527 10701 10788 10875\n11049 11223 11310 11397\n11571 11745 11832 11919\n12093 12267 ..."
},
{
"input": "3081 11",
"output": "203335\n11 33 44 55\n77 99 110 121\n143 165 176 187\n209 231 242 253\n275 297 308 319\n341 363 374 385\n407 429 440 451\n473 495 506 517\n539 561 572 583\n605 627 638 649\n671 693 704 715\n737 759 770 781\n803 825 836 847\n869 891 902 913\n935 957 968 979\n1001 1023 1034 1045\n1067 1089 1100 1111\n1133 1155 1166 1177\n1199 1221 1232 1243\n1265 1287 1298 1309\n1331 1353 1364 1375\n1397 1419 1430 1441\n1463 1485 1496 1507\n1529 1551 1562 1573\n1595 1617 1628 1639\n1661 1683 1694 1705\n1727 1749 1760 1771\n17..."
},
{
"input": "9221 39",
"output": "2157675\n39 117 156 195\n273 351 390 429\n507 585 624 663\n741 819 858 897\n975 1053 1092 1131\n1209 1287 1326 1365\n1443 1521 1560 1599\n1677 1755 1794 1833\n1911 1989 2028 2067\n2145 2223 2262 2301\n2379 2457 2496 2535\n2613 2691 2730 2769\n2847 2925 2964 3003\n3081 3159 3198 3237\n3315 3393 3432 3471\n3549 3627 3666 3705\n3783 3861 3900 3939\n4017 4095 4134 4173\n4251 4329 4368 4407\n4485 4563 4602 4641\n4719 4797 4836 4875\n4953 5031 5070 5109\n5187 5265 5304 5343\n5421 5499 5538 5577\n5655 5733 5772 5..."
},
{
"input": "770 59",
"output": "272521\n59 177 236 295\n413 531 590 649\n767 885 944 1003\n1121 1239 1298 1357\n1475 1593 1652 1711\n1829 1947 2006 2065\n2183 2301 2360 2419\n2537 2655 2714 2773\n2891 3009 3068 3127\n3245 3363 3422 3481\n3599 3717 3776 3835\n3953 4071 4130 4189\n4307 4425 4484 4543\n4661 4779 4838 4897\n5015 5133 5192 5251\n5369 5487 5546 5605\n5723 5841 5900 5959\n6077 6195 6254 6313\n6431 6549 6608 6667\n6785 6903 6962 7021\n7139 7257 7316 7375\n7493 7611 7670 7729\n7847 7965 8024 8083\n8201 8319 8378 8437\n8555 8673 8..."
},
{
"input": "5422 87",
"output": "2830197\n87 261 348 435\n609 783 870 957\n1131 1305 1392 1479\n1653 1827 1914 2001\n2175 2349 2436 2523\n2697 2871 2958 3045\n3219 3393 3480 3567\n3741 3915 4002 4089\n4263 4437 4524 4611\n4785 4959 5046 5133\n5307 5481 5568 5655\n5829 6003 6090 6177\n6351 6525 6612 6699\n6873 7047 7134 7221\n7395 7569 7656 7743\n7917 8091 8178 8265\n8439 8613 8700 8787\n8961 9135 9222 9309\n9483 9657 9744 9831\n10005 10179 10266 10353\n10527 10701 10788 10875\n11049 11223 11310 11397\n11571 11745 11832 11919\n12093 12267 ..."
},
{
"input": "1563 15",
"output": "140655\n15 45 60 75\n105 135 150 165\n195 225 240 255\n285 315 330 345\n375 405 420 435\n465 495 510 525\n555 585 600 615\n645 675 690 705\n735 765 780 795\n825 855 870 885\n915 945 960 975\n1005 1035 1050 1065\n1095 1125 1140 1155\n1185 1215 1230 1245\n1275 1305 1320 1335\n1365 1395 1410 1425\n1455 1485 1500 1515\n1545 1575 1590 1605\n1635 1665 1680 1695\n1725 1755 1770 1785\n1815 1845 1860 1875\n1905 1935 1950 1965\n1995 2025 2040 2055\n2085 2115 2130 2145\n2175 2205 2220 2235\n2265 2295 2310 2325\n2355 ..."
},
{
"input": "407 39",
"output": "95199\n39 117 156 195\n273 351 390 429\n507 585 624 663\n741 819 858 897\n975 1053 1092 1131\n1209 1287 1326 1365\n1443 1521 1560 1599\n1677 1755 1794 1833\n1911 1989 2028 2067\n2145 2223 2262 2301\n2379 2457 2496 2535\n2613 2691 2730 2769\n2847 2925 2964 3003\n3081 3159 3198 3237\n3315 3393 3432 3471\n3549 3627 3666 3705\n3783 3861 3900 3939\n4017 4095 4134 4173\n4251 4329 4368 4407\n4485 4563 4602 4641\n4719 4797 4836 4875\n4953 5031 5070 5109\n5187 5265 5304 5343\n5421 5499 5538 5577\n5655 5733 5772 581..."
},
{
"input": "6518 18",
"output": "703926\n18 54 72 90\n126 162 180 198\n234 270 288 306\n342 378 396 414\n450 486 504 522\n558 594 612 630\n666 702 720 738\n774 810 828 846\n882 918 936 954\n990 1026 1044 1062\n1098 1134 1152 1170\n1206 1242 1260 1278\n1314 1350 1368 1386\n1422 1458 1476 1494\n1530 1566 1584 1602\n1638 1674 1692 1710\n1746 1782 1800 1818\n1854 1890 1908 1926\n1962 1998 2016 2034\n2070 2106 2124 2142\n2178 2214 2232 2250\n2286 2322 2340 2358\n2394 2430 2448 2466\n2502 2538 2556 2574\n2610 2646 2664 2682\n2718 2754 2772 2790..."
},
{
"input": "1171 46",
"output": "323150\n46 138 184 230\n322 414 460 506\n598 690 736 782\n874 966 1012 1058\n1150 1242 1288 1334\n1426 1518 1564 1610\n1702 1794 1840 1886\n1978 2070 2116 2162\n2254 2346 2392 2438\n2530 2622 2668 2714\n2806 2898 2944 2990\n3082 3174 3220 3266\n3358 3450 3496 3542\n3634 3726 3772 3818\n3910 4002 4048 4094\n4186 4278 4324 4370\n4462 4554 4600 4646\n4738 4830 4876 4922\n5014 5106 5152 5198\n5290 5382 5428 5474\n5566 5658 5704 5750\n5842 5934 5980 6026\n6118 6210 6256 6302\n6394 6486 6532 6578\n6670 6762 6808..."
},
{
"input": "7311 70",
"output": "3070550\n70 210 280 350\n490 630 700 770\n910 1050 1120 1190\n1330 1470 1540 1610\n1750 1890 1960 2030\n2170 2310 2380 2450\n2590 2730 2800 2870\n3010 3150 3220 3290\n3430 3570 3640 3710\n3850 3990 4060 4130\n4270 4410 4480 4550\n4690 4830 4900 4970\n5110 5250 5320 5390\n5530 5670 5740 5810\n5950 6090 6160 6230\n6370 6510 6580 6650\n6790 6930 7000 7070\n7210 7350 7420 7490\n7630 7770 7840 7910\n8050 8190 8260 8330\n8470 8610 8680 8750\n8890 9030 9100 9170\n9310 9450 9520 9590\n9730 9870 9940 10010\n10150 1..."
},
{
"input": "6155 94",
"output": "3471326\n94 282 376 470\n658 846 940 1034\n1222 1410 1504 1598\n1786 1974 2068 2162\n2350 2538 2632 2726\n2914 3102 3196 3290\n3478 3666 3760 3854\n4042 4230 4324 4418\n4606 4794 4888 4982\n5170 5358 5452 5546\n5734 5922 6016 6110\n6298 6486 6580 6674\n6862 7050 7144 7238\n7426 7614 7708 7802\n7990 8178 8272 8366\n8554 8742 8836 8930\n9118 9306 9400 9494\n9682 9870 9964 10058\n10246 10434 10528 10622\n10810 10998 11092 11186\n11374 11562 11656 11750\n11938 12126 12220 12314\n12502 12690 12784 12878\n13066 ..."
},
{
"input": "7704 18",
"output": "832014\n18 54 72 90\n126 162 180 198\n234 270 288 306\n342 378 396 414\n450 486 504 522\n558 594 612 630\n666 702 720 738\n774 810 828 846\n882 918 936 954\n990 1026 1044 1062\n1098 1134 1152 1170\n1206 1242 1260 1278\n1314 1350 1368 1386\n1422 1458 1476 1494\n1530 1566 1584 1602\n1638 1674 1692 1710\n1746 1782 1800 1818\n1854 1890 1908 1926\n1962 1998 2016 2034\n2070 2106 2124 2142\n2178 2214 2232 2250\n2286 2322 2340 2358\n2394 2430 2448 2466\n2502 2538 2556 2574\n2610 2646 2664 2682\n2718 2754 2772 2790..."
},
{
"input": "3844 46",
"output": "1060898\n46 138 184 230\n322 414 460 506\n598 690 736 782\n874 966 1012 1058\n1150 1242 1288 1334\n1426 1518 1564 1610\n1702 1794 1840 1886\n1978 2070 2116 2162\n2254 2346 2392 2438\n2530 2622 2668 2714\n2806 2898 2944 2990\n3082 3174 3220 3266\n3358 3450 3496 3542\n3634 3726 3772 3818\n3910 4002 4048 4094\n4186 4278 4324 4370\n4462 4554 4600 4646\n4738 4830 4876 4922\n5014 5106 5152 5198\n5290 5382 5428 5474\n5566 5658 5704 5750\n5842 5934 5980 6026\n6118 6210 6256 6302\n6394 6486 6532 6578\n6670 6762 680..."
},
{
"input": "1 10",
"output": "50\n10 30 40 50"
}
] | 1,566,992,003 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 109 | 0 | class CodeforcesTask477BSolution:
def __init__(self):
self.result = ''
self.n_k = []
def read_input(self):
self.n_k = [int(x) for x in input().split(" ")]
def process_task(self):
m = (6 * self.n_k[0] - 1) * self.n_k[1]
results = []
for x in range(self.n_k[0]):
results.append([6 * x + 1, 6 * x + 2, 6 * x + 3, 6 * x + 5])
self.result = "{0}\n{1}".format(m, "\n".join([" ".join([str(x) for x in res]) for res in results]))
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask477BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dreamoon likes to play with sets, integers and . is defined as the largest positive integer that divides both *a* and *b*.
Let *S* be a set of exactly four distinct integers greater than 0. Define *S* to be of rank *k* if and only if for all pairs of distinct elements *s**i*, *s**j* from *S*, .
Given *k* and *n*, Dreamoon wants to make up *n* sets of rank *k* using integers from 1 to *m* such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum *m* that makes it possible and print one possible solution.
Input Specification:
The single line of the input contains two space separated integers *n*, *k* (1<=≤<=*n*<=≤<=10<=000,<=1<=≤<=*k*<=≤<=100).
Output Specification:
On the first line print a single integer — the minimal possible *m*.
On each of the next *n* lines print four space separated integers representing the *i*-th set.
Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal *m*, print any one of them.
Demo Input:
['1 1\n', '2 2\n']
Demo Output:
['5\n1 2 3 5\n', '22\n2 4 6 22\n14 18 10 16\n']
Note:
For the first example it's easy to see that set {1, 2, 3, 4} isn't a valid set of rank 1 since <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e2af04e5e60e1fe79a4d74bf22dfa575f0b0f7bb.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
class CodeforcesTask477BSolution:
def __init__(self):
self.result = ''
self.n_k = []
def read_input(self):
self.n_k = [int(x) for x in input().split(" ")]
def process_task(self):
m = (6 * self.n_k[0] - 1) * self.n_k[1]
results = []
for x in range(self.n_k[0]):
results.append([6 * x + 1, 6 * x + 2, 6 * x + 3, 6 * x + 5])
self.result = "{0}\n{1}".format(m, "\n".join([" ".join([str(x) for x in res]) for res in results]))
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask477BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | 0 | |
246 | B | Increase and Decrease | PROGRAMMING | 1,300 | [
"greedy",
"math"
] | null | null | Polycarpus has an array, consisting of *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
- he chooses two elements of the array *a**i*, *a**j* (*i*<=≠<=*j*); - he simultaneously increases number *a**i* by 1 and decreases number *a**j* by 1, that is, executes *a**i*<==<=*a**i*<=+<=1 and *a**j*<==<=*a**j*<=-<=1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the array size. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=104) — the original array. | Print a single integer — the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation. | [
"2\n2 1\n",
"3\n1 4 1\n"
] | [
"1\n",
"3\n"
] | none | 1,000 | [
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 4 1",
"output": "3"
},
{
"input": "4\n2 -7 -2 -6",
"output": "3"
},
{
"input": "4\n2 0 -2 -1",
"output": "3"
},
{
"input": "6\n-1 1 0 0 -1 -1",
"output": "5"
},
{
"input": "5\n0 0 0 0 0",
"output": "5"
},
{
"input": "100\n968 793 -628 -416 942 -308 977 168 728 -879 952 781 -425 -475 -480 738 -740 142 -319 -116 -701 -183 41 324 -918 -391 -176 781 763 888 475 -617 134 -802 -133 -211 855 -869 -236 503 550 387 137 -221 -957 -800 -56 -673 440 -791 -217 -959 -892 886 -593 427 890 223 -425 -342 -262 693 -137 924 860 156 -110 444 -12 703 421 -638 725 -551 597 -206 -80 393 -441 -759 617 424 -506 209 579 -685 -861 507 312 597 -81 -491 424 222 -892 629 512 743 -452 -504",
"output": "99"
}
] | 1,604,295,661 | 2,147,483,647 | PyPy 3 | OK | TESTS | 30 | 310 | 5,120,000 | n = int(input())
l=list(map(int,input().split()))
l=sum(l)
if l%n==0 :
print(n)
else:
print(n-1)
| Title: Increase and Decrease
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has an array, consisting of *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
- he chooses two elements of the array *a**i*, *a**j* (*i*<=≠<=*j*); - he simultaneously increases number *a**i* by 1 and decreases number *a**j* by 1, that is, executes *a**i*<==<=*a**i*<=+<=1 and *a**j*<==<=*a**j*<=-<=1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the array size. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=104) — the original array.
Output Specification:
Print a single integer — the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Demo Input:
['2\n2 1\n', '3\n1 4 1\n']
Demo Output:
['1\n', '3\n']
Note:
none | ```python
n = int(input())
l=list(map(int,input().split()))
l=sum(l)
if l%n==0 :
print(n)
else:
print(n-1)
``` | 3 | |
614 | B | Gena's Code | PROGRAMMING | 1,400 | [
"implementation",
"math"
] | null | null | It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly *n* distinct countries in the world and the *i*-th country added *a**i* tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case. | The first line of the input contains the number of countries *n* (1<=≤<=*n*<=≤<=100<=000). The second line contains *n* non-negative integers *a**i* without leading zeroes — the number of tanks of the *i*-th country.
It is guaranteed that the second line contains at least *n*<=-<=1 beautiful numbers and the total length of all these number's representations doesn't exceed 100<=000. | Print a single number without leading zeroes — the product of the number of tanks presented by each country. | [
"3\n5 10 1\n",
"4\n1 1 10 11\n",
"5\n0 3 1 100 1\n"
] | [
"50",
"110",
"0"
] | In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | 1,000 | [
{
"input": "3\n5 10 1",
"output": "50"
},
{
"input": "4\n1 1 10 11",
"output": "110"
},
{
"input": "5\n0 3 1 100 1",
"output": "0"
},
{
"input": "40\n10 100 10 1 10 10 100 10 10 100 10 100 100 10 1824868942 100 100 1 10 100 100 10 100 100 10 100 10 1 10 100 100 100 10 1 10 1 10 10 100 100",
"output": "1824868942000000000000000000000000000000000000000000000000000"
},
{
"input": "6\n1000000000000000000000000000000000000 6643573784 1000000000000000000000000000000000000 1000000000000000000000000000000000000 1000000000000000000000000000000000000 1000000000000000000000000000000000000",
"output": "6643573784000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n9",
"output": "9"
},
{
"input": "2\n10 50",
"output": "500"
},
{
"input": "3\n500 1 10",
"output": "5000"
}
] | 1,550,077,626 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 14 | 500 | 7,884,800 | import collections
n=int(input())
ss=""
l=list(map(int,input().split()))
d=collections.Counter(l)
if d[0]>0 :
print(0)
exit()
r=1
for x in d :
if x%10==0 and str(x)[0]=="1" :
ss+="0"*(len(str(x))-1)*d[x]
else :
if x!=1 :
r=x
print(str(r)+ss)
| Title: Gena's Code
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly *n* distinct countries in the world and the *i*-th country added *a**i* tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input Specification:
The first line of the input contains the number of countries *n* (1<=≤<=*n*<=≤<=100<=000). The second line contains *n* non-negative integers *a**i* without leading zeroes — the number of tanks of the *i*-th country.
It is guaranteed that the second line contains at least *n*<=-<=1 beautiful numbers and the total length of all these number's representations doesn't exceed 100<=000.
Output Specification:
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Demo Input:
['3\n5 10 1\n', '4\n1 1 10 11\n', '5\n0 3 1 100 1\n']
Demo Output:
['50', '110', '0']
Note:
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | ```python
import collections
n=int(input())
ss=""
l=list(map(int,input().split()))
d=collections.Counter(l)
if d[0]>0 :
print(0)
exit()
r=1
for x in d :
if x%10==0 and str(x)[0]=="1" :
ss+="0"*(len(str(x))-1)*d[x]
else :
if x!=1 :
r=x
print(str(r)+ss)
``` | 0 | |
567 | B | Berland National Library | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Berland National Library has recently been built in the capital of Berland. In addition, in the library you can take any of the collected works of Berland leaders, the library has a reading room.
Today was the pilot launch of an automated reading room visitors' accounting system! The scanner of the system is installed at the entrance to the reading room. It records the events of the form "reader entered room", "reader left room". Every reader is assigned a registration number during the registration procedure at the library — it's a unique integer from 1 to 106. Thus, the system logs events of two forms:
- "+ *r**i*" — the reader with registration number *r**i* entered the room; - "- *r**i*" — the reader with registration number *r**i* left the room.
The first launch of the system was a success, it functioned for some period of time, and, at the time of its launch and at the time of its shutdown, the reading room may already have visitors.
Significant funds of the budget of Berland have been spent on the design and installation of the system. Therefore, some of the citizens of the capital now demand to explain the need for this system and the benefits that its implementation will bring. Now, the developers of the system need to urgently come up with reasons for its existence.
Help the system developers to find the minimum possible capacity of the reading room (in visitors) using the log of the system available to you. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of records in the system log. Next follow *n* events from the system journal in the order in which the were made. Each event was written on a single line and looks as "+ *r**i*" or "- *r**i*", where *r**i* is an integer from 1 to 106, the registration number of the visitor (that is, distinct visitors always have distinct registration numbers).
It is guaranteed that the log is not contradictory, that is, for every visitor the types of any of his two consecutive events are distinct. Before starting the system, and after stopping the room may possibly contain visitors. | Print a single integer — the minimum possible capacity of the reading room. | [
"6\n+ 12001\n- 12001\n- 1\n- 1200\n+ 1\n+ 7\n",
"2\n- 1\n- 2\n",
"2\n+ 1\n- 1\n"
] | [
"3",
"2",
"1"
] | In the first sample test, the system log will ensure that at some point in the reading room were visitors with registration numbers 1, 1200 and 12001. More people were not in the room at the same time based on the log. Therefore, the answer to the test is 3. | 1,000 | [
{
"input": "6\n+ 12001\n- 12001\n- 1\n- 1200\n+ 1\n+ 7",
"output": "3"
},
{
"input": "2\n- 1\n- 2",
"output": "2"
},
{
"input": "2\n+ 1\n- 1",
"output": "1"
},
{
"input": "5\n+ 1\n- 1\n+ 2\n+ 3\n- 4",
"output": "3"
},
{
"input": "3\n- 1\n- 2\n- 3",
"output": "3"
},
{
"input": "4\n+ 1\n+ 2\n- 1\n+ 3",
"output": "2"
},
{
"input": "6\n+ 1\n+ 2\n- 1\n+ 3\n- 2\n+ 4",
"output": "2"
},
{
"input": "3\n+ 1\n+ 2\n- 3",
"output": "3"
},
{
"input": "3\n- 1\n+ 2\n- 2",
"output": "1"
},
{
"input": "4\n- 1\n- 2\n+ 3\n+ 4",
"output": "2"
},
{
"input": "1\n+ 1",
"output": "1"
},
{
"input": "1\n- 1",
"output": "1"
},
{
"input": "3\n- 1\n+ 1\n- 1",
"output": "1"
},
{
"input": "10\n+ 1\n+ 2\n+ 3\n+ 4\n+ 5\n+ 6\n+ 7\n+ 8\n+ 9\n+ 10",
"output": "10"
},
{
"input": "5\n+ 5\n+ 4\n- 4\n- 5\n+ 5",
"output": "2"
},
{
"input": "50\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100",
"output": "1"
},
{
"input": "10\n- 8\n- 4\n+ 8\n+ 10\n+ 6\n- 8\n+ 9\n- 2\n- 7\n+ 4",
"output": "5"
},
{
"input": "20\n+ 3\n- 3\n- 2\n+ 2\n+ 3\n- 5\n- 1\n+ 1\n- 3\n+ 4\n- 1\n+ 1\n+ 3\n- 3\n+ 5\n- 2\n- 1\n+ 2\n+ 1\n- 5",
"output": "4"
},
{
"input": "50\n+ 4\n+ 5\n+ 3\n+ 2\n- 2\n- 3\n- 4\n+ 3\n+ 2\n- 3\n+ 4\n- 2\n- 4\n+ 2\n+ 3\n- 3\n- 5\n- 1\n+ 4\n+ 5\n- 5\n+ 3\n- 4\n- 3\n- 2\n+ 4\n+ 3\n+ 2\n- 2\n- 4\n+ 5\n+ 1\n+ 4\n+ 2\n- 2\n+ 2\n- 3\n- 5\n- 4\n- 1\n+ 5\n- 2\n- 5\n+ 5\n+ 3\n- 3\n+ 1\n+ 3\n+ 2\n- 1",
"output": "5"
},
{
"input": "10\n- 2\n+ 1\n- 1\n+ 2\n- 2\n+ 2\n+ 1\n- 1\n- 2\n+ 1",
"output": "2"
},
{
"input": "50\n+ 1\n+ 2\n+ 3\n+ 4\n+ 5\n+ 6\n+ 7\n+ 8\n+ 9\n+ 10\n+ 11\n+ 12\n+ 13\n+ 14\n+ 15\n+ 16\n+ 17\n+ 18\n+ 19\n+ 20\n+ 21\n+ 22\n+ 23\n+ 24\n+ 25\n+ 26\n+ 27\n+ 28\n+ 29\n+ 30\n+ 31\n+ 32\n+ 33\n+ 34\n+ 35\n+ 36\n+ 37\n+ 38\n+ 39\n+ 40\n+ 41\n+ 42\n+ 43\n+ 44\n+ 45\n+ 46\n+ 47\n+ 48\n+ 49\n+ 50",
"output": "50"
},
{
"input": "50\n- 1\n- 2\n- 3\n- 4\n- 5\n- 6\n- 7\n- 8\n- 9\n- 10\n- 11\n- 12\n- 13\n- 14\n- 15\n- 16\n- 17\n- 18\n- 19\n- 20\n- 21\n- 22\n- 23\n- 24\n- 25\n- 26\n- 27\n- 28\n- 29\n- 30\n- 31\n- 32\n- 33\n- 34\n- 35\n- 36\n- 37\n- 38\n- 39\n- 40\n- 41\n- 42\n- 43\n- 44\n- 45\n- 46\n- 47\n- 48\n- 49\n- 50",
"output": "50"
}
] | 1,667,199,319 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 46 | 0 | n = int(input())
visitors = 0
best = 0
for _ in range(n):
op, id = input().split()
if op == "+":
visitors += 1
best = max(best, abs(visitors))
elif op == "-":
visitors -= 1
if visitors < 0:
visitors = 0
best += 1
print(best)
| Title: Berland National Library
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Berland National Library has recently been built in the capital of Berland. In addition, in the library you can take any of the collected works of Berland leaders, the library has a reading room.
Today was the pilot launch of an automated reading room visitors' accounting system! The scanner of the system is installed at the entrance to the reading room. It records the events of the form "reader entered room", "reader left room". Every reader is assigned a registration number during the registration procedure at the library — it's a unique integer from 1 to 106. Thus, the system logs events of two forms:
- "+ *r**i*" — the reader with registration number *r**i* entered the room; - "- *r**i*" — the reader with registration number *r**i* left the room.
The first launch of the system was a success, it functioned for some period of time, and, at the time of its launch and at the time of its shutdown, the reading room may already have visitors.
Significant funds of the budget of Berland have been spent on the design and installation of the system. Therefore, some of the citizens of the capital now demand to explain the need for this system and the benefits that its implementation will bring. Now, the developers of the system need to urgently come up with reasons for its existence.
Help the system developers to find the minimum possible capacity of the reading room (in visitors) using the log of the system available to you.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of records in the system log. Next follow *n* events from the system journal in the order in which the were made. Each event was written on a single line and looks as "+ *r**i*" or "- *r**i*", where *r**i* is an integer from 1 to 106, the registration number of the visitor (that is, distinct visitors always have distinct registration numbers).
It is guaranteed that the log is not contradictory, that is, for every visitor the types of any of his two consecutive events are distinct. Before starting the system, and after stopping the room may possibly contain visitors.
Output Specification:
Print a single integer — the minimum possible capacity of the reading room.
Demo Input:
['6\n+ 12001\n- 12001\n- 1\n- 1200\n+ 1\n+ 7\n', '2\n- 1\n- 2\n', '2\n+ 1\n- 1\n']
Demo Output:
['3', '2', '1']
Note:
In the first sample test, the system log will ensure that at some point in the reading room were visitors with registration numbers 1, 1200 and 12001. More people were not in the room at the same time based on the log. Therefore, the answer to the test is 3. | ```python
n = int(input())
visitors = 0
best = 0
for _ in range(n):
op, id = input().split()
if op == "+":
visitors += 1
best = max(best, abs(visitors))
elif op == "-":
visitors -= 1
if visitors < 0:
visitors = 0
best += 1
print(best)
``` | 0 | |
900 | A | Find Extra One | PROGRAMMING | 800 | [
"geometry",
"implementation"
] | null | null | You have *n* distinct points on a plane, none of them lie on *OY* axis. Check that there is a point after removal of which the remaining points are located on one side of the *OY* axis. | The first line contains a single positive integer *n* (2<=≤<=*n*<=≤<=105).
The following *n* lines contain coordinates of the points. The *i*-th of these lines contains two single integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=109, *x**i*<=≠<=0). No two points coincide. | Print "Yes" if there is such a point, "No" — otherwise.
You can print every letter in any case (upper or lower). | [
"3\n1 1\n-1 -1\n2 -1\n",
"4\n1 1\n2 2\n-1 1\n-2 2\n",
"3\n1 2\n2 1\n4 60\n"
] | [
"Yes",
"No",
"Yes"
] | In the first example the second point can be removed.
In the second example there is no suitable for the condition point.
In the third example any point can be removed. | 500 | [
{
"input": "3\n1 1\n-1 -1\n2 -1",
"output": "Yes"
},
{
"input": "4\n1 1\n2 2\n-1 1\n-2 2",
"output": "No"
},
{
"input": "3\n1 2\n2 1\n4 60",
"output": "Yes"
},
{
"input": "10\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n-1 -1",
"output": "Yes"
},
{
"input": "2\n1000000000 -1000000000\n1000000000 1000000000",
"output": "Yes"
},
{
"input": "23\n-1 1\n-1 2\n-2 4\n-7 -8\n-3 3\n-9 -14\n-5 3\n-6 2\n-7 11\n-4 4\n-8 5\n1 1\n-1 -1\n-1 -2\n-2 -4\n-7 8\n-3 -3\n-9 14\n-5 -3\n-6 -2\n-7 -11\n-4 -4\n-8 -5",
"output": "Yes"
},
{
"input": "4\n-1000000000 -1000000000\n1000000000 1000000000\n-1000000000 1000000000\n1000000000 -1000000000",
"output": "No"
},
{
"input": "2\n-1000000000 1000000000\n-1000000000 -1000000000",
"output": "Yes"
},
{
"input": "5\n-1 -1\n-2 2\n2 2\n2 -2\n3 2",
"output": "No"
},
{
"input": "2\n1 0\n-1 0",
"output": "Yes"
},
{
"input": "4\n-1 1\n-1 2\n-1 3\n-1 4",
"output": "Yes"
},
{
"input": "2\n-1 0\n1 0",
"output": "Yes"
},
{
"input": "2\n1 2\n-1 2",
"output": "Yes"
},
{
"input": "2\n8 0\n7 0",
"output": "Yes"
},
{
"input": "6\n-1 0\n-2 0\n-1 -1\n-1 5\n1 0\n1 1",
"output": "No"
},
{
"input": "4\n1 0\n2 0\n-1 0\n-2 0",
"output": "No"
},
{
"input": "4\n-2 0\n-1 0\n1 0\n2 0",
"output": "No"
},
{
"input": "2\n1 1\n-1 1",
"output": "Yes"
},
{
"input": "4\n-1 0\n-2 0\n1 0\n2 0",
"output": "No"
},
{
"input": "2\n4 3\n-4 -2",
"output": "Yes"
},
{
"input": "4\n1 0\n2 0\n-1 1\n-1 2",
"output": "No"
},
{
"input": "5\n1 1\n2 1\n3 1\n-1 1\n-2 1",
"output": "No"
},
{
"input": "2\n1 1\n-1 -1",
"output": "Yes"
},
{
"input": "4\n1 2\n1 0\n1 -2\n-1 2",
"output": "Yes"
},
{
"input": "5\n-2 3\n-3 3\n4 2\n3 2\n1 2",
"output": "No"
},
{
"input": "3\n2 0\n3 0\n4 0",
"output": "Yes"
},
{
"input": "5\n-3 1\n-2 1\n-1 1\n1 1\n2 1",
"output": "No"
},
{
"input": "4\n-3 0\n1 0\n2 0\n3 0",
"output": "Yes"
},
{
"input": "2\n1 0\n-1 1",
"output": "Yes"
},
{
"input": "3\n-1 0\n1 0\n2 0",
"output": "Yes"
},
{
"input": "5\n1 0\n3 0\n-1 0\n-6 0\n-4 1",
"output": "No"
},
{
"input": "5\n-1 2\n-2 2\n-3 1\n1 2\n2 3",
"output": "No"
},
{
"input": "3\n1 0\n-1 0\n-2 0",
"output": "Yes"
},
{
"input": "4\n1 0\n2 0\n3 1\n4 1",
"output": "Yes"
},
{
"input": "4\n1 0\n1 2\n1 3\n-1 5",
"output": "Yes"
},
{
"input": "4\n2 2\n2 5\n-2 3\n-2 0",
"output": "No"
},
{
"input": "4\n1 1\n-1 1\n-1 0\n-1 -1",
"output": "Yes"
},
{
"input": "4\n2 0\n3 0\n-3 -3\n-3 -4",
"output": "No"
},
{
"input": "4\n-1 0\n-2 0\n-3 0\n-4 0",
"output": "Yes"
},
{
"input": "2\n-1 1\n1 1",
"output": "Yes"
},
{
"input": "5\n1 1\n2 2\n3 3\n-4 -4\n-5 -5",
"output": "No"
},
{
"input": "5\n2 0\n3 0\n4 0\n5 0\n6 0",
"output": "Yes"
},
{
"input": "2\n-1 2\n1 2",
"output": "Yes"
},
{
"input": "4\n1 1\n2 1\n-3 0\n-4 0",
"output": "No"
},
{
"input": "4\n-1 0\n-2 0\n3 0\n4 0",
"output": "No"
},
{
"input": "3\n3 0\n2 0\n1 0",
"output": "Yes"
},
{
"input": "4\n-2 0\n-3 0\n1 -1\n3 1",
"output": "No"
},
{
"input": "3\n-1 -1\n1 1\n2 2",
"output": "Yes"
},
{
"input": "4\n-2 0\n-1 0\n2 0\n1 0",
"output": "No"
},
{
"input": "2\n-3 5\n3 5",
"output": "Yes"
},
{
"input": "2\n-1 5\n1 5",
"output": "Yes"
},
{
"input": "4\n2 0\n3 0\n-2 0\n-3 0",
"output": "No"
},
{
"input": "3\n-1 1\n1 1\n1 -1",
"output": "Yes"
},
{
"input": "2\n1 0\n2 0",
"output": "Yes"
},
{
"input": "4\n-1 1\n-2 1\n2 -1\n3 -1",
"output": "No"
},
{
"input": "5\n1 0\n2 0\n3 0\n-1 0\n-2 0",
"output": "No"
},
{
"input": "4\n-3 0\n-4 0\n-5 0\n-6 0",
"output": "Yes"
},
{
"input": "6\n-3 0\n-2 0\n-1 0\n1 0\n2 0\n3 0",
"output": "No"
},
{
"input": "4\n5 0\n5 1\n6 0\n6 1",
"output": "Yes"
}
] | 1,570,448,122 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 327 | 0 | n=int(input())
r=0
l=0
for _ in range(n):
x,y=map(int,input().split())
if (x>0 and y>=0) or (x>0 and y<=0):
r=r+1
else:
l=l+1
if min(r,l)>1:
print("No")
else:
print("Yes") | Title: Find Extra One
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have *n* distinct points on a plane, none of them lie on *OY* axis. Check that there is a point after removal of which the remaining points are located on one side of the *OY* axis.
Input Specification:
The first line contains a single positive integer *n* (2<=≤<=*n*<=≤<=105).
The following *n* lines contain coordinates of the points. The *i*-th of these lines contains two single integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=109, *x**i*<=≠<=0). No two points coincide.
Output Specification:
Print "Yes" if there is such a point, "No" — otherwise.
You can print every letter in any case (upper or lower).
Demo Input:
['3\n1 1\n-1 -1\n2 -1\n', '4\n1 1\n2 2\n-1 1\n-2 2\n', '3\n1 2\n2 1\n4 60\n']
Demo Output:
['Yes', 'No', 'Yes']
Note:
In the first example the second point can be removed.
In the second example there is no suitable for the condition point.
In the third example any point can be removed. | ```python
n=int(input())
r=0
l=0
for _ in range(n):
x,y=map(int,input().split())
if (x>0 and y>=0) or (x>0 and y<=0):
r=r+1
else:
l=l+1
if min(r,l)>1:
print("No")
else:
print("Yes")
``` | 3 | |
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,674,507,327 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 92 | 0 | n=input()
m=input()
sum=""
for i in range(-1,-len(n),-1):
sum+=n[i]
sum+=n[-len(n)]
if(sum==m):
print("YES")
elif(sum!=m):
print("NO")
| Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
n=input()
m=input()
sum=""
for i in range(-1,-len(n),-1):
sum+=n[i]
sum+=n[-len(n)]
if(sum==m):
print("YES")
elif(sum!=m):
print("NO")
``` | 3.977 |
237 | A | Free Cash | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period. | Print a single integer — the minimum number of cashes, needed to serve all clients next day. | [
"4\n8 0\n8 10\n8 10\n8 45\n",
"3\n0 12\n10 11\n22 22\n"
] | [
"2\n",
"1\n"
] | In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash. | 500 | [
{
"input": "4\n8 0\n8 10\n8 10\n8 45",
"output": "2"
},
{
"input": "3\n0 12\n10 11\n22 22",
"output": "1"
},
{
"input": "5\n12 8\n15 27\n15 27\n16 2\n19 52",
"output": "2"
},
{
"input": "7\n5 6\n7 34\n7 34\n7 34\n12 29\n15 19\n20 23",
"output": "3"
},
{
"input": "8\n0 36\n4 7\n4 7\n4 7\n11 46\n12 4\n15 39\n18 6",
"output": "3"
},
{
"input": "20\n4 12\n4 21\n4 27\n4 56\n5 55\n7 56\n11 28\n11 36\n14 58\n15 59\n16 8\n17 12\n17 23\n17 23\n17 23\n17 23\n17 23\n17 23\n20 50\n22 32",
"output": "6"
},
{
"input": "10\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30",
"output": "10"
},
{
"input": "50\n0 23\n1 21\n2 8\n2 45\n3 1\n4 19\n4 37\n7 7\n7 40\n8 43\n9 51\n10 13\n11 2\n11 19\n11 30\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 54\n13 32\n13 42\n14 29\n14 34\n14 48\n15 0\n15 27\n16 22\n16 31\n17 25\n17 26\n17 33\n18 14\n18 16\n18 20\n19 0\n19 5\n19 56\n20 22\n21 26\n22 0\n22 10\n22 11\n22 36\n23 17\n23 20",
"output": "8"
},
{
"input": "10\n0 39\n1 35\n1 49\n1 51\n5 24\n7 40\n7 56\n16 42\n23 33\n23 49",
"output": "1"
},
{
"input": "15\n0 16\n6 15\n8 2\n8 6\n8 7\n10 1\n10 1\n10 3\n10 12\n13 5\n14 16\n14 16\n14 16\n14 16\n14 16",
"output": "5"
},
{
"input": "2\n0 24\n1 0",
"output": "1"
},
{
"input": "1\n0 0",
"output": "1"
},
{
"input": "1\n1 5",
"output": "1"
},
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "3"
},
{
"input": "1\n5 0",
"output": "1"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "5"
},
{
"input": "1\n10 10",
"output": "1"
},
{
"input": "1\n8 0",
"output": "1"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "10"
},
{
"input": "2\n0 0\n0 1",
"output": "1"
},
{
"input": "2\n8 5\n8 5",
"output": "2"
}
] | 1,658,726,421 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 9 | 2,000 | 9,420,800 | def _input(): return map(int, input().split())
lst = []
for _ in range(int(input())):
a, b = _input()
lst.append((a,b))
res = 0; b = set(lst)
for i in b:
res = max(res, lst.count(i))
print(res) | Title: Free Cash
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
Output Specification:
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
Demo Input:
['4\n8 0\n8 10\n8 10\n8 45\n', '3\n0 12\n10 11\n22 22\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash. | ```python
def _input(): return map(int, input().split())
lst = []
for _ in range(int(input())):
a, b = _input()
lst.append((a,b))
res = 0; b = set(lst)
for i in b:
res = max(res, lst.count(i))
print(res)
``` | 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"
}
] | 1,607,667,322 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 154 | 307,200 | n=int(input())
lst=map(int,input().split())
min=99999
for i in range(0,n-1):
if abs(lst[i]-lst[i+1]<min):
min=abs(lst[i]-lst[i+1])
id1=i+1
id2=i+2
if(abs(lst[-1]-lst[0])<min):
min=abs(lst[-1]-lst[0])
id1=n
id2=1
print(id1,end=" ")
print(id2) | 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())
lst=map(int,input().split())
min=99999
for i in range(0,n-1):
if abs(lst[i]-lst[i+1]<min):
min=abs(lst[i]-lst[i+1])
id1=i+1
id2=i+2
if(abs(lst[-1]-lst[0])<min):
min=abs(lst[-1]-lst[0])
id1=n
id2=1
print(id1,end=" ")
print(id2)
``` | -1 |
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"
}
] | 1,615,559,191 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 80 | 186 | 0 | #!/usr/bin/env python
# coding: utf-8
# In[2]:
n = int(input())
summ=0
for i in range(0,n):
A = map(int, input().split())
summ+=sum(A)
if summ==0:
print('YES')
else:
print('NO')
# In[ ]:
| 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
#!/usr/bin/env python
# coding: utf-8
# In[2]:
n = int(input())
summ=0
for i in range(0,n):
A = map(int, input().split())
summ+=sum(A)
if summ==0:
print('YES')
else:
print('NO')
# In[ ]:
``` | 0 |
52 | C | Circular RMQ | PROGRAMMING | 2,200 | [
"data structures"
] | C. Circular RMQ | 1 | 256 | You are given circular array *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. There are two types of operations with it:
- *inc*(*lf*,<=*rg*,<=*v*) — this operation increases each element on the segment [*lf*,<=*rg*] (inclusively) by *v*; - *rmq*(*lf*,<=*rg*) — this operation returns minimal value on the segment [*lf*,<=*rg*] (inclusively).
Assume segments to be circular, so if *n*<==<=5 and *lf*<==<=3,<=*rg*<==<=1, it means the index sequence: 3,<=4,<=0,<=1.
Write program to process given sequence of operations. | The first line contains integer *n* (1<=≤<=*n*<=≤<=200000). The next line contains initial state of the array: *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (<=-<=106<=≤<=*a**i*<=≤<=106), *a**i* are integer. The third line contains integer *m* (0<=≤<=*m*<=≤<=200000), *m* — the number of operartons. Next *m* lines contain one operation each. If line contains two integer *lf*,<=*rg* (0<=≤<=*lf*,<=*rg*<=≤<=*n*<=-<=1) it means *rmq* operation, it contains three integers *lf*,<=*rg*,<=*v* (0<=≤<=*lf*,<=*rg*<=≤<=*n*<=-<=1;<=-<=106<=≤<=*v*<=≤<=106) — *inc* operation. | For each *rmq* operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d). | [
"4\n1 2 3 4\n4\n3 0\n3 0 -1\n0 1\n2 1\n"
] | [
"1\n0\n0\n"
] | none | 1,500 | [
{
"input": "4\n1 2 3 4\n4\n3 0\n3 0 -1\n0 1\n2 1",
"output": "1\n0\n0"
},
{
"input": "1\n-1\n10\n0 0 -1\n0 0\n0 0 1\n0 0\n0 0 1\n0 0\n0 0 0\n0 0\n0 0 -1\n0 0 1",
"output": "-2\n-1\n0\n0"
},
{
"input": "2\n-1 -1\n10\n0 0\n0 0\n0 0 1\n0 0\n1 1\n0 0 -1\n0 0 0\n0 0 1\n1 1 0\n0 0 -1",
"output": "-1\n-1\n0\n-1"
}
] | 1,591,197,418 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 124 | 0 | import sys
from math import inf, log2
class LazySegmentTree:
def __init__(self, array, func=min):
self.n = len(array)
self.size = 2**(int(log2(self.n-1))+1) if n != 1 else 1
self.func = func
self.default = 0 if self.func != min else inf
self.data = [self.default] * (2 * self.size)
self.lazy = [0] * (2 * self.size)
self.process(array)
def process(self, array):
self.data[self.size : self.size+self.n] = array
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):
self.lazy[2*index] += self.lazy[index]
self.lazy[2*index+1] += self.lazy[index]
self.data[2 * index] += self.lazy[index]
self.data[2 * index + 1] += self.lazy[index]
self.lazy[index] = 0
def build(self, index):
index >>= 1
while index:
self.data[index] = self.func(self.data[2*index], self.data[2*index+1]) + self.lazy[index]
index >>= 1
def query(self, alpha, omega):
res = self.default
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):
alpha += self.size
omega += self.size + 1
l, r = alpha, omega
while alpha < omega:
if alpha & 1:
self.data[alpha] += value
self.lazy[alpha] += value
alpha += 1
if omega & 1:
omega -= 1
self.data[omega] += value
self.lazy[omega] += value
alpha >>= 1
omega >>= 1
self.build(l)
self.build(r-1)
debug=False
lines = sys.stdin.read().split("\n")
lines.pop(-1)
print(lines)
| Title: Circular RMQ
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given circular array *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. There are two types of operations with it:
- *inc*(*lf*,<=*rg*,<=*v*) — this operation increases each element on the segment [*lf*,<=*rg*] (inclusively) by *v*; - *rmq*(*lf*,<=*rg*) — this operation returns minimal value on the segment [*lf*,<=*rg*] (inclusively).
Assume segments to be circular, so if *n*<==<=5 and *lf*<==<=3,<=*rg*<==<=1, it means the index sequence: 3,<=4,<=0,<=1.
Write program to process given sequence of operations.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=200000). The next line contains initial state of the array: *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (<=-<=106<=≤<=*a**i*<=≤<=106), *a**i* are integer. The third line contains integer *m* (0<=≤<=*m*<=≤<=200000), *m* — the number of operartons. Next *m* lines contain one operation each. If line contains two integer *lf*,<=*rg* (0<=≤<=*lf*,<=*rg*<=≤<=*n*<=-<=1) it means *rmq* operation, it contains three integers *lf*,<=*rg*,<=*v* (0<=≤<=*lf*,<=*rg*<=≤<=*n*<=-<=1;<=-<=106<=≤<=*v*<=≤<=106) — *inc* operation.
Output Specification:
For each *rmq* operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Demo Input:
['4\n1 2 3 4\n4\n3 0\n3 0 -1\n0 1\n2 1\n']
Demo Output:
['1\n0\n0\n']
Note:
none | ```python
import sys
from math import inf, log2
class LazySegmentTree:
def __init__(self, array, func=min):
self.n = len(array)
self.size = 2**(int(log2(self.n-1))+1) if n != 1 else 1
self.func = func
self.default = 0 if self.func != min else inf
self.data = [self.default] * (2 * self.size)
self.lazy = [0] * (2 * self.size)
self.process(array)
def process(self, array):
self.data[self.size : self.size+self.n] = array
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):
self.lazy[2*index] += self.lazy[index]
self.lazy[2*index+1] += self.lazy[index]
self.data[2 * index] += self.lazy[index]
self.data[2 * index + 1] += self.lazy[index]
self.lazy[index] = 0
def build(self, index):
index >>= 1
while index:
self.data[index] = self.func(self.data[2*index], self.data[2*index+1]) + self.lazy[index]
index >>= 1
def query(self, alpha, omega):
res = self.default
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):
alpha += self.size
omega += self.size + 1
l, r = alpha, omega
while alpha < omega:
if alpha & 1:
self.data[alpha] += value
self.lazy[alpha] += value
alpha += 1
if omega & 1:
omega -= 1
self.data[omega] += value
self.lazy[omega] += value
alpha >>= 1
omega >>= 1
self.build(l)
self.build(r-1)
debug=False
lines = sys.stdin.read().split("\n")
lines.pop(-1)
print(lines)
``` | 0 |
612 | E | Square Root of Permutation | PROGRAMMING | 2,200 | [
"combinatorics",
"constructive algorithms",
"dfs and similar",
"graphs",
"math"
] | null | null | A permutation of length *n* is an array containing each integer from 1 to *n* exactly once. For example, *q*<==<=[4,<=5,<=1,<=2,<=3] is a permutation. For the permutation *q* the square of permutation is the permutation *p* that *p*[*i*]<==<=*q*[*q*[*i*]] for each *i*<==<=1... *n*. For example, the square of *q*<==<=[4,<=5,<=1,<=2,<=3] is *p*<==<=*q*2<==<=[2,<=3,<=4,<=5,<=1].
This problem is about the inverse operation: given the permutation *p* you task is to find such permutation *q* that *q*2<==<=*p*. If there are several such *q* find any of them. | The first line contains integer *n* (1<=≤<=*n*<=≤<=106) — the number of elements in permutation *p*.
The second line contains *n* distinct integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the elements of permutation *p*. | If there is no permutation *q* such that *q*2<==<=*p* print the number "-1".
If the answer exists print it. The only line should contain *n* different integers *q**i* (1<=≤<=*q**i*<=≤<=*n*) — the elements of the permutation *q*. If there are several solutions print any of them. | [
"4\n2 1 4 3\n",
"4\n2 1 3 4\n",
"5\n2 3 4 5 1\n"
] | [
"3 4 2 1\n",
"-1\n",
"4 5 1 2 3\n"
] | none | 0 | [
{
"input": "4\n2 1 4 3",
"output": "3 4 2 1"
},
{
"input": "4\n2 1 3 4",
"output": "-1"
},
{
"input": "5\n2 3 4 5 1",
"output": "4 5 1 2 3"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n8 2 10 3 4 6 1 7 9 5",
"output": "-1"
},
{
"input": "10\n3 5 1 2 10 8 7 6 4 9",
"output": "6 9 8 10 4 3 7 1 5 2"
},
{
"input": "100\n11 9 35 34 51 74 16 67 26 21 14 80 84 79 7 61 28 3 53 43 42 5 56 36 69 30 22 88 1 27 65 91 46 31 59 50 17 96 25 18 64 55 78 2 63 24 95 48 93 13 38 76 89 94 15 90 45 81 52 87 83 73 44 49 23 82 85 75 86 33 47 19 58 97 37 20 40 10 92 4 6 68 77 54 71 12 62 60 100 39 41 99 72 29 57 8 70 32 66 98",
"output": "-1"
},
{
"input": "100\n94 22 24 99 58 97 20 29 67 30 38 64 77 50 15 44 92 88 39 42 25 70 2 76 84 6 37 49 17 71 31 19 26 79 10 35 65 63 32 95 5 8 52 27 83 18 53 93 13 81 48 68 54 82 34 60 87 23 16 86 55 40 61 45 28 7 74 41 14 91 3 72 33 11 98 89 90 69 78 36 80 59 56 21 43 1 75 46 47 12 96 73 57 51 4 85 9 100 66 62",
"output": "78 52 95 76 96 49 53 59 77 100 64 11 9 48 15 17 44 46 32 54 84 68 43 4 21 28 73 6 16 62 31 39 65 86 98 75 33 45 19 3 91 82 2 92 63 88 7 50 97 93 14 22 20 42 60 55 80 85 29 34 56 71 83 38 26 47 90 70 51 41 40 72 37 12 35 99 67 94 1 87 57 8 61 25 23 79 36 18 66 74 5 27 81 69 24 58 13 10 89 30"
}
] | 1,502,994,086 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 62 | 204,800 |
n = int(input())
c = input()
c = c.split(" ")
c = [int(p) - 1 for p in c]
visited = [0 for p in range(0,n,1)]
cycle_st = []
for i in range(0,n,1):
if visited[i] != 1:
cycle = []
cycle.append([i,c[i]])
visited[i] = 1
d = c[i]
while visited[d] != 1:
cycle.append([d,c[d]])
visited[d] = 1
d = c[d]
cycle_st.append(cycle)
used = [ 1 for p in range(0,len(cycle_st),1)]
posssible = True
have_cycle =True
i = 0
j = 0
for i in range(len(cycle_st)):
if used[i]:
have_cycle = False
if(len(cycle_st[i]) % 2 == 1):
for j in range(len(cycle_st[i])):
visited[cycle_st[i][j][0]] = cycle_st[i][(j+1)%len(cycle_st[i])][1]
else:
for j in range(i+1,len(cycle_st),1):
if used[j] and len(cycle_st[i]) == len(cycle_st[j]):
have_cycle = True
for k in range(len(cycle_st[i])):
visited[cycle_st[i][k][0]] = cycle_st[j][k-1][0]
visited[cycle_st[j][k-1][0]] = cycle_st[i][k-1][0]
used[j] = 0
break
if not have_cycle:
posssible = False
break
if not posssible:
print(-1)
else:
for p in visited:
print(str(p+1)) | Title: Square Root of Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation of length *n* is an array containing each integer from 1 to *n* exactly once. For example, *q*<==<=[4,<=5,<=1,<=2,<=3] is a permutation. For the permutation *q* the square of permutation is the permutation *p* that *p*[*i*]<==<=*q*[*q*[*i*]] for each *i*<==<=1... *n*. For example, the square of *q*<==<=[4,<=5,<=1,<=2,<=3] is *p*<==<=*q*2<==<=[2,<=3,<=4,<=5,<=1].
This problem is about the inverse operation: given the permutation *p* you task is to find such permutation *q* that *q*2<==<=*p*. If there are several such *q* find any of them.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=106) — the number of elements in permutation *p*.
The second line contains *n* distinct integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the elements of permutation *p*.
Output Specification:
If there is no permutation *q* such that *q*2<==<=*p* print the number "-1".
If the answer exists print it. The only line should contain *n* different integers *q**i* (1<=≤<=*q**i*<=≤<=*n*) — the elements of the permutation *q*. If there are several solutions print any of them.
Demo Input:
['4\n2 1 4 3\n', '4\n2 1 3 4\n', '5\n2 3 4 5 1\n']
Demo Output:
['3 4 2 1\n', '-1\n', '4 5 1 2 3\n']
Note:
none | ```python
n = int(input())
c = input()
c = c.split(" ")
c = [int(p) - 1 for p in c]
visited = [0 for p in range(0,n,1)]
cycle_st = []
for i in range(0,n,1):
if visited[i] != 1:
cycle = []
cycle.append([i,c[i]])
visited[i] = 1
d = c[i]
while visited[d] != 1:
cycle.append([d,c[d]])
visited[d] = 1
d = c[d]
cycle_st.append(cycle)
used = [ 1 for p in range(0,len(cycle_st),1)]
posssible = True
have_cycle =True
i = 0
j = 0
for i in range(len(cycle_st)):
if used[i]:
have_cycle = False
if(len(cycle_st[i]) % 2 == 1):
for j in range(len(cycle_st[i])):
visited[cycle_st[i][j][0]] = cycle_st[i][(j+1)%len(cycle_st[i])][1]
else:
for j in range(i+1,len(cycle_st),1):
if used[j] and len(cycle_st[i]) == len(cycle_st[j]):
have_cycle = True
for k in range(len(cycle_st[i])):
visited[cycle_st[i][k][0]] = cycle_st[j][k-1][0]
visited[cycle_st[j][k-1][0]] = cycle_st[i][k-1][0]
used[j] = 0
break
if not have_cycle:
posssible = False
break
if not posssible:
print(-1)
else:
for p in visited:
print(str(p+1))
``` | 0 | |
591 | A | Wizards' Duel | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length *l*. Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry's magic spell flies at a speed of *p* meters per second, and the impulse of You-Know-Who's magic spell flies at a speed of *q* meters per second.
The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it's caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse.
Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn't good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight. | The first line of the input contains a single integer *l* (1<=≤<=*l*<=≤<=1<=000) — the length of the corridor where the fight takes place.
The second line contains integer *p*, the third line contains integer *q* (1<=≤<=*p*,<=*q*<=≤<=500) — the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively. | Print a single real number — the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed 10<=-<=4.
Namely: let's assume that your answer equals *a*, and the answer of the jury is *b*. The checker program will consider your answer correct if . | [
"100\n50\n50\n",
"199\n60\n40\n"
] | [
"50\n",
"119.4\n"
] | In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor. | 500 | [
{
"input": "100\n50\n50",
"output": "50"
},
{
"input": "199\n60\n40",
"output": "119.4"
},
{
"input": "1\n1\n1",
"output": "0.5"
},
{
"input": "1\n1\n500",
"output": "0.001996007984"
},
{
"input": "1\n500\n1",
"output": "0.998003992"
},
{
"input": "1\n500\n500",
"output": "0.5"
},
{
"input": "1000\n1\n1",
"output": "500"
},
{
"input": "1000\n1\n500",
"output": "1.996007984"
},
{
"input": "1000\n500\n1",
"output": "998.003992"
},
{
"input": "1000\n500\n500",
"output": "500"
},
{
"input": "101\n11\n22",
"output": "33.66666667"
},
{
"input": "987\n1\n3",
"output": "246.75"
},
{
"input": "258\n25\n431",
"output": "14.14473684"
},
{
"input": "979\n39\n60",
"output": "385.6666667"
},
{
"input": "538\n479\n416",
"output": "287.9351955"
},
{
"input": "583\n112\n248",
"output": "181.3777778"
},
{
"input": "978\n467\n371",
"output": "545.0190931"
},
{
"input": "980\n322\n193",
"output": "612.7378641"
},
{
"input": "871\n401\n17",
"output": "835.576555"
},
{
"input": "349\n478\n378",
"output": "194.885514"
},
{
"input": "425\n458\n118",
"output": "337.9340278"
},
{
"input": "919\n323\n458",
"output": "380.0729834"
},
{
"input": "188\n59\n126",
"output": "59.95675676"
},
{
"input": "644\n428\n484",
"output": "302.2280702"
},
{
"input": "253\n80\n276",
"output": "56.85393258"
},
{
"input": "745\n152\n417",
"output": "199.0158172"
},
{
"input": "600\n221\n279",
"output": "265.2"
},
{
"input": "690\n499\n430",
"output": "370.6243272"
},
{
"input": "105\n68\n403",
"output": "15.15923567"
},
{
"input": "762\n462\n371",
"output": "422.6218487"
},
{
"input": "903\n460\n362",
"output": "505.3284672"
},
{
"input": "886\n235\n95",
"output": "630.9393939"
},
{
"input": "655\n203\n18",
"output": "601.6515837"
},
{
"input": "718\n29\n375",
"output": "51.53960396"
},
{
"input": "296\n467\n377",
"output": "163.7819905"
},
{
"input": "539\n61\n56",
"output": "281.017094"
},
{
"input": "133\n53\n124",
"output": "39.82485876"
},
{
"input": "998\n224\n65",
"output": "773.5363322"
},
{
"input": "961\n173\n47",
"output": "755.6954545"
},
{
"input": "285\n468\n62",
"output": "251.6603774"
},
{
"input": "496\n326\n429",
"output": "214.1668874"
},
{
"input": "627\n150\n285",
"output": "216.2068966"
},
{
"input": "961\n443\n50",
"output": "863.535497"
},
{
"input": "623\n422\n217",
"output": "411.4334898"
},
{
"input": "678\n295\n29",
"output": "617.3148148"
}
] | 1,445,765,046 | 1,446 | Python 3 | OK | TESTS | 45 | 62 | 0 | l, p, q = map(int, [input(), input(), input()])
print(l * p / float(p + q)) | Title: Wizards' Duel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length *l*. Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry's magic spell flies at a speed of *p* meters per second, and the impulse of You-Know-Who's magic spell flies at a speed of *q* meters per second.
The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it's caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse.
Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn't good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight.
Input Specification:
The first line of the input contains a single integer *l* (1<=≤<=*l*<=≤<=1<=000) — the length of the corridor where the fight takes place.
The second line contains integer *p*, the third line contains integer *q* (1<=≤<=*p*,<=*q*<=≤<=500) — the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively.
Output Specification:
Print a single real number — the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed 10<=-<=4.
Namely: let's assume that your answer equals *a*, and the answer of the jury is *b*. The checker program will consider your answer correct if .
Demo Input:
['100\n50\n50\n', '199\n60\n40\n']
Demo Output:
['50\n', '119.4\n']
Note:
In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor. | ```python
l, p, q = map(int, [input(), input(), input()])
print(l * p / float(p + q))
``` | 3 | |
834 | B | The Festive Evening | PROGRAMMING | 1,100 | [
"data structures",
"implementation"
] | null | null | It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in.
There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously.
For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed.
Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened. | Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26).
In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest. | Output «YES» if at least one door was unguarded during some time, and «NO» otherwise.
You can output each letter in arbitrary case (upper or lower). | [
"5 1\nAABBB\n",
"5 1\nABABB\n"
] | [
"NO\n",
"YES\n"
] | In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened.
In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door. | 1,000 | [
{
"input": "5 1\nAABBB",
"output": "NO"
},
{
"input": "5 1\nABABB",
"output": "YES"
},
{
"input": "26 1\nABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "NO"
},
{
"input": "27 1\nABCDEFGHIJKLMNOPQRSTUVWXYZA",
"output": "YES"
},
{
"input": "5 2\nABACA",
"output": "NO"
},
{
"input": "6 2\nABCABC",
"output": "YES"
},
{
"input": "8 3\nABCBCDCA",
"output": "NO"
},
{
"input": "73 2\nDEBECECBBADAADEAABEAEEEAEBEAEBCDDBABBAEBACCBEEBBAEADEECACEDEEDABACDCDBBBD",
"output": "YES"
},
{
"input": "44 15\nHGJIFCGGCDGIJDHBIBGAEABCIABIGBDEADBBBAGDFDHA",
"output": "NO"
},
{
"input": "41 19\nTMEYYIIELFDCMBDKWWKYNRNDUPRONYROXQCLVQALP",
"output": "NO"
},
{
"input": "377 3\nEADADBBBBDEAABBAEBABACDBDBBCACAADBEAEACDEAABACADEEDEACACDADABBBBDDEECBDABACACBAECBADAEBDEEBDBCDAEADBCDDACACDCCEEDBCCBBCEDBECBABCDDBBDEADEDAEACDECECBEBACBCCDCDBDAECDECADBCBEDBBDAAEBCAAECCDCCDBDDEBADEEBDCAEABBDEDBBDDEAECCBDDCDEACDAECCBDDABABEAEDCDEDBAECBDEACEBCECEACDCBABCBAAEAADACADBBBBABEADBCADEBCBECCABBDDDEEBCDEBADEBDAAABBEABADEDEAEABCEEBEEDEAEBEABCEDDBACBCCADEBAAAAAEABABBCE",
"output": "YES"
},
{
"input": "433 3\nFZDDHMJGBZCHFUXBBPIEBBEFDWOMXXEPOMDGSMPIUZOMRZQNSJAVNATGIWPDFISKFQXJNVFXPHOZDAEZFDAHDXXQKZMGNSGKQNWGNGJGJZVVITKNFLVCPMZSDMCHBTVAWYVZLIXXIADXNYILEYNIQHKMOGMVOCWGHCWIYMPEPADSJAAKEGTUSEDWAHMNYJDIHBKHVUHLYGNGZDBULRXLSAJHPCMNWCEAAPYMHDTYWPADOTJTXTXUKLCHWKUSZRHEKQEFPVJEJJHRWCKYOIWALRTIBUMNOCRXLSIKQCJVQXEPGOHRUDJDKMUUUDORURWXJNVRVMNOUNRFKSVMTMZGOIJLXEPAMVGESOADYIGZXRBJDIWKNOWTCSROAQTBECHTOZVSQUOOJRZIBAUHMKAXDCIMDZJFMABGRNTGPUJAUNFPFWCJG",
"output": "YES"
},
{
"input": "5 2\nABCAB",
"output": "YES"
},
{
"input": "5 1\nAZAZA",
"output": "YES"
},
{
"input": "7 2\nABCDBCD",
"output": "YES"
},
{
"input": "3 26\nAAB",
"output": "NO"
}
] | 1,501,427,069 | 1,769 | Python 3 | TIME_LIMIT_EXCEEDED | PRETESTS | 5 | 1,000 | 7,884,800 | n, k = map(int, input().strip().split(" "))
s = input().strip()
sum_hash = [0 for _ in range(26)]
cur_hash = [0 for _ in range(26)]
for c in s:
sum_hash[ord(c) - ord('A')] += 1
printyes = False
guard_need = 0
for c in s:
if cur_hash[ord(c) - ord('A')] == 0:
guard_need += 1
if guard_need > k:
printyes = True
cur_hash[ord(c) - ord('A')] += 1
if cur_hash[ord(c) - ord('A')] == sum_hash[ord(c) - ord('A')]:
guard_need -= 1
if printyes:
print("YES")
else:
print("NO") | Title: The Festive Evening
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in.
There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously.
For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed.
Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened.
Input Specification:
Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26).
In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest.
Output Specification:
Output «YES» if at least one door was unguarded during some time, and «NO» otherwise.
You can output each letter in arbitrary case (upper or lower).
Demo Input:
['5 1\nAABBB\n', '5 1\nABABB\n']
Demo Output:
['NO\n', 'YES\n']
Note:
In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened.
In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door. | ```python
n, k = map(int, input().strip().split(" "))
s = input().strip()
sum_hash = [0 for _ in range(26)]
cur_hash = [0 for _ in range(26)]
for c in s:
sum_hash[ord(c) - ord('A')] += 1
printyes = False
guard_need = 0
for c in s:
if cur_hash[ord(c) - ord('A')] == 0:
guard_need += 1
if guard_need > k:
printyes = True
cur_hash[ord(c) - ord('A')] += 1
if cur_hash[ord(c) - ord('A')] == sum_hash[ord(c) - ord('A')]:
guard_need -= 1
if printyes:
print("YES")
else:
print("NO")
``` | 0 | |
365 | A | Good Number | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*). | The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109). | Print a single integer — the number of *k*-good numbers in *a*. | [
"10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n",
"2 1\n1\n10\n"
] | [
"10\n",
"1\n"
] | none | 500 | [
{
"input": "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560",
"output": "10"
},
{
"input": "2 1\n1\n10",
"output": "1"
},
{
"input": "1 0\n1000000000",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "6 0\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "1 3\n1000000000",
"output": "0"
},
{
"input": "1 9\n1000000000",
"output": "0"
},
{
"input": "1 0\n1",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 3\n1",
"output": "0"
},
{
"input": "1 9\n1",
"output": "0"
},
{
"input": "2 8\n123456780\n123",
"output": "1"
},
{
"input": "2 8\n12345678\n1230",
"output": "0"
},
{
"input": "6 1\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "6 2\n10\n102\n120\n1032\n1212103\n1999999",
"output": "4"
},
{
"input": "6 3\n10\n102\n120\n1032\n1212103\n1999999",
"output": "2"
},
{
"input": "6 9\n10\n102\n120\n1032\n1212103\n1999999",
"output": "0"
},
{
"input": "1 2\n2220",
"output": "0"
}
] | 1,638,719,445 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 46 | 0 | #This was submitted to codeforces.com
n,k=map(int,input().split())
good=0
s=list(range(0,k+1))
for i in range(n):
lst=list(map(int,input()))
nums=[]
tem=list(sorted(list(set(lst)))) #Removing repeating occurences, and sorting
if k!=0:
nums=tem[0:len(s)]
else:
if s[0] in lst:
if lst[0]!=0:
good+=1
continue
elif lst[0]==0 and lst.count(0)>1:
good+=1
continue
else:
continue
if s==nums:
good+=1
print(good) | Title: Good Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
Input Specification:
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the number of *k*-good numbers in *a*.
Demo Input:
['10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n', '2 1\n1\n10\n']
Demo Output:
['10\n', '1\n']
Note:
none | ```python
#This was submitted to codeforces.com
n,k=map(int,input().split())
good=0
s=list(range(0,k+1))
for i in range(n):
lst=list(map(int,input()))
nums=[]
tem=list(sorted(list(set(lst)))) #Removing repeating occurences, and sorting
if k!=0:
nums=tem[0:len(s)]
else:
if s[0] in lst:
if lst[0]!=0:
good+=1
continue
elif lst[0]==0 and lst.count(0)>1:
good+=1
continue
else:
continue
if s==nums:
good+=1
print(good)
``` | 3 | |
653 | A | Bear and Three Balls | PROGRAMMING | 900 | [
"brute force",
"implementation",
"sortings"
] | null | null | Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
- No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above. | The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball. | Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes). | [
"4\n18 55 16 17\n",
"6\n40 41 43 44 44 44\n",
"8\n5 972 3 4 1 4 970 971\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971. | 500 | [
{
"input": "4\n18 55 16 17",
"output": "YES"
},
{
"input": "6\n40 41 43 44 44 44",
"output": "NO"
},
{
"input": "8\n5 972 3 4 1 4 970 971",
"output": "YES"
},
{
"input": "3\n959 747 656",
"output": "NO"
},
{
"input": "4\n1 2 2 3",
"output": "YES"
},
{
"input": "50\n998 30 384 289 505 340 872 223 663 31 929 625 864 699 735 589 676 399 745 635 963 381 75 97 324 612 597 797 103 382 25 894 219 458 337 572 201 355 294 275 278 311 586 573 965 704 936 237 715 543",
"output": "NO"
},
{
"input": "50\n941 877 987 982 966 979 984 810 811 909 872 980 957 897 845 995 924 905 984 914 824 840 868 910 815 808 872 858 883 952 823 835 860 874 959 972 931 867 866 987 982 837 800 921 887 910 982 980 828 869",
"output": "YES"
},
{
"input": "3\n408 410 409",
"output": "YES"
},
{
"input": "3\n903 902 904",
"output": "YES"
},
{
"input": "3\n399 400 398",
"output": "YES"
},
{
"input": "3\n450 448 449",
"output": "YES"
},
{
"input": "3\n390 389 388",
"output": "YES"
},
{
"input": "3\n438 439 440",
"output": "YES"
},
{
"input": "11\n488 688 490 94 564 615 641 170 489 517 669",
"output": "YES"
},
{
"input": "24\n102 672 983 82 720 501 81 721 982 312 207 897 159 964 611 956 118 984 37 271 596 403 772 954",
"output": "YES"
},
{
"input": "36\n175 551 70 479 875 480 979 32 465 402 640 116 76 687 874 678 359 785 753 401 978 629 162 963 886 641 39 845 132 930 2 372 478 947 407 318",
"output": "YES"
},
{
"input": "6\n10 79 306 334 304 305",
"output": "YES"
},
{
"input": "34\n787 62 26 683 486 364 684 891 846 801 969 837 359 800 836 359 471 637 732 91 841 836 7 799 959 405 416 841 737 803 615 483 323 365",
"output": "YES"
},
{
"input": "30\n860 238 14 543 669 100 428 789 576 484 754 274 849 850 586 377 711 386 510 408 520 693 23 477 266 851 728 711 964 73",
"output": "YES"
},
{
"input": "11\n325 325 324 324 324 325 325 324 324 324 324",
"output": "NO"
},
{
"input": "7\n517 517 518 517 518 518 518",
"output": "NO"
},
{
"input": "20\n710 710 711 711 711 711 710 710 710 710 711 710 710 710 710 710 710 711 711 710",
"output": "NO"
},
{
"input": "48\n29 30 29 29 29 30 29 30 30 30 30 29 30 30 30 29 29 30 30 29 30 29 29 30 29 30 29 30 30 29 30 29 29 30 30 29 29 30 30 29 29 30 30 30 29 29 30 29",
"output": "NO"
},
{
"input": "7\n880 880 514 536 881 881 879",
"output": "YES"
},
{
"input": "15\n377 432 262 376 261 375 377 262 263 263 261 376 262 262 375",
"output": "YES"
},
{
"input": "32\n305 426 404 961 426 425 614 304 404 425 615 403 303 304 615 303 305 405 427 614 403 303 425 615 404 304 427 403 206 616 405 404",
"output": "YES"
},
{
"input": "41\n115 686 988 744 762 519 745 519 518 83 85 115 520 44 687 686 685 596 988 687 989 988 114 745 84 519 519 746 988 84 745 744 115 114 85 115 520 746 745 116 987",
"output": "YES"
},
{
"input": "47\n1 2 483 28 7 109 270 651 464 162 353 521 224 989 721 499 56 69 197 716 313 446 580 645 828 197 100 138 789 499 147 677 384 711 783 937 300 543 540 93 669 604 739 122 632 822 116",
"output": "NO"
},
{
"input": "31\n1 2 1 373 355 692 750 920 578 666 615 232 141 129 663 929 414 704 422 559 568 731 354 811 532 618 39 879 292 602 995",
"output": "NO"
},
{
"input": "50\n5 38 41 4 15 40 27 39 20 3 44 47 30 6 36 29 35 12 19 26 10 2 21 50 11 46 48 49 17 16 33 13 32 28 31 18 23 34 7 14 24 45 9 37 1 8 42 25 43 22",
"output": "YES"
},
{
"input": "50\n967 999 972 990 969 978 963 987 954 955 973 970 959 981 995 983 986 994 979 957 965 982 992 977 953 975 956 961 993 997 998 958 980 962 960 951 996 991 1000 966 971 988 976 968 989 984 974 964 985 952",
"output": "YES"
},
{
"input": "50\n850 536 761 506 842 898 857 723 583 637 536 943 895 929 890 612 832 633 696 731 553 880 710 812 665 877 915 636 711 540 748 600 554 521 813 796 568 513 543 809 798 820 928 504 999 646 907 639 550 911",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "YES"
},
{
"input": "3\n500 999 1000",
"output": "NO"
},
{
"input": "10\n101 102 104 105 107 109 110 112 113 115",
"output": "NO"
},
{
"input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "NO"
},
{
"input": "50\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "3\n1000 999 998",
"output": "YES"
},
{
"input": "49\n343 322 248 477 53 156 245 493 209 141 370 66 229 184 434 137 276 472 216 456 147 180 140 114 493 323 393 262 380 314 222 124 98 441 129 346 48 401 347 460 122 125 114 106 189 260 374 165 456",
"output": "NO"
},
{
"input": "20\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3",
"output": "YES"
},
{
"input": "3\n999 999 1000",
"output": "NO"
},
{
"input": "9\n2 4 5 13 25 100 200 300 400",
"output": "NO"
},
{
"input": "9\n1 1 1 2 2 2 3 3 3",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "NO"
},
{
"input": "3\n998 999 1000",
"output": "YES"
},
{
"input": "12\n1 1 1 1 1 1 1 1 1 2 2 4",
"output": "NO"
},
{
"input": "4\n4 3 4 5",
"output": "YES"
},
{
"input": "6\n1 1 1 2 2 2",
"output": "NO"
},
{
"input": "3\n2 3 2",
"output": "NO"
},
{
"input": "5\n10 5 6 3 2",
"output": "NO"
},
{
"input": "3\n1 2 1",
"output": "NO"
},
{
"input": "3\n1 2 3",
"output": "YES"
},
{
"input": "4\n998 999 1000 1000",
"output": "YES"
},
{
"input": "5\n2 3 9 9 4",
"output": "YES"
},
{
"input": "4\n1 2 4 4",
"output": "NO"
},
{
"input": "3\n1 1 1",
"output": "NO"
},
{
"input": "3\n2 2 3",
"output": "NO"
},
{
"input": "7\n1 2 2 2 4 5 6",
"output": "YES"
},
{
"input": "5\n1 3 10 3 10",
"output": "NO"
},
{
"input": "3\n1 2 2",
"output": "NO"
},
{
"input": "4\n1000 1000 999 998",
"output": "YES"
},
{
"input": "3\n5 3 7",
"output": "NO"
},
{
"input": "6\n1 1 2 2 3 3",
"output": "YES"
},
{
"input": "9\n6 6 6 5 5 5 4 4 4",
"output": "YES"
},
{
"input": "7\n5 6 6 6 7 7 7",
"output": "YES"
},
{
"input": "5\n2 3 3 3 4",
"output": "YES"
},
{
"input": "5\n2 1 2 1 3",
"output": "YES"
},
{
"input": "3\n1 2 7",
"output": "NO"
},
{
"input": "3\n1000 1000 1000",
"output": "NO"
},
{
"input": "5\n1 100 2 100 3",
"output": "YES"
},
{
"input": "5\n5 4 6 5 5",
"output": "YES"
},
{
"input": "12\n1 1 1 1 2 2 2 2 3 3 3 3",
"output": "YES"
},
{
"input": "5\n9 9 1 2 3",
"output": "YES"
},
{
"input": "6\n1 2 3 1 2 3",
"output": "YES"
},
{
"input": "7\n1 1 1 1 2 3 3",
"output": "YES"
},
{
"input": "3\n13 13 13",
"output": "NO"
},
{
"input": "3\n42 42 42",
"output": "NO"
},
{
"input": "8\n1 1 1 1 2 2 2 2",
"output": "NO"
},
{
"input": "6\n1 1 1 1 2 3",
"output": "YES"
},
{
"input": "6\n1 1 2 2 6 6",
"output": "NO"
},
{
"input": "6\n1 2 5 5 5 5",
"output": "NO"
},
{
"input": "9\n1 2 3 1 2 3 1 2 3",
"output": "YES"
},
{
"input": "4\n1 2 1 100",
"output": "NO"
},
{
"input": "5\n1 1 2 2 3",
"output": "YES"
}
] | 1,459,282,600 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 61 | 4,608,000 | a = input("")
b = map(int, input("").split(" "))
b = sorted(b)
ans = "NO"
for i in range(len(b)-2):
if(b[i+2]-b[i]<=2 and b[i+1]!=b[i] and b[i+2]!=b[i] and b[i+2]!=b[i+1]):
ans = "YES"
print(ans)
| Title: Bear and Three Balls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a little polar bear. He has *n* balls, the *i*-th ball has size *t**i*.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
- No two friends can get balls of the same size. - No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
Input Specification:
The first line of the input contains one integer *n* (3<=≤<=*n*<=≤<=50) — the number of balls Limak has.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000) where *t**i* denotes the size of the *i*-th ball.
Output Specification:
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
Demo Input:
['4\n18 55 16 17\n', '6\n40 41 43 44 44 44\n', '8\n5 972 3 4 1 4 970 971\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5. 1. Choose balls with sizes 972, 970, 971. | ```python
a = input("")
b = map(int, input("").split(" "))
b = sorted(b)
ans = "NO"
for i in range(len(b)-2):
if(b[i+2]-b[i]<=2 and b[i+1]!=b[i] and b[i+2]!=b[i] and b[i+2]!=b[i+1]):
ans = "YES"
print(ans)
``` | 0 | |
978 | E | Bus Video System | PROGRAMMING | 1,400 | [
"combinatorics",
"math"
] | null | null | The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If $x$ is the number of passengers in a bus just before the current bus stop and $y$ is the number of passengers in the bus just after current bus stop, the system records the number $y-x$. So the system records show how number of passengers changed.
The test run was made for single bus and $n$ bus stops. Thus, the system recorded the sequence of integers $a_1, a_2, \dots, a_n$ (exactly one number for each bus stop), where $a_i$ is the record for the bus stop $i$. The bus stops are numbered from $1$ to $n$ in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to $w$ (that is, at any time in the bus there should be from $0$ to $w$ passengers inclusive). | The first line contains two integers $n$ and $w$ $(1 \le n \le 1\,000, 1 \le w \le 10^{9})$ — the number of bus stops and the capacity of the bus.
The second line contains a sequence $a_1, a_2, \dots, a_n$ $(-10^{6} \le a_i \le 10^{6})$, where $a_i$ equals to the number, which has been recorded by the video system after the $i$-th bus stop. | Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to $w$. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0. | [
"3 5\n2 1 -3\n",
"2 4\n-1 1\n",
"4 10\n2 4 1 2\n"
] | [
"3\n",
"4\n",
"2\n"
] | In the first example initially in the bus could be $0$, $1$ or $2$ passengers.
In the second example initially in the bus could be $1$, $2$, $3$ or $4$ passengers.
In the third example initially in the bus could be $0$ or $1$ passenger. | 0 | [
{
"input": "3 5\n2 1 -3",
"output": "3"
},
{
"input": "2 4\n-1 1",
"output": "4"
},
{
"input": "4 10\n2 4 1 2",
"output": "2"
},
{
"input": "2 10\n-1 2",
"output": "9"
},
{
"input": "3 4\n-3 -4 4",
"output": "0"
},
{
"input": "10 1\n-1 -1 3 -4 2 3 0 -3 3 2",
"output": "0"
},
{
"input": "5 21\n-3 2 -4 -1 -5",
"output": "11"
},
{
"input": "5 9\n-2 -1 2 -1 -2",
"output": "6"
},
{
"input": "8 7\n-5 0 -3 1 -1 5 0 -6",
"output": "0"
},
{
"input": "3 4\n-2 -1 0",
"output": "2"
},
{
"input": "1 1000000000\n0",
"output": "1000000001"
},
{
"input": "2 1000000000\n-1000000 -1000000",
"output": "998000001"
},
{
"input": "2 1000000000\n1000000 -1000000",
"output": "999000001"
},
{
"input": "2 1000000000\n-1000000 1000000",
"output": "999000001"
},
{
"input": "2 1000000000\n1000000 1000000",
"output": "998000001"
},
{
"input": "102 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 1 1 1 1 1 1",
"output": "0"
},
{
"input": "1 1234564\n1",
"output": "1234564"
},
{
"input": "3 4\n1 2 3",
"output": "0"
}
] | 1,691,952,405 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 61 | 0 | n,w=map(int,input().split())
a=list(map(int,input().split()))
l,r=0,w
for i in a[::-1]:
if i>0:
r-=i
if l>0:
l-=i
else:
l-=i
if r<w:
r-=i
#01234
#0123
#1234
#print(l,r)
print(abs(r-l)+1) | Title: Bus Video System
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.
If $x$ is the number of passengers in a bus just before the current bus stop and $y$ is the number of passengers in the bus just after current bus stop, the system records the number $y-x$. So the system records show how number of passengers changed.
The test run was made for single bus and $n$ bus stops. Thus, the system recorded the sequence of integers $a_1, a_2, \dots, a_n$ (exactly one number for each bus stop), where $a_i$ is the record for the bus stop $i$. The bus stops are numbered from $1$ to $n$ in chronological order.
Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to $w$ (that is, at any time in the bus there should be from $0$ to $w$ passengers inclusive).
Input Specification:
The first line contains two integers $n$ and $w$ $(1 \le n \le 1\,000, 1 \le w \le 10^{9})$ — the number of bus stops and the capacity of the bus.
The second line contains a sequence $a_1, a_2, \dots, a_n$ $(-10^{6} \le a_i \le 10^{6})$, where $a_i$ equals to the number, which has been recorded by the video system after the $i$-th bus stop.
Output Specification:
Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to $w$. If the situation is contradictory (i.e. for any initial number of passengers there will be a contradiction), print 0.
Demo Input:
['3 5\n2 1 -3\n', '2 4\n-1 1\n', '4 10\n2 4 1 2\n']
Demo Output:
['3\n', '4\n', '2\n']
Note:
In the first example initially in the bus could be $0$, $1$ or $2$ passengers.
In the second example initially in the bus could be $1$, $2$, $3$ or $4$ passengers.
In the third example initially in the bus could be $0$ or $1$ passenger. | ```python
n,w=map(int,input().split())
a=list(map(int,input().split()))
l,r=0,w
for i in a[::-1]:
if i>0:
r-=i
if l>0:
l-=i
else:
l-=i
if r<w:
r-=i
#01234
#0123
#1234
#print(l,r)
print(abs(r-l)+1)
``` | 0 | |
867 | A | Between the Offices | PROGRAMMING | 800 | [
"implementation"
] | null | null | As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not. | The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence. | Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower). | [
"4\nFSSF\n",
"2\nSF\n",
"10\nFFFFFFFFFF\n",
"10\nSSFFSFFSFF\n"
] | [
"NO\n",
"YES\n",
"NO\n",
"YES\n"
] | In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | 500 | [
{
"input": "4\nFSSF",
"output": "NO"
},
{
"input": "2\nSF",
"output": "YES"
},
{
"input": "10\nFFFFFFFFFF",
"output": "NO"
},
{
"input": "10\nSSFFSFFSFF",
"output": "YES"
},
{
"input": "20\nSFSFFFFSSFFFFSSSSFSS",
"output": "NO"
},
{
"input": "20\nSSFFFFFSFFFFFFFFFFFF",
"output": "YES"
},
{
"input": "20\nSSFSFSFSFSFSFSFSSFSF",
"output": "YES"
},
{
"input": "20\nSSSSFSFSSFSFSSSSSSFS",
"output": "NO"
},
{
"input": "100\nFFFSFSFSFSSFSFFSSFFFFFSSSSFSSFFFFSFFFFFSFFFSSFSSSFFFFSSFFSSFSFFSSFSSSFSFFSFSFFSFSFFSSFFSFSSSSFSFSFSS",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFFFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFSS",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFFSFFFFFFFFFSFSSFFFFFFFFFFFFFFFFFFFFFFSFFSFFFFFSFFFFFFFFSFFFFFFFFFFFFFSFFFFFFFFSFFFFFFFSF",
"output": "NO"
},
{
"input": "100\nSFFSSFFFFFFSSFFFSSFSFFFFFSSFFFSFFFFFFSFSSSFSFSFFFFSFSSFFFFFFFFSFFFFFSFFFFFSSFFFSFFSFSFFFFSFFSFFFFFFF",
"output": "YES"
},
{
"input": "100\nFFFFSSSSSFFSSSFFFSFFFFFSFSSFSFFSFFSSFFSSFSFFFFFSFSFSFSFFFFFFFFFSFSFFSFFFFSFSFFFFFFFFFFFFSFSSFFSSSSFF",
"output": "NO"
},
{
"input": "100\nFFFFFFFFFFFFSSFFFFSFSFFFSFSSSFSSSSSFSSSSFFSSFFFSFSFSSFFFSSSFFSFSFSSFSFSSFSFFFSFFFFFSSFSFFFSSSFSSSFFS",
"output": "NO"
},
{
"input": "100\nFFFSSSFSFSSSSFSSFSFFSSSFFSSFSSFFSSFFSFSSSSFFFSFFFSFSFSSSFSSFSFSFSFFSSSSSFSSSFSFSFFSSFSFSSFFSSFSFFSFS",
"output": "NO"
},
{
"input": "100\nFFSSSSFSSSFSSSSFSSSFFSFSSFFSSFSSSFSSSFFSFFSSSSSSSSSSSSFSSFSSSSFSFFFSSFFFFFFSFSFSSSSSSFSSSFSFSSFSSFSS",
"output": "NO"
},
{
"input": "100\nSSSFFFSSSSFFSSSSSFSSSSFSSSFSSSSSFSSSSSSSSFSFFSSSFFSSFSSSSFFSSSSSSFFSSSSFSSSSSSFSSSFSSSSSSSFSSSSFSSSS",
"output": "NO"
},
{
"input": "100\nFSSSSSSSSSSSFSSSSSSSSSSSSSSSSFSSSSSSFSSSSSSSSSSSSSFSSFSSSSSFSSFSSSSSSSSSFFSSSSSFSFSSSFFSSSSSSSSSSSSS",
"output": "NO"
},
{
"input": "100\nSSSSSSSSSSSSSFSSSSSSSSSSSSFSSSFSSSSSSSSSSSSSSSSSSSSSSSSSSSSSFSSSSSSSSSSSSSSSSFSFSSSSSSSSSSSSSSSSSSFS",
"output": "NO"
},
{
"input": "100\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS",
"output": "NO"
},
{
"input": "100\nSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF",
"output": "YES"
},
{
"input": "100\nSFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFSFSFFFFFFFFFFFSFSFFFFFFFFFFFFFSFFFFFFFFFFFFFFFFFFFFFFFFF",
"output": "YES"
},
{
"input": "100\nSFFFFFFFFFFFFSSFFFFSFFFFFFFFFFFFFFFFFFFSFFFSSFFFFSFSFFFSFFFFFFFFFFFFFFFSSFFFFFFFFSSFFFFFFFFFFFFFFSFF",
"output": "YES"
},
{
"input": "100\nSFFSSSFFSFSFSFFFFSSFFFFSFFFFFFFFSFSFFFSFFFSFFFSFFFFSFSFFFFFFFSFFFFFFFFFFSFFSSSFFSSFFFFSFFFFSFFFFSFFF",
"output": "YES"
},
{
"input": "100\nSFFFSFFFFSFFFSSFFFSFSFFFSFFFSSFSFFFFFSFFFFFFFFSFSFSFFSFFFSFSSFSFFFSFSFFSSFSFSSSFFFFFFSSFSFFSFFFFFFFF",
"output": "YES"
},
{
"input": "100\nSSSSFFFFSFFFFFFFSFFFFSFSFFFFSSFFFFFFFFFSFFSSFFFFFFSFSFSSFSSSFFFFFFFSFSFFFSSSFFFFFFFSFFFSSFFFFSSFFFSF",
"output": "YES"
},
{
"input": "100\nSSSFSSFFFSFSSSSFSSFSSSSFSSFFFFFSFFSSSSFFSSSFSSSFSSSSFSSSSFSSSSSSSFSFSSFFFSSFFSFFSSSSFSSFFSFSSFSFFFSF",
"output": "YES"
},
{
"input": "100\nSFFSFSSSSSSSFFSSSFSSSSFSFSSFFFSSSSSSFSSSSFSSFSSSFSSSSSSSFSSFSFFFSSFSSFSFSFSSSSSSSSSSSSFFFFSSSSSFSFFF",
"output": "YES"
},
{
"input": "100\nSSSFSFFSFSFFSSSSSFSSSFSSSFFFSSSSSSSSSFSFSFSSSSFSFSSFFFFFSSSSSSSSSSSSSSSSSSSFFSSSSSFSFSSSSFFSSSSFSSSF",
"output": "YES"
},
{
"input": "100\nSSSFSSSSSSSSSSFSSSSFSSSSSSFSSSSSSFSSSSSSSSSSSSSSFSSSFSSSFSSSSSSSSSSSFSSSSSSFSFSSSSFSSSSSSFSSSSSSSSFF",
"output": "YES"
},
{
"input": "100\nSSSSSSSSSSSSSSSFSFSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSFFSSSSSSSSSFSSSSSSSSSSSSSSSSSF",
"output": "YES"
},
{
"input": "100\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSF",
"output": "YES"
},
{
"input": "2\nSS",
"output": "NO"
}
] | 1,595,479,663 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 109 | 6,656,000 | n=int(input())
a=input()
sfs=0;
ssf=0;i=0;
while(i<n-1):
if(a[i]=='F' and a[i+1]=='S'):
sfs+=1;
elif(a[i]=='S' and a[i+1]=='F'):
ssf+=1;
i+=1;
if(ssf>sfs):
print("YES");
else:
print("NO") | Title: Between the Offices
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not.
Input Specification:
The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output Specification:
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['4\nFSSF\n', '2\nSF\n', '10\nFFFFFFFFFF\n', '10\nSSFFSFFSFF\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | ```python
n=int(input())
a=input()
sfs=0;
ssf=0;i=0;
while(i<n-1):
if(a[i]=='F' and a[i+1]=='S'):
sfs+=1;
elif(a[i]=='S' and a[i+1]=='F'):
ssf+=1;
i+=1;
if(ssf>sfs):
print("YES");
else:
print("NO")
``` | 3 | |
598 | B | Queries on a String | PROGRAMMING | 1,300 | [
"implementation",
"strings"
] | null | null | You are given a string *s* and should process *m* queries. Each query is described by two 1-based indices *l**i*, *r**i* and integer *k**i*. It means that you should cyclically shift the substring *s*[*l**i*... *r**i*] *k**i* times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string *s* is abacaba and the query is *l*1<==<=3,<=*r*1<==<=6,<=*k*1<==<=1 then the answer is abbacaa. If after that we would process the query *l*2<==<=1,<=*r*2<==<=4,<=*k*2<==<=2 then we would get the string baabcaa. | The first line of the input contains the string *s* (1<=≤<=|*s*|<=≤<=10<=000) in its initial state, where |*s*| stands for the length of *s*. It contains only lowercase English letters.
Second line contains a single integer *m* (1<=≤<=*m*<=≤<=300) — the number of queries.
The *i*-th of the next *m* lines contains three integers *l**i*, *r**i* and *k**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=|*s*|,<=1<=≤<=*k**i*<=≤<=1<=000<=000) — the description of the *i*-th query. | Print the resulting string *s* after processing all *m* queries. | [
"abacaba\n2\n3 6 1\n1 4 2\n"
] | [
"baabcaa\n"
] | The sample is described in problem statement. | 0 | [
{
"input": "abacaba\n2\n3 6 1\n1 4 2",
"output": "baabcaa"
},
{
"input": "u\n1\n1 1 1",
"output": "u"
},
{
"input": "p\n5\n1 1 5\n1 1 9\n1 1 10\n1 1 10\n1 1 4",
"output": "p"
},
{
"input": "ssssssssss\n5\n5 7 9\n3 9 3\n2 7 1\n7 7 10\n1 9 6",
"output": "ssssssssss"
},
{
"input": "tcpyzttcpo\n10\n2 3 6\n2 4 1\n2 6 9\n7 10 5\n2 3 5\n4 5 6\n3 4 5\n7 9 4\n9 10 7\n1 10 8",
"output": "zctycopttp"
},
{
"input": "yywlblbblw\n10\n4 7 2\n3 8 2\n4 10 6\n4 7 1\n3 9 6\n1 7 3\n3 7 3\n3 7 1\n1 8 7\n2 7 5",
"output": "bylwlwylbb"
},
{
"input": "thisisahacktest\n1\n1 2 1",
"output": "htisisahacktest"
},
{
"input": "ozozumuhackleyan\n1\n1 4 1",
"output": "zozoumuhackleyan"
},
{
"input": "lacongaithattuyet\n1\n1 1 1",
"output": "lacongaithattuyet"
}
] | 1,620,658,200 | 2,147,483,647 | PyPy 3 | OK | TESTS | 24 | 327 | 36,659,200 | import sys
import bisect
from bisect import bisect_left as lb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
#sys.setrecursionlimit(10**6)
mod=10**9+7
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs(root,par):
if par!=-1:
dp[root]=dp[par]+1
for i in range(1,20):
if kthpar[root][i-1]!=-1:
kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
for child in gp[root]:
if child==par:continue
kthpar[child][0]=root
dfs(child,root)
def hnbhai():
s=list(sa())
for i in range(sb()):
l,r,k=sd()
l-=1
r-=1
k=(k%(r-l+1))
s=s[:l]+s[r-k+1:r+1]+s[l:r-k+1]+s[r+1:]
print("".join(s))
for _ in range(1):
hnbhai()
| Title: Queries on a String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *s* and should process *m* queries. Each query is described by two 1-based indices *l**i*, *r**i* and integer *k**i*. It means that you should cyclically shift the substring *s*[*l**i*... *r**i*] *k**i* times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string *s* is abacaba and the query is *l*1<==<=3,<=*r*1<==<=6,<=*k*1<==<=1 then the answer is abbacaa. If after that we would process the query *l*2<==<=1,<=*r*2<==<=4,<=*k*2<==<=2 then we would get the string baabcaa.
Input Specification:
The first line of the input contains the string *s* (1<=≤<=|*s*|<=≤<=10<=000) in its initial state, where |*s*| stands for the length of *s*. It contains only lowercase English letters.
Second line contains a single integer *m* (1<=≤<=*m*<=≤<=300) — the number of queries.
The *i*-th of the next *m* lines contains three integers *l**i*, *r**i* and *k**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=|*s*|,<=1<=≤<=*k**i*<=≤<=1<=000<=000) — the description of the *i*-th query.
Output Specification:
Print the resulting string *s* after processing all *m* queries.
Demo Input:
['abacaba\n2\n3 6 1\n1 4 2\n']
Demo Output:
['baabcaa\n']
Note:
The sample is described in problem statement. | ```python
import sys
import bisect
from bisect import bisect_left as lb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
#sys.setrecursionlimit(10**6)
mod=10**9+7
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs(root,par):
if par!=-1:
dp[root]=dp[par]+1
for i in range(1,20):
if kthpar[root][i-1]!=-1:
kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
for child in gp[root]:
if child==par:continue
kthpar[child][0]=root
dfs(child,root)
def hnbhai():
s=list(sa())
for i in range(sb()):
l,r,k=sd()
l-=1
r-=1
k=(k%(r-l+1))
s=s[:l]+s[r-k+1:r+1]+s[l:r-k+1]+s[r+1:]
print("".join(s))
for _ in range(1):
hnbhai()
``` | 3 | |
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"
}
] | 1,601,751,155 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 218 | 102,400 | s = input()
upper, lower = 0, 0
for i in s:
if i >= 'A' and i <= 'Z':
upper += 1
elif i >= 'a' and i <= 'z':
lower +=1
if upper > lower:
print(s.upper())
else:
print(s.lower())
| 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()
upper, lower = 0, 0
for i in s:
if i >= 'A' and i <= 'Z':
upper += 1
elif i >= 'a' and i <= 'z':
lower +=1
if upper > lower:
print(s.upper())
else:
print(s.lower())
``` | 3.945309 |
156 | A | Message | PROGRAMMING | 1,700 | [
"brute force"
] | null | null | Dr. Moriarty is about to send a message to Sherlock Holmes. He has a string *s*.
String *p* is called a substring of string *s* if you can read it starting from some position in the string *s*. For example, string "aba" has six substrings: "a", "b", "a", "ab", "ba", "aba".
Dr. Moriarty plans to take string *s* and cut out some substring from it, let's call it *t*. Then he needs to change the substring *t* zero or more times. As a result, he should obtain a fixed string *u* (which is the string that should be sent to Sherlock Holmes). One change is defined as making one of the following actions:
- Insert one letter to any end of the string. - Delete one letter from any end of the string. - Change one letter into any other one.
Moriarty is very smart and after he chooses some substring *t*, he always makes the minimal number of changes to obtain *u*.
Help Moriarty choose the best substring *t* from all substrings of the string *s*. The substring *t* should minimize the number of changes Moriarty should make to obtain the string *u* from it. | The first line contains a non-empty string *s*, consisting of lowercase Latin letters. The second line contains a non-empty string *u*, consisting of lowercase Latin letters. The lengths of both strings are in the range from 1 to 2000, inclusive. | Print the only integer — the minimum number of changes that Dr. Moriarty has to make with the string that you choose. | [
"aaaaa\naaa\n",
"abcabc\nbcd\n",
"abcdef\nklmnopq\n"
] | [
"0\n",
"1\n",
"7\n"
] | In the first sample Moriarty can take any substring of length 3, and it will be equal to the required message *u*, so Moriarty won't have to make any changes.
In the second sample you should take a substring consisting of characters from second to fourth ("bca") or from fifth to sixth ("bc"). Then you will only have to make one change: to change or to add the last character.
In the third sample the initial string *s* doesn't contain any character that the message should contain, so, whatever string you choose, you will have to make at least 7 changes to obtain the required message. | 500 | [
{
"input": "aaaaa\naaa",
"output": "0"
},
{
"input": "abcabc\nbcd",
"output": "1"
},
{
"input": "abcdef\nklmnopq",
"output": "7"
},
{
"input": "aaabbbaaa\naba",
"output": "1"
},
{
"input": "a\na",
"output": "0"
},
{
"input": "z\nz",
"output": "0"
},
{
"input": "a\nz",
"output": "1"
},
{
"input": "d\nt",
"output": "1"
},
{
"input": "o\nu",
"output": "1"
},
{
"input": "a\nm",
"output": "1"
},
{
"input": "t\nv",
"output": "1"
},
{
"input": "n\ng",
"output": "1"
},
{
"input": "c\nh",
"output": "1"
},
{
"input": "r\ni",
"output": "1"
},
{
"input": "h\nb",
"output": "1"
},
{
"input": "r\na",
"output": "1"
},
{
"input": "c\np",
"output": "1"
},
{
"input": "wbdbzf\nfpvlerhsuf",
"output": "9"
},
{
"input": "zafsqbsu\nhl",
"output": "2"
},
{
"input": "juhlp\nycqugugk",
"output": "7"
},
{
"input": "ladfasxt\ncpvtd",
"output": "4"
},
{
"input": "ally\ncjidwuj",
"output": "7"
},
{
"input": "rgug\npgqwslo",
"output": "6"
},
{
"input": "wmjwu\nrfew",
"output": "3"
},
{
"input": "cpnwcdqff\nq",
"output": "0"
},
{
"input": "dkwh\nm",
"output": "1"
},
{
"input": "zfinrlju\nwiiegborjl",
"output": "9"
},
{
"input": "swconajiqpgziitbpwjsfcalqvmwbfed\nridfnsyumichlhpnurrnwkyjcdzchznpmno",
"output": "32"
},
{
"input": "vfjofvgkdwgqdlomtmcvedtmimdnxavhfirienxfdflldkbwjsynablhdvgaipvcghgaxipotwmmlzxekipgbvpfivlgzfwqz\njkdfjnessjfgcqpnxgtqdxtqimolbdlnipkoqht",
"output": "34"
},
{
"input": "dtvxepnxfkzcaoh\nkpdzbtwjitzlyzvsbwcsrfglaycrhzwsdtidrelndsq",
"output": "41"
},
{
"input": "sweaucynwsnduofyaqunoxttbipgrbfpssplfp\nuifmuxmczznobefdsyoclwzekewxmcwfqryuevnxxlgxsuhoytkaddorbdaygo",
"output": "57"
},
{
"input": "eohztfsxoyhirnzxgwaevfqstinlxeiyywmpmlbedkjihaxfdtsocof\nbloqrjbidxiqozvwregxxgmxuqcvhwzhytfckbafd",
"output": "37"
},
{
"input": "ybshzefoxkqdigcjafs\nnffvaxdmditsolfxbyquira",
"output": "19"
},
{
"input": "ytfqnuhqzbjjheejjbzcaorilcyvuxvviaiba\nxnhgkdfceialuujgcxmrhjbzvibcoknofafmdjnhij",
"output": "37"
},
{
"input": "ibdjtvgaveujdyidqldrxgwhsammmfpgxwljkptmeyejdvudhctmqjazalyzmzfgebetyqncu\nercdngwctdarcennbuqhsjlwfwrcqjbcjxqftycoulrhrimwhznogjmrrqdygtmllottpjgmkndraearezvxxmdhcuokhyngu",
"output": "90"
},
{
"input": "bwhvaanyxupicwobeevcwewhcriwowfovnylalpuhxzqxtzyjrzlxcmejujvliomdfedgtaioauwrcluhfxtzu\nplinvtsvytepojsecnjisxlmqkfhgknitvuw",
"output": "28"
},
{
"input": "sjxykdmrzpescabubcjflhnpckgytklc\nsxirpuqnmjqhlnvdeyvxvzzcygkpsujyifzgzmtvxsimddjahiephqlgfzngrzjtcrgrimewsxipczsgu",
"output": "76"
},
{
"input": "ksmbytfyhhnstlecripupiwdhbkhfpfmimrbqgszohcqnezcybvwasxmkxfupvuecsctcpadccnqexsglwaiyxcoyheefcjmdedesgjqdtqgrvfjonzesffousooutsjumrxl\nhgjqihcfbnmgufonaiudbjegexexthrzcdkuimwogpbyovemztzcmqnrbhabxyyxyfuzpyhjgnioexbezzupcxlyzuuncstiiqsjzdtqppqhxilvqimlpjejiqbwpeekzweeyvthvjffgfvqauqrugajjjzibgzhxphcvtncjzecbtupwkehcrgsgfgkvwwnifglyamjkzfvabybsstwrwugnmiwflhemgnfbrtskzfxcepqhtelgiowzeuujpkuzsfsipcvtfoeshawvryaubilcbwukdhlwamsqenzvr",
"output": "287"
},
{
"input": "abcd\ndabc",
"output": "1"
},
{
"input": "medxx\nahmed",
"output": "2"
},
{
"input": "ab\ndab",
"output": "1"
},
{
"input": "nasldkfnsb\nyyyynasld",
"output": "4"
},
{
"input": "abcde\ncabc",
"output": "1"
},
{
"input": "a\nzzzzzzzzzza",
"output": "10"
},
{
"input": "abcde\nabde",
"output": "2"
},
{
"input": "bac\ntbdca",
"output": "3"
},
{
"input": "abcdef\nxyzabc",
"output": "3"
},
{
"input": "abcdef\nbctsf",
"output": "2"
},
{
"input": "xxxabaxxx\nxxxaaxxx",
"output": "2"
},
{
"input": "bcd\nabc",
"output": "1"
},
{
"input": "d\nabcdef",
"output": "5"
}
] | 1,422,703,930 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 4 | 2,000 | 0 | # 156A
from sys import stdin
__author__ = 'artyom'
def read_line():
return stdin.readline().strip()
s = read_line()
u = read_line()
len_s = len(s)
len_u = len(u)
mn = 10000
for i in range(-len_u + 1, len_s):
count = max(0, len_u - len_s + i) + (-i if i < 0 else 0)
start = max(0, i)
finish = min(len_s, i + len_u)
k = min(0, start - i)
for j in range(start, finish):
if s[j] != u[k]:
count += 1
k += 1
if count < mn:
mn = count
print(mn) | Title: Message
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dr. Moriarty is about to send a message to Sherlock Holmes. He has a string *s*.
String *p* is called a substring of string *s* if you can read it starting from some position in the string *s*. For example, string "aba" has six substrings: "a", "b", "a", "ab", "ba", "aba".
Dr. Moriarty plans to take string *s* and cut out some substring from it, let's call it *t*. Then he needs to change the substring *t* zero or more times. As a result, he should obtain a fixed string *u* (which is the string that should be sent to Sherlock Holmes). One change is defined as making one of the following actions:
- Insert one letter to any end of the string. - Delete one letter from any end of the string. - Change one letter into any other one.
Moriarty is very smart and after he chooses some substring *t*, he always makes the minimal number of changes to obtain *u*.
Help Moriarty choose the best substring *t* from all substrings of the string *s*. The substring *t* should minimize the number of changes Moriarty should make to obtain the string *u* from it.
Input Specification:
The first line contains a non-empty string *s*, consisting of lowercase Latin letters. The second line contains a non-empty string *u*, consisting of lowercase Latin letters. The lengths of both strings are in the range from 1 to 2000, inclusive.
Output Specification:
Print the only integer — the minimum number of changes that Dr. Moriarty has to make with the string that you choose.
Demo Input:
['aaaaa\naaa\n', 'abcabc\nbcd\n', 'abcdef\nklmnopq\n']
Demo Output:
['0\n', '1\n', '7\n']
Note:
In the first sample Moriarty can take any substring of length 3, and it will be equal to the required message *u*, so Moriarty won't have to make any changes.
In the second sample you should take a substring consisting of characters from second to fourth ("bca") or from fifth to sixth ("bc"). Then you will only have to make one change: to change or to add the last character.
In the third sample the initial string *s* doesn't contain any character that the message should contain, so, whatever string you choose, you will have to make at least 7 changes to obtain the required message. | ```python
# 156A
from sys import stdin
__author__ = 'artyom'
def read_line():
return stdin.readline().strip()
s = read_line()
u = read_line()
len_s = len(s)
len_u = len(u)
mn = 10000
for i in range(-len_u + 1, len_s):
count = max(0, len_u - len_s + i) + (-i if i < 0 else 0)
start = max(0, i)
finish = min(len_s, i + len_u)
k = min(0, start - i)
for j in range(start, finish):
if s[j] != u[k]:
count += 1
k += 1
if count < mn:
mn = count
print(mn)
``` | 0 | |
680 | B | Bear and Finding Criminals | PROGRAMMING | 1,000 | [
"constructive algorithms",
"implementation"
] | null | null | There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD. | The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city. | Print the number of criminals Limak will catch. | [
"6 3\n1 1 1 0 1 0\n",
"5 2\n0 0 0 1 0\n"
] | [
"3\n",
"1\n"
] | In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is. | 1,000 | [
{
"input": "6 3\n1 1 1 0 1 0",
"output": "3"
},
{
"input": "5 2\n0 0 0 1 0",
"output": "1"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "9 3\n1 1 1 1 1 1 1 1 0",
"output": "8"
},
{
"input": "9 5\n1 0 1 0 1 0 1 0 1",
"output": "5"
},
{
"input": "20 17\n1 1 0 1 1 1 1 0 1 0 1 1 1 0 1 1 0 0 0 0",
"output": "10"
},
{
"input": "100 60\n1 1 1 1 1 1 0 1 0 0 1 1 0 1 1 1 1 1 0 0 1 1 0 0 0 0 0 1 0 1 1 0 1 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 0 0 1 0 0 0 1 1 1 0 1 0 0 1 1 1 0 1 1 1 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 1 1 1 0 1 0 0",
"output": "27"
},
{
"input": "8 1\n1 0 1 1 0 0 1 0",
"output": "4"
},
{
"input": "11 11\n0 1 0 0 1 1 1 0 0 0 0",
"output": "4"
},
{
"input": "19 10\n0 1 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 0 1",
"output": "4"
},
{
"input": "100 38\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100 38\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100 38\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "3"
},
{
"input": "99 38\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "25"
},
{
"input": "99 38\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "24"
},
{
"input": "99 38\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "24"
},
{
"input": "98 70\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "41"
},
{
"input": "99 70\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "9"
},
{
"input": "99 60\n0 0 0 1 1 0 1 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 0 1 1 0 1 1 1 1 0 0 0 0 0 0 1 1 1",
"output": "34"
},
{
"input": "98 24\n0 0 0 1 1 0 1 1 0 0 0 0 1 0 0 1 0 0 1 1 1 0 0 1 0 0 1 0 1 0 0 1 1 0 1 1 1 0 1 0 0 1 0 0 0 1 1 1 1 0 1 1 1 0 1 1 1 1 0 0 1 0 0 0 1 0 1 1 0 1 0 1 1 1 0 0 1 0 0 0 1 1 0 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1",
"output": "39"
},
{
"input": "100 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100"
},
{
"input": "100 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "2 1\n0 1",
"output": "1"
}
] | 1,686,605,902 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 6 | 62 | 0 | a, n = list(map(int, input().split()))
def valid(x):
if (x < 0 or x >= a):
return False
return True
criminals = list(map(int, input().split()))
left_idx = n-1
right_idx = n-1
criminals_caught = 0
while(left_idx > 0 or right_idx <a):
if (valid(left_idx) and valid(right_idx)):
if (criminals[left_idx] == criminals[right_idx] and criminals[left_idx] == 1):
criminals_caught +=2
if (left_idx == right_idx):
criminals_caught-=1
elif(valid(left_idx)):
criminals_caught +=criminals[left_idx]
elif(valid(right_idx)):
criminals_caught +=criminals[right_idx]
left_idx -=1
right_idx += 1
print (criminals_caught)
| Title: Bear and Finding Criminals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.
Input Specification:
The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city.
Output Specification:
Print the number of criminals Limak will catch.
Demo Input:
['6 3\n1 1 1 0 1 0\n', '5 2\n0 0 0 1 0\n']
Demo Output:
['3\n', '1\n']
Note:
In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is. | ```python
a, n = list(map(int, input().split()))
def valid(x):
if (x < 0 or x >= a):
return False
return True
criminals = list(map(int, input().split()))
left_idx = n-1
right_idx = n-1
criminals_caught = 0
while(left_idx > 0 or right_idx <a):
if (valid(left_idx) and valid(right_idx)):
if (criminals[left_idx] == criminals[right_idx] and criminals[left_idx] == 1):
criminals_caught +=2
if (left_idx == right_idx):
criminals_caught-=1
elif(valid(left_idx)):
criminals_caught +=criminals[left_idx]
elif(valid(right_idx)):
criminals_caught +=criminals[right_idx]
left_idx -=1
right_idx += 1
print (criminals_caught)
``` | 0 | |
258 | A | Little Elephant and Bits | PROGRAMMING | 1,100 | [
"greedy",
"math"
] | null | null | The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation. | The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits. | In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem. | [
"101\n",
"110010\n"
] | [
"11\n",
"11010\n"
] | In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>. | 500 | [
{
"input": "101",
"output": "11"
},
{
"input": "110010",
"output": "11010"
},
{
"input": "10000",
"output": "1000"
},
{
"input": "1111111110",
"output": "111111111"
},
{
"input": "10100101011110101",
"output": "1100101011110101"
},
{
"input": "111010010111",
"output": "11110010111"
},
{
"input": "11110111011100000000",
"output": "1111111011100000000"
},
{
"input": "11110010010100001110110101110011110110100111101",
"output": "1111010010100001110110101110011110110100111101"
},
{
"input": "1001011111010010100111111",
"output": "101011111010010100111111"
},
{
"input": "1111111111",
"output": "111111111"
},
{
"input": "1111111111111111111100111101001110110111111000001111110101001101001110011000001011001111111000110101",
"output": "111111111111111111110111101001110110111111000001111110101001101001110011000001011001111111000110101"
},
{
"input": "11010110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100",
"output": "1110110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100"
},
{
"input": "11111111111111111111111110110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011",
"output": "1111111111111111111111111110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011"
},
{
"input": "11100010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011",
"output": "1110010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011"
},
{
"input": "11",
"output": "1"
},
{
"input": "111",
"output": "11"
},
{
"input": "111111",
"output": "11111"
},
{
"input": "11111",
"output": "1111"
},
{
"input": "1111",
"output": "111"
}
] | 1,567,328,350 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 9 | 218 | 0 | t = input()
if t.count("1") == len(t):
print(t[:len(t)-1])
n = t.index("0")
print(t[:n]+t[n+1:]) | Title: Little Elephant and Bits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
Input Specification:
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
Output Specification:
In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.
Demo Input:
['101\n', '110010\n']
Demo Output:
['11\n', '11010\n']
Note:
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>. | ```python
t = input()
if t.count("1") == len(t):
print(t[:len(t)-1])
n = t.index("0")
print(t[:n]+t[n+1:])
``` | -1 | |
31 | D | Chocolate | PROGRAMMING | 2,000 | [
"dfs and similar",
"implementation"
] | D. Chocolate | 2 | 256 | Bob has a rectangular chocolate bar of the size *W*<=×<=*H*. He introduced a cartesian coordinate system so that the point (0,<=0) corresponds to the lower-left corner of the bar, and the point (*W*,<=*H*) corresponds to the upper-right corner. Bob decided to split the bar into pieces by breaking it. Each break is a segment parallel to one of the coordinate axes, which connects the edges of the bar. More formally, each break goes along the line *x*<==<=*x**c* or *y*<==<=*y**c*, where *x**c* and *y**c* are integers. It should divide one part of the bar into two non-empty parts. After Bob breaks some part into two parts, he breaks the resulting parts separately and independently from each other. Also he doesn't move the parts of the bar. Bob made *n* breaks and wrote them down in his notebook in arbitrary order. At the end he got *n*<=+<=1 parts. Now he wants to calculate their areas. Bob is lazy, so he asks you to do this task. | The first line contains 3 integers *W*, *H* and *n* (1<=≤<=*W*,<=*H*,<=*n*<=≤<=100) — width of the bar, height of the bar and amount of breaks. Each of the following *n* lines contains four integers *x**i*,<=1,<=*y**i*,<=1,<=*x**i*,<=2,<=*y**i*,<=2 — coordinates of the endpoints of the *i*-th break (0<=≤<=*x**i*,<=1<=≤<=*x**i*,<=2<=≤<=*W*,<=0<=≤<=*y**i*,<=1<=≤<=*y**i*,<=2<=≤<=*H*, or *x**i*,<=1<==<=*x**i*,<=2, or *y**i*,<=1<==<=*y**i*,<=2). Breaks are given in arbitrary order.
It is guaranteed that the set of breaks is correct, i.e. there is some order of the given breaks that each next break divides exactly one part of the bar into two non-empty parts. | Output *n*<=+<=1 numbers — areas of the resulting parts in the increasing order. | [
"2 2 2\n1 0 1 2\n0 1 1 1\n",
"2 2 3\n1 0 1 2\n0 1 1 1\n1 1 2 1\n",
"2 4 2\n0 1 2 1\n0 3 2 3\n"
] | [
"1 1 2 ",
"1 1 1 1 ",
"2 2 4 "
] | none | 2,000 | [
{
"input": "2 2 2\n1 0 1 2\n0 1 1 1",
"output": "1 1 2 "
},
{
"input": "2 2 3\n1 0 1 2\n0 1 1 1\n1 1 2 1",
"output": "1 1 1 1 "
},
{
"input": "2 4 2\n0 1 2 1\n0 3 2 3",
"output": "2 2 4 "
},
{
"input": "5 5 3\n2 1 2 5\n0 1 5 1\n4 0 4 1",
"output": "1 4 8 12 "
},
{
"input": "10 10 4\n9 0 9 10\n4 0 4 10\n1 0 1 10\n1 4 4 4",
"output": "10 10 12 18 50 "
},
{
"input": "9 8 5\n4 3 4 4\n0 4 9 4\n5 4 5 8\n0 3 9 3\n1 4 1 8",
"output": "4 4 5 16 16 27 "
},
{
"input": "100 100 1\n0 14 100 14",
"output": "1400 8600 "
}
] | 1,674,897,784 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 3 | 92 | 0 | w,h,n=map(int,input().split())
N=5
x=[[0]*N for _ in range(N)]
y=[[0]*N for _ in range(N)]
x[h][0],y[0][w],p=2,2,[]
for a,b,c,d in [[*map(int,input().split())] for i in range(n)]:
m,n=sorted([(a,b),(c,d)])
x[m[1]][m[0]]=1
y[m[1]][m[0]]=1
if m[0]==n[0]:
x[n[1]][n[0]]=2
if m[1]==n[1]:
y[n[1]][n[0]]=2
def v(a,b):
s,t=1,1
while y[b][a+s]==0:
s+=1
if y[b][a+s]==1:
y[b][a+s]=x[b][a+s]=2
v(a+s,b)
while x[b+t][a]==0:
t+=1
if x[b+t][a]==1:
x[b+t][a]=y[b+t][a]=2
v(a,b+t)
p.append(s*t)
v(0,0)
print(*sorted(p))
| Title: Chocolate
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob has a rectangular chocolate bar of the size *W*<=×<=*H*. He introduced a cartesian coordinate system so that the point (0,<=0) corresponds to the lower-left corner of the bar, and the point (*W*,<=*H*) corresponds to the upper-right corner. Bob decided to split the bar into pieces by breaking it. Each break is a segment parallel to one of the coordinate axes, which connects the edges of the bar. More formally, each break goes along the line *x*<==<=*x**c* or *y*<==<=*y**c*, where *x**c* and *y**c* are integers. It should divide one part of the bar into two non-empty parts. After Bob breaks some part into two parts, he breaks the resulting parts separately and independently from each other. Also he doesn't move the parts of the bar. Bob made *n* breaks and wrote them down in his notebook in arbitrary order. At the end he got *n*<=+<=1 parts. Now he wants to calculate their areas. Bob is lazy, so he asks you to do this task.
Input Specification:
The first line contains 3 integers *W*, *H* and *n* (1<=≤<=*W*,<=*H*,<=*n*<=≤<=100) — width of the bar, height of the bar and amount of breaks. Each of the following *n* lines contains four integers *x**i*,<=1,<=*y**i*,<=1,<=*x**i*,<=2,<=*y**i*,<=2 — coordinates of the endpoints of the *i*-th break (0<=≤<=*x**i*,<=1<=≤<=*x**i*,<=2<=≤<=*W*,<=0<=≤<=*y**i*,<=1<=≤<=*y**i*,<=2<=≤<=*H*, or *x**i*,<=1<==<=*x**i*,<=2, or *y**i*,<=1<==<=*y**i*,<=2). Breaks are given in arbitrary order.
It is guaranteed that the set of breaks is correct, i.e. there is some order of the given breaks that each next break divides exactly one part of the bar into two non-empty parts.
Output Specification:
Output *n*<=+<=1 numbers — areas of the resulting parts in the increasing order.
Demo Input:
['2 2 2\n1 0 1 2\n0 1 1 1\n', '2 2 3\n1 0 1 2\n0 1 1 1\n1 1 2 1\n', '2 4 2\n0 1 2 1\n0 3 2 3\n']
Demo Output:
['1 1 2 ', '1 1 1 1 ', '2 2 4 ']
Note:
none | ```python
w,h,n=map(int,input().split())
N=5
x=[[0]*N for _ in range(N)]
y=[[0]*N for _ in range(N)]
x[h][0],y[0][w],p=2,2,[]
for a,b,c,d in [[*map(int,input().split())] for i in range(n)]:
m,n=sorted([(a,b),(c,d)])
x[m[1]][m[0]]=1
y[m[1]][m[0]]=1
if m[0]==n[0]:
x[n[1]][n[0]]=2
if m[1]==n[1]:
y[n[1]][n[0]]=2
def v(a,b):
s,t=1,1
while y[b][a+s]==0:
s+=1
if y[b][a+s]==1:
y[b][a+s]=x[b][a+s]=2
v(a+s,b)
while x[b+t][a]==0:
t+=1
if x[b+t][a]==1:
x[b+t][a]=y[b+t][a]=2
v(a,b+t)
p.append(s*t)
v(0,0)
print(*sorted(p))
``` | -1 |
920 | E | Connected Components? | PROGRAMMING | 2,100 | [
"data structures",
"dfs and similar",
"dsu",
"graphs"
] | null | null | You are given an undirected graph consisting of *n* vertices and edges. Instead of giving you the edges that exist in the graph, we give you *m* unordered pairs (*x*,<=*y*) such that there is no edge between *x* and *y*, and if some pair of vertices is not listed in the input, then there is an edge between these vertices.
You have to find the number of connected components in the graph and the size of each component. A connected component is a set of vertices *X* such that for every two vertices from this set there exists at least one path in the graph connecting these vertices, but adding any other vertex to *X* violates this rule. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=200000, ).
Then *m* lines follow, each containing a pair of integers *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*, *x*<=≠<=*y*) denoting that there is no edge between *x* and *y*. Each pair is listed at most once; (*x*,<=*y*) and (*y*,<=*x*) are considered the same (so they are never listed in the same test). If some pair of vertices is not listed in the input, then there exists an edge between those vertices. | Firstly print *k* — the number of connected components in this graph.
Then print *k* integers — the sizes of components. You should output these integers in non-descending order. | [
"5 5\n1 2\n3 4\n3 2\n4 2\n2 5\n"
] | [
"2\n1 4 "
] | none | 0 | [
{
"input": "5 5\n1 2\n3 4\n3 2\n4 2\n2 5",
"output": "2\n1 4 "
},
{
"input": "8 15\n2 1\n4 5\n2 4\n3 4\n2 5\n3 5\n2 6\n3 6\n5 6\n4 6\n2 7\n3 8\n2 8\n3 7\n6 7",
"output": "1\n8 "
},
{
"input": "12 58\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 10\n1 11\n1 12\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n3 4\n3 5\n3 6\n3 7\n3 8\n3 9\n3 10\n3 11\n3 12\n4 5\n4 6\n4 8\n4 11\n4 12\n5 6\n5 7\n5 8\n5 9\n5 10\n5 11\n6 7\n6 8\n6 9\n6 10\n6 11\n6 12\n7 8\n7 9\n7 10\n7 11\n7 12\n8 9\n8 10\n8 11\n9 10\n9 11\n9 12\n10 12",
"output": "4\n1 1 1 9 "
},
{
"input": "5 7\n1 2\n2 3\n3 4\n1 5\n2 5\n3 5\n4 5",
"output": "2\n1 4 "
},
{
"input": "6 10\n1 2\n1 3\n1 4\n1 6\n2 3\n2 4\n2 5\n3 5\n3 6\n4 6",
"output": "1\n6 "
},
{
"input": "8 23\n1 2\n1 4\n1 6\n1 8\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n3 4\n3 5\n3 6\n3 7\n3 8\n4 5\n4 6\n4 7\n5 6\n5 7\n5 8\n6 8\n7 8",
"output": "3\n1 2 5 "
},
{
"input": "4 3\n2 1\n3 1\n4 2",
"output": "1\n4 "
},
{
"input": "6 9\n1 2\n1 4\n1 5\n2 3\n2 5\n2 6\n3 5\n4 6\n5 6",
"output": "1\n6 "
},
{
"input": "2 0",
"output": "1\n2 "
},
{
"input": "8 18\n1 4\n1 6\n1 7\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n3 4\n3 8\n4 7\n5 6\n5 7\n5 8\n6 7\n6 8\n7 8",
"output": "1\n8 "
},
{
"input": "4 3\n1 2\n3 1\n4 3",
"output": "1\n4 "
},
{
"input": "8 23\n2 7\n7 5\n8 6\n8 2\n6 3\n3 5\n8 1\n8 4\n8 3\n3 4\n1 2\n2 6\n5 2\n6 4\n7 6\n6 5\n7 8\n7 1\n5 4\n3 7\n1 4\n3 1\n3 2",
"output": "3\n1 3 4 "
},
{
"input": "4 4\n2 1\n3 1\n1 4\n3 2",
"output": "2\n1 3 "
},
{
"input": "2 1\n1 2",
"output": "2\n1 1 "
},
{
"input": "4 3\n1 3\n1 4\n2 3",
"output": "1\n4 "
},
{
"input": "3 1\n2 3",
"output": "1\n3 "
},
{
"input": "5 4\n1 4\n2 3\n4 3\n4 2",
"output": "1\n5 "
},
{
"input": "10 36\n7 8\n7 9\n2 3\n2 4\n2 5\n9 10\n2 7\n2 8\n2 9\n2 10\n4 5\n4 6\n4 7\n4 8\n4 10\n6 7\n6 9\n6 10\n1 2\n1 3\n1 4\n8 9\n1 5\n8 10\n1 7\n1 8\n1 9\n1 10\n3 4\n3 6\n3 7\n3 9\n5 6\n5 7\n5 9\n5 10",
"output": "2\n2 8 "
},
{
"input": "10 34\n7 10\n2 3\n2 4\n2 5\n9 10\n2 7\n2 8\n2 10\n4 5\n4 6\n4 7\n4 8\n4 9\n6 7\n6 8\n6 9\n6 10\n1 2\n1 3\n1 5\n8 9\n1 6\n1 7\n1 8\n1 9\n1 10\n3 4\n3 5\n3 6\n3 8\n3 10\n5 6\n5 9\n5 10",
"output": "1\n10 "
},
{
"input": "12 56\n9 5\n2 6\n9 8\n5 4\n1 11\n1 6\n4 1\n1 10\n10 3\n8 4\n5 1\n9 1\n5 10\n2 7\n11 5\n6 11\n5 8\n7 6\n3 2\n12 7\n8 6\n12 3\n1 2\n8 1\n2 11\n10 12\n4 6\n5 12\n2 4\n10 2\n7 3\n12 11\n7 10\n7 1\n9 2\n11 9\n9 10\n8 7\n11 3\n7 9\n5 7\n4 12\n3 5\n12 2\n4 10\n9 12\n5 2\n9 4\n11 8\n8 2\n3 6\n4 11\n8 10\n6 10\n3 9\n3 4",
"output": "3\n1 4 7 "
},
{
"input": "11 49\n10 3\n6 4\n11 3\n7 6\n10 6\n6 1\n4 3\n10 2\n4 5\n9 2\n10 1\n5 7\n1 5\n9 7\n2 11\n8 6\n3 9\n2 5\n9 5\n6 5\n1 4\n11 9\n1 7\n8 10\n3 6\n3 7\n11 5\n6 9\n4 10\n8 7\n4 9\n8 2\n4 2\n8 11\n7 4\n9 10\n8 1\n10 7\n3 2\n5 8\n8 9\n1 3\n2 7\n10 11\n5 3\n10 5\n4 11\n1 11\n8 3",
"output": "5\n1 1 1 2 6 "
}
] | 1,599,096,373 | 2,147,483,647 | PyPy 3 | MEMORY_LIMIT_EXCEEDED | TESTS | 8 | 1,559 | 268,390,400 | class Node:
def __init__(self, val):
self.val = val
self.parent = self
self.size = 1
class UnionFind:
def __init__(self):
self.map = {}
def find(self, node):
if node.val not in self.map:
self.map[node.val] = node
elif node.parent != node:
node = self.find(node.parent)
return node
def union(self, node1, node2):
parent1, parent2 = self.find(node1), self.find(node2)
if parent1 != parent2:
if parent1.size > parent2.size:
parent2.parent = parent1
elif parent2.size > parent1.size:
parent1.parent = parent2
else:
parent1.parent = parent2
parent2.size += 1
import collections
n, m = map(int, input().split())
graph = [[1] * n for _ in range(n)]
for _ in range(m):
r, c = map(int, input().split())
graph[r - 1][c - 1] = 0
graph[c - 1][r - 1] = 0
def solve():
uf = UnionFind()
for i in range(1, n + 1):
node1 = uf.find(Node(i) if i not in uf.map else uf.map[i])
for j in range(1, n + 1):
if i == j or not graph[i - 1][j - 1]: continue
node2 = Node(j) if j not in uf.map else uf.map[j]
uf.union(node1, node2)
ans = collections.Counter()
for i in range(1, n + 1):
ans[uf.find(uf.map[i])] += 1
print(len(ans))
ans = sorted(list(ans.values()))
print(*ans, ' ')
solve()
| Title: Connected Components?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an undirected graph consisting of *n* vertices and edges. Instead of giving you the edges that exist in the graph, we give you *m* unordered pairs (*x*,<=*y*) such that there is no edge between *x* and *y*, and if some pair of vertices is not listed in the input, then there is an edge between these vertices.
You have to find the number of connected components in the graph and the size of each component. A connected component is a set of vertices *X* such that for every two vertices from this set there exists at least one path in the graph connecting these vertices, but adding any other vertex to *X* violates this rule.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=200000, ).
Then *m* lines follow, each containing a pair of integers *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*, *x*<=≠<=*y*) denoting that there is no edge between *x* and *y*. Each pair is listed at most once; (*x*,<=*y*) and (*y*,<=*x*) are considered the same (so they are never listed in the same test). If some pair of vertices is not listed in the input, then there exists an edge between those vertices.
Output Specification:
Firstly print *k* — the number of connected components in this graph.
Then print *k* integers — the sizes of components. You should output these integers in non-descending order.
Demo Input:
['5 5\n1 2\n3 4\n3 2\n4 2\n2 5\n']
Demo Output:
['2\n1 4 ']
Note:
none | ```python
class Node:
def __init__(self, val):
self.val = val
self.parent = self
self.size = 1
class UnionFind:
def __init__(self):
self.map = {}
def find(self, node):
if node.val not in self.map:
self.map[node.val] = node
elif node.parent != node:
node = self.find(node.parent)
return node
def union(self, node1, node2):
parent1, parent2 = self.find(node1), self.find(node2)
if parent1 != parent2:
if parent1.size > parent2.size:
parent2.parent = parent1
elif parent2.size > parent1.size:
parent1.parent = parent2
else:
parent1.parent = parent2
parent2.size += 1
import collections
n, m = map(int, input().split())
graph = [[1] * n for _ in range(n)]
for _ in range(m):
r, c = map(int, input().split())
graph[r - 1][c - 1] = 0
graph[c - 1][r - 1] = 0
def solve():
uf = UnionFind()
for i in range(1, n + 1):
node1 = uf.find(Node(i) if i not in uf.map else uf.map[i])
for j in range(1, n + 1):
if i == j or not graph[i - 1][j - 1]: continue
node2 = Node(j) if j not in uf.map else uf.map[j]
uf.union(node1, node2)
ans = collections.Counter()
for i in range(1, n + 1):
ans[uf.find(uf.map[i])] += 1
print(len(ans))
ans = sorted(list(ans.values()))
print(*ans, ' ')
solve()
``` | 0 | |
208 | A | Dubstep | PROGRAMMING | 900 | [
"strings"
] | null | null | Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song. | The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word. | Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space. | [
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] | [
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] | In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | 500 | [
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "ZJWUBWUBWUBJWUBWUBWUBL",
"output": "ZJ J L "
},
{
"input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB",
"output": "C B E Q "
},
{
"input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB",
"output": "JKD WBIRAQKF YE WV "
},
{
"input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB",
"output": "KSDHEMIXUJ R S H "
},
{
"input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB",
"output": "OG X I KO "
},
{
"input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH",
"output": "Q QQ I WW JOPJPBRH "
},
{
"input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB",
"output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C "
},
{
"input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV",
"output": "E IQMJNIQ GZZBQZAUHYP PMR DCV "
},
{
"input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB",
"output": "FV BPS RXNETCJ JDMBH B V B "
},
{
"input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL",
"output": "FBQ IDFSY CTWDM SXO QI L "
},
{
"input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL",
"output": "I QLHD YIIKZDFQ CX U K NL "
},
{
"input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE",
"output": "K UPDYXGOKU AGOAH IZD IY V P E "
},
{
"input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB",
"output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ "
},
{
"input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB",
"output": "PAMJGY XGPQM TKGSXUY E N H E "
},
{
"input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB",
"output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB "
},
{
"input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM",
"output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M "
},
{
"input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW",
"output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W "
},
{
"input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG",
"output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G "
},
{
"input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN",
"output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N "
},
{
"input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG",
"output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG "
},
{
"input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB",
"output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L "
},
{
"input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB",
"output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U "
},
{
"input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB",
"output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ "
},
{
"input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB",
"output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J "
},
{
"input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO",
"output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O "
},
{
"input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR",
"output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R "
},
{
"input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB",
"output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM "
},
{
"input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB",
"output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H "
},
{
"input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB",
"output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S "
},
{
"input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB",
"output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J "
},
{
"input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF",
"output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F "
},
{
"input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY",
"output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y "
},
{
"input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB",
"output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO "
},
{
"input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW",
"output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W "
},
{
"input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD",
"output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D "
},
{
"input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB",
"output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A "
},
{
"input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB",
"output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y "
},
{
"input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB",
"output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R "
},
{
"input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY",
"output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y "
},
{
"input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE",
"output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E "
},
{
"input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS",
"output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S "
},
{
"input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP",
"output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP "
},
{
"input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB",
"output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y "
},
{
"input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL",
"output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL "
},
{
"input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO",
"output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O "
},
{
"input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU",
"output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU "
},
{
"input": "A",
"output": "A "
},
{
"input": "WUBA",
"output": "A "
},
{
"input": "WUBWUBA",
"output": "A "
},
{
"input": "AWUBWUBWUB",
"output": "A "
},
{
"input": "AWUBBWUBCWUBD",
"output": "A B C D "
},
{
"input": "WUBWWUBWUBWUBUWUBWUBBWUB",
"output": "W U B "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "WUWUBBWWUBUB",
"output": "WU BW UB "
},
{
"input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB",
"output": "WUAB "
},
{
"input": "ZWUB",
"output": "Z "
},
{
"input": "WU",
"output": "WU "
},
{
"input": "UB",
"output": "UB "
},
{
"input": "U",
"output": "U "
},
{
"input": "WUBW",
"output": "W "
},
{
"input": "WUBWU",
"output": "WU "
},
{
"input": "WUWUB",
"output": "WU "
},
{
"input": "UBWUB",
"output": "UB "
},
{
"input": "WUWUBUBWUBUWUB",
"output": "WU UB U "
},
{
"input": "WUBWWUBAWUB",
"output": "W A "
},
{
"input": "WUUUUU",
"output": "WUUUUU "
}
] | 1,693,161,233 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 92 | 0 | remix = input()
# remix = "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB"
remix_list = list(remix.split("WUB"))
org_list = []
# print(remix_list)
for i in remix_list:
# print(f"- step one: {i}")
if i != '':
org_list.append(i)
else:
pass
org = ' '.join(org_list)
print(org) | Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | ```python
remix = input()
# remix = "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB"
remix_list = list(remix.split("WUB"))
org_list = []
# print(remix_list)
for i in remix_list:
# print(f"- step one: {i}")
if i != '':
org_list.append(i)
else:
pass
org = ' '.join(org_list)
print(org)
``` | 3 | |
817 | B | Makes And The Product | PROGRAMMING | 1,500 | [
"combinatorics",
"implementation",
"math",
"sortings"
] | null | null | After returning from the army Makes received a gift — an array *a* consisting of *n* positive integer numbers. He hadn't been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (*i*,<= *j*,<= *k*) (*i*<=<<=*j*<=<<=*k*), such that *a**i*·*a**j*·*a**k* is minimum possible, are there in the array? Help him with it! | The first line of input contains a positive integer number *n* (3<=≤<=*n*<=≤<=105) — the number of elements in array *a*. The second line contains *n* positive integer numbers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of a given array. | Print one number — the quantity of triples (*i*,<= *j*,<= *k*) such that *i*,<= *j* and *k* are pairwise distinct and *a**i*·*a**j*·*a**k* is minimum possible. | [
"4\n1 1 1 1\n",
"5\n1 3 2 3 4\n",
"6\n1 3 3 1 3 2\n"
] | [
"4\n",
"2\n",
"1\n"
] | In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.
In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.
In the third example a triple of numbers (1, 1, 2) is chosen, and there's only one way to choose indices. | 0 | [
{
"input": "4\n1 1 1 1",
"output": "4"
},
{
"input": "5\n1 3 2 3 4",
"output": "2"
},
{
"input": "6\n1 3 3 1 3 2",
"output": "1"
},
{
"input": "3\n1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "4\n1 1 2 2",
"output": "2"
},
{
"input": "3\n1 3 1",
"output": "1"
},
{
"input": "11\n1 2 2 2 2 2 2 2 2 2 2",
"output": "45"
},
{
"input": "5\n1 2 2 2 2",
"output": "6"
},
{
"input": "6\n1 2 2 3 3 4",
"output": "1"
},
{
"input": "8\n1 1 2 2 2 3 3 3",
"output": "3"
},
{
"input": "6\n1 2 2 2 2 3",
"output": "6"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "6\n1 2 2 2 3 3",
"output": "3"
},
{
"input": "6\n1 2 2 2 2 2",
"output": "10"
},
{
"input": "4\n1 2 2 2",
"output": "3"
},
{
"input": "5\n1 2 3 2 3",
"output": "1"
},
{
"input": "6\n2 2 3 3 3 3",
"output": "4"
},
{
"input": "6\n1 2 2 2 5 6",
"output": "3"
},
{
"input": "10\n1 2 2 2 2 2 2 2 2 2",
"output": "36"
},
{
"input": "3\n2 1 2",
"output": "1"
},
{
"input": "5\n1 2 3 3 3",
"output": "3"
},
{
"input": "6\n1 2 2 2 4 5",
"output": "3"
},
{
"input": "4\n1 2 2 3",
"output": "1"
},
{
"input": "10\n2 2 2 2 2 1 2 2 2 2",
"output": "36"
},
{
"input": "7\n2 2 2 3 3 3 1",
"output": "3"
},
{
"input": "3\n1 1 2",
"output": "1"
},
{
"input": "5\n1 1 2 2 2",
"output": "3"
},
{
"input": "3\n1 2 3",
"output": "1"
},
{
"input": "9\n2 2 3 3 3 3 3 3 3",
"output": "7"
},
{
"input": "5\n1 1 2 2 3",
"output": "2"
},
{
"input": "4\n1 1 3 3",
"output": "2"
},
{
"input": "4\n33554432 33554432 67108864 33554432",
"output": "1"
},
{
"input": "6\n2 2 2 1 2 2",
"output": "10"
},
{
"input": "10\n1 2 1 2 3 2 3 2 2 2",
"output": "6"
},
{
"input": "10\n9 6 4 7 1 8 9 5 9 4",
"output": "1"
},
{
"input": "4\n5 7 2 7",
"output": "2"
},
{
"input": "3\n7 6 7",
"output": "1"
},
{
"input": "6\n3 2 8 2 5 3",
"output": "2"
},
{
"input": "3\n5 9 5",
"output": "1"
},
{
"input": "5\n6 3 7 6 3",
"output": "2"
},
{
"input": "9\n10 10 4 10 7 9 6 7 3",
"output": "1"
},
{
"input": "5\n9 10 10 3 8",
"output": "1"
},
{
"input": "5\n2 9 5 10 5",
"output": "1"
},
{
"input": "9\n7 1 9 6 6 8 3 1 3",
"output": "2"
},
{
"input": "5\n3 4 4 4 5",
"output": "3"
},
{
"input": "3\n3 1 3",
"output": "1"
},
{
"input": "8\n3 2 2 5 2 2 1 2",
"output": "10"
}
] | 1,554,962,587 | 2,147,483,647 | PyPy 3 | OK | TESTS | 80 | 202 | 15,667,200 | from math import factorial
n=int(input())
a=[int(i) for i in input().split()]
a.sort()
c1 = c2 = c3 = 0
for i in range(n):
if a[i] == a[0]:
c1 += 1
if a[i] == a[1]:
c2 += 1
if a[i] == a[2]:
c3 += 1
if a[0] == a[1]:
if a[1] == a[2]:
print((c1*(c1-1)*(c1-2))//6)
exit()
else:
print((c1*(c1-1))//2*c3)
exit()
if a[1] == a[2]:
print((c1*c2*(c2-1))//2)
exit()
print(c1*c2*c3) | Title: Makes And The Product
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After returning from the army Makes received a gift — an array *a* consisting of *n* positive integer numbers. He hadn't been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (*i*,<= *j*,<= *k*) (*i*<=<<=*j*<=<<=*k*), such that *a**i*·*a**j*·*a**k* is minimum possible, are there in the array? Help him with it!
Input Specification:
The first line of input contains a positive integer number *n* (3<=≤<=*n*<=≤<=105) — the number of elements in array *a*. The second line contains *n* positive integer numbers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of a given array.
Output Specification:
Print one number — the quantity of triples (*i*,<= *j*,<= *k*) such that *i*,<= *j* and *k* are pairwise distinct and *a**i*·*a**j*·*a**k* is minimum possible.
Demo Input:
['4\n1 1 1 1\n', '5\n1 3 2 3 4\n', '6\n1 3 3 1 3 2\n']
Demo Output:
['4\n', '2\n', '1\n']
Note:
In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.
In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.
In the third example a triple of numbers (1, 1, 2) is chosen, and there's only one way to choose indices. | ```python
from math import factorial
n=int(input())
a=[int(i) for i in input().split()]
a.sort()
c1 = c2 = c3 = 0
for i in range(n):
if a[i] == a[0]:
c1 += 1
if a[i] == a[1]:
c2 += 1
if a[i] == a[2]:
c3 += 1
if a[0] == a[1]:
if a[1] == a[2]:
print((c1*(c1-1)*(c1-2))//6)
exit()
else:
print((c1*(c1-1))//2*c3)
exit()
if a[1] == a[2]:
print((c1*c2*(c2-1))//2)
exit()
print(c1*c2*c3)
``` | 3 | |
387 | A | George and Sleep | PROGRAMMING | 900 | [
"implementation"
] | null | null | George woke up and saw the current time *s* on the digital clock. Besides, George knows that he has slept for time *t*.
Help George! Write a program that will, given time *s* and *t*, determine the time *p* when George went to bed. Note that George could have gone to bed yesterday relatively to the current time (see the second test sample). | The first line contains current time *s* as a string in the format "hh:mm". The second line contains time *t* in the format "hh:mm" — the duration of George's sleep. It is guaranteed that the input contains the correct time in the 24-hour format, that is, 00<=≤<=*hh*<=≤<=23, 00<=≤<=*mm*<=≤<=59. | In the single line print time *p* — the time George went to bed in the format similar to the format of the time in the input. | [
"05:50\n05:44\n",
"00:00\n01:00\n",
"00:01\n00:00\n"
] | [
"00:06\n",
"23:00\n",
"00:01\n"
] | In the first sample George went to bed at "00:06". Note that you should print the time only in the format "00:06". That's why answers "0:06", "00:6" and others will be considered incorrect.
In the second sample, George went to bed yesterday.
In the third sample, George didn't do to bed at all. | 500 | [
{
"input": "05:50\n05:44",
"output": "00:06"
},
{
"input": "00:00\n01:00",
"output": "23:00"
},
{
"input": "00:01\n00:00",
"output": "00:01"
},
{
"input": "23:59\n23:59",
"output": "00:00"
},
{
"input": "23:44\n23:55",
"output": "23:49"
},
{
"input": "00:00\n13:12",
"output": "10:48"
},
{
"input": "12:00\n23:59",
"output": "12:01"
},
{
"input": "12:44\n12:44",
"output": "00:00"
},
{
"input": "05:55\n07:12",
"output": "22:43"
},
{
"input": "07:12\n05:55",
"output": "01:17"
},
{
"input": "22:22\n22:22",
"output": "00:00"
},
{
"input": "22:22\n22:23",
"output": "23:59"
},
{
"input": "23:24\n23:23",
"output": "00:01"
},
{
"input": "00:00\n00:00",
"output": "00:00"
},
{
"input": "23:30\n00:00",
"output": "23:30"
},
{
"input": "01:00\n00:00",
"output": "01:00"
},
{
"input": "05:44\n06:00",
"output": "23:44"
},
{
"input": "00:00\n23:59",
"output": "00:01"
},
{
"input": "21:00\n01:00",
"output": "20:00"
},
{
"input": "21:21\n12:21",
"output": "09:00"
},
{
"input": "12:21\n21:12",
"output": "15:09"
},
{
"input": "12:33\n23:33",
"output": "13:00"
},
{
"input": "07:55\n05:53",
"output": "02:02"
},
{
"input": "19:30\n02:00",
"output": "17:30"
},
{
"input": "21:30\n02:00",
"output": "19:30"
},
{
"input": "19:30\n09:30",
"output": "10:00"
},
{
"input": "13:08\n00:42",
"output": "12:26"
},
{
"input": "13:04\n09:58",
"output": "03:06"
},
{
"input": "21:21\n23:06",
"output": "22:15"
},
{
"input": "20:53\n10:23",
"output": "10:30"
},
{
"input": "12:59\n00:45",
"output": "12:14"
},
{
"input": "12:39\n22:21",
"output": "14:18"
},
{
"input": "21:10\n13:50",
"output": "07:20"
},
{
"input": "03:38\n23:46",
"output": "03:52"
},
{
"input": "03:48\n00:41",
"output": "03:07"
},
{
"input": "07:43\n12:27",
"output": "19:16"
},
{
"input": "03:23\n08:52",
"output": "18:31"
},
{
"input": "16:04\n10:28",
"output": "05:36"
},
{
"input": "12:53\n08:37",
"output": "04:16"
},
{
"input": "13:43\n17:23",
"output": "20:20"
},
{
"input": "00:00\n00:01",
"output": "23:59"
},
{
"input": "10:10\n01:01",
"output": "09:09"
},
{
"input": "10:05\n00:00",
"output": "10:05"
},
{
"input": "09:09\n00:00",
"output": "09:09"
},
{
"input": "09:10\n00:01",
"output": "09:09"
},
{
"input": "23:24\n00:28",
"output": "22:56"
},
{
"input": "10:00\n01:00",
"output": "09:00"
}
] | 1,546,694,832 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<iostream>
#include<cstring>
using namespace std;
int main()
{
string a,b;
int h1=0,h2=0,m1=0,m2=0,a1,a2;
bool flag=false;
cin>>a>>b;
for(int i=0;i<a.size();i++)
if(flag==false&&a[i]!=':')
h1*=10,h1+=(a[i]-48);
else if(a[i]==':')
flag=true;
else
m1*=10,m1+=(a[i]-48);
flag=false;
for(int i=0;i<a.size();i++)
if(flag==false&&b[i]!=':')
h2*=10,h2+=(b[i]-48);
else if(b[i]==':')
flag=true;
else
m2*=10,m2+=(b[i]-48);
if(m1<m2)
m1+=60,h1--;
if(h1<h2)
h1+=24;
a1=h1-h2;
a2=m1-m2;
if(a1<10)
cout<<0;
cout<<a1<<':';
if(a2<10)
cout<<0;
cout<<a2;
return 0;
}
| Title: George and Sleep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George woke up and saw the current time *s* on the digital clock. Besides, George knows that he has slept for time *t*.
Help George! Write a program that will, given time *s* and *t*, determine the time *p* when George went to bed. Note that George could have gone to bed yesterday relatively to the current time (see the second test sample).
Input Specification:
The first line contains current time *s* as a string in the format "hh:mm". The second line contains time *t* in the format "hh:mm" — the duration of George's sleep. It is guaranteed that the input contains the correct time in the 24-hour format, that is, 00<=≤<=*hh*<=≤<=23, 00<=≤<=*mm*<=≤<=59.
Output Specification:
In the single line print time *p* — the time George went to bed in the format similar to the format of the time in the input.
Demo Input:
['05:50\n05:44\n', '00:00\n01:00\n', '00:01\n00:00\n']
Demo Output:
['00:06\n', '23:00\n', '00:01\n']
Note:
In the first sample George went to bed at "00:06". Note that you should print the time only in the format "00:06". That's why answers "0:06", "00:6" and others will be considered incorrect.
In the second sample, George went to bed yesterday.
In the third sample, George didn't do to bed at all. | ```python
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
string a,b;
int h1=0,h2=0,m1=0,m2=0,a1,a2;
bool flag=false;
cin>>a>>b;
for(int i=0;i<a.size();i++)
if(flag==false&&a[i]!=':')
h1*=10,h1+=(a[i]-48);
else if(a[i]==':')
flag=true;
else
m1*=10,m1+=(a[i]-48);
flag=false;
for(int i=0;i<a.size();i++)
if(flag==false&&b[i]!=':')
h2*=10,h2+=(b[i]-48);
else if(b[i]==':')
flag=true;
else
m2*=10,m2+=(b[i]-48);
if(m1<m2)
m1+=60,h1--;
if(h1<h2)
h1+=24;
a1=h1-h2;
a2=m1-m2;
if(a1<10)
cout<<0;
cout<<a1<<':';
if(a2<10)
cout<<0;
cout<<a2;
return 0;
}
``` | -1 | |
929 | C | Красивая команда | PROGRAMMING | 1,700 | [
"*special",
"combinatorics",
"math"
] | null | null | Завтра у хоккейной команды, которой руководит Евгений, важный матч. Евгению нужно выбрать шесть игроков, которые выйдут на лед в стартовом составе: один вратарь, два защитника и три нападающих.
Так как это стартовый состав, Евгения больше волнует, насколько красива будет команда на льду, чем способности игроков. А именно, Евгений хочет выбрать такой стартовый состав, чтобы номера любых двух игроков из стартового состава отличались не более, чем в два раза. Например, игроки с номерами 13, 14, 10, 18, 15 и 20 устроят Евгения, а если, например, на лед выйдут игроки с номерами 8 и 17, то это не устроит Евгения.
Про каждого из игроков вам известно, на какой позиции он играет (вратарь, защитник или нападающий), а также его номер. В хоккее номера игроков не обязательно идут подряд. Посчитайте число различных стартовых составов из одного вратаря, двух защитников и трех нападающих, которые может выбрать Евгений, чтобы выполнялось его условие красоты. | Первая строка содержит три целых числа *g*, *d* и *f* (1<=≤<=*g*<=≤<=1<=000, 1<=≤<=*d*<=≤<=1<=000, 1<=≤<=*f*<=≤<=1<=000) — число вратарей, защитников и нападающих в команде Евгения.
Вторая строка содержит *g* целых чисел, каждое в пределах от 1 до 100<=000 — номера вратарей.
Третья строка содержит *d* целых чисел, каждое в пределах от 1 до 100<=000 — номера защитников.
Четвертая строка содержит *f* целых чисел, каждое в пределах от 1 до 100<=000 — номера нападающих.
Гарантируется, что общее количество игроков не превосходит 1<=000, т. е. *g*<=+<=*d*<=+<=*f*<=≤<=1<=000. Все *g*<=+<=*d*<=+<=*f* номеров игроков различны. | Выведите одно целое число — количество возможных стартовых составов. | [
"1 2 3\n15\n10 19\n20 11 13\n",
"2 3 4\n16 40\n20 12 19\n13 21 11 10\n"
] | [
"1\n",
"6\n"
] | В первом примере всего один вариант для выбора состава, который удовлетворяет описанным условиям, поэтому ответ 1.
Во втором примере подходят следующие игровые сочетания (в порядке вратарь-защитник-защитник-нападающий-нападающий-нападающий):
- 16 20 12 13 21 11 - 16 20 12 13 11 10 - 16 20 19 13 21 11 - 16 20 19 13 11 10 - 16 12 19 13 21 11 - 16 12 19 13 11 10
Таким образом, ответ на этот пример — 6. | 1,750 | [
{
"input": "1 2 3\n15\n10 19\n20 11 13",
"output": "1"
},
{
"input": "2 3 4\n16 40\n20 12 19\n13 21 11 10",
"output": "6"
},
{
"input": "4 4 5\n15 16 19 6\n8 11 9 18\n5 3 1 12 14",
"output": "0"
},
{
"input": "6 7 7\n32 35 26 33 16 23\n4 40 36 12 28 24 3\n39 11 31 37 1 25 6",
"output": "120"
},
{
"input": "9 10 7\n935 433 848 137 548 958 758 576 592\n780 129 631 991 575 421 245 944 487 771\n430 34 276 8 165 188 727",
"output": "0"
},
{
"input": "17 15 17\n598 1369 806 247 1570 361 1650 1250 1269 1744 1400 1074 947 115 863 1392 1044\n1252 1797 1574 1445 1274 246 1483 1814 231 804 543 1142 1425 125 1280\n1276 1724 512 1975 1215 1205 1415 1141 993 199 1318 855 389 376 1386 146 1297",
"output": "108025"
},
{
"input": "29 20 26\n250 44 142 149 3 84 85 267 191 144 100 164 66 125 278 37 244 288 124 50 47 16 141 93 9 242 78 238 59\n176 276 33 91 248 234 205 60 8 80 81 88 4 213 53 175 290 206 168 185\n10 56 225 193 73 209 246 296 152 146 221 294 275 83 42 192 23 24 82 226 70 222 189 20 210 265",
"output": "360518"
},
{
"input": "30 24 30\n61 189 108 126 2 180 15 141 75 67 115 107 144 196 4 135 38 106 146 136 31 114 14 49 158 54 173 69 91 98\n151 109 46 182 23 94 39 168 165 30 103 66 179 70 40 198 8 152 163 87 176 187 55 3\n65 140 21 142 80 185 125 19 190 157 73 186 58 188 105 93 83 1 7 79 52 82 113 13 10 164 174 119 96 78",
"output": "670920"
},
{
"input": "29 42 50\n605 254 369 842 889 103 937 235 135 698 482 883 738 467 848 70 1000 129 970 58 94 873 140 363 133 913 834 727 185\n17 676 703 245 149 296 800 106 153 111 285 382 12 704 830 664 30 533 604 380 469 155 216 466 36 347 270 170 10 349 86 5 164 599 517 593 373 461 908 34 569 573\n614 439 78 172 109 217 85 463 720 176 571 486 503 318 977 501 910 196 882 107 584 940 928 249 537 962 333 477 897 875 500 915 227 256 194 808 193 759 934 324 525 174 792 425 449 843 824 261 654 868",
"output": "7743753"
},
{
"input": "1 2 3\n1\n100 200\n300 400 500",
"output": "0"
},
{
"input": "40 40 40\n1 118 100 19 91 115 34 22 28 55 43 109 13 94 7 4 31 79 10 52 16 88 37 112 97 76 70 25 64 103 61 106 58 85 67 40 82 49 46 73\n59 80 23 113 35 56 95 116 107 44 65 26 38 98 47 14 86 11 50 89 29 119 41 5 17 71 92 110 2 32 20 104 83 8 53 77 62 101 74 68\n63 78 54 90 75 3 99 6 93 42 111 9 51 102 57 81 66 48 21 87 12 84 117 24 69 120 15 45 33 108 39 72 18 60 105 114 96 36 30 27",
"output": "9339317"
},
{
"input": "40 40 40\n100 73 109 115 40 88 58 76 22 31 34 7 97 61 70 16 10 64 103 94 79 106 67 13 118 43 85 46 19 112 1 55 4 91 28 49 37 82 52 25\n9 72 102 21 51 90 69 114 27 60 75 18 42 78 120 84 57 39 93 3 6 63 117 48 99 111 24 45 108 54 33 12 30 81 87 36 15 96 105 66\n119 98 113 23 116 71 83 56 68 65 44 50 29 107 26 38 5 35 14 101 86 77 62 80 89 92 104 2 59 20 11 74 53 47 17 32 95 41 8 110",
"output": "9166683"
},
{
"input": "40 40 40\n116 101 80 62 38 11 20 50 65 41 110 119 68 56 5 53 83 14 107 98 104 92 32 2 113 95 71 59 89 23 74 86 29 35 47 17 77 8 26 44\n67 97 22 37 4 55 46 100 40 16 64 79 43 19 82 109 34 10 52 7 88 85 1 13 73 94 25 106 91 115 58 31 61 28 70 112 76 49 118 103\n39 6 57 120 87 51 81 99 90 15 33 21 12 66 3 48 114 111 75 9 27 117 105 72 42 102 60 108 18 84 93 69 63 30 78 54 24 36 45 96",
"output": "9199268"
},
{
"input": "40 40 40\n86 41 89 2 32 29 11 107 20 101 35 8 59 47 104 74 56 50 95 92 53 119 68 113 14 77 71 23 38 5 62 44 65 83 110 98 116 80 17 26\n96 75 60 30 57 78 108 12 36 93 111 66 6 48 72 33 3 84 90 45 9 117 42 18 21 120 114 24 51 15 39 63 69 87 27 102 105 54 81 99\n94 10 1 112 22 103 109 46 82 25 40 34 61 79 19 85 13 70 106 28 31 118 97 67 76 16 91 115 58 4 88 49 73 52 55 7 100 64 43 37",
"output": "8979951"
},
{
"input": "40 40 40\n33 69 27 30 72 108 57 75 99 42 66 84 15 24 120 54 9 87 60 18 117 93 6 39 111 81 21 48 96 12 102 78 3 105 90 45 114 36 51 63\n61 40 4 7 34 55 94 46 112 19 85 97 28 100 115 79 103 82 67 109 73 91 64 16 106 22 70 1 25 49 37 76 88 43 13 118 31 52 10 58\n50 59 8 56 14 86 89 110 47 104 68 95 107 77 62 17 20 38 92 83 71 53 23 113 32 101 98 11 29 65 80 74 119 116 5 35 41 2 44 26",
"output": "9067332"
},
{
"input": "40 40 40\n93 90 27 120 108 21 12 114 66 45 48 57 9 81 18 75 111 39 6 102 117 15 30 3 51 96 99 33 72 24 78 54 36 87 105 69 42 63 84 60\n107 83 77 104 95 65 113 35 8 86 89 119 29 98 68 38 92 110 14 5 23 56 50 59 2 47 41 26 11 116 44 74 80 101 53 17 71 20 62 32\n22 7 43 40 85 49 79 31 46 61 118 82 115 67 112 34 28 13 88 91 73 16 25 4 19 70 37 1 103 10 55 76 97 94 58 64 52 106 100 109",
"output": "9020649"
},
{
"input": "2 1 3\n10 20\n15\n13 14 16",
"output": "0"
},
{
"input": "2 2 2\n10 11\n12 13\n14 15",
"output": "0"
},
{
"input": "1 2 1\n10\n11 12\n13",
"output": "0"
}
] | 1,520,495,004 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 1,000 | 63,488,000 | from itertools import product,permutations,combinations
a,b,c=[int(s) for s in input().split()]
k=0
lists=[]
masa=[int(s) for s in input().split()]
masb=[int(s) for s in input().split()]
masc=[int(s) for s in input().split()]
comb1=list(combinations(masa,1))
comb2=list(combinations(masb,2))
comb3=list(combinations(masc,3))
comb=list(product(comb1,comb2,comb3))
for x in comb:
for i in x:
lists+=list(i)
if min(lists)*2>=max(lists):
k+=1
lists=[]
print(k) | Title: Красивая команда
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Завтра у хоккейной команды, которой руководит Евгений, важный матч. Евгению нужно выбрать шесть игроков, которые выйдут на лед в стартовом составе: один вратарь, два защитника и три нападающих.
Так как это стартовый состав, Евгения больше волнует, насколько красива будет команда на льду, чем способности игроков. А именно, Евгений хочет выбрать такой стартовый состав, чтобы номера любых двух игроков из стартового состава отличались не более, чем в два раза. Например, игроки с номерами 13, 14, 10, 18, 15 и 20 устроят Евгения, а если, например, на лед выйдут игроки с номерами 8 и 17, то это не устроит Евгения.
Про каждого из игроков вам известно, на какой позиции он играет (вратарь, защитник или нападающий), а также его номер. В хоккее номера игроков не обязательно идут подряд. Посчитайте число различных стартовых составов из одного вратаря, двух защитников и трех нападающих, которые может выбрать Евгений, чтобы выполнялось его условие красоты.
Input Specification:
Первая строка содержит три целых числа *g*, *d* и *f* (1<=≤<=*g*<=≤<=1<=000, 1<=≤<=*d*<=≤<=1<=000, 1<=≤<=*f*<=≤<=1<=000) — число вратарей, защитников и нападающих в команде Евгения.
Вторая строка содержит *g* целых чисел, каждое в пределах от 1 до 100<=000 — номера вратарей.
Третья строка содержит *d* целых чисел, каждое в пределах от 1 до 100<=000 — номера защитников.
Четвертая строка содержит *f* целых чисел, каждое в пределах от 1 до 100<=000 — номера нападающих.
Гарантируется, что общее количество игроков не превосходит 1<=000, т. е. *g*<=+<=*d*<=+<=*f*<=≤<=1<=000. Все *g*<=+<=*d*<=+<=*f* номеров игроков различны.
Output Specification:
Выведите одно целое число — количество возможных стартовых составов.
Demo Input:
['1 2 3\n15\n10 19\n20 11 13\n', '2 3 4\n16 40\n20 12 19\n13 21 11 10\n']
Demo Output:
['1\n', '6\n']
Note:
В первом примере всего один вариант для выбора состава, который удовлетворяет описанным условиям, поэтому ответ 1.
Во втором примере подходят следующие игровые сочетания (в порядке вратарь-защитник-защитник-нападающий-нападающий-нападающий):
- 16 20 12 13 21 11 - 16 20 12 13 11 10 - 16 20 19 13 21 11 - 16 20 19 13 11 10 - 16 12 19 13 21 11 - 16 12 19 13 11 10
Таким образом, ответ на этот пример — 6. | ```python
from itertools import product,permutations,combinations
a,b,c=[int(s) for s in input().split()]
k=0
lists=[]
masa=[int(s) for s in input().split()]
masb=[int(s) for s in input().split()]
masc=[int(s) for s in input().split()]
comb1=list(combinations(masa,1))
comb2=list(combinations(masb,2))
comb3=list(combinations(masc,3))
comb=list(product(comb1,comb2,comb3))
for x in comb:
for i in x:
lists+=list(i)
if min(lists)*2>=max(lists):
k+=1
lists=[]
print(k)
``` | 0 | |
984 | A | Game | PROGRAMMING | 800 | [
"sortings"
] | null | null | Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves. | The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$). | Print one number that will be left on the board. | [
"3\n2 1 3\n",
"3\n2 2 2\n"
] | [
"2",
"2"
] | In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players. | 500 | [
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 2 2",
"output": "2"
},
{
"input": "9\n44 53 51 80 5 27 74 79 94",
"output": "53"
},
{
"input": "10\n38 82 23 37 96 4 81 60 67 86",
"output": "60"
},
{
"input": "10\n58 26 77 15 53 81 68 48 22 65",
"output": "53"
},
{
"input": "1\n124",
"output": "124"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 1 1000",
"output": "1"
},
{
"input": "2\n322 322",
"output": "322"
},
{
"input": "3\n9 92 12",
"output": "12"
},
{
"input": "3\n1 2 2",
"output": "2"
}
] | 1,686,311,506 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 30 | 0 | import sys
n = sys.stdin.readline()
y = list(sys.stdin.readline().split())
y.remove(max(y))
y.remove(min(y))
print(*y) | Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
Input Specification:
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
Output Specification:
Print one number that will be left on the board.
Demo Input:
['3\n2 1 3\n', '3\n2 2 2\n']
Demo Output:
['2', '2']
Note:
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players. | ```python
import sys
n = sys.stdin.readline()
y = list(sys.stdin.readline().split())
y.remove(max(y))
y.remove(min(y))
print(*y)
``` | 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"
}
] | 1,549,295,023 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 218 | 0 | n = int(input())
t_x = 0
t_y = 0
t_z = 0
for i in range(0,n):
s = str(input())
l = s.split()
t_x = t_x + int(l[0])
t_y = t_y + int(l[1])
t_z = t_z + int(l[2])
if t_x == 0 and t_y == 0 and t_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())
t_x = 0
t_y = 0
t_z = 0
for i in range(0,n):
s = str(input())
l = s.split()
t_x = t_x + int(l[0])
t_y = t_y + int(l[1])
t_z = t_z + int(l[2])
if t_x == 0 and t_y == 0 and t_z == 0 :
print("YES")
else:
print("NO")
``` | 3.9455 |
913 | A | Modular Exponentiation | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | The following problem is well-known: given integers *n* and *m*, calculate
where 2*n*<==<=2·2·...·2 (*n* factors), and denotes the remainder of division of *x* by *y*.
You are asked to solve the "reverse" problem. Given integers *n* and *m*, calculate | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=108).
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=108). | Output a single integer — the value of . | [
"4\n42\n",
"1\n58\n",
"98765432\n23456789\n"
] | [
"10\n",
"0\n",
"23456789\n"
] | In the first example, the remainder of division of 42 by 2<sup class="upper-index">4</sup> = 16 is equal to 10.
In the second example, 58 is divisible by 2<sup class="upper-index">1</sup> = 2 without remainder, and the answer is 0. | 500 | [
{
"input": "4\n42",
"output": "10"
},
{
"input": "1\n58",
"output": "0"
},
{
"input": "98765432\n23456789",
"output": "23456789"
},
{
"input": "8\n88127381",
"output": "149"
},
{
"input": "32\n92831989",
"output": "92831989"
},
{
"input": "92831989\n25",
"output": "25"
},
{
"input": "100000000\n100000000",
"output": "100000000"
},
{
"input": "7\n1234",
"output": "82"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n100000000",
"output": "0"
},
{
"input": "100000000\n1",
"output": "1"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "2\n1",
"output": "1"
},
{
"input": "2\n2",
"output": "2"
},
{
"input": "2\n3",
"output": "3"
},
{
"input": "2\n4",
"output": "0"
},
{
"input": "2\n5",
"output": "1"
},
{
"input": "25\n33554432",
"output": "0"
},
{
"input": "26\n33554432",
"output": "33554432"
},
{
"input": "25\n67108864",
"output": "0"
},
{
"input": "26\n67108864",
"output": "0"
},
{
"input": "25\n92831989",
"output": "25723125"
},
{
"input": "26\n92831989",
"output": "25723125"
},
{
"input": "27\n92831989",
"output": "92831989"
},
{
"input": "29\n92831989",
"output": "92831989"
},
{
"input": "30\n92831989",
"output": "92831989"
},
{
"input": "31\n92831989",
"output": "92831989"
},
{
"input": "99812399\n93172892",
"output": "93172892"
},
{
"input": "67108864\n12345678",
"output": "12345678"
},
{
"input": "3\n98391849",
"output": "1"
},
{
"input": "13\n1231232",
"output": "2432"
},
{
"input": "18\n9842194",
"output": "142866"
},
{
"input": "24\n71237177",
"output": "4128313"
},
{
"input": "11\n65535",
"output": "2047"
},
{
"input": "11\n65536",
"output": "0"
},
{
"input": "12\n123",
"output": "123"
},
{
"input": "123456\n12",
"output": "12"
},
{
"input": "88\n888",
"output": "888"
},
{
"input": "888\n88",
"output": "88"
},
{
"input": "8888\n8888",
"output": "8888"
},
{
"input": "64\n128",
"output": "128"
},
{
"input": "128\n64",
"output": "64"
},
{
"input": "60\n9999999",
"output": "9999999"
},
{
"input": "200\n100000",
"output": "100000"
},
{
"input": "3\n8",
"output": "0"
},
{
"input": "10000000\n25024886",
"output": "25024886"
},
{
"input": "100\n100000",
"output": "100000"
},
{
"input": "99999999\n100000000",
"output": "100000000"
},
{
"input": "70\n12345678",
"output": "12345678"
}
] | 1,594,404,806 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 1,000 | 22,425,600 | from sys import stdin
def binpow(a,b):
res = 1
while b > 0:
if b & 1:
res *= a
a *= a
b >>= 1
return res
n = int(stdin.readline())
m = int(stdin.readline())
print(m % binpow(2,n)) | Title: Modular Exponentiation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The following problem is well-known: given integers *n* and *m*, calculate
where 2*n*<==<=2·2·...·2 (*n* factors), and denotes the remainder of division of *x* by *y*.
You are asked to solve the "reverse" problem. Given integers *n* and *m*, calculate
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=108).
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=108).
Output Specification:
Output a single integer — the value of .
Demo Input:
['4\n42\n', '1\n58\n', '98765432\n23456789\n']
Demo Output:
['10\n', '0\n', '23456789\n']
Note:
In the first example, the remainder of division of 42 by 2<sup class="upper-index">4</sup> = 16 is equal to 10.
In the second example, 58 is divisible by 2<sup class="upper-index">1</sup> = 2 without remainder, and the answer is 0. | ```python
from sys import stdin
def binpow(a,b):
res = 1
while b > 0:
if b & 1:
res *= a
a *= a
b >>= 1
return res
n = int(stdin.readline())
m = int(stdin.readline())
print(m % binpow(2,n))
``` | 0 | |
476 | B | Dreamoon and WiFi | PROGRAMMING | 1,300 | [
"bitmasks",
"brute force",
"combinatorics",
"dp",
"math",
"probabilities"
] | null | 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*<sub class="lower-index">1</sub> and *s*<sub class="lower-index">2</sub> will lead Dreamoon to finish at the same position + 1.
For the second sample, *s*<sub class="lower-index">1</sub> will lead Dreamoon to finish at position 0, while there are four possibilites for *s*<sub class="lower-index">2</sub>: {"+-++", "+-+-", "+--+", "+---"} 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*<sub class="lower-index">2</sub> 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"
}
] | 1,656,087,788 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 46 | 0 | f = input()
s = input()
t = f.count('+') - f.count('-')
known = s.count('+') - s.count('-')
sp = s.count('?')
if sp == 0:
if known == t:
print(1.0)
else:
print(0.0)
exit(0)
diff = abs(t-known)
if diff - sp > 0:
print(0.0)
exit(0)
print(1/sp) | 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*<sub class="lower-index">1</sub> and *s*<sub class="lower-index">2</sub> will lead Dreamoon to finish at the same position + 1.
For the second sample, *s*<sub class="lower-index">1</sub> will lead Dreamoon to finish at position 0, while there are four possibilites for *s*<sub class="lower-index">2</sub>: {"+-++", "+-+-", "+--+", "+---"} 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*<sub class="lower-index">2</sub> could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position + 3 is 0. | ```python
f = input()
s = input()
t = f.count('+') - f.count('-')
known = s.count('+') - s.count('-')
sp = s.count('?')
if sp == 0:
if known == t:
print(1.0)
else:
print(0.0)
exit(0)
diff = abs(t-known)
if diff - sp > 0:
print(0.0)
exit(0)
print(1/sp)
``` | 0 | |
1,010 | C | Border | PROGRAMMING | 1,800 | [
"number theory"
] | null | null | Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are $n$ banknote denominations on Mars: the value of $i$-th banknote is $a_i$. Natasha has an infinite number of banknotes of each denomination.
Martians have $k$ fingers on their hands, so they use a number system with base $k$. In addition, the Martians consider the digit $d$ (in the number system with base $k$) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base $k$ is $d$, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values $d$ Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change. | The first line contains two integers $n$ and $k$ ($1 \le n \le 100\,000$, $2 \le k \le 100\,000$) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — denominations of banknotes on Mars.
All numbers are given in decimal notation. | On the first line output the number of values $d$ for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation. | [
"2 8\n12 20\n",
"3 10\n10 20 30\n"
] | [
"2\n0 4 ",
"1\n0 "
] | Consider the first test case. It uses the octal number system.
If you take one banknote with the value of $12$, you will get $14_8$ in octal system. The last digit is $4_8$.
If you take one banknote with the value of $12$ and one banknote with the value of $20$, the total value will be $32$. In the octal system, it is $40_8$. The last digit is $0_8$.
If you take two banknotes with the value of $20$, the total value will be $40$, this is $50_8$ in the octal system. The last digit is $0_8$.
No other digits other than $0_8$ and $4_8$ can be obtained. Digits $0_8$ and $4_8$ could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero. | 1,000 | [
{
"input": "2 8\n12 20",
"output": "2\n0 4 "
},
{
"input": "3 10\n10 20 30",
"output": "1\n0 "
},
{
"input": "5 10\n20 16 4 16 2",
"output": "5\n0 2 4 6 8 "
},
{
"input": "10 5\n4 6 8 6 4 10 2 10 8 6",
"output": "5\n0 1 2 3 4 "
},
{
"input": "20 25\n15 10 5 20 10 5 15 5 15 10 15 5 5 5 5 10 15 20 20 20",
"output": "5\n0 5 10 15 20 "
},
{
"input": "30 30\n11 23 7 30 13 6 25 29 1 15 20 5 28 15 19 22 21 5 27 25 29 10 1 4 12 19 1 5 8 10",
"output": "30\n0 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 "
},
{
"input": "40 30\n16 12 12 22 18 28 32 24 36 26 12 30 22 16 32 30 36 18 20 16 12 24 28 20 16 28 8 34 18 18 18 4 4 36 18 10 30 38 18 10",
"output": "15\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 "
},
{
"input": "50 30\n15 9 21 39 42 39 3 42 42 39 6 48 39 30 12 39 27 45 30 48 18 33 18 36 27 3 48 12 36 27 15 12 42 39 18 21 48 39 15 42 24 36 33 48 6 48 15 12 30 18",
"output": "10\n0 3 6 9 12 15 18 21 24 27 "
},
{
"input": "1 10\n1",
"output": "10\n0 1 2 3 4 5 6 7 8 9 "
},
{
"input": "1 2\n1",
"output": "2\n0 1 "
},
{
"input": "60 30\n10 30 45 15 25 60 10 40 35 25 5 40 35 40 15 5 15 35 10 60 25 15 60 10 30 10 5 25 10 15 60 20 30 5 50 50 40 20 55 40 35 15 15 10 60 40 50 50 30 15 25 45 35 40 15 5 5 20 60 45",
"output": "6\n0 5 10 15 20 25 "
},
{
"input": "70 30\n54 30 12 48 42 24 42 60 54 6 36 42 54 66 12 48 54 42 24 54 30 18 30 54 18 60 24 30 54 48 48 60 18 60 60 66 54 18 54 30 24 30 60 54 36 36 60 48 12 60 6 60 42 66 6 42 18 60 54 48 42 18 48 66 18 42 48 30 12 66",
"output": "5\n0 6 12 18 24 "
},
{
"input": "80 30\n30 80 40 40 60 60 40 80 70 80 30 30 60 80 30 70 60 10 10 30 70 60 70 20 40 20 30 10 60 70 70 50 60 70 70 30 70 60 60 70 20 60 10 60 70 80 20 30 30 20 60 50 40 40 80 70 70 20 40 80 30 50 40 10 40 20 70 10 80 10 50 40 50 70 40 80 10 40 60 60",
"output": "3\n0 10 20 "
},
{
"input": "90 30\n90 45 75 75 90 90 45 30 90 15 45 90 15 30 45 60 30 15 30 45 45 45 45 15 45 60 15 60 45 75 45 75 90 60 30 15 60 30 90 75 15 60 15 30 45 30 45 15 30 15 45 30 15 75 90 15 45 15 75 15 75 30 75 45 60 75 15 45 30 75 45 90 45 60 90 45 75 30 30 30 15 15 75 60 75 90 75 60 90 45",
"output": "2\n0 15 "
},
{
"input": "100 30\n30 30 30 90 30 30 90 90 30 90 30 90 90 30 30 30 60 60 60 30 30 60 90 90 90 60 30 90 60 60 90 60 60 60 30 60 60 60 60 90 60 30 60 90 90 90 60 60 90 60 60 60 60 30 30 60 30 60 60 90 30 60 60 60 90 60 90 30 30 60 30 90 90 90 90 60 90 30 30 60 60 30 60 60 60 30 90 60 60 60 90 60 30 90 60 30 30 60 90 90",
"output": "1\n0 "
},
{
"input": "1 10\n2",
"output": "5\n0 2 4 6 8 "
},
{
"input": "1 10\n3",
"output": "10\n0 1 2 3 4 5 6 7 8 9 "
},
{
"input": "5 2\n1 1 1 1 1",
"output": "2\n0 1 "
},
{
"input": "2 30\n6 10",
"output": "15\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 "
},
{
"input": "1 10\n10",
"output": "1\n0 "
},
{
"input": "1 10\n20",
"output": "1\n0 "
},
{
"input": "1 2\n1000000000",
"output": "1\n0 "
},
{
"input": "2 6\n2 3",
"output": "6\n0 1 2 3 4 5 "
},
{
"input": "1 5\n4",
"output": "5\n0 1 2 3 4 "
},
{
"input": "2 5\n2 4",
"output": "5\n0 1 2 3 4 "
},
{
"input": "3 30\n6 10 15",
"output": "30\n0 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 "
},
{
"input": "2 7\n3 6",
"output": "7\n0 1 2 3 4 5 6 "
},
{
"input": "2 15\n3 5",
"output": "15\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 "
},
{
"input": "2 12\n4 6",
"output": "6\n0 2 4 6 8 10 "
},
{
"input": "2 10\n3 6",
"output": "10\n0 1 2 3 4 5 6 7 8 9 "
},
{
"input": "2 100000\n2 4",
"output": "50000\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120 122 124 126 128 130 132 134 136 138 140 142 144 146 148 150 152 154 156 158 160 162 164 166 168 170 172 174 176 178 180 182 184 186 188 190 192 194 196 198 200 202 204 206 208 210 212 214 216 218 220 222 224 226 228 230 232 234 236 238 240 242 244 246 248 250 252 254 256 258 260 262 264 266 268 270 272 274 276 278..."
},
{
"input": "1 14\n4",
"output": "7\n0 2 4 6 8 10 12 "
},
{
"input": "1 13\n5",
"output": "13\n0 1 2 3 4 5 6 7 8 9 10 11 12 "
},
{
"input": "2 420\n412 363",
"output": "420\n0 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 ..."
},
{
"input": "2 30\n10 6",
"output": "15\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 "
},
{
"input": "1 5\n3",
"output": "5\n0 1 2 3 4 "
},
{
"input": "1 8\n6",
"output": "4\n0 2 4 6 "
},
{
"input": "3 10\n6 6 6",
"output": "5\n0 2 4 6 8 "
},
{
"input": "1 7\n6",
"output": "7\n0 1 2 3 4 5 6 "
},
{
"input": "1 9\n6",
"output": "3\n0 3 6 "
},
{
"input": "2 10\n7 7",
"output": "10\n0 1 2 3 4 5 6 7 8 9 "
},
{
"input": "1 8\n5",
"output": "8\n0 1 2 3 4 5 6 7 "
},
{
"input": "2 50\n3 15",
"output": "50\n0 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 "
},
{
"input": "1 6\n4",
"output": "3\n0 2 4 "
},
{
"input": "1 7\n4",
"output": "7\n0 1 2 3 4 5 6 "
},
{
"input": "1 10\n7",
"output": "10\n0 1 2 3 4 5 6 7 8 9 "
},
{
"input": "1 3\n2",
"output": "3\n0 1 2 "
},
{
"input": "2 10\n2 5",
"output": "10\n0 1 2 3 4 5 6 7 8 9 "
},
{
"input": "2 50\n15 3",
"output": "50\n0 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 "
},
{
"input": "1 5\n2",
"output": "5\n0 1 2 3 4 "
},
{
"input": "2 10\n6 5",
"output": "10\n0 1 2 3 4 5 6 7 8 9 "
},
{
"input": "2 35\n5 7",
"output": "35\n0 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 "
},
{
"input": "2 300\n3 10",
"output": "300\n0 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 ..."
},
{
"input": "1 100000\n47",
"output": "100000\n0 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 1..."
},
{
"input": "2 54\n6 9",
"output": "18\n0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 "
}
] | 1,578,300,511 | 2,147,483,647 | PyPy 3 | OK | TESTS | 107 | 249 | 13,824,000 | n,k=map(int,input().split())
g=lambda x:int(x)%k
a=list(map(g,input().split()))
def nod(a,b):
while b>0:
a%=b
a,b=b,a
return a
c=k
for x in a:
if x!=0:
c=nod(c,x)
print(k//c)
print(' '.join(map(str,list(range(0,k,c)))))
| Title: Border
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are $n$ banknote denominations on Mars: the value of $i$-th banknote is $a_i$. Natasha has an infinite number of banknotes of each denomination.
Martians have $k$ fingers on their hands, so they use a number system with base $k$. In addition, the Martians consider the digit $d$ (in the number system with base $k$) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base $k$ is $d$, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values $d$ Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input Specification:
The first line contains two integers $n$ and $k$ ($1 \le n \le 100\,000$, $2 \le k \le 100\,000$) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output Specification:
On the first line output the number of values $d$ for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Demo Input:
['2 8\n12 20\n', '3 10\n10 20 30\n']
Demo Output:
['2\n0 4 ', '1\n0 ']
Note:
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of $12$, you will get $14_8$ in octal system. The last digit is $4_8$.
If you take one banknote with the value of $12$ and one banknote with the value of $20$, the total value will be $32$. In the octal system, it is $40_8$. The last digit is $0_8$.
If you take two banknotes with the value of $20$, the total value will be $40$, this is $50_8$ in the octal system. The last digit is $0_8$.
No other digits other than $0_8$ and $4_8$ can be obtained. Digits $0_8$ and $4_8$ could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero. | ```python
n,k=map(int,input().split())
g=lambda x:int(x)%k
a=list(map(g,input().split()))
def nod(a,b):
while b>0:
a%=b
a,b=b,a
return a
c=k
for x in a:
if x!=0:
c=nod(c,x)
print(k//c)
print(' '.join(map(str,list(range(0,k,c)))))
``` | 3 | |
32 | B | Borze | PROGRAMMING | 800 | [
"expression parsing",
"implementation"
] | B. Borze | 2 | 256 | Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet. | The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes). | Output the decoded ternary number. It can have leading zeroes. | [
".-.--\n",
"--.\n",
"-..-.--\n"
] | [
"012",
"20",
"1012"
] | none | 1,000 | [
{
"input": ".-.--",
"output": "012"
},
{
"input": "--.",
"output": "20"
},
{
"input": "-..-.--",
"output": "1012"
},
{
"input": "---..",
"output": "210"
},
{
"input": "..--.---..",
"output": "0020210"
},
{
"input": "-.....----.",
"output": "10000220"
},
{
"input": ".",
"output": "0"
},
{
"input": "-.",
"output": "1"
},
{
"input": "--",
"output": "2"
},
{
"input": "..",
"output": "00"
},
{
"input": "--.",
"output": "20"
},
{
"input": ".--.",
"output": "020"
},
{
"input": ".-.-..",
"output": "0110"
},
{
"input": "----.-.",
"output": "2201"
},
{
"input": "-..--.-.",
"output": "10201"
},
{
"input": "..--..--.",
"output": "0020020"
},
{
"input": "-.-.---.--..-..-.-.-..-..-.--.",
"output": "112120010111010120"
},
{
"input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.",
"output": "21112220010101011012011011221011"
},
{
"input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------",
"output": "11020111110111222212021020002022111100201121222020012022110010222"
},
{
"input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..",
"output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110"
},
{
"input": ".-..-.-.---.-----.--.---...-.--.-.-....-..",
"output": "01011212212021001201100010"
},
{
"input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..",
"output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210"
},
{
"input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.",
"output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220"
},
{
"input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----",
"output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122"
},
{
"input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..",
"output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110"
},
{
"input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.",
"output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020"
},
{
"input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..",
"output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200"
},
{
"input": "-.---...----...--.--...-.--.----",
"output": "121002200020200012022"
},
{
"input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.",
"output": "202020212012111010201021100001021221200011222112012201012220"
}
] | 1,563,435,320 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 216 | 0 | lis = input()
i = 0
ln = len(lis)
while i< ln-1:
if lis[i] == '-':
i+=1
if lis[i] == '-':
print(2)
else:
print(1)
elif lis[i]== '.':
print(0)
i+=1
| Title: Borze
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
Input Specification:
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
Output Specification:
Output the decoded ternary number. It can have leading zeroes.
Demo Input:
['.-.--\n', '--.\n', '-..-.--\n']
Demo Output:
['012', '20', '1012']
Note:
none | ```python
lis = input()
i = 0
ln = len(lis)
while i< ln-1:
if lis[i] == '-':
i+=1
if lis[i] == '-':
print(2)
else:
print(1)
elif lis[i]== '.':
print(0)
i+=1
``` | 0 |
368 | B | Sereja and Suffixes | PROGRAMMING | 1,100 | [
"data structures",
"dp"
] | null | null | Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.?
Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements.
Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*). | Print *m* lines — on the *i*-th line print the answer to the number *l**i*. | [
"10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n"
] | [
"6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n"
] | none | 1,000 | [
{
"input": "10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10",
"output": "6\n6\n6\n6\n6\n5\n4\n3\n2\n1"
},
{
"input": "8 3\n8 6 4 3 4 2 4 8\n6\n4\n2",
"output": "3\n4\n5"
},
{
"input": "7 10\n1 3 8 6 2 2 7\n4\n2\n6\n3\n4\n4\n6\n2\n7\n4",
"output": "3\n5\n2\n4\n3\n3\n2\n5\n1\n3"
},
{
"input": "10 2\n2 6 5 7 2 2 3 2 4 8\n1\n2",
"output": "7\n7"
},
{
"input": "7 1\n68346 10956 76708 23018 84063 34833 80407\n1",
"output": "7"
},
{
"input": "2 2\n8 4\n1\n1",
"output": "2\n2"
},
{
"input": "1 5\n5\n1\n1\n1\n1\n1",
"output": "1\n1\n1\n1\n1"
},
{
"input": "4 7\n3 1 4 2\n4\n1\n2\n3\n2\n4\n4",
"output": "1\n4\n3\n2\n3\n1\n1"
},
{
"input": "4 3\n9 1 7 1\n1\n4\n2",
"output": "3\n1\n2"
},
{
"input": "8 3\n9280 6676 2720 6172 8329 10413 3975 1394\n5\n6\n7",
"output": "4\n3\n2"
},
{
"input": "1 1\n1\n1",
"output": "1"
}
] | 1,667,962,596 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 1,000 | 4,915,200 | [n,m]=[int(j) for j in input().split()]
a=list(int(j) for j in input().split())
b=[]
c=0
ans=[0]*n
for i in range(n):
if (a[n-1-i] in b):
ans[n-1-i]=int(c)
else:
c+=1
b.append(a[n-1-i])
ans[n-1-i]=int(c)
for i in range(m):
c=int(input())
print(ans[c-1]) | Title: Sereja and Suffixes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.?
Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements.
Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*).
Output Specification:
Print *m* lines — on the *i*-th line print the answer to the number *l**i*.
Demo Input:
['10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n']
Demo Output:
['6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n']
Note:
none | ```python
[n,m]=[int(j) for j in input().split()]
a=list(int(j) for j in input().split())
b=[]
c=0
ans=[0]*n
for i in range(n):
if (a[n-1-i] in b):
ans[n-1-i]=int(c)
else:
c+=1
b.append(a[n-1-i])
ans[n-1-i]=int(c)
for i in range(m):
c=int(input())
print(ans[c-1])
``` | 0 | |
496 | A | Minimum Difficulty | PROGRAMMING | 900 | [
"brute force",
"implementation",
"math"
] | null | null | Mike is trying rock climbing but he is awful at it.
There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=<<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.
Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.
Help Mike determine the minimum difficulty of the track after removing one hold. | The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds.
The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one). | Print a single number — the minimum difficulty of the track after removing a single hold. | [
"3\n1 4 6\n",
"5\n1 2 3 4 5\n",
"5\n1 2 3 7 8\n"
] | [
"5\n",
"2\n",
"4\n"
] | In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.
In the second test after removing every hold the difficulty equals 2.
In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4. | 500 | [
{
"input": "3\n1 4 6",
"output": "5"
},
{
"input": "5\n1 2 3 4 5",
"output": "2"
},
{
"input": "5\n1 2 3 7 8",
"output": "4"
},
{
"input": "3\n1 500 1000",
"output": "999"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "2"
},
{
"input": "10\n1 4 9 16 25 36 49 64 81 100",
"output": "19"
},
{
"input": "10\n300 315 325 338 350 365 379 391 404 416",
"output": "23"
},
{
"input": "15\n87 89 91 92 93 95 97 99 101 103 105 107 109 111 112",
"output": "2"
},
{
"input": "60\n3 5 7 8 15 16 18 21 24 26 40 41 43 47 48 49 50 51 52 54 55 60 62 71 74 84 85 89 91 96 406 407 409 412 417 420 423 424 428 431 432 433 436 441 445 446 447 455 458 467 469 471 472 475 480 485 492 493 497 500",
"output": "310"
},
{
"input": "3\n159 282 405",
"output": "246"
},
{
"input": "81\n6 7 22 23 27 38 40 56 59 71 72 78 80 83 86 92 95 96 101 122 125 127 130 134 154 169 170 171 172 174 177 182 184 187 195 197 210 211 217 223 241 249 252 253 256 261 265 269 274 277 291 292 297 298 299 300 302 318 338 348 351 353 381 386 387 397 409 410 419 420 428 430 453 460 461 473 478 493 494 500 741",
"output": "241"
},
{
"input": "10\n218 300 388 448 535 629 680 740 836 925",
"output": "111"
},
{
"input": "100\n6 16 26 36 46 56 66 76 86 96 106 116 126 136 146 156 166 176 186 196 206 216 226 236 246 256 266 276 286 296 306 316 326 336 346 356 366 376 386 396 406 416 426 436 446 456 466 476 486 496 506 516 526 536 546 556 566 576 586 596 606 616 626 636 646 656 666 676 686 696 706 716 726 736 746 756 766 776 786 796 806 816 826 836 846 856 866 876 886 896 906 916 926 936 946 956 966 976 986 996",
"output": "20"
},
{
"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 50 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000",
"output": "901"
},
{
"input": "100\n1 9 15 17 28 29 30 31 32 46 48 49 52 56 62 77 82 85 90 91 94 101 102 109 111 113 116 118 124 125 131 132 136 138 139 143 145 158 161 162 165 167 171 173 175 177 179 183 189 196 801 802 804 806 817 819 827 830 837 840 842 846 850 855 858 862 863 866 869 870 878 881 883 884 896 898 899 901 904 906 908 909 910 911 912 917 923 924 925 935 939 943 945 956 963 964 965 972 976 978",
"output": "605"
},
{
"input": "100\n2 43 47 49 50 57 59 67 74 98 901 903 904 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 938 939 940 942 943 944 945 946 947 948 949 950 952 953 954 956 957 958 959 960 961 962 963 965 966 967 968 969 970 971 972 973 974 975 976 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 998 999",
"output": "803"
},
{
"input": "72\n178 186 196 209 217 226 236 248 260 273 281 291 300 309 322 331 343 357 366 377 389 399 409 419 429 442 450 459 469 477 491 501 512 524 534 548 557 568 582 593 602 616 630 643 652 660 670 679 693 707 715 728 737 750 759 768 776 789 797 807 815 827 837 849 863 873 881 890 901 910 920 932",
"output": "17"
},
{
"input": "38\n1 28 55 82 109 136 163 190 217 244 271 298 325 352 379 406 433 460 487 514 541 568 595 622 649 676 703 730 757 784 811 838 865 892 919 946 973 1000",
"output": "54"
},
{
"input": "28\n1 38 75 112 149 186 223 260 297 334 371 408 445 482 519 556 593 630 667 704 741 778 815 852 889 926 963 1000",
"output": "74"
}
] | 1,633,493,453 | 2,147,483,647 | PyPy 3 | OK | TESTS | 19 | 93 | 20,172,800 | numnum = int(input())
nums = [int(x) for x in input().split()]
oned = nums[1] - nums[0]
twod = nums[2] - nums[0]
for i in range(2, numnum):
oned = max(oned, nums[i] - nums[i - 1])
twod = min(twod, nums[i] - nums[i - 2])
print(max(oned, twod)) | Title: Minimum Difficulty
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike is trying rock climbing but he is awful at it.
There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=<<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.
Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.
Help Mike determine the minimum difficulty of the track after removing one hold.
Input Specification:
The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds.
The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one).
Output Specification:
Print a single number — the minimum difficulty of the track after removing a single hold.
Demo Input:
['3\n1 4 6\n', '5\n1 2 3 4 5\n', '5\n1 2 3 7 8\n']
Demo Output:
['5\n', '2\n', '4\n']
Note:
In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.
In the second test after removing every hold the difficulty equals 2.
In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4. | ```python
numnum = int(input())
nums = [int(x) for x in input().split()]
oned = nums[1] - nums[0]
twod = nums[2] - nums[0]
for i in range(2, numnum):
oned = max(oned, nums[i] - nums[i - 1])
twod = min(twod, nums[i] - nums[i - 2])
print(max(oned, twod))
``` | 3 | |
615 | A | Bulbs | PROGRAMMING | 800 | [
"implementation"
] | null | null | Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on. | The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively.
Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs. | If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO". | [
"3 4\n2 1 4\n3 1 3 1\n1 2\n",
"3 3\n1 1\n1 2\n1 1\n"
] | [
"YES\n",
"NO\n"
] | In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | 500 | [
{
"input": "3 4\n2 1 4\n3 1 3 1\n1 2",
"output": "YES"
},
{
"input": "3 3\n1 1\n1 2\n1 1",
"output": "NO"
},
{
"input": "3 4\n1 1\n1 2\n1 3",
"output": "NO"
},
{
"input": "1 5\n5 1 2 3 4 5",
"output": "YES"
},
{
"input": "1 5\n5 4 4 1 2 3",
"output": "NO"
},
{
"input": "1 5\n5 1 1 1 1 5",
"output": "NO"
},
{
"input": "2 5\n4 3 1 4 2\n4 2 3 4 5",
"output": "YES"
},
{
"input": "5 7\n2 6 7\n5 1 1 1 1 1\n3 6 5 4\n0\n4 4 3 2 1",
"output": "YES"
},
{
"input": "100 100\n0\n0\n0\n1 53\n0\n0\n1 34\n1 54\n0\n1 14\n0\n1 33\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 82\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n1 26\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n0\n0\n0\n1 3\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 40\n0\n0\n0\n1 26\n0\n0\n0\n0\n0\n1 97\n0\n1 5\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "100 100\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "5 6\n3 1 2 6\n3 1 2 6\n1 1\n2 3 4\n3 1 5 6",
"output": "YES"
},
{
"input": "5 2\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "NO"
},
{
"input": "1 4\n3 1 2 3",
"output": "NO"
},
{
"input": "1 4\n3 2 3 4",
"output": "NO"
},
{
"input": "2 4\n3 2 3 4\n1 1",
"output": "YES"
},
{
"input": "2 4\n3 1 2 3\n1 4",
"output": "YES"
},
{
"input": "5 1\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "1 1\n0",
"output": "NO"
},
{
"input": "1 10\n10 1 2 3 4 5 6 7 8 9 10",
"output": "YES"
},
{
"input": "1 1\n1 1",
"output": "YES"
},
{
"input": "1 100\n99 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",
"output": "NO"
},
{
"input": "1 3\n3 1 2 1",
"output": "NO"
},
{
"input": "1 100\n100 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",
"output": "YES"
}
] | 1,665,477,709 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 46 | 0 | a,b = [int(x) for x in input().split()]
list1 = [0]*b
for i in range(a):
list2 = [int(x) for x in input().split()]
for j in list2[1:]:
list1[j-1] += 1
if list1.count(0)==0:
print("YES")
else:
print("NO") | Title: Bulbs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input Specification:
The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively.
Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs.
Output Specification:
If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO".
Demo Input:
['3 4\n2 1 4\n3 1 3 1\n1 2\n', '3 3\n1 1\n1 2\n1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | ```python
a,b = [int(x) for x in input().split()]
list1 = [0]*b
for i in range(a):
list2 = [int(x) for x in input().split()]
for j in list2[1:]:
list1[j-1] += 1
if list1.count(0)==0:
print("YES")
else:
print("NO")
``` | 3 | |
235 | A | LCM Challenge | PROGRAMMING | 1,600 | [
"number theory"
] | null | null | Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than *n*. Can you help me to find the maximum possible least common multiple of these three integers? | The first line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the *n* mentioned in the statement. | Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than *n*. | [
"9\n",
"7\n"
] | [
"504\n",
"210\n"
] | The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | 500 | [
{
"input": "9",
"output": "504"
},
{
"input": "7",
"output": "210"
},
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "60"
},
{
"input": "6",
"output": "60"
},
{
"input": "33",
"output": "32736"
},
{
"input": "21",
"output": "7980"
},
{
"input": "2",
"output": "2"
},
{
"input": "41",
"output": "63960"
},
{
"input": "29",
"output": "21924"
},
{
"input": "117",
"output": "1560780"
},
{
"input": "149",
"output": "3241644"
},
{
"input": "733",
"output": "392222436"
},
{
"input": "925",
"output": "788888100"
},
{
"input": "509",
"output": "131096004"
},
{
"input": "829",
"output": "567662724"
},
{
"input": "117",
"output": "1560780"
},
{
"input": "605",
"output": "220348260"
},
{
"input": "245",
"output": "14526540"
},
{
"input": "925",
"output": "788888100"
},
{
"input": "213",
"output": "9527916"
},
{
"input": "53",
"output": "140556"
},
{
"input": "341",
"output": "39303660"
},
{
"input": "21",
"output": "7980"
},
{
"input": "605",
"output": "220348260"
},
{
"input": "149",
"output": "3241644"
},
{
"input": "733",
"output": "392222436"
},
{
"input": "117",
"output": "1560780"
},
{
"input": "53",
"output": "140556"
},
{
"input": "245",
"output": "14526540"
},
{
"input": "829",
"output": "567662724"
},
{
"input": "924",
"output": "783776526"
},
{
"input": "508",
"output": "130065780"
},
{
"input": "700",
"output": "341042100"
},
{
"input": "636",
"output": "254839470"
},
{
"input": "20",
"output": "6460"
},
{
"input": "604",
"output": "218891412"
},
{
"input": "796",
"output": "501826260"
},
{
"input": "732",
"output": "389016270"
},
{
"input": "412",
"output": "69256788"
},
{
"input": "700",
"output": "341042100"
},
{
"input": "244",
"output": "14289372"
},
{
"input": "828",
"output": "563559150"
},
{
"input": "508",
"output": "130065780"
},
{
"input": "796",
"output": "501826260"
},
{
"input": "636",
"output": "254839470"
},
{
"input": "924",
"output": "783776526"
},
{
"input": "245",
"output": "14526540"
},
{
"input": "828",
"output": "563559150"
},
{
"input": "21",
"output": "7980"
},
{
"input": "605",
"output": "220348260"
},
{
"input": "636",
"output": "254839470"
},
{
"input": "924",
"output": "783776526"
},
{
"input": "116",
"output": "1507420"
},
{
"input": "700",
"output": "341042100"
},
{
"input": "732",
"output": "389016270"
},
{
"input": "20",
"output": "6460"
},
{
"input": "508",
"output": "130065780"
},
{
"input": "148",
"output": "3154620"
},
{
"input": "828",
"output": "563559150"
},
{
"input": "763116",
"output": "444394078546562430"
},
{
"input": "756604",
"output": "433115377058855412"
},
{
"input": "447244",
"output": "89460162932862372"
},
{
"input": "372636",
"output": "51742503205363470"
},
{
"input": "546924",
"output": "163597318076822526"
},
{
"input": "540412",
"output": "157823524476316788"
},
{
"input": "714700",
"output": "365063922340784100"
},
{
"input": "520731",
"output": "141201007712496270"
},
{
"input": "695019",
"output": "335728459024850814"
},
{
"input": "688507",
"output": "326379736779169710"
},
{
"input": "862795",
"output": "642275489615199390"
},
{
"input": "668827",
"output": "299184742915995150"
},
{
"input": "810411",
"output": "532248411551110590"
},
{
"input": "836603",
"output": "585540171302562606"
},
{
"input": "978187",
"output": "935975171582120670"
},
{
"input": "816923",
"output": "545182335484592526"
},
{
"input": "958507",
"output": "880611813728059710"
},
{
"input": "984699",
"output": "954792870629291694"
},
{
"input": "642635",
"output": "265393998349453470"
},
{
"input": "296604",
"output": "26092892528622606"
},
{
"input": "1000000",
"output": "999996000003000000"
},
{
"input": "8",
"output": "280"
},
{
"input": "3",
"output": "6"
},
{
"input": "4",
"output": "12"
},
{
"input": "30",
"output": "21924"
},
{
"input": "18",
"output": "4080"
},
{
"input": "12",
"output": "990"
}
] | 1,632,393,674 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 186 | 20,172,800 | from math import gcd
n = int(input())
arr = [n, n-1, n-2]
res = arr[0]
for i in range(1, 3):
res = (res*arr[i])//gcd(res, arr[i])
print(res) | Title: LCM Challenge
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than *n*. Can you help me to find the maximum possible least common multiple of these three integers?
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the *n* mentioned in the statement.
Output Specification:
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than *n*.
Demo Input:
['9\n', '7\n']
Demo Output:
['504\n', '210\n']
Note:
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | ```python
from math import gcd
n = int(input())
arr = [n, n-1, n-2]
res = arr[0]
for i in range(1, 3):
res = (res*arr[i])//gcd(res, arr[i])
print(res)
``` | 0 | |
767 | A | Snacktower | PROGRAMMING | 1,100 | [
"data structures",
"implementation"
] | null | null | According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*. | Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty. | [
"3\n3 1 2\n",
"5\n4 5 1 2 3\n"
] | [
"3\n \n2 1",
"5 4\n \n \n3 2 1\n"
] | In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before. | 500 | [
{
"input": "3\n3 1 2",
"output": "3 \n\n2 1 "
},
{
"input": "5\n4 5 1 2 3",
"output": "5 4 \n\n\n3 2 1 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n1 2",
"output": "2 1 "
},
{
"input": "10\n5 1 6 2 8 3 4 10 9 7",
"output": "10 \n9 8 \n7 6 5 4 3 2 1 "
},
{
"input": "30\n16 10 4 29 5 28 12 21 11 30 18 6 14 3 17 22 20 15 9 1 27 19 24 26 13 25 2 23 8 7",
"output": "30 29 28 \n\n\n\n\n\n\n\n\n\n\n27 \n\n\n26 \n\n25 24 \n\n23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 \n8 \n7 6 5 4 3 2 1 "
},
{
"input": "100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 4 80 72 39",
"output": "100 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n99 98 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 \n\n\n\n\n\n\n\n\n80 79 78 77 76 75 74 73 \n72 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 \n39 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 "
},
{
"input": "2\n2 1",
"output": "2 \n1 "
}
] | 1,683,032,803 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 2,000 | 13,516,800 | n = int(input())
arr = list(map(int,input().split()))
pack=[]
for i in range(n):
pack.append(arr[i])
while n in pack:
print(n,end=" ")
n-=1
else:
print("") | Title: Snacktower
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
Output Specification:
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
Demo Input:
['3\n3 1 2\n', '5\n4 5 1 2 3\n']
Demo Output:
['3\n\xa0\n2 1', '5 4\n\xa0\n\xa0\n3 2 1\n']
Note:
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before. | ```python
n = int(input())
arr = list(map(int,input().split()))
pack=[]
for i in range(n):
pack.append(arr[i])
while n in pack:
print(n,end=" ")
n-=1
else:
print("")
``` | 0 | |
599 | A | Patrick and Shopping | PROGRAMMING | 800 | [
"implementation"
] | null | null | Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled. | The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops. | Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house. | [
"10 20 30\n",
"1 1 5\n"
] | [
"60\n",
"4\n"
] | The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. | 500 | [
{
"input": "10 20 30",
"output": "60"
},
{
"input": "1 1 5",
"output": "4"
},
{
"input": "100 33 34",
"output": "134"
},
{
"input": "777 777 777",
"output": "2331"
},
{
"input": "2 2 8",
"output": "8"
},
{
"input": "12 34 56",
"output": "92"
},
{
"input": "789 101112 131415",
"output": "203802"
},
{
"input": "27485716 99999999 35182",
"output": "55041796"
},
{
"input": "1 293548 5",
"output": "12"
},
{
"input": "12059 259855 5874875",
"output": "543828"
},
{
"input": "46981 105809 585858",
"output": "305580"
},
{
"input": "9889 1221 2442",
"output": "7326"
},
{
"input": "100500 200600 300700",
"output": "601800"
},
{
"input": "318476 318476 318476",
"output": "955428"
},
{
"input": "23985 3353 75633",
"output": "54676"
},
{
"input": "120 1298 2222",
"output": "2836"
},
{
"input": "98437 23487 666672",
"output": "243848"
},
{
"input": "100000000 100000000 100000000",
"output": "300000000"
},
{
"input": "2 5 2",
"output": "8"
},
{
"input": "1 1000 1",
"output": "4"
},
{
"input": "1 100000000 1",
"output": "4"
}
] | 1,598,972,699 | 2,147,483,647 | PyPy 3 | OK | TESTS | 21 | 140 | 0 | a,b,c=map(int,input().split())
m=2*a+2*b
v=a+b+c
g=2*a+2*c
h=2*b+2*c
print(min(m,v,g,h)) | Title: Patrick and Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
Input Specification:
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
Output Specification:
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
Demo Input:
['10 20 30\n', '1 1 5\n']
Demo Output:
['60\n', '4\n']
Note:
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. | ```python
a,b,c=map(int,input().split())
m=2*a+2*b
v=a+b+c
g=2*a+2*c
h=2*b+2*c
print(min(m,v,g,h))
``` | 3 | |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,662,112,968 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | m,n,a=map(int,input().split())
if n%a == 0:
b = 0
else:
b = 1
if m%a == 0:
c = 0
else:
c = 1
t = (n/a)+b
t1 = (m/a)+c
print(t*t1) | Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
m,n,a=map(int,input().split())
if n%a == 0:
b = 0
else:
b = 1
if m%a == 0:
c = 0
else:
c = 1
t = (n/a)+b
t1 = (m/a)+c
print(t*t1)
``` | 0 |
742 | A | Arpa’s hard exam and Mehrdad’s naive cheat | PROGRAMMING | 1,000 | [
"implementation",
"math",
"number theory"
] | null | null | There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given *n*, print the last digit of 1378*n*.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat. | The single line of input contains one integer *n* (0<=<=≤<=<=*n*<=<=≤<=<=109). | Print single integer — the last digit of 1378*n*. | [
"1\n",
"2\n"
] | [
"8",
"4"
] | In the first example, last digit of 1378<sup class="upper-index">1</sup> = 1378 is 8.
In the second example, last digit of 1378<sup class="upper-index">2</sup> = 1378·1378 = 1898884 is 4. | 500 | [
{
"input": "1",
"output": "8"
},
{
"input": "2",
"output": "4"
},
{
"input": "1000",
"output": "6"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "6"
},
{
"input": "1000000000",
"output": "6"
},
{
"input": "5",
"output": "8"
},
{
"input": "6",
"output": "4"
},
{
"input": "999999999",
"output": "2"
},
{
"input": "1378",
"output": "4"
},
{
"input": "13781378",
"output": "4"
},
{
"input": "51202278",
"output": "4"
},
{
"input": "999999998",
"output": "4"
},
{
"input": "999999997",
"output": "8"
},
{
"input": "12193721",
"output": "8"
},
{
"input": "0",
"output": "1"
},
{
"input": "989898989",
"output": "8"
},
{
"input": "7",
"output": "2"
},
{
"input": "8",
"output": "6"
},
{
"input": "9",
"output": "8"
},
{
"input": "10",
"output": "4"
},
{
"input": "11",
"output": "2"
},
{
"input": "12",
"output": "6"
},
{
"input": "13",
"output": "8"
},
{
"input": "14",
"output": "4"
},
{
"input": "15",
"output": "2"
},
{
"input": "16",
"output": "6"
},
{
"input": "999999996",
"output": "6"
},
{
"input": "999999995",
"output": "2"
},
{
"input": "999999994",
"output": "4"
}
] | 1,694,875,404 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 15 | 46 | 0 | import sys
import math
for line in sys.stdin:
numbers = [int(x) for x in line.strip().split()]
n= numbers[0]
if (n%4)==0:
print(6)
elif (n%4==1):
print(8)
elif (n%4==2):
print(4)
elif (n%4==3):
print(2) | Title: Arpa’s hard exam and Mehrdad’s naive cheat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given *n*, print the last digit of 1378*n*.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.
Input Specification:
The single line of input contains one integer *n* (0<=<=≤<=<=*n*<=<=≤<=<=109).
Output Specification:
Print single integer — the last digit of 1378*n*.
Demo Input:
['1\n', '2\n']
Demo Output:
['8', '4']
Note:
In the first example, last digit of 1378<sup class="upper-index">1</sup> = 1378 is 8.
In the second example, last digit of 1378<sup class="upper-index">2</sup> = 1378·1378 = 1898884 is 4. | ```python
import sys
import math
for line in sys.stdin:
numbers = [int(x) for x in line.strip().split()]
n= numbers[0]
if (n%4)==0:
print(6)
elif (n%4==1):
print(8)
elif (n%4==2):
print(4)
elif (n%4==3):
print(2)
``` | 0 | |
984 | A | Game | PROGRAMMING | 800 | [
"sortings"
] | null | null | Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves. | The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$). | Print one number that will be left on the board. | [
"3\n2 1 3\n",
"3\n2 2 2\n"
] | [
"2",
"2"
] | In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players. | 500 | [
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 2 2",
"output": "2"
},
{
"input": "9\n44 53 51 80 5 27 74 79 94",
"output": "53"
},
{
"input": "10\n38 82 23 37 96 4 81 60 67 86",
"output": "60"
},
{
"input": "10\n58 26 77 15 53 81 68 48 22 65",
"output": "53"
},
{
"input": "1\n124",
"output": "124"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 1 1000",
"output": "1"
},
{
"input": "2\n322 322",
"output": "322"
},
{
"input": "3\n9 92 12",
"output": "12"
},
{
"input": "3\n1 2 2",
"output": "2"
}
] | 1,560,800,018 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 124 | 0 | a=int(input())
b=input().split()
if a%2==1:
for i in range(a//2):
b.remove(max(b))
b.remove(min(b))
else:
for i in range(a//2-1):
b.remove(max(b))
b.remove(min(b))
b.remove(max(b))
for i in range(len(b)):
print(b[i])
| Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
Input Specification:
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
Output Specification:
Print one number that will be left on the board.
Demo Input:
['3\n2 1 3\n', '3\n2 2 2\n']
Demo Output:
['2', '2']
Note:
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players. | ```python
a=int(input())
b=input().split()
if a%2==1:
for i in range(a//2):
b.remove(max(b))
b.remove(min(b))
else:
for i in range(a//2-1):
b.remove(max(b))
b.remove(min(b))
b.remove(max(b))
for i in range(len(b)):
print(b[i])
``` | 0 | |
859 | A | Declined Finalists | PROGRAMMING | 800 | [
"greedy",
"implementation"
] | null | null | This year, as in previous years, MemSQL is inviting the top 25 competitors from the Start[c]up qualification round to compete onsite for the final round. Not everyone who is eligible to compete onsite can afford to travel to the office, though. Initially the top 25 contestants are invited to come onsite. Each eligible contestant must either accept or decline the invitation. Whenever a contestant declines, the highest ranked contestant not yet invited is invited to take the place of the one that declined. This continues until 25 contestants have accepted invitations.
After the qualifying round completes, you know *K* of the onsite finalists, as well as their qualifying ranks (which start at 1, there are no ties). Determine the minimum possible number of contestants that declined the invitation to compete onsite in the final round. | The first line of input contains *K* (1<=≤<=*K*<=≤<=25), the number of onsite finalists you know. The second line of input contains *r*1,<=*r*2,<=...,<=*r**K* (1<=≤<=*r**i*<=≤<=106), the qualifying ranks of the finalists you know. All these ranks are distinct. | Print the minimum possible number of contestants that declined the invitation to compete onsite. | [
"25\n2 3 4 5 6 7 8 9 10 11 12 14 15 16 17 18 19 20 21 22 23 24 25 26 28\n",
"5\n16 23 8 15 4\n",
"3\n14 15 92\n"
] | [
"3\n",
"0\n",
"67\n"
] | In the first example, you know all 25 onsite finalists. The contestants who ranked 1-st, 13-th, and 27-th must have declined, so the answer is 3. | 500 | [
{
"input": "25\n2 3 4 5 6 7 8 9 10 11 12 14 15 16 17 18 19 20 21 22 23 24 25 26 28",
"output": "3"
},
{
"input": "5\n16 23 8 15 4",
"output": "0"
},
{
"input": "3\n14 15 92",
"output": "67"
},
{
"input": "1\n1000000",
"output": "999975"
},
{
"input": "25\n1000000 999999 999998 999997 999996 999995 999994 999993 999992 999991 999990 999989 999988 999987 999986 999985 999984 999983 999982 999981 999980 999979 999978 999977 999976",
"output": "999975"
},
{
"input": "25\n13 15 24 2 21 18 9 4 16 6 10 25 20 11 23 17 8 3 1 12 5 19 22 14 7",
"output": "0"
},
{
"input": "10\n17 11 7 13 18 12 14 5 16 2",
"output": "0"
},
{
"input": "22\n22 14 23 20 11 21 4 12 3 8 7 9 19 10 13 17 15 1 5 18 16 2",
"output": "0"
},
{
"input": "21\n6 21 24 3 10 23 14 2 26 12 8 1 15 13 9 5 19 20 4 16 22",
"output": "1"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "2\n100 60",
"output": "75"
},
{
"input": "4\n999 581 787 236",
"output": "974"
},
{
"input": "6\n198 397 732 1234 309 827",
"output": "1209"
},
{
"input": "11\n6494 3961 1858 4351 8056 780 7720 6211 1961 8192 3621",
"output": "8167"
},
{
"input": "14\n18809 9534 11652 6493 8929 9370 4125 23888 16403 3559 23649 19243 14289 17852",
"output": "23863"
},
{
"input": "18\n24939 35558 47058 70307 26221 12866 3453 40422 47557 36322 40698 64060 10825 77777 48645 26124 4859 64222",
"output": "77752"
},
{
"input": "24\n633483 654321 122445 481150 347578 37803 525083 151084 211073 358699 339420 452023 219553 119727 74852 66750 371279 405099 618894 649977 235337 607819 81649 649804",
"output": "654296"
},
{
"input": "25\n58115 794098 753382 484882 238434 674285 690118 858677 196185 173301 349729 918792 600745 636016 122678 366783 137179 377098 917081 369620 449039 379412 503678 1000000 292099",
"output": "999975"
},
{
"input": "2\n26 27",
"output": "2"
},
{
"input": "3\n40 30 35",
"output": "15"
},
{
"input": "2\n46 45",
"output": "21"
},
{
"input": "3\n1 25 90",
"output": "65"
},
{
"input": "5\n14 15 16 30 92",
"output": "67"
},
{
"input": "2\n1000 1001",
"output": "976"
},
{
"input": "25\n3 4 5 6 7 8 9 10 11 12 14 15 16 17 18 19 20 21 22 23 24 25 26 28 2",
"output": "3"
}
] | 1,505,632,550 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 62 | 0 | n = int(input())
L = input().split()
mx = 0
for i in range(n):
L[i] = int(L[i])
mx = max(mx, L[i])
print(max(0, mx - 25)) | Title: Declined Finalists
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This year, as in previous years, MemSQL is inviting the top 25 competitors from the Start[c]up qualification round to compete onsite for the final round. Not everyone who is eligible to compete onsite can afford to travel to the office, though. Initially the top 25 contestants are invited to come onsite. Each eligible contestant must either accept or decline the invitation. Whenever a contestant declines, the highest ranked contestant not yet invited is invited to take the place of the one that declined. This continues until 25 contestants have accepted invitations.
After the qualifying round completes, you know *K* of the onsite finalists, as well as their qualifying ranks (which start at 1, there are no ties). Determine the minimum possible number of contestants that declined the invitation to compete onsite in the final round.
Input Specification:
The first line of input contains *K* (1<=≤<=*K*<=≤<=25), the number of onsite finalists you know. The second line of input contains *r*1,<=*r*2,<=...,<=*r**K* (1<=≤<=*r**i*<=≤<=106), the qualifying ranks of the finalists you know. All these ranks are distinct.
Output Specification:
Print the minimum possible number of contestants that declined the invitation to compete onsite.
Demo Input:
['25\n2 3 4 5 6 7 8 9 10 11 12 14 15 16 17 18 19 20 21 22 23 24 25 26 28\n', '5\n16 23 8 15 4\n', '3\n14 15 92\n']
Demo Output:
['3\n', '0\n', '67\n']
Note:
In the first example, you know all 25 onsite finalists. The contestants who ranked 1-st, 13-th, and 27-th must have declined, so the answer is 3. | ```python
n = int(input())
L = input().split()
mx = 0
for i in range(n):
L[i] = int(L[i])
mx = max(mx, L[i])
print(max(0, mx - 25))
``` | 3 |
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