evanellis/Codeforces-Python-Submissions_correct

432

A

Choosing Teams

PROGRAMMING

800

[ "greedy", "implementation", "sortings" ]

null

The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times. The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?

The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.

Print a single number — the answer to the problem.

[ "5 2\n0 4 5 1 0\n", "6 4\n0 1 2 3 4 5\n", "6 5\n0 0 0 0 0 0\n" ]

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

In the first sample only one team could be made: the first, the fourth and the fifth participants. In the second sample no teams could be created. In the third sample two teams could be created. Any partition into two teams fits.

500

[ { "input": "5 2\n0 4 5 1 0", "output": "1" }, { "input": "6 4\n0 1 2 3 4 5", "output": "0" }, { "input": "6 5\n0 0 0 0 0 0", "output": "2" }, { "input": "3 4\n0 1 0", "output": "1" }, { "input": "3 4\n0 2 0", "output": "0" }, { "input": "6 5\n0 0 0 0 0 0", "output": "2" }, { "input": "12 2\n0 1 2 3 4 0 1 2 3 4 0 1", "output": "3" }, { "input": "15 2\n0 1 2 3 4 0 1 2 3 4 0 1 2 3 4", "output": "4" }, { "input": "13 1\n5 0 5 0 1 5 0 4 1 1 1 4 1", "output": "3" }, { "input": "20 1\n5 0 4 2 2 3 2 1 2 4 1 3 5 5 5 4 4 1 3 0", "output": "5" }, { "input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4", "output": "4" }, { "input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4", "output": "4" }, { "input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4", "output": "4" }, { "input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4", "output": "4" }, { "input": "6 3\n4 4 4 4 4 4", "output": "0" }, { "input": "1 4\n1", "output": "0" }, { "input": "1 1\n0", "output": "0" }, { "input": "2 3\n0 0", "output": "0" } ]

1,673,721,814

2,147,483,647

Python 3

OK

TESTS

35

46

0

x,y = map(int,input().split()) z = list(map(int,input().split())) n = 5-y c = 0 for i in z: if i <= n: c+=1 print(c//3)

Title: Choosing Teams Time Limit: None seconds Memory Limit: None megabytes Problem Description: The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times. The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times? Input Specification: The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship. Output Specification: Print a single number — the answer to the problem. Demo Input: ['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n'] Demo Output: ['1\n', '0\n', '2\n'] Note: In the first sample only one team could be made: the first, the fourth and the fifth participants. In the second sample no teams could be created. In the third sample two teams could be created. Any partition into two teams fits.

```python x,y = map(int,input().split()) z = list(map(int,input().split())) n = 5-y c = 0 for i in z: if i <= n: c+=1 print(c//3) ```

3

237

A

Free Cash

PROGRAMMING

1,000

[ "implementation" ]

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" } ]

1,697,522,828

2,147,483,647

Python 3

OK

TESTS

67

404

8,601,600

l = [] T = int(input()) for i in range(0, T): a, b = map(int, input().split()) l.append((a, b)) d = {} for item in l: if item not in d: d[item] = 1 else: d[item] += 1 # print(d) mx = 1 for key, value in d.items(): # print(key, value) mx = max(mx, value) print(mx)

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 l = [] T = int(input()) for i in range(0, T): a, b = map(int, input().split()) l.append((a, b)) d = {} for item in l: if item not in d: d[item] = 1 else: d[item] += 1 # print(d) mx = 1 for key, value in d.items(): # print(key, value) mx = max(mx, value) print(mx) ```

3

200

B

Drinks

PROGRAMMING

800

[ "implementation", "math" ]

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" } ]

1,686,982,773

2,147,483,647

Python 3

OK

TESTS

31

92

0

n = int(input()) s = input().split() c = 0 for j in range(n): c+=(int(s[j])/100) print ((c/n)*100)

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()) s = input().split() c = 0 for j in range(n): c+=(int(s[j])/100) print ((c/n)*100) ```

3

959

A

Mahmoud and Ehab and the even-odd game

PROGRAMMING

800

[ "games", "math" ]

null

Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that: - 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd. If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?

The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game.

Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.

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

[ "Ehab", "Mahmoud" ]

In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins. In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.

500

[ { "input": "1", "output": "Ehab" }, { "input": "2", "output": "Mahmoud" }, { "input": "10000", "output": "Mahmoud" }, { "input": "33333", "output": "Ehab" }, { "input": "5", "output": "Ehab" }, { "input": "1000000000", "output": "Mahmoud" }, { "input": "999999999", "output": "Ehab" }, { "input": "123123123", "output": "Ehab" }, { "input": "22222221", "output": "Ehab" }, { "input": "22222220", "output": "Mahmoud" }, { "input": "3", "output": "Ehab" }, { "input": "4", "output": "Mahmoud" }, { "input": "6", "output": "Mahmoud" }, { "input": "7", "output": "Ehab" }, { "input": "8", "output": "Mahmoud" }, { "input": "9", "output": "Ehab" }, { "input": "10", "output": "Mahmoud" }, { "input": "536870912", "output": "Mahmoud" }, { "input": "536870913", "output": "Ehab" }, { "input": "536870911", "output": "Ehab" } ]

1,677,521,382

2,147,483,647

PyPy 3-64

OK

TESTS

20

46

0

# k=int(input()) # for i in range(k): e=int(input()) if e%2!=0 or e==0 : print ("Ehab") else: print("Mahmoud")

Title: Mahmoud and Ehab and the even-odd game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that: - 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd. If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally? Input Specification: The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game. Output Specification: Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise. Demo Input: ['1\n', '2\n'] Demo Output: ['Ehab', 'Mahmoud'] Note: In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins. In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.

```python # k=int(input()) # for i in range(k): e=int(input()) if e%2!=0 or e==0 : print ("Ehab") else: print("Mahmoud") ```

3

990

B

Micro-World

PROGRAMMING

1,200

[ "greedy", "sortings" ]

null

You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them. You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$. The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i > a_j$ and $a_i \le a_j + K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i > a_j$ and $a_i \le a_j + K$. The swallow operations go one after another. For example, the sequence of bacteria sizes $a=[101, 53, 42, 102, 101, 55, 54]$ and $K=1$. The one of possible sequences of swallows is: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$. In total there are $3$ bacteria remained in the Petri dish. Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.

The first line contains two space separated positive integers $n$ and $K$ ($1 \le n \le 2 \cdot 10^5$, $1 \le K \le 10^6$) — number of bacteria and intergalactic constant $K$. The second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — sizes of bacteria you have.

Print the only integer — minimal possible number of bacteria can remain.

[ "7 1\n101 53 42 102 101 55 54\n", "6 5\n20 15 10 15 20 25\n", "7 1000000\n1 1 1 1 1 1 1\n" ]

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

The first example is clarified in the problem statement. In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, \underline{20}, 25]$ $\to$ $[20, 15, 10, \underline{15}, 25]$ $\to$ $[20, 15, \underline{10}, 25]$ $\to$ $[20, \underline{15}, 25]$ $\to$ $[\underline{20}, 25]$ $\to$ $[25]$. In the third example no bacteria can swallow any other bacteria.

0

[ { "input": "7 1\n101 53 42 102 101 55 54", "output": "3" }, { "input": "6 5\n20 15 10 15 20 25", "output": "1" }, { "input": "7 1000000\n1 1 1 1 1 1 1", "output": "7" }, { "input": "1 1\n1", "output": "1" }, { "input": "1 4\n8", "output": "1" }, { "input": "10 1\n1 2 3 5 6 8 10 11 9 4", "output": "2" }, { "input": "9 2\n1 6 1 5 5 8 6 8 7", "output": "4" }, { "input": "15 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "15" }, { "input": "2 1000000\n1 1000000", "output": "1" }, { "input": "7 2\n1 5 5 8 9 8 8", "output": "4" }, { "input": "10 1\n2 6 3 4 2 4 4 3 2 1", "output": "4" }, { "input": "4 1\n2 2 1 1", "output": "2" }, { "input": "10 1\n6 3 1 3 6 4 1 3 6 4", "output": "7" }, { "input": "2 1\n1 1", "output": "2" }, { "input": "2 1\n1 2", "output": "1" }, { "input": "8 2\n3 13 9 8 3 13 9 14", "output": "5" }, { "input": "8 1000000\n1 1 5 1000000 1000000 2 2 2", "output": "2" }, { "input": "2 1\n999152 999153", "output": "1" } ]

1,528,818,226

2,147,483,647

Python 3

OK

TESTS

36

373

14,233,600

n, k = [int(a) for a in input().split()] s = sorted([int(a) for a in input().split()], reverse=1) ans = 1 q = s[0] for i in range(1, n): if not(abs(s[i] - s[i - 1]) <= k and s[i] < q): ans += 1 q = s[i] print(ans)

Title: Micro-World Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them. You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$. The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i > a_j$ and $a_i \le a_j + K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i > a_j$ and $a_i \le a_j + K$. The swallow operations go one after another. For example, the sequence of bacteria sizes $a=[101, 53, 42, 102, 101, 55, 54]$ and $K=1$. The one of possible sequences of swallows is: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$. In total there are $3$ bacteria remained in the Petri dish. Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope. Input Specification: The first line contains two space separated positive integers $n$ and $K$ ($1 \le n \le 2 \cdot 10^5$, $1 \le K \le 10^6$) — number of bacteria and intergalactic constant $K$. The second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — sizes of bacteria you have. Output Specification: Print the only integer — minimal possible number of bacteria can remain. Demo Input: ['7 1\n101 53 42 102 101 55 54\n', '6 5\n20 15 10 15 20 25\n', '7 1000000\n1 1 1 1 1 1 1\n'] Demo Output: ['3\n', '1\n', '7\n'] Note: The first example is clarified in the problem statement. In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, \underline{20}, 25]$ $\to$ $[20, 15, 10, \underline{15}, 25]$ $\to$ $[20, 15, \underline{10}, 25]$ $\to$ $[20, \underline{15}, 25]$ $\to$ $[\underline{20}, 25]$ $\to$ $[25]$. In the third example no bacteria can swallow any other bacteria.

```python n, k = [int(a) for a in input().split()] s = sorted([int(a) for a in input().split()], reverse=1) ans = 1 q = s[0] for i in range(1, n): if not(abs(s[i] - s[i - 1]) <= k and s[i] < q): ans += 1 q = s[i] print(ans) ```

3

18

C

Stripe

PROGRAMMING

1,200

[ "data structures", "implementation" ]

C. Stripe

2

64

Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?

The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.

Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.

[ "9\n1 5 -6 7 9 -16 0 -2 2\n", "3\n1 1 1\n", "2\n0 0\n" ]

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

none

0

[ { "input": "9\n1 5 -6 7 9 -16 0 -2 2", "output": "3" }, { "input": "3\n1 1 1", "output": "0" }, { "input": "2\n0 0", "output": "1" }, { "input": "4\n100 1 10 111", "output": "1" }, { "input": "10\n0 4 -3 0 -2 2 -3 -3 2 5", "output": "3" }, { "input": "10\n0 -1 2 2 -1 1 0 0 0 2", "output": "0" }, { "input": "10\n-1 -1 1 -1 0 1 0 1 1 1", "output": "1" }, { "input": "10\n0 0 0 0 0 0 0 0 0 0", "output": "9" }, { "input": "50\n-4 -3 3 4 -1 0 2 -4 -3 -4 1 4 3 0 4 1 0 -3 4 -3 -2 2 2 1 0 -4 -4 -5 3 2 -1 4 5 -3 -3 4 4 -5 2 -3 4 -5 2 5 -4 4 1 -2 -4 3", "output": "3" }, { "input": "15\n0 4 0 3 -1 4 -2 -2 -4 -4 3 2 4 -1 -3", "output": "0" }, { "input": "10\n3 -1 -3 -1 3 -2 0 3 1 -2", "output": "0" }, { "input": "100\n-4 2 4 4 1 3 -3 -3 2 1 -4 0 0 2 3 -1 -4 -3 4 -2 -3 -3 -3 -1 -2 -3 -1 -4 0 4 0 -1 4 0 -4 -4 4 -4 -2 1 -4 1 -3 -2 3 -4 4 0 -1 3 -1 4 -1 4 -1 3 -3 -3 -2 -2 4 -3 -3 4 -3 -2 -1 0 -2 4 0 -3 -1 -2 -3 1 -4 1 -3 -3 -3 -2 -3 0 1 -2 -2 -4 -3 -1 2 3 -1 1 1 0 3 -3 -1 -2", "output": "1" }, { "input": "100\n-2 -1 1 0 -2 -1 2 2 0 0 2 1 0 2 0 2 1 0 -1 -1 -1 0 -2 -1 2 -1 -2 2 -2 2 -2 -2 2 1 1 1 -2 2 0 0 2 -1 2 2 2 0 -1 -1 -1 1 -2 2 2 2 -2 0 0 -2 0 -2 -2 0 -1 -1 -2 -1 1 2 -2 -1 1 -2 -1 0 -2 2 2 -1 -2 2 0 0 0 1 0 1 2 1 -2 -2 -1 -1 -2 -2 -2 -1 2 2 2 -2", "output": "6" }, { "input": "100\n2 7 6 0 8 -2 0 1 8 -1 7 -4 -1 1 0 3 4 -4 3 7 8 -4 -6 6 4 -2 -5 7 4 6 1 6 3 8 -2 6 -6 -3 0 -1 -7 -8 -2 5 8 6 -2 -2 -5 -4 -7 -3 -5 -3 -3 -1 5 8 4 0 -7 -6 8 -4 -1 7 2 6 6 4 -5 -4 -5 -2 2 -2 -7 -1 5 -8 -6 -2 -5 4 2 8 -6 7 -8 -1 -5 8 6 -3 -1 7 -1 2 -8 -8", "output": "0" }, { "input": "1\n0", "output": "0" }, { "input": "1\n10000", "output": "0" }, { "input": "1\n-10000", "output": "0" } ]

1,528,064,580

2,147,483,647

Python 3

OK

TESTS

45

310

14,438,400

def main(): n = int(input()) arr = list(map(int, input().split())) right = sum(arr) left = 0 result = 0 for i in range(0, n - 1): left += arr[i] right -= arr[i] if left == right: result += 1 print(result) if __name__ == "__main__": main()

Title: Stripe Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem? Input Specification: The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value. Output Specification: Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only. Demo Input: ['9\n1 5 -6 7 9 -16 0 -2 2\n', '3\n1 1 1\n', '2\n0 0\n'] Demo Output: ['3\n', '0\n', '1\n'] Note: none

```python def main(): n = int(input()) arr = list(map(int, input().split())) right = sum(arr) left = 0 result = 0 for i in range(0, n - 1): left += arr[i] right -= arr[i] if left == right: result += 1 print(result) if __name__ == "__main__": main() ```

3.814926

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" } ]

1,677,529,194

2,147,483,647

Python 3

OK

TESTS

20

46

0

for i in range(0, int(input())): string = input() if len(string) > 10: answer = string[0] + str(len(string) - 2) + string[-1] print(answer) else: print(string)

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 for i in range(0, int(input())): string = input() if len(string) > 10: answer = string[0] + str(len(string) - 2) + string[-1] print(answer) else: print(string) ```

3.977

248

B

Chilly Willy

PROGRAMMING

1,400

[ "math", "number theory" ]

null

Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them. Chilly Willy wants to find the minimum number of length *n*, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that. A number's length is the number of digits in its decimal representation without leading zeros.

A single input line contains a single integer *n* (1<=≤<=*n*<=≤<=105).

Print a single integer — the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.

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

[ "-1\n", "10080" ]

none

1,000

[ { "input": "1", "output": "-1" }, { "input": "5", "output": "10080" }, { "input": "6", "output": "100170" }, { "input": "4", "output": "1050" }, { "input": "15", "output": "100000000000110" }, { "input": "16", "output": "1000000000000050" }, { "input": "17", "output": "10000000000000080" }, { "input": "7", "output": "1000020" }, { "input": "120", "output": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000170" }, { "input": "8", "output": "10000200" }, { "input": "3", "output": "210" }, { "input": "2", "output": "-1" }, { "input": "9", "output": "100000110" }, { "input": "10", "output": "1000000050" }, { "input": "11", "output": "10000000080" }, { "input": "12", "output": "100000000170" }, { "input": "13", "output": "1000000000020" }, { "input": "14", "output": "10000000000200" }, { "input": "100000", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99999", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99998", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99997", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99996", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99995", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99994", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99993", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99992", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99991", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99990", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99989", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99988", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99987", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99988", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99987", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "99986", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "10000", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "5000", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "5001", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "5002", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "121", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000020" }, { "input": "122", "output": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000200" }, { "input": "123", "output": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000110" }, { "input": "18", "output": "100000000000000170" }, { "input": "19", "output": "1000000000000000020" }, { "input": "20", "output": "10000000000000000200" }, { "input": "21", "output": "100000000000000000110" }, { "input": "22", "output": "1000000000000000000050" }, { "input": "23", "output": "10000000000000000000080" }, { "input": "24", "output": "100000000000000000000170" }, { "input": "25", "output": "1000000000000000000000020" }, { "input": "31", "output": "1000000000000000000000000000020" }, { "input": "33", "output": "100000000000000000000000000000110" }, { "input": "65", "output": "10000000000000000000000000000000000000000000000000000000000000080" }, { "input": "2345", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "5522", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "8824", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "9003", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "88888", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "77777", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "66666", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "55553", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "34532", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "27324", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "45332", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "1000", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "12398", "output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." } ]

1,578,129,578

2,147,483,647

PyPy 3

OK

TESTS

66

280

512,000

n = int(input()) if n < 3: print(-1) elif n == 3: print(210) else: suff = "" if n % 6 == 4: suff = "50" elif n % 6 == 5: suff = "80" elif n % 6 == 0: suff = "170" elif n % 6 == 1: suff = "20" elif n % 6 == 2: suff = "200" elif n % 6 == 3: suff = "110" main = n - 1 - len(suff) print("1" + "0" * main + suff)

Title: Chilly Willy Time Limit: None seconds Memory Limit: None megabytes Problem Description: Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them. Chilly Willy wants to find the minimum number of length *n*, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that. A number's length is the number of digits in its decimal representation without leading zeros. Input Specification: A single input line contains a single integer *n* (1<=≤<=*n*<=≤<=105). Output Specification: Print a single integer — the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist. Demo Input: ['1\n', '5\n'] Demo Output: ['-1\n', '10080'] Note: none

```python n = int(input()) if n < 3: print(-1) elif n == 3: print(210) else: suff = "" if n % 6 == 4: suff = "50" elif n % 6 == 5: suff = "80" elif n % 6 == 0: suff = "170" elif n % 6 == 1: suff = "20" elif n % 6 == 2: suff = "200" elif n % 6 == 3: suff = "110" main = n - 1 - len(suff) print("1" + "0" * main + suff) ```

3

55

B

Smallest number

PROGRAMMING

1,600

[ "brute force" ]

B. Smallest number

2

256

Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers *a*, *b*, *c*, *d* on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.

First line contains four integers separated by space: 0<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)

Output one integer number — the minimal result which can be obtained. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).

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

[ "3\n", "8\n", "9\n" ]

none

1,000

[ { "input": "1 1 1 1\n+ + *", "output": "3" }, { "input": "2 2 2 2\n* * +", "output": "8" }, { "input": "1 2 3 4\n* + +", "output": "9" }, { "input": "15 1 3 1\n* * +", "output": "18" }, { "input": "8 1 7 14\n+ + +", "output": "30" }, { "input": "7 17 3 25\n+ * +", "output": "63" }, { "input": "13 87 4 17\n* * *", "output": "76908" }, { "input": "7 0 8 15\n+ + *", "output": "0" }, { "input": "52 0 43 239\n+ + +", "output": "334" }, { "input": "1000 1000 999 1000\n* * *", "output": "999000000000" }, { "input": "720 903 589 804\n* * *", "output": "307887168960" }, { "input": "631 149 496 892\n* * +", "output": "445884" }, { "input": "220 127 597 394\n* + +", "output": "28931" }, { "input": "214 862 466 795\n+ + +", "output": "2337" }, { "input": "346 290 587 525\n* * *", "output": "30922279500" }, { "input": "323 771 559 347\n+ * *", "output": "149067730" }, { "input": "633 941 836 254\n* + +", "output": "162559" }, { "input": "735 111 769 553\n+ * *", "output": "92320032" }, { "input": "622 919 896 120\n* * +", "output": "667592" }, { "input": "652 651 142 661\n+ + +", "output": "2106" }, { "input": "450 457 975 35\n* * *", "output": "7017806250" }, { "input": "883 954 804 352\n* * +", "output": "1045740" }, { "input": "847 206 949 358\n* + *", "output": "62660050" }, { "input": "663 163 339 76\n+ + +", "output": "1241" }, { "input": "990 330 253 553\n+ * +", "output": "85033" }, { "input": "179 346 525 784\n* * *", "output": "25492034400" }, { "input": "780 418 829 778\n+ + *", "output": "997766" }, { "input": "573 598 791 124\n* * *", "output": "33608874936" }, { "input": "112 823 202 223\n* * +", "output": "137222" }, { "input": "901 166 994 315\n* + *", "output": "47278294" }, { "input": "393 342 840 486\n+ * *", "output": "178222356" }, { "input": "609 275 153 598\n+ + *", "output": "226746" }, { "input": "56 828 386 57\n+ * *", "output": "3875088" }, { "input": "944 398 288 986\n+ + *", "output": "670464" }, { "input": "544 177 162 21\n+ + *", "output": "18543" }, { "input": "105 238 316 265\n+ + +", "output": "924" }, { "input": "31 353 300 911\n* * *", "output": "2990721900" }, { "input": "46 378 310 194\n* * +", "output": "77528" }, { "input": "702 534 357 657\n+ * *", "output": "259077042" }, { "input": "492 596 219 470\n+ + *", "output": "341202" }, { "input": "482 842 982 902\n+ * +", "output": "407728" }, { "input": "827 578 394 351\n* * *", "output": "66105361764" }, { "input": "901 884 426 451\n* + *", "output": "170223210" }, { "input": "210 295 12 795\n* * +", "output": "71490" }, { "input": "40 734 948 202\n+ * *", "output": "13590560" }, { "input": "136 611 963 195\n+ + *", "output": "240584" }, { "input": "695 74 871 760\n+ * +", "output": "53061" }, { "input": "666 884 772 54\n* + +", "output": "37620" }, { "input": "975 785 753 224\n+ * +", "output": "170432" }, { "input": "35 187 126 596\n+ + +", "output": "944" }, { "input": "243 386 431 35\n* + *", "output": "3298015" }, { "input": "229 602 133 635\n* * +", "output": "222313" }, { "input": "916 207 238 891\n+ + *", "output": "423315" }, { "input": "922 145 883 357\n+ + *", "output": "313490" }, { "input": "69 355 762 111\n* + +", "output": "8776" }, { "input": "209 206 34 67\n* + *", "output": "476374" }, { "input": "693 824 375 361\n* * +", "output": "557339" }, { "input": "45 712 635 467\n* + +", "output": "22362" }, { "input": "426 283 179 211\n+ + +", "output": "1099" }, { "input": "802 387 686 12\n+ + +", "output": "1887" } ]

1,638,214,266

2,147,483,647

Python 3

OK

TESTS

60

124

0

class Solution: def find_Smallest(self,a,s): global smallest if len(a)==1: if smallest > a[0]: smallest = a[0] return for i in range(len(a)): for j in range(i+1,len(a)): tmp = [0] for k in range(len(a)): if k!=i and k!=j: tmp.append(a[k]) tmp[0] = a[i]*a[j] if s[0]=="*" else a[i]+a[j] self.find_Smallest(tmp,s[1:]) if __name__=="__main__": Sol = Solution() num = list(map(int,input().split())) signs = list(input().split()) smallest = float('inf') Sol.find_Smallest(num,signs) print(smallest)

Title: Smallest number Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers *a*, *b*, *c*, *d* on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations. Input Specification: First line contains four integers separated by space: 0<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication) Output Specification: Output one integer number — the minimal result which can be obtained. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d). Demo Input: ['1 1 1 1\n+ + *\n', '2 2 2 2\n* * +\n', '1 2 3 4\n* + +\n'] Demo Output: ['3\n', '8\n', '9\n'] Note: none

```python class Solution: def find_Smallest(self,a,s): global smallest if len(a)==1: if smallest > a[0]: smallest = a[0] return for i in range(len(a)): for j in range(i+1,len(a)): tmp = [0] for k in range(len(a)): if k!=i and k!=j: tmp.append(a[k]) tmp[0] = a[i]*a[j] if s[0]=="*" else a[i]+a[j] self.find_Smallest(tmp,s[1:]) if __name__=="__main__": Sol = Solution() num = list(map(int,input().split())) signs = list(input().split()) smallest = float('inf') Sol.find_Smallest(num,signs) print(smallest) ```

3.969

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" } ]

1,513,659,505

2,147,483,647

Python 3

OK

TESTS

32

124

5,734,400

n=int(input()) line=[int(i) for i in input().split()] d=0 s=0 for i in range(n): if line[i]%2==0: s+=1 else: d+=1 if s>d: a=1 if d>s: a=0 for i in range(n): if line[i]%2==a: x=i print(x+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()) line=[int(i) for i in input().split()] d=0 s=0 for i in range(n): if line[i]%2==0: s+=1 else: d+=1 if s>d: a=1 if d>s: a=0 for i in range(n): if line[i]%2==a: x=i print(x+1) ```

3.958319

697

A

Pineapple Incident

PROGRAMMING

900

[ "implementation", "math" ]

null

Ted has a pineapple. This pineapple is able to bark like a bulldog! At time *t* (in seconds) it barks for the first time. Then every *s* seconds after it, it barks twice with 1 second interval. Thus it barks at times *t*, *t*<=+<=*s*, *t*<=+<=*s*<=+<=1, *t*<=+<=2*s*, *t*<=+<=2*s*<=+<=1, etc. Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time *x* (in seconds), so he asked you to tell him if it's gonna bark at that time.

The first and only line of input contains three integers *t*, *s* and *x* (0<=≤<=*t*,<=*x*<=≤<=109, 2<=≤<=*s*<=≤<=109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively.

Print a single "YES" (without quotes) if the pineapple will bark at time *x* or a single "NO" (without quotes) otherwise in the only line of output.

[ "3 10 4\n", "3 10 3\n", "3 8 51\n", "3 8 52\n" ]

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

In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3. In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52.

500

[ { "input": "3 10 4", "output": "NO" }, { "input": "3 10 3", "output": "YES" }, { "input": "3 8 51", "output": "YES" }, { "input": "3 8 52", "output": "YES" }, { "input": "456947336 740144 45", "output": "NO" }, { "input": "33 232603 599417964", "output": "YES" }, { "input": "4363010 696782227 701145238", "output": "YES" }, { "input": "9295078 2 6", "output": "NO" }, { "input": "76079 281367 119938421", "output": "YES" }, { "input": "93647 7 451664565", "output": "YES" }, { "input": "5 18553 10908", "output": "NO" }, { "input": "6 52 30", "output": "NO" }, { "input": "6431 855039 352662", "output": "NO" }, { "input": "749399100 103031711 761562532", "output": "NO" }, { "input": "21 65767 55245", "output": "NO" }, { "input": "4796601 66897 4860613", "output": "NO" }, { "input": "8 6728951 860676", "output": "NO" }, { "input": "914016 6 914019", "output": "NO" }, { "input": "60686899 78474 60704617", "output": "NO" }, { "input": "3 743604 201724", "output": "NO" }, { "input": "571128 973448796 10", "output": "NO" }, { "input": "688051712 67 51", "output": "NO" }, { "input": "74619 213344 6432326", "output": "NO" }, { "input": "6947541 698167 6", "output": "NO" }, { "input": "83 6 6772861", "output": "NO" }, { "input": "251132 67561 135026988", "output": "NO" }, { "input": "8897216 734348516 743245732", "output": "YES" }, { "input": "50 64536 153660266", "output": "YES" }, { "input": "876884 55420 971613604", "output": "YES" }, { "input": "0 6906451 366041903", "output": "YES" }, { "input": "11750 8 446010134", "output": "YES" }, { "input": "582692707 66997 925047377", "output": "YES" }, { "input": "11 957526890 957526901", "output": "YES" }, { "input": "556888 514614196 515171084", "output": "YES" }, { "input": "6 328006 584834704", "output": "YES" }, { "input": "4567998 4 204966403", "output": "YES" }, { "input": "60 317278 109460971", "output": "YES" }, { "input": "906385 342131991 685170368", "output": "YES" }, { "input": "1 38 902410512", "output": "YES" }, { "input": "29318 787017 587931018", "output": "YES" }, { "input": "351416375 243431 368213115", "output": "YES" }, { "input": "54 197366062 197366117", "output": "YES" }, { "input": "586389 79039 850729874", "output": "YES" }, { "input": "723634470 2814619 940360134", "output": "YES" }, { "input": "0 2 0", "output": "YES" }, { "input": "0 2 1", "output": "NO" }, { "input": "0 2 2", "output": "YES" }, { "input": "0 2 3", "output": "YES" }, { "input": "0 2 1000000000", "output": "YES" }, { "input": "0 10 23", "output": "NO" }, { "input": "0 2 999999999", "output": "YES" }, { "input": "10 5 11", "output": "NO" }, { "input": "1 2 1000000000", "output": "YES" }, { "input": "1 10 20", "output": "NO" }, { "input": "1 2 999999937", "output": "YES" }, { "input": "10 3 5", "output": "NO" }, { "input": "3 2 5", "output": "YES" }, { "input": "0 4 0", "output": "YES" }, { "input": "0 215 403", "output": "NO" }, { "input": "5 2 10", "output": "YES" }, { "input": "0 2 900000000", "output": "YES" }, { "input": "0 79 4000", "output": "NO" }, { "input": "5 1000 1000", "output": "NO" }, { "input": "1 5 103", "output": "NO" }, { "input": "5 2 6", "output": "NO" }, { "input": "120 2 1000000000", "output": "YES" }, { "input": "2 2 1000000000", "output": "YES" }, { "input": "5 5 13", "output": "NO" }, { "input": "10 5 15", "output": "YES" }, { "input": "11 2 0", "output": "NO" }, { "input": "3 8 53", "output": "NO" }, { "input": "2 2 4", "output": "YES" }, { "input": "4 4 0", "output": "NO" }, { "input": "1 2 3", "output": "YES" }, { "input": "5 3 9", "output": "YES" }, { "input": "5 6 19", "output": "NO" }, { "input": "3 10 125", "output": "NO" }, { "input": "5 3 8", "output": "YES" }, { "input": "6 3 9", "output": "YES" }, { "input": "0 3 5", "output": "NO" }, { "input": "5 3 300000035", "output": "YES" }, { "input": "5 2 7", "output": "YES" }, { "input": "1 5 6", "output": "YES" }, { "input": "4 2 6", "output": "YES" }, { "input": "0 3 999999998", "output": "NO" }, { "input": "0 10001 0", "output": "YES" }, { "input": "6 5 3", "output": "NO" }, { "input": "1 5 1000000000", "output": "NO" }, { "input": "1 3 6", "output": "NO" }, { "input": "3 3 1000000000", "output": "YES" }, { "input": "3 3 4", "output": "NO" }, { "input": "3 3 5", "output": "NO" }, { "input": "3 3 0", "output": "NO" }, { "input": "1 2 4", "output": "YES" }, { "input": "5 5 10", "output": "YES" } ]

1,581,848,955

2,147,483,647

PyPy 3

OK

TESTS

95

156

0

t, s, x = map(int, input().split()) c1 = x - t - 1 - s c2 = x - t if (c1%s==0 and c1 >= 0) or (c2%s==0 and c2 >= 0): print('YES') else: print('NO')

Title: Pineapple Incident Time Limit: None seconds Memory Limit: None megabytes Problem Description: Ted has a pineapple. This pineapple is able to bark like a bulldog! At time *t* (in seconds) it barks for the first time. Then every *s* seconds after it, it barks twice with 1 second interval. Thus it barks at times *t*, *t*<=+<=*s*, *t*<=+<=*s*<=+<=1, *t*<=+<=2*s*, *t*<=+<=2*s*<=+<=1, etc. Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time *x* (in seconds), so he asked you to tell him if it's gonna bark at that time. Input Specification: The first and only line of input contains three integers *t*, *s* and *x* (0<=≤<=*t*,<=*x*<=≤<=109, 2<=≤<=*s*<=≤<=109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively. Output Specification: Print a single "YES" (without quotes) if the pineapple will bark at time *x* or a single "NO" (without quotes) otherwise in the only line of output. Demo Input: ['3 10 4\n', '3 10 3\n', '3 8 51\n', '3 8 52\n'] Demo Output: ['NO\n', 'YES\n', 'YES\n', 'YES\n'] Note: In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3. In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52.

```python t, s, x = map(int, input().split()) c1 = x - t - 1 - s c2 = x - t if (c1%s==0 and c1 >= 0) or (c2%s==0 and c2 >= 0): print('YES') else: print('NO') ```

3

743

B

Chloe and the sequence

PROGRAMMING

1,200

[ "binary search", "bitmasks", "constructive algorithms", "implementation" ]

null

Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad. Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step. The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps. Please help Chloe to solve the problem!

The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1).

Print single integer — the integer at the *k*-th position in the obtained sequence.

[ "3 2\n", "4 8\n" ]

[ "2", "4" ]

In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2. In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.

1,000

[ { "input": "3 2", "output": "2" }, { "input": "4 8", "output": "4" }, { "input": "5 27", "output": "1" }, { "input": "7 44", "output": "3" }, { "input": "15 18432", "output": "12" }, { "input": "20 259676", "output": "3" }, { "input": "30 671088640", "output": "28" }, { "input": "38 137438953472", "output": "38" }, { "input": "1 1", "output": "1" }, { "input": "3 3", "output": "1" }, { "input": "4 12", "output": "3" }, { "input": "8 224", "output": "6" }, { "input": "13 1368", "output": "4" }, { "input": "16 49152", "output": "15" }, { "input": "19 1024", "output": "11" }, { "input": "24 15204352", "output": "20" }, { "input": "27 6586544", "output": "5" }, { "input": "31 536870912", "output": "30" }, { "input": "38 94489280512", "output": "34" }, { "input": "42 215268308020", "output": "3" }, { "input": "42 3452074721280", "output": "21" }, { "input": "46 34394312982528", "output": "15" }, { "input": "48 133865540681728", "output": "39" }, { "input": "49 76541041985542", "output": "2" }, { "input": "49 104822971826176", "output": "34" }, { "input": "49 351843720888320", "output": "47" }, { "input": "50 743042492421629", "output": "1" }, { "input": "50 666266740436818", "output": "2" }, { "input": "50 704239287953456", "output": "5" }, { "input": "50 116938486513664", "output": "17" }, { "input": "50 806423059496960", "output": "37" }, { "input": "50 985162418487296", "output": "48" }, { "input": "50 844424930131968", "output": "49" }, { "input": "50 562949953421312", "output": "50" }, { "input": "50 1", "output": "1" }, { "input": "50 1125899906842623", "output": "1" }, { "input": "50 1125899906842620", "output": "3" }, { "input": "39 549755813887", "output": "1" }, { "input": "50 100000000000", "output": "12" } ]

1,670,259,542

2,147,483,647

PyPy 3-64

OK

TESTS

39

62

0

def main(): n, k = get_list_int() if k % 2 == 1: print(1) return first_occurance_idx = [0] for i in range(n): step_size = 2**i prev = first_occurance_idx[-1] first_occurance_idx.append(prev + step_size) for i in range(len(first_occurance_idx)): start_idx = first_occurance_idx[i] repeat_freq = 2 ** (i + 1) if (k - 1 - start_idx) % repeat_freq == 0: print(i + 1) def get_int() -> int: return int(input()) def get_list_int() -> list[int]: return [int(x) for x in input().split(" ")] def get_list_str() -> list[str]: return input().split(" ") if __name__ == "__main__": main()

Title: Chloe and the sequence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad. Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step. The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps. Please help Chloe to solve the problem! Input Specification: The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1). Output Specification: Print single integer — the integer at the *k*-th position in the obtained sequence. Demo Input: ['3 2\n', '4 8\n'] Demo Output: ['2', '4'] Note: In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2. In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.

```python def main(): n, k = get_list_int() if k % 2 == 1: print(1) return first_occurance_idx = [0] for i in range(n): step_size = 2**i prev = first_occurance_idx[-1] first_occurance_idx.append(prev + step_size) for i in range(len(first_occurance_idx)): start_idx = first_occurance_idx[i] repeat_freq = 2 ** (i + 1) if (k - 1 - start_idx) % repeat_freq == 0: print(i + 1) def get_int() -> int: return int(input()) def get_list_int() -> list[int]: return [int(x) for x in input().split(" ")] def get_list_str() -> list[str]: return input().split(" ") if __name__ == "__main__": main() ```

3

501

B

Misha and Changing Handles

PROGRAMMING

1,100

[ "data structures", "dsu", "strings" ]

null

Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point. Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.

The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests. Next *q* lines contain the descriptions of the requests, one per line. Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20. The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone.

In the first line output the integer *n* — the number of users that changed their handles at least once. In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order. Each user who changes the handle must occur exactly once in this description.

[ "5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n" ]

[ "3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n" ]

none

500

[ { "input": "5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov", "output": "3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123" }, { "input": "1\nMisha Vasya", "output": "1\nMisha Vasya" }, { "input": "10\na b\nb c\nc d\nd e\ne f\nf g\ng h\nh i\ni j\nj k", "output": "1\na k" }, { "input": "5\n123abc abc123\nabc123 a1b2c3\na1b2c3 1A2B3C\n1 2\n2 Misha", "output": "2\n123abc 1A2B3C\n1 Misha" }, { "input": "8\nM F\nS D\n1 2\nF G\n2 R\nD Q\nQ W\nW e", "output": "3\nM G\n1 R\nS e" }, { "input": "17\nn5WhQ VCczxtxKwFio5U\nVCczxtxKwFio5U 1WMVGA17cd1LRcp4r\n1WMVGA17cd1LRcp4r SJl\nSJl D8bPUoIft5v1\nNAvvUgunbPZNCL9ZY2 jnLkarKYsotz\nD8bPUoIft5v1 DnDkHi7\njnLkarKYsotz GfjX109HSQ81gFEBJc\nGfjX109HSQ81gFEBJc kBJ0zrH78mveJ\nkBJ0zrH78mveJ 9DrAypYW\nDnDkHi7 3Wkho2PglMDaFQw\n3Wkho2PglMDaFQw pOqW\n9DrAypYW G3y0cXXGsWAh\npOqW yr1Ec\nG3y0cXXGsWAh HrmWWg5u4Hsy\nyr1Ec GkFeivXjQ01\nGkFeivXjQ01 mSsWgbCCZcotV4goiA\nHrmWWg5u4Hsy zkCmEV", "output": "2\nn5WhQ mSsWgbCCZcotV4goiA\nNAvvUgunbPZNCL9ZY2 zkCmEV" }, { "input": "10\nH1nauWCJOImtVqXk gWPMQ9DHv5CtkYp9lwm9\nSEj 2knOMLyzr\n0v69ijnAc S7d7zGTjmlku01Gv\n2knOMLyzr otGmEd\nacwr3TfMV7oCIp RUSVFa9TIWlLsd7SB\nS7d7zGTjmlku01Gv Gd6ZufVmQnBpi\nS1 WOJLpk\nWOJLpk Gu\nRUSVFa9TIWlLsd7SB RFawatGnbVB\notGmEd OTB1zKiOI", "output": "5\n0v69ijnAc Gd6ZufVmQnBpi\nS1 Gu\nSEj OTB1zKiOI\nacwr3TfMV7oCIp RFawatGnbVB\nH1nauWCJOImtVqXk gWPMQ9DHv5CtkYp9lwm9" }, { "input": "14\nTPdoztSZROpjZe z6F8bYFvnER4V5SP0n\n8Aa3PQY3hzHZTPEUz fhrZZPJ3iUS\nm9p888KaZAoQaO KNmdRSAlUVn8zXOM0\nAO s1VGWTCbHzM\ni 4F\nfhrZZPJ3iUS j0OVZQF6MvNcKN9xDZFJ\nDnlkXtaKNlYEI2ApBuwu DMA9i8ScKRxwhe72a3\nj0OVZQF6MvNcKN9xDZFJ DzjmeNqN0H4Teq0Awr\n4F wJcdxt1kwqfDeJ\nqxXlsa5t RHCL1K6aUyns\nr6WYbDaXt hEHw\nJ0Usg DKdKMFJ6tK8XA\nz6F8bYFvnER4V5SP0n 0alJ\nMijh2O6 qic8kXWuR6", "output": "10\nTPdoztSZROpjZe 0alJ\nJ0Usg DKdKMFJ6tK8XA\nDnlkXtaKNlYEI2ApBuwu DMA9i8ScKRxwhe72a3\n8Aa3PQY3hzHZTPEUz DzjmeNqN0H4Teq0Awr\nm9p888KaZAoQaO KNmdRSAlUVn8zXOM0\nqxXlsa5t RHCL1K6aUyns\nr6WYbDaXt hEHw\nMijh2O6 qic8kXWuR6\nAO s1VGWTCbHzM\ni wJcdxt1kwqfDeJ" }, { "input": "14\nHAXRxayyf1Dj1F0mT hjR4A8IQMb0nyBtqG\nWNuMJa5Jg05qkqZOrL noNkWXrSidHGwxgbQ\nmOitVy6W52s0FENMz6 oLUkLNfojssvLvb1t\nhjR4A8IQMb0nyBtqG oA7uBFu4Oo\noA7uBFu4Oo M450\nM450 LXEzO4\noLUkLNfojssvLvb1t YG5\nnoNkWXrSidHGwxgbQ L\nL YBWzu4W\nYBWzu4W ML\nML scVZE9m8JnH\nLXEzO4 Ne0oBPY0Iy\nscVZE9m8JnH GXhznv\nYG5 UY08abilYF1LaXj49hQ", "output": "3\nWNuMJa5Jg05qkqZOrL GXhznv\nHAXRxayyf1Dj1F0mT Ne0oBPY0Iy\nmOitVy6W52s0FENMz6 UY08abilYF1LaXj49hQ" } ]

1,658,992,524

2,147,483,647

PyPy 3

OK

TESTS

49

140

3,686,400

import sys import math n = int(input()) lst = dict() for _ in range(n): s = input().split(' ') if s[0] in lst.values(): k = list(lst.keys())[list(lst.values()).index(s[0])] lst[k] = s[1] else: lst.update({s[0]: s[1]}) print(len(lst)) for x, y in lst.items(): print(x,y)

Title: Misha and Changing Handles Time Limit: None seconds Memory Limit: None megabytes Problem Description: Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point. Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that. Input Specification: The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests. Next *q* lines contain the descriptions of the requests, one per line. Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20. The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone. Output Specification: In the first line output the integer *n* — the number of users that changed their handles at least once. In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order. Each user who changes the handle must occur exactly once in this description. Demo Input: ['5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n'] Demo Output: ['3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n'] Note: none

```python import sys import math n = int(input()) lst = dict() for _ in range(n): s = input().split(' ') if s[0] in lst.values(): k = list(lst.keys())[list(lst.values()).index(s[0])] lst[k] = s[1] else: lst.update({s[0]: s[1]}) print(len(lst)) for x, y in lst.items(): print(x,y) ```

3

69

A

Young Physicist

PROGRAMMING

1,000

[ "implementation", "math" ]

A. Young Physicist

2

256

A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.

The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).

Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.

[ "3\n4 1 7\n-2 4 -1\n1 -5 -3\n", "3\n3 -1 7\n-5 2 -4\n2 -1 -3\n" ]

[ "NO", "YES" ]

none

500

[ { "input": "3\n4 1 7\n-2 4 -1\n1 -5 -3", "output": "NO" }, { "input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3", "output": "YES" }, { "input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41", "output": "NO" }, { "input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33", "output": "NO" }, { "input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7", "output": "NO" }, { "input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38", "output": "NO" }, { "input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6", "output": "YES" }, { "input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0", "output": "YES" }, { "input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0", "output": "YES" }, { "input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0", "output": "YES" }, { "input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0", "output": "YES" }, { "input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0", "output": "YES" }, { "input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87", "output": "YES" }, { "input": "3\n1 2 3\n3 2 1\n0 0 0", "output": "NO" }, { "input": "2\n5 -23 12\n0 0 0", "output": "NO" }, { "input": "1\n0 0 0", "output": "YES" }, { "input": "1\n1 -2 0", "output": "NO" }, { "input": "2\n-23 77 -86\n23 -77 86", "output": "YES" }, { "input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82", "output": "YES" }, { "input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0", "output": "YES" }, { "input": "3\n96 49 -12\n2 -66 28\n-98 17 -16", "output": "YES" }, { "input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0", "output": "YES" }, { "input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53", "output": "YES" }, { "input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1", "output": "YES" }, { "input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0", "output": "YES" }, { "input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0", "output": "YES" }, { "input": "3\n0 2 -2\n1 -1 3\n-3 0 0", "output": "NO" } ]

1,593,851,715

2,147,483,647

Python 3

OK

TESTS

81

218

0

from sys import stdin input = stdin.readline n = int(input()) force = [list(map(int,input().split())) for _ in range(n)] nx, ny, nz = 0, 0, 0 for f in force: x, y, z = f nx += x ny += y nz += z if not nx and not ny and not nz: 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 from sys import stdin input = stdin.readline n = int(input()) force = [list(map(int,input().split())) for _ in range(n)] nx, ny, nz = 0, 0, 0 for f in force: x, y, z = f nx += x ny += y nz += z if not nx and not ny and not nz: print("YES") else: print("NO") ```

3.9455

793

A

Oleg and shares

PROGRAMMING

900

[ "implementation", "math" ]

null

Oleg the bank client checks share prices every day. There are *n* share prices he is interested in. Today he observed that each second exactly one of these prices decreases by *k* rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg found this process interesting, and he asked Igor the financial analyst, what is the minimum time needed for all *n* prices to become equal, or it is impossible at all? Igor is busy right now, so he asked you to help Oleg. Can you answer this question?

The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109) — the number of share prices, and the amount of rubles some price decreases each second. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the initial prices.

Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible.

[ "3 3\n12 9 15\n", "2 2\n10 9\n", "4 1\n1 1000000000 1000000000 1000000000\n" ]

[ "3", "-1", "2999999997" ]

Consider the first example. Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds. There could be other possibilities, but this minimizes the time needed for all prices to become equal. Thus the answer is 3. In the second example we can notice that parity of first and second price is different and never changes within described process. Thus prices never can become equal. In the third example following scenario can take place: firstly, the second price drops, then the third price, and then fourth price. It happens 999999999 times, and, since in one second only one price can drop, the whole process takes 999999999 * 3 = 2999999997 seconds. We can note that this is the minimum possible time.

500

[ { "input": "3 3\n12 9 15", "output": "3" }, { "input": "2 2\n10 9", "output": "-1" }, { "input": "4 1\n1 1000000000 1000000000 1000000000", "output": "2999999997" }, { "input": "1 11\n123", "output": "0" }, { "input": "20 6\n38 86 86 50 98 62 32 2 14 62 98 50 2 50 32 38 62 62 8 14", "output": "151" }, { "input": "20 5\n59 54 19 88 55 100 54 3 6 13 99 38 36 71 59 6 64 85 45 54", "output": "-1" }, { "input": "100 10\n340 70 440 330 130 120 340 210 440 110 410 120 180 40 50 230 70 110 310 360 480 70 230 120 230 310 470 60 210 60 210 480 290 250 450 440 150 40 500 230 280 250 30 50 310 50 230 360 420 260 330 80 50 160 70 470 140 180 380 190 250 30 220 410 80 310 280 50 20 430 440 180 310 190 190 330 90 190 320 390 170 460 230 30 80 500 470 370 80 500 400 120 220 150 70 120 70 320 260 260", "output": "2157" }, { "input": "100 18\n489 42 300 366 473 105 220 448 70 488 201 396 168 281 67 235 324 291 313 387 407 223 39 144 224 233 72 318 229 377 62 171 448 119 354 282 147 447 260 384 172 199 67 326 311 431 337 142 281 202 404 468 38 120 90 437 33 420 249 372 367 253 255 411 309 333 103 176 162 120 203 41 352 478 216 498 224 31 261 493 277 99 375 370 394 229 71 488 246 194 233 13 66 111 366 456 277 360 116 354", "output": "-1" }, { "input": "4 2\n1 2 3 4", "output": "-1" }, { "input": "3 4\n3 5 5", "output": "-1" }, { "input": "3 2\n88888884 88888886 88888888", "output": "3" }, { "input": "2 1\n1000000000 1000000000", "output": "0" }, { "input": "4 2\n1000000000 100000000 100000000 100000000", "output": "450000000" }, { "input": "2 2\n1000000000 1000000000", "output": "0" }, { "input": "3 3\n3 2 1", "output": "-1" }, { "input": "3 4\n3 5 3", "output": "-1" }, { "input": "3 2\n1 2 2", "output": "-1" }, { "input": "4 2\n2 3 3 2", "output": "-1" }, { "input": "3 2\n1 2 4", "output": "-1" }, { "input": "3 2\n3 4 4", "output": "-1" }, { "input": "3 3\n4 7 10", "output": "3" }, { "input": "4 3\n2 2 5 1", "output": "-1" }, { "input": "3 3\n1 3 5", "output": "-1" }, { "input": "2 5\n5 9", "output": "-1" }, { "input": "2 3\n5 7", "output": "-1" }, { "input": "3 137\n1000000000 1000000000 1000000000", "output": "0" }, { "input": "5 1000000000\n1000000000 1000000000 1000000000 1000000000 1000000000", "output": "0" }, { "input": "3 5\n1 2 5", "output": "-1" }, { "input": "3 3\n1000000000 1000000000 999999997", "output": "2" }, { "input": "2 4\n5 6", "output": "-1" }, { "input": "4 1\n1000000000 1000000000 1000000000 1000000000", "output": "0" }, { "input": "2 3\n5 8", "output": "1" }, { "input": "2 6\n8 16", "output": "-1" }, { "input": "5 3\n15 14 9 12 18", "output": "-1" }, { "input": "3 3\n1 2 3", "output": "-1" }, { "input": "3 3\n3 4 5", "output": "-1" }, { "input": "2 5\n8 17", "output": "-1" }, { "input": "2 1\n1 2", "output": "1" }, { "input": "1 1\n1000000000", "output": "0" }, { "input": "3 3\n5 3 4", "output": "-1" }, { "input": "3 6\n10 14 12", "output": "-1" }, { "input": "2 2\n3 5", "output": "1" }, { "input": "3 5\n1 3 4", "output": "-1" }, { "input": "4 3\n1 6 6 6", "output": "-1" }, { "input": "2 3\n1 8", "output": "-1" }, { "input": "3 5\n6 11 17", "output": "-1" }, { "input": "2 2\n1 4", "output": "-1" }, { "input": "2 4\n6 8", "output": "-1" }, { "input": "2 1\n2 3", "output": "1" }, { "input": "4 4\n1 5 8 14", "output": "-1" }, { "input": "3 3\n1 5 3", "output": "-1" }, { "input": "4 3\n1 2 2 5", "output": "-1" }, { "input": "3 2\n1 4 6", "output": "-1" }, { "input": "2 3\n6 9", "output": "1" }, { "input": "3 3\n2 3 4", "output": "-1" }, { "input": "3 2\n9 10 10", "output": "-1" }, { "input": "2 2\n9 12", "output": "-1" }, { "input": "2 2\n100000003 100000005", "output": "1" }, { "input": "2 3\n2 4", "output": "-1" }, { "input": "3 2\n2 3 5", "output": "-1" }, { "input": "3 3\n1 3 4", "output": "-1" }, { "input": "10 2\n2 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "4499999991" }, { "input": "3 5\n2 4 5", "output": "-1" }, { "input": "2 3\n7 10", "output": "1" }, { "input": "3 10\n10 13 17", "output": "-1" }, { "input": "2 3\n1 6", "output": "-1" }, { "input": "1 7\n1000000000", "output": "0" }, { "input": "2 4\n3 7", "output": "1" }, { "input": "2 3\n2 5", "output": "1" }, { "input": "20 1\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "0" }, { "input": "3 3\n7 8 8", "output": "-1" }, { "input": "4 10\n1 11 100 11", "output": "-1" } ]

1,607,511,184

2,147,483,647

Python 3

OK

TESTS

88

218

9,216,000

n, k = map(int, input().split()) a = sorted(list(map(int, input().split()))) result = 0 for i in a: if (i-a[0])%k!=0: print(-1) exit() else: result+=(i-a[0])//k print(result)

Title: Oleg and shares Time Limit: None seconds Memory Limit: None megabytes Problem Description: Oleg the bank client checks share prices every day. There are *n* share prices he is interested in. Today he observed that each second exactly one of these prices decreases by *k* rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg found this process interesting, and he asked Igor the financial analyst, what is the minimum time needed for all *n* prices to become equal, or it is impossible at all? Igor is busy right now, so he asked you to help Oleg. Can you answer this question? Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109) — the number of share prices, and the amount of rubles some price decreases each second. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the initial prices. Output Specification: Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible. Demo Input: ['3 3\n12 9 15\n', '2 2\n10 9\n', '4 1\n1 1000000000 1000000000 1000000000\n'] Demo Output: ['3', '-1', '2999999997'] Note: Consider the first example. Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds. There could be other possibilities, but this minimizes the time needed for all prices to become equal. Thus the answer is 3. In the second example we can notice that parity of first and second price is different and never changes within described process. Thus prices never can become equal. In the third example following scenario can take place: firstly, the second price drops, then the third price, and then fourth price. It happens 999999999 times, and, since in one second only one price can drop, the whole process takes 999999999 * 3 = 2999999997 seconds. We can note that this is the minimum possible time.

```python n, k = map(int, input().split()) a = sorted(list(map(int, input().split()))) result = 0 for i in a: if (i-a[0])%k!=0: print(-1) exit() else: result+=(i-a[0])//k print(result) ```

3

513

B1

Permutations

PROGRAMMING

1,400

[ "brute force" ]

null

You are given a permutation *p* of numbers 1,<=2,<=...,<=*n*. Let's define *f*(*p*) as the following sum: Find the lexicographically *m*-th permutation of length *n* in the set of permutations having the maximum possible value of *f*(*p*).

The single line of input contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*cnt**n*), where *cnt**n* is the number of permutations of length *n* with maximum possible value of *f*(*p*). The problem consists of two subproblems. The subproblems have different constraints on the input. You will get some score for the correct submission of the subproblem. The description of the subproblems follows. - In subproblem B1 (3 points), the constraint 1<=≤<=*n*<=≤<=8 will hold. - In subproblem B2 (4 points), the constraint 1<=≤<=*n*<=≤<=50 will hold.

Output *n* number forming the required permutation.

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

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

In the first example, both permutations of numbers {1, 2} yield maximum possible *f*(*p*) which is equal to 4. Among them, (2, 1) comes second in lexicographical order.

3

[ { "input": "2 2", "output": "2 1 " }, { "input": "3 2", "output": "1 3 2 " }, { "input": "1 1", "output": "1 " }, { "input": "3 1", "output": "1 2 3 " }, { "input": "3 3", "output": "2 3 1 " }, { "input": "3 4", "output": "3 2 1 " }, { "input": "4 1", "output": "1 2 3 4 " }, { "input": "4 3", "output": "1 3 4 2 " }, { "input": "4 4", "output": "1 4 3 2 " }, { "input": "4 8", "output": "4 3 2 1 " }, { "input": "5 2", "output": "1 2 3 5 4 " }, { "input": "5 7", "output": "1 4 5 3 2 " }, { "input": "5 15", "output": "4 5 3 2 1 " }, { "input": "6 23", "output": "2 5 6 4 3 1 " }, { "input": "7 7", "output": "1 2 3 6 7 5 4 " }, { "input": "7 44", "output": "2 4 7 6 5 3 1 " }, { "input": "8 1", "output": "1 2 3 4 5 6 7 8 " }, { "input": "8 127", "output": "7 8 6 5 4 3 2 1 " }, { "input": "8 128", "output": "8 7 6 5 4 3 2 1 " } ]

1,423,331,254

2,854

Python 3

OK

TESTS1

19

1,045

0

__author__ = 'yushchenko' def countf(f): sum = 0 for i in range(len(f)): for j in range(len(f))[i:]: # print(i, j) # print(f[i:j + 1]) sum += min(f[i:j + 1]) return sum import itertools n,m = input().split() n = int(n) m = int(m) maxf = 0; count = 0; save = () for x in itertools.permutations(range(n + 1)[1:]): t = countf(x) if t > maxf: maxf = t count = 0 if t == maxf: # print (x , '-' , t) count = count + 1 if count == m: save = x print(' '.join(str(e) for e in save))

Title: Permutations Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a permutation *p* of numbers 1,<=2,<=...,<=*n*. Let's define *f*(*p*) as the following sum: Find the lexicographically *m*-th permutation of length *n* in the set of permutations having the maximum possible value of *f*(*p*). Input Specification: The single line of input contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*cnt**n*), where *cnt**n* is the number of permutations of length *n* with maximum possible value of *f*(*p*). The problem consists of two subproblems. The subproblems have different constraints on the input. You will get some score for the correct submission of the subproblem. The description of the subproblems follows. - In subproblem B1 (3 points), the constraint 1<=≤<=*n*<=≤<=8 will hold. - In subproblem B2 (4 points), the constraint 1<=≤<=*n*<=≤<=50 will hold. Output Specification: Output *n* number forming the required permutation. Demo Input: ['2 2\n', '3 2\n'] Demo Output: ['2 1 \n', '1 3 2 \n'] Note: In the first example, both permutations of numbers {1, 2} yield maximum possible *f*(*p*) which is equal to 4. Among them, (2, 1) comes second in lexicographical order.

```python __author__ = 'yushchenko' def countf(f): sum = 0 for i in range(len(f)): for j in range(len(f))[i:]: # print(i, j) # print(f[i:j + 1]) sum += min(f[i:j + 1]) return sum import itertools n,m = input().split() n = int(n) m = int(m) maxf = 0; count = 0; save = () for x in itertools.permutations(range(n + 1)[1:]): t = countf(x) if t > maxf: maxf = t count = 0 if t == maxf: # print (x , '-' , t) count = count + 1 if count == m: save = x print(' '.join(str(e) for e in save)) ```

3

18

C

Stripe

PROGRAMMING

1,200

[ "data structures", "implementation" ]

C. Stripe

2

64

Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?

The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.

Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.

[ "9\n1 5 -6 7 9 -16 0 -2 2\n", "3\n1 1 1\n", "2\n0 0\n" ]

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

none

0

[ { "input": "9\n1 5 -6 7 9 -16 0 -2 2", "output": "3" }, { "input": "3\n1 1 1", "output": "0" }, { "input": "2\n0 0", "output": "1" }, { "input": "4\n100 1 10 111", "output": "1" }, { "input": "10\n0 4 -3 0 -2 2 -3 -3 2 5", "output": "3" }, { "input": "10\n0 -1 2 2 -1 1 0 0 0 2", "output": "0" }, { "input": "10\n-1 -1 1 -1 0 1 0 1 1 1", "output": "1" }, { "input": "10\n0 0 0 0 0 0 0 0 0 0", "output": "9" }, { "input": "50\n-4 -3 3 4 -1 0 2 -4 -3 -4 1 4 3 0 4 1 0 -3 4 -3 -2 2 2 1 0 -4 -4 -5 3 2 -1 4 5 -3 -3 4 4 -5 2 -3 4 -5 2 5 -4 4 1 -2 -4 3", "output": "3" }, { "input": "15\n0 4 0 3 -1 4 -2 -2 -4 -4 3 2 4 -1 -3", "output": "0" }, { "input": "10\n3 -1 -3 -1 3 -2 0 3 1 -2", "output": "0" }, { "input": "100\n-4 2 4 4 1 3 -3 -3 2 1 -4 0 0 2 3 -1 -4 -3 4 -2 -3 -3 -3 -1 -2 -3 -1 -4 0 4 0 -1 4 0 -4 -4 4 -4 -2 1 -4 1 -3 -2 3 -4 4 0 -1 3 -1 4 -1 4 -1 3 -3 -3 -2 -2 4 -3 -3 4 -3 -2 -1 0 -2 4 0 -3 -1 -2 -3 1 -4 1 -3 -3 -3 -2 -3 0 1 -2 -2 -4 -3 -1 2 3 -1 1 1 0 3 -3 -1 -2", "output": "1" }, { "input": "100\n-2 -1 1 0 -2 -1 2 2 0 0 2 1 0 2 0 2 1 0 -1 -1 -1 0 -2 -1 2 -1 -2 2 -2 2 -2 -2 2 1 1 1 -2 2 0 0 2 -1 2 2 2 0 -1 -1 -1 1 -2 2 2 2 -2 0 0 -2 0 -2 -2 0 -1 -1 -2 -1 1 2 -2 -1 1 -2 -1 0 -2 2 2 -1 -2 2 0 0 0 1 0 1 2 1 -2 -2 -1 -1 -2 -2 -2 -1 2 2 2 -2", "output": "6" }, { "input": "100\n2 7 6 0 8 -2 0 1 8 -1 7 -4 -1 1 0 3 4 -4 3 7 8 -4 -6 6 4 -2 -5 7 4 6 1 6 3 8 -2 6 -6 -3 0 -1 -7 -8 -2 5 8 6 -2 -2 -5 -4 -7 -3 -5 -3 -3 -1 5 8 4 0 -7 -6 8 -4 -1 7 2 6 6 4 -5 -4 -5 -2 2 -2 -7 -1 5 -8 -6 -2 -5 4 2 8 -6 7 -8 -1 -5 8 6 -3 -1 7 -1 2 -8 -8", "output": "0" }, { "input": "1\n0", "output": "0" }, { "input": "1\n10000", "output": "0" }, { "input": "1\n-10000", "output": "0" } ]

1,608,399,223

2,147,483,647

Python 3

OK

TESTS

45

436

7,782,400

if __name__ == "__main__": stripe_length = int(input()) square_Values = list(map(int, input().split())) # create a prefix sum array for i in range(1, stripe_length): square_Values[i] += square_Values[i-1] flag = 0 for i in range(stripe_length): if i != stripe_length - 1 and square_Values[i] == square_Values[stripe_length-1] - square_Values[i]: flag += 1 print(flag) # basic logic >> time exceed when it comes to large array # flag = 0 # i = 1 # while i < len(square_Values): # if sum(square_Values[:i]) == sum(square_Values[i:]): # flag+=1 # i += 1

Title: Stripe Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem? Input Specification: The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value. Output Specification: Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only. Demo Input: ['9\n1 5 -6 7 9 -16 0 -2 2\n', '3\n1 1 1\n', '2\n0 0\n'] Demo Output: ['3\n', '0\n', '1\n'] Note: none

```python if __name__ == "__main__": stripe_length = int(input()) square_Values = list(map(int, input().split())) # create a prefix sum array for i in range(1, stripe_length): square_Values[i] += square_Values[i-1] flag = 0 for i in range(stripe_length): if i != stripe_length - 1 and square_Values[i] == square_Values[stripe_length-1] - square_Values[i]: flag += 1 print(flag) # basic logic >> time exceed when it comes to large array # flag = 0 # i = 1 # while i < len(square_Values): # if sum(square_Values[:i]) == sum(square_Values[i:]): # flag+=1 # i += 1 ```

3.833017

576

A

Vasya and Petya's Game

PROGRAMMING

1,500

[ "math", "number theory" ]

null

Vasya and Petya are playing a simple game. Vasya thought of number *x* between 1 and *n*, and Petya tries to guess the number. Petya can ask questions like: "Is the unknown number divisible by number *y*?". The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of. Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers *y**i*, he should ask the questions about.

A single line contains number *n* (1<=≤<=*n*<=≤<=103).

Print the length of the sequence of questions *k* (0<=≤<=*k*<=≤<=*n*), followed by *k* numbers — the questions *y**i* (1<=≤<=*y**i*<=≤<=*n*). If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.

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

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

The sequence from the answer to the first sample test is actually correct. If the unknown number is not divisible by one of the sequence numbers, it is equal to 1. If the unknown number is divisible by 4, it is 4. If the unknown number is divisible by 3, then the unknown number is 3. Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter.

500

[ { "input": "4", "output": "3\n2 4 3 " }, { "input": "6", "output": "4\n2 4 3 5 " }, { "input": "1", "output": "0" }, { "input": "15", "output": "9\n2 4 8 3 9 5 7 11 13 " }, { "input": "19", "output": "12\n2 4 8 16 3 9 5 7 11 13 17 19 " }, { "input": "20", "output": "12\n2 4 8 16 3 9 5 7 11 13 17 19 " }, { "input": "37", "output": "19\n2 4 8 16 32 3 9 27 5 25 7 11 13 17 19 23 29 31 37 " }, { "input": "211", "output": "61\n2 4 8 16 32 64 128 3 9 27 81 5 25 125 7 49 11 121 13 169 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 " }, { "input": "557", "output": "123\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 5 25 125 7 49 343 11 121 13 169 17 289 19 361 23 529 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 " }, { "input": "907", "output": "179\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 ..." }, { "input": "953", "output": "186\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 ..." }, { "input": "289", "output": "78\n2 4 8 16 32 64 128 256 3 9 27 81 243 5 25 125 7 49 11 121 13 169 17 289 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 " }, { "input": "400", "output": "97\n2 4 8 16 32 64 128 256 3 9 27 81 243 5 25 125 7 49 343 11 121 13 169 17 289 19 361 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 " }, { "input": "900", "output": "178\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 ..." }, { "input": "625", "output": "136\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 " }, { "input": "729", "output": "152\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 ..." }, { "input": "784", "output": "160\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 ..." }, { "input": "31", "output": "17\n2 4 8 16 3 9 27 5 25 7 11 13 17 19 23 29 31 " }, { "input": "44", "output": "21\n2 4 8 16 32 3 9 27 5 25 7 11 13 17 19 23 29 31 37 41 43 " }, { "input": "160", "output": "50\n2 4 8 16 32 64 128 3 9 27 81 5 25 125 7 49 11 121 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 " }, { "input": "322", "output": "83\n2 4 8 16 32 64 128 256 3 9 27 81 243 5 25 125 7 49 11 121 13 169 17 289 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 " }, { "input": "894", "output": "178\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 ..." }, { "input": "998", "output": "193\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 961 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 ..." }, { "input": "1000", "output": "193\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 961 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 ..." }, { "input": "1000", "output": "193\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 961 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 ..." }, { "input": "30", "output": "16\n2 4 8 16 3 9 27 5 25 7 11 13 17 19 23 29 " } ]

1,662,257,529

2,147,483,647

Python 3

OK

TESTS

26

93

0

n = int(input()) a = " ".join([" ".join([str(i ** k) for k in range(1, 30) if i ** k <= n]) for i in range(2, n + 1) if all([i % j != 0 for j in range(2, i)])]) print((len(a.split())), "\n" + a) num_inp=lambda: int(input()) arr_inp=lambda: list(map(int,input().split())) sp_inp=lambda: map(int,input().split()) str_inp=lambda:input()

Title: Vasya and Petya's Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya and Petya are playing a simple game. Vasya thought of number *x* between 1 and *n*, and Petya tries to guess the number. Petya can ask questions like: "Is the unknown number divisible by number *y*?". The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of. Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers *y**i*, he should ask the questions about. Input Specification: A single line contains number *n* (1<=≤<=*n*<=≤<=103). Output Specification: Print the length of the sequence of questions *k* (0<=≤<=*k*<=≤<=*n*), followed by *k* numbers — the questions *y**i* (1<=≤<=*y**i*<=≤<=*n*). If there are several correct sequences of questions of the minimum length, you are allowed to print any of them. Demo Input: ['4\n', '6\n'] Demo Output: ['3\n2 4 3 \n', '4\n2 4 3 5 \n'] Note: The sequence from the answer to the first sample test is actually correct. If the unknown number is not divisible by one of the sequence numbers, it is equal to 1. If the unknown number is divisible by 4, it is 4. If the unknown number is divisible by 3, then the unknown number is 3. Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter.

```python n = int(input()) a = " ".join([" ".join([str(i ** k) for k in range(1, 30) if i ** k <= n]) for i in range(2, n + 1) if all([i % j != 0 for j in range(2, i)])]) print((len(a.split())), "\n" + a) num_inp=lambda: int(input()) arr_inp=lambda: list(map(int,input().split())) sp_inp=lambda: map(int,input().split()) str_inp=lambda:input() ```

3

165

B

Burning Midnight Oil

PROGRAMMING

1,500

[ "binary search", "implementation" ]

null

One day a highly important task was commissioned to Vasya — writing a program in a night. The program consists of *n* lines of code. Vasya is already exhausted, so he works like that: first he writes *v* lines of code, drinks a cup of tea, then he writes as much as lines, drinks another cup of tea, then he writes lines and so on: , , , ... The expression is regarded as the integral part from dividing number *a* by number *b*. The moment the current value equals 0, Vasya immediately falls asleep and he wakes up only in the morning, when the program should already be finished. Vasya is wondering, what minimum allowable value *v* can take to let him write not less than *n* lines of code before he falls asleep.

The input consists of two integers *n* and *k*, separated by spaces — the size of the program in lines and the productivity reduction coefficient, 1<=≤<=*n*<=≤<=109, 2<=≤<=*k*<=≤<=10.

Print the only integer — the minimum value of *v* that lets Vasya write the program in one night.

[ "7 2\n", "59 9\n" ]

[ "4\n", "54\n" ]

In the first sample the answer is *v* = 4. Vasya writes the code in the following portions: first 4 lines, then 2, then 1, and then Vasya falls asleep. Thus, he manages to write 4 + 2 + 1 = 7 lines in a night and complete the task. In the second sample the answer is *v* = 54. Vasya writes the code in the following portions: 54, 6. The total sum is 54 + 6 = 60, that's even more than *n* = 59.

1,000

[ { "input": "7 2", "output": "4" }, { "input": "59 9", "output": "54" }, { "input": "1 9", "output": "1" }, { "input": "11 2", "output": "7" }, { "input": "747 2", "output": "376" }, { "input": "6578 2", "output": "3293" }, { "input": "37212 2", "output": "18609" }, { "input": "12357 2", "output": "6181" }, { "input": "7998332 2", "output": "3999172" }, { "input": "86275251 2", "output": "43137632" }, { "input": "75584551 2", "output": "37792280" }, { "input": "6 3", "output": "5" }, { "input": "43 4", "output": "33" }, { "input": "811 3", "output": "543" }, { "input": "3410 4", "output": "2560" }, { "input": "21341 4", "output": "16009" }, { "input": "696485 4", "output": "522368" }, { "input": "8856748 3", "output": "5904504" }, { "input": "2959379 4", "output": "2219538" }, { "input": "831410263 3", "output": "554273516" }, { "input": "2 5", "output": "2" }, { "input": "19 6", "output": "17" }, { "input": "715 7", "output": "615" }, { "input": "9122 5", "output": "7300" }, { "input": "89117 6", "output": "74268" }, { "input": "689973 7", "output": "591408" }, { "input": "3024524 5", "output": "2419624" }, { "input": "67127156 6", "output": "55939302" }, { "input": "412262167 7", "output": "353367574" }, { "input": "6 8", "output": "6" }, { "input": "59 9", "output": "54" }, { "input": "246 10", "output": "222" }, { "input": "5314 8", "output": "4651" }, { "input": "15309 9", "output": "13609" }, { "input": "35648 10", "output": "32085" }, { "input": "3018012 8", "output": "2640764" }, { "input": "92153348 9", "output": "81914089" }, { "input": "177583558 10", "output": "159825206" }, { "input": "1000000000 2", "output": "500000008" }, { "input": "1000000000 3", "output": "666666672" }, { "input": "1000000000 4", "output": "750000005" }, { "input": "1000000000 5", "output": "800000003" }, { "input": "1000000000 6", "output": "833333338" }, { "input": "1000000000 7", "output": "857142861" }, { "input": "1000000000 8", "output": "875000004" }, { "input": "1000000000 9", "output": "888888894" }, { "input": "1000000000 10", "output": "900000001" }, { "input": "1 4", "output": "1" }, { "input": "2 10", "output": "2" }, { "input": "1 2", "output": "1" }, { "input": "6 8", "output": "6" }, { "input": "987862820 9", "output": "878100288" }, { "input": "979591791 9", "output": "870748262" }, { "input": "948889213 9", "output": "843457081" }, { "input": "8 9", "output": "8" }, { "input": "999999999 10", "output": "900000000" } ]

1,657,861,995

2,147,483,647

PyPy 3-64

OK

TESTS

56

124

0

def check(x): val=0 div=0 while val<n and x//(k**div)>0: val=val+x//(k**(div)) div=div+1 # print(val,div) # print(x,val) if val<n: return False return True n,k=list(map(int,input().split())) # print(check(4)) i=-1 j=10**10 while i+1<j: mid=(i+j)//2 # print(check(mid),mid) if check(mid): j=mid else: i=mid print(j)

Title: Burning Midnight Oil Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day a highly important task was commissioned to Vasya — writing a program in a night. The program consists of *n* lines of code. Vasya is already exhausted, so he works like that: first he writes *v* lines of code, drinks a cup of tea, then he writes as much as lines, drinks another cup of tea, then he writes lines and so on: , , , ... The expression is regarded as the integral part from dividing number *a* by number *b*. The moment the current value equals 0, Vasya immediately falls asleep and he wakes up only in the morning, when the program should already be finished. Vasya is wondering, what minimum allowable value *v* can take to let him write not less than *n* lines of code before he falls asleep. Input Specification: The input consists of two integers *n* and *k*, separated by spaces — the size of the program in lines and the productivity reduction coefficient, 1<=≤<=*n*<=≤<=109, 2<=≤<=*k*<=≤<=10. Output Specification: Print the only integer — the minimum value of *v* that lets Vasya write the program in one night. Demo Input: ['7 2\n', '59 9\n'] Demo Output: ['4\n', '54\n'] Note: In the first sample the answer is *v* = 4. Vasya writes the code in the following portions: first 4 lines, then 2, then 1, and then Vasya falls asleep. Thus, he manages to write 4 + 2 + 1 = 7 lines in a night and complete the task. In the second sample the answer is *v* = 54. Vasya writes the code in the following portions: 54, 6. The total sum is 54 + 6 = 60, that's even more than *n* = 59.

```python def check(x): val=0 div=0 while val<n and x//(k**div)>0: val=val+x//(k**(div)) div=div+1 # print(val,div) # print(x,val) if val<n: return False return True n,k=list(map(int,input().split())) # print(check(4)) i=-1 j=10**10 while i+1<j: mid=(i+j)//2 # print(check(mid),mid) if check(mid): j=mid else: i=mid print(j) ```

3

805

B

3-palindrome

PROGRAMMING

1,000

[ "constructive algorithms" ]

null

In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick. He is too selfish, so for a given *n* he wants to obtain a string of *n* characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible.

The first line contains single integer *n* (1<=≤<=*n*<=≤<=2·105) — the length of the string.

Print the string that satisfies all the constraints. If there are multiple answers, print any of them.

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

[ "aa\n", "bba\n" ]

A palindrome is a sequence of characters which reads the same backward and forward.

1,000

[ { "input": "2", "output": "aa" }, { "input": "3", "output": "aab" }, { "input": "38", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaa" }, { "input": "47", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab" }, { "input": "59", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab" }, { "input": "67", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab" }, { "input": "77", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba" }, { "input": "89", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba" }, { "input": "98", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaa" }, { "input": "109", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba" }, { "input": "117", "output": 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"aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..." }, { "input": "60131", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..." }, { "input": "146883", "output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..." }, { "input": "6", "output": "aabbaa" }, { "input": "4", "output": "aabb" } ]

1,591,127,448

2,147,483,647

Python 3

OK

TESTS

42

124

819,200

n=int(input()) s='' c='a' for i in range(n//2+1): if c=='a': s=s+'aa' c='b' else: s=s+'bb' c='a' print(s[0:n])

Title: 3-palindrome Time Limit: None seconds Memory Limit: None megabytes Problem Description: In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick. He is too selfish, so for a given *n* he wants to obtain a string of *n* characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible. Input Specification: The first line contains single integer *n* (1<=≤<=*n*<=≤<=2·105) — the length of the string. Output Specification: Print the string that satisfies all the constraints. If there are multiple answers, print any of them. Demo Input: ['2\n', '3\n'] Demo Output: ['aa\n', 'bba\n'] Note: A palindrome is a sequence of characters which reads the same backward and forward.

```python n=int(input()) s='' c='a' for i in range(n//2+1): if c=='a': s=s+'aa' c='b' else: s=s+'bb' c='a' print(s[0:n]) ```

3

430

B

Balls Game

PROGRAMMING

1,400

[ "brute force", "two pointers" ]

null

Iahub is training for the IOI. What is a better way to train than playing a Zuma-like game? There are *n* balls put in a row. Each ball is colored in one of *k* colors. Initially the row doesn't contain three or more contiguous balls with the same color. Iahub has a single ball of color *x*. He can insert his ball at any position in the row (probably, between two other balls). If at any moment there are three or more contiguous balls of the same color in the row, they are destroyed immediately. This rule is applied multiple times, until there are no more sets of 3 or more contiguous balls of the same color. For example, if Iahub has the row of balls [black, black, white, white, black, black] and a white ball, he can insert the ball between two white balls. Thus three white balls are destroyed, and then four black balls become contiguous, so all four balls are destroyed. The row will not contain any ball in the end, so Iahub can destroy all 6 balls. Iahub wants to destroy as many balls as possible. You are given the description of the row of balls, and the color of Iahub's ball. Help Iahub train for the IOI by telling him the maximum number of balls from the row he can destroy.

The first line of input contains three integers: *n* (1<=≤<=*n*<=≤<=100), *k* (1<=≤<=*k*<=≤<=100) and *x* (1<=≤<=*x*<=≤<=*k*). The next line contains *n* space-separated integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=*k*). Number *c**i* means that the *i*-th ball in the row has color *c**i*. It is guaranteed that the initial row of balls will never contain three or more contiguous balls of the same color.

Print a single integer — the maximum number of balls Iahub can destroy.

[ "6 2 2\n1 1 2 2 1 1\n", "1 1 1\n1\n" ]

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

none

1,000

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

1,427,117,765

2,147,483,647

Python 3

OK

TESTS

18

62

204,800

R = lambda: list(map(int, input().split())) def dfs(a): cnt = 0 n = len(a) a.append(10000000) for i in range(1, n): if a[i] != a[i - 1]: cnt = 0 else: cnt += 1 if cnt >= 2: j = i while a[j] == a[i]: j += 1 return j - i + 2 + dfs(a[:i - 2] + a[j:n]) return 0 n, k, x = R() a = R() ans=1 for i in range(n+1): ans=max(ans, dfs(a[:i]+[x]+a[i:])) print(ans-1)

Title: Balls Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Iahub is training for the IOI. What is a better way to train than playing a Zuma-like game? There are *n* balls put in a row. Each ball is colored in one of *k* colors. Initially the row doesn't contain three or more contiguous balls with the same color. Iahub has a single ball of color *x*. He can insert his ball at any position in the row (probably, between two other balls). If at any moment there are three or more contiguous balls of the same color in the row, they are destroyed immediately. This rule is applied multiple times, until there are no more sets of 3 or more contiguous balls of the same color. For example, if Iahub has the row of balls [black, black, white, white, black, black] and a white ball, he can insert the ball between two white balls. Thus three white balls are destroyed, and then four black balls become contiguous, so all four balls are destroyed. The row will not contain any ball in the end, so Iahub can destroy all 6 balls. Iahub wants to destroy as many balls as possible. You are given the description of the row of balls, and the color of Iahub's ball. Help Iahub train for the IOI by telling him the maximum number of balls from the row he can destroy. Input Specification: The first line of input contains three integers: *n* (1<=≤<=*n*<=≤<=100), *k* (1<=≤<=*k*<=≤<=100) and *x* (1<=≤<=*x*<=≤<=*k*). The next line contains *n* space-separated integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=*k*). Number *c**i* means that the *i*-th ball in the row has color *c**i*. It is guaranteed that the initial row of balls will never contain three or more contiguous balls of the same color. Output Specification: Print a single integer — the maximum number of balls Iahub can destroy. Demo Input: ['6 2 2\n1 1 2 2 1 1\n', '1 1 1\n1\n'] Demo Output: ['6\n', '0\n'] Note: none

```python R = lambda: list(map(int, input().split())) def dfs(a): cnt = 0 n = len(a) a.append(10000000) for i in range(1, n): if a[i] != a[i - 1]: cnt = 0 else: cnt += 1 if cnt >= 2: j = i while a[j] == a[i]: j += 1 return j - i + 2 + dfs(a[:i - 2] + a[j:n]) return 0 n, k, x = R() a = R() ans=1 for i in range(n+1): ans=max(ans, dfs(a[:i]+[x]+a[i:])) print(ans-1) ```

3

488

A

Giga Tower

PROGRAMMING

1,100

[ "brute force" ]

null

Giga Tower is the tallest and deepest building in Cyberland. There are 17<=777<=777<=777 floors, numbered from <=-<=8<=888<=888<=888 to 8<=888<=888<=888. In particular, there is floor 0 between floor <=-<=1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view. In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8<=888<=888<=888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8,<=<=-<=180,<=808 are all lucky while 42,<=<=-<=10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?). Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered *a*. He wants to find the minimum positive integer *b*, such that, if he walks *b* floors higher, he will arrive at a floor with a lucky number.

The only line of input contains an integer *a* (<=-<=109<=≤<=*a*<=≤<=109).

Print the minimum *b* in a line.

[ "179\n", "-1\n", "18\n" ]

[ "1\n", "9\n", "10\n" ]

For the first sample, he has to arrive at the floor numbered 180. For the second sample, he will arrive at 8. Note that *b* should be positive, so the answer for the third sample is 10, not 0.

500

[ { "input": "179", "output": "1" }, { "input": "-1", "output": "9" }, { "input": "18", "output": "10" }, { "input": "-410058385", "output": "1" }, { "input": "-586825624", "output": "1" }, { "input": "852318890", "output": "1" }, { "input": "919067153", "output": "5" }, { "input": "690422411", "output": "7" }, { "input": "-408490162", "output": "1" }, { "input": "-8", "output": "16" }, { "input": "-6", "output": "14" }, { "input": "-4", "output": "12" }, { "input": "-2", "output": "10" }, { "input": "0", "output": "8" }, { "input": "2", "output": "6" }, { "input": "4", "output": "4" }, { "input": "6", "output": "2" }, { "input": "8", "output": "10" }, { "input": "1000000000", "output": "8" }, { "input": "-1000000000", "output": "2" }, { "input": "88888", "output": "1" }, { "input": "89", "output": "9" }, { "input": "-80000000", "output": "2" }, { "input": "-8888", "output": "1" }, { "input": "-17", "output": "9" }, { "input": "78", "output": "2" }, { "input": "-19", "output": "1" }, { "input": "-999999998", "output": "9" }, { "input": "-999999997", "output": "8" }, { "input": "999999997", "output": "1" }, { "input": "811111111", "output": "1" }, { "input": "-8", "output": "16" }, { "input": "-5", "output": "13" }, { "input": "-7", "output": "15" }, { "input": "1000000000", "output": "8" } ]

1,416,594,741

4,341

Python 3

OK

TESTS

35

93

0

n = input() for i in range(1,20): if "8" in str(int(n)+i): print(i) break

Title: Giga Tower Time Limit: None seconds Memory Limit: None megabytes Problem Description: Giga Tower is the tallest and deepest building in Cyberland. There are 17<=777<=777<=777 floors, numbered from <=-<=8<=888<=888<=888 to 8<=888<=888<=888. In particular, there is floor 0 between floor <=-<=1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view. In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8<=888<=888<=888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8,<=<=-<=180,<=808 are all lucky while 42,<=<=-<=10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?). Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered *a*. He wants to find the minimum positive integer *b*, such that, if he walks *b* floors higher, he will arrive at a floor with a lucky number. Input Specification: The only line of input contains an integer *a* (<=-<=109<=≤<=*a*<=≤<=109). Output Specification: Print the minimum *b* in a line. Demo Input: ['179\n', '-1\n', '18\n'] Demo Output: ['1\n', '9\n', '10\n'] Note: For the first sample, he has to arrive at the floor numbered 180. For the second sample, he will arrive at 8. Note that *b* should be positive, so the answer for the third sample is 10, not 0.

```python n = input() for i in range(1,20): if "8" in str(int(n)+i): print(i) break ```

3

841

A

Generous Kefa

PROGRAMMING

900

[ "brute force", "implementation" ]

null

One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.

The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends. Next line contains string *s* — colors of baloons.

Answer to the task — «YES» or «NO» in a single line. You can choose the case (lower or upper) for each letter arbitrary.

[ "4 2\naabb\n", "6 3\naacaab\n" ]

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

In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second. In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».

500

[ { "input": "4 2\naabb", "output": "YES" }, { "input": "6 3\naacaab", "output": "NO" }, { "input": "2 2\nlu", "output": "YES" }, { "input": "5 3\novvoo", "output": "YES" }, { "input": "36 13\nbzbzcffczzcbcbzzfzbbfzfzzbfbbcbfccbf", "output": "YES" }, { "input": "81 3\nooycgmvvrophvcvpoupepqllqttwcocuilvyxbyumdmmfapvpnxhjhxfuagpnntonibicaqjvwfhwxhbv", "output": "NO" }, { "input": "100 100\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx", "output": "YES" }, { "input": "100 1\nnubcvvjvbjgnjsdkajimdcxvewbcytvfkihunycdrlconddlwgzjasjlsrttlrzsumzpyumpveglfqzmaofbshbojmwuwoxxvrod", "output": "NO" }, { "input": "100 13\nvyldolgryldqrvoldvzvrdrgorlorszddtgqvrlisxxrxdxlqtvtgsrqlzixoyrozxzogqxlsgzdddzqrgitxxritoolzolgrtvl", "output": "YES" }, { "input": "18 6\njzwtnkvmscqhmdlsxy", "output": "YES" }, { "input": "21 2\nfscegcqgzesefghhwcexs", "output": "NO" }, { "input": "32 22\ncduamsptaklqtxlyoutlzepxgyfkvngc", "output": "YES" }, { "input": "49 27\noxyorfnkzwsfllnyvdhdanppuzrnbxehugvmlkgeymqjlmfxd", "output": "YES" }, { "input": "50 24\nxxutzjwbggcwvxztttkmzovtmuwttzcbwoztttohzzxghuuthv", "output": "YES" }, { "input": "57 35\nglxshztrqqfyxthqamagvtmrdparhelnzrqvcwqxjytkbuitovkdxueul", "output": "YES" }, { "input": "75 23\nittttiiuitutuiiuuututiuttiuiuutuuuiuiuuuuttuuttuutuiiuiuiiuiitttuututuiuuii", "output": "NO" }, { "input": "81 66\nfeqevfqfebhvubhuuvfuqheuqhbeeuebehuvhffvbqvqvfbqqvvhevqffbqqhvvqhfeehuhqeqhueuqqq", "output": "YES" }, { "input": "93 42\npqeiafraiavfcteumflpcbpozcomlvpovlzdbldvoopnhdoeqaopzthiuzbzmeieiatthdeqovaqfipqlddllmfcrrnhb", "output": "YES" }, { "input": "100 53\nizszyqyndzwzyzgsdagdwdazadiawizinagqqgczaqqnawgijziziawzszdjdcqjdjqiwgadydcnqisaayjiqqsscwwzjzaycwwc", "output": "YES" }, { "input": "100 14\nvkrdcqbvkwuckpmnbydmczdxoagdsgtqxvhaxntdcxhjcrjyvukhugoglbmyoaqexgtcfdgemmizoniwtmisqqwcwfusmygollab", "output": "YES" }, { "input": "100 42\naaaaaiiiiaiiiaaiaiiaaiiiiiaaaaaiaiiiaiiiiaiiiaaaaaiiiaaaiiaaiiiaiiiaiaaaiaiiiiaaiiiaiiaiaiiaiiiaaaia", "output": "NO" }, { "input": "100 89\ntjbkmydejporbqhcbztkcumxjjgsrvxpuulbhzeeckkbchpbxwhedrlhjsabcexcohgdzouvsgphjdthpuqrlkgzxvqbuhqxdsmf", "output": "YES" }, { "input": "100 100\njhpyiuuzizhubhhpxbbhpyxzhbpjphzppuhiahihiappbhuypyauhizpbibzixjbzxzpbphuiaypyujappuxiyuyaajaxjupbahb", "output": "YES" }, { "input": "100 3\nsszoovvzysavsvzsozzvoozvysozsaszayaszasaysszzzysosyayyvzozovavzoyavsooaoyvoozvvozsaosvayyovazzszzssa", "output": "NO" }, { "input": "100 44\ndluthkxwnorabqsukgnxnvhmsmzilyulpursnxkdsavgemiuizbyzebhyjejgqrvuckhaqtuvdmpziesmpmewpvozdanjyvwcdgo", "output": "YES" }, { "input": "100 90\ntljonbnwnqounictqqctgonktiqoqlocgoblngijqokuquoolciqwnctgoggcbojtwjlculoikbggquqncittwnjbkgkgubnioib", "output": "YES" }, { "input": "100 79\nykxptzgvbqxlregvkvucewtydvnhqhuggdsyqlvcfiuaiddnrrnstityyehiamrggftsqyduwxpuldztyzgmfkehprrneyvtknmf", "output": "YES" }, { "input": "100 79\naagwekyovbviiqeuakbqbqifwavkfkutoriovgfmittulhwojaptacekdirgqoovlleeoqkkdukpadygfwavppohgdrmymmulgci", "output": "YES" }, { "input": "100 93\nearrehrehenaddhdnrdddhdahnadndheeennrearrhraharddreaeraddhehhhrdnredanndneheddrraaneerreedhnadnerhdn", "output": "YES" }, { "input": "100 48\nbmmaebaebmmmbbmxvmammbvvebvaemvbbaxvbvmaxvvmveaxmbbxaaemxmxvxxxvxbmmxaaaevvaxmvamvvmaxaxavexbmmbmmev", "output": "YES" }, { "input": "100 55\nhsavbkehaaesffaeeffakhkhfehbbvbeasahbbbvkesbfvkefeesesevbsvfkbffakvshsbkahfkfakebsvafkbvsskfhfvaasss", "output": "YES" }, { "input": "100 2\ncscffcffsccffsfsfffccssfsscfsfsssffcffsscfccssfffcfscfsscsccccfsssffffcfcfsfffcsfsccffscffcfccccfffs", "output": "NO" }, { "input": "100 3\nzrgznxgdpgfoiifrrrsjfuhvtqxjlgochhyemismjnanfvvpzzvsgajcbsulxyeoepjfwvhkqogiiwqxjkrpsyaqdlwffoockxnc", "output": "NO" }, { "input": "100 5\njbltyyfjakrjeodqepxpkjideulofbhqzxjwlarufwzwsoxhaexpydpqjvhybmvjvntuvhvflokhshpicbnfgsqsmrkrfzcrswwi", "output": "NO" }, { "input": "100 1\nfnslnqktlbmxqpvcvnemxcutebdwepoxikifkzaaixzzydffpdxodmsxjribmxuqhueifdlwzytxkklwhljswqvlejedyrgguvah", "output": "NO" }, { "input": "100 21\nddjenetwgwmdtjbpzssyoqrtirvoygkjlqhhdcjgeurqpunxpupwaepcqkbjjfhnvgpyqnozhhrmhfwararmlcvpgtnopvjqsrka", "output": "YES" }, { "input": "100 100\nnjrhiauqlgkkpkuvciwzivjbbplipvhslqgdkfnmqrxuxnycmpheenmnrglotzuyxycosfediqcuadklsnzjqzfxnbjwvfljnlvq", "output": "YES" }, { "input": "100 100\nbbbbbbbtbbttbtbbbttbttbtbbttttbbbtbttbbbtbttbtbbttttbbbbbtbbttbtbbtbttbbbtbtbtbtbtbtbbbttbbtbtbtbbtb", "output": "YES" }, { "input": "14 5\nfssmmsfffmfmmm", "output": "NO" }, { "input": "2 1\nff", "output": "NO" }, { "input": "2 1\nhw", "output": "YES" }, { "input": "2 2\nss", "output": "YES" }, { "input": "1 1\nl", "output": "YES" }, { "input": "100 50\nfffffttttttjjjuuuvvvvvdddxxxxwwwwgggbsssncccczzyyyyyhhhhhkrreeeeeeaaaaaiiillllllllooooqqqqqqmmpppppp", "output": "YES" }, { "input": "100 50\nbbbbbbbbgggggggggggaaaaaaaahhhhhhhhhhpppppppppsssssssrrrrrrrrllzzzzzzzeeeeeeekkkkkkkwwwwwwwwjjjjjjjj", "output": "YES" }, { "input": "100 50\nwwwwwwwwwwwwwwxxxxxxxxxxxxxxxxxxxxxxxxzzzzzzzzzzzzzzzzzzbbbbbbbbbbbbbbbbbbbbjjjjjjjjjjjjjjjjjjjjjjjj", "output": "YES" }, { "input": "100 80\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm", "output": "YES" }, { "input": "100 10\nbbttthhhhiiiiiiijjjjjvvvvpppssssseeeeeeewwwwgggkkkkkkkkmmmddddduuuzzzzllllnnnnnxxyyyffffccraaaaooooq", "output": "YES" }, { "input": "100 20\nssssssssssbbbbbbbhhhhhhhyyyyyyyzzzzzzzzzzzzcccccxxxxxxxxxxddddmmmmmmmeeeeeeejjjjjjjjjwwwwwwwtttttttt", "output": "YES" }, { "input": "1 2\na", "output": "YES" }, { "input": "3 1\nabb", "output": "NO" }, { "input": "2 1\naa", "output": "NO" }, { "input": "2 1\nab", "output": "YES" }, { "input": "6 2\naaaaaa", "output": "NO" }, { "input": "8 4\naaaaaaaa", "output": "NO" }, { "input": "4 2\naaaa", "output": "NO" }, { "input": "4 3\naaaa", "output": "NO" }, { "input": "1 3\na", "output": "YES" }, { "input": "4 3\nzzzz", "output": "NO" }, { "input": "4 1\naaaa", "output": "NO" }, { "input": "3 4\nabc", "output": "YES" }, { "input": "2 5\nab", "output": "YES" }, { "input": "2 4\nab", "output": "YES" }, { "input": "1 10\na", "output": "YES" }, { "input": "5 2\nzzzzz", "output": "NO" }, { "input": "53 26\naaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbb", "output": "NO" }, { "input": "4 1\nabab", "output": "NO" }, { "input": "4 1\nabcb", "output": "NO" }, { "input": "4 2\nabbb", "output": "NO" }, { "input": "5 2\nabccc", "output": "NO" }, { "input": "2 3\nab", "output": "YES" }, { "input": "4 3\nbbbs", "output": "YES" }, { "input": "10 2\nazzzzzzzzz", "output": "NO" }, { "input": "1 2\nb", "output": "YES" }, { "input": "1 3\nb", "output": "YES" }, { "input": "4 5\nabcd", "output": "YES" }, { "input": "4 6\naabb", "output": "YES" }, { "input": "5 2\naaaab", "output": "NO" }, { "input": "3 5\naaa", "output": "YES" }, { "input": "5 3\nazzzz", "output": "NO" }, { "input": "4 100\naabb", "output": "YES" }, { "input": "3 10\naaa", "output": "YES" }, { "input": "3 4\naaa", "output": "YES" }, { "input": "12 5\naaaaabbbbbbb", "output": "NO" }, { "input": "5 2\naabbb", "output": "NO" }, { "input": "10 5\nzzzzzzzzzz", "output": "NO" }, { "input": "2 4\naa", "output": "YES" }, { "input": "1 5\na", "output": "YES" }, { "input": "10 5\naaaaaaaaaa", "output": "NO" }, { "input": "6 3\naaaaaa", "output": "NO" }, { "input": "7 1\nabcdeee", "output": "NO" }, { "input": "18 3\naaaaaabbbbbbcccccc", "output": "NO" }, { "input": "8 2\naabbccdd", "output": "YES" }, { "input": "4 2\nzzzz", "output": "NO" }, { "input": "4 2\nabaa", "output": "NO" }, { "input": "3 2\naaa", "output": "NO" }, { "input": "3 1\nzzz", "output": "NO" }, { "input": "5 4\nzzzzz", "output": "NO" }, { "input": "6 2\naabbbc", "output": "NO" }, { "input": "3 6\naaa", "output": "YES" }, { "input": "2 1\nzz", "output": "NO" }, { "input": "10 3\naaaeeeeeee", "output": "NO" }, { "input": "4 5\naabb", "output": "YES" }, { "input": "3 1\naaa", "output": "NO" }, { "input": "5 2\naazzz", "output": "NO" }, { "input": "6 2\nabbbbc", "output": "NO" }, { "input": "4 2\nxxxx", "output": "NO" }, { "input": "6 3\nzzzzzz", "output": "NO" }, { "input": "3 2\nabb", "output": "YES" }, { "input": "3 2\nzzz", "output": "NO" }, { "input": "6 5\nzzzzzz", "output": "NO" }, { "input": "6 3\nbcaaaa", "output": "NO" }, { "input": "100 100\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "YES" }, { "input": "3 6\nabc", "output": "YES" } ]

1,601,864,071

2,147,483,647

Python 3

OK

TESTS

114

124

0

n,k = map(int,input().split()) l = input() if k >= n: print("YES") else: for x in l: t = l.count(x) if t > k: print("NO") break else: print("YES")

Title: Generous Kefa Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends. Next line contains string *s* — colors of baloons. Output Specification: Answer to the task — «YES» or «NO» in a single line. You can choose the case (lower or upper) for each letter arbitrary. Demo Input: ['4 2\naabb\n', '6 3\naacaab\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second. In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».

```python n,k = map(int,input().split()) l = input() if k >= n: print("YES") else: for x in l: t = l.count(x) if t > k: print("NO") break else: print("YES") ```

3

0

none

0

[ "none" ]

null

Santa Claus has Robot which lives on the infinite grid and can move along its lines. He can also, having a sequence of *m* points *p*1,<=*p*2,<=...,<=*p**m* with integer coordinates, do the following: denote its initial location by *p*0. First, the robot will move from *p*0 to *p*1 along one of the shortest paths between them (please notice that since the robot moves only along the grid lines, there can be several shortest paths). Then, after it reaches *p*1, it'll move to *p*2, again, choosing one of the shortest ways, then to *p*3, and so on, until he has visited all points in the given order. Some of the points in the sequence may coincide, in that case Robot will visit that point several times according to the sequence order. While Santa was away, someone gave a sequence of points to Robot. This sequence is now lost, but Robot saved the protocol of its unit movements. Please, find the minimum possible length of the sequence.

The first line of input contains the only positive integer *n* (1<=≤<=*n*<=≤<=2·105) which equals the number of unit segments the robot traveled. The second line contains the movements protocol, which consists of *n* letters, each being equal either L, or R, or U, or D. *k*-th letter stands for the direction which Robot traveled the *k*-th unit segment in: L means that it moved to the left, R — to the right, U — to the top and D — to the bottom. Have a look at the illustrations for better explanation.

The only line of input should contain the minimum possible length of the sequence.

[ "4\nRURD\n", "6\nRRULDD\n", "26\nRRRULURURUULULLLDLDDRDRDLD\n", "3\nRLL\n", "4\nLRLR\n" ]

[ "2\n", "2\n", "7\n", "2\n", "4\n" ]

The illustrations to the first three tests are given below. <img class="tex-graphics" src="https://espresso.codeforces.com/832fb8f97a482be815e0f87edde26c9791a0d330.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/119a8ba68772b2c2bf76f2acdc58027f6c5cde1f.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/c7b4534f24cbad48148bcba24bc44f37bf7a2dbf.png" style="max-width: 100.0%;max-height: 100.0%;"/> The last example illustrates that each point in the sequence should be counted as many times as it is presented in the sequence.

0

[ { "input": "4\nRURD", "output": "2" }, { "input": "6\nRRULDD", "output": "2" }, { "input": "26\nRRRULURURUULULLLDLDDRDRDLD", "output": "7" }, { "input": "3\nRLL", "output": "2" }, { "input": "4\nLRLR", "output": "4" }, { "input": "5\nLRDLR", "output": "4" }, { "input": "10\nDDRDUULUDD", "output": "3" }, { "input": "1\nD", "output": "1" } ]

1,482,659,151

2,451

PyPy 3

OK

TESTS

32

187

27,955,200

from sys import stdin n = int(stdin.readline()) s = stdin.readline().strip() x = 0 y = 0 xs = 0 ys = 0 ans = 0 l = 0 for ch in s: xp = x yp = y if ch == 'L': x -= 1 if ch == 'R': x += 1 if ch == 'U': y += 1 if ch == 'D': y -= 1 l += 1 if l > abs(x - xs) + abs(y - ys): xs = xp ys = yp l = 0 ans += 1 print(ans+1)

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Santa Claus has Robot which lives on the infinite grid and can move along its lines. He can also, having a sequence of *m* points *p*1,<=*p*2,<=...,<=*p**m* with integer coordinates, do the following: denote its initial location by *p*0. First, the robot will move from *p*0 to *p*1 along one of the shortest paths between them (please notice that since the robot moves only along the grid lines, there can be several shortest paths). Then, after it reaches *p*1, it'll move to *p*2, again, choosing one of the shortest ways, then to *p*3, and so on, until he has visited all points in the given order. Some of the points in the sequence may coincide, in that case Robot will visit that point several times according to the sequence order. While Santa was away, someone gave a sequence of points to Robot. This sequence is now lost, but Robot saved the protocol of its unit movements. Please, find the minimum possible length of the sequence. Input Specification: The first line of input contains the only positive integer *n* (1<=≤<=*n*<=≤<=2·105) which equals the number of unit segments the robot traveled. The second line contains the movements protocol, which consists of *n* letters, each being equal either L, or R, or U, or D. *k*-th letter stands for the direction which Robot traveled the *k*-th unit segment in: L means that it moved to the left, R — to the right, U — to the top and D — to the bottom. Have a look at the illustrations for better explanation. Output Specification: The only line of input should contain the minimum possible length of the sequence. Demo Input: ['4\nRURD\n', '6\nRRULDD\n', '26\nRRRULURURUULULLLDLDDRDRDLD\n', '3\nRLL\n', '4\nLRLR\n'] Demo Output: ['2\n', '2\n', '7\n', '2\n', '4\n'] Note: The illustrations to the first three tests are given below. <img class="tex-graphics" src="https://espresso.codeforces.com/832fb8f97a482be815e0f87edde26c9791a0d330.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/119a8ba68772b2c2bf76f2acdc58027f6c5cde1f.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/c7b4534f24cbad48148bcba24bc44f37bf7a2dbf.png" style="max-width: 100.0%;max-height: 100.0%;"/> The last example illustrates that each point in the sequence should be counted as many times as it is presented in the sequence.

```python from sys import stdin n = int(stdin.readline()) s = stdin.readline().strip() x = 0 y = 0 xs = 0 ys = 0 ans = 0 l = 0 for ch in s: xp = x yp = y if ch == 'L': x -= 1 if ch == 'R': x += 1 if ch == 'U': y += 1 if ch == 'D': y -= 1 l += 1 if l > abs(x - xs) + abs(y - ys): xs = xp ys = yp l = 0 ans += 1 print(ans+1) ```

3

981

A

Antipalindrome

PROGRAMMING

900

[ "brute force", "implementation", "strings" ]

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" } ]

1,567,688,810

2,147,483,647

Python 3

OK

TESTS

133

187

0

n = input() s = {x for x in n} r = n[::-1] print(0) if len(s) == 1 else print(len(n)-1) if r == n else print(len(n))

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 n = input() s = {x for x in n} r = n[::-1] print(0) if len(s) == 1 else print(len(n)-1) if r == n else print(len(n)) ```

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" } ]

1,670,427,662

2,147,483,647

PyPy 3-64

OK

TESTS

32

124

0

input() a = list(map(int, input().split())) j = list() t = list() for i in a: if i % 2 == 0: j.append(i) else: t.append(i) if len(j) == 1: print(a.index(j[0]) + 1) elif len(t) == 1: print(a.index(t[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 input() a = list(map(int, input().split())) j = list() t = list() for i in a: if i % 2 == 0: j.append(i) else: t.append(i) if len(j) == 1: print(a.index(j[0]) + 1) elif len(t) == 1: print(a.index(t[0]) + 1) ```

3.969

820

A

Mister B and Book Reading

PROGRAMMING

900

[ "implementation" ]

null

Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages. At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day. Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time. Help Mister B to calculate how many days he needed to finish the book.

First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading.

Print one integer — the number of days Mister B needed to finish the book.

[ "5 5 10 5 4\n", "12 4 12 4 1\n", "15 1 100 0 0\n" ]

[ "1\n", "3\n", "15\n" ]

In the first sample test the book contains 5 pages, so Mister B read it right at the first day. In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book. In third sample test every day Mister B read 1 page of the book, so he finished in 15 days.

500

[ { "input": "5 5 10 5 4", "output": "1" }, { "input": "12 4 12 4 1", "output": "3" }, { "input": "15 1 100 0 0", "output": "15" }, { "input": "1 1 1 0 0", "output": "1" }, { "input": "1000 999 1000 1000 998", "output": "2" }, { "input": "1000 2 2 5 1", "output": "999" }, { "input": "1000 1 1 1000 0", "output": "1000" }, { "input": "737 41 74 12 11", "output": "13" }, { "input": "1000 1000 1000 0 999", "output": "1" }, { "input": "765 12 105 5 7", "output": "17" }, { "input": "15 2 2 1000 0", "output": "8" }, { "input": "1000 1 1000 1000 0", "output": "2" }, { "input": "20 3 7 1 2", "output": "6" }, { "input": "1000 500 500 1000 499", "output": "501" }, { "input": "1 1000 1000 1000 0", "output": "1" }, { "input": "1000 2 1000 56 0", "output": "7" }, { "input": "1000 2 1000 802 0", "output": "3" }, { "input": "16 1 8 2 0", "output": "4" }, { "input": "20 6 10 2 2", "output": "3" }, { "input": "8 2 12 4 1", "output": "3" }, { "input": "8 6 13 2 5", "output": "2" }, { "input": "70 4 20 87 0", "output": "5" }, { "input": "97 8 13 234 5", "output": "13" }, { "input": "16 4 23 8 3", "output": "3" }, { "input": "65 7 22 7 4", "output": "5" }, { "input": "93 10 18 11 7", "output": "9" }, { "input": "86 13 19 15 9", "output": "9" }, { "input": "333 17 50 10 16", "output": "12" }, { "input": "881 16 55 10 12", "output": "23" }, { "input": "528 11 84 3 9", "output": "19" }, { "input": "896 2 184 8 1", "output": "16" }, { "input": "236 10 930 9 8", "output": "8" }, { "input": "784 1 550 14 0", "output": "12" }, { "input": "506 1 10 4 0", "output": "53" }, { "input": "460 1 3 2 0", "output": "154" }, { "input": "701 1 3 1 0", "output": "235" }, { "input": "100 49 50 1000 2", "output": "3" }, { "input": "100 1 100 100 0", "output": "2" }, { "input": "12 1 4 2 0", "output": "4" }, { "input": "22 10 12 0 0", "output": "3" }, { "input": "20 10 15 1 4", "output": "3" }, { "input": "1000 5 10 1 4", "output": "169" }, { "input": "1000 1 1000 1 0", "output": "45" }, { "input": "4 1 2 2 0", "output": "3" }, { "input": "1 5 5 1 1", "output": "1" }, { "input": "19 10 11 0 2", "output": "3" }, { "input": "1 2 3 0 0", "output": "1" }, { "input": "10 1 4 10 0", "output": "4" }, { "input": "20 3 100 1 1", "output": "5" }, { "input": "1000 5 9 5 0", "output": "112" }, { "input": "1 11 12 0 10", "output": "1" }, { "input": "1 1 1 1 0", "output": "1" }, { "input": "1000 1 20 1 0", "output": "60" }, { "input": "9 1 4 2 0", "output": "4" }, { "input": "129 2 3 4 0", "output": "44" }, { "input": "4 2 2 0 1", "output": "3" }, { "input": "1000 1 10 100 0", "output": "101" }, { "input": "100 1 100 1 0", "output": "14" }, { "input": "8 3 4 2 0", "output": "3" }, { "input": "20 1 6 4 0", "output": "5" }, { "input": "8 2 4 2 0", "output": "3" }, { "input": "11 5 6 7 2", "output": "3" }, { "input": "100 120 130 120 0", "output": "1" }, { "input": "7 1 4 1 0", "output": "4" }, { "input": "5 3 10 0 2", "output": "3" }, { "input": "5 2 2 0 0", "output": "3" }, { "input": "1000 10 1000 10 0", "output": "14" }, { "input": "25 3 50 4 2", "output": "4" }, { "input": "9 10 10 10 9", "output": "1" }, { "input": "17 10 12 6 5", "output": "2" }, { "input": "15 5 10 3 0", "output": "3" }, { "input": "8 3 5 1 0", "output": "3" }, { "input": "19 1 12 5 0", "output": "4" }, { "input": "1000 10 1000 1 0", "output": "37" }, { "input": "100 1 2 1000 0", "output": "51" }, { "input": "20 10 11 1000 9", "output": "6" }, { "input": "16 2 100 1 1", "output": "5" }, { "input": "18 10 13 2 5", "output": "3" }, { "input": "12 3 5 3 1", "output": "4" }, { "input": "17 3 11 2 0", "output": "4" }, { "input": "4 2 100 1 1", "output": "2" }, { "input": "7 4 5 2 3", "output": "3" }, { "input": "100 1 2 2 0", "output": "51" }, { "input": "50 4 5 5 0", "output": "11" }, { "input": "1 2 2 0 1", "output": "1" }, { "input": "1000 2 3 10 1", "output": "500" }, { "input": "500 10 500 1000 0", "output": "2" }, { "input": "1000 4 12 1 0", "output": "87" }, { "input": "18 10 13 1 5", "output": "3" }, { "input": "7 3 6 2 2", "output": "3" }, { "input": "15 5 100 1 2", "output": "4" }, { "input": "100 1 10 1 0", "output": "15" }, { "input": "8 2 7 5 1", "output": "2" }, { "input": "11 2 4 1 1", "output": "5" }, { "input": "1000 500 900 100 300", "output": "3" }, { "input": "7 1 2 5 0", "output": "4" }, { "input": "7 3 5 3 2", "output": "3" }, { "input": "7 3 10 2 1", "output": "2" }, { "input": "1000 501 510 1 499", "output": "50" }, { "input": "1000 1 1000 2 0", "output": "32" }, { "input": "1 5 5 0 0", "output": "1" }, { "input": "18 10 15 1 5", "output": "3" }, { "input": "100 4 1000 1 2", "output": "13" }, { "input": "20 2 40 1 1", "output": "6" }, { "input": "1 11 1000 100 1", "output": "1" }, { "input": "6 4 4 1 2", "output": "2" }, { "input": "8 3 5 3 1", "output": "3" }, { "input": "10 5 7 1 2", "output": "3" }, { "input": "400 100 198 1 99", "output": "25" }, { "input": "3 1 2 5 0", "output": "2" } ]

1,499,530,406

1,106

Python 3

OK

TESTS

110

62

5,529,600

a = input() b = a.split(" ") c, v0, v1, a, l = int(b[0]), int(b[1]), int(b[2]), int(b[3]), int(b[4]) day = 1 while c > 0: if day > 1: c = c + l c = c - min(v0 + a * (day - 1), v1) if c <= 0: break day += 1 print(day)

Title: Mister B and Book Reading Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages. At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day. Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time. Help Mister B to calculate how many days he needed to finish the book. Input Specification: First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading. Output Specification: Print one integer — the number of days Mister B needed to finish the book. Demo Input: ['5 5 10 5 4\n', '12 4 12 4 1\n', '15 1 100 0 0\n'] Demo Output: ['1\n', '3\n', '15\n'] Note: In the first sample test the book contains 5 pages, so Mister B read it right at the first day. In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book. In third sample test every day Mister B read 1 page of the book, so he finished in 15 days.

```python a = input() b = a.split(" ") c, v0, v1, a, l = int(b[0]), int(b[1]), int(b[2]), int(b[3]), int(b[4]) day = 1 while c > 0: if day > 1: c = c + l c = c - min(v0 + a * (day - 1), v1) if c <= 0: break day += 1 print(day) ```

3

753

A

Santa Claus and Candies

PROGRAMMING

1,000

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

null

Santa Claus has *n* candies, he dreams to give them as gifts to children. What is the maximal number of children for whose he can give candies if Santa Claus want each kid should get distinct positive integer number of candies. Santa Class wants to give all *n* candies he has.

The only line contains positive integer number *n* (1<=≤<=*n*<=≤<=1000) — number of candies Santa Claus has.

Print to the first line integer number *k* — maximal number of kids which can get candies. Print to the second line *k* distinct integer numbers: number of candies for each of *k* kid. The sum of *k* printed numbers should be exactly *n*. If there are many solutions, print any of them.

[ "5\n", "9\n", "2\n" ]

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

none

500

[ { "input": "5", "output": "2\n1 4 " }, { "input": "9", "output": "3\n1 2 6 " }, { "input": "2", "output": "1\n2 " }, { "input": "1", "output": "1\n1 " }, { "input": "3", "output": "2\n1 2 " }, { "input": "1000", "output": "44\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 54 " }, { "input": "4", "output": "2\n1 3 " }, { "input": "6", "output": "3\n1 2 3 " }, { "input": "7", "output": "3\n1 2 4 " }, { "input": "8", "output": "3\n1 2 5 " }, { "input": "10", "output": "4\n1 2 3 4 " }, { "input": "11", "output": "4\n1 2 3 5 " }, { "input": "12", "output": "4\n1 2 3 6 " }, { "input": "13", "output": "4\n1 2 3 7 " }, { "input": "14", "output": "4\n1 2 3 8 " }, { "input": "15", "output": "5\n1 2 3 4 5 " }, { "input": "16", "output": "5\n1 2 3 4 6 " }, { "input": "20", "output": "5\n1 2 3 4 10 " }, { "input": "21", "output": "6\n1 2 3 4 5 6 " }, { "input": "22", "output": "6\n1 2 3 4 5 7 " }, { "input": "27", "output": "6\n1 2 3 4 5 12 " }, { "input": "28", "output": "7\n1 2 3 4 5 6 7 " }, { "input": "29", "output": "7\n1 2 3 4 5 6 8 " }, { "input": "35", "output": "7\n1 2 3 4 5 6 14 " }, { "input": "36", "output": "8\n1 2 3 4 5 6 7 8 " }, { "input": "37", "output": "8\n1 2 3 4 5 6 7 9 " }, { "input": "44", "output": "8\n1 2 3 4 5 6 7 16 " }, { "input": "45", "output": "9\n1 2 3 4 5 6 7 8 9 " }, { "input": "46", "output": "9\n1 2 3 4 5 6 7 8 10 " }, { "input": "230", "output": "20\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 40 " }, { "input": "231", "output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 " }, { "input": "232", "output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 " }, { "input": "239", "output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 29 " }, { "input": "629", "output": "34\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 68 " }, { "input": "630", "output": "35\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 " }, { "input": "631", "output": "35\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 36 " }, { "input": "945", "output": "42\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 84 " }, { "input": "946", "output": "43\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 " }, { "input": "947", "output": "43\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 44 " }, { "input": "989", "output": "43\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 86 " }, { "input": "990", "output": "44\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 " }, { "input": "991", "output": "44\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 45 " }, { "input": "956", "output": "43\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 53 " }, { "input": "981", "output": "43\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 78 " }, { "input": "867", "output": "41\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 47 " }, { "input": "906", "output": "42\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 45 " }, { "input": "999", "output": "44\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 53 " }, { "input": "100", "output": "13\n1 2 3 4 5 6 7 8 9 10 11 12 22 " }, { "input": "126", "output": "15\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 21 " } ]

1,568,563,150

2,147,483,647

PyPy 3

OK

TESTS

49

155

0

n = int(input()) answer = [] s, i = 0, 1 while s < n: answer.append(i) s += i i += 1 d = s - n answer = list(filter(lambda x: x != d, answer)) print(len(answer)) print(' '.join(map(str, answer)))

Title: Santa Claus and Candies Time Limit: None seconds Memory Limit: None megabytes Problem Description: Santa Claus has *n* candies, he dreams to give them as gifts to children. What is the maximal number of children for whose he can give candies if Santa Claus want each kid should get distinct positive integer number of candies. Santa Class wants to give all *n* candies he has. Input Specification: The only line contains positive integer number *n* (1<=≤<=*n*<=≤<=1000) — number of candies Santa Claus has. Output Specification: Print to the first line integer number *k* — maximal number of kids which can get candies. Print to the second line *k* distinct integer numbers: number of candies for each of *k* kid. The sum of *k* printed numbers should be exactly *n*. If there are many solutions, print any of them. Demo Input: ['5\n', '9\n', '2\n'] Demo Output: ['2\n2 3\n', '3\n3 5 1\n', '1\n2 \n'] Note: none

```python n = int(input()) answer = [] s, i = 0, 1 while s < n: answer.append(i) s += i i += 1 d = s - n answer = list(filter(lambda x: x != d, answer)) print(len(answer)) print(' '.join(map(str, answer))) ```

3

352

A

Jeff and Digits

PROGRAMMING

1,000

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

null

Jeff's got *n* cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got? Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards.

The first line contains integer *n* (1<=≤<=*n*<=≤<=103). The next line contains *n* integers *a*1, *a*2, ..., *a**n* (*a**i*<==<=0 or *a**i*<==<=5). Number *a**i* represents the digit that is written on the *i*-th card.

In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1.

[ "4\n5 0 5 0\n", "11\n5 5 5 5 5 5 5 5 0 5 5\n" ]

[ "0\n", "5555555550\n" ]

In the first test you can make only one number that is a multiple of 90 — 0. In the second test you can make number 5555555550, it is a multiple of 90.

500

[ { "input": "4\n5 0 5 0", "output": "0" }, { "input": "11\n5 5 5 5 5 5 5 5 0 5 5", "output": "5555555550" }, { "input": "7\n5 5 5 5 5 5 5", "output": "-1" }, { "input": "1\n5", "output": "-1" }, { "input": "1\n0", "output": "0" }, { "input": "11\n5 0 5 5 5 0 0 5 5 5 5", "output": "0" }, { "input": "23\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 0 0 0 0 0", "output": "55555555555555555500000" }, { "input": "9\n5 5 5 5 5 5 5 5 5", "output": "-1" }, { "input": "24\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 0 0 0 0 0", "output": "55555555555555555500000" }, { "input": "10\n0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "10\n5 5 5 5 5 0 0 5 0 5", "output": "0" }, { "input": "3\n5 5 0", "output": "0" }, { "input": "5\n5 5 0 5 5", "output": "0" }, { "input": "14\n0 5 5 0 0 0 0 0 0 5 5 5 5 5", "output": "0" }, { "input": "3\n5 5 5", "output": "-1" }, { "input": "3\n0 5 5", "output": "0" }, { "input": "13\n0 0 5 0 5 0 5 5 0 0 0 0 0", "output": "0" }, { "input": "9\n5 5 0 5 5 5 5 5 5", "output": "0" }, { "input": "8\n0 0 0 0 0 0 0 0", "output": "0" }, { "input": "101\n5 0 0 0 0 0 0 0 5 0 0 0 0 5 0 0 5 0 0 0 0 0 5 0 0 0 0 0 0 0 0 5 0 0 5 0 0 0 0 0 0 0 5 0 0 5 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 5 0 0 0 0 0 0 0 0 0 5 0 0 5 0 0 0 0 5 0 0", "output": "5555555550000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "214\n5 0 5 0 5 0 0 0 5 5 0 5 0 5 5 0 5 0 0 0 0 5 5 0 0 5 5 0 0 0 0 5 5 5 5 0 5 0 0 0 0 0 0 5 0 0 0 5 0 0 5 0 0 5 5 0 0 5 5 0 0 0 0 0 5 0 5 0 5 5 0 5 0 0 5 5 5 0 5 0 5 0 5 5 0 5 0 0 0 5 5 0 5 0 5 5 5 5 5 0 0 0 0 0 0 5 0 5 5 0 5 0 5 0 5 5 0 0 0 0 5 0 5 0 5 0 0 5 0 0 5 5 5 5 5 0 0 5 0 0 5 0 0 5 0 0 5 0 0 5 0 5 0 0 0 5 0 0 5 5 5 0 0 5 5 5 0 0 5 5 0 0 0 5 0 0 5 5 5 5 5 5 0 5 0 0 5 5 5 5 0 5 5 0 0 0 5 5 5 5 0 0 0 0 5 0 0 5 0 0 5 5 0 0", "output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555550000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "80\n0 0 0 0 5 0 5 5 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 5 0 0 0 0 0 0 0 0 0 5 5 0 5 0 0 0 0 0 0 5 0 0 0 0 0 0 0 5 0 0 0 0 5 0 5 5 0 0 0", "output": "555555555000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "2\n0 0", "output": "0" }, { "input": "3\n5 0 0", "output": "0" }, { "input": "4\n5 5 5 5", "output": "-1" }, { "input": "2\n0 5", "output": "0" }, { "input": "14\n5 5 5 5 5 5 5 5 5 5 5 5 5 0", "output": "5555555550" }, { "input": "18\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5", "output": "-1" }, { "input": "10\n5 5 5 5 5 5 5 5 5 0", "output": "5555555550" }, { "input": "10\n5 5 5 5 5 5 5 5 5 5", "output": "-1" }, { "input": "20\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5", "output": "-1" } ]

1,650,095,461

2,147,483,647

PyPy 3

OK

TESTS

42

184

0

input();a=input().split() print('5'*(a.count('5')//9*9)+'0'*a.count('0') if a.count('5')>8 and a.count('0') else -(a.count('0')==0))

Title: Jeff and Digits Time Limit: None seconds Memory Limit: None megabytes Problem Description: Jeff's got *n* cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got? Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=103). The next line contains *n* integers *a*1, *a*2, ..., *a**n* (*a**i*<==<=0 or *a**i*<==<=5). Number *a**i* represents the digit that is written on the *i*-th card. Output Specification: In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1. Demo Input: ['4\n5 0 5 0\n', '11\n5 5 5 5 5 5 5 5 0 5 5\n'] Demo Output: ['0\n', '5555555550\n'] Note: In the first test you can make only one number that is a multiple of 90 — 0. In the second test you can make number 5555555550, it is a multiple of 90.

```python input();a=input().split() print('5'*(a.count('5')//9*9)+'0'*a.count('0') if a.count('5')>8 and a.count('0') else -(a.count('0')==0)) ```

3

659

C

Tanya and Toys

PROGRAMMING

1,200

[ "greedy", "implementation" ]

null

In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new collection of the *i*-th type costs *i* bourles. Tania has managed to collect *n* different types of toys *a*1,<=*a*2,<=...,<=*a**n* from the new collection. Today is Tanya's birthday, and her mother decided to spend no more than *m* bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has. Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this.

The first line contains two integers *n* (1<=≤<=*n*<=≤<=100<=000) and *m* (1<=≤<=*m*<=≤<=109) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys. The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the types of toys that Tanya already has.

In the first line print a single integer *k* — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed *m*. In the second line print *k* distinct space-separated integers *t*1,<=*t*2,<=...,<=*t**k* (1<=≤<=*t**i*<=≤<=109) — the types of toys that Tanya should choose. If there are multiple answers, you may print any of them. Values of *t**i* can be printed in any order.

[ "3 7\n1 3 4\n", "4 14\n4 6 12 8\n" ]

[ "2\n2 5 \n", "4\n7 2 3 1\n" ]

In the first sample mom should buy two toys: one toy of the 2-nd type and one toy of the 5-th type. At any other purchase for 7 bourles (assuming that the toys of types 1, 3 and 4 have already been bought), it is impossible to buy two and more toys.

1,000

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135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "1 999999119\n34421", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "1 1000000000\n1", "output": "44719\n2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..." }, { "input": "1 1000000000\n44720", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "1 1000000000\n44719", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "1 1000000000\n44721", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "3 1000000000\n123456789 234567891 345678912", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 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126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..." }, { "input": "3 100000\n1000000000 100000000 1", "output": "445\n2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 ..." }, { "input": "5 5\n100000000 200000000 300000000 400000000 1000000000", "output": "2\n1 2 " }, { "input": "6 3\n1 2 3 4 5 6", "output": "0" }, { "input": "2 1\n1 2", "output": "0" }, { "input": "1 1000000000\n1000000000", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "5 1000000\n1000000000 100000000 10000000 99999999 123456789", "output": "1413\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..." }, { "input": "2 10000000\n1234567 123456", "output": "4471\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..." }, { "input": "1 1\n1000000000", "output": "1\n1 " }, { "input": "1 1000000000\n9999999", "output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..." }, { "input": "5 10000\n1000000000 888888888 777777777 666666666 959595959", "output": "140\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 " }, { "input": "3 1\n1000000000 999999999 999999998", "output": "1\n1 " }, { "input": "5 100000000\n100000000 999999999 1 2 3", "output": "14138\n4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 15..." }, { "input": "3 55\n100000000 1000000000 999999999", "output": "10\n1 2 3 4 5 6 7 8 9 10 " }, { "input": "2 10\n5 10000009", "output": "4\n1 2 3 4 " }, { "input": "3 10000000\n999999999 999999998 999999997", "output": "4471\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..." }, { "input": "1 1100\n1000000000", "output": "46\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 " }, { "input": "1 40\n1000000000", "output": "8\n1 2 3 4 5 6 7 8 " } ]

1,645,810,400

2,147,483,647

Python 3

OK

TESTS

61

139

10,240,000

def main(): n, k = map(int, input().split()) arr = set(map(int, input().split())) i = 1 s = 0 answer_arr = [] while True: if i not in arr: if s + i > k: break else: s += i answer_arr.append(i) i += 1 else: i += 1 print(len(answer_arr)) print(*answer_arr) if __name__ == "__main__": main()

Title: Tanya and Toys Time Limit: None seconds Memory Limit: None megabytes Problem Description: In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new collection of the *i*-th type costs *i* bourles. Tania has managed to collect *n* different types of toys *a*1,<=*a*2,<=...,<=*a**n* from the new collection. Today is Tanya's birthday, and her mother decided to spend no more than *m* bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has. Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this. Input Specification: The first line contains two integers *n* (1<=≤<=*n*<=≤<=100<=000) and *m* (1<=≤<=*m*<=≤<=109) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys. The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the types of toys that Tanya already has. Output Specification: In the first line print a single integer *k* — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed *m*. In the second line print *k* distinct space-separated integers *t*1,<=*t*2,<=...,<=*t**k* (1<=≤<=*t**i*<=≤<=109) — the types of toys that Tanya should choose. If there are multiple answers, you may print any of them. Values of *t**i* can be printed in any order. Demo Input: ['3 7\n1 3 4\n', '4 14\n4 6 12 8\n'] Demo Output: ['2\n2 5 \n', '4\n7 2 3 1\n'] Note: In the first sample mom should buy two toys: one toy of the 2-nd type and one toy of the 5-th type. At any other purchase for 7 bourles (assuming that the toys of types 1, 3 and 4 have already been bought), it is impossible to buy two and more toys.

```python def main(): n, k = map(int, input().split()) arr = set(map(int, input().split())) i = 1 s = 0 answer_arr = [] while True: if i not in arr: if s + i > k: break else: s += i answer_arr.append(i) i += 1 else: i += 1 print(len(answer_arr)) print(*answer_arr) if __name__ == "__main__": main() ```

3

69

A

Young Physicist

PROGRAMMING

1,000

[ "implementation", "math" ]

A. Young Physicist

2

256

A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.

The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).

Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.

[ "3\n4 1 7\n-2 4 -1\n1 -5 -3\n", "3\n3 -1 7\n-5 2 -4\n2 -1 -3\n" ]

[ "NO", "YES" ]

none

500

[ { "input": "3\n4 1 7\n-2 4 -1\n1 -5 -3", "output": "NO" }, { "input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3", "output": "YES" }, { "input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41", "output": "NO" }, { "input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33", "output": "NO" }, { "input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7", "output": "NO" }, { "input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38", "output": "NO" }, { "input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6", "output": "YES" }, { "input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0", "output": "YES" }, { "input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0", "output": "YES" }, { "input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0", "output": "YES" }, { "input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0", "output": "YES" }, { "input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0", "output": "YES" }, { "input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87", "output": "YES" }, { "input": "3\n1 2 3\n3 2 1\n0 0 0", "output": "NO" }, { "input": "2\n5 -23 12\n0 0 0", "output": "NO" }, { "input": "1\n0 0 0", "output": "YES" }, { "input": "1\n1 -2 0", "output": "NO" }, { "input": "2\n-23 77 -86\n23 -77 86", "output": "YES" }, { "input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82", "output": "YES" }, { "input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0", "output": "YES" }, { "input": "3\n96 49 -12\n2 -66 28\n-98 17 -16", "output": "YES" }, { "input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0", "output": "YES" }, { "input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53", "output": "YES" }, { "input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1", "output": "YES" }, { "input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0", "output": "YES" }, { "input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0", "output": "YES" }, { "input": "3\n0 2 -2\n1 -1 3\n-3 0 0", "output": "NO" } ]

1,601,626,233

2,147,483,647

Python 3

OK

TESTS

81

218

307,200

x=int(input()) a,b,c=0,0,0 l=[] for i in range(x): l.append(input().split(' ')) for i in range(x): a+=int(l[i][0]) b+=int(l[i][1]) c+=int(l[i][2]) print(['YES','NO'][a!=0 or b!=0 or c!=0])

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 x=int(input()) a,b,c=0,0,0 l=[] for i in range(x): l.append(input().split(' ')) for i in range(x): a+=int(l[i][0]) b+=int(l[i][1]) c+=int(l[i][2]) print(['YES','NO'][a!=0 or b!=0 or c!=0]) ```

3.944928

66

A

Petya and Java

PROGRAMMING

1,300

[ "implementation", "strings" ]

A. Petya and Java

2

256

Little Petya has recently started attending a programming club. Naturally he is facing the problem of choosing a programming language. After long considerations he realized that Java is the best choice. The main argument in favor of choosing Java was that it has a very large integer data type, called BigInteger. But having attended several classes of the club, Petya realized that not all tasks require using the BigInteger type. It turned out that in some tasks it is much easier to use small data types. That's why a question arises: "Which integer type to use if one wants to store a positive integer *n*?" Petya knows only 5 integer types: 1) byte occupies 1 byte and allows you to store numbers from <=-<=128 to 127 2) short occupies 2 bytes and allows you to store numbers from <=-<=32768 to 32767 3) int occupies 4 bytes and allows you to store numbers from <=-<=2147483648 to 2147483647 4) long occupies 8 bytes and allows you to store numbers from <=-<=9223372036854775808 to 9223372036854775807 5) BigInteger can store any integer number, but at that it is not a primitive type, and operations with it are much slower. For all the types given above the boundary values are included in the value range. From this list, Petya wants to choose the smallest type that can store a positive integer *n*. Since BigInteger works much slower, Peter regards it last. Help him.

The first line contains a positive number *n*. It consists of no more than 100 digits and doesn't contain any leading zeros. The number *n* can't be represented as an empty string. 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).

Print the first type from the list "byte, short, int, long, BigInteger", that can store the natural number *n*, in accordance with the data given above.

[ "127\n", "130\n", "123456789101112131415161718192021222324\n" ]

[ "byte\n", "short\n", "BigInteger\n" ]

none

500

[ { "input": "127", "output": "byte" }, { "input": "130", "output": "short" }, { "input": "123456789101112131415161718192021222324", "output": "BigInteger" }, { "input": "6", "output": "byte" }, { "input": "16", "output": "byte" }, { "input": "126", "output": "byte" }, { "input": "128", "output": "short" }, { "input": "32766", "output": "short" }, { "input": "111111", "output": "int" }, { "input": "22222", "output": "short" }, { "input": "32767", "output": "short" }, { "input": "32768", "output": "int" }, { "input": "32769", "output": "int" }, { "input": "2147483645", "output": "int" }, { "input": "2147483646", "output": "int" }, { "input": "2147483647", "output": "int" }, { "input": "2147483648", "output": "long" }, { "input": "2147483649", "output": "long" }, { "input": "9223372036854775805", "output": "long" }, { "input": "9223372036854775806", "output": "long" }, { "input": "9223372036854775807", "output": "long" }, { "input": "9223372036854775808", "output": "BigInteger" }, { "input": "9223372036854775809", "output": "BigInteger" }, { "input": "1111111111111111111111111111111111111111111111", "output": "BigInteger" }, { "input": "232", "output": "short" }, { "input": "241796563564014133460267652699", "output": "BigInteger" }, { "input": "29360359146807441660707083821018832188095237636414144034857851003419752010124705615779249", "output": "BigInteger" }, { "input": "337300529263821789926982715723773719445001702036602052198530564", "output": "BigInteger" }, { "input": "381127467969689863953686682245136076127159921", "output": "BigInteger" }, { "input": "2158324958633591462", "output": "long" }, { "input": "268659422768117401499491767189496733446324586965055954729177892248858259490346", "output": "BigInteger" }, { "input": "3023764505449745844381036446038799100004717936344985", "output": "BigInteger" }, { "input": "13408349824892484976400774", "output": "BigInteger" }, { "input": "18880842614378213198381172973704766723997934818440985546083314104481253291692101136681", "output": "BigInteger" }, { "input": "1180990956946757129733650596194933741", "output": "BigInteger" }, { "input": "73795216631038776655609800540262114612084443385902708034055020082090470662930545328551", "output": "BigInteger" }, { "input": "1658370691480968202384509492140362150472696196949", "output": "BigInteger" }, { "input": "59662093286671707493190399502717308574459619342109544431740791973099298641871347858082458491958703", "output": "BigInteger" }, { "input": "205505005582428018613354752739589866670902346355933720701937", "output": "BigInteger" }, { "input": "53348890623013817139699", "output": "BigInteger" }, { "input": "262373979958859125198440634134122707574734706745701184688685117904709744", "output": "BigInteger" }, { "input": "69113784278456828987289369893745977", "output": "BigInteger" }, { "input": "2210209454022702335652564247406666491086662454147967686455330365147159266087", "output": "BigInteger" }, { "input": "630105816139991597267787581532092408135", "output": "BigInteger" }, { "input": "800461429306907809762708270", "output": "BigInteger" }, { "input": "7685166910821197056344900917707673568669808490600751439157", "output": "BigInteger" }, { "input": "713549841568602590705962611607726022334779480510421458817648621376683672722573289661127894", "output": "BigInteger" }, { "input": "680504312323996476676434432", "output": "BigInteger" }, { "input": "3376595620091080825479292544658464163405755746884100218035", "output": "BigInteger" }, { "input": "303681723783491968617491075591006152690484825330764215796396316561122383310011589365655481", "output": "BigInteger" }, { "input": "4868659422768117401499491767189496733446324586965055954729177892248858259490346614099717639491763430", "output": "BigInteger" }, { "input": "3502376450544974584438103644603879910000471793634498544789130945841846713263971487355748226237288709", "output": "BigInteger" }, { "input": "2334083498248924849764007740114454487565621932425948046430072197452845278935316358800789014185793377", "output": "BigInteger" }, { "input": "1988808426143782131983811729737047667239979348184409855460833141044812532916921011366813880911319644", "output": "BigInteger" }, { "input": "1018099095694675712973365059619493374113337270925179793757322992466016001294122941535439492265169131", "output": "BigInteger" }, { "input": "8437952166310387766556098005402621146120844433859027080340550200820904706629305453285512716464931911", "output": "BigInteger" }, { "input": "6965837069148096820238450949214036215047269619694967357734070611376013382163559966747678150791825071", "output": "BigInteger" }, { "input": "4596620932866717074931903995027173085744596193421095444317407919730992986418713478580824584919587030", "output": "BigInteger" }, { "input": "1905505005582428018613354752739589866670902346355933720701937408006000562951996789032987808118459990", "output": "BigInteger" }, { "input": "8433488906230138171396997888322916936677429522910871017295155818340819168386140293774243244435122950", "output": "BigInteger" }, { "input": "6862373979958859125198440634134122707574734706745701184688685117904709744830303784215298687654884810", "output": "BigInteger" }, { "input": "4491137842784568289872893698937459777201151060689848471272003426250808340375567208957554901863756992", "output": "BigInteger" }, { "input": "9721020945402270233565256424740666649108666245414796768645533036514715926608741510409618545180420952", "output": "BigInteger" }, { "input": "7330105816139991597267787581532092408135003429259616955239761315950805521264994021242873979309182812", "output": "BigInteger" }, { "input": "2000461429306907809762708270752707617318091579531521957022940951538737203583768926365382290530636885", "output": "BigInteger" }, { "input": "9868516691082119705634490091770767356866980849060075143915700796802700437762260163478754592094654326", "output": "BigInteger" }, { "input": "8713549841568602590705962611607726022334779480510421458817648621376683672722573289661127894678771177", "output": "BigInteger" }, { "input": "4580504312323996476676434432646986768872786931159974634901608445720467716981185426771899006352697916", "output": "BigInteger" }, { "input": "2537659562009108082547929254465846416340575574688410021803548570097340949141688442074263189944614467", "output": "BigInteger" }, { "input": "1403681723783491968617491075591006152690484825330764215796396316561122383310011589365655481428540208", "output": "BigInteger" }, { "input": "26", "output": "byte" }, { "input": "302376450544", "output": "long" }, { "input": "13", "output": "byte" }, { "input": "188808426143", "output": "long" }, { "input": "118099095694675", "output": "long" }, { "input": "73795216631038", "output": "long" }, { "input": "1658370691480", "output": "long" }, { "input": "596620932866", "output": "long" }, { "input": "2055050055", "output": "int" }, { "input": "533488906", "output": "int" }, { "input": "26237397", "output": "int" }, { "input": "6911378", "output": "int" }, { "input": "221020945402270233", "output": "long" }, { "input": "63010581613999159", "output": "long" }, { "input": "80046142930", "output": "long" }, { "input": "7685166910821197", "output": "long" }, { "input": "71", "output": "byte" }, { "input": "6805043123239964766", "output": "long" }, { "input": "3376", "output": "short" }, { "input": "3036817237", "output": "long" } ]

1,643,272,038

2,147,483,647

Python 3

OK

TESTS

90

154

0

n=int(input()) if n<=127: print("byte") elif n<=32767: print("short") elif n<=2147483647: print("int") elif n<=9223372036854775807: print("long") else: print("BigInteger")

Title: Petya and Java Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Little Petya has recently started attending a programming club. Naturally he is facing the problem of choosing a programming language. After long considerations he realized that Java is the best choice. The main argument in favor of choosing Java was that it has a very large integer data type, called BigInteger. But having attended several classes of the club, Petya realized that not all tasks require using the BigInteger type. It turned out that in some tasks it is much easier to use small data types. That's why a question arises: "Which integer type to use if one wants to store a positive integer *n*?" Petya knows only 5 integer types: 1) byte occupies 1 byte and allows you to store numbers from <=-<=128 to 127 2) short occupies 2 bytes and allows you to store numbers from <=-<=32768 to 32767 3) int occupies 4 bytes and allows you to store numbers from <=-<=2147483648 to 2147483647 4) long occupies 8 bytes and allows you to store numbers from <=-<=9223372036854775808 to 9223372036854775807 5) BigInteger can store any integer number, but at that it is not a primitive type, and operations with it are much slower. For all the types given above the boundary values are included in the value range. From this list, Petya wants to choose the smallest type that can store a positive integer *n*. Since BigInteger works much slower, Peter regards it last. Help him. Input Specification: The first line contains a positive number *n*. It consists of no more than 100 digits and doesn't contain any leading zeros. The number *n* can't be represented as an empty string. 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). Output Specification: Print the first type from the list "byte, short, int, long, BigInteger", that can store the natural number *n*, in accordance with the data given above. Demo Input: ['127\n', '130\n', '123456789101112131415161718192021222324\n'] Demo Output: ['byte\n', 'short\n', 'BigInteger\n'] Note: none

```python n=int(input()) if n<=127: print("byte") elif n<=32767: print("short") elif n<=2147483647: print("int") elif n<=9223372036854775807: print("long") else: print("BigInteger") ```

3.9615

768

A

Oath of the Night's Watch

PROGRAMMING

900

[ "constructive algorithms", "sortings" ]

null

"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath. With that begins the watch of Jon Snow. He is assigned the task to support the stewards. This time he has *n* stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him. Can you find how many stewards will Jon support?

First line consists of a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of stewards with Jon Snow. Second line consists of *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) representing the values assigned to the stewards.

Output a single integer representing the number of stewards which Jon will feed.

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

[ "0", "1" ]

In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5. In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.

500

[ { "input": "2\n1 5", "output": "0" }, { "input": "3\n1 2 5", "output": "1" }, { "input": "4\n1 2 3 4", "output": "2" }, { "input": "8\n7 8 9 4 5 6 1 2", "output": "6" }, { "input": "1\n1", "output": "0" }, { "input": "1\n100", "output": "0" }, { "input": "205\n5 5 3 3 6 2 9 3 8 9 6 6 10 8 1 5 3 3 1 2 9 9 9 3 9 10 3 9 8 3 5 6 6 4 6 9 2 9 10 9 5 6 6 7 4 2 6 3 4 1 10 1 7 2 7 7 3 2 6 5 5 2 9 3 8 8 7 6 6 4 2 2 6 2 3 5 7 2 2 10 1 4 6 9 2 3 7 2 2 7 4 4 9 10 7 5 8 6 5 3 6 10 2 7 5 6 6 8 3 3 9 4 3 5 7 9 3 2 1 1 3 2 1 9 3 1 4 4 10 2 5 5 8 1 4 8 5 3 1 10 8 6 5 8 3 5 4 5 4 4 6 7 2 8 10 8 7 6 6 9 6 7 1 10 3 2 5 10 4 4 5 4 3 4 8 5 3 8 10 3 10 9 7 2 1 8 6 4 6 5 8 10 2 6 7 4 9 4 5 1 8 7 10 3 1", "output": "174" }, { "input": "4\n1000000000 99999999 1000000000 1000000000", "output": "0" }, { "input": "3\n2 2 2", "output": "0" }, { "input": "5\n1 1 1 1 1", "output": "0" }, { "input": "3\n1 1 1", "output": "0" }, { "input": "6\n1 1 3 3 2 2", "output": "2" }, { "input": "7\n1 1 1 1 1 1 1", "output": "0" }, { "input": "4\n1 1 2 5", "output": "1" }, { "input": "3\n0 0 0", "output": "0" }, { "input": "5\n0 0 0 0 0", "output": "0" }, { "input": "5\n1 1 1 1 5", "output": "0" }, { "input": "5\n1 1 2 3 3", "output": "1" }, { "input": "3\n1 1 3", "output": "0" }, { "input": "3\n2 2 3", "output": "0" }, { "input": "1\n6", "output": "0" }, { "input": "5\n1 5 3 5 1", "output": "1" }, { "input": "7\n1 2 2 2 2 2 3", "output": "5" }, { "input": "4\n2 2 2 2", "output": "0" }, { "input": "9\n2 2 2 3 4 5 6 6 6", "output": "3" }, { "input": "10\n1 1 1 2 3 3 3 3 3 3", "output": "1" }, { "input": "6\n1 1 1 1 1 1", "output": "0" }, { "input": "3\n0 0 1", "output": "0" }, { "input": "9\n1 1 1 2 2 2 3 3 3", "output": "3" }, { "input": "3\n1 2 2", "output": "0" }, { "input": "6\n2 2 2 2 2 2", "output": "0" }, { "input": "5\n2 2 2 2 2", "output": "0" }, { "input": "5\n5 5 5 5 5", "output": "0" }, { "input": "1\n0", "output": "0" }, { "input": "6\n1 2 5 5 5 5", "output": "1" }, { "input": "5\n1 2 3 3 3", "output": "1" }, { "input": "3\n1 1 2", "output": "0" }, { "input": "6\n1 1 1 1 1 2", "output": "0" }, { "input": "5\n1 1 2 4 4", "output": "1" }, { "input": "3\n999999 5999999 9999999", "output": "1" }, { "input": "4\n1 1 5 5", "output": "0" }, { "input": "9\n1 1 1 2 2 2 4 4 4", "output": "3" }, { "input": "5\n1 3 4 5 1", "output": "2" }, { "input": "5\n3 3 3 3 3", "output": "0" }, { "input": "5\n1 1 2 2 2", "output": "0" }, { "input": "5\n2 1 1 1 3", "output": "1" }, { "input": "5\n0 0 0 1 2", "output": "1" }, { "input": "4\n2 2 2 3", "output": "0" }, { "input": "7\n1 1 1 1 5 5 5", "output": "0" }, { "input": "5\n1 2 3 4 4", "output": "2" }, { "input": "2\n5 4", "output": "0" }, { "input": "4\n5 5 5 5", "output": "0" }, { "input": "5\n1 1 1 5 5", "output": "0" }, { "input": "2\n1 1", "output": "0" }, { "input": "1\n3", "output": "0" }, { "input": "3\n2 1 2", "output": "0" }, { "input": "4\n1 2 2 2", "output": "0" }, { "input": "8\n1000000000 1000000000 1000000000 999999999 999999999 999999999 999999998 999999998", "output": "3" }, { "input": "5\n1 1 3 4 4", "output": "1" }, { "input": "6\n1 1 2 2 3 3", "output": "2" }, { "input": "4\n1 1 1 1", "output": "0" }, { "input": "9\n1 2 3 4 1 5 6 7 8", "output": "6" }, { "input": "8\n5 4 4 6 6 4 4 3", "output": "5" }, { "input": "8\n4 3 3 3 3 3 3 3", "output": "0" }, { "input": "7\n4 3 3 3 3 3 3", "output": "0" }, { "input": "6\n4 3 3 3 3 3", "output": "0" }, { "input": "5\n4 3 3 3 3", "output": "0" } ]

1,684,008,681

2,147,483,647

PyPy 3-64

OK

TESTS

88

92

13,619,200

n = int(input()) lst = list(map(int, input().split())) a = min(lst) b = max(lst) a1 = lst.count(a) b1 = lst.count(b) if n - a1-b1>0: print(n - a1 - b1) else: print(0)

Title: Oath of the Night's Watch Time Limit: None seconds Memory Limit: None megabytes Problem Description: "Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath. With that begins the watch of Jon Snow. He is assigned the task to support the stewards. This time he has *n* stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him. Can you find how many stewards will Jon support? Input Specification: First line consists of a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of stewards with Jon Snow. Second line consists of *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) representing the values assigned to the stewards. Output Specification: Output a single integer representing the number of stewards which Jon will feed. Demo Input: ['2\n1 5\n', '3\n1 2 5\n'] Demo Output: ['0', '1'] Note: In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5. In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.

```python n = int(input()) lst = list(map(int, input().split())) a = min(lst) b = max(lst) a1 = lst.count(a) b1 = lst.count(b) if n - a1-b1>0: print(n - a1 - b1) else: print(0) ```

3

216

A

Tiling with Hexagons

PROGRAMMING

1,200

[ "implementation", "math" ]

null

Several ages ago Berland was a kingdom. The King of Berland adored math. That's why, when he first visited one of his many palaces, he first of all paid attention to the floor in one hall. The floor was tiled with hexagonal tiles. The hall also turned out hexagonal in its shape. The King walked along the perimeter of the hall and concluded that each of the six sides has *a*, *b*, *c*, *a*, *b* and *c* adjacent tiles, correspondingly. To better visualize the situation, look at the picture showing a similar hexagon for *a*<==<=2, *b*<==<=3 and *c*<==<=4. According to the legend, as the King of Berland obtained the values *a*, *b* and *c*, he almost immediately calculated the total number of tiles on the hall floor. Can you do the same?

The first line contains three integers: *a*, *b* and *c* (2<=≤<=*a*,<=*b*,<=*c*<=≤<=1000).

Print a single number — the total number of tiles on the hall floor.

[ "2 3 4\n" ]

[ "18" ]

none

500

[ { "input": "2 3 4", "output": "18" }, { "input": "2 2 2", "output": "7" }, { "input": "7 8 13", "output": "224" }, { "input": "14 7 75", "output": "1578" }, { "input": "201 108 304", "output": "115032" }, { "input": "999 998 996", "output": "2983022" }, { "input": "2 2 3", "output": "10" }, { "input": "2 3 2", "output": "10" }, { "input": "3 2 2", "output": "10" }, { "input": "2 3 3", "output": "14" }, { "input": "3 2 3", "output": "14" }, { "input": "3 3 2", "output": "14" }, { "input": "3 3 3", "output": "19" }, { "input": "4 5 3", "output": "36" }, { "input": "2 2 856", "output": "2569" }, { "input": "2 986 2", "output": "2959" }, { "input": "985 2 2", "output": "2956" }, { "input": "2 958 983", "output": "943654" }, { "input": "992 2 912", "output": "906607" }, { "input": "789 894 2", "output": "707048" }, { "input": "1000 1000 1000", "output": "2997001" }, { "input": "384 458 284", "output": "413875" }, { "input": "709 14 290", "output": "218584" }, { "input": "485 117 521", "output": "369265" }, { "input": "849 333 102", "output": "401998" }, { "input": "998 999 1000", "output": "2991006" }, { "input": "2 2 1000", "output": "3001" }, { "input": "2 1000 2", "output": "3001" }, { "input": "1000 2 2", "output": "3001" }, { "input": "1000 2 1000", "output": "1001999" }, { "input": "865 291 383", "output": "692925" }, { "input": "41 49 28", "output": "4412" }, { "input": "34 86 90", "output": "13515" }, { "input": "39 23 56", "output": "4252" }, { "input": "14 99 81", "output": "10346" }, { "input": "48 38 193", "output": "18144" }, { "input": "395 85 22", "output": "43634" }, { "input": "38 291 89", "output": "39922" }, { "input": "7 23 595", "output": "17387" }, { "input": "948 48 3", "output": "47494" } ]

1,518,303,624

2,147,483,647

Python 3

OK

TESTS

40

124

5,632,000

def readInts(): return map(int, input().split()) a, b, c = readInts() ans = 0 while min(a,b,c) > 1: ans += 2*(a+b+c) - 6 a -= 1 b -= 1 c -= 1 ans += a*b*c print(ans)

Title: Tiling with Hexagons Time Limit: None seconds Memory Limit: None megabytes Problem Description: Several ages ago Berland was a kingdom. The King of Berland adored math. That's why, when he first visited one of his many palaces, he first of all paid attention to the floor in one hall. The floor was tiled with hexagonal tiles. The hall also turned out hexagonal in its shape. The King walked along the perimeter of the hall and concluded that each of the six sides has *a*, *b*, *c*, *a*, *b* and *c* adjacent tiles, correspondingly. To better visualize the situation, look at the picture showing a similar hexagon for *a*<==<=2, *b*<==<=3 and *c*<==<=4. According to the legend, as the King of Berland obtained the values *a*, *b* and *c*, he almost immediately calculated the total number of tiles on the hall floor. Can you do the same? Input Specification: The first line contains three integers: *a*, *b* and *c* (2<=≤<=*a*,<=*b*,<=*c*<=≤<=1000). Output Specification: Print a single number — the total number of tiles on the hall floor. Demo Input: ['2 3 4\n'] Demo Output: ['18'] Note: none

```python def readInts(): return map(int, input().split()) a, b, c = readInts() ans = 0 while min(a,b,c) > 1: ans += 2*(a+b+c) - 6 a -= 1 b -= 1 c -= 1 ans += a*b*c print(ans) ```

3

82

A

Double Cola

PROGRAMMING

1,100

[ "implementation", "math" ]

A. Double Cola

1

256

Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum. For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny. Write a program that will print the name of a man who will drink the *n*-th can. Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.

The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109). It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.

Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.

[ "1\n", "6\n", "1802\n" ]

[ "Sheldon\n", "Sheldon\n", "Penny\n" ]

none

500

[ { "input": "1", "output": "Sheldon" }, { "input": "6", "output": "Sheldon" }, { "input": "1802", "output": "Penny" }, { "input": "1", "output": "Sheldon" }, { "input": "2", "output": "Leonard" }, { "input": "3", "output": "Penny" }, { "input": "4", "output": "Rajesh" }, { "input": "5", "output": "Howard" }, { "input": "10", "output": "Penny" }, { "input": "534", "output": "Rajesh" }, { "input": "5033", "output": "Howard" }, { "input": "10010", "output": "Howard" }, { "input": "500000000", "output": "Penny" }, { "input": "63", "output": "Rajesh" }, { "input": "841", "output": "Leonard" }, { "input": "3667", "output": "Penny" }, { "input": "38614", "output": "Howard" }, { "input": "282798", "output": "Rajesh" }, { "input": "9266286", "output": "Rajesh" }, { "input": "27385966", "output": "Leonard" }, { "input": "121580142", "output": "Penny" }, { "input": "5", "output": "Howard" }, { "input": "300", "output": "Howard" }, { "input": "1745", "output": "Leonard" }, { "input": "8302", "output": "Rajesh" }, { "input": "184518", "output": "Sheldon" }, { "input": "1154414", "output": "Rajesh" }, { "input": "28643950", "output": "Leonard" }, { "input": "159222638", "output": "Howard" }, { "input": "24", "output": "Penny" }, { "input": "505", "output": "Penny" }, { "input": "4425", "output": "Rajesh" }, { "input": "12079", "output": "Sheldon" }, { "input": "469726", "output": "Penny" }, { "input": "3961838", "output": "Penny" }, { "input": "57710446", "output": "Leonard" }, { "input": "80719470", "output": "Howard" }, { "input": "1000000000", "output": "Penny" }, { "input": "999999999", "output": "Penny" }, { "input": "999999998", "output": "Penny" }, { "input": "5", "output": "Howard" } ]

1,683,954,261

2,147,483,647

Python 3

OK

TESTS

41

31

0

a = int(input()) i = 0 s = 5 while s < a: i += 1 s += 5 * 2 ** i n = ['Sheldon'] + ['Leonard'] + ['Penny'] + ['Rajesh'] + ['Howard'] print(n[(a - (s - 5 * 2 ** i) - 1) // 2 ** i])

Title: Double Cola Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum. For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny. Write a program that will print the name of a man who will drink the *n*-th can. Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon. Input Specification: The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109). It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers. Output Specification: Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially. Demo Input: ['1\n', '6\n', '1802\n'] Demo Output: ['Sheldon\n', 'Sheldon\n', 'Penny\n'] Note: none

```python a = int(input()) i = 0 s = 5 while s < a: i += 1 s += 5 * 2 ** i n = ['Sheldon'] + ['Leonard'] + ['Penny'] + ['Rajesh'] + ['Howard'] print(n[(a - (s - 5 * 2 ** i) - 1) // 2 ** i]) ```

3.9845

282

A

Bit++

PROGRAMMING

800

[ "implementation" ]

null

The classic programming language of Bitland is Bit++. This language is so peculiar and complicated. The language is that peculiar as it has exactly one variable, called *x*. Also, there are two operations: - Operation ++ increases the value of variable *x* by 1. - Operation -- decreases the value of variable *x* by 1. A statement in language Bit++ is a sequence, consisting of exactly one operation and one variable *x*. The statement is written without spaces, that is, it can only contain characters "+", "-", "X". Executing a statement means applying the operation it contains. A programme in Bit++ is a sequence of statements, each of them needs to be executed. Executing a programme means executing all the statements it contains. You're given a programme in language Bit++. The initial value of *x* is 0. Execute the programme and find its final value (the value of the variable when this programme is executed).

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=150) — the number of statements in the programme. Next *n* lines contain a statement each. Each statement contains exactly one operation (++ or --) and exactly one variable *x* (denoted as letter «X»). Thus, there are no empty statements. The operation and the variable can be written in any order.

Print a single integer — the final value of *x*.

[ "1\n++X\n", "2\nX++\n--X\n" ]

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

none

500

[ { "input": "1\n++X", "output": "1" }, { "input": "2\nX++\n--X", "output": "0" }, { "input": "3\n++X\n++X\n++X", "output": "3" }, { "input": "2\n--X\n--X", "output": "-2" }, { "input": "5\n++X\n--X\n++X\n--X\n--X", "output": "-1" }, { "input": "28\nX--\n++X\nX++\nX++\nX++\n--X\n--X\nX++\nX--\n++X\nX++\n--X\nX--\nX++\nX--\n++X\n++X\nX++\nX++\nX++\nX++\n--X\n++X\n--X\n--X\n--X\n--X\nX++", "output": "4" }, { "input": "94\nX++\nX++\n++X\n++X\nX--\n--X\nX++\n--X\nX++\n++X\nX++\n++X\n--X\n--X\n++X\nX++\n--X\nX--\nX--\n--X\nX--\nX--\n--X\n++X\n--X\nX--\nX--\nX++\n++X\n--X\nX--\n++X\n--X\n--X\nX--\nX--\nX++\nX++\nX--\nX++\nX--\nX--\nX--\n--X\nX--\nX--\nX--\nX++\n++X\nX--\n++X\nX++\n--X\n--X\n--X\n--X\n++X\nX--\n--X\n--X\n++X\nX--\nX--\nX++\n++X\nX++\n++X\n--X\n--X\nX--\n++X\nX--\nX--\n++X\n++X\n++X\n++X\nX++\n++X\n--X\nX++\n--X\n--X\n++X\n--X\nX++\n++X\nX++\n--X\nX--\nX--\n--X\n++X\nX++", "output": "-10" }, { "input": "56\n--X\nX--\n--X\n--X\nX--\nX--\n--X\nX++\n++X\n--X\nX++\nX--\n--X\n++X\n--X\nX--\nX--\n++X\nX--\nX--\n--X\n++X\n--X\n++X\n--X\nX++\n++X\nX++\n--X\n++X\nX++\nX++\n--X\nX++\nX--\n--X\nX--\n--X\nX++\n++X\n--X\n++X\nX++\nX--\n--X\n--X\n++X\nX--\nX--\n--X\nX--\n--X\nX++\n--X\n++X\n--X", "output": "-14" }, { "input": "59\nX--\n--X\nX++\n++X\nX--\n--X\n--X\n++X\n++X\n++X\n++X\nX++\n++X\n++X\nX++\n--X\nX--\nX++\n++X\n--X\nX++\n--X\n++X\nX++\n--X\n--X\nX++\nX++\n--X\nX++\nX++\nX++\nX--\nX--\n--X\nX++\nX--\nX--\n++X\nX--\nX++\n--X\nX++\nX--\nX--\nX--\nX--\n++X\n--X\nX++\nX++\nX--\nX++\n++X\nX--\nX++\nX--\nX--\n++X", "output": "3" }, { "input": "87\n--X\n++X\n--X\nX++\n--X\nX--\n--X\n++X\nX--\n++X\n--X\n--X\nX++\n--X\nX--\nX++\n++X\n--X\n++X\n++X\n--X\n++X\n--X\nX--\n++X\n++X\nX--\nX++\nX++\n--X\n--X\n++X\nX--\n--X\n++X\n--X\nX++\n--X\n--X\nX--\n++X\n++X\n--X\nX--\nX--\nX--\nX--\nX--\nX++\n--X\n++X\n--X\nX++\n++X\nX++\n++X\n--X\nX++\n++X\nX--\n--X\nX++\n++X\nX++\nX++\n--X\n--X\n++X\n--X\nX++\nX++\n++X\nX++\nX++\nX++\nX++\n--X\n--X\n--X\n--X\n--X\n--X\n--X\nX--\n--X\n++X\n++X", "output": "-5" }, { "input": "101\nX++\nX++\nX++\n++X\n--X\nX--\nX++\nX--\nX--\n--X\n--X\n++X\nX++\n++X\n++X\nX--\n--X\n++X\nX++\nX--\n++X\n--X\n--X\n--X\n++X\n--X\n++X\nX++\nX++\n++X\n--X\nX++\nX--\nX++\n++X\n++X\nX--\nX--\nX--\nX++\nX++\nX--\nX--\nX++\n++X\n++X\n++X\n--X\n--X\n++X\nX--\nX--\n--X\n++X\nX--\n++X\nX++\n++X\nX--\nX--\n--X\n++X\n--X\n++X\n++X\n--X\nX++\n++X\nX--\n++X\nX--\n++X\nX++\nX--\n++X\nX++\n--X\nX++\nX++\n++X\n--X\n++X\n--X\nX++\n--X\nX--\n--X\n++X\n++X\n++X\n--X\nX--\nX--\nX--\nX--\n--X\n--X\n--X\n++X\n--X\n--X", "output": "1" }, { "input": "63\n--X\nX--\n++X\n--X\n++X\nX++\n--X\n--X\nX++\n--X\n--X\nX++\nX--\nX--\n--X\n++X\nX--\nX--\nX++\n++X\nX++\nX++\n--X\n--X\n++X\nX--\nX--\nX--\n++X\nX++\nX--\n--X\nX--\n++X\n++X\nX++\n++X\nX++\nX++\n--X\nX--\n++X\nX--\n--X\nX--\nX--\nX--\n++X\n++X\n++X\n++X\nX++\nX++\n++X\n--X\n--X\n++X\n++X\n++X\nX--\n++X\n++X\nX--", "output": "1" }, { "input": "45\n--X\n++X\nX--\n++X\n++X\nX++\n--X\n--X\n--X\n--X\n--X\n--X\n--X\nX++\n++X\nX--\n++X\n++X\nX--\nX++\nX--\n--X\nX--\n++X\n++X\n--X\n--X\nX--\nX--\n--X\n++X\nX--\n--X\n++X\n++X\n--X\n--X\nX--\n++X\n++X\nX++\nX++\n++X\n++X\nX++", "output": "-3" }, { "input": "21\n++X\nX++\n--X\nX--\nX++\n++X\n--X\nX--\nX++\nX--\nX--\nX--\nX++\n++X\nX++\n++X\n--X\nX--\n--X\nX++\n++X", "output": "1" }, { "input": "100\n--X\n++X\nX++\n++X\nX--\n++X\nX--\nX++\n--X\nX++\nX--\nX--\nX--\n++X\nX--\nX++\nX++\n++X\nX++\nX++\nX++\nX++\n++X\nX++\n++X\nX--\n--X\n++X\nX--\n--X\n++X\n++X\nX--\nX++\nX++\nX++\n++X\n--X\n++X\nX++\nX--\n++X\n++X\n--X\n++X\nX--\nX--\nX--\nX++\nX--\nX--\nX++\nX++\n--X\nX++\nX++\n--X\nX--\n--X\n++X\n--X\n++X\n++X\nX--\n--X\n++X\n++X\n--X\n--X\n++X\nX++\nX--\nX++\nX--\nX++\nX++\n--X\nX--\nX--\n++X\nX--\n--X\n--X\nX++\n--X\n--X\nX--\nX--\n++X\n++X\nX--\n++X\nX++\n--X\n++X\n++X\nX++\n--X\n--X\nX++", "output": "8" }, { "input": "17\nX++\nX++\n++X\n--X\n--X\n++X\n++X\n--X\nX--\nX++\nX--\n--X\n--X\nX--\n++X\nX--\nX++", "output": "-1" }, { "input": "77\n++X\nX++\n--X\nX--\n--X\n--X\nX--\nX++\nX--\nX++\nX--\n++X\n--X\n--X\n--X\n--X\n++X\nX--\nX++\nX--\n--X\nX--\n--X\nX--\n++X\n--X\n++X\n++X\nX++\nX++\nX--\n--X\nX--\nX--\nX++\n--X\n--X\n++X\nX--\nX--\n++X\nX++\nX--\n++X\n--X\nX++\nX--\n++X\n++X\n++X\nX--\nX--\nX--\n--X\n++X\n++X\n++X\nX++\n--X\n--X\n++X\n--X\nX--\nX++\n++X\nX++\n++X\nX--\nX++\nX++\n--X\nX++\nX++\nX++\n--X\nX++\nX--", "output": "-5" }, { "input": "21\nX--\n++X\n--X\nX--\n++X\nX--\n++X\nX--\n--X\n++X\nX++\n++X\nX++\n++X\nX--\n--X\nX++\nX++\nX--\n++X\nX--", "output": "1" }, { "input": "1\nX--", "output": "-1" } ]

1,696,758,283

2,147,483,647

Python 3

OK

TESTS

36

46

0

n = int(input()) cou = 0 for i in range(n): a = str(input()) if a[0] == '+' or a[len(a) - 1] == '+': cou += 1 elif a[0] == '-' or a[len(a) - 1] == '-': cou -= 1 print(cou)

Title: Bit++ Time Limit: None seconds Memory Limit: None megabytes Problem Description: The classic programming language of Bitland is Bit++. This language is so peculiar and complicated. The language is that peculiar as it has exactly one variable, called *x*. Also, there are two operations: - Operation ++ increases the value of variable *x* by 1. - Operation -- decreases the value of variable *x* by 1. A statement in language Bit++ is a sequence, consisting of exactly one operation and one variable *x*. The statement is written without spaces, that is, it can only contain characters "+", "-", "X". Executing a statement means applying the operation it contains. A programme in Bit++ is a sequence of statements, each of them needs to be executed. Executing a programme means executing all the statements it contains. You're given a programme in language Bit++. The initial value of *x* is 0. Execute the programme and find its final value (the value of the variable when this programme is executed). Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=150) — the number of statements in the programme. Next *n* lines contain a statement each. Each statement contains exactly one operation (++ or --) and exactly one variable *x* (denoted as letter «X»). Thus, there are no empty statements. The operation and the variable can be written in any order. Output Specification: Print a single integer — the final value of *x*. Demo Input: ['1\n++X\n', '2\nX++\n--X\n'] Demo Output: ['1\n', '0\n'] Note: none

```python n = int(input()) cou = 0 for i in range(n): a = str(input()) if a[0] == '+' or a[len(a) - 1] == '+': cou += 1 elif a[0] == '-' or a[len(a) - 1] == '-': cou -= 1 print(cou) ```

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" } ]

1,687,269,890

2,147,483,647

Python 3

OK

TESTS

40

92

0

n1 = input() n2 = input() t = len(n2) - 1 count = 0 for i in range(len(n1)): if (n1[i] == n2[t]): count += 1 t -= 1 if count == len(n1): print("YES") else: 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 n1 = input() n2 = input() t = len(n2) - 1 count = 0 for i in range(len(n1)): if (n1[i] == n2[t]): count += 1 t -= 1 if count == len(n1): print("YES") else: print("NO") ```

3.977

703

A

Mishka and Game

PROGRAMMING

800

[ "implementation" ]

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!^^" } ]

1,688,762,715

2,147,483,647

Python 3

OK

TESTS

69

46

0

# это задача 703A n = int(input()) m = 0 c = 0 for i in range(n): a, b = map(int, input().split()) if a > b: m = m + 1 elif b > a: c = c + 1 if m > c: print("Mishka") if m < c: print("Chris") if m == c: 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 # это задача 703A n = int(input()) m = 0 c = 0 for i in range(n): a, b = map(int, input().split()) if a > b: m = m + 1 elif b > a: c = c + 1 if m > c: print("Mishka") if m < c: print("Chris") if m == c: print("Friendship is magic!^^") ```

3

10

A

Power Consumption Calculation

PROGRAMMING

900

[ "implementation" ]

A. Power Consumption Calculation

1

256

Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*].

The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=≤<=*n*<=≤<=100,<=0<=≤<=*P*1,<=*P*2,<=*P*3<=≤<=100,<=1<=≤<=*T*1,<=*T*2<=≤<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=1440, *r**i*<=<<=*l**i*<=+<=1 for *i*<=<<=*n*), which stand for the start and the end of the *i*-th period of work.

Output the answer to the problem.

[ "1 3 2 1 5 10\n0 10\n", "2 8 4 2 5 10\n20 30\n50 100\n" ]

[ "30", "570" ]

none

0

[ { "input": "1 3 2 1 5 10\n0 10", "output": "30" }, { "input": "2 8 4 2 5 10\n20 30\n50 100", "output": "570" }, { "input": "3 15 9 95 39 19\n873 989\n1003 1137\n1172 1436", "output": "8445" }, { "input": "4 73 2 53 58 16\n51 52\n209 242\n281 407\n904 945", "output": "52870" }, { "input": "5 41 20 33 43 4\n46 465\n598 875\n967 980\n1135 1151\n1194 1245", "output": "46995" }, { "input": "6 88 28 100 53 36\n440 445\n525 614\n644 844\n1238 1261\n1305 1307\n1425 1434", "output": "85540" }, { "input": "7 46 61 55 28 59\n24 26\n31 61\n66 133\n161 612\n741 746\n771 849\n1345 1357", "output": "67147" }, { "input": "8 83 18 30 28 5\n196 249\n313 544\n585 630\n718 843\n1040 1194\n1207 1246\n1268 1370\n1414 1422", "output": "85876" }, { "input": "9 31 65 27 53 54\n164 176\n194 210\n485 538\n617 690\n875 886\n888 902\n955 957\n1020 1200\n1205 1282", "output": "38570" }, { "input": "30 3 1 58 44 7\n11 13\n14 32\n37 50\n70 74\n101 106\n113 129\n184 195\n197 205\n213 228\n370 394\n443 446\n457 460\n461 492\n499 585\n602 627\n709 776\n812 818\n859 864\n910 913\n918 964\n1000 1010\n1051 1056\n1063 1075\n1106 1145\n1152 1189\n1211 1212\n1251 1259\n1272 1375\n1412 1417\n1430 1431", "output": "11134" }, { "input": "30 42 3 76 28 26\n38 44\n55 66\n80 81\n84 283\n298 314\n331 345\n491 531\n569 579\n597 606\n612 617\n623 701\n723 740\n747 752\n766 791\n801 827\n842 846\n853 891\n915 934\n945 949\n955 964\n991 1026\n1051 1059\n1067 1179\n1181 1191\n1214 1226\n1228 1233\n1294 1306\n1321 1340\n1371 1374\n1375 1424", "output": "59043" }, { "input": "30 46 5 93 20 46\n12 34\n40 41\n54 58\n100 121\n162 182\n220 349\n358 383\n390 398\n401 403\n408 409\n431 444\n466 470\n471 535\n556 568\n641 671\n699 709\n767 777\n786 859\n862 885\n912 978\n985 997\n1013 1017\n1032 1038\n1047 1048\n1062 1080\n1094 1097\n1102 1113\n1122 1181\n1239 1280\n1320 1369", "output": "53608" }, { "input": "30 50 74 77 4 57\n17 23\n24 61\n67 68\n79 87\n93 101\n104 123\n150 192\n375 377\n398 414\n461 566\n600 633\n642 646\n657 701\n771 808\n812 819\n823 826\n827 833\n862 875\n880 891\n919 920\n928 959\n970 1038\n1057 1072\n1074 1130\n1165 1169\n1171 1230\n1265 1276\n1279 1302\n1313 1353\n1354 1438", "output": "84067" }, { "input": "30 54 76 95 48 16\n9 11\n23 97\n112 116\n126 185\n214 223\n224 271\n278 282\n283 348\n359 368\n373 376\n452 463\n488 512\n532 552\n646 665\n681 685\n699 718\n735 736\n750 777\n791 810\n828 838\n841 858\n874 1079\n1136 1171\n1197 1203\n1210 1219\n1230 1248\n1280 1292\n1324 1374\n1397 1435\n1438 1439", "output": "79844" }, { "input": "30 58 78 12 41 28\n20 26\n27 31\n35 36\n38 99\n103 104\n106 112\n133 143\n181 246\n248 251\n265 323\n350 357\n378 426\n430 443\n466 476\n510 515\n517 540\n542 554\n562 603\n664 810\n819 823\n826 845\n869 895\n921 973\n1002 1023\n1102 1136\n1143 1148\n1155 1288\n1316 1388\n1394 1403\n1434 1437", "output": "82686" }, { "input": "30 62 80 97 25 47\n19 20\n43 75\n185 188\n199 242\n252 258\n277 310\n316 322\n336 357\n398 399\n404 436\n443 520\n549 617\n637 649\n679 694\n705 715\n725 730\n731 756\n768 793\n806 833\n834 967\n1003 1079\n1088 1097\n1100 1104\n1108 1121\n1127 1164\n1240 1263\n1274 1307\n1367 1407\n1419 1425\n1433 1437", "output": "92356" }, { "input": "30 100 48 14 9 7\n26 55\n75 107\n145 146\n158 167\n197 199\n201 213\n222 238\n257 265\n271 338\n357 439\n454 462\n463 562\n633 660\n670 691\n696 829\n844 864\n880 888\n891 894\n895 900\n950 973\n1004 1007\n1018 1021\n1045 1049\n1062 1073\n1138 1184\n1240 1272\n1278 1315\n1403 1410\n1412 1418\n1426 1434", "output": "107490" }, { "input": "30 3 50 32 1 18\n46 60\n66 94\n99 141\n162 239\n309 334\n400 550\n594 630\n659 663\n664 701\n735 736\n741 755\n756 801\n823 854\n865 915\n937 964\n984 986\n990 1007\n1008 1031\n1052 1064\n1073 1080\n1190 1191\n1192 1254\n1265 1266\n1272 1273\n1274 1277\n1284 1302\n1308 1312\n1358 1383\n1388 1389\n1390 1401", "output": "25258" }, { "input": "30 7 52 49 46 37\n2 73\n129 130\n131 140\n143 150\n187 190\n217 238\n248 271\n375 426\n465 470\n475 503\n506 541\n569 571\n578 585\n680 683\n690 738\n742 761\n868 877\n890 892\n904 915\n935 961\n1076 1100\n1103 1114\n1232 1244\n1246 1247\n1257 1261\n1271 1276\n1280 1298\n1346 1360\n1373 1382\n1407 1419", "output": "23992" } ]

1,588,366,710

2,147,483,647

PyPy 3

OK

TESTS

30

280

0

#10A - Power Consumption from sys import stdin #Get information about the working periods #P - how much power the computer uses in the current state #P1 = working mode power consumption #P2 = screensaving mode power consumption #P3 = sleep mode power consumption #T - the time that passed from the last time Tom touched the mouse #T1 - the time period: from the last time Tom touched the laptop to # till it enters the screensaving mode #T2 - the time period: from the start of screensaving mode to # till it enters the sleeping mode n, P1, P2, P3, T1, T2 = map(int, input().split()) #How much the computer has rested before it started screensaving last = list(map(int, input().split())) rest = (last[1]-last[0]) * P1 #The following n-1 lines contain information about each working period for i in range(1, n): current = list(map(int, stdin.readline().split())) d = current[0] - last[1] rest += min(d, T1) * P1 rest += min(max(d - T1, 0), T2) * P2 rest += max(d - (T1+T2), 0) * P3 rest += (current[1]-current[0]) * P1 last = current print(rest)

Title: Power Consumption Calculation Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*]. Input Specification: The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=≤<=*n*<=≤<=100,<=0<=≤<=*P*1,<=*P*2,<=*P*3<=≤<=100,<=1<=≤<=*T*1,<=*T*2<=≤<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=1440, *r**i*<=<<=*l**i*<=+<=1 for *i*<=<<=*n*), which stand for the start and the end of the *i*-th period of work. Output Specification: Output the answer to the problem. Demo Input: ['1 3 2 1 5 10\n0 10\n', '2 8 4 2 5 10\n20 30\n50 100\n'] Demo Output: ['30', '570'] Note: none

```python #10A - Power Consumption from sys import stdin #Get information about the working periods #P - how much power the computer uses in the current state #P1 = working mode power consumption #P2 = screensaving mode power consumption #P3 = sleep mode power consumption #T - the time that passed from the last time Tom touched the mouse #T1 - the time period: from the last time Tom touched the laptop to # till it enters the screensaving mode #T2 - the time period: from the start of screensaving mode to # till it enters the sleeping mode n, P1, P2, P3, T1, T2 = map(int, input().split()) #How much the computer has rested before it started screensaving last = list(map(int, input().split())) rest = (last[1]-last[0]) * P1 #The following n-1 lines contain information about each working period for i in range(1, n): current = list(map(int, stdin.readline().split())) d = current[0] - last[1] rest += min(d, T1) * P1 rest += min(max(d - T1, 0), T2) * P2 rest += max(d - (T1+T2), 0) * P3 rest += (current[1]-current[0]) * P1 last = current print(rest) ```

3.86

893

C

Rumor

PROGRAMMING

1,300

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

null

Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it. Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it. Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on. The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest? Take a look at the notes if you think you haven't understood the problem completely.

The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends. The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor. Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once.

Print one number — the minimum amount of gold Vova has to spend in order to finish the quest.

[ "5 2\n2 5 3 4 8\n1 4\n4 5\n", "10 0\n1 2 3 4 5 6 7 8 9 10\n", "10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n" ]

[ "10\n", "55\n", "15\n" ]

In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor. In the second example Vova has to bribe everyone. In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters.

0

[ { "input": "5 2\n2 5 3 4 8\n1 4\n4 5", "output": "10" }, { "input": "10 0\n1 2 3 4 5 6 7 8 9 10", "output": "55" }, { "input": "10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10", "output": "15" }, { "input": "1 0\n0", "output": "0" }, { "input": "1 0\n1000000000", "output": "1000000000" }, { "input": "2 0\n0 0", "output": "0" }, { "input": "2 0\n1000000000 0", "output": "1000000000" }, { "input": "2 0\n0 1000000000", "output": "1000000000" }, { "input": "2 0\n1000000000 1000000000", "output": "2000000000" }, { "input": "2 1\n0 0\n1 2", "output": "0" } ]

1,666,893,790

2,147,483,647

PyPy 3

OK

TESTS

46

982

19,456,000

# funtion of bfs def dfs(index, visited, neighbours, cost): stack = [index] min = cost[index] while len(stack) > 0: u = stack.pop() if visited[u] == 0: if cost[u] < min: min = cost[u] visited[u] = 1 for i in neighbours[u]: if visited[i] == 0: stack.append(i) return min n, m = input().split() n = int(n) m = int(m) cost_str = list(input().split()) cost = [0] neighbours = [[]] visited = [0] for i in range(n): cost.append(int(cost_str[i])) for i in range(n): neighbours.append([]) visited.append(0) for i in range(m): a, b = input().split() a = int(a) b = int(b) neighbours[a].append(b) neighbours[b].append(a) ans = 0 for i in range(1, n+1): if visited[i] == 0: ans += dfs(i, visited, neighbours, cost) print(ans)

Title: Rumor Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it. Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it. Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on. The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest? Take a look at the notes if you think you haven't understood the problem completely. Input Specification: The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends. The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor. Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once. Output Specification: Print one number — the minimum amount of gold Vova has to spend in order to finish the quest. Demo Input: ['5 2\n2 5 3 4 8\n1 4\n4 5\n', '10 0\n1 2 3 4 5 6 7 8 9 10\n', '10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n'] Demo Output: ['10\n', '55\n', '15\n'] Note: In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor. In the second example Vova has to bribe everyone. In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters.

```python # funtion of bfs def dfs(index, visited, neighbours, cost): stack = [index] min = cost[index] while len(stack) > 0: u = stack.pop() if visited[u] == 0: if cost[u] < min: min = cost[u] visited[u] = 1 for i in neighbours[u]: if visited[i] == 0: stack.append(i) return min n, m = input().split() n = int(n) m = int(m) cost_str = list(input().split()) cost = [0] neighbours = [[]] visited = [0] for i in range(n): cost.append(int(cost_str[i])) for i in range(n): neighbours.append([]) visited.append(0) for i in range(m): a, b = input().split() a = int(a) b = int(b) neighbours[a].append(b) neighbours[b].append(a) ans = 0 for i in range(1, n+1): if visited[i] == 0: ans += dfs(i, visited, neighbours, cost) print(ans) ```

3

807

A

Is it rated?

PROGRAMMING

900

[ "implementation", "sortings" ]

null

Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it. Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known. It's known that if at least one participant's rating has changed, then the round was rated for sure. It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed. In this problem, you should not make any other assumptions about the rating system. Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.

The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants. Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.

If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".

[ "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n", "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n", "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n" ]

[ "rated\n", "unrated\n", "maybe\n" ]

In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated. In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure. In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.

500

[ { "input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated" }, { "input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe" }, { "input": "2\n1 1\n1 1", "output": "maybe" }, { "input": "2\n4126 4126\n4126 4126", "output": "maybe" }, { "input": "10\n446 446\n1331 1331\n3594 3594\n1346 1902\n91 91\n3590 3590\n2437 2437\n4007 3871\n2797 699\n1423 1423", "output": "rated" }, { "input": "10\n4078 4078\n2876 2876\n1061 1061\n3721 3721\n143 143\n2992 2992\n3279 3279\n3389 3389\n1702 1702\n1110 1110", "output": "unrated" }, { "input": "10\n4078 4078\n3721 3721\n3389 3389\n3279 3279\n2992 2992\n2876 2876\n1702 1702\n1110 1110\n1061 1061\n143 143", "output": "maybe" }, { "input": "2\n3936 3936\n2967 2967", "output": "maybe" }, { "input": "2\n1 1\n2 2", "output": "unrated" }, { "input": "2\n2 2\n1 1", "output": "maybe" }, { "input": "2\n2 1\n1 2", "output": "rated" }, { "input": "2\n2967 2967\n3936 3936", "output": "unrated" }, { "input": "3\n1200 1200\n1200 1200\n1300 1300", "output": "unrated" }, { "input": "3\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "3\n1 1\n1 1\n2 2", "output": "unrated" }, { "input": "2\n3 2\n3 2", "output": "rated" }, { "input": "3\n5 5\n4 4\n3 4", "output": "rated" }, { "input": "3\n200 200\n200 200\n300 300", "output": "unrated" }, { "input": "3\n1 1\n2 2\n3 3", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2245 2245\n1699 1699", "output": "maybe" }, { "input": "2\n10 10\n8 8", "output": "maybe" }, { "input": "3\n1500 1500\n1500 1500\n1600 1600", "output": "unrated" }, { "input": "3\n1500 1500\n1500 1500\n1700 1700", "output": "unrated" }, { "input": "4\n100 100\n100 100\n70 70\n80 80", "output": "unrated" }, { "input": "2\n1 2\n2 1", "output": "rated" }, { "input": "3\n5 5\n4 3\n3 3", "output": "rated" }, { "input": "3\n1600 1650\n1500 1550\n1400 1450", "output": "rated" }, { "input": "4\n2000 2000\n1500 1500\n1500 1500\n1700 1700", "output": "unrated" }, { "input": "4\n1500 1500\n1400 1400\n1400 1400\n1700 1700", "output": "unrated" }, { "input": "2\n1600 1600\n1400 1400", "output": "maybe" }, { "input": "2\n3 1\n9 8", "output": "rated" }, { "input": "2\n2 1\n1 1", "output": "rated" }, { "input": "4\n4123 4123\n4123 4123\n2670 2670\n3670 3670", "output": "unrated" }, { "input": "2\n2 2\n3 3", "output": "unrated" }, { "input": "2\n10 11\n5 4", "output": "rated" }, { "input": "2\n15 14\n13 12", "output": "rated" }, { "input": "2\n2 1\n2 2", "output": "rated" }, { "input": "3\n2670 2670\n3670 3670\n4106 4106", "output": "unrated" }, { "input": "3\n4 5\n3 3\n2 2", "output": "rated" }, { "input": "2\n10 9\n10 10", "output": "rated" }, { "input": "3\n1011 1011\n1011 999\n2200 2100", "output": "rated" }, { "input": "2\n3 3\n5 5", "output": "unrated" }, { "input": "2\n1500 1500\n3000 2000", "output": "rated" }, { "input": "2\n5 6\n5 5", "output": "rated" }, { "input": "3\n2000 2000\n1500 1501\n500 500", "output": "rated" }, { "input": "2\n2 3\n2 2", "output": "rated" }, { "input": "2\n3 3\n2 2", "output": "maybe" }, { "input": "2\n1 2\n1 1", "output": "rated" }, { "input": "4\n3123 3123\n2777 2777\n2246 2246\n1699 1699", "output": "maybe" }, { "input": "2\n15 14\n14 13", "output": "rated" }, { "input": "4\n3000 3000\n2900 2900\n3000 3000\n2900 2900", "output": "unrated" }, { "input": "6\n30 3060\n24 2194\n26 2903\n24 2624\n37 2991\n24 2884", "output": "rated" }, { "input": "2\n100 99\n100 100", "output": "rated" }, { "input": "4\n2 2\n1 1\n1 1\n2 2", "output": "unrated" }, { "input": "3\n100 101\n100 100\n100 100", "output": "rated" }, { "input": "4\n1000 1001\n900 900\n950 950\n890 890", "output": "rated" }, { "input": "2\n2 3\n1 1", "output": "rated" }, { "input": "2\n2 2\n1 1", "output": "maybe" }, { "input": "2\n3 2\n2 2", "output": "rated" }, { "input": "2\n3 2\n3 3", "output": "rated" }, { "input": "2\n1 1\n2 2", "output": "unrated" }, { "input": "3\n3 2\n3 3\n3 3", "output": "rated" }, { "input": "4\n1500 1501\n1300 1300\n1200 1200\n1400 1400", "output": "rated" }, { "input": "3\n1000 1000\n500 500\n400 300", "output": "rated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n3000 3000", "output": "unrated" }, { "input": "2\n1 1\n2 3", "output": "rated" }, { "input": "2\n6 2\n6 2", "output": "rated" }, { "input": "5\n3123 3123\n1699 1699\n2777 2777\n2246 2246\n2246 2246", "output": "unrated" }, { "input": "2\n1500 1500\n1600 1600", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2241 2241\n1699 1699", "output": "maybe" }, { "input": "2\n20 30\n10 5", "output": "rated" }, { "input": "3\n1 1\n2 2\n1 1", "output": "unrated" }, { "input": "2\n1 2\n3 3", "output": "rated" }, { "input": "5\n5 5\n4 4\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "2\n2 2\n2 1", "output": "rated" }, { "input": "2\n100 100\n90 89", "output": "rated" }, { "input": "2\n1000 900\n2000 2000", "output": "rated" }, { "input": "2\n50 10\n10 50", "output": "rated" }, { "input": "2\n200 200\n100 100", "output": "maybe" }, { "input": "3\n2 2\n2 2\n3 3", "output": "unrated" }, { "input": "3\n1000 1000\n300 300\n100 100", "output": "maybe" }, { "input": "4\n2 2\n2 2\n3 3\n4 4", "output": "unrated" }, { "input": "2\n5 3\n6 3", "output": "rated" }, { "input": "2\n1200 1100\n1200 1000", "output": "rated" }, { "input": "2\n5 5\n4 4", "output": "maybe" }, { "input": "2\n5 5\n3 3", "output": "maybe" }, { "input": "5\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n1100 1100", "output": "unrated" }, { "input": "5\n10 10\n9 9\n8 8\n7 7\n6 6", "output": "maybe" }, { "input": "3\n1000 1000\n300 300\n10 10", "output": "maybe" }, { "input": "5\n6 6\n5 5\n4 4\n3 3\n2 2", "output": "maybe" }, { "input": "2\n3 3\n1 1", "output": "maybe" }, { "input": "4\n2 2\n2 2\n2 2\n3 3", "output": "unrated" }, { "input": "2\n1000 1000\n700 700", "output": "maybe" }, { "input": "2\n4 3\n5 3", "output": "rated" }, { "input": "2\n1000 1000\n1100 1100", "output": "unrated" }, { "input": "4\n5 5\n4 4\n3 3\n2 2", "output": "maybe" }, { "input": "3\n1 1\n2 3\n2 2", "output": "rated" }, { "input": "2\n1 2\n1 3", "output": "rated" }, { "input": "2\n3 3\n1 2", "output": "rated" }, { "input": "4\n1501 1500\n1300 1300\n1200 1200\n1400 1400", "output": "rated" }, { "input": "5\n1 1\n2 2\n3 3\n4 4\n5 5", "output": "unrated" }, { "input": "2\n10 10\n1 2", "output": "rated" }, { "input": "6\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n1900 1900", "output": "unrated" }, { "input": "6\n3123 3123\n2777 2777\n3000 3000\n2246 2246\n2246 2246\n1699 1699", "output": "unrated" }, { "input": "2\n100 100\n110 110", "output": "unrated" }, { "input": "3\n3 3\n3 3\n4 4", "output": "unrated" }, { "input": "3\n3 3\n3 2\n4 4", "output": "rated" }, { "input": "3\n5 2\n4 4\n3 3", "output": "rated" }, { "input": "4\n4 4\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "2\n1 1\n3 2", "output": "rated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n2699 2699", "output": "unrated" }, { "input": "3\n3 3\n3 3\n3 4", "output": "rated" }, { "input": "3\n1 2\n2 2\n3 3", "output": "rated" }, { "input": "3\n1 2\n1 2\n1 2", "output": "rated" }, { "input": "2\n2 1\n2 1", "output": "rated" }, { "input": "2\n1 2\n3 4", "output": "rated" }, { "input": "2\n3 2\n2 3", "output": "rated" }, { "input": "3\n1500 1500\n1600 1600\n1600 1600", "output": "unrated" }, { "input": "3\n1 1\n3 3\n4 4", "output": "unrated" }, { "input": "3\n1 1\n2 2\n2 2", "output": "unrated" }, { "input": "2\n10 12\n8 8", "output": "rated" }, { "input": "5\n1200 1200\n1500 1500\n1500 1500\n1500 1500\n1500 1500", "output": "unrated" }, { "input": "2\n1 2\n2 2", "output": "rated" }, { "input": "3\n1500 1400\n1200 1200\n1100 1100", "output": "rated" }, { "input": "2\n10 12\n10 10", "output": "rated" }, { "input": "3\n1500 1500\n1400 1400\n1300 1300", "output": "maybe" }, { "input": "3\n3 3\n4 4\n5 5", "output": "unrated" }, { "input": "3\n2 6\n3 5\n4 4", "output": "rated" }, { "input": "2\n5 6\n4 6", "output": "rated" }, { "input": "4\n10 10\n10 10\n7 7\n8 8", "output": "unrated" }, { "input": "2\n4 4\n3 3", "output": "maybe" } ]

1,659,183,925

2,147,483,647

Python 3

OK

TESTS

150

46

0

def isSorted(arr:list,ascending:bool): old = arr[0] for i in arr: if ascending and old>i: return False if not ascending and old < i: return False old = i return True n = int(input()) arr = [] while n: n-=1 inn = input().split() if(inn[0]!=inn[1]): print('rated') exit(0) arr.append(int(inn[0])) if(isSorted(arr,ascending=False)): print('maybe') else: print('unrated')

Title: Is it rated? Time Limit: None seconds Memory Limit: None megabytes Problem Description: Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it. Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known. It's known that if at least one participant's rating has changed, then the round was rated for sure. It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed. In this problem, you should not make any other assumptions about the rating system. Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants. Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings. Output Specification: If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". Demo Input: ['6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n', '4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n', '5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n'] Demo Output: ['rated\n', 'unrated\n', 'maybe\n'] Note: In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated. In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure. In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.

```python def isSorted(arr:list,ascending:bool): old = arr[0] for i in arr: if ascending and old>i: return False if not ascending and old < i: return False old = i return True n = int(input()) arr = [] while n: n-=1 inn = input().split() if(inn[0]!=inn[1]): print('rated') exit(0) arr.append(int(inn[0])) if(isSorted(arr,ascending=False)): print('maybe') else: print('unrated') ```

3

9

A

Die Roll

PROGRAMMING

800

[ "math", "probabilities" ]

A. Die Roll

1

64

Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place. But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams. Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania. It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.

The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.

Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».

[ "4 2\n" ]

[ "1/2\n" ]

Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.

0

[ { "input": "4 2", "output": "1/2" }, { "input": "1 1", "output": "1/1" }, { "input": "1 2", "output": "5/6" }, { "input": "1 3", "output": "2/3" }, { "input": "1 4", "output": "1/2" }, { "input": "1 5", "output": "1/3" }, { "input": "1 6", "output": "1/6" }, { "input": "2 1", "output": "5/6" }, { "input": "2 2", "output": "5/6" }, { "input": "2 3", "output": "2/3" }, { "input": "2 4", "output": "1/2" }, { "input": "2 5", "output": "1/3" }, { "input": "2 6", "output": "1/6" }, { "input": "3 1", "output": "2/3" }, { "input": "3 2", "output": "2/3" }, { "input": "3 3", "output": "2/3" }, { "input": "3 4", "output": "1/2" }, { "input": "3 5", "output": "1/3" }, { "input": "3 6", "output": "1/6" }, { "input": "4 1", "output": "1/2" }, { "input": "4 3", "output": "1/2" }, { "input": "4 4", "output": "1/2" }, { "input": "4 5", "output": "1/3" }, { "input": "4 6", "output": "1/6" }, { "input": "5 1", "output": "1/3" }, { "input": "5 2", "output": "1/3" }, { "input": "5 3", "output": "1/3" }, { "input": "5 4", "output": "1/3" }, { "input": "5 5", "output": "1/3" }, { "input": "5 6", "output": "1/6" }, { "input": "6 1", "output": "1/6" }, { "input": "6 2", "output": "1/6" }, { "input": "6 3", "output": "1/6" }, { "input": "6 4", "output": "1/6" }, { "input": "6 5", "output": "1/6" }, { "input": "6 6", "output": "1/6" } ]

1,671,487,223

2,147,483,647

PyPy 3-64

OK

TESTS

36

122

0

y,w = map(int,input().split()) r = (7-max(y,w))/6 if r == 1/6 : print("1/6") elif r == 2/6 : print("1/3") elif r == 3/6 : print("1/2") elif r == 4/6 : print("2/3") elif r == 5/6 : print("5/6") else: print("1/1")

Title: Die Roll Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place. But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams. Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania. It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. Input Specification: The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. Output Specification: Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». Demo Input: ['4 2\n'] Demo Output: ['1/2\n'] Note: Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.

```python y,w = map(int,input().split()) r = (7-max(y,w))/6 if r == 1/6 : print("1/6") elif r == 2/6 : print("1/3") elif r == 3/6 : print("1/2") elif r == 4/6 : print("2/3") elif r == 5/6 : print("5/6") else: print("1/1") ```

3.939

224

A

Parallelepiped

PROGRAMMING

1,100

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

null

You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.

The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.

Print a single number — the sum of all edges of the parallelepiped.

[ "1 1 1\n", "4 6 6\n" ]

[ "12\n", "28\n" ]

In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3.

500

[ { "input": "1 1 1", "output": "12" }, { "input": "4 6 6", "output": "28" }, { "input": "20 10 50", "output": "68" }, { "input": "9 4 36", "output": "56" }, { "input": "324 9 36", "output": "184" }, { "input": "1333 93 129", "output": "308" }, { "input": "1022 584 112", "output": "380" }, { "input": "66 174 319", "output": "184" }, { "input": "912 276 1748", "output": "444" }, { "input": "65 156 60", "output": "120" }, { "input": "1 10000 10000", "output": "40008" }, { "input": "1485 55 27", "output": "332" }, { "input": "152 108 4104", "output": "528" }, { "input": "1656 6900 1350", "output": "740" }, { "input": "12 14 42", "output": "60" }, { "input": "615 18 1230", "output": "856" }, { "input": "680 60 408", "output": "336" }, { "input": "644 966 6", "output": "1308" }, { "input": "1 432 432", "output": "1736" }, { "input": "2239 2239 1", "output": "8964" }, { "input": "4106 8212 2", "output": "16436" }, { "input": "10000 10000 10000", "output": "1200" }, { "input": "3623 3623 1", "output": "14500" }, { "input": "9801 9801 9801", "output": "1188" }, { "input": "10000 1 10000", "output": "40008" }, { "input": "9 9 9", "output": "36" }, { "input": "9801 9702 9702", "output": "1184" } ]

1,631,842,899

2,147,483,647

PyPy 3

OK

TESTS

27

218

20,172,800

x,y,z = tuple(map(int,input().split())) a = int(((x*z)//y)**(1/2)) b = int(x//a) c = int(z//a) print(4*(a+b+c))

Title: Parallelepiped Time Limit: None seconds Memory Limit: None megabytes Problem Description: You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped. Input Specification: The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement. Output Specification: Print a single number — the sum of all edges of the parallelepiped. Demo Input: ['1 1 1\n', '4 6 6\n'] Demo Output: ['12\n', '28\n'] Note: In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3.

```python x,y,z = tuple(map(int,input().split())) a = int(((x*z)//y)**(1/2)) b = int(x//a) c = int(z//a) print(4*(a+b+c)) ```

3

669

A

Little Artem and Presents

PROGRAMMING

800

[ "math" ]

null

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

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

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

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

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

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

500

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

1,461,515,868

168

Python 3

OK

TESTS

26

62

4,915,200

a = int(input()) print(a//3*2+[0,1][a%3!=0])

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

```python a = int(input()) print(a//3*2+[0,1][a%3!=0]) ```

3

686

A

Free Ice Cream

PROGRAMMING

800

[ "constructive algorithms", "implementation" ]

null

After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer. At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue). If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress. Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.

The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109). Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.

Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.

[ "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n", "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n" ]

[ "22 1\n", "3 2\n" ]

Consider the first sample. 1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.

500

[ { "input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20", "output": "22 1" }, { "input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98", "output": "3 2" }, { "input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000", "output": "7000000000 0" }, { "input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46", "output": "46 0" }, { "input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100", "output": "700 0" }, { "input": "1 0\n- 526403222", "output": "0 1" }, { "input": "1 897986543\n- 371188251", "output": "526798292 0" }, { "input": "1 0\n+ 1", "output": "1 0" }, { "input": "1 0\n- 1", "output": "0 1" }, { "input": "1 10\n+ 10", "output": "20 0" }, { "input": "1 3\n- 5", "output": "3 1" }, { "input": "1 0\n- 5", "output": "0 1" }, { "input": "1 0\n+ 5", "output": "5 0" } ]

1,651,674,717

2,147,483,647

PyPy 3-64

OK

TESTS

34

77

2,355,200

n,x = (map(int,input().split())) d = 0 for _ in range(n) : a = input().split() if a[0]=="+" : x+=int(a[1]) else : if x<int(a[1]) : d+=1 else : x-=int(a[1]) print(str(x)+" "+str(d))

Title: Free Ice Cream Time Limit: None seconds Memory Limit: None megabytes Problem Description: After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer. At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue). If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress. Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids. Input Specification: The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109). Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place. Output Specification: Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress. Demo Input: ['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n'] Demo Output: ['22 1\n', '3 2\n'] Note: Consider the first sample. 1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.

```python n,x = (map(int,input().split())) d = 0 for _ in range(n) : a = input().split() if a[0]=="+" : x+=int(a[1]) else : if x<int(a[1]) : d+=1 else : x-=int(a[1]) print(str(x)+" "+str(d)) ```

3

672

B

Different is Good

PROGRAMMING

1,000

[ "constructive algorithms", "implementation", "strings" ]

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" } ]

1,546,662,367

2,147,483,647

Python 3

OK

TESTS

47

124

204,800

def main(): n = int(input()) s = input() arr = [0] * 26 for char in s: arr[ord(char) - ord('a')] += 1 ans = 0 count = 0 for c in range(26): if arr[c] > 0: count += 1 ans += arr[c] - 1 if 26 - count < ans: print(-1) else: print(ans) main()

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 def main(): n = int(input()) s = input() arr = [0] * 26 for char in s: arr[ord(char) - ord('a')] += 1 ans = 0 count = 0 for c in range(26): if arr[c] > 0: count += 1 ans += arr[c] - 1 if 26 - count < ans: print(-1) else: print(ans) main() ```

3

265

B

Roadside Trees (Simplified Edition)

PROGRAMMING

1,000

[ "greedy", "implementation" ]

null

Squirrel Liss loves nuts. There are *n* trees (numbered 1 to *n* from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree *i* is *h**i*. Liss wants to eat all nuts. Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions: - Walk up or down one unit on a tree. - Eat a nut on the top of the current tree. - Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height *h* of the tree *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), she jumps to height *h* of the tree *i*<=+<=1. This action can't be performed if *h*<=><=*h**i*<=+<=1. Compute the minimal time (in seconds) required to eat all nuts.

The first line contains an integer *n* (1<=<=≤<=<=*n*<=≤<=105) — the number of trees. Next *n* lines contains the height of trees: *i*-th line contains an integer *h**i* (1<=≤<=*h**i*<=≤<=104) — the height of the tree with the number *i*.

Print a single integer — the minimal time required to eat all nuts in seconds.

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

[ "5\n", "14\n" ]

none

1,000

[ { "input": "2\n1\n2", "output": "5" }, { "input": "5\n2\n1\n2\n1\n1", "output": "14" }, { "input": "1\n1", "output": "2" } ]

1,611,901,019

2,147,483,647

Python 3

OK

TESTS

15

404

0

i=0 ans=0 past=0 for x in range(int(input())): n=int(input()) if i==0: ans=n+1 past=n ##print(ans) else: ans+=abs(past-n)+2 ##print(ans) i+=1 past=n print(ans)

Title: Roadside Trees (Simplified Edition) Time Limit: None seconds Memory Limit: None megabytes Problem Description: Squirrel Liss loves nuts. There are *n* trees (numbered 1 to *n* from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree *i* is *h**i*. Liss wants to eat all nuts. Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions: - Walk up or down one unit on a tree. - Eat a nut on the top of the current tree. - Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height *h* of the tree *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), she jumps to height *h* of the tree *i*<=+<=1. This action can't be performed if *h*<=><=*h**i*<=+<=1. Compute the minimal time (in seconds) required to eat all nuts. Input Specification: The first line contains an integer *n* (1<=<=≤<=<=*n*<=≤<=105) — the number of trees. Next *n* lines contains the height of trees: *i*-th line contains an integer *h**i* (1<=≤<=*h**i*<=≤<=104) — the height of the tree with the number *i*. Output Specification: Print a single integer — the minimal time required to eat all nuts in seconds. Demo Input: ['2\n1\n2\n', '5\n2\n1\n2\n1\n1\n'] Demo Output: ['5\n', '14\n'] Note: none

```python i=0 ans=0 past=0 for x in range(int(input())): n=int(input()) if i==0: ans=n+1 past=n ##print(ans) else: ans+=abs(past-n)+2 ##print(ans) i+=1 past=n print(ans) ```

3

18

C

Stripe

PROGRAMMING

1,200

[ "data structures", "implementation" ]

C. Stripe

2

64

Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?

The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.

Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.

[ "9\n1 5 -6 7 9 -16 0 -2 2\n", "3\n1 1 1\n", "2\n0 0\n" ]

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

none

0

[ { "input": "9\n1 5 -6 7 9 -16 0 -2 2", "output": "3" }, { "input": "3\n1 1 1", "output": "0" }, { "input": "2\n0 0", "output": "1" }, { "input": "4\n100 1 10 111", "output": "1" }, { "input": "10\n0 4 -3 0 -2 2 -3 -3 2 5", "output": "3" }, { "input": "10\n0 -1 2 2 -1 1 0 0 0 2", "output": "0" }, { "input": "10\n-1 -1 1 -1 0 1 0 1 1 1", "output": "1" }, { "input": "10\n0 0 0 0 0 0 0 0 0 0", "output": "9" }, { "input": "50\n-4 -3 3 4 -1 0 2 -4 -3 -4 1 4 3 0 4 1 0 -3 4 -3 -2 2 2 1 0 -4 -4 -5 3 2 -1 4 5 -3 -3 4 4 -5 2 -3 4 -5 2 5 -4 4 1 -2 -4 3", "output": "3" }, { "input": "15\n0 4 0 3 -1 4 -2 -2 -4 -4 3 2 4 -1 -3", "output": "0" }, { "input": "10\n3 -1 -3 -1 3 -2 0 3 1 -2", "output": "0" }, { "input": "100\n-4 2 4 4 1 3 -3 -3 2 1 -4 0 0 2 3 -1 -4 -3 4 -2 -3 -3 -3 -1 -2 -3 -1 -4 0 4 0 -1 4 0 -4 -4 4 -4 -2 1 -4 1 -3 -2 3 -4 4 0 -1 3 -1 4 -1 4 -1 3 -3 -3 -2 -2 4 -3 -3 4 -3 -2 -1 0 -2 4 0 -3 -1 -2 -3 1 -4 1 -3 -3 -3 -2 -3 0 1 -2 -2 -4 -3 -1 2 3 -1 1 1 0 3 -3 -1 -2", "output": "1" }, { "input": "100\n-2 -1 1 0 -2 -1 2 2 0 0 2 1 0 2 0 2 1 0 -1 -1 -1 0 -2 -1 2 -1 -2 2 -2 2 -2 -2 2 1 1 1 -2 2 0 0 2 -1 2 2 2 0 -1 -1 -1 1 -2 2 2 2 -2 0 0 -2 0 -2 -2 0 -1 -1 -2 -1 1 2 -2 -1 1 -2 -1 0 -2 2 2 -1 -2 2 0 0 0 1 0 1 2 1 -2 -2 -1 -1 -2 -2 -2 -1 2 2 2 -2", "output": "6" }, { "input": "100\n2 7 6 0 8 -2 0 1 8 -1 7 -4 -1 1 0 3 4 -4 3 7 8 -4 -6 6 4 -2 -5 7 4 6 1 6 3 8 -2 6 -6 -3 0 -1 -7 -8 -2 5 8 6 -2 -2 -5 -4 -7 -3 -5 -3 -3 -1 5 8 4 0 -7 -6 8 -4 -1 7 2 6 6 4 -5 -4 -5 -2 2 -2 -7 -1 5 -8 -6 -2 -5 4 2 8 -6 7 -8 -1 -5 8 6 -3 -1 7 -1 2 -8 -8", "output": "0" }, { "input": "1\n0", "output": "0" }, { "input": "1\n10000", "output": "0" }, { "input": "1\n-10000", "output": "0" } ]

1,584,862,260

2,147,483,647

PyPy 3

OK

TESTS

45

404

11,878,400

def main(): n = int(input()) a = list(map(int,input().split())) b = [a[0]] for i in range(1,n): b.append(b[i-1]+a[i]) poss_count = 0 for i in range(n-1): if b[i] == (b[n-1] - b[i]): poss_count += 1 #print(i) print(poss_count) if __name__ == '__main__': main()

Title: Stripe Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem? Input Specification: The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value. Output Specification: Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only. Demo Input: ['9\n1 5 -6 7 9 -16 0 -2 2\n', '3\n1 1 1\n', '2\n0 0\n'] Demo Output: ['3\n', '0\n', '1\n'] Note: none

```python def main(): n = int(input()) a = list(map(int,input().split())) b = [a[0]] for i in range(1,n): b.append(b[i-1]+a[i]) poss_count = 0 for i in range(n-1): if b[i] == (b[n-1] - b[i]): poss_count += 1 #print(i) print(poss_count) if __name__ == '__main__': main() ```

3.810499

14

A

Letter

PROGRAMMING

800

[ "implementation" ]

A. Letter

1

64

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

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

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

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

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

none

0

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

1,587,635,518

2,147,483,647

PyPy 3

OK

TESTS

30

342

2,560,000

n, m = map(int, input().split()) str_arr = list() for i in range(n): string = input() str_arr.append(string) i_arr = [] for i in range(n): if "*" in str_arr[i]: i_arr.append(i) i_min = min(i_arr) i_max = max(i_arr) min_j = str_arr[i_min].find("*") tmp_index = 0 for i in range(n): tmp_index = str_arr[i].find("*") if tmp_index < min_j and tmp_index != -1: min_j = tmp_index max_j = str_arr[i_max].rfind("*") tmp_index = 0 for i in range(n): tmp_index = str_arr[i].rfind("*") if tmp_index != -1 and tmp_index > max_j: max_j = tmp_index for i in range(i_min, i_max+1): for j in range(min_j, max_j+1): print(str_arr[i][j], end='') print()

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

```python n, m = map(int, input().split()) str_arr = list() for i in range(n): string = input() str_arr.append(string) i_arr = [] for i in range(n): if "*" in str_arr[i]: i_arr.append(i) i_min = min(i_arr) i_max = max(i_arr) min_j = str_arr[i_min].find("*") tmp_index = 0 for i in range(n): tmp_index = str_arr[i].find("*") if tmp_index < min_j and tmp_index != -1: min_j = tmp_index max_j = str_arr[i_max].rfind("*") tmp_index = 0 for i in range(n): tmp_index = str_arr[i].rfind("*") if tmp_index != -1 and tmp_index > max_j: max_j = tmp_index for i in range(i_min, i_max+1): for j in range(min_j, max_j+1): print(str_arr[i][j], end='') print() ```

3.809927

386

A

Second-Price Auction

PROGRAMMING

800

[ "implementation" ]

null

In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction). Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.

The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.

The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.

[ "2\n5 7\n", "3\n10 2 8\n", "6\n3 8 2 9 4 14\n" ]

[ "2 5\n", "1 8\n", "6 9\n" ]

none

500

[ { "input": "2\n5 7", "output": "2 5" }, { "input": "3\n10 2 8", "output": "1 8" }, { "input": "6\n3 8 2 9 4 14", "output": "6 9" }, { "input": "4\n4707 7586 4221 5842", "output": "2 5842" }, { "input": "5\n3304 4227 4869 6937 6002", "output": "4 6002" }, { "input": "6\n5083 3289 7708 5362 9031 7458", "output": "5 7708" }, { "input": "7\n9038 6222 3392 1706 3778 1807 2657", "output": "1 6222" }, { "input": "8\n7062 2194 4481 3864 7470 1814 8091 733", "output": "7 7470" }, { "input": "9\n2678 5659 9199 2628 7906 7496 4524 2663 3408", "output": "3 7906" }, { "input": "2\n3458 1504", "output": "1 1504" }, { "input": "50\n9237 3904 407 9052 6657 9229 9752 3888 7732 2512 4614 1055 2355 7108 6506 6849 2529 8862 159 8630 7906 7941 960 8470 333 8659 54 9475 3163 5625 6393 6814 2656 3388 169 7918 4881 8468 9983 6281 6340 280 5108 2996 101 7617 3313 8172 326 1991", "output": "39 9752" }, { "input": "100\n2515 3324 7975 6171 4240 1217 4829 5203 8603 6900 3031 4699 4732 6070 4221 3228 6497 7359 9130 4346 4619 1109 3945 5442 3271 16 9711 2045 6410 2301 3406 8125 6003 1892 1260 9661 3940 6692 4708 7027 4930 6925 1979 5361 4263 3144 867 8639 6230 5562 9714 3676 4231 3347 4733 4920 4881 3431 1059 7313 8912 3038 9308 72 9583 7009 3034 7425 2398 6141 3245 2495 2933 6710 8289 9806 1226 8393 7349 6462 1618 9613 3546 6012 2964 9995 1578 210 2123 4874 1252 8625 348 8020 803 7244 9080 5088 706 2602", "output": "86 9806" }, { "input": "2\n2 1", "output": "1 1" }, { "input": "2\n1 2", "output": "2 1" }, { "input": "3\n10 20 30", "output": "3 20" }, { "input": "3\n10 30 20", "output": "2 20" }, { "input": "3\n20 10 30", "output": "3 20" }, { "input": "3\n20 30 10", "output": "2 20" }, { "input": "3\n30 10 20", "output": "1 20" }, { "input": "3\n30 20 10", "output": "1 20" }, { "input": "2\n1 10000", "output": "2 1" }, { "input": "2\n10000 999", "output": "1 999" }, { "input": "3\n3 4 1", "output": "2 3" }, { "input": "6\n1 2 24 6 5 7", "output": "3 7" } ]

1,630,998,838

2,147,483,647

Python 3

OK

TESTS

42

77

6,963,200

n=int(input()) sbid=input().split() bid=[0]*n for i in range(n): sbid[i]=int(sbid[i]) bid[i]=sbid[i] sbid.sort() winner=sbid[len(sbid)-1] price=sbid[len(sbid)-2] print(bid.index(winner)+1,price)

Title: Second-Price Auction Time Limit: None seconds Memory Limit: None megabytes Problem Description: In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction). Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different. Input Specification: The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder. Output Specification: The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based. Demo Input: ['2\n5 7\n', '3\n10 2 8\n', '6\n3 8 2 9 4 14\n'] Demo Output: ['2 5\n', '1 8\n', '6 9\n'] Note: none

```python n=int(input()) sbid=input().split() bid=[0]*n for i in range(n): sbid[i]=int(sbid[i]) bid[i]=sbid[i] sbid.sort() winner=sbid[len(sbid)-1] price=sbid[len(sbid)-2] print(bid.index(winner)+1,price) ```

3

0

none

0

[ "none" ]

null

Sereja loves all sorts of algorithms. He has recently come up with a new algorithm, which receives a string as an input. Let's represent the input string of the algorithm as *q*<==<=*q*1*q*2... *q**k*. The algorithm consists of two steps: 1. Find any continuous subsequence (substring) of three characters of string *q*, which doesn't equal to either string "zyx", "xzy", "yxz". If *q* doesn't contain any such subsequence, terminate the algorithm, otherwise go to step 2. 1. Rearrange the letters of the found subsequence randomly and go to step 1. Sereja thinks that the algorithm works correctly on string *q* if there is a non-zero probability that the algorithm will be terminated. But if the algorithm anyway will work for infinitely long on a string, then we consider the algorithm to work incorrectly on this string. Sereja wants to test his algorithm. For that, he has string *s*<==<=*s*1*s*2... *s**n*, consisting of *n* characters. The boy conducts a series of *m* tests. As the *i*-th test, he sends substring *s**l**i**s**l**i*<=+<=1... *s**r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*) to the algorithm input. Unfortunately, the implementation of his algorithm works too long, so Sereja asked you to help. For each test (*l**i*,<=*r**i*) determine if the algorithm works correctly on this test or not.

The first line contains non-empty string *s*, its length (*n*) doesn't exceed 105. It is guaranteed that string *s* only contains characters: 'x', 'y', 'z'. The second line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of tests. Next *m* lines contain the tests. The *i*-th line contains a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*).

For each test, print "YES" (without the quotes) if the algorithm works correctly on the corresponding test and "NO" (without the quotes) otherwise.

[ "zyxxxxxxyyz\n5\n5 5\n1 3\n1 11\n1 4\n3 6\n" ]

[ "YES\nYES\nNO\nYES\nNO\n" ]

In the first example, in test one and two the algorithm will always be terminated in one step. In the fourth test you can get string "xzyx" on which the algorithm will terminate. In all other tests the algorithm doesn't work correctly.

0

[ { "input": "zyxxxxxxyyz\n5\n5 5\n1 3\n1 11\n1 4\n3 6", "output": "YES\nYES\nNO\nYES\nNO" }, { "input": "yxzyzxzzxyyzzxxxzyyzzyzxxzxyzyyzxyzxyxxyzxyxzyzxyzxyyxzzzyzxyyxyzxxy\n10\n17 67\n6 35\n12 45\n56 56\n14 30\n25 54\n1 1\n46 54\n3 33\n19 40", "output": "NO\nNO\nNO\nYES\nYES\nNO\nYES\nNO\nNO\nYES" }, { "input": "xxxxyyxyyzzyxyxzxyzyxzyyyzyzzxxxxzyyzzzzyxxxxzzyzzyzx\n5\n4 4\n3 3\n1 24\n3 28\n18 39", "output": "YES\nYES\nNO\nNO\nNO" }, { "input": "yzxyzxyzxzxzyzxyzyzzzyxzyz\n9\n4 6\n2 7\n3 5\n14 24\n3 13\n2 24\n2 5\n2 14\n3 15", "output": "YES\nYES\nYES\nNO\nYES\nNO\nYES\nNO\nNO" }, { "input": "zxyzxyzyyzxzzxyzxyzx\n15\n7 10\n17 17\n6 7\n8 14\n4 7\n11 18\n12 13\n1 1\n3 8\n1 1\n9 17\n4 4\n5 11\n3 15\n1 1", "output": "NO\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nYES\nNO\nYES" }, { "input": "x\n1\n1 1", "output": "YES" } ]

1,684,755,733

2,147,483,647

PyPy 3-64

OK

TESTS

38

296

14,028,800

# https://codeforces.com/contest/368 import sys input = lambda: sys.stdin.readline().rstrip() # faster! s = input() m = int(input()) n = len(s) cx, cy, cz = [0] * (n + 1), [0] * (n + 1), [0] * (n + 1) for i in range(n): cx[i + 1] = cx[i] + int(s[i] == "x") cy[i + 1] = cy[i] + int(s[i] == "y") cz[i + 1] = cz[i] + int(s[i] == "z") for _ in range(m): l, r = map(int, input().split()) if r - l < 2: print("YES") else: x = cx[r] - cx[l - 1] y = cy[r] - cy[l - 1] z = cz[r] - cz[l - 1] print("YES" if max(x, y, z) - min(x, y, z) <= 1 else "NO")

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sereja loves all sorts of algorithms. He has recently come up with a new algorithm, which receives a string as an input. Let's represent the input string of the algorithm as *q*<==<=*q*1*q*2... *q**k*. The algorithm consists of two steps: 1. Find any continuous subsequence (substring) of three characters of string *q*, which doesn't equal to either string "zyx", "xzy", "yxz". If *q* doesn't contain any such subsequence, terminate the algorithm, otherwise go to step 2. 1. Rearrange the letters of the found subsequence randomly and go to step 1. Sereja thinks that the algorithm works correctly on string *q* if there is a non-zero probability that the algorithm will be terminated. But if the algorithm anyway will work for infinitely long on a string, then we consider the algorithm to work incorrectly on this string. Sereja wants to test his algorithm. For that, he has string *s*<==<=*s*1*s*2... *s**n*, consisting of *n* characters. The boy conducts a series of *m* tests. As the *i*-th test, he sends substring *s**l**i**s**l**i*<=+<=1... *s**r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*) to the algorithm input. Unfortunately, the implementation of his algorithm works too long, so Sereja asked you to help. For each test (*l**i*,<=*r**i*) determine if the algorithm works correctly on this test or not. Input Specification: The first line contains non-empty string *s*, its length (*n*) doesn't exceed 105. It is guaranteed that string *s* only contains characters: 'x', 'y', 'z'. The second line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of tests. Next *m* lines contain the tests. The *i*-th line contains a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). Output Specification: For each test, print "YES" (without the quotes) if the algorithm works correctly on the corresponding test and "NO" (without the quotes) otherwise. Demo Input: ['zyxxxxxxyyz\n5\n5 5\n1 3\n1 11\n1 4\n3 6\n'] Demo Output: ['YES\nYES\nNO\nYES\nNO\n'] Note: In the first example, in test one and two the algorithm will always be terminated in one step. In the fourth test you can get string "xzyx" on which the algorithm will terminate. In all other tests the algorithm doesn't work correctly.

```python # https://codeforces.com/contest/368 import sys input = lambda: sys.stdin.readline().rstrip() # faster! s = input() m = int(input()) n = len(s) cx, cy, cz = [0] * (n + 1), [0] * (n + 1), [0] * (n + 1) for i in range(n): cx[i + 1] = cx[i] + int(s[i] == "x") cy[i + 1] = cy[i] + int(s[i] == "y") cz[i + 1] = cz[i] + int(s[i] == "z") for _ in range(m): l, r = map(int, input().split()) if r - l < 2: print("YES") else: x = cx[r] - cx[l - 1] y = cy[r] - cy[l - 1] z = cz[r] - cz[l - 1] print("YES" if max(x, y, z) - min(x, y, z) <= 1 else "NO") ```

3

522

B

Photo to Remember

PROGRAMMING

1,100

[ "*special", "data structures", "dp", "implementation" ]

null

One day *n* friends met at a party, they hadn't seen each other for a long time and so they decided to make a group photo together. Simply speaking, the process of taking photos can be described as follows. On the photo, each photographed friend occupies a rectangle of pixels: the *i*-th of them occupies the rectangle of width *w**i* pixels and height *h**i* pixels. On the group photo everybody stands in a line, thus the minimum pixel size of the photo including all the photographed friends, is *W*<=×<=*H*, where *W* is the total sum of all widths and *H* is the maximum height of all the photographed friends. As is usually the case, the friends made *n* photos — the *j*-th (1<=≤<=*j*<=≤<=*n*) photo had everybody except for the *j*-th friend as he was the photographer. Print the minimum size of each made photo in pixels.

The first line contains integer *n* (2<=≤<=*n*<=≤<=200<=000) — the number of friends. Then *n* lines follow: the *i*-th line contains information about the *i*-th friend. The line contains a pair of integers *w**i*,<=*h**i* (1<=≤<=*w**i*<=≤<=10,<=1<=≤<=*h**i*<=≤<=1000) — the width and height in pixels of the corresponding rectangle.

Print *n* space-separated numbers *b*1,<=*b*2,<=...,<=*b**n*, where *b**i* — the total number of pixels on the minimum photo containing all friends expect for the *i*-th one.

[ "3\n1 10\n5 5\n10 1\n", "3\n2 1\n1 2\n2 1\n" ]

[ "75 110 60 ", "6 4 6 " ]

none

1,000

[ { "input": "3\n1 10\n5 5\n10 1", "output": "75 110 60 " }, { "input": "3\n2 1\n1 2\n2 1", "output": "6 4 6 " }, { "input": "2\n1 5\n2 3", "output": "6 5 " }, { "input": "2\n2 3\n1 1", "output": "1 6 " }, { "input": "3\n1 10\n2 10\n3 10", "output": "50 40 30 " }, { "input": "3\n2 10\n1 9\n3 7", "output": "36 50 30 " }, { "input": "3\n1 1\n3 2\n2 3", "output": "15 9 8 " }, { "input": "3\n3 123\n1 456\n2 789", "output": "2367 3945 1824 " }, { "input": "3\n2 987\n3 654\n1 321", "output": "2616 2961 4935 " }, { "input": "3\n3 143\n2 543\n1 893", "output": "2679 3572 2715 " }, { "input": "2\n1 1\n1 2", "output": "2 1 " }, { "input": "3\n2 22\n1 11\n2 22", "output": "66 88 66 " }, { "input": "3\n1 11\n1 12\n1 13", "output": "26 26 24 " }, { "input": "3\n1 11\n1 12\n2 10", "output": "36 33 24 " }, { "input": "10\n6 20\n1 175\n1 758\n1 169\n2 490\n2 600\n4 463\n7 377\n9 40\n4 961", "output": "29791 34596 34596 34596 33635 33635 31713 28830 26908 25014 " }, { "input": "10\n8 158\n1 709\n6 766\n4 335\n5 356\n2 972\n1 108\n4 235\n3 631\n1 414", "output": "26244 33048 28188 30132 29160 25278 33048 30132 31104 33048 " }, { "input": "10\n7 549\n9 115\n8 141\n3 650\n5 730\n3 841\n7 18\n9 170\n2 217\n1 155", "output": "39527 37845 38686 42891 41209 37230 39527 37845 43732 44573 " }, { "input": "10\n6 386\n9 816\n9 268\n9 481\n8 284\n10 715\n9 351\n7 580\n4 327\n7 392", "output": "58752 49335 56304 56304 57120 55488 56304 57936 60384 57936 " }, { "input": "10\n9 292\n4 6\n6 638\n8 461\n10 970\n10 488\n9 769\n10 644\n8 280\n5 334", "output": "67900 72750 70810 68870 53061 66930 67900 66930 68870 71780 " }, { "input": "10\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000", "output": "90000 90000 90000 90000 90000 90000 90000 90000 90000 90000 " } ]

1,626,110,889

2,147,483,647

Python 3

OK

TESTS

44

795

19,763,200

n=int(input()); a=[];b=[]; for i in range(n): k=list(map(int,input().split())); a.append(k[0]); b.append(k[1]); M=max(b); m=b.index(M); MM=0; for i in range(n): if(i!=m and b[i]>MM):MM=b[i]; k=sum(a); c=[]; for i in range(n): if(i==m):c.append((k-a[i])*MM); else:c.append((k-a[i])*M); print(*c);

Title: Photo to Remember Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day *n* friends met at a party, they hadn't seen each other for a long time and so they decided to make a group photo together. Simply speaking, the process of taking photos can be described as follows. On the photo, each photographed friend occupies a rectangle of pixels: the *i*-th of them occupies the rectangle of width *w**i* pixels and height *h**i* pixels. On the group photo everybody stands in a line, thus the minimum pixel size of the photo including all the photographed friends, is *W*<=×<=*H*, where *W* is the total sum of all widths and *H* is the maximum height of all the photographed friends. As is usually the case, the friends made *n* photos — the *j*-th (1<=≤<=*j*<=≤<=*n*) photo had everybody except for the *j*-th friend as he was the photographer. Print the minimum size of each made photo in pixels. Input Specification: The first line contains integer *n* (2<=≤<=*n*<=≤<=200<=000) — the number of friends. Then *n* lines follow: the *i*-th line contains information about the *i*-th friend. The line contains a pair of integers *w**i*,<=*h**i* (1<=≤<=*w**i*<=≤<=10,<=1<=≤<=*h**i*<=≤<=1000) — the width and height in pixels of the corresponding rectangle. Output Specification: Print *n* space-separated numbers *b*1,<=*b*2,<=...,<=*b**n*, where *b**i* — the total number of pixels on the minimum photo containing all friends expect for the *i*-th one. Demo Input: ['3\n1 10\n5 5\n10 1\n', '3\n2 1\n1 2\n2 1\n'] Demo Output: ['75 110 60 ', '6 4 6 '] Note: none

```python n=int(input()); a=[];b=[]; for i in range(n): k=list(map(int,input().split())); a.append(k[0]); b.append(k[1]); M=max(b); m=b.index(M); MM=0; for i in range(n): if(i!=m and b[i]>MM):MM=b[i]; k=sum(a); c=[]; for i in range(n): if(i==m):c.append((k-a[i])*MM); else:c.append((k-a[i])*M); print(*c); ```

3

928

C

Dependency management

PROGRAMMING

1,900

[ "*special", "graphs", "implementation" ]

null

Polycarp is currently developing a project in Vaja language and using a popular dependency management system called Vamen. From Vamen's point of view both Vaja project and libraries are treated projects for simplicity. A project in Vaja has its own uniqie non-empty name consisting of lowercase latin letters with length not exceeding 10 and version — positive integer from 1 to 106. Each project (keep in mind that it is determined by both its name and version) might depend on other projects. For sure, there are no cyclic dependencies. You're given a list of project descriptions. The first of the given projects is the one being developed by Polycarp at this moment. Help Polycarp determine all projects that his project depends on (directly or via a certain chain). It's possible that Polycarp's project depends on two different versions of some project. In this case collision resolving is applied, i.e. for each such project the system chooses the version that minimizes the distance from it to Polycarp's project. If there are several options, the newer (with the maximum version) is preferred. This version is considered actual; other versions and their dependencies are ignored. More formal, choose such a set of projects of minimum possible size that the following conditions hold: - Polycarp's project is chosen; - Polycarp's project depends (directly or indirectly) on all other projects in the set; - no two projects share the name; - for each project *x* that some other project in the set depends on we have either *x* or some *y* with other version and shorter chain to Polycarp's project chosen. In case of ties the newer one is chosen. Output all Polycarp's project's dependencies (Polycarp's project itself should't be printed) in lexicographical order.

The first line contains an only integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of projects in Vaja. The following lines contain the project descriptions. Each project is described by a line consisting of its name and version separated by space. The next line gives the number of direct dependencies (from 0 to *n*<=-<=1) and the dependencies themselves (one in a line) in arbitrary order. Each dependency is specified by its name and version. The projects are also given in arbitrary order, but the first of them is always Polycarp's. Project descriptions are separated by one empty line. Refer to samples for better understanding. It's guaranteed that there are no cyclic dependencies.

Output all Polycarp's project's dependencies in lexicographical order.

[ "4\na 3\n2\nb 1\nc 1\n \nb 2\n0\n \nb 1\n1\nb 2\n \nc 1\n1\nb 2\n", "9\ncodehorses 5\n3\nwebfrmk 6\nmashadb 1\nmashadb 2\n \ncommons 2\n0\n \nmashadb 3\n0\n \nwebfrmk 6\n2\nmashadb 3\ncommons 2\n \nextra 4\n1\nextra 3\n \nextra 3\n0\n \nextra 1\n0\n \nmashadb 1\n1\nextra 3\n \nmashadb 2\n1\nextra 1\n", "3\nabc 1\n2\nabc 3\ncba 2\n\nabc 3\n0\n\ncba 2\n0\n" ]

[ "2\nb 1\nc 1\n", "4\ncommons 2\nextra 1\nmashadb 2\nwebfrmk 6\n", "1\ncba 2\n" ]

The first sample is given in the pic below. Arrow from *A* to *B* means that *B* directly depends on *A*. Projects that Polycarp's project «a» (version 3) depends on are painted black. The second sample is again given in the pic below. Arrow from *A* to *B* means that *B* directly depends on *A*. Projects that Polycarp's project «codehorses» (version 5) depends on are paint it black. Note that «extra 1» is chosen instead of «extra 3» since «mashadb 1» and all of its dependencies are ignored due to «mashadb 2».

2,000

[ { "input": "4\na 3\n2\nb 1\nc 1\n\nb 2\n0\n\nb 1\n1\nb 2\n\nc 1\n1\nb 2", "output": "2\nb 1\nc 1" }, { "input": "9\ncodehorses 5\n3\nwebfrmk 6\nmashadb 1\nmashadb 2\n\ncommons 2\n0\n\nmashadb 3\n0\n\nwebfrmk 6\n2\nmashadb 3\ncommons 2\n\nextra 4\n1\nextra 3\n\nextra 3\n0\n\nextra 1\n0\n\nmashadb 1\n1\nextra 3\n\nmashadb 2\n1\nextra 1", "output": "4\ncommons 2\nextra 1\nmashadb 2\nwebfrmk 6" }, { "input": "3\nabc 1\n2\nabc 3\ncba 2\n\nabc 3\n0\n\ncba 2\n0", "output": "1\ncba 2" }, { "input": "1\nabc 1000000\n0", "output": "0" }, { "input": "3\nppdpd 283157\n1\npddpdpp 424025\n\nppdpd 529292\n1\nppdpd 283157\n\npddpdpp 424025\n0", "output": "1\npddpdpp 424025" }, { "input": "5\nabbzzz 646068\n0\n\nzabza 468048\n2\nbb 902619\nzabza 550912\n\nzabza 217401\n2\nabbzzz 646068\nbb 902619\n\nzabza 550912\n1\nzabza 217401\n\nbb 902619\n1\nabbzzz 646068", "output": "0" }, { "input": "5\nyyyy 223967\n1\nyyyyyyy 254197\n\nyyyyyyy 254197\n0\n\ny 442213\n0\n\ny 965022\n1\nyyyyyyy 254197\n\nyyyy 766922\n4\nyyyyyyy 254197\ny 442213\nyyyy 223967\ny 965022", "output": "1\nyyyyyyy 254197" }, { "input": "3\nvvgvvgv 991444\n1\ngvgvgvgvgg 206648\n\nvvgvvgv 296188\n0\n\ngvgvgvgvgg 206648\n1\nvvgvvgv 296188", "output": "1\ngvgvgvgvgg 206648" }, { "input": "5\ntctocototo 984516\n1\ncttocottt 486791\n\ntctocototo 688522\n2\ncttocottt 486791\ntctocototo 984516\n\ncttocottt 486791\n0\n\ntctocototo 676435\n1\ntctocototo 394244\n\ntctocototo 394244\n2\ntctocototo 688522\ncttocottt 486791", "output": "1\ncttocottt 486791" }, { "input": "5\nggggggggg 202537\n4\ngggggggg 868552\ngggg 234633\ngggg 402994\ngggggggg 86863\n\ngggg 402994\n0\n\ngggggggg 868552\n1\ngggg 234633\n\ngggg 234633\n1\ngggg 402994\n\ngggggggg 86863\n0", "output": "2\ngggg 402994\ngggggggg 868552" }, { "input": "4\ncf 1\n2\ndb 1\ndb 2\n\ndb 2\n0\n\ndb 1\n1\nold 1\n\nold 1\n0", "output": "1\ndb 2" }, { "input": "5\ncf 1\n2\ndb 1\ndb 2\n\ndb 2\n1\nold 1\n\ndb 1\n1\nold 2\n\nold 1\n0\n\nold 2\n0", "output": "2\ndb 2\nold 1" } ]

1,519,588,713

2,147,483,647

Python 3

OK

TESTS

49

1,247

56,934,400

def bfs(polycarp, all_projects, existed_projects): # print(polycarp) # print(all_projects) # print(existed_projects) queue0 = {polycarp[0]: polycarp[1]} queue1 = {} while queue0: for p in queue0.items(): existed_projects[p[0]] = p[1] for pp in queue0.items(): for p in all_projects[pp]: if p[0] not in existed_projects: if p[0] not in queue1: queue1[p[0]] = p[1] else: queue1[p[0]] = max(queue1[p[0]], p[1]) else: continue queue0, queue1 = queue1, {} def main(): n = int(input()) all_projects = {} for _ in range(n): name, version = input().split() version = int(version) if _ == 0: polycarp = (name, version) nn = int(input()) all_projects[(name, version)] = [] for _2 in range(nn): name2, version2 = input().split() version2 = int(version2) all_projects[(name, version)].append((name2, version2)) if _ != n - 1: input() existed_projects = {} bfs(polycarp, all_projects, existed_projects) print(len(existed_projects) - 1) for name in sorted(existed_projects.keys()): if name != polycarp[0]: print(name, existed_projects[name]) if __name__ == '__main__': main()

Title: Dependency management Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp is currently developing a project in Vaja language and using a popular dependency management system called Vamen. From Vamen's point of view both Vaja project and libraries are treated projects for simplicity. A project in Vaja has its own uniqie non-empty name consisting of lowercase latin letters with length not exceeding 10 and version — positive integer from 1 to 106. Each project (keep in mind that it is determined by both its name and version) might depend on other projects. For sure, there are no cyclic dependencies. You're given a list of project descriptions. The first of the given projects is the one being developed by Polycarp at this moment. Help Polycarp determine all projects that his project depends on (directly or via a certain chain). It's possible that Polycarp's project depends on two different versions of some project. In this case collision resolving is applied, i.e. for each such project the system chooses the version that minimizes the distance from it to Polycarp's project. If there are several options, the newer (with the maximum version) is preferred. This version is considered actual; other versions and their dependencies are ignored. More formal, choose such a set of projects of minimum possible size that the following conditions hold: - Polycarp's project is chosen; - Polycarp's project depends (directly or indirectly) on all other projects in the set; - no two projects share the name; - for each project *x* that some other project in the set depends on we have either *x* or some *y* with other version and shorter chain to Polycarp's project chosen. In case of ties the newer one is chosen. Output all Polycarp's project's dependencies (Polycarp's project itself should't be printed) in lexicographical order. Input Specification: The first line contains an only integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of projects in Vaja. The following lines contain the project descriptions. Each project is described by a line consisting of its name and version separated by space. The next line gives the number of direct dependencies (from 0 to *n*<=-<=1) and the dependencies themselves (one in a line) in arbitrary order. Each dependency is specified by its name and version. The projects are also given in arbitrary order, but the first of them is always Polycarp's. Project descriptions are separated by one empty line. Refer to samples for better understanding. It's guaranteed that there are no cyclic dependencies. Output Specification: Output all Polycarp's project's dependencies in lexicographical order. Demo Input: ['4\na 3\n2\nb 1\nc 1\n\xa0\nb 2\n0\n\xa0\nb 1\n1\nb 2\n\xa0\nc 1\n1\nb 2\n', '9\ncodehorses 5\n3\nwebfrmk 6\nmashadb 1\nmashadb 2\n\xa0\ncommons 2\n0\n\xa0\nmashadb 3\n0\n\xa0\nwebfrmk 6\n2\nmashadb 3\ncommons 2\n\xa0\nextra 4\n1\nextra 3\n\xa0\nextra 3\n0\n\xa0\nextra 1\n0\n\xa0\nmashadb 1\n1\nextra 3\n\xa0\nmashadb 2\n1\nextra 1\n', '3\nabc 1\n2\nabc 3\ncba 2\n\nabc 3\n0\n\ncba 2\n0\n'] Demo Output: ['2\nb 1\nc 1\n', '4\ncommons 2\nextra 1\nmashadb 2\nwebfrmk 6\n', '1\ncba 2\n'] Note: The first sample is given in the pic below. Arrow from *A* to *B* means that *B* directly depends on *A*. Projects that Polycarp's project «a» (version 3) depends on are painted black. The second sample is again given in the pic below. Arrow from *A* to *B* means that *B* directly depends on *A*. Projects that Polycarp's project «codehorses» (version 5) depends on are paint it black. Note that «extra 1» is chosen instead of «extra 3» since «mashadb 1» and all of its dependencies are ignored due to «mashadb 2».

```python def bfs(polycarp, all_projects, existed_projects): # print(polycarp) # print(all_projects) # print(existed_projects) queue0 = {polycarp[0]: polycarp[1]} queue1 = {} while queue0: for p in queue0.items(): existed_projects[p[0]] = p[1] for pp in queue0.items(): for p in all_projects[pp]: if p[0] not in existed_projects: if p[0] not in queue1: queue1[p[0]] = p[1] else: queue1[p[0]] = max(queue1[p[0]], p[1]) else: continue queue0, queue1 = queue1, {} def main(): n = int(input()) all_projects = {} for _ in range(n): name, version = input().split() version = int(version) if _ == 0: polycarp = (name, version) nn = int(input()) all_projects[(name, version)] = [] for _2 in range(nn): name2, version2 = input().split() version2 = int(version2) all_projects[(name, version)].append((name2, version2)) if _ != n - 1: input() existed_projects = {} bfs(polycarp, all_projects, existed_projects) print(len(existed_projects) - 1) for name in sorted(existed_projects.keys()): if name != polycarp[0]: print(name, existed_projects[name]) if __name__ == '__main__': main() ```

3

543

B

Destroying Roads

PROGRAMMING

2,100

[ "constructive algorithms", "graphs", "shortest paths" ]

null

In some country there are exactly *n* cities and *m* bidirectional roads connecting the cities. Cities are numbered with integers from 1 to *n*. If cities *a* and *b* are connected by a road, then in an hour you can go along this road either from city *a* to city *b*, or from city *b* to city *a*. The road network is such that from any city you can get to any other one by moving along the roads. You want to destroy the largest possible number of roads in the country so that the remaining roads would allow you to get from city *s*1 to city *t*1 in at most *l*1 hours and get from city *s*2 to city *t*2 in at most *l*2 hours. Determine what maximum number of roads you need to destroy in order to meet the condition of your plan. If it is impossible to reach the desired result, print -1.

The first line contains two integers *n*, *m* (1<=≤<=*n*<=≤<=3000, ) — the number of cities and roads in the country, respectively. Next *m* lines contain the descriptions of the roads as pairs of integers *a**i*, *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*). It is guaranteed that the roads that are given in the description can transport you from any city to any other one. It is guaranteed that each pair of cities has at most one road between them. The last two lines contains three integers each, *s*1, *t*1, *l*1 and *s*2, *t*2, *l*2, respectively (1<=≤<=*s**i*,<=*t**i*<=≤<=*n*, 0<=≤<=*l**i*<=≤<=*n*).

Print a single number — the answer to the problem. If the it is impossible to meet the conditions, print -1.

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

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

none

1,000

[ { "input": "5 4\n1 2\n2 3\n3 4\n4 5\n1 3 2\n3 5 2", "output": "0" }, { "input": "5 4\n1 2\n2 3\n3 4\n4 5\n1 3 2\n2 4 2", "output": "1" }, { "input": "5 4\n1 2\n2 3\n3 4\n4 5\n1 3 2\n3 5 1", "output": "-1" }, { "input": "9 9\n1 2\n2 3\n2 4\n4 5\n5 7\n5 6\n3 8\n8 9\n9 6\n1 7 4\n3 6 3", "output": "2" }, { "input": "9 9\n1 2\n2 3\n2 4\n4 5\n5 7\n5 6\n3 8\n8 9\n9 6\n1 7 4\n3 6 4", "output": "3" }, { "input": "10 11\n1 3\n2 3\n3 4\n4 5\n4 6\n3 7\n3 8\n4 9\n4 10\n7 9\n8 10\n1 5 3\n6 2 3", "output": "6" }, { "input": "1 0\n1 1 0\n1 1 0", "output": "0" }, { "input": "2 1\n1 2\n1 1 0\n1 2 1", "output": "0" }, { "input": "2 1\n1 2\n1 1 0\n1 2 0", "output": "-1" }, { "input": "6 5\n1 3\n2 3\n3 4\n4 5\n4 6\n1 6 3\n5 2 3", "output": "0" }, { "input": "6 5\n1 2\n2 3\n3 4\n3 5\n2 6\n1 4 3\n5 6 3", "output": "0" }, { "input": "5 4\n1 2\n2 3\n3 4\n4 5\n1 3 2\n4 2 2", "output": "1" } ]

1,665,074,435

2,147,483,647

PyPy 3-64

OK

TESTS

63

1,731

91,955,200

from collections import deque n,m=map(int,input().split()) graph=[[] for _ in range(n)] for i in range(m): a,b=map(int,input().split()) a-=1 b-=1 graph[a].append(b) graph[b].append(a) d=[[-1]*n for _ in range(n)] for i in range(n): q=deque() q.append(i) d[i][i]=0 while q: node=q.popleft() for j in range(len(graph[node])): to=graph[node][j] if d[i][to]==-1: d[i][to]=d[i][node]+1 q.append(to) a,b,c=map(int,input().split()) o,e,f=map(int,input().split()) a-=1 b-=1 e-=1 o-=1 ans=m+1 for _ in range(2): a,b=b,a for i in range(n): for j in range(n): v=[d[a][i]+d[i][j]+d[j][b],d[o][i]+d[i][j]+d[j][e]] if v[0]<=c and v[1]<=f: ans=min(ans,v[0]+v[1]-d[i][j]) if d[a][b]<=c and d[o][e]<=f: ans=min(ans,d[a][b]+d[o][e]) if ans>m: print(-1) else : ans=m-ans print(ans)

Title: Destroying Roads Time Limit: None seconds Memory Limit: None megabytes Problem Description: In some country there are exactly *n* cities and *m* bidirectional roads connecting the cities. Cities are numbered with integers from 1 to *n*. If cities *a* and *b* are connected by a road, then in an hour you can go along this road either from city *a* to city *b*, or from city *b* to city *a*. The road network is such that from any city you can get to any other one by moving along the roads. You want to destroy the largest possible number of roads in the country so that the remaining roads would allow you to get from city *s*1 to city *t*1 in at most *l*1 hours and get from city *s*2 to city *t*2 in at most *l*2 hours. Determine what maximum number of roads you need to destroy in order to meet the condition of your plan. If it is impossible to reach the desired result, print -1. Input Specification: The first line contains two integers *n*, *m* (1<=≤<=*n*<=≤<=3000, ) — the number of cities and roads in the country, respectively. Next *m* lines contain the descriptions of the roads as pairs of integers *a**i*, *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*). It is guaranteed that the roads that are given in the description can transport you from any city to any other one. It is guaranteed that each pair of cities has at most one road between them. The last two lines contains three integers each, *s*1, *t*1, *l*1 and *s*2, *t*2, *l*2, respectively (1<=≤<=*s**i*,<=*t**i*<=≤<=*n*, 0<=≤<=*l**i*<=≤<=*n*). Output Specification: Print a single number — the answer to the problem. If the it is impossible to meet the conditions, print -1. Demo Input: ['5 4\n1 2\n2 3\n3 4\n4 5\n1 3 2\n3 5 2\n', '5 4\n1 2\n2 3\n3 4\n4 5\n1 3 2\n2 4 2\n', '5 4\n1 2\n2 3\n3 4\n4 5\n1 3 2\n3 5 1\n'] Demo Output: ['0\n', '1\n', '-1\n'] Note: none

```python from collections import deque n,m=map(int,input().split()) graph=[[] for _ in range(n)] for i in range(m): a,b=map(int,input().split()) a-=1 b-=1 graph[a].append(b) graph[b].append(a) d=[[-1]*n for _ in range(n)] for i in range(n): q=deque() q.append(i) d[i][i]=0 while q: node=q.popleft() for j in range(len(graph[node])): to=graph[node][j] if d[i][to]==-1: d[i][to]=d[i][node]+1 q.append(to) a,b,c=map(int,input().split()) o,e,f=map(int,input().split()) a-=1 b-=1 e-=1 o-=1 ans=m+1 for _ in range(2): a,b=b,a for i in range(n): for j in range(n): v=[d[a][i]+d[i][j]+d[j][b],d[o][i]+d[i][j]+d[j][e]] if v[0]<=c and v[1]<=f: ans=min(ans,v[0]+v[1]-d[i][j]) if d[a][b]<=c and d[o][e]<=f: ans=min(ans,d[a][b]+d[o][e]) if ans>m: print(-1) else : ans=m-ans print(ans) ```

3

148

A

Insomnia cure

PROGRAMMING

800

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

null

«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine. However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic. How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?

Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).

Output the number of damaged dragons.

[ "1\n2\n3\n4\n12\n", "2\n3\n4\n5\n24\n" ]

[ "12\n", "17\n" ]

In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough. In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.

1,000

[ { "input": "1\n2\n3\n4\n12", "output": "12" }, { "input": "2\n3\n4\n5\n24", "output": "17" }, { "input": "1\n1\n1\n1\n100000", "output": "100000" }, { "input": "10\n9\n8\n7\n6", "output": "0" }, { "input": "8\n4\n4\n3\n65437", "output": "32718" }, { "input": "8\n4\n1\n10\n59392", "output": "59392" }, { "input": "4\n1\n8\n7\n44835", "output": "44835" }, { "input": "6\n1\n7\n2\n62982", "output": "62982" }, { "input": "2\n7\n4\n9\n56937", "output": "35246" }, { "input": "2\n9\n8\n1\n75083", "output": "75083" }, { "input": "8\n7\n7\n6\n69038", "output": "24656" }, { "input": "4\n4\n2\n3\n54481", "output": "36320" }, { "input": "6\n4\n9\n8\n72628", "output": "28244" }, { "input": "9\n7\n8\n10\n42357", "output": "16540" }, { "input": "5\n6\n4\n3\n60504", "output": "36302" }, { "input": "7\n2\n3\n8\n21754", "output": "15539" }, { "input": "1\n2\n10\n4\n39901", "output": "39901" }, { "input": "3\n4\n7\n1\n58048", "output": "58048" }, { "input": "9\n10\n4\n6\n52003", "output": "21956" }, { "input": "5\n10\n9\n3\n70149", "output": "32736" }, { "input": "5\n5\n5\n10\n55592", "output": "11118" }, { "input": "1\n5\n2\n6\n49547", "output": "49547" }, { "input": "3\n7\n7\n7\n84046", "output": "36019" }, { "input": "10\n2\n1\n8\n63537", "output": "63537" }, { "input": "7\n2\n10\n5\n81684", "output": "53678" }, { "input": "7\n1\n6\n8\n99831", "output": "99831" }, { "input": "3\n9\n3\n3\n61082", "output": "20360" }, { "input": "5\n5\n8\n9\n79228", "output": "29931" }, { "input": "1\n5\n5\n6\n89535", "output": "89535" }, { "input": "5\n5\n2\n1\n7682", "output": "7682" }, { "input": "7\n4\n8\n8\n25829", "output": "9224" }, { "input": "3\n4\n3\n3\n87079", "output": "43539" }, { "input": "5\n8\n4\n9\n5226", "output": "2438" }, { "input": "4\n9\n6\n6\n1070", "output": "415" }, { "input": "10\n10\n10\n10\n100000", "output": "10000" }, { "input": "1\n1\n1\n1\n1", "output": "1" }, { "input": "10\n10\n10\n10\n1", "output": "0" }, { "input": "10\n10\n1\n10\n10", "output": "10" }, { "input": "10\n10\n1\n10\n100", "output": "100" }, { "input": "2\n2\n2\n2\n1", "output": "0" } ]

1,691,280,748

2,147,483,647

PyPy 3

OK

TESTS

40

186

2,969,600

import sys *A,d = map(int,sys.stdin.read().split()) print(sum(any((i%p==0) for p in A) for i in range(1,d+1)))

Title: Insomnia cure Time Limit: None seconds Memory Limit: None megabytes Problem Description: «One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine. However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic. How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons? Input Specification: Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105). Output Specification: Output the number of damaged dragons. Demo Input: ['1\n2\n3\n4\n12\n', '2\n3\n4\n5\n24\n'] Demo Output: ['12\n', '17\n'] Note: In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough. In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.

```python import sys *A,d = map(int,sys.stdin.read().split()) print(sum(any((i%p==0) for p in A) for i in range(1,d+1))) ```

3

119

A

Epic Game

PROGRAMMING

800

[ "implementation" ]

null

Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game.

The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.

If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).

[ "3 5 9\n", "1 1 100\n" ]

[ "0", "1" ]

The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses. In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.

500

[ { "input": "3 5 9", "output": "0" }, { "input": "1 1 100", "output": "1" }, { "input": "23 12 16", "output": "1" }, { "input": "95 26 29", "output": "1" }, { "input": "73 32 99", "output": "1" }, { "input": "1 1 1", "output": "0" }, { "input": "41 12 65", "output": "1" }, { "input": "13 61 100", "output": "1" }, { "input": "100 100 10", "output": "0" }, { "input": "12 24 26", "output": "1" }, { "input": "73 21 96", "output": "1" }, { "input": "17 22 81", "output": "1" }, { "input": "14 88 97", "output": "1" }, { "input": "42 81 17", "output": "0" }, { "input": "34 36 19", "output": "1" }, { "input": "21 87 12", "output": "1" }, { "input": "97 92 98", "output": "1" }, { "input": "49 25 62", "output": "1" }, { "input": "29 17 44", "output": "1" }, { "input": "10 1 10", "output": "0" }, { "input": "99 98 100", "output": "1" }, { "input": "100 100 1", "output": "0" }, { "input": "11 99 55", "output": "0" }, { "input": "8 13 21", "output": "0" }, { "input": "34 55 89", "output": "0" }, { "input": "21 34 55", "output": "1" }, { "input": "13 21 34", "output": "1" }, { "input": "34 21 55", "output": "0" }, { "input": "10 20 84", "output": "1" }, { "input": "4 2 5", "output": "0" }, { "input": "3 2 3", "output": "0" }, { "input": "31 23 1", "output": "0" }, { "input": "100 100 100", "output": "0" }, { "input": "100 1 100", "output": "0" }, { "input": "1 100 100", "output": "1" }, { "input": "1 1 100", "output": "1" }, { "input": "17 23 100", "output": "1" }, { "input": "99 98 100", "output": "1" }, { "input": "1 1 2", "output": "1" }, { "input": "2 1 1", "output": "0" }, { "input": "1 2 1", "output": "0" }, { "input": "2 1 2", "output": "0" }, { "input": "10 10 3", "output": "1" }, { "input": "2 2 1", "output": "0" }, { "input": "9 5 1", "output": "0" } ]

1,622,114,598

2,147,483,647

Python 3

OK

TESTS

45

124

0

import math a,b,n=input().split() a=int(a) b=int(b) n=int(n) i=1 while(n>0): m=0 if i%2!=0: m=math.gcd(a,n) n=n-m elif i%2==0: m=math.gcd(b,n) n=n-m i+=1 if i%2==0: print(0) else: print(1)

Title: Epic Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game. Input Specification: The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. Output Specification: If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). Demo Input: ['3 5 9\n', '1 1 100\n'] Demo Output: ['0', '1'] Note: The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses. In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.

```python import math a,b,n=input().split() a=int(a) b=int(b) n=int(n) i=1 while(n>0): m=0 if i%2!=0: m=math.gcd(a,n) n=n-m elif i%2==0: m=math.gcd(b,n) n=n-m i+=1 if i%2==0: print(0) else: print(1) ```

3

119

A

Epic Game

PROGRAMMING

800

[ "implementation" ]

null

Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game.

The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.

If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).

[ "3 5 9\n", "1 1 100\n" ]

[ "0", "1" ]

The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses. In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.

500

[ { "input": "3 5 9", "output": "0" }, { "input": "1 1 100", "output": "1" }, { "input": "23 12 16", "output": "1" }, { "input": "95 26 29", "output": "1" }, { "input": "73 32 99", "output": "1" }, { "input": "1 1 1", "output": "0" }, { "input": "41 12 65", "output": "1" }, { "input": "13 61 100", "output": "1" }, { "input": "100 100 10", "output": "0" }, { "input": "12 24 26", "output": "1" }, { "input": "73 21 96", "output": "1" }, { "input": "17 22 81", "output": "1" }, { "input": "14 88 97", "output": "1" }, { "input": "42 81 17", "output": "0" }, { "input": "34 36 19", "output": "1" }, { "input": "21 87 12", "output": "1" }, { "input": "97 92 98", "output": "1" }, { "input": "49 25 62", "output": "1" }, { "input": "29 17 44", "output": "1" }, { "input": "10 1 10", "output": "0" }, { "input": "99 98 100", "output": "1" }, { "input": "100 100 1", "output": "0" }, { "input": "11 99 55", "output": "0" }, { "input": "8 13 21", "output": "0" }, { "input": "34 55 89", "output": "0" }, { "input": "21 34 55", "output": "1" }, { "input": "13 21 34", "output": "1" }, { "input": "34 21 55", "output": "0" }, { "input": "10 20 84", "output": "1" }, { "input": "4 2 5", "output": "0" }, { "input": "3 2 3", "output": "0" }, { "input": "31 23 1", "output": "0" }, { "input": "100 100 100", "output": "0" }, { "input": "100 1 100", "output": "0" }, { "input": "1 100 100", "output": "1" }, { "input": "1 1 100", "output": "1" }, { "input": "17 23 100", "output": "1" }, { "input": "99 98 100", "output": "1" }, { "input": "1 1 2", "output": "1" }, { "input": "2 1 1", "output": "0" }, { "input": "1 2 1", "output": "0" }, { "input": "2 1 2", "output": "0" }, { "input": "10 10 3", "output": "1" }, { "input": "2 2 1", "output": "0" }, { "input": "9 5 1", "output": "0" } ]

1,659,523,890

2,147,483,647

PyPy 3-64

OK

TESTS

45

122

0

import math x = (input().split()) a = int(x[0]) b = int(x[1]) c = int(x[2]) for i in range(100000000): if i%2==0: c-=math.gcd(c,a) else: c-=math.gcd(c,b) if c==0: print(i%2) break

Title: Epic Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take). Your task is to determine by the given *a*, *b* and *n* who wins the game. Input Specification: The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. Output Specification: If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). Demo Input: ['3 5 9\n', '1 1 100\n'] Demo Output: ['0', '1'] Note: The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*. In the first sample the game will go like that: - Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses. In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.

```python import math x = (input().split()) a = int(x[0]) b = int(x[1]) c = int(x[2]) for i in range(100000000): if i%2==0: c-=math.gcd(c,a) else: c-=math.gcd(c,b) if c==0: print(i%2) break ```

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" } ]

1,676,770,823

2,147,483,647

Python 3

OK

TESTS

30

92

0

cadena=input() contador=0 n=len(cadena) for i in cadena: if ord(i)<97: contador+=1 j=n/2 if (contador>j): print(cadena.upper()) elif (contador<j): print(cadena.lower()) else: print(cadena.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 cadena=input() contador=0 n=len(cadena) for i in cadena: if ord(i)<97: contador+=1 j=n/2 if (contador>j): print(cadena.upper()) elif (contador<j): print(cadena.lower()) else: print(cadena.lower()) ```

3.977

801

A

Vicious Keyboard

PROGRAMMING

1,100

[ "brute force" ]

null

Tonio has a keyboard with only two letters, "V" and "K". One day, he has typed out a string *s* with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i. e. a letter "K" right after a letter "V") in the resulting string.

The first line will contain a string *s* consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100.

Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character.

[ "VK\n", "VV\n", "V\n", "VKKKKKKKKKVVVVVVVVVK\n", "KVKV\n" ]

[ "1\n", "1\n", "0\n", "3\n", "1\n" ]

For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear. For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring. For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.

500

[ { "input": "VK", "output": "1" }, { "input": "VV", "output": "1" }, { "input": "V", "output": "0" }, { "input": "VKKKKKKKKKVVVVVVVVVK", "output": "3" }, { "input": "KVKV", "output": "1" }, { "input": "VKKVVVKVKVK", "output": "5" }, { "input": "VKVVKVKVVKVKKKKVVVVVVVVKVKVVVVVVKKVKKVKVVKVKKVVVVKV", "output": "14" }, { "input": "VVKKVKKVVKKVKKVKVVKKVKKVVKKVKVVKKVKKVKVVKKVVKKVKVVKKVKVVKKVVKVVKKVKKVKKVKKVKKVKVVKKVKKVKKVKKVKKVVKVK", "output": "32" }, { "input": "KVVKKVKVKVKVKVKKVKVKVVKVKVVKVVKVKKVKVKVKVKVKVKVKVKVKVKVKVKVKVKVVKVKVVKKVKVKK", "output": "32" }, { "input": "KVVVVVKKVKVVKVVVKVVVKKKVKKKVVKVKKKVKKKKVKVVVVVKKKVVVVKKVVVVKKKVKVVVVVVVKKVKVKKKVVKVVVKVVKK", "output": "21" }, { "input": "VVVVVKKVKVKVKVVKVVKKVVKVKKKKKKKVKKKVVVVVVKKVVVKVKVVKVKKVVKVVVKKKKKVVVVVKVVVVKVVVKKVKKVKKKVKKVKKVVKKV", "output": "25" }, { "input": "KKVVKVVKVVKKVVKKVKVVKKV", "output": "7" }, { "input": "KKVVKKVVVKKVKKVKKVVVKVVVKKVKKVVVKKVVVKVVVKVVVKKVVVKKVVVKVVVKKVVVKVVKKVVVKKVVVKKVVKVVVKKVVKKVKKVVVKKV", "output": "24" }, { "input": "KVKVKVKVKVKVKVKVKVKVVKVKVKVKVKVKVKVVKVKVKKVKVKVKVKVVKVKVKVKVKVKVKVKVKKVKVKVV", "output": "35" }, { "input": "VKVVVKKKVKVVKVKVKVKVKVV", "output": "9" }, { "input": "KKKKVKKVKVKVKKKVVVVKK", "output": "6" }, { "input": "KVKVKKVVVVVVKKKVKKKKVVVVKVKKVKVVK", "output": "9" }, { "input": "KKVKKVKKKVKKKVKKKVKVVVKKVVVVKKKVKKVVKVKKVKVKVKVVVKKKVKKKKKVVKVVKVVVKKVVKVVKKKKKVK", "output": "22" }, { "input": "VVVKVKVKVVVVVKVVVKKVVVKVVVVVKKVVKVVVKVVVKVKKKVVKVVVVVKVVVVKKVVKVKKVVKKKVKVVKVKKKKVVKVVVKKKVKVKKKKKK", "output": "25" }, { "input": "VKVVKVVKKKVVKVKKKVVKKKVVKVVKVVKKVKKKVKVKKKVVKVKKKVVKVVKKKVVKKKVKKKVVKKVVKKKVKVKKKVKKKVKKKVKVKKKVVKVK", "output": "29" }, { "input": "KKVKVVVKKVV", "output": "3" }, { "input": "VKVKVKVKVKVKVKVKVKVKVVKVKVKVKVKVK", "output": "16" }, { "input": "VVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVV", "output": "13" }, { "input": "VVKKVKVKKKVVVKVVVKVKKVKKKVVVKVVKVKKVKKVKVKVVKKVVKKVKVVKKKVVKKVVVKVKVVVKVKVVKVKKVKKV", "output": "26" }, { "input": "VVKVKKVVKKVVKKVVKKVVKKVKKVVKVKKVVKKVVKKVVKKVVKKVVKVVKKVVKVVKKVVKVVKKVVKKVKKVVKVVKKVVKVVKKVV", "output": "26" }, { "input": "K", "output": "0" }, { "input": "VKVK", "output": "2" }, { "input": "VKVV", "output": "2" }, { "input": "KV", "output": "0" }, { "input": "KK", "output": "1" }, { "input": "KKVK", "output": "2" }, { "input": "KKKK", "output": "1" }, { "input": "KKV", "output": "1" }, { "input": "KKVKVK", "output": "3" }, { "input": "VKKVK", "output": "2" }, { "input": "VKKK", "output": "2" }, { "input": "KKK", "output": "1" }, { "input": "KVV", "output": "1" }, { "input": "KKVKV", "output": "2" }, { "input": "VVK", "output": "1" }, { "input": "VVVKVKVKVKVKVKVK", "output": "8" }, { "input": "KVVVK", "output": "2" }, { "input": "VVVKK", "output": "2" }, { "input": "KKVV", "output": "1" }, { "input": "KKKKKKK", "output": "1" }, { "input": "VKKKVK", "output": "3" }, { "input": "KKVVV", "output": "1" }, { "input": "VVVVVV", "output": "1" }, { "input": "KKKV", "output": "1" }, { "input": "VVKVV", "output": "2" }, { "input": "VKVKKK", "output": "3" }, { "input": "VKKV", "output": "1" }, { "input": "VKKVV", "output": "2" }, { "input": "VVKKVV", "output": "2" }, { "input": "KKVVKKV", "output": "2" }, { "input": "KKKKK", "output": "1" }, { "input": "VKVVKKVKKVVKVKKVKKKVKKVKVKK", "output": "10" }, { "input": "VKVKVV", "output": "3" }, { "input": "VKVVKVV", "output": "3" }, { "input": "VVV", "output": "1" }, { "input": "VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV", "output": "1" }, { "input": "VVKKKKKKVKK", "output": "3" }, { "input": "KVKVKVV", "output": "3" } ]

1,492,357,542

642

Python 3

OK

TESTS

63

62

5,529,600

#l=[int(i)for i in input().split()] s=input() i=0 a=s.find("VK") ans=0 while a!=-1: ans+=1 s=s[:a]+"."+s[a+2:] a=s.find("VK") if s.count("VV")>0 or s.count("KK")>0:ans+=1 print(ans)

Title: Vicious Keyboard Time Limit: None seconds Memory Limit: None megabytes Problem Description: Tonio has a keyboard with only two letters, "V" and "K". One day, he has typed out a string *s* with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i. e. a letter "K" right after a letter "V") in the resulting string. Input Specification: The first line will contain a string *s* consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100. Output Specification: Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character. Demo Input: ['VK\n', 'VV\n', 'V\n', 'VKKKKKKKKKVVVVVVVVVK\n', 'KVKV\n'] Demo Output: ['1\n', '1\n', '0\n', '3\n', '1\n'] Note: For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear. For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring. For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.

```python #l=[int(i)for i in input().split()] s=input() i=0 a=s.find("VK") ans=0 while a!=-1: ans+=1 s=s[:a]+"."+s[a+2:] a=s.find("VK") if s.count("VV")>0 or s.count("KK")>0:ans+=1 print(ans) ```

3

791

A

Bear and Big Brother

PROGRAMMING

800

[ "implementation" ]

null

Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob. Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight. Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year. After how many full years will Limak become strictly larger (strictly heavier) than Bob?

The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively.

Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob.

[ "4 7\n", "4 9\n", "1 1\n" ]

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

In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2. In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights. In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then.

500

[ { "input": "4 7", "output": "2" }, { "input": "4 9", "output": "3" }, { "input": "1 1", "output": "1" }, { "input": "4 6", "output": "2" }, { "input": "1 10", "output": "6" }, { "input": "1 1", "output": "1" }, { "input": "1 2", "output": "2" }, { "input": "1 3", "output": "3" }, { "input": "1 4", "output": "4" }, { "input": "1 5", "output": "4" }, { "input": "1 6", "output": "5" }, { "input": "1 7", "output": "5" }, { "input": "1 8", "output": "6" }, { "input": "1 9", "output": "6" }, { "input": "1 10", "output": "6" }, { "input": "2 2", "output": "1" }, { "input": "2 3", "output": "2" }, { "input": "2 4", "output": "2" }, { "input": "2 5", "output": "3" }, { "input": "2 6", "output": "3" }, { "input": "2 7", "output": "4" }, { "input": "2 8", "output": "4" }, { "input": "2 9", "output": "4" }, { "input": "2 10", "output": "4" }, { "input": "3 3", "output": "1" }, { "input": "3 4", "output": "1" }, { "input": "3 5", "output": "2" }, { "input": "3 6", "output": "2" }, { "input": "3 7", "output": "3" }, { "input": "3 8", "output": "3" }, { "input": "3 9", "output": "3" }, { "input": "3 10", "output": "3" }, { "input": "4 4", "output": "1" }, { "input": "4 5", "output": "1" }, { "input": "4 6", "output": "2" }, { "input": "4 7", "output": "2" }, { "input": "4 8", "output": "2" }, { "input": "4 9", "output": "3" }, { "input": "4 10", "output": "3" }, { "input": "5 5", "output": "1" }, { "input": "5 6", "output": "1" }, { "input": "5 7", "output": "1" }, { "input": "5 8", "output": "2" }, { "input": "5 9", "output": "2" }, { "input": "5 10", "output": "2" }, { "input": "6 6", "output": "1" }, { "input": "6 7", "output": "1" }, { "input": "6 8", "output": "1" }, { "input": "6 9", "output": "2" }, { "input": "6 10", "output": "2" }, { "input": "7 7", "output": "1" }, { "input": "7 8", "output": "1" }, { "input": "7 9", "output": "1" }, { "input": "7 10", "output": "1" }, { "input": "8 8", "output": "1" }, { "input": "8 9", "output": "1" }, { "input": "8 10", "output": "1" }, { "input": "9 9", "output": "1" }, { "input": "9 10", "output": "1" }, { "input": "10 10", "output": "1" }, { "input": "10 10", "output": "1" }, { "input": "1 2", "output": "2" } ]

1,693,757,144

2,147,483,647

Python 3

OK

TESTS

62

46

0

x,y=map(int,input().split()) c=0 while x<=y: c=c+1 x*=3 y*=2 print(c)

Title: Bear and Big Brother Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob. Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight. Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year. After how many full years will Limak become strictly larger (strictly heavier) than Bob? Input Specification: The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively. Output Specification: Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob. Demo Input: ['4 7\n', '4 9\n', '1 1\n'] Demo Output: ['2\n', '3\n', '1\n'] Note: In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2. In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights. In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then.

```python x,y=map(int,input().split()) c=0 while x<=y: c=c+1 x*=3 y*=2 print(c) ```

3

996

A

Hit the Lottery

PROGRAMMING

800

[ "dp", "greedy" ]

null

Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?

The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).

Output the minimum number of bills that Allen could receive.

[ "125\n", "43\n", "1000000000\n" ]

[ "3\n", "5\n", "10000000\n" ]

In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills. In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills. In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.

500

[ { "input": "125", "output": "3" }, { "input": "43", "output": "5" }, { "input": "1000000000", "output": "10000000" }, { "input": "4", "output": "4" }, { "input": "5", "output": "1" }, { "input": "1", "output": "1" }, { "input": "74", "output": "8" }, { "input": "31", "output": "3" }, { "input": "59", "output": "8" }, { "input": "79", "output": "9" }, { "input": "7", "output": "3" }, { "input": "55", "output": "4" }, { "input": "40", "output": "2" }, { "input": "719", "output": "13" }, { "input": "847", "output": "13" }, { "input": "225", "output": "4" }, { "input": "4704", "output": "51" }, { "input": "1132", "output": "15" }, { "input": "7811", "output": "80" }, { "input": "7981", "output": "84" }, { "input": "82655", "output": "830" }, { "input": "6364", "output": "70" }, { "input": "74611", "output": "748" }, { "input": "45391", "output": "459" }, { "input": "620448", "output": "6210" }, { "input": "265145", "output": "2654" }, { "input": "671704", "output": "6721" }, { "input": "365173", "output": "3658" }, { "input": "7130872", "output": "71314" }, { "input": "9628747", "output": "96292" }, { "input": "8898399", "output": "88993" }, { "input": "9497953", "output": "94985" }, { "input": "19070947", "output": "190714" }, { "input": "20185520", "output": "201856" }, { "input": "91402248", "output": "914028" }, { "input": "27035533", "output": "270360" }, { "input": "22717528", "output": "227180" }, { "input": "24403439", "output": "244041" }, { "input": "163565555", "output": "1635659" }, { "input": "152541908", "output": "1525423" }, { "input": "668439837", "output": "6684403" }, { "input": "20", "output": "1" }, { "input": "999999999", "output": "10000009" }, { "input": "10", "output": "1" }, { "input": "7", "output": "3" }, { "input": "6", "output": "2" }, { "input": "99999999", "output": "1000009" }, { "input": "1000001", "output": "10001" }, { "input": "2521", "output": "27" } ]

1,694,426,516

2,147,483,647

Python 3

OK

TESTS

49

46

0

n = int(input()) ct = 0 while(n>0): if n >= 100: ct += n//100 n = n%100 elif n >= 20: ct += n//20 n = n%20 elif n >= 10: ct += n//10 n = n%10 elif n >= 5: ct += n//5 n = n%5 else: ct += n n = 0 print(ct)

Title: Hit the Lottery Time Limit: None seconds Memory Limit: None megabytes Problem Description: Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance? Input Specification: The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$). Output Specification: Output the minimum number of bills that Allen could receive. Demo Input: ['125\n', '43\n', '1000000000\n'] Demo Output: ['3\n', '5\n', '10000000\n'] Note: In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills. In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills. In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.

```python n = int(input()) ct = 0 while(n>0): if n >= 100: ct += n//100 n = n%100 elif n >= 20: ct += n//20 n = n%20 elif n >= 10: ct += n//10 n = n%10 elif n >= 5: ct += n//5 n = n%5 else: ct += n n = 0 print(ct) ```

3

734

A

Anton and Danik

PROGRAMMING

800

[ "implementation", "strings" ]

null

Anton likes to play chess, and so does his friend Danik. Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie. Now Anton wonders, who won more games, he or Danik? Help him determine this.

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played. The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game.

If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output. If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output. If Anton and Danik won the same number of games, print "Friendship" (without quotes).

[ "6\nADAAAA\n", "7\nDDDAADA\n", "6\nDADADA\n" ]

[ "Anton\n", "Danik\n", "Friendship\n" ]

In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton". In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik". In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".

500

[ { "input": "6\nADAAAA", "output": "Anton" }, { "input": "7\nDDDAADA", "output": "Danik" }, { "input": "6\nDADADA", "output": "Friendship" }, { "input": "10\nDDDDADDADD", "output": "Danik" }, { "input": "40\nAAAAAAAAADDAAAAAAAAAAADADDAAAAAAAAAAADAA", "output": "Anton" }, { "input": "200\nDDDDDDDADDDDDDAADADAADAAADAADADAAADDDADDDDDDADDDAADDDAADADDDDDADDDAAAADAAADDDDDAAADAADDDAAAADDADADDDAADDAADAAADAADAAAADDAADDADAAAADADDDAAAAAADDAADAADAADADDDAAADAAAADADDADAAAAAADADADDDADDDAADDADDDAAAAD", "output": "Friendship" }, { "input": "1\nA", "output": "Anton" }, { "input": "1\nD", "output": "Danik" }, { "input": "2\nDA", "output": "Friendship" }, { "input": "4\nDADA", "output": "Friendship" }, { "input": "4\nDAAD", "output": "Friendship" }, { "input": "3\nADD", "output": "Danik" }, { "input": "3\nDAD", "output": "Danik" }, { "input": "2\nDA", "output": "Friendship" }, { "input": "379\nAADAAAAAADDAAAAAADAADADADDAAAAADADDAADAAAADDDADAAAAAAADAADAAAAAAADAAAAAAAAADAAAAAAADAAAAAAAAAAADDDADAAAAAAAADAADADAAAADAAAAAAAAAAAAAAAAADAAAADDDAADAAAAAAADAAADAAADAADDDADDAAADAAAAAADDDADDDAAADAAAADAAAAAAAAADAAADAAAAAAAAADAAAAAAAAAAAAAAAAAADADAAAAAAAAAAADAAAAADAAAADAAAAAAAAAAAAADADAADAAAAAAAADAADAAAAAAAADAAAAAAAADDDAAAAAADAAADAAAAAADAADAAAAAADAAAADADAADAAAAAADAAAADAADDAADAADAAA", "output": "Anton" } ]

1,694,275,547

2,147,483,647

PyPy 3-64

OK

TESTS

25

62

4,915,200

num = input() results = input() winners = list(results) if winners.count("A") > winners.count("D"): print("Anton") elif winners.count("A") < winners.count("D"): print("Danik") else: print("Friendship")

Title: Anton and Danik Time Limit: None seconds Memory Limit: None megabytes Problem Description: Anton likes to play chess, and so does his friend Danik. Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie. Now Anton wonders, who won more games, he or Danik? Help him determine this. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played. The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game. Output Specification: If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output. If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output. If Anton and Danik won the same number of games, print "Friendship" (without quotes). Demo Input: ['6\nADAAAA\n', '7\nDDDAADA\n', '6\nDADADA\n'] Demo Output: ['Anton\n', 'Danik\n', 'Friendship\n'] Note: In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton". In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik". In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".

```python num = input() results = input() winners = list(results) if winners.count("A") > winners.count("D"): print("Anton") elif winners.count("A") < winners.count("D"): print("Danik") else: print("Friendship") ```

3

469

A

I Wanna Be the Guy

PROGRAMMING

800

[ "greedy", "implementation" ]

null

There is a game called "I Wanna Be the Guy", consisting of *n* levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game. Little X can pass only *p* levels of the game. And Little Y can pass only *q* levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?

The first line contains a single integer *n* (1<=≤<=<=*n*<=≤<=100). The next line contains an integer *p* (0<=≤<=*p*<=≤<=*n*) at first, then follows *p* distinct integers *a*1,<=*a*2,<=...,<=*a**p* (1<=≤<=*a**i*<=≤<=*n*). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to *n*.

If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).

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

[ "I become the guy.\n", "Oh, my keyboard!\n" ]

In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both. In the second sample, no one can pass level 4.

500

[ { "input": "4\n3 1 2 3\n2 2 4", "output": "I become the guy." }, { "input": "4\n3 1 2 3\n2 2 3", "output": "Oh, my keyboard!" }, { "input": "10\n5 8 6 1 5 4\n6 1 3 2 9 4 6", "output": "Oh, my keyboard!" }, { "input": "10\n8 8 10 7 3 1 4 2 6\n8 9 5 10 3 7 2 4 8", "output": "I become the guy." }, { "input": "10\n9 6 1 8 3 9 7 5 10 4\n7 1 3 2 7 6 9 5", "output": "I become the guy." }, { "input": "100\n75 83 69 73 30 76 37 48 14 41 42 21 35 15 50 61 86 85 46 3 31 13 78 10 2 44 80 95 56 82 38 75 77 4 99 9 84 53 12 11 36 74 39 72 43 89 57 28 54 1 51 66 27 22 93 59 68 88 91 29 7 20 63 8 52 23 64 58 100 79 65 49 96 71 33 45\n83 50 89 73 34 28 99 67 77 44 19 60 68 42 8 27 94 85 14 39 17 78 24 21 29 63 92 32 86 22 71 81 31 82 65 48 80 59 98 3 70 55 37 12 15 72 47 9 11 33 16 7 91 74 13 64 38 84 6 61 93 90 45 69 1 54 52 100 57 10 35 49 53 75 76 43 62 5 4 18 36 96 79 23", "output": "Oh, my keyboard!" }, { "input": "1\n1 1\n1 1", "output": "I become the guy." }, { "input": "1\n0\n1 1", "output": "I become the guy." }, { "input": "1\n1 1\n0", "output": "I become the guy." }, { "input": "1\n0\n0", "output": "Oh, my keyboard!" }, { "input": "100\n0\n0", "output": "Oh, my keyboard!" }, { "input": "100\n44 71 70 55 49 43 16 53 7 95 58 56 38 76 67 94 20 73 29 90 25 30 8 84 5 14 77 52 99 91 66 24 39 37 22 44 78 12 63 59 32 51 15 82 34\n56 17 10 96 80 69 13 81 31 57 4 48 68 89 50 45 3 33 36 2 72 100 64 87 21 75 54 74 92 65 23 40 97 61 18 28 98 93 35 83 9 79 46 27 41 62 88 6 47 60 86 26 42 85 19 1 11", "output": "I become the guy." }, { "input": "100\n78 63 59 39 11 58 4 2 80 69 22 95 90 26 65 16 30 100 66 99 67 79 54 12 23 28 45 56 70 74 60 82 73 91 68 43 92 75 51 21 17 97 86 44 62 47 85 78 72 64 50 81 71 5 57 13 31 76 87 9 49 96 25 42 19 35 88 53 7 83 38 27 29 41 89 93 10 84 18\n78 1 16 53 72 99 9 36 59 49 75 77 94 79 35 4 92 42 82 83 76 97 20 68 55 47 65 50 14 30 13 67 98 8 7 40 64 32 87 10 33 90 93 18 26 71 17 46 24 28 89 58 37 91 39 34 25 48 84 31 96 95 80 88 3 51 62 52 85 61 12 15 27 6 45 38 2 22 60", "output": "I become the guy." }, { "input": "2\n2 2 1\n0", "output": "I become the guy." }, { "input": "2\n1 2\n2 1 2", "output": "I become the guy." }, { "input": "80\n57 40 1 47 36 69 24 76 5 72 26 4 29 62 6 60 3 70 8 64 18 37 16 14 13 21 25 7 66 68 44 74 61 39 38 33 15 63 34 65 10 23 56 51 80 58 49 75 71 12 50 57 2 30 54 27 17 52\n61 22 67 15 28 41 26 1 80 44 3 38 18 37 79 57 11 7 65 34 9 36 40 5 48 29 64 31 51 63 27 4 50 13 24 32 58 23 19 46 8 73 39 2 21 56 77 53 59 78 43 12 55 45 30 74 33 68 42 47 17 54", "output": "Oh, my keyboard!" }, { "input": "100\n78 87 96 18 73 32 38 44 29 64 40 70 47 91 60 69 24 1 5 34 92 94 99 22 83 65 14 68 15 20 74 31 39 100 42 4 97 46 25 6 8 56 79 9 71 35 54 19 59 93 58 62 10 85 57 45 33 7 86 81 30 98 26 61 84 41 23 28 88 36 66 51 80 53 37 63 43 95 75\n76 81 53 15 26 37 31 62 24 87 41 39 75 86 46 76 34 4 51 5 45 65 67 48 68 23 71 27 94 47 16 17 9 96 84 89 88 100 18 52 69 42 6 92 7 64 49 12 98 28 21 99 25 55 44 40 82 19 36 30 77 90 14 43 50 3 13 95 78 35 20 54 58 11 2 1 33", "output": "Oh, my keyboard!" }, { "input": "100\n77 55 26 98 13 91 78 60 23 76 12 11 36 62 84 80 18 1 68 92 81 67 19 4 2 10 17 77 96 63 15 69 46 97 82 42 83 59 50 72 14 40 89 9 52 29 56 31 74 39 45 85 22 99 44 65 95 6 90 38 54 32 49 34 3 70 75 33 94 53 21 71 5 66 73 41 100 24\n69 76 93 5 24 57 59 6 81 4 30 12 44 15 67 45 73 3 16 8 47 95 20 64 68 85 54 17 90 86 66 58 13 37 42 51 35 32 1 28 43 80 7 14 48 19 62 55 2 91 25 49 27 26 38 79 89 99 22 60 75 53 88 82 34 21 87 71 72 61", "output": "I become the guy." }, { "input": "100\n74 96 32 63 12 69 72 99 15 22 1 41 79 77 71 31 20 28 75 73 85 37 38 59 42 100 86 89 55 87 68 4 24 57 52 8 92 27 56 98 95 58 34 9 45 14 11 36 66 76 61 19 25 23 78 49 90 26 80 43 70 13 65 10 5 74 81 21 44 60 97 3 47 93 6\n64 68 21 27 16 91 23 22 33 12 71 88 90 50 62 43 28 29 57 59 5 74 10 95 35 1 67 93 36 32 86 40 6 64 78 46 89 15 84 53 18 30 17 85 2 3 47 92 25 48 76 51 20 82 52 83 99 63 80 11 94 54 39 7 58", "output": "I become the guy." }, { "input": "100\n75 11 98 44 47 88 94 23 78 59 70 2 43 39 34 63 71 19 42 61 30 74 14 77 97 53 92 60 67 36 37 13 6 86 62 46 41 3 25 93 7 12 27 48 55 49 31 35 51 10 57 54 95 82 28 90 73 26 17 50 81 56 20 87 40 85 72 64 99 29 91 5 80 18 24 52\n72 93 59 5 88 47 9 58 48 1 43 50 100 87 61 91 45 98 99 56 25 84 53 73 78 54 63 38 37 2 77 95 89 85 4 90 10 33 12 22 74 32 34 70 71 52 96 57 15 66 31 27 75 8 21 39 62 44 67 94 81 68 14 19 36 28 11 79 16 65 46 83 76", "output": "Oh, my keyboard!" }, { "input": "3\n1 2\n2 2 3", "output": "Oh, my keyboard!" }, { "input": "4\n1 2\n3 1 3 4", "output": "I become the guy." }, { "input": "6\n2 1 2\n3 4 5 6", "output": "Oh, my keyboard!" } ]

1,695,950,118

2,147,483,647

PyPy 3-64

OK

TESTS

29

62

0

uniq = [] num = int(input()) litx = list(input().split(" ")) lity = list(input().split(" ")) del litx[0] del lity[0] for x in litx: if x not in uniq: uniq.append(x) for y in lity: if y not in uniq: uniq.append(y) if len(uniq) == num: print("I become the guy.") else: print("Oh, my keyboard!")

Title: I Wanna Be the Guy Time Limit: None seconds Memory Limit: None megabytes Problem Description: There is a game called "I Wanna Be the Guy", consisting of *n* levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game. Little X can pass only *p* levels of the game. And Little Y can pass only *q* levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other? Input Specification: The first line contains a single integer *n* (1<=≤<=<=*n*<=≤<=100). The next line contains an integer *p* (0<=≤<=*p*<=≤<=*n*) at first, then follows *p* distinct integers *a*1,<=*a*2,<=...,<=*a**p* (1<=≤<=*a**i*<=≤<=*n*). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to *n*. Output Specification: If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes). Demo Input: ['4\n3 1 2 3\n2 2 4\n', '4\n3 1 2 3\n2 2 3\n'] Demo Output: ['I become the guy.\n', 'Oh, my keyboard!\n'] Note: In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both. In the second sample, no one can pass level 4.

```python uniq = [] num = int(input()) litx = list(input().split(" ")) lity = list(input().split(" ")) del litx[0] del lity[0] for x in litx: if x not in uniq: uniq.append(x) for y in lity: if y not in uniq: uniq.append(y) if len(uniq) == num: print("I become the guy.") else: print("Oh, my keyboard!") ```

3

69

A

Young Physicist

PROGRAMMING

1,000

[ "implementation", "math" ]

A. Young Physicist

2

256

A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.

The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).

Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.

[ "3\n4 1 7\n-2 4 -1\n1 -5 -3\n", "3\n3 -1 7\n-5 2 -4\n2 -1 -3\n" ]

[ "NO", "YES" ]

none

500

[ { "input": "3\n4 1 7\n-2 4 -1\n1 -5 -3", "output": "NO" }, { "input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3", "output": "YES" }, { "input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41", "output": "NO" }, { "input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33", "output": "NO" }, { "input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7", "output": "NO" }, { "input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38", "output": "NO" }, { "input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6", "output": "YES" }, { "input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0", "output": "YES" }, { "input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0", "output": "YES" }, { "input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0", "output": "YES" }, { "input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0", "output": "YES" }, { "input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0", "output": "YES" }, { "input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87", "output": "YES" }, { "input": "3\n1 2 3\n3 2 1\n0 0 0", "output": "NO" }, { "input": "2\n5 -23 12\n0 0 0", "output": "NO" }, { "input": "1\n0 0 0", "output": "YES" }, { "input": "1\n1 -2 0", "output": "NO" }, { "input": "2\n-23 77 -86\n23 -77 86", "output": "YES" }, { "input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82", "output": "YES" }, { "input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0", "output": "YES" }, { "input": "3\n96 49 -12\n2 -66 28\n-98 17 -16", "output": "YES" }, { "input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0", "output": "YES" }, { "input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53", "output": "YES" }, { "input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1", "output": "YES" }, { "input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0", "output": "YES" }, { "input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0", "output": "YES" }, { "input": "3\n0 2 -2\n1 -1 3\n-3 0 0", "output": "NO" } ]

1,561,377,275

2,147,483,647

PyPy 3

OK

TESTS

81

248

0

a=0 b=0 c=0 n=int(input()) for x in range (n): x,y,z=map(int,input().split()) a=a+x b=b+y c=c+z if a==b==c==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 a=0 b=0 c=0 n=int(input()) for x in range (n): x,y,z=map(int,input().split()) a=a+x b=b+y c=c+z if a==b==c==0: print("YES") else: print("NO") ```

3.938

965

C

Greedy Arkady

PROGRAMMING

2,000

[ "math" ]

null

$k$ people want to split $n$ candies between them. Each candy should be given to exactly one of them or be thrown away. The people are numbered from $1$ to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$) will be thrown away. Arkady can't choose $x$ greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$ times, as it is considered a slow splitting. Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$.

The only line contains four integers $n$, $k$, $M$ and $D$ ($2 \le n \le 10^{18}$, $2 \le k \le n$, $1 \le M \le n$, $1 \le D \le \min{(n, 1000)}$, $M \cdot D \cdot k \ge n$) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies.

Print a single integer — the maximum possible number of candies Arkady can give to himself. Note that it is always possible to choose some valid $x$.

[ "20 4 5 2\n", "30 9 4 1\n" ]

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

In the first example Arkady should choose $x = 4$. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$ candies in total. Note that if Arkady chooses $x = 5$, he will receive only $5$ candies, and if he chooses $x = 3$, he will receive only $3 + 3 = 6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x = 1$ nor $x = 2$ because in these cases he will receive candies more than $2$ times. In the second example Arkady has to choose $x = 4$, because any smaller value leads to him receiving candies more than $1$ time.

1,500

[ { "input": "20 4 5 2", "output": "8" }, { "input": "30 9 4 1", "output": "4" }, { "input": "2 2 1 1", "output": "1" }, { "input": "42 20 5 29", "output": "5" }, { "input": "1000000000000000000 135 1000000000000000 1000", "output": "8325624421831635" }, { "input": "100 33 100 100", "output": "100" }, { "input": "1000000000 1000000000 1000000000 1000", "output": "1000000000" }, { "input": "1000000000 32428 1000000000 1000", "output": "1000000000" }, { "input": "1000000000 324934 1000 1000", "output": "4000" }, { "input": "1000000000000000000 32400093004 10000000 1000", "output": "40000000" }, { "input": "885 2 160 842", "output": "504" }, { "input": "216 137 202 208", "output": "202" }, { "input": "72 66 28 9", "output": "28" }, { "input": "294 4 13 8", "output": "80" }, { "input": "9 2 2 3", "output": "4" }, { "input": "31 3 2 8", "output": "10" }, { "input": "104 2 5 11", "output": "50" }, { "input": "1000000000000000000 1000000000000000000 1000 1000", "output": "1000" }, { "input": "1000000000000000000 100000000000000000 1 1000", "output": "10" }, { "input": "23925738098196565 23925738098196565 23925738098196565 1000", "output": "23925738098196565" }, { "input": "576460752303423488 576460752303423488 351082447248993993 1000", "output": "351082447248993993" }, { "input": "962768465676381898 72057594037927936 586039918340257175 256", "output": "586039918340257175" }, { "input": "1000000000000000000 1000000000000000000 10 1000", "output": "10" }, { "input": "23925738098196565 23925738098196565 1 1000", "output": "1" }, { "input": "1000000000000000000 1000000000000000000 1 1000", "output": "1" } ]

1,530,231,318

2,147,483,647

Python 3

OK

TESTS

25

124

0

n,k,m,d=map(int,input().split()) ans=0 for i in range(1,d+1): s=1 e=m while s<e: mid=(s+e+1)//2 if mid*(k*(i-1)+1)<=n: s=mid else: e=mid-1 if s*(k*(i-1)+1)<=n: ans=max(ans,i*s) """print("%d %d %d ",ans,i,s)""" print(ans)

Title: Greedy Arkady Time Limit: None seconds Memory Limit: None megabytes Problem Description: $k$ people want to split $n$ candies between them. Each candy should be given to exactly one of them or be thrown away. The people are numbered from $1$ to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$) will be thrown away. Arkady can't choose $x$ greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$ times, as it is considered a slow splitting. Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$. Input Specification: The only line contains four integers $n$, $k$, $M$ and $D$ ($2 \le n \le 10^{18}$, $2 \le k \le n$, $1 \le M \le n$, $1 \le D \le \min{(n, 1000)}$, $M \cdot D \cdot k \ge n$) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies. Output Specification: Print a single integer — the maximum possible number of candies Arkady can give to himself. Note that it is always possible to choose some valid $x$. Demo Input: ['20 4 5 2\n', '30 9 4 1\n'] Demo Output: ['8\n', '4\n'] Note: In the first example Arkady should choose $x = 4$. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$ candies in total. Note that if Arkady chooses $x = 5$, he will receive only $5$ candies, and if he chooses $x = 3$, he will receive only $3 + 3 = 6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x = 1$ nor $x = 2$ because in these cases he will receive candies more than $2$ times. In the second example Arkady has to choose $x = 4$, because any smaller value leads to him receiving candies more than $1$ time.

```python n,k,m,d=map(int,input().split()) ans=0 for i in range(1,d+1): s=1 e=m while s<e: mid=(s+e+1)//2 if mid*(k*(i-1)+1)<=n: s=mid else: e=mid-1 if s*(k*(i-1)+1)<=n: ans=max(ans,i*s) """print("%d %d %d ",ans,i,s)""" print(ans) ```

3

527

A

Playing with Paper

PROGRAMMING

1,100

[ "implementation", "math" ]

null

One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part. After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop. Can you determine how many ships Vasya will make during the lesson?

The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper.

Print a single integer — the number of ships that Vasya will make.

[ "2 1\n", "10 7\n", "1000000000000 1\n" ]

[ "2\n", "6\n", "1000000000000\n" ]

Pictures to the first and second sample test.

500

[ { "input": "2 1", "output": "2" }, { "input": "10 7", "output": "6" }, { "input": "1000000000000 1", "output": "1000000000000" }, { "input": "3 1", "output": "3" }, { "input": "4 1", "output": "4" }, { "input": "3 2", "output": "3" }, { "input": "4 2", "output": "2" }, { "input": "1000 700", "output": "6" }, { "input": "959986566087 524054155168", "output": "90" }, { "input": "4 3", "output": "4" }, { "input": "7 6", "output": "7" }, { "input": "1000 999", "output": "1000" }, { "input": "1000 998", "output": "500" }, { "input": "1000 997", "output": "336" }, { "input": "42 1", "output": "42" }, { "input": "1000 1", "output": "1000" }, { "input": "8 5", "output": "5" }, { "input": "13 8", "output": "6" }, { "input": "987 610", "output": "15" }, { "input": "442 42", "output": "22" }, { "input": "754 466", "output": "13" }, { "input": "1000000000000 999999999999", "output": "1000000000000" }, { "input": "1000000000000 999999999998", "output": "500000000000" }, { "input": "941 14", "output": "74" }, { "input": "998 2", "output": "499" }, { "input": "1000 42", "output": "32" }, { "input": "1000 17", "output": "66" }, { "input": "5 1", "output": "5" }, { "input": "5 2", "output": "4" }, { "input": "5 3", "output": "4" }, { "input": "5 4", "output": "5" }, { "input": "293 210", "output": "17" }, { "input": "787878787878 424242424242", "output": "8" }, { "input": "956722026041 591286729879", "output": "58" }, { "input": "956722026041 365435296162", "output": "58" }, { "input": "628625247282 464807889701", "output": "102" }, { "input": "695928431619 424778620208", "output": "167" }, { "input": "1000000000000 42", "output": "23809523821" }, { "input": "987654345678 23", "output": "42941493300" }, { "input": "10000000001 2", "output": "5000000002" }, { "input": "1000000000000 2", "output": "500000000000" }, { "input": "1000000000000 3", "output": "333333333336" }, { "input": "100000000000 3", "output": "33333333336" }, { "input": "100000000000 23", "output": "4347826109" }, { "input": "999999999997 7", "output": "142857142861" }, { "input": "8589934592 4294967296", "output": "2" } ]

1,511,514,508

2,147,483,647

Python 3

OK

TESTS

46

93

6,041,600

# -*- coding: utf-8 -*- import math import collections import bisect import heapq import time import random import itertools import sys """ created by shhuan at 2017/11/24 17:02 """ # a, b # a-b, b # ... => b, a-kb # def cal(a, b): if a < b: return cal(b, a) if a == b: return 1 if a == 0 or b == 0: return 0 k = a//b return k + cal(b, a-k*b) a, b = map(int, input().split()) print(cal(a, b))

Title: Playing with Paper Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part. After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop. Can you determine how many ships Vasya will make during the lesson? Input Specification: The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper. Output Specification: Print a single integer — the number of ships that Vasya will make. Demo Input: ['2 1\n', '10 7\n', '1000000000000 1\n'] Demo Output: ['2\n', '6\n', '1000000000000\n'] Note: Pictures to the first and second sample test.

```python # -*- coding: utf-8 -*- import math import collections import bisect import heapq import time import random import itertools import sys """ created by shhuan at 2017/11/24 17:02 """ # a, b # a-b, b # ... => b, a-kb # def cal(a, b): if a < b: return cal(b, a) if a == b: return 1 if a == 0 or b == 0: return 0 k = a//b return k + cal(b, a-k*b) a, b = map(int, input().split()) print(cal(a, b)) ```

3

918

A

Eleven

PROGRAMMING

800

[ "brute force", "implementation" ]

null

Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly *n* characters. Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the *i*-th letter of her name should be 'O' (uppercase) if *i* is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to *n*. Fibonacci sequence is the sequence *f* where - *f*1<==<=1, - *f*2<==<=1, - *f**n*<==<=*f**n*<=-<=2<=+<=*f**n*<=-<=1 (*n*<=><=2). As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name.

The first and only line of input contains an integer *n* (1<=≤<=*n*<=≤<=1000).

Print Eleven's new name on the first and only line of output.

[ "8\n", "15\n" ]

[ "OOOoOooO\n", "OOOoOooOooooOoo\n" ]

none

500

[ { "input": "8", "output": "OOOoOooO" }, { "input": "15", "output": "OOOoOooOooooOoo" }, { "input": "85", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooo" }, { "input": "381", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooo" }, { "input": "805", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..." }, { "input": "1000", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..." }, { "input": "1", "output": "O" }, { "input": "2", "output": "OO" }, { "input": "3", "output": "OOO" }, { "input": "5", "output": "OOOoO" }, { "input": "17", "output": "OOOoOooOooooOoooo" }, { "input": "49", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooo" }, { "input": "256", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooo" }, { "input": "512", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..." }, { "input": "933", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..." }, { "input": "61", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooo" }, { "input": "781", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..." }, { "input": "999", "output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..." } ]

1,617,009,995

2,147,483,647

Python 3

OK

TESTS

18

77

307,200

n=int(input()) f1=0 f2=1 count=0 l=[] if n==1: l.append(n) else: l.append(f1) while count<n: f=f1+f2 l.append(f) f1=f2 f2=f count+=1 # print(l) for i in range(1,n+1): if i in l: print('O',end="") else: print('o',end="")

Title: Eleven Time Limit: None seconds Memory Limit: None megabytes Problem Description: Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly *n* characters. Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the *i*-th letter of her name should be 'O' (uppercase) if *i* is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to *n*. Fibonacci sequence is the sequence *f* where - *f*1<==<=1, - *f*2<==<=1, - *f**n*<==<=*f**n*<=-<=2<=+<=*f**n*<=-<=1 (*n*<=><=2). As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name. Input Specification: The first and only line of input contains an integer *n* (1<=≤<=*n*<=≤<=1000). Output Specification: Print Eleven's new name on the first and only line of output. Demo Input: ['8\n', '15\n'] Demo Output: ['OOOoOooO\n', 'OOOoOooOooooOoo\n'] Note: none

```python n=int(input()) f1=0 f2=1 count=0 l=[] if n==1: l.append(n) else: l.append(f1) while count<n: f=f1+f2 l.append(f) f1=f2 f2=f count+=1 # print(l) for i in range(1,n+1): if i in l: print('O',end="") else: print('o',end="") ```

3

991

B

Getting an A

PROGRAMMING

900

[ "greedy", "sortings" ]

null

Translator's note: in Russia's most widespread grading system, there are four grades: 5, 4, 3, 2, the higher the better, roughly corresponding to A, B, C and F respectively in American grading system. The term is coming to an end and students start thinking about their grades. Today, a professor told his students that the grades for his course would be given out automatically — he would calculate the simple average (arithmetic mean) of all grades given out for lab works this term and round to the nearest integer. The rounding would be done in favour of the student — $4.5$ would be rounded up to $5$ (as in example 3), but $4.4$ would be rounded down to $4$. This does not bode well for Vasya who didn't think those lab works would influence anything, so he may receive a grade worse than $5$ (maybe even the dreaded $2$). However, the professor allowed him to redo some of his works of Vasya's choosing to increase his average grade. Vasya wants to redo as as few lab works as possible in order to get $5$ for the course. Of course, Vasya will get $5$ for the lab works he chooses to redo. Help Vasya — calculate the minimum amount of lab works Vasya has to redo.

The first line contains a single integer $n$ — the number of Vasya's grades ($1 \leq n \leq 100$). The second line contains $n$ integers from $2$ to $5$ — Vasya's grades for his lab works.

Output a single integer — the minimum amount of lab works that Vasya has to redo. It can be shown that Vasya can always redo enough lab works to get a $5$.

[ "3\n4 4 4\n", "4\n5 4 5 5\n", "4\n5 3 3 5\n" ]

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

In the first sample, it is enough to redo two lab works to make two $4$s into $5$s. In the second sample, Vasya's average is already $4.75$ so he doesn't have to redo anything to get a $5$. In the second sample Vasya has to redo one lab work to get rid of one of the $3$s, that will make the average exactly $4.5$ so the final grade would be $5$.

1,000

[ { "input": "3\n4 4 4", "output": "2" }, { "input": "4\n5 4 5 5", "output": "0" }, { "input": "4\n5 3 3 5", "output": "1" }, { "input": "1\n5", "output": "0" }, { "input": "4\n3 2 5 4", "output": "2" }, { "input": "5\n5 4 3 2 5", "output": "2" }, { "input": "8\n5 4 2 5 5 2 5 5", "output": "1" }, { "input": "5\n5 5 2 5 5", "output": "1" }, { "input": "6\n5 5 5 5 5 2", "output": "0" }, { "input": "6\n2 2 2 2 2 2", "output": "5" }, { "input": "100\n3 2 4 3 3 3 4 2 3 5 5 2 5 2 3 2 4 4 4 5 5 4 2 5 4 3 2 5 3 4 3 4 2 4 5 4 2 4 3 4 5 2 5 3 3 4 2 2 4 4 4 5 4 3 3 3 2 5 2 2 2 3 5 4 3 2 4 5 5 5 2 2 4 2 3 3 3 5 3 2 2 4 5 5 4 5 5 4 2 3 2 2 2 2 5 3 5 2 3 4", "output": "40" }, { "input": "1\n2", "output": "1" }, { "input": "1\n3", "output": "1" }, { "input": "1\n4", "output": "1" }, { "input": "4\n3 2 5 5", "output": "1" }, { "input": "6\n4 3 3 3 3 4", "output": "4" }, { "input": "8\n3 3 5 3 3 3 5 5", "output": "3" }, { "input": "10\n2 4 5 5 5 5 2 3 3 2", "output": "3" }, { "input": "20\n5 2 5 2 2 2 2 2 5 2 2 5 2 5 5 2 2 5 2 2", "output": "10" }, { "input": "25\n4 4 4 4 3 4 3 3 3 3 3 4 4 3 4 4 4 4 4 3 3 3 4 3 4", "output": "13" }, { "input": "30\n4 2 4 2 4 2 2 4 4 4 4 2 4 4 4 2 2 2 2 4 2 4 4 4 2 4 2 4 2 2", "output": "15" }, { "input": "52\n5 3 4 4 4 3 5 3 4 5 3 4 4 3 5 5 4 3 3 3 4 5 4 4 5 3 5 3 5 4 5 5 4 3 4 5 3 4 3 3 4 4 4 3 5 3 4 5 3 5 4 5", "output": "14" }, { "input": "77\n5 3 2 3 2 3 2 3 5 2 2 3 3 3 3 5 3 3 2 2 2 5 5 5 5 3 2 2 5 2 3 2 2 5 2 5 3 3 2 2 5 5 2 3 3 2 3 3 3 2 5 5 2 2 3 3 5 5 2 2 5 5 3 3 5 5 2 2 5 2 2 5 5 5 2 5 2", "output": "33" }, { "input": "55\n3 4 2 3 3 2 4 4 3 3 4 2 4 4 3 3 2 3 2 2 3 3 2 3 2 3 2 4 4 3 2 3 2 3 3 2 2 4 2 4 4 3 4 3 2 4 3 2 4 2 2 3 2 3 4", "output": "34" }, { "input": "66\n5 4 5 5 4 4 4 4 4 2 5 5 2 4 2 2 2 5 4 4 4 4 5 2 2 5 5 2 2 4 4 2 4 2 2 5 2 5 4 5 4 5 4 4 2 5 2 4 4 4 2 2 5 5 5 5 4 4 4 4 4 2 4 5 5 5", "output": "16" }, { "input": "99\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2", "output": "83" }, { "input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2", "output": "84" }, { "input": "99\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3", "output": "75" }, { "input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3", "output": "75" }, { "input": "99\n2 2 3 3 3 3 3 2 2 3 2 3 2 3 2 2 3 2 3 2 3 3 3 3 2 2 2 2 3 2 3 3 3 3 3 2 3 3 3 3 2 3 2 3 3 3 2 3 2 3 3 3 3 2 2 3 2 3 2 3 2 3 2 2 2 3 3 2 3 2 2 2 2 2 2 2 2 3 3 3 3 2 3 2 3 3 2 3 2 3 2 3 3 2 2 2 3 2 3", "output": "75" }, { "input": "100\n3 2 3 3 2 2 3 2 2 3 3 2 3 2 2 2 2 2 3 2 2 2 3 2 3 3 2 2 3 2 2 2 2 3 2 3 3 2 2 3 2 2 3 2 3 2 2 3 2 3 2 2 3 2 2 3 3 3 3 3 2 2 3 2 3 3 2 2 3 2 2 2 3 2 2 3 3 2 2 3 3 3 3 2 3 2 2 2 3 3 2 2 3 2 2 2 2 3 2 2", "output": "75" }, { "input": "99\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", "output": "50" }, { "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": "50" }, { "input": "99\n2 2 2 2 4 2 2 2 2 4 4 4 4 2 4 4 2 2 4 4 2 2 2 4 4 2 4 4 2 4 4 2 2 2 4 4 2 2 2 2 4 4 4 2 2 2 4 4 2 4 2 4 2 2 4 2 4 4 4 4 4 2 2 4 4 4 2 2 2 2 4 2 4 2 2 2 2 2 2 4 4 2 4 2 2 4 2 2 2 2 2 4 2 4 2 2 4 4 4", "output": "54" }, { "input": "100\n4 2 4 4 2 4 2 2 4 4 4 4 4 4 4 4 4 2 4 4 2 2 4 4 2 2 4 4 2 2 2 4 4 2 4 4 2 4 2 2 4 4 2 4 2 4 4 4 2 2 2 2 2 2 2 4 2 2 2 4 4 4 2 2 2 2 4 2 2 2 2 2 2 2 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2 2 2 4 4 4 4 2 4 2 2 4", "output": "50" }, { "input": "99\n4 3 4 4 4 4 4 3 4 3 3 4 3 3 4 4 3 3 3 4 3 4 3 3 4 3 3 3 3 4 3 4 4 3 4 4 3 3 4 4 4 3 3 3 4 4 3 3 4 3 4 3 4 3 4 3 3 3 3 4 3 4 4 4 4 4 4 3 4 4 3 3 3 3 3 3 3 3 4 3 3 3 4 4 4 4 4 4 3 3 3 3 4 4 4 3 3 4 3", "output": "51" }, { "input": "100\n3 3 4 4 4 4 4 3 4 4 3 3 3 3 4 4 4 4 4 4 3 3 3 4 3 4 3 4 3 3 4 3 3 3 3 3 3 3 3 4 3 4 3 3 4 3 3 3 4 4 3 4 4 3 3 4 4 4 4 4 4 3 4 4 3 4 3 3 3 4 4 3 3 4 4 3 4 4 4 3 3 4 3 3 4 3 4 3 4 3 3 4 4 4 3 3 4 3 3 4", "output": "51" }, { "input": "99\n3 3 4 4 4 2 4 4 3 2 3 4 4 4 2 2 2 3 2 4 4 2 4 3 2 2 2 4 2 3 4 3 4 2 3 3 4 2 3 3 2 3 4 4 3 2 4 3 4 3 3 3 3 3 4 4 3 3 4 4 2 4 3 4 3 2 3 3 3 4 4 2 4 4 2 3 4 2 3 3 3 4 2 2 3 2 4 3 2 3 3 2 3 4 2 3 3 2 3", "output": "58" }, { "input": "100\n2 2 4 2 2 3 2 3 4 4 3 3 4 4 4 2 3 2 2 3 4 2 3 2 4 3 4 2 3 3 3 2 4 3 3 2 2 3 2 4 4 2 4 3 4 4 3 3 3 2 4 2 2 2 2 2 2 3 2 3 2 3 4 4 4 2 2 3 4 4 3 4 3 3 2 3 3 3 4 3 2 3 3 2 4 2 3 3 4 4 3 3 4 3 4 3 3 4 3 3", "output": "61" }, { "input": "99\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", "output": "0" }, { "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": "0" }, { "input": "99\n2 2 2 2 2 5 2 2 5 2 5 2 5 2 2 2 2 2 5 2 2 2 5 2 2 5 2 2 2 5 5 2 5 2 2 5 2 5 2 2 5 5 2 2 2 2 5 5 2 2 2 5 2 2 5 2 2 2 2 2 5 5 5 5 2 2 5 2 5 2 2 2 2 2 5 2 2 5 5 2 2 2 2 2 5 5 2 2 5 5 2 2 2 2 5 5 5 2 5", "output": "48" }, { "input": "100\n5 5 2 2 2 2 2 2 5 5 2 5 2 2 2 2 5 2 5 2 5 5 2 5 5 2 2 2 2 2 2 5 2 2 2 5 2 2 5 2 2 5 5 5 2 5 5 5 5 5 5 2 2 5 2 2 5 5 5 5 5 2 5 2 5 2 2 2 5 2 5 2 5 5 2 5 5 2 2 5 2 5 5 2 5 2 2 5 2 2 2 5 2 2 2 2 5 5 2 5", "output": "38" }, { "input": "99\n5 3 3 3 5 3 3 3 3 3 3 3 3 5 3 3 3 3 3 3 3 3 5 3 3 3 5 5 3 5 5 3 3 5 5 5 3 5 3 3 3 3 5 3 3 5 5 3 5 5 5 3 5 3 5 3 5 5 5 5 3 3 3 5 3 5 3 3 3 5 5 5 5 5 3 5 5 3 3 5 5 3 5 5 3 5 5 3 3 5 5 5 3 3 3 5 3 3 3", "output": "32" }, { "input": "100\n3 3 3 5 3 3 3 3 3 3 5 5 5 5 3 3 3 3 5 3 3 3 3 3 5 3 5 3 3 5 5 5 5 5 5 3 3 5 3 3 5 3 5 5 5 3 5 3 3 3 3 3 3 3 3 3 3 3 5 5 3 5 3 5 5 3 5 3 3 5 3 5 5 5 5 3 5 3 3 3 5 5 5 3 3 3 5 3 5 5 5 3 3 3 5 3 5 5 3 5", "output": "32" }, { "input": "99\n5 3 5 5 3 3 3 2 2 5 2 5 3 2 5 2 5 2 3 5 3 2 3 2 5 5 2 2 3 3 5 5 3 5 5 2 3 3 5 2 2 5 3 2 5 2 3 5 5 2 5 2 2 5 3 3 5 3 3 5 3 2 3 5 3 2 3 2 3 2 2 2 2 5 2 2 3 2 5 5 5 3 3 2 5 3 5 5 5 2 3 2 5 5 2 5 2 5 3", "output": "39" }, { "input": "100\n3 5 3 3 5 5 3 3 2 5 5 3 3 3 2 2 3 2 5 3 2 2 3 3 3 3 2 5 3 2 3 3 5 2 2 2 3 2 3 5 5 3 2 5 2 2 5 5 3 5 5 5 2 2 5 5 3 3 2 2 2 5 3 3 2 2 3 5 3 2 3 5 5 3 2 3 5 5 3 3 2 3 5 2 5 5 5 5 5 5 3 5 3 2 3 3 2 5 2 2", "output": "42" }, { "input": "99\n4 4 4 5 4 4 5 5 4 4 5 5 5 4 5 4 5 5 5 4 4 5 5 5 5 4 5 5 5 4 4 5 5 4 5 4 4 4 5 5 5 5 4 4 5 4 4 5 4 4 4 4 5 5 5 4 5 4 5 5 5 5 5 4 5 4 5 4 4 4 4 5 5 5 4 5 5 4 4 5 5 5 4 5 4 4 5 5 4 5 5 5 5 4 5 5 4 4 4", "output": "0" }, { "input": "100\n4 4 5 5 5 5 5 5 4 4 5 5 4 4 5 5 4 5 4 4 4 4 4 4 4 4 5 5 5 5 5 4 4 4 4 4 5 4 4 5 4 4 4 5 5 5 4 5 5 5 5 5 5 4 4 4 4 4 4 5 5 4 5 4 4 5 4 4 4 4 5 5 4 5 5 4 4 4 5 5 5 5 4 5 5 5 4 4 5 5 5 4 5 4 5 4 4 5 5 4", "output": "1" }, { "input": "99\n2 2 2 5 2 2 2 2 2 4 4 5 5 2 2 4 2 5 2 2 2 5 2 2 5 5 5 4 5 5 4 4 2 2 5 2 2 2 2 5 5 2 2 4 4 4 2 2 2 5 2 4 4 2 4 2 4 2 5 4 2 2 5 2 4 4 4 2 5 2 2 5 4 2 2 5 5 5 2 4 5 4 5 5 4 4 4 5 4 5 4 5 4 2 5 2 2 2 4", "output": "37" }, { "input": "100\n4 4 5 2 2 5 4 5 2 2 2 4 2 5 4 4 2 2 4 5 2 4 2 5 5 4 2 4 4 2 2 5 4 2 5 4 5 2 5 2 4 2 5 4 5 2 2 2 5 2 5 2 5 2 2 4 4 5 5 5 5 5 5 5 4 2 2 2 4 2 2 4 5 5 4 5 4 2 2 2 2 4 2 2 5 5 4 2 2 5 4 5 5 5 4 5 5 5 2 2", "output": "31" }, { "input": "99\n5 3 4 4 5 4 4 4 3 5 4 3 3 4 3 5 5 5 5 4 3 3 5 3 4 5 3 5 4 4 3 5 5 4 4 4 4 3 5 3 3 5 5 5 5 5 4 3 4 4 3 5 5 3 3 4 4 4 5 4 4 5 4 4 4 4 5 5 4 3 3 4 3 5 3 3 3 3 4 4 4 4 3 4 5 4 4 5 5 5 3 4 5 3 4 5 4 3 3", "output": "24" }, { "input": "100\n5 4 4 4 5 5 5 4 5 4 4 3 3 4 4 4 5 4 5 5 3 5 5 4 5 5 5 4 4 5 3 5 3 5 3 3 5 4 4 5 5 4 5 5 3 4 5 4 4 3 4 4 3 3 5 4 5 4 5 3 4 5 3 4 5 4 3 5 4 5 4 4 4 3 4 5 3 4 3 5 3 4 4 4 3 4 4 5 3 3 4 4 5 5 4 3 4 4 3 5", "output": "19" }, { "input": "99\n2 2 5 2 5 3 4 2 3 5 4 3 4 2 5 3 2 2 4 2 4 4 5 4 4 5 2 5 5 3 2 3 2 2 3 4 5 3 5 2 5 4 4 5 4 2 2 3 2 3 3 3 4 4 3 2 2 4 4 2 5 3 5 3 5 4 4 4 5 4 5 2 2 5 4 4 4 3 3 2 5 2 5 2 3 2 5 2 2 5 5 3 4 5 3 4 4 4 4", "output": "37" }, { "input": "2\n5 2", "output": "1" }, { "input": "5\n2 2 2 2 2", "output": "5" }, { "input": "100\n2 3 2 2 2 3 2 3 3 3 3 3 2 3 3 2 2 3 3 2 3 2 3 2 3 4 4 4 3 3 3 3 3 4 4 3 3 4 3 2 3 4 3 3 3 3 2 3 4 3 4 3 3 2 4 4 2 4 4 3 3 3 3 4 3 2 3 4 3 4 4 4 4 4 3 2 2 3 4 2 4 4 4 2 2 4 2 2 3 2 2 4 4 3 4 2 3 3 2 2", "output": "61" }, { "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 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": "1" }, { "input": "100\n5 4 3 5 3 5 4 2 3 3 4 5 4 5 5 4 2 4 2 2 5 2 5 3 4 4 4 5 5 5 3 4 4 4 3 5 3 2 5 4 3 3 3 5 2 3 4 2 5 4 3 4 5 2 2 3 4 4 2 3 3 3 2 5 2 3 4 3 3 3 2 5 4 3 4 5 4 2 5 4 5 2 2 4 2 2 5 5 4 5 2 2 2 2 5 2 4 4 4 5", "output": "35" }, { "input": "2\n2 2", "output": "2" }, { "input": "20\n4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5", "output": "1" } ]

1,611,608,277

777

PyPy 3

OK

TESTS

61

109

1,331,200

def crb(a): return sum(a)/len(a) n=int(input()) a=list(map(int,input().split())) a.sort() i=0 ans=0 if crb(a)>=4.5: print(0) else: while crb(a)<4.5: a[i]=5 i +=1 ans +=1 print(ans)

Title: Getting an A Time Limit: None seconds Memory Limit: None megabytes Problem Description: Translator's note: in Russia's most widespread grading system, there are four grades: 5, 4, 3, 2, the higher the better, roughly corresponding to A, B, C and F respectively in American grading system. The term is coming to an end and students start thinking about their grades. Today, a professor told his students that the grades for his course would be given out automatically — he would calculate the simple average (arithmetic mean) of all grades given out for lab works this term and round to the nearest integer. The rounding would be done in favour of the student — $4.5$ would be rounded up to $5$ (as in example 3), but $4.4$ would be rounded down to $4$. This does not bode well for Vasya who didn't think those lab works would influence anything, so he may receive a grade worse than $5$ (maybe even the dreaded $2$). However, the professor allowed him to redo some of his works of Vasya's choosing to increase his average grade. Vasya wants to redo as as few lab works as possible in order to get $5$ for the course. Of course, Vasya will get $5$ for the lab works he chooses to redo. Help Vasya — calculate the minimum amount of lab works Vasya has to redo. Input Specification: The first line contains a single integer $n$ — the number of Vasya's grades ($1 \leq n \leq 100$). The second line contains $n$ integers from $2$ to $5$ — Vasya's grades for his lab works. Output Specification: Output a single integer — the minimum amount of lab works that Vasya has to redo. It can be shown that Vasya can always redo enough lab works to get a $5$. Demo Input: ['3\n4 4 4\n', '4\n5 4 5 5\n', '4\n5 3 3 5\n'] Demo Output: ['2\n', '0\n', '1\n'] Note: In the first sample, it is enough to redo two lab works to make two $4$s into $5$s. In the second sample, Vasya's average is already $4.75$ so he doesn't have to redo anything to get a $5$. In the second sample Vasya has to redo one lab work to get rid of one of the $3$s, that will make the average exactly $4.5$ so the final grade would be $5$.

```python def crb(a): return sum(a)/len(a) n=int(input()) a=list(map(int,input().split())) a.sort() i=0 ans=0 if crb(a)>=4.5: print(0) else: while crb(a)<4.5: a[i]=5 i +=1 ans +=1 print(ans) ```

3

686

A

Free Ice Cream

PROGRAMMING

800

[ "constructive algorithms", "implementation" ]

null

After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer. At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue). If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress. Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.

The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109). Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.

Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.

[ "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n", "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n" ]

[ "22 1\n", "3 2\n" ]

Consider the first sample. 1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.

500

[ { "input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20", "output": "22 1" }, { "input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98", "output": "3 2" }, { "input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000", "output": "7000000000 0" }, { "input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46", "output": "46 0" }, { "input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100", "output": "700 0" }, { "input": "1 0\n- 526403222", "output": "0 1" }, { "input": "1 897986543\n- 371188251", "output": "526798292 0" }, { "input": "1 0\n+ 1", "output": "1 0" }, { "input": "1 0\n- 1", "output": "0 1" }, { "input": "1 10\n+ 10", "output": "20 0" }, { "input": "1 3\n- 5", "output": "3 1" }, { "input": "1 0\n- 5", "output": "0 1" }, { "input": "1 0\n+ 5", "output": "5 0" } ]

1,688,604,953

2,147,483,647

Python 3

OK

TESTS

34

46

0

n, girls_pack = input().split() n = int(n) girls_pack = int(girls_pack) nbr_depressed = 0 counter = girls_pack my_list = [] for i in range(n): packs = input() my_list.append(packs) for j in range(len(my_list)): sign, packs = my_list[j].split() packs = int(packs) if sign == "+": counter += packs continue elif sign == "-": if counter >= packs: counter -= packs else: nbr_depressed += 1 print(str(counter) + " " + str(nbr_depressed))

Title: Free Ice Cream Time Limit: None seconds Memory Limit: None megabytes Problem Description: After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer. At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue). If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress. Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids. Input Specification: The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109). Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place. Output Specification: Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress. Demo Input: ['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n'] Demo Output: ['22 1\n', '3 2\n'] Note: Consider the first sample. 1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.

```python n, girls_pack = input().split() n = int(n) girls_pack = int(girls_pack) nbr_depressed = 0 counter = girls_pack my_list = [] for i in range(n): packs = input() my_list.append(packs) for j in range(len(my_list)): sign, packs = my_list[j].split() packs = int(packs) if sign == "+": counter += packs continue elif sign == "-": if counter >= packs: counter -= packs else: nbr_depressed += 1 print(str(counter) + " " + str(nbr_depressed)) ```

3

63

A

Sinking Ship

PROGRAMMING

900

[ "implementation", "sortings", "strings" ]

A. Sinking Ship

2

256

The ship crashed into a reef and is sinking. Now the entire crew must be evacuated. All *n* crew members have already lined up in a row (for convenience let's label them all from left to right with positive integers from 1 to *n*) and await further instructions. However, one should evacuate the crew properly, in a strict order. Specifically: The first crew members to leave the ship are rats. Then women and children (both groups have the same priority) leave the ship. After that all men are evacuated from the ship. The captain leaves the sinking ship last. If we cannot determine exactly who should leave the ship first for any two members of the crew by the rules from the previous paragraph, then the one who stands to the left in the line leaves the ship first (or in other words, the one whose number in the line is less). For each crew member we know his status as a crew member, and also his name. All crew members have different names. Determine the order in which to evacuate the crew.

The first line contains an integer *n*, which is the number of people in the crew (1<=≤<=*n*<=≤<=100). Then follow *n* lines. The *i*-th of those lines contains two words — the name of the crew member who is *i*-th in line, and his status on the ship. The words are separated by exactly one space. There are no other spaces in the line. The names consist of Latin letters, the first letter is uppercase, the rest are lowercase. The length of any name is from 1 to 10 characters. The status can have the following values: rat for a rat, woman for a woman, child for a child, man for a man, captain for the captain. The crew contains exactly one captain.

Print *n* lines. The *i*-th of them should contain the name of the crew member who must be the *i*-th one to leave the ship.

[ "6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman\n" ]

[ "Teddy\nAlice\nBob\nJulia\nCharlie\nJack\n" ]

none

500

[ { "input": "6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman", "output": "Teddy\nAlice\nBob\nJulia\nCharlie\nJack" }, { "input": "1\nA captain", "output": "A" }, { "input": "1\nAbcdefjhij captain", "output": "Abcdefjhij" }, { "input": "5\nA captain\nB man\nD woman\nC child\nE rat", "output": "E\nD\nC\nB\nA" }, { "input": "10\nCap captain\nD child\nC woman\nA woman\nE child\nMan man\nB child\nF woman\nRat rat\nRatt rat", "output": "Rat\nRatt\nD\nC\nA\nE\nB\nF\nMan\nCap" }, { "input": "5\nJoyxnkypf captain\nDxssgr woman\nKeojmnpd rat\nGdv man\nHnw man", "output": "Keojmnpd\nDxssgr\nGdv\nHnw\nJoyxnkypf" }, { "input": "11\nJue rat\nWyglbyphk rat\nGjlgu child\nGi man\nAttx rat\nTheorpkgx man\nYm rat\nX child\nB captain\nEnualf rat\nKktsgyuyv woman", "output": "Jue\nWyglbyphk\nAttx\nYm\nEnualf\nGjlgu\nX\nKktsgyuyv\nGi\nTheorpkgx\nB" }, { "input": "22\nWswwcvvm woman\nBtmfats rat\nI rat\nOcmtsnwx man\nUrcqv rat\nYghnogt woman\nWtyfc man\nWqle child\nUjfrelpu rat\nDstixj man\nAhksnio woman\nKhkvaap woman\nSjppvwm rat\nEgdmsv rat\nDank rat\nNquicjnw rat\nLh captain\nTdyaqaqln rat\nQtj rat\nTfgwijvq rat\nNbiso child\nNqthvbf woman", "output": "Btmfats\nI\nUrcqv\nUjfrelpu\nSjppvwm\nEgdmsv\nDank\nNquicjnw\nTdyaqaqln\nQtj\nTfgwijvq\nWswwcvvm\nYghnogt\nWqle\nAhksnio\nKhkvaap\nNbiso\nNqthvbf\nOcmtsnwx\nWtyfc\nDstixj\nLh" }, { "input": "36\nKqxmtwmsf child\nIze woman\nDlpr child\nK woman\nF captain\nRjwfeuhba rat\nBbv rat\nS rat\nMnmg woman\nSmzyx woman\nSr man\nQmhroracn rat\nSoqpuqock rat\nPibdq man\nIlrkrptx rat\nZaecfyqka man\nMmersfs child\nVvvocqi man\nHjeqxvq rat\nMpmb woman\nWmgu woman\nCerelmhoxi child\nA man\nDylv man\nXrdgmmtcpq woman\nXj woman\nCeh child\nOfccguo man\nNevr man\nAknkwdkq child\nBmo man\nLqo woman\nVej child\nT child\nRlijc man\nNsdtpdg woman", "output": "Rjwfeuhba\nBbv\nS\nQmhroracn\nSoqpuqock\nIlrkrptx\nHjeqxvq\nKqxmtwmsf\nIze\nDlpr\nK\nMnmg\nSmzyx\nMmersfs\nMpmb\nWmgu\nCerelmhoxi\nXrdgmmtcpq\nXj\nCeh\nAknkwdkq\nLqo\nVej\nT\nNsdtpdg\nSr\nPibdq\nZaecfyqka\nVvvocqi\nA\nDylv\nOfccguo\nNevr\nBmo\nRlijc\nF" }, { "input": "2\nA captain\nB man", "output": "B\nA" }, { "input": "2\nB woman\nA captain", "output": "B\nA" }, { "input": "2\nX child\nY captain", "output": "X\nY" }, { "input": "2\nX captain\nY rat", "output": "Y\nX" }, { "input": "6\nAbdelrhman captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman", "output": "Teddy\nAlice\nBob\nJulia\nCharlie\nAbdelrhman" }, { "input": "4\nAratos captain\nMind child\nAman woman\nCarrot man", "output": "Mind\nAman\nCarrot\nAratos" } ]

1,619,062,242

2,147,483,647

Python 3

OK

TESTS

26

124

0

n=int(input()) l=[] for i in range(n): l.append(input().split()) for i,j in l: if j=='rat': print(i) for i,j in l: if j=='woman': print(i) if j=='child': print(i) for i,j in l: if j=='man': print(i) for i,j in l: if j=='captain': print(i)

Title: Sinking Ship Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The ship crashed into a reef and is sinking. Now the entire crew must be evacuated. All *n* crew members have already lined up in a row (for convenience let's label them all from left to right with positive integers from 1 to *n*) and await further instructions. However, one should evacuate the crew properly, in a strict order. Specifically: The first crew members to leave the ship are rats. Then women and children (both groups have the same priority) leave the ship. After that all men are evacuated from the ship. The captain leaves the sinking ship last. If we cannot determine exactly who should leave the ship first for any two members of the crew by the rules from the previous paragraph, then the one who stands to the left in the line leaves the ship first (or in other words, the one whose number in the line is less). For each crew member we know his status as a crew member, and also his name. All crew members have different names. Determine the order in which to evacuate the crew. Input Specification: The first line contains an integer *n*, which is the number of people in the crew (1<=≤<=*n*<=≤<=100). Then follow *n* lines. The *i*-th of those lines contains two words — the name of the crew member who is *i*-th in line, and his status on the ship. The words are separated by exactly one space. There are no other spaces in the line. The names consist of Latin letters, the first letter is uppercase, the rest are lowercase. The length of any name is from 1 to 10 characters. The status can have the following values: rat for a rat, woman for a woman, child for a child, man for a man, captain for the captain. The crew contains exactly one captain. Output Specification: Print *n* lines. The *i*-th of them should contain the name of the crew member who must be the *i*-th one to leave the ship. Demo Input: ['6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman\n'] Demo Output: ['Teddy\nAlice\nBob\nJulia\nCharlie\nJack\n'] Note: none

```python n=int(input()) l=[] for i in range(n): l.append(input().split()) for i,j in l: if j=='rat': print(i) for i,j in l: if j=='woman': print(i) if j=='child': print(i) for i,j in l: if j=='man': print(i) for i,j in l: if j=='captain': print(i) ```

3.969

712

B

Memory and Trident

PROGRAMMING

1,100

[ "implementation", "strings" ]

null

Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string *s* with his directions for motion: - An 'L' indicates he should move one unit left. - An 'R' indicates he should move one unit right. - A 'U' indicates he should move one unit up. - A 'D' indicates he should move one unit down. But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in *s* with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string.

The first and only line contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) — the instructions Memory is given.

If there is a string satisfying the conditions, output a single integer — the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1.

[ "RRU\n", "UDUR\n", "RUUR\n" ]

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

In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk. In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change *s* to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin.

1,000

[ { "input": "RRU", "output": "-1" }, { "input": "UDUR", "output": "1" }, { "input": "RUUR", "output": "2" }, { "input": "DDDD", "output": "2" }, { "input": "RRRR", "output": "2" }, { "input": "RRRUUD", "output": "2" }, { "input": "UDURLRDURLRD", "output": "1" }, { "input": "RLRU", "output": "1" }, { "input": "RDDLLDLUUUDDRDRURLUUURLLDDLRLUURRLLRRLDRLLUDRLRULLDLRRLRLRLRUDUUDLULURLLDUURULURLLRRRURRRDRUUDLDRLRDRLRRDDLDLDLLUDRUDRLLLLDRDUULRUURRDLULLULDUDULRURRDDDLLUDRLUDDLDDDRRDDDULLLLDLDRLRRLRRDDRLULURRUDRDUUUULDURUDRDLDDUDUDRRURDULRRUDRLRRDLUURURDLDRLRDUDDDLDDDURURLUULRDUUULRURUDUDRRUDULLLUUUDRLLDRRDDLRUDRDRDLLRURURRRULURURRRLUUULRRRUURUUDURUDDLLDLDRLRDLDRLLDLDRDRRLRRRURUUUDRDLRLRUDRLULUUULUDDLULDLRLLRDUULLRLRURLRURULLLUDUDDLRULRDUURURLDLUURRRDURRLLDRUUDRDLLDUUDLURUDDUUUULRLLURLUDDRLRRDRURLRUDRLDDRLLL", "output": "-1" }, { "input": "LDLDLDLDLDRULD", "output": "5" }, { "input": "LULULURULLLU", "output": "5" }, { "input": "DRDRDDRR", "output": "4" }, { "input": "LR", "output": "0" }, { "input": "UL", "output": "1" }, { "input": "UD", "output": "0" }, { "input": "R", "output": "-1" }, { "input": "LU", "output": "1" }, { "input": "RR", "output": "1" }, { "input": "UDLR", "output": "0" }, { "input": "RRRRRLLUUUUDD", "output": "-1" }, { "input": "UUUUDLLLLR", "output": "3" }, { "input": "LLRUD", "output": "-1" }, { "input": "LLRDDU", "output": "1" } ]

1,518,955,951

2,147,483,647

Python 3

OK

TESTS

75

62

5,836,800

s = input() u,d,l,r = [s.count(x) for x in 'UDLR'] ud = abs(u-d) lr = abs(l-r) if len(s)%2: print (-1) else: print (ud//2 + lr//2 + ud%2)

Title: Memory and Trident Time Limit: None seconds Memory Limit: None megabytes Problem Description: Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string *s* with his directions for motion: - An 'L' indicates he should move one unit left. - An 'R' indicates he should move one unit right. - A 'U' indicates he should move one unit up. - A 'D' indicates he should move one unit down. But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in *s* with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string. Input Specification: The first and only line contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) — the instructions Memory is given. Output Specification: If there is a string satisfying the conditions, output a single integer — the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1. Demo Input: ['RRU\n', 'UDUR\n', 'RUUR\n'] Demo Output: ['-1\n', '1\n', '2\n'] Note: In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk. In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change *s* to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin.

```python s = input() u,d,l,r = [s.count(x) for x in 'UDLR'] ud = abs(u-d) lr = abs(l-r) if len(s)%2: print (-1) else: print (ud//2 + lr//2 + ud%2) ```

3

991

D

Bishwock

PROGRAMMING

1,500

[ "dp", "greedy" ]

null

Bishwock is a chess figure that consists of three squares resembling an "L-bar". This figure can be rotated by 90, 180 and 270 degrees so it can have four possible states: Bishwocks don't attack any squares and can even occupy on the adjacent squares as long as they don't occupy the same square. Vasya has a board with $2\times n$ squares onto which he wants to put some bishwocks. To his dismay, several squares on this board are already occupied by pawns and Vasya can't put bishwocks there. However, pawns also don't attack bishwocks and they can occupy adjacent squares peacefully. Knowing the positions of pawns on the board, help Vasya to determine the maximum amount of bishwocks he can put onto the board so that they wouldn't occupy the same squares and wouldn't occupy squares with pawns.

The input contains two nonempty strings that describe Vasya's board. Those strings contain only symbols "0" (zero) that denote the empty squares and symbols "X" (uppercase English letter) that denote the squares occupied by pawns. Strings are nonempty and are of the same length that does not exceed $100$.

Output a single integer — the maximum amount of bishwocks that can be placed onto the given board.

[ "00\n00\n", "00X00X0XXX0\n0XXX0X00X00\n", "0X0X0\n0X0X0\n", "0XXX0\n00000\n" ]

[ "1", "4", "0", "2" ]

none

1,500

[ { "input": "00\n00", "output": "1" }, { "input": "00X00X0XXX0\n0XXX0X00X00", "output": "4" }, { "input": "0X0X0\n0X0X0", "output": "0" }, { "input": "0XXX0\n00000", "output": "2" }, { "input": "0\n0", "output": "0" }, { "input": "0\nX", "output": "0" }, { "input": "X\n0", "output": "0" }, { "input": "XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX", "output": "0" }, { "input": "0000X0XX000X0XXXX0X0XXXX000X0X0XX000XXX0X00XX00XX00X0000XX0XX00X0X00X0X00X0XX000XX00XXXXXXXXXXXXXXX0\nX00XX0XX00XXXX00XXXX00XX0000000000XXX0X00XX0XX00XXX00X00X0XX0000X00XXXXXXX00X00000XXX00XXX00XXX0X0XX", "output": "18" }, { "input": "X\nX", "output": "0" }, { "input": "X0\n00", "output": "1" }, { "input": "0X\n00", "output": "1" }, { "input": "00\nX0", "output": "1" }, { "input": "00\n0X", "output": "1" }, { "input": "XX\nXX", "output": "0" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "66" }, { "input": "00000\n00000", "output": "3" }, { "input": "00000000\nXXXXXXXX", "output": "0" }, { "input": "X00X0XXXX0\nX0XXX0XX00", "output": "2" }, { "input": "00000XX0000000000000\n0X00000XX0000X00X000", "output": "10" }, { "input": "XXX00XXX0XXX0X0XXXXX\nXXX00XXX0XXX0X0XXXXX", "output": "1" }, { "input": "000X00000X00000X00000000000000\n000X00000X00000X00000000000000", "output": "17" }, { "input": "00X0X00000X0X0X00X0X0XXX0000X0\n0000000X00X000X000000000X00000", "output": "12" }, { "input": "000000000000000000000000000000000000000000\n00X000X00X00X0000X0XX000000000X000X0000000", "output": "23" }, { "input": "X0XXX00XX00X0XXXXXXXX0X0X0XX0X0X0XXXXX00X0XXXX00XX000XX0X000XX000XX\n0000000000000000000000000000000000000000000000000000000000000000000", "output": "24" }, { "input": "0000000000000000000000000000X00000000000000XX0X00000X0000000000000000000000000000000000000\n0000000000000000000000000X0000000000000000000000000000000000000000000000000000000000000000", "output": "57" }, { "input": "0000000000000000000000000000000000000X000000000000000000000X0X00000000000000000000000000000\n000000000000000000000000000X0X0000000000000000000000000000000000000000000000000000000000000", "output": "58" }, { "input": "00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\nX0X00000000000000000000000000X000000000X0000X00X000000XX000000X0X00000000X000X000000X0000X00", "output": "55" }, { "input": "000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX0XXXXXXXXXXXXXXXX0XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX", "output": "2" }, { "input": "XXXXXXXXXXXXXXXXXXXXXXX0XXX000XXXX0XXXXXXXXXXXXXXXXXXXXXXXXX0XXXXXXXXXXXX0X0XXXXXXXXXXXXXXXXXX\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "7" }, { "input": "00000XX0000000000000000000000000000000000000000000X0000000X0000000000000X0000000000000000X00000\n00000XX0000000000000000000000000000000000000000000X0000000X0000000000000X0000000000000000X00000", "output": "56" }, { "input": "000000000000000X0000000000000000000000000XX0000000000000000X00000000000000000000000X000000000000\n000000000000000X0000000000000000000000000XX0000000000000000X00000000000000000000000X000000000000", "output": "59" }, { "input": "000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "64" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000X000X0000000000X00000000X00000000000000000000000000000000000000000000000000000000", "output": "65" }, { "input": "000000000000000000X00X000000000000000000000000000000000000000X00000000X0000000X0000000000000000000X0\n000000000000000000X00X000000000000000000000000000000000000000X00000000X0000000X0000000000000000000X0", "output": "60" }, { "input": "XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX0XX0XXXXXXXXXXXXXXXX0XXXXXXXXXXXXXXXXXXXXXXX0XXXXXXXXXXXX\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX0XX0XXXXXXXXXXXXXXXX0XXXXXXXXXXXXXXXXXXXXXXX0XXXXXXXXXXXX", "output": "0" }, { "input": "XXXXXXXXXXX0X00XXXXXXXXXXXXXXXXXXXX0XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX0XXXXXXXXXXX00XXXXXXXXX0X0XXX0XX\nXXXXXXXXXXX0X00XXXXXXXXXXXXXXXXXXXX0XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX0XXXXXXXXXXX00XXXXXXXXX0X0XXX0XX", "output": "2" }, { "input": "0X0X0\nX0X0X", "output": "0" }, { "input": "X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0\n0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X", "output": "0" }, { "input": "X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0\n0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X", "output": "0" }, { "input": "X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X\n0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0", "output": "0" }, { "input": "0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X\nX0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0X0", "output": "0" }, { "input": "00000000000000X0000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "66" }, { "input": "XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX0XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX00XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX", "output": "1" }, { "input": "XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX00\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX0", "output": "1" }, { "input": "00XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX\nX0XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX", "output": "1" }, { "input": "XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX0XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX00XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX", "output": "0" }, { "input": "0000000000000000000000000000000000000000000000000000000000X0000000000000000000000000000000000000X000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "66" }, { "input": "00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000XX\n000000000000000000000000000000000X00000000000000000X000000000000000000000000000000000000000000000000", "output": "65" }, { "input": "0000X00X000000X0000X00X00X0000000000X0000000X000X00000X0X000XXX00000000XX0XX000000000000X00000000000\n000000000XX000000X00000X00X00X00000000000000000X0X000XX0000000000000X0X00X0000X0000X000000X0000000XX", "output": "49" }, { "input": "0000000000000000000000000000000000X0000000000000000000000000000000000000000000000000000000000000000\n000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "65" }, { "input": "00000000000000000000000000X000000000000000000000000000000000000000000X00000X0000X000000000000000000\n000X0000000000X000000000000000000000X0000000000X0X0000000000000000000X00000000000000000000000000000", "output": "62" }, { "input": "000X00XX0XX0X00X0XX0XXXX00XXX0X00000000XXX0XXXXXXX0X00X00XX00X0XXX00000XXXX0XX00X00XXX00X0X0XXXX000\nXXXXX000X0XXX000XXXXX0XX0000XX0XXX0XXX000XXX00XXXXX00X00XX0000X0XX0XX0XX000X0XX00X00XX00XX00X00XX0X", "output": "16" }, { "input": "X0X0XXXX0XXXXXXXXXX00XXXXXXXXXXXXXXXXXX0XXXXX0XX0X00X0XXXXXXX0X0XXXXXXXXX0X0X0XX0XX0000XXXX00XXX0XX\nXX0XXXXXXX0X0X00XXXX0X000X0XXXXXX0XXX0X0XXXX0XXXXXXXXXXXXXX00XX00XXX0XXXXXXXXXX00XXXX0XXXX0XXXXXXXX", "output": "4" }, { "input": "000\n000", "output": "2" }, { "input": "000000000000000000000\n000000000000000000000", "output": "14" }, { "input": "00000000000000000000000000000000000000\n00000000000000000000000000000000000000", "output": "25" }, { "input": "0000\n00X0", "output": "2" }, { "input": "000\n00X", "output": "1" }, { "input": "X000\nX000", "output": "2" }, { "input": "X0000X000XX00000000000000000000XX000X0000X00X0X00XX000000000000X0000X0X0XX00\n00X00XX00000000X0000000X000X000X0X0X00X000X0X0X0000X0000000X000XX000XX000X00", "output": "33" }, { "input": "0X00\n0000", "output": "2" }, { "input": "0X0\n000", "output": "1" }, { "input": "000000\n000000", "output": "4" }, { "input": "X00X0\n00000", "output": "2" }, { "input": "000\n0X0", "output": "1" }, { "input": "X000\n00X0", "output": "2" }, { "input": "X0000\n00000", "output": "3" }, { "input": "X000X\nX000X", "output": "2" }, { "input": "X0000X\n00000X", "output": "3" }, { "input": "000000000000\n000000000000", "output": "8" }, { "input": "00000\n0000X", "output": "3" } ]

1,689,043,217

2,147,483,647

PyPy 3-64

OK

TESTS

72

62

0

A = input() B = input() res = 0 p0,p1=1,1 for a,b in zip(A,B): a, b = a=='X',b=='X' w = a+b+p0+p1 if w<=1: res += 1 p0,p1 = 1,w else: p0,p1 = a,b print(res)

Title: Bishwock Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bishwock is a chess figure that consists of three squares resembling an "L-bar". This figure can be rotated by 90, 180 and 270 degrees so it can have four possible states: Bishwocks don't attack any squares and can even occupy on the adjacent squares as long as they don't occupy the same square. Vasya has a board with $2\times n$ squares onto which he wants to put some bishwocks. To his dismay, several squares on this board are already occupied by pawns and Vasya can't put bishwocks there. However, pawns also don't attack bishwocks and they can occupy adjacent squares peacefully. Knowing the positions of pawns on the board, help Vasya to determine the maximum amount of bishwocks he can put onto the board so that they wouldn't occupy the same squares and wouldn't occupy squares with pawns. Input Specification: The input contains two nonempty strings that describe Vasya's board. Those strings contain only symbols "0" (zero) that denote the empty squares and symbols "X" (uppercase English letter) that denote the squares occupied by pawns. Strings are nonempty and are of the same length that does not exceed $100$. Output Specification: Output a single integer — the maximum amount of bishwocks that can be placed onto the given board. Demo Input: ['00\n00\n', '00X00X0XXX0\n0XXX0X00X00\n', '0X0X0\n0X0X0\n', '0XXX0\n00000\n'] Demo Output: ['1', '4', '0', '2'] Note: none

```python A = input() B = input() res = 0 p0,p1=1,1 for a,b in zip(A,B): a, b = a=='X',b=='X' w = a+b+p0+p1 if w<=1: res += 1 p0,p1 = 1,w else: p0,p1 = a,b print(res) ```

3

468

C

Hack it!

PROGRAMMING

2,500

[ "binary search", "constructive algorithms", "math" ]

null

Little X has met the following problem recently. Let's define *f*(*x*) as the sum of digits in decimal representation of number *x* (for example, *f*(1234)<==<=1<=+<=2<=+<=3<=+<=4). You are to calculate Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:

The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1018).

Print two integers: *l*,<=*r* (1<=≤<=*l*<=≤<=*r*<=<<=10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.

[ "46\n", "126444381000032\n" ]

[ "1 10\n", "2333333 2333333333333\n" ]

none

1,500

[ { "input": "46", "output": "1 10" }, { "input": "126444381000032", "output": "2333333 2333333333333" }, { "input": "69645082595", "output": "613752823618441225798858488535 713259406474207764329704856394" }, { "input": "70602205995", "output": "11 249221334020432074498656960922" }, { "input": "33898130785", "output": "9 558855506346909386939077840182" }, { "input": "58929554039", "output": "22 855783114773435710171914224422" }, { "input": "81696185182", "output": "499118531974994927425925323518 956291458400902769638235161661" }, { "input": "1", "output": "149268802942315027273202513064 277551734280589260570057105889" }, { "input": "2", "output": "119692200833686078608961312319 629363568954685219494592939495" }, { "input": "3", "output": "2 302254410562920936884653943506" }, { "input": "4", "output": "284378053387469023431537894255 317250990020830090421009164911" }, { "input": "5", "output": "2 62668056583245293799710157951" }, { "input": "6", "output": "3 93810188780011787541394067841" }, { "input": "7", "output": "2 834286447477504059026206246185" }, { "input": "8", "output": "3 257583347960907690857477857197" }, { "input": "10", "output": "3 163048811987317819669274448265" }, { "input": "11", "output": "3 919618203693907154039906935669" }, { "input": "12", "output": "448221703341269567451520778454 698029790336105644790102859494" }, { "input": "43", "output": "9 172412961300207091437973214327" }, { "input": "36", "output": "8 619355518777647869838990701242" }, { "input": "65", "output": "6 709024330418134127413755925068" }, { "input": "43", "output": "7 669540448846929747909766131221" }, { "input": "23", "output": "2 104579054315773428039906118259" }, { "input": "100", "output": "15 324437778467489559125023403167" }, { "input": "10000", "output": "2 936791129091842315790163514642" }, { "input": "1000000", "output": "18 369591628030718549289473454545" }, { "input": "100000000", "output": "7 870405265198051697453938746950" }, { "input": "10000000000", "output": "20 972749766921651560604778558599" }, { "input": "1000000000000", "output": "6 68997070398311657294228230677" }, { "input": "100000000000000", "output": "249537318528661282822184562278 397003438246047829818181818181" }, { "input": "10000000000000000", "output": "10 778165727326620883431915444624" }, { "input": "1000000000000000000", "output": "408256298986776744812953390000 824018301451167837914299999999" }, { "input": "450000000000000000", "output": "2 357722688084551093593033993033" }, { "input": "432022", "output": "3 333556238531076799985515487090" }, { "input": "428033", "output": "22 730314748425770554502599499142" }, { "input": "776930", "output": "20 521232359366297130685112811874" }, { "input": "329824", "output": "308969571112207311167474021348 745620588073413831210052337999" }, { "input": "85058261498638", "output": "16 931187081941564769324316582547" }, { "input": "2130909834463", "output": "21 895378349209612431051450316022" }, { "input": "3427089130241", "output": "10 676758114393938690602742889714" }, { "input": "22881472397923", "output": "174523915446146844994793303441 429392837423394397373605399524" }, { "input": "756499070280135900", "output": "17 414254565210363110239866979636" }, { "input": "348919756917359012", "output": "15 148711424846749040017925495268" }, { "input": "789136710974630947", "output": "13 214740524663440316532540770911" }, { "input": "32851678509090784", "output": "15 728787006985867306455350840047" }, { "input": "219760524792138559", "output": "8 51530430692061277616906746431" }, { "input": "1000000007", "output": "19 96093944063249823335491170999" }, { "input": "999999874000003969", "output": "306259936121099507549028331741 983025349899931537481252156006" }, { "input": "999999999999999989", "output": "140766388309929586494194507367 705096656792775954903252198639" }, { "input": "999999999999999983", "output": "4 365946710276937286091309116340" }, { "input": "999999999999999999", "output": "26 983966655749419565978291923474" } ]

1,606,406,545

1,945

PyPy 3

OK

TESTS

51

140

0

A = int(input()) l = 1 r = 10**100 cnt = 10**99*100*45 + 1 cnt = -cnt % A l += cnt r += cnt print(l, r)

Title: Hack it! Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little X has met the following problem recently. Let's define *f*(*x*) as the sum of digits in decimal representation of number *x* (for example, *f*(1234)<==<=1<=+<=2<=+<=3<=+<=4). You are to calculate Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code: Input Specification: The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1018). Output Specification: Print two integers: *l*,<=*r* (1<=≤<=*l*<=≤<=*r*<=<<=10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists. Demo Input: ['46\n', '126444381000032\n'] Demo Output: ['1 10\n', '2333333 2333333333333\n'] Note: none

```python A = int(input()) l = 1 r = 10**100 cnt = 10**99*100*45 + 1 cnt = -cnt % A l += cnt r += cnt print(l, r) ```

3

651

A

Joysticks

PROGRAMMING

1,100

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

null

Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger). Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops. Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.

The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively.

Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.

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

[ "6\n", "5\n" ]

In the first sample game lasts for 6 minute by using the following algorithm: - at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%. After that the first joystick is completely discharged and the game is stopped.

500

[ { "input": "3 5", "output": "6" }, { "input": "4 4", "output": "5" }, { "input": "100 100", "output": "197" }, { "input": "1 100", "output": "98" }, { "input": "100 1", "output": "98" }, { "input": "1 4", "output": "2" }, { "input": "1 1", "output": "0" }, { "input": "8 8", "output": "13" }, { "input": "7 2", "output": "7" }, { "input": "24 15", "output": "36" }, { "input": "19 30", "output": "47" }, { "input": "15 31", "output": "44" }, { "input": "14 15", "output": "27" }, { "input": "58 33", "output": "89" }, { "input": "15 25", "output": "38" }, { "input": "59 45", "output": "102" }, { "input": "3 73", "output": "74" }, { "input": "48 1", "output": "47" }, { "input": "100 25", "output": "122" }, { "input": "40 49", "output": "86" }, { "input": "85 73", "output": "155" }, { "input": "29 1", "output": "28" }, { "input": "74 25", "output": "97" }, { "input": "24 57", "output": "78" }, { "input": "23 12", "output": "33" }, { "input": "2 99", "output": "99" }, { "input": "98 2", "output": "97" }, { "input": "2 97", "output": "97" }, { "input": "30 54", "output": "81" }, { "input": "32 53", "output": "82" }, { "input": "32 54", "output": "84" }, { "input": "1 2", "output": "1" }, { "input": "2 1", "output": "1" }, { "input": "2 2", "output": "1" }, { "input": "1 3", "output": "2" }, { "input": "3 1", "output": "2" }, { "input": "1 4", "output": "2" }, { "input": "2 3", "output": "3" }, { "input": "3 2", "output": "3" } ]

1,658,801,454

2,147,483,647

PyPy 3

OK

TESTS

39

93

0

def solve(p,q): ans=0 while p>0 and q>0: ans+=1 if p>q: p,q=q,p p+=1 q-=2 if p<0 or q<0: ans-=1 break return ans p,q=map(int,input().split()) print(solve(p,q))

Title: Joysticks Time Limit: None seconds Memory Limit: None megabytes Problem Description: Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger). Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops. Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent. Input Specification: The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively. Output Specification: Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged. Demo Input: ['3 5\n', '4 4\n'] Demo Output: ['6\n', '5\n'] Note: In the first sample game lasts for 6 minute by using the following algorithm: - at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%. After that the first joystick is completely discharged and the game is stopped.

```python def solve(p,q): ans=0 while p>0 and q>0: ans+=1 if p>q: p,q=q,p p+=1 q-=2 if p<0 or q<0: ans-=1 break return ans p,q=map(int,input().split()) print(solve(p,q)) ```

3

40

A

Find Color

PROGRAMMING

1,300

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

A. Find Color

2

256

Not so long ago as a result of combat operations the main Berland place of interest — the magic clock — was damaged. The cannon's balls made several holes in the clock, that's why the residents are concerned about the repair. The magic clock can be represented as an infinite Cartesian plane, where the origin corresponds to the clock center. The clock was painted two colors as is shown in the picture: The picture shows only the central part of the clock. This coloring naturally extends to infinity. The balls can be taken to be points on the plane. Your task is to find the color of the area, damaged by the given ball. All the points located on the border of one of the areas have to be considered painted black.

The first and single line contains two integers *x* and *y* — the coordinates of the hole made in the clock by the ball. Each of the numbers *x* and *y* has an absolute value that does not exceed 1000.

Find the required color. All the points between which and the origin of coordinates the distance is integral-value are painted black.

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

[ "white\n", "black\n", "black\n" ]

none

500

[ { "input": "-2 1", "output": "white" }, { "input": "2 1", "output": "black" }, { "input": "4 3", "output": "black" }, { "input": "3 3", "output": "black" }, { "input": "4 4", "output": "white" }, { "input": "-4 4", "output": "black" }, { "input": "4 -4", "output": "black" }, { "input": "-4 -4", "output": "white" }, { "input": "0 0", "output": "black" }, { "input": "0 1", "output": "black" }, { "input": "0 2", "output": "black" }, { "input": "0 1000", "output": "black" }, { "input": "1000 0", "output": "black" }, { "input": "-1000 0", "output": "black" }, { "input": "0 -1000", "output": "black" }, { "input": "1000 -1000", "output": "white" }, { "input": "12 5", "output": "black" }, { "input": "12 -5", "output": "black" }, { "input": "-12 -35", "output": "black" }, { "input": "20 -21", "output": "black" }, { "input": "-677 492", "output": "white" }, { "input": "-673 -270", "output": "white" }, { "input": "-668 970", "output": "black" }, { "input": "-220 208", "output": "white" }, { "input": "-215 -996", "output": "black" }, { "input": "-211 243", "output": "black" }, { "input": "-206 -518", "output": "white" }, { "input": "-201 278", "output": "black" }, { "input": "-196 -484", "output": "black" }, { "input": "902 479", "output": "white" }, { "input": "-441 572", "output": "white" }, { "input": "217 221", "output": "white" }, { "input": "875 -129", "output": "white" }, { "input": "-469 -36", "output": "black" }, { "input": "189 -387", "output": "white" }, { "input": "847 -294", "output": "white" }, { "input": "-496 -644", "output": "black" }, { "input": "-281 -552", "output": "white" }, { "input": "377 -902", "output": "black" }, { "input": "165 -738", "output": "white" }, { "input": "61 -175", "output": "black" }, { "input": "-42 389", "output": "black" }, { "input": "-589 952", "output": "black" }, { "input": "-693 -929", "output": "white" }, { "input": "-796 -365", "output": "white" }, { "input": "658 198", "output": "white" }, { "input": "555 319", "output": "black" }, { "input": "8 882", "output": "black" }, { "input": "-96 -556", "output": "black" }, { "input": "-129 489", "output": "black" }, { "input": "207 -224", "output": "black" }, { "input": "64 0", "output": "black" }, { "input": "17 144", "output": "black" }, { "input": "60 -448", "output": "black" }, { "input": "-399 -40", "output": "black" }, { "input": "128 -504", "output": "black" }, { "input": "0 72", "output": "black" }, { "input": "168 -26", "output": "black" }, { "input": "72 -154", "output": "black" }, { "input": "117 -44", "output": "black" }, { "input": "-72 -646", "output": "black" }, { "input": "253 -204", "output": "black" }, { "input": "-40 198", "output": "black" }, { "input": "-216 -90", "output": "black" }, { "input": "15 -8", "output": "black" }, { "input": "-180 -432", "output": "black" }, { "input": "280 342", "output": "black" }, { "input": "132 224", "output": "black" }, { "input": "-192 -256", "output": "black" }, { "input": "351 -280", "output": "black" } ]

1,597,499,367

2,147,483,647

Python 3

OK

TESTS

70

218

0

# from decimal import * # getcontext().prec=16 # from math import sqrt # from scipy.special import binom # from collections import defaultdict from math import sin,pi,sqrt x,y=list(map(int,input().split(" "))) if sqrt(x**2+y**2)%1==0: color=1 else: if x*y>=0: color=1 else: color=0 int_r=sqrt(x**2+y**2)//1 if int_r%2==1: color=1-color if color==1: print("black") else: print("white")

Title: Find Color Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Not so long ago as a result of combat operations the main Berland place of interest — the magic clock — was damaged. The cannon's balls made several holes in the clock, that's why the residents are concerned about the repair. The magic clock can be represented as an infinite Cartesian plane, where the origin corresponds to the clock center. The clock was painted two colors as is shown in the picture: The picture shows only the central part of the clock. This coloring naturally extends to infinity. The balls can be taken to be points on the plane. Your task is to find the color of the area, damaged by the given ball. All the points located on the border of one of the areas have to be considered painted black. Input Specification: The first and single line contains two integers *x* and *y* — the coordinates of the hole made in the clock by the ball. Each of the numbers *x* and *y* has an absolute value that does not exceed 1000. Output Specification: Find the required color. All the points between which and the origin of coordinates the distance is integral-value are painted black. Demo Input: ['-2 1\n', '2 1\n', '4 3\n'] Demo Output: ['white\n', 'black\n', 'black\n'] Note: none

```python # from decimal import * # getcontext().prec=16 # from math import sqrt # from scipy.special import binom # from collections import defaultdict from math import sin,pi,sqrt x,y=list(map(int,input().split(" "))) if sqrt(x**2+y**2)%1==0: color=1 else: if x*y>=0: color=1 else: color=0 int_r=sqrt(x**2+y**2)//1 if int_r%2==1: color=1-color if color==1: print("black") else: print("white") ```

3.9455

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" } ]

1,692,986,760

2,147,483,647

Python 3

OK

TESTS

30

92

0

def main(): s=input() i=0;ans='' while i<=len(s)-1: if s[i]=='-' and s[i+1]=='-': ans+='2';i+=2; elif s[i]=='-' and s[i+1]=='.': ans+='1';i+=2; else: ans+='0'; i+=1 return ans print(main())

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 def main(): s=input() i=0;ans='' while i<=len(s)-1: if s[i]=='-' and s[i+1]=='-': ans+='2';i+=2; elif s[i]=='-' and s[i+1]=='.': ans+='1';i+=2; else: ans+='0'; i+=1 return ans print(main()) ```

3.977

71

A

Way Too Long Words

PROGRAMMING

800

[ "strings" ]

A. Way Too Long Words

1

256

Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.

Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.

[ "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n" ]

[ "word\nl10n\ni18n\np43s\n" ]

none

500

[ { "input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis", "output": "word\nl10n\ni18n\np43s" }, { "input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm", "output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m" }, { "input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd", "output": "n20n\nj27b\nk15d" }, { "input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt", "output": "t41t" }, { "input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings", "output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings" }, { "input": "1\na", "output": "a" }, { "input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz", "output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz" }, { "input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij", "output": "a98j" }, { "input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco", "output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco" }, { "input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt", "output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t" }, { "input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr", "output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr" } ]

1,696,421,419

2,147,483,647

PyPy 3-64

OK

TESTS

20

62

0

loops = int(input()) for i in range(loops): str = input() ln = len(str) if ln > 10: print('{}{}{}'.format(str[0], len(str[1:ln-1]), str[ln-1])) else: print(str)

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 loops = int(input()) for i in range(loops): str = input() ln = len(str) if ln > 10: print('{}{}{}'.format(str[0], len(str[1:ln-1]), str[ln-1])) else: print(str) ```

3.969

378

A

Playing with Dice

PROGRAMMING

800

[ "brute force" ]

null

Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw. The first player wrote number *a*, the second player wrote number *b*. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins?

The single line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=6) — the numbers written on the paper by the first and second player, correspondingly.

Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly.

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

[ "3 0 3\n", "2 1 3\n" ]

The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct. You can assume that number *a* is closer to number *x* than number *b*, if |*a* - *x*| < |*b* - *x*|.

500

[ { "input": "2 5", "output": "3 0 3" }, { "input": "2 4", "output": "2 1 3" }, { "input": "5 3", "output": "2 1 3" }, { "input": "1 6", "output": "3 0 3" }, { "input": "5 1", "output": "3 1 2" }, { "input": "6 3", "output": "2 0 4" }, { "input": "2 3", "output": "2 0 4" }, { "input": "5 6", "output": "5 0 1" }, { "input": "4 4", "output": "0 6 0" }, { "input": "1 1", "output": "0 6 0" }, { "input": "6 4", "output": "1 1 4" }, { "input": "1 4", "output": "2 0 4" }, { "input": "5 5", "output": "0 6 0" }, { "input": "4 5", "output": "4 0 2" }, { "input": "4 3", "output": "3 0 3" }, { "input": "1 5", "output": "2 1 3" }, { "input": "6 5", "output": "1 0 5" }, { "input": "2 2", "output": "0 6 0" }, { "input": "1 3", "output": "1 1 4" }, { "input": "3 6", "output": "4 0 2" }, { "input": "3 1", "output": "4 1 1" }, { "input": "3 2", "output": "4 0 2" }, { "input": "3 5", "output": "3 1 2" }, { "input": "3 3", "output": "0 6 0" }, { "input": "6 2", "output": "2 1 3" }, { "input": "4 1", "output": "4 0 2" }, { "input": "5 2", "output": "3 0 3" }, { "input": "4 2", "output": "3 1 2" }, { "input": "2 1", "output": "5 0 1" }, { "input": "6 1", "output": "3 0 3" }, { "input": "4 6", "output": "4 1 1" }, { "input": "2 6", "output": "3 1 2" }, { "input": "3 4", "output": "3 0 3" }, { "input": "1 2", "output": "1 0 5" }, { "input": "6 6", "output": "0 6 0" }, { "input": "5 4", "output": "2 0 4" }, { "input": "3 3", "output": "0 6 0" }, { "input": "1 1", "output": "0 6 0" } ]

1,622,820,401

2,147,483,647

PyPy 3

OK

TESTS

38

202

0

a,b=map(int,input().split()) c1=c2=c3=0 for i in range(1,7): if abs(a-i)<abs(i-b): c1+=1 elif abs(a-i)==abs(b-i): c2+=1 else: c3+=1 print(c1,c2,c3)

Title: Playing with Dice Time Limit: None seconds Memory Limit: None megabytes Problem Description: Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw. The first player wrote number *a*, the second player wrote number *b*. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins? Input Specification: The single line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=6) — the numbers written on the paper by the first and second player, correspondingly. Output Specification: Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly. Demo Input: ['2 5\n', '2 4\n'] Demo Output: ['3 0 3\n', '2 1 3\n'] Note: The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct. You can assume that number *a* is closer to number *x* than number *b*, if |*a* - *x*| < |*b* - *x*|.

```python a,b=map(int,input().split()) c1=c2=c3=0 for i in range(1,7): if abs(a-i)<abs(i-b): c1+=1 elif abs(a-i)==abs(b-i): c2+=1 else: c3+=1 print(c1,c2,c3) ```

3

303

A

Lucky Permutation Triple

PROGRAMMING

1,300

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

null

Bike is interested in permutations. A permutation of length *n* is an integer sequence such that each integer from 0 to (*n*<=-<=1) appears exactly once in it. For example, [0,<=2,<=1] is a permutation of length 3 while both [0,<=2,<=2] and [1,<=2,<=3] is not. A permutation triple of permutations of length *n* (*a*,<=*b*,<=*c*) is called a Lucky Permutation Triple if and only if . The sign *a**i* denotes the *i*-th element of permutation *a*. The modular equality described above denotes that the remainders after dividing *a**i*<=+<=*b**i* by *n* and dividing *c**i* by *n* are equal. Now, he has an integer *n* and wants to find a Lucky Permutation Triple. Could you please help him?

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105).

If no Lucky Permutation Triple of length *n* exists print -1. Otherwise, you need to print three lines. Each line contains *n* space-seperated integers. The first line must contain permutation *a*, the second line — permutation *b*, the third — permutation *c*. If there are multiple solutions, print any of them.

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

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

In Sample 1, the permutation triple ([1, 4, 3, 2, 0], [1, 0, 2, 4, 3], [2, 4, 0, 1, 3]) is Lucky Permutation Triple, as following holds: - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a6bf1b9b57809dbec5021f65f89616f259587c07.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/48cc13134296b68f459f69d78e0240859aaec702.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ac44412de7b46833e90348a6b3298f9796e3977c.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3825b0bb758208dda2ead1c5224c05d89ad9ab55.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0a72e2da40048a507839927a211267ac01c9bf89.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In Sample 2, you can easily notice that no lucky permutation triple exists.

500

[ { "input": "5", "output": "1 4 3 2 0\n1 0 2 4 3\n2 4 0 1 3" }, { "input": "2", "output": "-1" }, { "input": "8", "output": "-1" }, { "input": "9", "output": "0 1 2 3 4 5 6 7 8 \n0 1 2 3 4 5 6 7 8 \n0 2 4 6 8 1 3 5 7 " }, { "input": "2", "output": "-1" }, { "input": "77", "output": "0 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 \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 \n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 4..." }, { "input": "6", "output": "-1" }, { "input": "87", "output": "0 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 \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 \n0 2 4..." }, { "input": "72", "output": "-1" }, { "input": "1", "output": "0 \n0 \n0 " }, { "input": "23", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n0 2 4 6 8 10 12 14 16 18 20 22 1 3 5 7 9 11 13 15 17 19 21 " }, { "input": "52", "output": "-1" }, { "input": "32", "output": "-1" }, { "input": "25", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 \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 \n0 2 4 6 8 10 12 14 16 18 20 22 24 1 3 5 7 9 11 13 15 17 19 21 23 " }, { "input": "54", "output": "-1" }, { "input": "39", "output": "0 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 \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 \n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 " }, { "input": "20", "output": "-1" }, { "input": "53", "output": "0 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 \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 \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 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 " }, { "input": "34", "output": "-1" }, { "input": "23", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n0 2 4 6 8 10 12 14 16 18 20 22 1 3 5 7 9 11 13 15 17 19 21 " }, { "input": "37123", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "41904", "output": "-1" }, { "input": "46684", "output": "-1" }, { "input": "67817", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "72598", "output": "-1" }, { "input": "85891", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "74320", "output": "-1" }, { "input": "11805", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "16586", "output": "-1" }, { "input": "5014", "output": "-1" }, { "input": "73268", "output": "-1" }, { "input": "61697", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "99182", "output": "-1" }, { "input": "79771", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "68199", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "5684", "output": "-1" }, { "input": "10465", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "31598", "output": "-1" }, { "input": "36379", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "16968", "output": "-1" }, { "input": "93061", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "73650", "output": "-1" }, { "input": "94783", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "99564", "output": "-1" }, { "input": "37049", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "25478", "output": "-1" }, { "input": "30259", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "43551", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "31980", "output": "-1" }, { "input": "69465", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "1", "output": "0 \n0 \n0 " }, { "input": "100000", "output": "-1" }, { "input": "99999", "output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..." }, { "input": "99998", "output": "-1" } ]

1,619,279,320

2,147,483,647

Python 3

OK

TESTS

54

934

0

n = int(input()) if n%2: j = 2 while j: for i in range(n): print(i,end=' ') j -= 1 print('') for i in range(n): print((2*i)%n,end=' ') else: print(-1)

Title: Lucky Permutation Triple Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bike is interested in permutations. A permutation of length *n* is an integer sequence such that each integer from 0 to (*n*<=-<=1) appears exactly once in it. For example, [0,<=2,<=1] is a permutation of length 3 while both [0,<=2,<=2] and [1,<=2,<=3] is not. A permutation triple of permutations of length *n* (*a*,<=*b*,<=*c*) is called a Lucky Permutation Triple if and only if . The sign *a**i* denotes the *i*-th element of permutation *a*. The modular equality described above denotes that the remainders after dividing *a**i*<=+<=*b**i* by *n* and dividing *c**i* by *n* are equal. Now, he has an integer *n* and wants to find a Lucky Permutation Triple. Could you please help him? Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105). Output Specification: If no Lucky Permutation Triple of length *n* exists print -1. Otherwise, you need to print three lines. Each line contains *n* space-seperated integers. The first line must contain permutation *a*, the second line — permutation *b*, the third — permutation *c*. If there are multiple solutions, print any of them. Demo Input: ['5\n', '2\n'] Demo Output: ['1 4 3 2 0\n1 0 2 4 3\n2 4 0 1 3\n', '-1\n'] Note: In Sample 1, the permutation triple ([1, 4, 3, 2, 0], [1, 0, 2, 4, 3], [2, 4, 0, 1, 3]) is Lucky Permutation Triple, as following holds: - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a6bf1b9b57809dbec5021f65f89616f259587c07.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/48cc13134296b68f459f69d78e0240859aaec702.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ac44412de7b46833e90348a6b3298f9796e3977c.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3825b0bb758208dda2ead1c5224c05d89ad9ab55.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0a72e2da40048a507839927a211267ac01c9bf89.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In Sample 2, you can easily notice that no lucky permutation triple exists.

```python n = int(input()) if n%2: j = 2 while j: for i in range(n): print(i,end=' ') j -= 1 print('') for i in range(n): print((2*i)%n,end=' ') else: print(-1) ```

3

368

B

Sereja and Suffixes

PROGRAMMING

1,100

[ "data structures", "dp" ]

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" } ]

1,668,520,672

2,147,483,647

PyPy 3-64

OK

TESTS

30

998

16,896,000

n, m = map(int, input().split()) array = list(map(int, input().split())) test = set() dp = [] out = 0 for a_j in reversed(array): if a_j not in test: test.add(a_j) out += 1 dp.append(out) for i in range(m): rev_index = len(array) - int(input()) print(dp[rev_index])

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 = map(int, input().split()) array = list(map(int, input().split())) test = set() dp = [] out = 0 for a_j in reversed(array): if a_j not in test: test.add(a_j) out += 1 dp.append(out) for i in range(m): rev_index = len(array) - int(input()) print(dp[rev_index]) ```

3

264

B

Good Sequences

PROGRAMMING

1,500

[ "dp", "number theory" ]

null

Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks *n* integers *a*1,<=*a*2,<=...,<=*a**n* are good. Now she is interested in good sequences. A sequence *x*1,<=*x*2,<=...,<=*x**k* is called good if it satisfies the following three conditions: - The sequence is strictly increasing, i.e. *x**i*<=<<=*x**i*<=+<=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1). - No two adjacent elements are coprime, i.e. *gcd*(*x**i*,<=*x**i*<=+<=1)<=><=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1) (where *gcd*(*p*,<=*q*) denotes the greatest common divisor of the integers *p* and *q*). - All elements of the sequence are good integers. Find the length of the longest good sequence.

The input consists of two lines. The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of good integers. The second line contains a single-space separated list of good integers *a*1,<=*a*2,<=...,<=*a**n* in strictly increasing order (1<=≤<=*a**i*<=≤<=105; *a**i*<=<<=*a**i*<=+<=1).

Print a single integer — the length of the longest good sequence.

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

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

In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4.

1,000

[ { "input": "5\n2 3 4 6 9", "output": "4" }, { "input": "9\n1 2 3 5 6 7 8 9 10", "output": "4" }, { "input": "4\n1 2 4 6", "output": "3" }, { "input": "7\n1 2 3 4 7 9 10", "output": "3" }, { "input": "1\n1", "output": "1" }, { "input": "8\n3 4 5 6 7 8 9 10", "output": "4" }, { "input": "5\n2 3 7 9 10", "output": "2" }, { "input": "3\n1 4 7", "output": "1" }, { "input": "1\n4", "output": "1" }, { "input": "9\n1 2 3 4 5 6 7 9 10", "output": "4" }, { "input": "49\n10 34 58 72 126 166 176 180 198 200 208 228 238 248 302 332 340 344 350 354 380 406 418 428 438 442 482 532 536 544 546 554 596 626 642 682 684 704 714 792 804 820 862 880 906 946 954 966 970", "output": "49" }, { "input": "44\n1 5 37 97 107 147 185 187 195 241 249 295 311 323 341 345 363 391 425 431 473 525 539 541 555 577 595 611 647 695 757 759 775 779 869 877 927 935 963 965 967 969 973 975", "output": "15" }, { "input": "3\n21 67 243", "output": "2" }, { "input": "3\n150 358 382", "output": "3" }, { "input": "10\n13 2187 2197 4567 5200 29873 67866 98798 99999 100000", "output": "6" }, { "input": "10\n1 2 4 8 16 32 33 64 128 256", "output": "8" }, { "input": "10\n2 4 8 67 128 324 789 1296 39877 98383", "output": "7" }, { "input": "10\n2 3 4 5 6 8 9 10 17 92", "output": "6" }, { "input": "10\n2 3 10 40 478 3877 28787 88888 99999 100000", "output": "6" }, { "input": "10\n2 3 45 67 89 101 234 567 890 1234", "output": "5" }, { "input": "10\n2 3 4 5 6 7 8 9 10 11", "output": "5" }, { "input": "2\n5101 10202", "output": "2" }, { "input": "1\n99991", "output": "1" }, { "input": "2\n1009 2018", "output": "2" }, { "input": "2\n601 1202", "output": "2" }, { "input": "3\n3 14 22", "output": "2" } ]

1,601,999,870

2,147,483,647

PyPy 3

OK

TESTS

86

498

16,076,800

n=10**5+1 p=[0]*n t=[[] for _ in range(n)] t[1]=[1] for i in range(2,n): if not t[i]: t[i]=[i] for j in range(2*i,n,i): t[j].append(i) m=input() arr=list(map(int,input().split())) x=0 #print(t[:10]) for i in arr: x=0 for j in t[i]: x=max(p[j],x) x+=1 for j in t[i]: p[j]=x #print(p[0:12]) print(max(p))

Title: Good Sequences Time Limit: None seconds Memory Limit: None megabytes Problem Description: Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks *n* integers *a*1,<=*a*2,<=...,<=*a**n* are good. Now she is interested in good sequences. A sequence *x*1,<=*x*2,<=...,<=*x**k* is called good if it satisfies the following three conditions: - The sequence is strictly increasing, i.e. *x**i*<=<<=*x**i*<=+<=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1). - No two adjacent elements are coprime, i.e. *gcd*(*x**i*,<=*x**i*<=+<=1)<=><=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1) (where *gcd*(*p*,<=*q*) denotes the greatest common divisor of the integers *p* and *q*). - All elements of the sequence are good integers. Find the length of the longest good sequence. Input Specification: The input consists of two lines. The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of good integers. The second line contains a single-space separated list of good integers *a*1,<=*a*2,<=...,<=*a**n* in strictly increasing order (1<=≤<=*a**i*<=≤<=105; *a**i*<=<<=*a**i*<=+<=1). Output Specification: Print a single integer — the length of the longest good sequence. Demo Input: ['5\n2 3 4 6 9\n', '9\n1 2 3 5 6 7 8 9 10\n'] Demo Output: ['4\n', '4\n'] Note: In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4.

```python n=10**5+1 p=[0]*n t=[[] for _ in range(n)] t[1]=[1] for i in range(2,n): if not t[i]: t[i]=[i] for j in range(2*i,n,i): t[j].append(i) m=input() arr=list(map(int,input().split())) x=0 #print(t[:10]) for i in arr: x=0 for j in t[i]: x=max(p[j],x) x+=1 for j in t[i]: p[j]=x #print(p[0:12]) print(max(p)) ```

3

987

A

Infinity Gauntlet

PROGRAMMING

800

[ "implementation" ]

null

You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems: - the Power Gem of purple color, - the Time Gem of green color, - the Space Gem of blue color, - the Soul Gem of orange color, - the Reality Gem of red color, - the Mind Gem of yellow color. Using colors of Gems you saw in the Gauntlet determine the names of absent Gems.

In the first line of input there is one integer $n$ ($0 \le n \le 6$) — the number of Gems in Infinity Gauntlet. In next $n$ lines there are colors of Gems you saw. Words used for colors are: purple, green, blue, orange, red, yellow. It is guaranteed that all the colors are distinct. All colors are given in lowercase English letters.

In the first line output one integer $m$ ($0 \le m \le 6$) — the number of absent Gems. Then in $m$ lines print the names of absent Gems, each on its own line. Words used for names are: Power, Time, Space, Soul, Reality, Mind. Names can be printed in any order. Keep the first letter uppercase, others lowercase.

[ "4\nred\npurple\nyellow\norange\n", "0\n" ]

[ "2\nSpace\nTime\n", "6\nTime\nMind\nSoul\nPower\nReality\nSpace\n" ]

In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space. In the second sample Thanos doesn't have any Gems, so he needs all six.

500

[ { "input": "4\nred\npurple\nyellow\norange", "output": "2\nSpace\nTime" }, { "input": "0", "output": "6\nMind\nSpace\nPower\nTime\nReality\nSoul" }, { "input": "6\npurple\nblue\nyellow\nred\ngreen\norange", "output": "0" }, { "input": "1\npurple", "output": "5\nTime\nReality\nSoul\nSpace\nMind" }, { "input": "3\nblue\norange\npurple", "output": "3\nTime\nReality\nMind" }, { "input": "2\nyellow\nred", "output": "4\nPower\nSoul\nSpace\nTime" }, { "input": "1\ngreen", "output": "5\nReality\nSpace\nPower\nSoul\nMind" }, { "input": "2\npurple\ngreen", "output": "4\nReality\nMind\nSpace\nSoul" }, { "input": "1\nblue", "output": "5\nPower\nReality\nSoul\nTime\nMind" }, { "input": "2\npurple\nblue", "output": "4\nMind\nSoul\nTime\nReality" }, { "input": "2\ngreen\nblue", "output": "4\nReality\nMind\nPower\nSoul" }, { "input": "3\npurple\ngreen\nblue", "output": "3\nMind\nReality\nSoul" }, { "input": "1\norange", "output": "5\nReality\nTime\nPower\nSpace\nMind" }, { "input": "2\npurple\norange", "output": "4\nReality\nMind\nTime\nSpace" }, { "input": "2\norange\ngreen", "output": "4\nSpace\nMind\nReality\nPower" }, { "input": "3\norange\npurple\ngreen", "output": "3\nReality\nSpace\nMind" }, { "input": "2\norange\nblue", "output": "4\nTime\nMind\nReality\nPower" }, { "input": "3\nblue\ngreen\norange", "output": "3\nPower\nMind\nReality" }, { "input": "4\nblue\norange\ngreen\npurple", "output": "2\nMind\nReality" }, { "input": "1\nred", "output": "5\nTime\nSoul\nMind\nPower\nSpace" }, { "input": "2\nred\npurple", "output": "4\nMind\nSpace\nTime\nSoul" }, { "input": "2\nred\ngreen", "output": "4\nMind\nSpace\nPower\nSoul" }, { "input": "3\nred\npurple\ngreen", "output": "3\nSoul\nSpace\nMind" }, { "input": "2\nblue\nred", "output": "4\nMind\nTime\nPower\nSoul" }, { "input": "3\nred\nblue\npurple", "output": "3\nTime\nMind\nSoul" }, { "input": "3\nred\nblue\ngreen", "output": "3\nSoul\nPower\nMind" }, { "input": "4\npurple\nblue\ngreen\nred", "output": "2\nMind\nSoul" }, { "input": "2\norange\nred", "output": "4\nPower\nMind\nTime\nSpace" }, { "input": "3\nred\norange\npurple", "output": "3\nMind\nSpace\nTime" }, { "input": "3\nred\norange\ngreen", "output": "3\nMind\nSpace\nPower" }, { "input": "4\nred\norange\ngreen\npurple", "output": "2\nSpace\nMind" }, { "input": "3\nblue\norange\nred", "output": "3\nPower\nMind\nTime" }, { "input": "4\norange\nblue\npurple\nred", "output": "2\nTime\nMind" }, { "input": "4\ngreen\norange\nred\nblue", "output": "2\nMind\nPower" }, { "input": "5\npurple\norange\nblue\nred\ngreen", "output": "1\nMind" }, { "input": "1\nyellow", "output": "5\nPower\nSoul\nReality\nSpace\nTime" }, { "input": "2\npurple\nyellow", "output": "4\nTime\nReality\nSpace\nSoul" }, { "input": "2\ngreen\nyellow", "output": "4\nSpace\nReality\nPower\nSoul" }, { "input": "3\npurple\nyellow\ngreen", "output": "3\nSoul\nReality\nSpace" }, { "input": "2\nblue\nyellow", "output": "4\nTime\nReality\nPower\nSoul" }, { "input": "3\nyellow\nblue\npurple", "output": "3\nSoul\nReality\nTime" }, { "input": "3\ngreen\nyellow\nblue", "output": "3\nSoul\nReality\nPower" }, { "input": "4\nyellow\nblue\ngreen\npurple", "output": "2\nReality\nSoul" }, { "input": "2\nyellow\norange", "output": "4\nTime\nSpace\nReality\nPower" }, { "input": "3\nyellow\npurple\norange", "output": "3\nSpace\nReality\nTime" }, { "input": "3\norange\nyellow\ngreen", "output": "3\nSpace\nReality\nPower" }, { "input": "4\ngreen\nyellow\norange\npurple", "output": "2\nSpace\nReality" }, { "input": "3\nyellow\nblue\norange", "output": "3\nTime\nReality\nPower" }, { "input": "4\norange\npurple\nblue\nyellow", "output": "2\nReality\nTime" }, { "input": "4\nblue\norange\nyellow\ngreen", "output": "2\nReality\nPower" }, { "input": "5\ngreen\nyellow\norange\nblue\npurple", "output": "1\nReality" }, { "input": "3\nyellow\npurple\nred", "output": "3\nTime\nSoul\nSpace" }, { "input": "3\nred\ngreen\nyellow", "output": "3\nPower\nSoul\nSpace" }, { "input": "4\nred\npurple\ngreen\nyellow", "output": "2\nSpace\nSoul" }, { "input": "3\nred\nyellow\nblue", "output": "3\nPower\nSoul\nTime" }, { "input": "4\nblue\nyellow\nred\npurple", "output": "2\nTime\nSoul" }, { "input": "4\nblue\nyellow\nred\ngreen", "output": "2\nSoul\nPower" }, { "input": "5\nred\nyellow\ngreen\nblue\npurple", "output": "1\nSoul" }, { "input": "3\nred\nyellow\norange", "output": "3\nPower\nSpace\nTime" }, { "input": "4\norange\ngreen\nyellow\nred", "output": "2\nPower\nSpace" }, { "input": "5\norange\nred\ngreen\nyellow\npurple", "output": "1\nSpace" }, { "input": "4\nyellow\nred\norange\nblue", "output": "2\nTime\nPower" }, { "input": "5\npurple\nblue\norange\nyellow\nred", "output": "1\nTime" }, { "input": "5\norange\nblue\nyellow\nred\ngreen", "output": "1\nPower" } ]

1,626,954,335

2,147,483,647

Python 3

OK

TESTS

64

77

6,963,200

stones=int(input()) if stones==6: print(0) else: d={'purple':'Power','green':'Time','blue':'Space','orange':'Soul','red':'Reality','yellow':'Mind'} z=[] for i in range(stones): s=input() z+=s, print(6-stones) for i in d: if i not in z: print(d[i])

Title: Infinity Gauntlet Time Limit: None seconds Memory Limit: None megabytes Problem Description: You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems: - the Power Gem of purple color, - the Time Gem of green color, - the Space Gem of blue color, - the Soul Gem of orange color, - the Reality Gem of red color, - the Mind Gem of yellow color. Using colors of Gems you saw in the Gauntlet determine the names of absent Gems. Input Specification: In the first line of input there is one integer $n$ ($0 \le n \le 6$) — the number of Gems in Infinity Gauntlet. In next $n$ lines there are colors of Gems you saw. Words used for colors are: purple, green, blue, orange, red, yellow. It is guaranteed that all the colors are distinct. All colors are given in lowercase English letters. Output Specification: In the first line output one integer $m$ ($0 \le m \le 6$) — the number of absent Gems. Then in $m$ lines print the names of absent Gems, each on its own line. Words used for names are: Power, Time, Space, Soul, Reality, Mind. Names can be printed in any order. Keep the first letter uppercase, others lowercase. Demo Input: ['4\nred\npurple\nyellow\norange\n', '0\n'] Demo Output: ['2\nSpace\nTime\n', '6\nTime\nMind\nSoul\nPower\nReality\nSpace\n'] Note: In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space. In the second sample Thanos doesn't have any Gems, so he needs all six.

```python stones=int(input()) if stones==6: print(0) else: d={'purple':'Power','green':'Time','blue':'Space','orange':'Soul','red':'Reality','yellow':'Mind'} z=[] for i in range(stones): s=input() z+=s, print(6-stones) for i in d: if i not in z: print(d[i]) ```

3

870

C

Maximum splitting

PROGRAMMING

1,300

[ "dp", "greedy", "math", "number theory" ]

null

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

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

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

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

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

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

1,500

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

1,683,726,912

2,147,483,647

PyPy 3-64

OK

TESTS

20

982

9,523,200

nums=[4,9,6,15] t=int(input()) for _ in range(t): n=int(input()) if n < nums[n%4]: print(-1) else: ans = (n-nums[n%4])//4+1 if(n%4==3): ans+=1 print(ans)

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

```python nums=[4,9,6,15] t=int(input()) for _ in range(t): n=int(input()) if n < nums[n%4]: print(-1) else: ans = (n-nums[n%4])//4+1 if(n%4==3): ans+=1 print(ans) ```

3

520

A

Pangram

PROGRAMMING

800

[ "implementation", "strings" ]

null

A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices. You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string. The second line contains the string. The string consists only of uppercase and lowercase Latin letters.

Output "YES", if the string is a pangram and "NO" otherwise.

[ "12\ntoosmallword\n", "35\nTheQuickBrownFoxJumpsOverTheLazyDog\n" ]

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

none

500

[ { "input": "12\ntoosmallword", "output": "NO" }, { "input": "35\nTheQuickBrownFoxJumpsOverTheLazyDog", "output": "YES" }, { "input": "1\na", "output": "NO" }, { "input": "26\nqwertyuiopasdfghjklzxcvbnm", "output": "YES" }, { "input": "26\nABCDEFGHIJKLMNOPQRSTUVWXYZ", "output": "YES" }, { "input": "48\nthereisasyetinsufficientdataforameaningfulanswer", "output": "NO" }, { "input": "30\nToBeOrNotToBeThatIsTheQuestion", "output": "NO" }, { "input": "30\njackdawslovemybigsphinxofquarz", "output": "NO" }, { "input": "31\nTHEFIVEBOXINGWIZARDSJUMPQUICKLY", "output": "YES" }, { "input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "NO" }, { "input": "26\nMGJYIZDKsbhpVeNFlquRTcWoAx", "output": "YES" }, { "input": "26\nfWMOhAPsbIVtyUEZrGNQXDklCJ", "output": "YES" }, { "input": "26\nngPMVFSThiRCwLEuyOAbKxQzDJ", "output": "YES" }, { "input": "25\nnxYTzLFwzNolAumjgcAboyxAj", "output": "NO" }, { "input": "26\npRWdodGdxUESvcScPGbUoooZsC", "output": "NO" }, { "input": "66\nBovdMlDzTaqKllZILFVfxbLGsRnzmtVVTmqiIDTYrossLEPlmsPrkUYtWEsGHVOnFj", "output": "NO" }, { "input": "100\nmKtsiDRJypUieHIkvJaMFkwaKxcCIbBszZQLIyPpCDCjhNpAnYFngLjRpnKWpKWtGnwoSteeZXuFHWQxxxOpFlNeYTwKocsXuCoa", "output": "YES" }, { "input": "26\nEoqxUbsLjPytUHMiFnvcGWZdRK", "output": "NO" }, { "input": "26\nvCUFRKElZOnjmXGylWQaHDiPst", "output": "NO" }, { "input": "26\nWtrPuaHdXLKJMsnvQfgOiJZBEY", "output": "NO" }, { "input": "26\npGiFluRteQwkaVoPszJyNBChxM", "output": "NO" }, { "input": "26\ncTUpqjPmANrdbzSFhlWIoKxgVY", "output": "NO" }, { "input": "26\nLndjgvAEuICHKxPwqYztosrmBN", "output": "NO" }, { "input": "26\nMdaXJrCipnOZLykfqHWEStevbU", "output": "NO" }, { "input": "26\nEjDWsVxfKTqGXRnUMOLYcIzPba", "output": "NO" }, { "input": "26\nxKwzRMpunYaqsdfaBgJcVElTHo", "output": "NO" }, { "input": "26\nnRYUQsTwCPLZkgshfEXvBdoiMa", "output": "NO" }, { "input": "26\nHNCQPfJutyAlDGsvRxZWMEbIdO", "output": "NO" }, { "input": "26\nDaHJIpvKznQcmUyWsTGObXRFDe", "output": "NO" }, { "input": "26\nkqvAnFAiRhzlJbtyuWedXSPcOG", "output": "NO" }, { "input": "26\nhlrvgdwsIOyjcmUZXtAKEqoBpF", "output": "NO" }, { "input": "26\njLfXXiMhBTcAwQVReGnpKzdsYu", "output": "NO" }, { "input": "26\nlNMcVuwItjxRBGAekjhyDsQOzf", "output": "NO" }, { "input": "26\nRkSwbNoYldUGtAZvpFMcxhIJFE", "output": "NO" }, { "input": "26\nDqspXZJTuONYieKgaHLMBwfVSC", "output": "NO" }, { "input": "26\necOyUkqNljFHRVXtIpWabGMLDz", "output": "NO" }, { "input": "26\nEKAvqZhBnPmVCDRlgWJfOusxYI", "output": "NO" }, { "input": "26\naLbgqeYchKdMrsZxIPFvTOWNjA", "output": "NO" }, { "input": "26\nxfpBLsndiqtacOCHGmeWUjRkYz", "output": "NO" }, { "input": "26\nXsbRKtqleZPNIVCdfUhyagAomJ", "output": "NO" }, { "input": "26\nAmVtbrwquEthZcjKPLiyDgSoNF", "output": "NO" }, { "input": "26\nOhvXDcwqAUmSEPRZGnjFLiKtNB", "output": "NO" }, { "input": "26\nEKWJqCFLRmstxVBdYuinpbhaOg", "output": "NO" }, { "input": "26\nmnbvcxxlkjhgfdsapoiuytrewq", "output": "NO" }, { "input": "26\naAbcdefghijklmnopqrstuvwxy", "output": "NO" }, { "input": "30\nABCDEFGHTYRIOPLabcdefghtyriopl", "output": "NO" }, { "input": "25\nabcdefghijklmnopqrstuvwxy", "output": "NO" }, { "input": "26\nabcdefhijklmnopqrstVxyzABC", "output": "NO" }, { "input": "25\nqwertyuiopasdfghjklxcvbnm", "output": "NO" }, { "input": "34\nTheQuickBrownFoxJumpsOverTheLayDog", "output": "NO" }, { "input": "26\nabcdefghigklmnopqrstuvwxyz", "output": "NO" }, { "input": "26\nabcdefghijklmnopqrstuvwxyA", "output": "NO" }, { "input": "50\nqazwsxedcrfvtgbyhnujmikolQWERTYUIOASDFGHJKLZXCVBNM", "output": "NO" }, { "input": "35\nTheQuickBrownFoxJumpsOverTheLasyDog", "output": "NO" }, { "input": "25\nbcdefghijklmnopqrstuvwxyz", "output": "NO" }, { "input": "38\nAbCdEfGhIjKlMnOpQrStVwXyZzzzzzzaaaaaaa", "output": "NO" }, { "input": "26\nabcdefghiklmnopqrstvxyzABC", "output": "NO" }, { "input": "26\nabcdefghijklmnopqrstuvwxzZ", "output": "NO" }, { "input": "50\nabcdefghijklmnopqrstuvwxyABCDEFGHIJKLMNOPQRSTUVWXY", "output": "NO" } ]

1,699,113,088

2,147,483,647

Python 3

OK

TESTS

59

46

0

n = int(input()) d = {} s = input() count = 0 for i in s: x = i.lower() if x not in d: d[x] = 1 count = count + 1 else: d[x] = d[x] + 1 if count==26: print('YES') else: print('NO')

Title: Pangram Time Limit: None seconds Memory Limit: None megabytes Problem Description: A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices. You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string. The second line contains the string. The string consists only of uppercase and lowercase Latin letters. Output Specification: Output "YES", if the string is a pangram and "NO" otherwise. Demo Input: ['12\ntoosmallword\n', '35\nTheQuickBrownFoxJumpsOverTheLazyDog\n'] Demo Output: ['NO\n', 'YES\n'] Note: none

```python n = int(input()) d = {} s = input() count = 0 for i in s: x = i.lower() if x not in d: d[x] = 1 count = count + 1 else: d[x] = d[x] + 1 if count==26: print('YES') else: print('NO') ```

3

556

A

Case of the Zeros and Ones

PROGRAMMING

900

[ "greedy" ]

null

Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones. Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result. Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.

First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has. The second line contains the string of length *n* consisting only from zeros and ones.

Output the minimum length of the string that may remain after applying the described operations several times.

[ "4\n1100\n", "5\n01010\n", "8\n11101111\n" ]

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

In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

250

[ { "input": "4\n1100", "output": "0" }, { "input": "5\n01010", "output": "1" }, { "input": "8\n11101111", "output": "6" }, { "input": "1\n0", "output": "1" }, { "input": "1\n1", "output": "1" }, { "input": "2\n00", "output": "2" }, { "input": "2\n01", "output": "0" }, { "input": "2\n10", "output": "0" }, { "input": "2\n11", "output": "2" }, { "input": "3\n001", "output": "1" }, { "input": "6\n110110", "output": "2" }, { "input": "7\n0000011", "output": "3" }, { "input": "6\n110010", "output": "0" }, { "input": "6\n110100", "output": "0" }, { "input": "3\n100", "output": "1" }, { "input": "6\n010111", "output": "2" }, { "input": "8\n01011100", "output": "0" }, { "input": "6\n001011", "output": "0" }, { "input": "7\n1110000", "output": "1" }, { "input": "9\n011111101", "output": "5" } ]

1,694,640,089

2,147,483,647

Python 3

OK

TESTS

49

46

409,600

n = int(input()) text_input = input() a = text_input.count("0") b = text_input.count("1") c = min(a, b) print(len(text_input) - (2 * c))

Title: Case of the Zeros and Ones Time Limit: None seconds Memory Limit: None megabytes Problem Description: Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones. Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result. Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number. Input Specification: First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has. The second line contains the string of length *n* consisting only from zeros and ones. Output Specification: Output the minimum length of the string that may remain after applying the described operations several times. Demo Input: ['4\n1100\n', '5\n01010\n', '8\n11101111\n'] Demo Output: ['0\n', '1\n', '6\n'] Note: In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

```python n = int(input()) text_input = input() a = text_input.count("0") b = text_input.count("1") c = min(a, b) print(len(text_input) - (2 * c)) ```

3

264

B

Good Sequences

PROGRAMMING

1,500

[ "dp", "number theory" ]

null

Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks *n* integers *a*1,<=*a*2,<=...,<=*a**n* are good. Now she is interested in good sequences. A sequence *x*1,<=*x*2,<=...,<=*x**k* is called good if it satisfies the following three conditions: - The sequence is strictly increasing, i.e. *x**i*<=<<=*x**i*<=+<=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1). - No two adjacent elements are coprime, i.e. *gcd*(*x**i*,<=*x**i*<=+<=1)<=><=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1) (where *gcd*(*p*,<=*q*) denotes the greatest common divisor of the integers *p* and *q*). - All elements of the sequence are good integers. Find the length of the longest good sequence.

The input consists of two lines. The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of good integers. The second line contains a single-space separated list of good integers *a*1,<=*a*2,<=...,<=*a**n* in strictly increasing order (1<=≤<=*a**i*<=≤<=105; *a**i*<=<<=*a**i*<=+<=1).

Print a single integer — the length of the longest good sequence.

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

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

In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4.

1,000

[ { "input": "5\n2 3 4 6 9", "output": "4" }, { "input": "9\n1 2 3 5 6 7 8 9 10", "output": "4" }, { "input": "4\n1 2 4 6", "output": "3" }, { "input": "7\n1 2 3 4 7 9 10", "output": "3" }, { "input": "1\n1", "output": "1" }, { "input": "8\n3 4 5 6 7 8 9 10", "output": "4" }, { "input": "5\n2 3 7 9 10", "output": "2" }, { "input": "3\n1 4 7", "output": "1" }, { "input": "1\n4", "output": "1" }, { "input": "9\n1 2 3 4 5 6 7 9 10", "output": "4" }, { "input": "49\n10 34 58 72 126 166 176 180 198 200 208 228 238 248 302 332 340 344 350 354 380 406 418 428 438 442 482 532 536 544 546 554 596 626 642 682 684 704 714 792 804 820 862 880 906 946 954 966 970", "output": "49" }, { "input": "44\n1 5 37 97 107 147 185 187 195 241 249 295 311 323 341 345 363 391 425 431 473 525 539 541 555 577 595 611 647 695 757 759 775 779 869 877 927 935 963 965 967 969 973 975", "output": "15" }, { "input": "3\n21 67 243", "output": "2" }, { "input": "3\n150 358 382", "output": "3" }, { "input": "10\n13 2187 2197 4567 5200 29873 67866 98798 99999 100000", "output": "6" }, { "input": "10\n1 2 4 8 16 32 33 64 128 256", "output": "8" }, { "input": "10\n2 4 8 67 128 324 789 1296 39877 98383", "output": "7" }, { "input": "10\n2 3 4 5 6 8 9 10 17 92", "output": "6" }, { "input": "10\n2 3 10 40 478 3877 28787 88888 99999 100000", "output": "6" }, { "input": "10\n2 3 45 67 89 101 234 567 890 1234", "output": "5" }, { "input": "10\n2 3 4 5 6 7 8 9 10 11", "output": "5" }, { "input": "2\n5101 10202", "output": "2" }, { "input": "1\n99991", "output": "1" }, { "input": "2\n1009 2018", "output": "2" }, { "input": "2\n601 1202", "output": "2" }, { "input": "3\n3 14 22", "output": "2" } ]

1,644,234,615

2,147,483,647

PyPy 3

OK

TESTS

86

1,870

75,264,000

import math def solve(): n = int(input()) # l, r = list(map(int, input().split(" "))) a = list(map(int, input().split(" "))) # b = list(map(int, input().split(" "))) # s = input() N = int(10e5 + 5) dp = [0]*N fac = [[] for _ in range(N)] seive = [1] * N for i in range(2, N): if seive[i]: for j in range(i, N, i): seive[j] = 0 fac[j].append(i) ans = 1 for i in range(n): z = 0 for x in fac[a[i]]: z = max(dp[x], z) z += 1 ans = max(ans, z) for x in fac[a[i]]: dp[x] = max(z, dp[x]) print(ans) return # for _ in range(int(input())): # solve() solve()

Title: Good Sequences Time Limit: None seconds Memory Limit: None megabytes Problem Description: Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks *n* integers *a*1,<=*a*2,<=...,<=*a**n* are good. Now she is interested in good sequences. A sequence *x*1,<=*x*2,<=...,<=*x**k* is called good if it satisfies the following three conditions: - The sequence is strictly increasing, i.e. *x**i*<=<<=*x**i*<=+<=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1). - No two adjacent elements are coprime, i.e. *gcd*(*x**i*,<=*x**i*<=+<=1)<=><=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1) (where *gcd*(*p*,<=*q*) denotes the greatest common divisor of the integers *p* and *q*). - All elements of the sequence are good integers. Find the length of the longest good sequence. Input Specification: The input consists of two lines. The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of good integers. The second line contains a single-space separated list of good integers *a*1,<=*a*2,<=...,<=*a**n* in strictly increasing order (1<=≤<=*a**i*<=≤<=105; *a**i*<=<<=*a**i*<=+<=1). Output Specification: Print a single integer — the length of the longest good sequence. Demo Input: ['5\n2 3 4 6 9\n', '9\n1 2 3 5 6 7 8 9 10\n'] Demo Output: ['4\n', '4\n'] Note: In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4.

```python import math def solve(): n = int(input()) # l, r = list(map(int, input().split(" "))) a = list(map(int, input().split(" "))) # b = list(map(int, input().split(" "))) # s = input() N = int(10e5 + 5) dp = [0]*N fac = [[] for _ in range(N)] seive = [1] * N for i in range(2, N): if seive[i]: for j in range(i, N, i): seive[j] = 0 fac[j].append(i) ans = 1 for i in range(n): z = 0 for x in fac[a[i]]: z = max(dp[x], z) z += 1 ans = max(ans, z) for x in fac[a[i]]: dp[x] = max(z, dp[x]) print(ans) return # for _ in range(int(input())): # solve() solve() ```

3

665

B

Shopping

PROGRAMMING

1,400

[ "brute force" ]

null

Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online. The store contains *k* items. *n* customers have already used the above service. Each user paid for *m* items. Let *a**ij* denote the *j*-th item in the *i*-th person's order. Due to the space limitations all the items are arranged in one single row. When Ayush receives the *i*-th order he will find one by one all the items *a**ij* (1<=≤<=*j*<=≤<=*m*) in the row. Let *pos*(*x*) denote the position of the item *x* in the row at the moment of its collection. Then Ayush takes time equal to *pos*(*a**i*1)<=+<=*pos*(*a**i*2)<=+<=...<=+<=*pos*(*a**im*) for the *i*-th customer. When Ayush accesses the *x*-th element he keeps a new stock in the front of the row and takes away the *x*-th element. Thus the values are updating. Your task is to calculate the total time it takes for Ayush to process all the orders. You can assume that the market has endless stock.

The first line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*k*<=≤<=100,<=1<=≤<=*m*<=≤<=*k*) — the number of users, the number of items each user wants to buy and the total number of items at the market. The next line contains *k* distinct integers *p**l* (1<=≤<=*p**l*<=≤<=*k*) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to *k*. Each of the next *n* lines contains *m* distinct integers *a**ij* (1<=≤<=*a**ij*<=≤<=*k*) — the order of the *i*-th person.

Print the only integer *t* — the total time needed for Ayush to process all the orders.

[ "2 2 5\n3 4 1 2 5\n1 5\n3 1\n" ]

[ "14\n" ]

Customer 1 wants the items 1 and 5. *pos*(1) = 3, so the new positions are: [1, 3, 4, 2, 5]. *pos*(5) = 5, so the new positions are: [5, 1, 3, 4, 2]. Time taken for the first customer is 3 + 5 = 8. Customer 2 wants the items 3 and 1. *pos*(3) = 3, so the new positions are: [3, 5, 1, 4, 2]. *pos*(1) = 3, so the new positions are: [1, 3, 5, 4, 2]. Time taken for the second customer is 3 + 3 = 6. Total time is 8 + 6 = 14. Formally *pos*(*x*) is the index of *x* in the current row.

0

[ { "input": "2 2 5\n3 4 1 2 5\n1 5\n3 1", "output": "14" }, { "input": "4 4 4\n1 2 3 4\n3 4 2 1\n4 3 2 1\n4 1 2 3\n4 1 2 3", "output": "59" }, { "input": "1 1 1\n1\n1", "output": "1" }, { "input": "10 1 100\n1 55 67 75 40 86 24 84 82 26 81 23 70 79 51 54 21 78 31 98 68 93 66 88 99 65 20 52 35 85 16 12 94 100 59 56 18 33 47 46 71 8 38 57 2 92 3 95 6 4 87 22 48 80 15 29 11 45 72 76 44 60 91 90 39 74 41 36 13 27 53 83 32 5 30 63 89 64 49 17 9 97 69 14 50 77 37 96 10 42 28 34 61 19 73 7 62 43 58 25\n33\n69\n51\n7\n68\n70\n1\n35\n24\n7", "output": "335" }, { "input": "100 1 1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1", "output": "100" }, { "input": "3 2 3\n3 1 2\n1 2\n2 1\n2 3", "output": "13" }, { "input": "10 10 10\n3 4 1 2 8 9 5 10 6 7\n9 10 7 8 6 1 2 3 4 5\n2 5 3 6 1 4 9 7 8 10\n2 9 1 8 4 7 5 10 6 3\n10 9 7 1 3 6 2 8 5 4\n2 5 1 3 7 10 4 9 8 6\n6 1 8 7 9 2 3 5 4 10\n1 3 2 8 6 9 4 10 5 7\n5 2 4 8 6 1 10 9 3 7\n5 1 7 10 4 6 2 8 9 3\n2 1 3 9 7 10 6 4 8 5", "output": "771" } ]

1,547,023,657

2,147,483,647

Python 3

OK

TESTS

10

109

0

n,m,k=map(int,input().split()) q=list(map(int,input().split())) ans=0 for i in range(n): lst=list(map(int,input().split())) for j in lst: #for x in input().split(): ind=q.index(j) ans+=(ind+1) q.pop(ind) q.insert(0,j) print(ans)

Title: Shopping Time Limit: None seconds Memory Limit: None megabytes Problem Description: Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online. The store contains *k* items. *n* customers have already used the above service. Each user paid for *m* items. Let *a**ij* denote the *j*-th item in the *i*-th person's order. Due to the space limitations all the items are arranged in one single row. When Ayush receives the *i*-th order he will find one by one all the items *a**ij* (1<=≤<=*j*<=≤<=*m*) in the row. Let *pos*(*x*) denote the position of the item *x* in the row at the moment of its collection. Then Ayush takes time equal to *pos*(*a**i*1)<=+<=*pos*(*a**i*2)<=+<=...<=+<=*pos*(*a**im*) for the *i*-th customer. When Ayush accesses the *x*-th element he keeps a new stock in the front of the row and takes away the *x*-th element. Thus the values are updating. Your task is to calculate the total time it takes for Ayush to process all the orders. You can assume that the market has endless stock. Input Specification: The first line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*k*<=≤<=100,<=1<=≤<=*m*<=≤<=*k*) — the number of users, the number of items each user wants to buy and the total number of items at the market. The next line contains *k* distinct integers *p**l* (1<=≤<=*p**l*<=≤<=*k*) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to *k*. Each of the next *n* lines contains *m* distinct integers *a**ij* (1<=≤<=*a**ij*<=≤<=*k*) — the order of the *i*-th person. Output Specification: Print the only integer *t* — the total time needed for Ayush to process all the orders. Demo Input: ['2 2 5\n3 4 1 2 5\n1 5\n3 1\n'] Demo Output: ['14\n'] Note: Customer 1 wants the items 1 and 5. *pos*(1) = 3, so the new positions are: [1, 3, 4, 2, 5]. *pos*(5) = 5, so the new positions are: [5, 1, 3, 4, 2]. Time taken for the first customer is 3 + 5 = 8. Customer 2 wants the items 3 and 1. *pos*(3) = 3, so the new positions are: [3, 5, 1, 4, 2]. *pos*(1) = 3, so the new positions are: [1, 3, 5, 4, 2]. Time taken for the second customer is 3 + 3 = 6. Total time is 8 + 6 = 14. Formally *pos*(*x*) is the index of *x* in the current row.

```python n,m,k=map(int,input().split()) q=list(map(int,input().split())) ans=0 for i in range(n): lst=list(map(int,input().split())) for j in lst: #for x in input().split(): ind=q.index(j) ans+=(ind+1) q.pop(ind) q.insert(0,j) print(ans) ```

3

610

A

Pasha and Stick

PROGRAMMING

1,000

[ "combinatorics", "math" ]

null

Pasha has a wooden stick of some positive integer length *n*. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be *n*. Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square. Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer *x*, such that the number of parts of length *x* in the first way differ from the number of parts of length *x* in the second way.

The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=2·109) — the length of Pasha's stick.

The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.

[ "6\n", "20\n" ]

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

There is only one way to divide the stick in the first sample {1, 1, 2, 2}. Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work.

500

[ { "input": "6", "output": "1" }, { "input": "20", "output": "4" }, { "input": "1", "output": "0" }, { "input": "2", "output": "0" }, { "input": "3", "output": "0" }, { "input": "4", "output": "0" }, { "input": "2000000000", "output": "499999999" }, { "input": "1924704072", "output": "481176017" }, { "input": "73740586", "output": "18435146" }, { "input": "1925088820", "output": "481272204" }, { "input": "593070992", "output": "148267747" }, { "input": "1925473570", "output": "481368392" }, { "input": "629490186", "output": "157372546" }, { "input": "1980649112", "output": "495162277" }, { "input": "36661322", "output": "9165330" }, { "input": "1943590793", "output": "0" }, { "input": "71207034", "output": "17801758" }, { "input": "1757577394", "output": "439394348" }, { "input": "168305294", "output": "42076323" }, { "input": "1934896224", "output": "483724055" }, { "input": "297149088", "output": "74287271" }, { "input": "1898001634", "output": "474500408" }, { "input": "176409698", "output": "44102424" }, { "input": "1873025522", "output": "468256380" }, { "input": "5714762", "output": "1428690" }, { "input": "1829551192", "output": "457387797" }, { "input": "16269438", "output": "4067359" }, { "input": "1663283390", "output": "415820847" }, { "input": "42549941", "output": "0" }, { "input": "1967345604", "output": "491836400" }, { "input": "854000", "output": "213499" }, { "input": "1995886626", "output": "498971656" }, { "input": "10330019", "output": "0" }, { "input": "1996193634", "output": "499048408" }, { "input": "9605180", "output": "2401294" }, { "input": "1996459740", "output": "499114934" }, { "input": "32691948", "output": "8172986" }, { "input": "1975903308", "output": "493975826" }, { "input": "1976637136", "output": "494159283" }, { "input": "29803038", "output": "7450759" }, { "input": "1977979692", "output": "494494922" }, { "input": "1978595336", "output": "494648833" }, { "input": "27379344", "output": "6844835" }, { "input": "1979729912", "output": "494932477" }, { "input": "1980253780", "output": "495063444" }, { "input": "1980751584", "output": "495187895" }, { "input": "53224878", "output": "13306219" }, { "input": "5", "output": "0" }, { "input": "7", "output": "0" }, { "input": "8", "output": "1" }, { "input": "9", "output": "0" }, { "input": "10", "output": "2" }, { "input": "11", "output": "0" }, { "input": "12", "output": "2" }, { "input": "13", "output": "0" }, { "input": "14", "output": "3" }, { "input": "15", "output": "0" }, { "input": "16", "output": "3" }, { "input": "17", "output": "0" }, { "input": "18", "output": "4" }, { "input": "19", "output": "0" }, { "input": "21", "output": "0" }, { "input": "22", "output": "5" }, { "input": "23", "output": "0" }, { "input": "24", "output": "5" }, { "input": "25", "output": "0" }, { "input": "26", "output": "6" }, { "input": "27", "output": "0" }, { "input": "28", "output": "6" }, { "input": "29", "output": "0" }, { "input": "30", "output": "7" }, { "input": "111", "output": "0" }, { "input": "55", "output": "0" }, { "input": "105", "output": "0" }, { "input": "199", "output": "0" }, { "input": "151", "output": "0" } ]

1,559,034,017

2,147,483,647

Python 3

OK

TESTS

76

109

0

x = int(input()) if x%2 == 1: print(0) else: k = x//2 if k%2 == 1: print((k-1)//2) else: print((k)//2-1)

Title: Pasha and Stick Time Limit: None seconds Memory Limit: None megabytes Problem Description: Pasha has a wooden stick of some positive integer length *n*. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be *n*. Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square. Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer *x*, such that the number of parts of length *x* in the first way differ from the number of parts of length *x* in the second way. Input Specification: The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=2·109) — the length of Pasha's stick. Output Specification: The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square. Demo Input: ['6\n', '20\n'] Demo Output: ['1\n', '4\n'] Note: There is only one way to divide the stick in the first sample {1, 1, 2, 2}. Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work.

```python x = int(input()) if x%2 == 1: print(0) else: k = x//2 if k%2 == 1: print((k-1)//2) else: print((k)//2-1) ```

3

743

B

Chloe and the sequence

PROGRAMMING

1,200

[ "binary search", "bitmasks", "constructive algorithms", "implementation" ]

null

Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad. Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step. The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps. Please help Chloe to solve the problem!

The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1).

Print single integer — the integer at the *k*-th position in the obtained sequence.

[ "3 2\n", "4 8\n" ]

[ "2", "4" ]

In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2. In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.

1,000

[ { "input": "3 2", "output": "2" }, { "input": "4 8", "output": "4" }, { "input": "5 27", "output": "1" }, { "input": "7 44", "output": "3" }, { "input": "15 18432", "output": "12" }, { "input": "20 259676", "output": "3" }, { "input": "30 671088640", "output": "28" }, { "input": "38 137438953472", "output": "38" }, { "input": "1 1", "output": "1" }, { "input": "3 3", "output": "1" }, { "input": "4 12", "output": "3" }, { "input": "8 224", "output": "6" }, { "input": "13 1368", "output": "4" }, { "input": "16 49152", "output": "15" }, { "input": "19 1024", "output": "11" }, { "input": "24 15204352", "output": "20" }, { "input": "27 6586544", "output": "5" }, { "input": "31 536870912", "output": "30" }, { "input": "38 94489280512", "output": "34" }, { "input": "42 215268308020", "output": "3" }, { "input": "42 3452074721280", "output": "21" }, { "input": "46 34394312982528", "output": "15" }, { "input": "48 133865540681728", "output": "39" }, { "input": "49 76541041985542", "output": "2" }, { "input": "49 104822971826176", "output": "34" }, { "input": "49 351843720888320", "output": "47" }, { "input": "50 743042492421629", "output": "1" }, { "input": "50 666266740436818", "output": "2" }, { "input": "50 704239287953456", "output": "5" }, { "input": "50 116938486513664", "output": "17" }, { "input": "50 806423059496960", "output": "37" }, { "input": "50 985162418487296", "output": "48" }, { "input": "50 844424930131968", "output": "49" }, { "input": "50 562949953421312", "output": "50" }, { "input": "50 1", "output": "1" }, { "input": "50 1125899906842623", "output": "1" }, { "input": "50 1125899906842620", "output": "3" }, { "input": "39 549755813887", "output": "1" }, { "input": "50 100000000000", "output": "12" } ]

1,600,361,671

2,147,483,647

PyPy 3

OK

TESTS

39

156

0

import sys def input(): return sys.stdin.readline().rstrip("\r\n") def List(): return list(map(int, input().split())) def Num(): return int(input()) def solve(): n, k = List() return (2 * k - bin(k).count('1')) - (2 * (k - 1) - bin(k - 1).count('1')) print(solve())

Title: Chloe and the sequence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad. Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step. The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps. Please help Chloe to solve the problem! Input Specification: The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1). Output Specification: Print single integer — the integer at the *k*-th position in the obtained sequence. Demo Input: ['3 2\n', '4 8\n'] Demo Output: ['2', '4'] Note: In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2. In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.

```python import sys def input(): return sys.stdin.readline().rstrip("\r\n") def List(): return list(map(int, input().split())) def Num(): return int(input()) def solve(): n, k = List() return (2 * k - bin(k).count('1')) - (2 * (k - 1) - bin(k - 1).count('1')) print(solve()) ```

3

431

A

Black Square

PROGRAMMING

800

[ "implementation" ]

null

Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone. In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip. You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares?

The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=≤<=*a*1,<=*a*2,<=*a*3,<=*a*4<=≤<=104). The second line contains string *s* (1<=≤<=|*s*|<=≤<=105), where the *і*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip.

Print a single integer — the total number of calories that Jury wastes.

[ "1 2 3 4\n123214\n", "1 5 3 2\n11221\n" ]

[ "13\n", "13\n" ]

none

500

[ { "input": "1 2 3 4\n123214", "output": "13" }, { "input": "1 5 3 2\n11221", "output": "13" }, { "input": "5 5 5 1\n3422", "output": "16" }, { "input": "4 3 2 1\n2", "output": "3" }, { "input": "5651 6882 6954 4733\n2442313421", "output": "60055" }, { "input": "0 0 0 0\n4132", "output": "0" }, { "input": "3163 5778 83 7640\n11141442444", "output": "64270" }, { "input": "1809 1302 7164 6122\n3144121413113111223311232232114144321414421243443243422322144324121433444342231344234443332241322442", "output": "420780" }, { "input": "0 0 0 0\n1", "output": "0" }, { "input": "1 2 3 4\n4", "output": "4" }, { "input": "2343 7653 1242 5432\n1", "output": "2343" }, { "input": "2343 7653 1242 5432\n2", "output": "7653" }, { "input": "2343 7653 1242 5432\n3", "output": "1242" }, { "input": "2343 7653 1242 5432\n4", "output": "5432" }, { "input": "1 2 3 4\n123412", "output": "13" }, { "input": "50 50 50 50\n11111111111111111111111111111111111111111111111111111", "output": "2650" }, { "input": "1 2 3 4\n11111111111111111111111111111111111111111111111111", "output": "50" }, { "input": "1 2 3 4\n23123231321231231231231231221232123121312321", "output": "87" }, { "input": "1 2 3 4\n1111111111111222222222233333333333444444444444444", "output": "126" }, { "input": "2 3 1 4\n121321232412342112312313213123123412131231231232", "output": "105" } ]

1,687,290,791

2,147,483,647

PyPy 3-64

OK

TESTS

49

62

2,048,000

a,b,c,d=map(int, input().split()) s=input() cnt=0 for i in range(len(s)): if s[i]=='1': cnt+=a elif s[i]=='2': cnt+=b elif s[i]=='3': cnt+=c elif s[i]=='4': cnt+=d print(cnt)

Title: Black Square Time Limit: None seconds Memory Limit: None megabytes Problem Description: Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone. In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip. You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares? Input Specification: The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=≤<=*a*1,<=*a*2,<=*a*3,<=*a*4<=≤<=104). The second line contains string *s* (1<=≤<=|*s*|<=≤<=105), where the *і*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip. Output Specification: Print a single integer — the total number of calories that Jury wastes. Demo Input: ['1 2 3 4\n123214\n', '1 5 3 2\n11221\n'] Demo Output: ['13\n', '13\n'] Note: none

```python a,b,c,d=map(int, input().split()) s=input() cnt=0 for i in range(len(s)): if s[i]=='1': cnt+=a elif s[i]=='2': cnt+=b elif s[i]=='3': cnt+=c elif s[i]=='4': cnt+=d print(cnt) ```

3

40

A

Find Color

PROGRAMMING

1,300

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

A. Find Color

2

256

Not so long ago as a result of combat operations the main Berland place of interest — the magic clock — was damaged. The cannon's balls made several holes in the clock, that's why the residents are concerned about the repair. The magic clock can be represented as an infinite Cartesian plane, where the origin corresponds to the clock center. The clock was painted two colors as is shown in the picture: The picture shows only the central part of the clock. This coloring naturally extends to infinity. The balls can be taken to be points on the plane. Your task is to find the color of the area, damaged by the given ball. All the points located on the border of one of the areas have to be considered painted black.

The first and single line contains two integers *x* and *y* — the coordinates of the hole made in the clock by the ball. Each of the numbers *x* and *y* has an absolute value that does not exceed 1000.

Find the required color. All the points between which and the origin of coordinates the distance is integral-value are painted black.

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

[ "white\n", "black\n", "black\n" ]

none

500

[ { "input": "-2 1", "output": "white" }, { "input": "2 1", "output": "black" }, { "input": "4 3", "output": "black" }, { "input": "3 3", "output": "black" }, { "input": "4 4", "output": "white" }, { "input": "-4 4", "output": "black" }, { "input": "4 -4", "output": "black" }, { "input": "-4 -4", "output": "white" }, { "input": "0 0", "output": "black" }, { "input": "0 1", "output": "black" }, { "input": "0 2", "output": "black" }, { "input": "0 1000", "output": "black" }, { "input": "1000 0", "output": "black" }, { "input": "-1000 0", "output": "black" }, { "input": "0 -1000", "output": "black" }, { "input": "1000 -1000", "output": "white" }, { "input": "12 5", "output": "black" }, { "input": "12 -5", "output": "black" }, { "input": "-12 -35", "output": "black" }, { "input": "20 -21", "output": "black" }, { "input": "-677 492", "output": "white" }, { "input": "-673 -270", "output": "white" }, { "input": "-668 970", "output": "black" }, { "input": "-220 208", "output": "white" }, { "input": "-215 -996", "output": "black" }, { "input": "-211 243", "output": "black" }, { "input": "-206 -518", "output": "white" }, { "input": "-201 278", "output": "black" }, { "input": "-196 -484", "output": "black" }, { "input": "902 479", "output": "white" }, { "input": "-441 572", "output": "white" }, { "input": "217 221", "output": "white" }, { "input": "875 -129", "output": "white" }, { "input": "-469 -36", "output": "black" }, { "input": "189 -387", "output": "white" }, { "input": "847 -294", "output": "white" }, { "input": "-496 -644", "output": "black" }, { "input": "-281 -552", "output": "white" }, { "input": "377 -902", "output": "black" }, { "input": "165 -738", "output": "white" }, { "input": "61 -175", "output": "black" }, { "input": "-42 389", "output": "black" }, { "input": "-589 952", "output": "black" }, { "input": "-693 -929", "output": "white" }, { "input": "-796 -365", "output": "white" }, { "input": "658 198", "output": "white" }, { "input": "555 319", "output": "black" }, { "input": "8 882", "output": "black" }, { "input": "-96 -556", "output": "black" }, { "input": "-129 489", "output": "black" }, { "input": "207 -224", "output": "black" }, { "input": "64 0", "output": "black" }, { "input": "17 144", "output": "black" }, { "input": "60 -448", "output": "black" }, { "input": "-399 -40", "output": "black" }, { "input": "128 -504", "output": "black" }, { "input": "0 72", "output": "black" }, { "input": "168 -26", "output": "black" }, { "input": "72 -154", "output": "black" }, { "input": "117 -44", "output": "black" }, { "input": "-72 -646", "output": "black" }, { "input": "253 -204", "output": "black" }, { "input": "-40 198", "output": "black" }, { "input": "-216 -90", "output": "black" }, { "input": "15 -8", "output": "black" }, { "input": "-180 -432", "output": "black" }, { "input": "280 342", "output": "black" }, { "input": "132 224", "output": "black" }, { "input": "-192 -256", "output": "black" }, { "input": "351 -280", "output": "black" } ]

1,653,695,714

2,147,483,647

Python 3

OK

TESTS

70

92

0

from sys import stdin from math import sqrt, floor x, y = stdin.readline().split() x = int(x) y = int(y) r2 = (x * x) + (y * y) r = floor(sqrt(r2)) if (r * r) < r2 and ((r + 1) * (r + 1)) > r2 and (r % 2 == 1 and x * y > 0 or r % 2 == 0 and x * y < 0): print("white") else: print("black");

Title: Find Color Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Not so long ago as a result of combat operations the main Berland place of interest — the magic clock — was damaged. The cannon's balls made several holes in the clock, that's why the residents are concerned about the repair. The magic clock can be represented as an infinite Cartesian plane, where the origin corresponds to the clock center. The clock was painted two colors as is shown in the picture: The picture shows only the central part of the clock. This coloring naturally extends to infinity. The balls can be taken to be points on the plane. Your task is to find the color of the area, damaged by the given ball. All the points located on the border of one of the areas have to be considered painted black. Input Specification: The first and single line contains two integers *x* and *y* — the coordinates of the hole made in the clock by the ball. Each of the numbers *x* and *y* has an absolute value that does not exceed 1000. Output Specification: Find the required color. All the points between which and the origin of coordinates the distance is integral-value are painted black. Demo Input: ['-2 1\n', '2 1\n', '4 3\n'] Demo Output: ['white\n', 'black\n', 'black\n'] Note: none

```python from sys import stdin from math import sqrt, floor x, y = stdin.readline().split() x = int(x) y = int(y) r2 = (x * x) + (y * y) r = floor(sqrt(r2)) if (r * r) < r2 and ((r + 1) * (r + 1)) > r2 and (r % 2 == 1 and x * y > 0 or r % 2 == 0 and x * y < 0): print("white") else: print("black"); ```

3.977

714

B

Filya and Homework

PROGRAMMING

1,200

[ "implementation", "sortings" ]

null

Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help. Filya is given an array of non-negative integers *a*1,<=*a*2,<=...,<=*a**n*. First, he pick an integer *x* and then he adds *x* to some elements of the array (no more than once), subtract *x* from some other elements (also, no more than once) and do no change other elements. He wants all elements of the array to be equal. Now he wonders if it's possible to pick such integer *x* and change some elements of the array using this *x* in order to make all elements equal.

The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of integers in the Filya's array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — elements of the array.

If it's impossible to make all elements of the array equal using the process given in the problem statement, then print "NO" (without quotes) in the only line of the output. Otherwise print "YES" (without quotes).

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

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

In the first sample Filya should select *x* = 1, then add it to the first and the last elements of the array and subtract from the second and the third elements.

1,000

[ { "input": "5\n1 3 3 2 1", "output": "YES" }, { "input": "5\n1 2 3 4 5", "output": "NO" }, { "input": "2\n1 2", "output": "YES" }, { "input": "3\n1 2 3", "output": "YES" }, { "input": "3\n1 1 1", "output": "YES" }, { "input": "2\n1 1000000000", "output": "YES" }, { "input": "4\n1 2 3 4", "output": "NO" }, { "input": "10\n1 1 1 1 1 2 2 2 2 2", "output": "YES" }, { "input": "2\n4 2", "output": "YES" }, { "input": "4\n1 1 4 7", "output": "YES" }, { "input": "3\n99999999 1 50000000", "output": "YES" }, { "input": "1\n0", "output": "YES" }, { "input": "5\n0 0 0 0 0", "output": "YES" }, { "input": "4\n4 2 2 1", "output": "NO" }, { "input": "3\n1 4 2", "output": "NO" }, { "input": "3\n1 4 100", "output": "NO" }, { "input": "3\n2 5 11", "output": "NO" }, { "input": "3\n1 4 6", "output": "NO" }, { "input": "3\n1 2 4", "output": "NO" }, { "input": "3\n1 2 7", "output": "NO" }, { "input": "5\n1 1 1 4 5", "output": "NO" }, { "input": "2\n100000001 100000003", "output": "YES" }, { "input": "3\n7 4 5", "output": "NO" }, { "input": "3\n2 3 5", "output": "NO" }, { "input": "3\n1 2 5", "output": "NO" }, { "input": "2\n2 3", "output": "YES" }, { "input": "3\n2 100 29", "output": "NO" }, { "input": "3\n0 1 5", "output": "NO" }, { "input": "3\n1 3 6", "output": "NO" }, { "input": "3\n2 1 3", "output": "YES" }, { "input": "3\n1 5 100", "output": "NO" }, { "input": "3\n1 4 8", "output": "NO" }, { "input": "3\n1 7 10", "output": "NO" }, { "input": "3\n5 4 1", "output": "NO" }, { "input": "3\n1 6 10", "output": "NO" }, { "input": "4\n1 3 4 5", "output": "NO" }, { "input": "3\n1 5 4", "output": "NO" }, { "input": "5\n1 2 3 3 5", "output": "NO" }, { "input": "3\n2 3 1", "output": "YES" }, { "input": "3\n2 3 8", "output": "NO" }, { "input": "3\n0 3 5", "output": "NO" }, { "input": "3\n1 5 10", "output": "NO" }, { "input": "3\n1 7 2", "output": "NO" }, { "input": "3\n1 3 9", "output": "NO" }, { "input": "3\n1 1 2", "output": "YES" }, { "input": "7\n1 1 1 1 1 2 4", "output": "NO" }, { "input": "5\n1 4 4 4 6", "output": "NO" }, { "input": "5\n1 2 2 4 4", "output": "NO" }, { "input": "3\n1 9 10", "output": "NO" }, { "input": "8\n1 1 1 1 1 1 2 3", "output": "YES" }, { "input": "3\n1 2 100", "output": "NO" }, { "input": "3\n1 3 4", "output": "NO" }, { "input": "3\n1 15 14", "output": "NO" }, { "input": "3\n1 3 8", "output": "NO" }, { "input": "3\n1 2 10", "output": "NO" }, { "input": "4\n2 2 4 5", "output": "NO" }, { "input": "3\n1 3 5", "output": "YES" }, { "input": "5\n3 6 7 8 9", "output": "NO" }, { "input": "3\n7 6 8", "output": "YES" }, { "input": "3\n3 2 1", "output": "YES" }, { "input": "5\n1 2 2 2 3", "output": "YES" }, { "input": "3\n4 6 7", "output": "NO" }, { "input": "3\n2 0 4", "output": "YES" }, { "input": "4\n10 20 21 30", "output": "NO" }, { "input": "4\n0 2 3 4", "output": "NO" }, { "input": "3\n3 6 12", "output": "NO" }, { "input": "5\n0 0 1 3 5", "output": "NO" }, { "input": "3\n3 5 8", "output": "NO" }, { "input": "3\n1 4 4", "output": "YES" }, { "input": "4\n2 4 5 6", "output": "NO" } ]

1,484,331,565

2,147,483,647

Python 3

OK

TESTS

79

93

13,824,000

a=int(input()) b=list(map(int,input().split())) c=set(b) d=sorted(list(c)) if len(c)>3: print('NO') elif len(c)==3: if (d[2]+d[0])==2*d[1]: print('YES') else: print('NO') else: print('YES')

Title: Filya and Homework Time Limit: None seconds Memory Limit: None megabytes Problem Description: Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help. Filya is given an array of non-negative integers *a*1,<=*a*2,<=...,<=*a**n*. First, he pick an integer *x* and then he adds *x* to some elements of the array (no more than once), subtract *x* from some other elements (also, no more than once) and do no change other elements. He wants all elements of the array to be equal. Now he wonders if it's possible to pick such integer *x* and change some elements of the array using this *x* in order to make all elements equal. Input Specification: The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of integers in the Filya's array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — elements of the array. Output Specification: If it's impossible to make all elements of the array equal using the process given in the problem statement, then print "NO" (without quotes) in the only line of the output. Otherwise print "YES" (without quotes). Demo Input: ['5\n1 3 3 2 1\n', '5\n1 2 3 4 5\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first sample Filya should select *x* = 1, then add it to the first and the last elements of the array and subtract from the second and the third elements.

```python a=int(input()) b=list(map(int,input().split())) c=set(b) d=sorted(list(c)) if len(c)>3: print('NO') elif len(c)==3: if (d[2]+d[0])==2*d[1]: print('YES') else: print('NO') else: print('YES') ```

3

337

A

Puzzles

PROGRAMMING

900

[ "greedy" ]

null

The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces). The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on. Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.

The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.

Print a single integer — the least possible difference the teacher can obtain.

[ "4 6\n10 12 10 7 5 22\n" ]

[ "5\n" ]

Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.

500

[ { "input": "4 6\n10 12 10 7 5 22", "output": "5" }, { "input": "2 2\n4 4", "output": "0" }, { "input": "2 10\n4 5 6 7 8 9 10 11 12 12", "output": "0" }, { "input": "4 5\n818 136 713 59 946", "output": "759" }, { "input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903", "output": "13" }, { "input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94", "output": "3" }, { "input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190", "output": "31" }, { "input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997", "output": "0" }, { "input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176", "output": "412" }, { "input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886", "output": "938" }, { "input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993", "output": "937" }, { "input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7", "output": "0" }, { "input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19", "output": "12" }, { "input": "40 50\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", "output": "0" }, { "input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17", "output": "31" }, { "input": "2 2\n1000 4", "output": "996" }, { "input": "2 3\n4 502 1000", "output": "498" }, { "input": "3 3\n4 1000 4", "output": "996" } ]

1,693,844,697

2,147,483,647

Python 3

OK

TESTS

18

62

0

n, m = map(int, input().split(" ")) f = list(map(int, input().split(" "))) f.sort() # Sort the list in ascending order min_diff = float('inf') # Initialize with a large value for right in range(n, m + 1): min_diff = min(min_diff, f[right - 1] - f[right - n]) print(min_diff)

Title: Puzzles Time Limit: None seconds Memory Limit: None megabytes Problem Description: The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces). The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on. Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*. Input Specification: The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop. Output Specification: Print a single integer — the least possible difference the teacher can obtain. Demo Input: ['4 6\n10 12 10 7 5 22\n'] Demo Output: ['5\n'] Note: Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.

```python n, m = map(int, input().split(" ")) f = list(map(int, input().split(" "))) f.sort() # Sort the list in ascending order min_diff = float('inf') # Initialize with a large value for right in range(n, m + 1): min_diff = min(min_diff, f[right - 1] - f[right - n]) print(min_diff) ```

3

900

A

Find Extra One

PROGRAMMING

800

[ "geometry", "implementation" ]

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" } ]

1,570,742,357

2,147,483,647

PyPy 3

OK

TESTS

71

967

9,830,400

n = int(input()) d = 0 e = 0 for i in range(n): x, y = map(int, input().split()) if x > 0: d += 1 elif x < 0: e += 1 YES = 'YES' NO = 'NO' if d == 0 or d == 1 or e == 0 or e == 1: print(YES) else: print(NO)

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()) d = 0 e = 0 for i in range(n): x, y = map(int, input().split()) if x > 0: d += 1 elif x < 0: e += 1 YES = 'YES' NO = 'NO' if d == 0 or d == 1 or e == 0 or e == 1: print(YES) else: print(NO) ```

3

361

A

Levko and Table

PROGRAMMING

800

[ "constructive algorithms", "implementation" ]

null

Levko loves tables that consist of *n* rows and *n* columns very much. He especially loves beautiful tables. A table is beautiful to Levko if the sum of elements in each row and column of the table equals *k*. Unfortunately, he doesn't know any such table. Your task is to help him to find at least one of them.

The single line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1000).

Print any beautiful table. Levko doesn't like too big numbers, so all elements of the table mustn't exceed 1000 in their absolute value. If there are multiple suitable tables, you are allowed to print any of them.

[ "2 4\n", "4 7\n" ]

[ "1 3\n3 1\n", "2 1 0 4\n4 0 2 1\n1 3 3 0\n0 3 2 2\n" ]

In the first sample the sum in the first row is 1 + 3 = 4, in the second row — 3 + 1 = 4, in the first column — 1 + 3 = 4 and in the second column — 3 + 1 = 4. There are other beautiful tables for this sample. In the second sample the sum of elements in each row and each column equals 7. Besides, there are other tables that meet the statement requirements.

500

[ { "input": "2 4", "output": "4 0 \n0 4 " }, { "input": "4 7", "output": "7 0 0 0 \n0 7 0 0 \n0 0 7 0 \n0 0 0 7 " }, { "input": "1 8", "output": "8 " }, { "input": "9 3", "output": "3 0 0 0 0 0 0 0 0 \n0 3 0 0 0 0 0 0 0 \n0 0 3 0 0 0 0 0 0 \n0 0 0 3 0 0 0 0 0 \n0 0 0 0 3 0 0 0 0 \n0 0 0 0 0 3 0 0 0 \n0 0 0 0 0 0 3 0 0 \n0 0 0 0 0 0 0 3 0 \n0 0 0 0 0 0 0 0 3 " }, { "input": "31 581", "output": "581 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 \n0 581 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 \n0 0 581 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 \n0 0 0 581 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 \n0 0 0 0 581 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 \n0 0 0 0 0 581 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 \n0 0 0 0 0 0 581 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 \n0 0 0 0 0 0 0 581 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0..." }, { "input": "100 1000", "output": "1000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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}, { "input": "3 2", "output": "2 0 0 \n0 2 0 \n0 0 2 " } ]

1,622,198,053

553

PyPy 3

OK

TESTS

22

233

3,276,800

n,k=map(int,input().split()) numberofones=n-1 req=k-(n-1) cnt=n-1 for i in range(n): for j in range(n): if j==cnt: print(req,end=" ") else: print(1,end=" ") cnt-=1 print()

Title: Levko and Table Time Limit: None seconds Memory Limit: None megabytes Problem Description: Levko loves tables that consist of *n* rows and *n* columns very much. He especially loves beautiful tables. A table is beautiful to Levko if the sum of elements in each row and column of the table equals *k*. Unfortunately, he doesn't know any such table. Your task is to help him to find at least one of them. Input Specification: The single line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1000). Output Specification: Print any beautiful table. Levko doesn't like too big numbers, so all elements of the table mustn't exceed 1000 in their absolute value. If there are multiple suitable tables, you are allowed to print any of them. Demo Input: ['2 4\n', '4 7\n'] Demo Output: ['1 3\n3 1\n', '2 1 0 4\n4 0 2 1\n1 3 3 0\n0 3 2 2\n'] Note: In the first sample the sum in the first row is 1 + 3 = 4, in the second row — 3 + 1 = 4, in the first column — 1 + 3 = 4 and in the second column — 3 + 1 = 4. There are other beautiful tables for this sample. In the second sample the sum of elements in each row and each column equals 7. Besides, there are other tables that meet the statement requirements.

```python n,k=map(int,input().split()) numberofones=n-1 req=k-(n-1) cnt=n-1 for i in range(n): for j in range(n): if j==cnt: print(req,end=" ") else: print(1,end=" ") cnt-=1 print() ```

3

236

A

Boy or Girl

PROGRAMMING

800

[ "brute force", "implementation", "strings" ]

null

Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network. But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names. This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.

The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.

If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).

[ "wjmzbmr\n", "xiaodao\n", "sevenkplus\n" ]

[ "CHAT WITH HER!\n", "IGNORE HIM!\n", "CHAT WITH HER!\n" ]

For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".

500

[ { "input": "wjmzbmr", "output": "CHAT WITH HER!" }, { "input": "xiaodao", "output": "IGNORE HIM!" }, { "input": "sevenkplus", "output": "CHAT WITH HER!" }, { "input": "pezu", "output": "CHAT WITH HER!" }, { "input": "wnemlgppy", "output": "CHAT WITH HER!" }, { "input": "zcinitufxoldnokacdvtmdohsfdjepyfioyvclhmujiqwvmudbfjzxjfqqxjmoiyxrfsbvseawwoyynn", "output": "IGNORE HIM!" }, { "input": "qsxxuoynwtebujwpxwpajitiwxaxwgbcylxneqiebzfphugwkftpaikixmumkhfbjiswmvzbtiyifbx", "output": "CHAT WITH HER!" }, { "input": "qwbdfzfylckctudyjlyrtmvbidfatdoqfmrfshsqqmhzohhsczscvwzpwyoyswhktjlykumhvaounpzwpxcspxwlgt", "output": "IGNORE HIM!" }, { "input": "nuezoadauueermoeaabjrkxttkatspjsjegjcjcdmcxgodowzbwuqncfbeqlhkk", "output": "IGNORE HIM!" }, { "input": "lggvdmulrsvtuagoavstuyufhypdxfomjlzpnduulukszqnnwfvxbvxyzmleocmofwclmzz", "output": "IGNORE HIM!" }, { "input": "tgcdptnkc", "output": "IGNORE HIM!" }, { "input": "wvfgnfrzabgibzxhzsojskmnlmrokydjoexnvi", "output": "IGNORE HIM!" }, { "input": "sxtburpzskucowowebgrbovhadrrayamuwypmmxhscrujkmcgvyinp", "output": "IGNORE HIM!" }, { "input": "pjqxhvxkyeqqvyuujxhmbspatvrckhhkfloottuybjivkkhpyivcighxumavrxzxslfpggnwbtalmhysyfllznphzia", "output": "IGNORE HIM!" }, { "input": "fpellxwskyekoyvrfnuf", "output": "CHAT WITH HER!" }, { "input": "xninyvkuvakfbs", "output": "IGNORE HIM!" }, { "input": "vnxhrweyvhqufpfywdwftoyrfgrhxuamqhblkvdpxmgvphcbeeqbqssresjifwyzgfhurmamhkwupymuomak", "output": "CHAT WITH HER!" }, { "input": "kmsk", "output": "IGNORE HIM!" }, { "input": "lqonogasrkzhryjxppjyriyfxmdfubieglthyswz", "output": "CHAT WITH HER!" }, { "input": "ndormkufcrkxlihdhmcehzoimcfhqsmombnfjrlcalffq", "output": "CHAT WITH HER!" }, { "input": "zqzlnnuwcfufwujygtczfakhcpqbtxtejrbgoodychepzdphdahtxyfpmlrycyicqthsgm", "output": "IGNORE HIM!" }, { "input": "ppcpbnhwoizajrl", "output": "IGNORE HIM!" }, { "input": "sgubujztzwkzvztitssxxxwzanfmddfqvv", "output": "CHAT WITH HER!" }, { "input": "ptkyaxycecpbrjnvxcjtbqiocqcswnmicxbvhdsptbxyxswbw", "output": "IGNORE HIM!" }, { "input": "yhbtzfppwcycxqjpqdfmjnhwaogyuaxamwxpnrdrnqsgdyfvxu", "output": "CHAT WITH HER!" }, { "input": "ojjvpnkrxibyevxk", "output": "CHAT WITH HER!" }, { "input": "wjweqcrqfuollfvfbiyriijovweg", "output": "IGNORE HIM!" }, { "input": "hkdbykboclchfdsuovvpknwqr", "output": "IGNORE HIM!" }, { "input": "stjvyfrfowopwfjdveduedqylerqugykyu", "output": "IGNORE HIM!" }, { "input": "rafcaanqytfclvfdegak", "output": "CHAT WITH HER!" }, { "input": "xczn", "output": "CHAT WITH HER!" }, { "input": "arcoaeozyeawbveoxpmafxxzdjldsielp", "output": "IGNORE HIM!" }, { "input": "smdfafbyehdylhaleevhoggiurdgeleaxkeqdixyfztkuqsculgslheqfafxyghyuibdgiuwrdxfcitojxika", "output": "CHAT WITH HER!" }, { "input": "vbpfgjqnhfazmvtkpjrdasfhsuxnpiepxfrzvoh", "output": "CHAT WITH HER!" }, { "input": "dbdokywnpqnotfrhdbrzmuyoxfdtrgrzcccninbtmoqvxfatcqg", "output": "CHAT WITH HER!" }, { "input": "udlpagtpq", "output": "CHAT WITH HER!" }, { "input": "zjurevbytijifnpfuyswfchdzelxheboruwjqijxcucylysmwtiqsqqhktexcynquvcwhbjsipy", "output": "CHAT WITH HER!" }, { "input": "qagzrqjomdwhagkhrjahhxkieijyten", "output": "CHAT WITH HER!" }, { "input": "achhcfjnnfwgoufxamcqrsontgjjhgyfzuhklkmiwybnrlsvblnsrjqdytglipxsulpnphpjpoewvlusalsgovwnsngb", "output": "CHAT WITH HER!" }, { "input": "qbkjsdwpahdbbohggbclfcufqelnojoehsxxkr", "output": "CHAT WITH HER!" }, { "input": "cpvftiwgyvnlmbkadiafddpgfpvhqqvuehkypqjsoibpiudfvpkhzlfrykc", "output": "IGNORE HIM!" }, { "input": "lnpdosnceumubvk", "output": "IGNORE HIM!" }, { "input": "efrk", "output": "CHAT WITH HER!" }, { "input": "temnownneghnrujforif", "output": "IGNORE HIM!" }, { "input": "ottnneymszwbumgobazfjyxewkjakglbfflsajuzescplpcxqta", "output": "IGNORE HIM!" }, { "input": "eswpaclodzcwhgixhpyzvhdwsgneqidanbzdzszquefh", "output": "IGNORE HIM!" }, { "input": "gwntwbpj", "output": "IGNORE HIM!" }, { "input": "wuqvlbblkddeindiiswsinkfrnkxghhwunzmmvyovpqapdfbolyim", "output": "IGNORE HIM!" }, { "input": "swdqsnzmzmsyvktukaoyqsqzgfmbzhezbfaqeywgwizrwjyzquaahucjchegknqaioliqd", "output": "CHAT WITH HER!" }, { "input": "vlhrpzezawyolhbmvxbwhtjustdbqggexmzxyieihjlelvwjosmkwesfjmramsikhkupzvfgezmrqzudjcalpjacmhykhgfhrjx", "output": "IGNORE HIM!" }, { "input": "lxxwbkrjgnqjwsnflfnsdyxihmlspgivirazsbveztnkuzpaxtygidniflyjheejelnjyjvgkgvdqks", "output": "CHAT WITH HER!" }, { "input": "wpxbxzfhtdecetpljcrvpjjnllosdqirnkzesiqeukbedkayqx", "output": "CHAT WITH HER!" }, { "input": "vmzxgacicvweclaodrunmjnfwtimceetsaoickarqyrkdghcmyjgmtgsqastcktyrjgvjqimdc", "output": "CHAT WITH HER!" }, { "input": "yzlzmesxdttfcztooypjztlgxwcr", "output": "IGNORE HIM!" }, { "input": "qpbjwzwgdzmeluheirjrvzrhbmagfsjdgvzgwumjtjzecsfkrfqjasssrhhtgdqqfydlmrktlgfc", "output": "IGNORE HIM!" }, { "input": "aqzftsvezdgouyrirsxpbuvdjupnzvbhguyayeqozfzymfnepvwgblqzvmxxkxcilmsjvcgyqykpoaktjvsxbygfgsalbjoq", "output": "CHAT WITH HER!" }, { "input": "znicjjgijhrbdlnwmtjgtdgziollrfxroabfhadygnomodaembllreorlyhnehijfyjbfxucazellblegyfrzuraogadj", "output": "IGNORE HIM!" }, { "input": "qordzrdiknsympdrkgapjxokbldorpnmnpucmwakklmqenpmkom", "output": "CHAT WITH HER!" }, { "input": "wqfldgihuxfktzanyycluzhtewmwvnawqlfoavuguhygqrrxtstxwouuzzsryjqtfqo", "output": "CHAT WITH HER!" }, { "input": "vujtrrpshinkskgyknlcfckmqdrwtklkzlyipmetjvaqxdsslkskschbalmdhzsdrrjmxdltbtnxbh", "output": "IGNORE HIM!" }, { "input": "zioixjibuhrzyrbzqcdjbbhhdmpgmqykixcxoqupggaqajuzonrpzihbsogjfsrrypbiphehonyhohsbybnnukqebopppa", "output": "CHAT WITH HER!" }, { "input": "oh", "output": "CHAT WITH HER!" }, { "input": "kxqthadqesbpgpsvpbcbznxpecqrzjoilpauttzlnxvaczcqwuri", "output": "IGNORE HIM!" }, { "input": "zwlunigqnhrwirkvufqwrnwcnkqqonebrwzcshcbqqwkjxhymjjeakuzjettebciadjlkbfp", "output": "CHAT WITH HER!" }, { "input": "fjuldpuejgmggvvigkwdyzytfxzwdlofrpifqpdnhfyroginqaufwgjcbgshyyruwhofctsdaisqpjxqjmtpp", "output": "CHAT WITH HER!" }, { "input": "xiwntnheuitbtqxrmzvxmieldudakogealwrpygbxsbluhsqhtwmdlpjwzyafckrqrdduonkgo", "output": "CHAT WITH HER!" }, { "input": "mnmbupgo", "output": "IGNORE HIM!" }, { "input": "mcjehdiygkbmrbfjqwpwxidbdfelifwhstaxdapigbymmsgrhnzsdjhsqchl", "output": "IGNORE HIM!" }, { "input": "yocxrzspinchmhtmqo", "output": "CHAT WITH HER!" }, { "input": "vasvvnpymtgjirnzuynluluvmgpquskuaafwogeztfnvybblajvuuvfomtifeuzpikjrolzeeoftv", "output": "CHAT WITH HER!" }, { "input": "ecsdicrznvglwggrdbrvehwzaenzjutjydhvimtqegweurpxtjkmpcznshtrvotkvrghxhacjkedidqqzrduzad", "output": "IGNORE HIM!" }, { "input": "ubvhyaebyxoghakajqrpqpctwbrfqzli", "output": "CHAT WITH HER!" }, { "input": "gogbxfeqylxoummvgxpkoqzsmobasesxbqjjktqbwqxeiaagnnhbvepbpy", "output": "IGNORE HIM!" }, { "input": "nheihhxkbbrmlpxpxbhnpofcjmxemyvqqdbanwd", "output": "IGNORE HIM!" }, { "input": "acrzbavz", "output": "CHAT WITH HER!" }, { "input": "drvzznznvrzskftnrhvvzxcalwutxmdza", "output": "IGNORE HIM!" }, { "input": "oacwxipdfcoabhkwxqdbtowiekpnflnqhlrkustgzryvws", "output": "CHAT WITH HER!" }, { "input": "tpnwfmfsibnccvdwjvzviyvjfljupinfigfunyff", "output": "CHAT WITH HER!" }, { "input": "gavaihhamfolcndgytcsgucqdqngxkrlovpthvteacmmthoglxu", "output": "CHAT WITH HER!" }, { "input": "hsfcfvameeupldgvchmogrvwxrvsmnwxxkxoawwodtsahqvehlcw", "output": "IGNORE HIM!" }, { "input": "sbkydrscoojychxchqsuciperfroumenelgiyiwlqfwximrgdbyvkmacy", "output": "CHAT WITH HER!" }, { "input": "rhh", "output": "CHAT WITH HER!" }, { "input": "zhdouqfmlkenjzdijxdfxnlegxeuvhelo", "output": "IGNORE HIM!" }, { "input": "yufkkfwyhhvcjntsgsvpzbhqtmtgyxifqoewmuplphykmptfdebjxuaxigomjtwgtljwdjhjernkitifbomifbhysnmadtnyn", "output": "CHAT WITH HER!" }, { "input": "urigreuzpxnej", "output": "CHAT WITH HER!" } ]

1,696,543,878

2,147,483,647

Python 3

OK

TESTS

85

92

0

username = list(input()) username = list(set(username)) if len(username)%2 == 1: print("IGNORE HIM!") else: print("CHAT WITH HER!")

Title: Boy or Girl Time Limit: None seconds Memory Limit: None megabytes Problem Description: Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network. But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names. This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method. Input Specification: The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters. Output Specification: If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes). Demo Input: ['wjmzbmr\n', 'xiaodao\n', 'sevenkplus\n'] Demo Output: ['CHAT WITH HER!\n', 'IGNORE HIM!\n', 'CHAT WITH HER!\n'] Note: For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".

```python username = list(input()) username = list(set(username)) if len(username)%2 == 1: print("IGNORE HIM!") else: print("CHAT WITH HER!") ```

3

580

A

Kefa and First Steps

PROGRAMMING

900

[ "brute force", "dp", "implementation" ]

null

Kefa decided to make some money doing business on the Internet for exactly *n* days. He knows that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) he makes *a**i* money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence *a**i*. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order. Help Kefa cope with this task!

The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (1<=≤<=*a**i*<=≤<=109).

Print a single integer — the length of the maximum non-decreasing subsegment of sequence *a*.

[ "6\n2 2 1 3 4 1\n", "3\n2 2 9\n" ]

[ "3", "3" ]

In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one. In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.

750

[ { "input": "6\n2 2 1 3 4 1", "output": "3" }, { "input": "3\n2 2 9", "output": "3" }, { "input": "5\n10 100 111 1 2", "output": "3" }, { "input": "10\n1 2 3 4 1 2 3 4 5 6", "output": "6" }, { "input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "50" }, { "input": "100\n1 838 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 605 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 27 533 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 835 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 992", "output": "42" }, { "input": "1\n1", "output": "1" }, { "input": "2\n1 1", "output": "2" }, { "input": "6\n5 4 3 2 1 2", "output": "2" }, { "input": "9\n1 2 3 4 5 6 7 8 9", "output": "9" }, { "input": "7\n99 100 1 2 3 4 5", "output": "5" }, { "input": "5\n3 3 1 2 3", "output": "3" }, { "input": "1\n100", "output": "1" }, { "input": "1\n5", "output": "1" } ]

1,699,968,819

2,147,483,647

PyPy 3

OK

TESTS

28

155

10,956,800

n = int(input()) niz = [int(x) for x in input().split()] def najduzi(niz): najveci = 0 ukupno = 0 prvi = niz[0] for i in range(1, len(niz)): if prvi<=niz[i]: najveci+=1 else: najveci=0 ukupno=max(ukupno,najveci) prvi= niz[i] return ukupno+1 print(najduzi(niz))

Title: Kefa and First Steps Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kefa decided to make some money doing business on the Internet for exactly *n* days. He knows that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) he makes *a**i* money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence *a**i*. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order. Help Kefa cope with this task! Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (1<=≤<=*a**i*<=≤<=109). Output Specification: Print a single integer — the length of the maximum non-decreasing subsegment of sequence *a*. Demo Input: ['6\n2 2 1 3 4 1\n', '3\n2 2 9\n'] Demo Output: ['3', '3'] Note: In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one. In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.

```python n = int(input()) niz = [int(x) for x in input().split()] def najduzi(niz): najveci = 0 ukupno = 0 prvi = niz[0] for i in range(1, len(niz)): if prvi<=niz[i]: najveci+=1 else: najveci=0 ukupno=max(ukupno,najveci) prvi= niz[i] return ukupno+1 print(najduzi(niz)) ```

3

1,003

A

Polycarp's Pockets

PROGRAMMING

800

[ "implementation" ]

null

Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket. For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$. Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that.

The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins. The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins.

Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket.

[ "6\n1 2 4 3 3 2\n", "1\n100\n" ]

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

none

0

[ { "input": "6\n1 2 4 3 3 2", "output": "2" }, { "input": "1\n100", "output": "1" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "100" }, { "input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "100" }, { "input": "100\n59 47 39 47 47 71 47 28 58 47 35 79 58 47 38 47 47 47 47 27 47 43 29 95 47 49 46 71 47 74 79 47 47 32 45 67 47 47 30 37 47 47 16 67 22 76 47 86 84 10 5 47 47 47 47 47 1 51 47 54 47 8 47 47 9 47 47 47 47 28 47 47 26 47 47 47 47 47 47 92 47 47 77 47 47 24 45 47 10 47 47 89 47 27 47 89 47 67 24 71", "output": "51" }, { "input": "100\n45 99 10 27 16 85 39 38 17 32 15 23 67 48 50 97 42 70 62 30 44 81 64 73 34 22 46 5 83 52 58 60 33 74 47 88 18 61 78 53 25 95 94 31 3 75 1 57 20 54 59 9 68 7 77 43 21 87 86 24 4 80 11 49 2 72 36 84 71 8 65 55 79 100 41 14 35 89 66 69 93 37 56 82 90 91 51 19 26 92 6 96 13 98 12 28 76 40 63 29", "output": "1" }, { "input": "100\n45 29 5 2 6 50 22 36 14 15 9 48 46 20 8 37 7 47 12 50 21 38 18 27 33 19 40 10 5 49 38 42 34 37 27 30 35 24 10 3 40 49 41 3 4 44 13 25 28 31 46 36 23 1 1 23 7 22 35 26 21 16 48 42 32 8 11 16 34 11 39 32 47 28 43 41 39 4 14 19 26 45 13 18 15 25 2 44 17 29 17 33 43 6 12 30 9 20 31 24", "output": "2" }, { "input": "50\n7 7 3 3 7 4 5 6 4 3 7 5 6 4 5 4 4 5 6 7 7 7 4 5 5 5 3 7 6 3 4 6 3 6 4 4 5 4 6 6 3 5 6 3 5 3 3 7 7 6", "output": "10" }, { "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 99 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "99" }, { "input": "7\n1 2 3 3 3 1 2", "output": "3" }, { "input": "5\n1 2 3 4 5", "output": "1" }, { "input": "7\n1 2 3 4 5 6 7", "output": "1" }, { "input": "8\n1 2 3 4 5 6 7 8", "output": "1" }, { "input": "9\n1 2 3 4 5 6 7 8 9", "output": "1" }, { "input": "10\n1 2 3 4 5 6 7 8 9 10", "output": "1" }, { "input": "3\n2 1 1", "output": "2" }, { "input": "11\n1 2 3 4 5 6 7 8 9 1 1", "output": "3" }, { "input": "12\n1 2 1 1 1 1 1 1 1 1 1 1", "output": "11" }, { "input": "13\n1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "13" }, { "input": "14\n1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "14" }, { "input": "15\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "15" }, { "input": "16\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "16" }, { "input": "3\n1 1 1", "output": "3" }, { "input": "3\n1 2 3", "output": "1" }, { "input": "10\n1 1 1 1 2 2 1 1 9 10", "output": "6" }, { "input": "2\n1 1", "output": "2" }, { "input": "56\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", "output": "56" }, { "input": "99\n35 96 73 72 70 83 22 93 98 75 45 32 81 82 45 54 25 7 53 72 29 2 94 19 21 98 34 28 39 99 55 85 44 23 6 47 98 2 33 34 19 57 49 35 67 4 60 4 4 23 55 6 57 66 16 68 34 45 84 79 48 63 4 9 46 88 98 13 19 27 83 12 4 63 57 22 44 77 44 62 28 52 44 64 9 24 55 22 48 4 2 9 80 76 45 1 56 22 92", "output": "6" }, { "input": "10\n1 2 2 3 3 3 4 4 4 4", "output": "4" }, { "input": "99\n97 44 33 56 42 10 61 85 64 26 40 39 82 34 75 9 51 51 39 73 58 38 74 31 13 99 58 1 28 89 76 19 52 7 40 56 12 27 72 72 67 75 62 46 22 55 35 16 18 39 60 63 92 42 85 69 34 61 73 50 57 95 30 4 45 63 76 58 32 35 48 81 10 78 95 79 55 97 21 21 22 94 30 17 78 57 89 93 100 44 16 89 68 55 19 46 42 73 21", "output": "3" }, { "input": "5\n5 5 5 5 1", "output": "4" }, { "input": "6\n2 3 2 5 2 6", "output": "3" }, { "input": "3\n58 59 58", "output": "2" }, { "input": "9\n1 2 3 4 5 6 7 8 8", "output": "2" }, { "input": "97\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", "output": "97" }, { "input": "3\n95 95 4", "output": "2" }, { "input": "3\n2 2 5", "output": "2" } ]

1,694,004,011

2,147,483,647

Python 3

OK

TESTS

37

46

0

freq = [0 for _ in range(101)] n = int(input()) for x in [int(x) for x in input().split()]: freq[x] += 1 print(max(freq))

Title: Polycarp's Pockets Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket. For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$. Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that. Input Specification: The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins. The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins. Output Specification: Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket. Demo Input: ['6\n1 2 4 3 3 2\n', '1\n100\n'] Demo Output: ['2\n', '1\n'] Note: none

```python freq = [0 for _ in range(101)] n = int(input()) for x in [int(x) for x in input().split()]: freq[x] += 1 print(max(freq)) ```

3

75

A

Life Without Zeros

PROGRAMMING

1,000

[ "implementation" ]

A. Life Without Zeros

2

256

Can you imagine our life if we removed all zeros from it? For sure we will have many problems. In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros? For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation. But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.

The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.

The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.

[ "101\n102\n", "105\n106\n" ]

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

none

500

[ { "input": "101\n102", "output": "YES" }, { "input": "105\n106", "output": "NO" }, { "input": "544\n397", "output": "YES" }, { "input": "822\n280", "output": "NO" }, { "input": "101\n413", "output": "NO" }, { "input": "309\n139", "output": "NO" }, { "input": "693\n970", "output": "NO" }, { "input": "981\n1", "output": "YES" }, { "input": "352\n276", "output": "YES" }, { "input": "164\n691", "output": "YES" }, { "input": "110036\n43", "output": "YES" }, { "input": "100\n1000", "output": "NO" }, { "input": "1000000000\n1000000000", "output": "YES" }, { "input": "999999999\n999999999", "output": "YES" }, { "input": "6\n4", "output": "NO" }, { "input": "123456\n876543", "output": "YES" }, { "input": "1234567\n9876543", "output": "NO" }, { "input": "1111111\n1119111", "output": "NO" }, { "input": "10502405\n40403403", "output": "YES" }, { "input": "1\n1", "output": "YES" }, { "input": "5\n4", "output": "YES" }, { "input": "1\n6", "output": "YES" }, { "input": "40\n17", "output": "NO" }, { "input": "35\n81", "output": "YES" }, { "input": "347\n908", "output": "NO" }, { "input": "2081\n7450", "output": "NO" }, { "input": "84355694\n373629503", "output": "NO" }, { "input": "885992048\n510468672", "output": "NO" }, { "input": "540144752\n647307841", "output": "NO" }, { "input": "341781108\n784147010", "output": "NO" }, { "input": "960440938\n572334653", "output": "NO" }, { "input": "657414510\n768771903", "output": "NO" }, { "input": "501871728\n725074574", "output": "NO" }, { "input": "968\n851", "output": "YES" }, { "input": "943\n939", "output": "YES" }, { "input": "347\n908", "output": "NO" }, { "input": "611\n41", "output": "YES" }, { "input": "737\n289", "output": "NO" }, { "input": "168\n571", "output": "YES" }, { "input": "868\n463", "output": "YES" }, { "input": "815732007\n913882008", "output": "NO" }, { "input": "459039901\n453033850", "output": "NO" }, { "input": "975305404\n484601256", "output": "NO" } ]

1,582,465,394

2,147,483,647

PyPy 3

OK

TESTS

43

310

0

a = input() b = input() c = int(a) d = int(b) ans1 = c+d e = a.replace('0', '') f = b.replace('0', '') g = int(e) f = int(f) ans2 = g+f temp = str(ans1) s1 = temp.replace('0', '') s2 = str(ans2) if s1 == s2: print('YES') else: print('NO')

Title: Life Without Zeros Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Can you imagine our life if we removed all zeros from it? For sure we will have many problems. In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros? For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation. But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation. Input Specification: The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*. Output Specification: The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise. Demo Input: ['101\n102\n', '105\n106\n'] Demo Output: ['YES\n', 'NO\n'] Note: none

```python a = input() b = input() c = int(a) d = int(b) ans1 = c+d e = a.replace('0', '') f = b.replace('0', '') g = int(e) f = int(f) ans2 = g+f temp = str(ans1) s1 = temp.replace('0', '') s2 = str(ans2) if s1 == s2: print('YES') else: print('NO') ```

3.9225

199

A

Hexadecimal's theorem

PROGRAMMING

900

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

null

Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers. Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1. So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ... If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible.

The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number.

Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes. If there are multiple answers, print any of them.

[ "3\n", "13\n" ]

[ "1 1 1\n", "2 3 8\n" ]

none

500

[ { "input": "3", "output": "1 1 1" }, { "input": "13", "output": "2 3 8" }, { "input": "0", "output": "0 0 0" }, { "input": "1", "output": "1 0 0" }, { "input": "2", "output": "1 1 0" }, { "input": "1597", "output": "233 377 987" }, { "input": "0", "output": "0 0 0" }, { "input": "1", "output": "1 0 0" }, { "input": "1", "output": "1 0 0" }, { "input": "2", "output": "1 1 0" }, { "input": "3", "output": "1 1 1" }, { "input": "5", "output": "1 1 3" }, { "input": "8", "output": "1 2 5" }, { "input": "13", "output": "2 3 8" }, { "input": "21", "output": "3 5 13" }, { "input": "34", "output": "5 8 21" }, { "input": "55", "output": "8 13 34" }, { "input": "89", "output": "13 21 55" }, { "input": "144", "output": "21 34 89" }, { "input": "233", "output": "34 55 144" }, { "input": "377", "output": "55 89 233" }, { "input": "610", "output": "89 144 377" }, { "input": "987", "output": "144 233 610" }, { "input": "1597", "output": "233 377 987" }, { "input": "2584", "output": "377 610 1597" }, { "input": "4181", "output": "610 987 2584" }, { "input": "6765", "output": "987 1597 4181" }, { "input": "10946", "output": "1597 2584 6765" }, { "input": "17711", "output": "2584 4181 10946" }, { "input": "28657", "output": "4181 6765 17711" }, { "input": "46368", "output": "6765 10946 28657" }, { "input": "75025", "output": "10946 17711 46368" }, { "input": "121393", "output": "17711 28657 75025" }, { "input": "196418", "output": "28657 46368 121393" }, { "input": "317811", "output": "46368 75025 196418" }, { "input": "514229", "output": "75025 121393 317811" }, { "input": "832040", "output": "121393 196418 514229" }, { "input": "1346269", "output": "196418 317811 832040" }, { "input": "2178309", "output": "317811 514229 1346269" }, { "input": "3524578", "output": "514229 832040 2178309" }, { "input": "5702887", "output": "832040 1346269 3524578" }, { "input": "9227465", "output": "1346269 2178309 5702887" }, { "input": "14930352", "output": "2178309 3524578 9227465" }, { "input": "24157817", "output": "3524578 5702887 14930352" }, { "input": "39088169", "output": "5702887 9227465 24157817" }, { "input": "63245986", "output": "9227465 14930352 39088169" }, { "input": "102334155", "output": "14930352 24157817 63245986" }, { "input": "165580141", "output": "24157817 39088169 102334155" }, { "input": "267914296", "output": "39088169 63245986 165580141" }, { "input": "433494437", "output": "63245986 102334155 267914296" }, { "input": "701408733", "output": "102334155 165580141 433494437" }, { "input": "701408733", "output": "102334155 165580141 433494437" }, { "input": "102334155", "output": "14930352 24157817 63245986" }, { "input": "63245986", "output": "9227465 14930352 39088169" } ]

1,595,983,492

2,147,483,647

PyPy 3

OK

TESTS

54

310

20,172,800

m = {} fibo = [0]*47 fibo[1] = 1 m[0] = 0 for i in range(2,47): fibo[i] = fibo[i-2] + fibo[i-1] m[fibo[i]] = i n = int(input()) if n == 0: print(0,0,0) exit(0) print(fibo[m[n]-1],fibo[m[n]-2],0)

Title: Hexadecimal's theorem Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers. Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1. So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ... If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible. Input Specification: The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number. Output Specification: Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes. If there are multiple answers, print any of them. Demo Input: ['3\n', '13\n'] Demo Output: ['1 1 1\n', '2 3 8\n'] Note: none

```python m = {} fibo = [0]*47 fibo[1] = 1 m[0] = 0 for i in range(2,47): fibo[i] = fibo[i-2] + fibo[i-1] m[fibo[i]] = i n = int(input()) if n == 0: print(0,0,0) exit(0) print(fibo[m[n]-1],fibo[m[n]-2],0) ```

3