MatrixStudio/Codeforces-Python-Submissions

727

A

Transformation: from A to B

PROGRAMMING

1,000

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

null

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

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

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

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

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

none

1,000

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

1,644,278,298

498

Python 3

RUNTIME_ERROR

TESTS

0

31

0

""" import sys import math sys.stdin = open('inputf.in', 'r') sys.stdout = open('outputf.in', 'w') """ a, b = [int(x) for x in input().split(' ')] #a = [int(x) for x in input().split()] res = [] trig = 1 while b > a: res.append(b) if b%2 ==0: b = b//2 elif ((b-1)%10) == 0: b = b//10 else: trig = 0 break if b != a: trig = 0 else: res.append(a) if not ok: print("N0") else: print('YES') print(len(res)) print(list(reversed(res)))

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

```python """ import sys import math sys.stdin = open('inputf.in', 'r') sys.stdout = open('outputf.in', 'w') """ a, b = [int(x) for x in input().split(' ')] #a = [int(x) for x in input().split()] res = [] trig = 1 while b > a: res.append(b) if b%2 ==0: b = b//2 elif ((b-1)%10) == 0: b = b//10 else: trig = 0 break if b != a: trig = 0 else: res.append(a) if not ok: print("N0") else: print('YES') print(len(res)) print(list(reversed(res))) ```

-1

327

A

Flipping Game

PROGRAMMING

1,200

[ "brute force", "dp", "implementation" ]

null

Iahub got bored, so he invented a game to be played on paper. He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*. The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.

The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1.

Print an integer — the maximal number of 1s that can be obtained after exactly one move.

[ "5\n1 0 0 1 0\n", "4\n1 0 0 1\n" ]

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

In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1]. In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1.

500

[ { "input": "5\n1 0 0 1 0", "output": "4" }, { "input": "4\n1 0 0 1", "output": "4" }, { "input": "1\n1", "output": "0" }, { "input": "1\n0", "output": "1" }, { "input": "8\n1 0 0 0 1 0 0 0", "output": "7" }, { "input": "18\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "18" }, { "input": "23\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "22" }, { "input": "100\n0 1 0 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 0 1 1 1 0 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1", "output": "70" }, { "input": "100\n0 1 1 0 1 0 0 1 0 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 0 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 0 0 1 0 0 0 0 1 1 1 1", "output": "60" }, { "input": "18\n0 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 1 0", "output": "11" }, { "input": "25\n0 1 0 0 0 0 0 1 0 1 0 1 0 0 0 0 1 1 1 0 0 1 1 0 1", "output": "18" }, { "input": "55\n0 0 1 1 0 0 0 1 0 1 1 0 1 1 1 0 1 1 1 1 1 0 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 1 1 1 0 1 1 0 0 0 0 1 1 0 1 1 1 1", "output": "36" }, { "input": "75\n1 1 0 1 0 1 1 0 0 0 0 0 1 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 1 0 1 0 0 0 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 0 1 1 0 0 0 1 0 0 0", "output": "44" }, { "input": "100\n0 0 1 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 0 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 1 1 1 0 0 0 0 1 0 1 1 1 0 0 1 0 1 1 1 1 1 1 1 0 1 0 1 0 0 1 0 1 1 1 0 0 0 0 1 0 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 0 1 0 1", "output": "61" }, { "input": "100\n0 0 0 1 0 0 0 1 0 1 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 1 1 0 0 0 1 1 1 1 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 1 1 0 1 1 0", "output": "61" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "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": "99" }, { "input": "100\n0 0 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 1 1 1 1 1 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 0 1 1 1 0 1 1 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 1 1 0 0 0 0 1 0", "output": "61" }, { "input": "100\n0 1 1 1 1 1 1 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 0 1 0 0 0 0 1 0 1 0 1 0 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 1", "output": "59" }, { "input": "99\n1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 0 1 1 0 0 1 0 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 1 1", "output": "61" }, { "input": "2\n1 1", "output": "1" } ]

1,676,812,721

2,147,483,647

Python 3

OK

TESTS

21

92

0

## n = int(input()) l = list(map(int, input().split())) m = [] for i in range(n): for j in range(i, n): m.append(l[i:j+1].count(0) + (l[:i] + l[j+1:]).count(1)) print(max(m))

Title: Flipping Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Iahub got bored, so he invented a game to be played on paper. He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*. The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub. Input Specification: The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1. Output Specification: Print an integer — the maximal number of 1s that can be obtained after exactly one move. Demo Input: ['5\n1 0 0 1 0\n', '4\n1 0 0 1\n'] Demo Output: ['4\n', '4\n'] Note: In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1]. In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1.

```python ## n = int(input()) l = list(map(int, input().split())) m = [] for i in range(n): for j in range(i, n): m.append(l[i:j+1].count(0) + (l[:i] + l[j+1:]).count(1)) print(max(m)) ```

3

287

B

Pipeline

PROGRAMMING

1,700

[ "binary search", "math" ]

null

Vova, the Ultimate Thule new shaman, wants to build a pipeline. As there are exactly *n* houses in Ultimate Thule, Vova wants the city to have exactly *n* pipes, each such pipe should be connected to the water supply. A pipe can be connected to the water supply if there's water flowing out of it. Initially Vova has only one pipe with flowing water. Besides, Vova has several splitters. A splitter is a construction that consists of one input (it can be connected to a water pipe) and *x* output pipes. When a splitter is connected to a water pipe, water flows from each output pipe. You can assume that the output pipes are ordinary pipes. For example, you can connect water supply to such pipe if there's water flowing out from it. At most one splitter can be connected to any water pipe. Vova has one splitter of each kind: with 2, 3, 4, ..., *k* outputs. Help Vova use the minimum number of splitters to build the required pipeline or otherwise state that it's impossible. Vova needs the pipeline to have exactly *n* pipes with flowing out water. Note that some of those pipes can be the output pipes of the splitters.

The first line contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=1018, 2<=≤<=*k*<=≤<=109). Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Print a single integer — the minimum number of splitters needed to build the pipeline. If it is impossible to build a pipeline with the given splitters, print -1.

[ "4 3\n", "5 5\n", "8 4\n" ]

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

none

1,500

[ { "input": "4 3", "output": "2" }, { "input": "5 5", "output": "1" }, { "input": "8 4", "output": "-1" }, { "input": "1000000000000000000 1000000000", "output": "-1" }, { "input": "499999998500000001 1000000000", "output": "999955279" }, { "input": "499999998500000000 1000000000", "output": "999955279" }, { "input": "499999999500000000 1000000000", "output": "999999998" }, { "input": "499999999500000001 1000000000", "output": "999999999" }, { "input": "525 34", "output": "25" }, { "input": "223265034477 726990", "output": "440662" }, { "input": "15597035789572051 185473109", "output": "128849771" }, { "input": "499999999500000002 1000000000", "output": "-1" }, { "input": "1 1000000000", "output": "0" }, { "input": "1000000000 2", "output": "-1" }, { "input": "462498979 204468265", "output": "3" }, { "input": "2107921 542531", "output": "4" }, { "input": "131 49", "output": "3" }, { "input": "20171878992939541 200857557", "output": "200853401" }, { "input": "399812655947 894219", "output": "893030" }, { "input": "93 17", "output": "8" }, { "input": "1000000000 999999999", "output": "2" }, { "input": "100000000000000000 1000000000", "output": "105572810" } ]

1,688,697,179

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

3

400

0

def main(): n, k = map(int, input().split()) print(solve(n, k)) def solve(n, k): low = 0 high = k minSplitters = -1 while low <= high: splittersTrial = low + ((high - low) // 2) # print("Splitters Trying:", splittersTrial) if is_valid(splittersTrial, n, k): minSplitters = splittersTrial high = splittersTrial - 1 # print("--Possible--") else: low = splittersTrial + 1 # print("--Not Possible--") return minSplitters def is_valid(x, totalPipes, maxSize): high = maxSize low = maxSize while x > 0: sum = ((high - low + 1) * (low + high)) // 2 pipeFlow = sum - (high - low) if pipeFlow >= totalPipes: return True low -= 1 x -= 1 return False if __name__ == "__main__": main()

Title: Pipeline Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vova, the Ultimate Thule new shaman, wants to build a pipeline. As there are exactly *n* houses in Ultimate Thule, Vova wants the city to have exactly *n* pipes, each such pipe should be connected to the water supply. A pipe can be connected to the water supply if there's water flowing out of it. Initially Vova has only one pipe with flowing water. Besides, Vova has several splitters. A splitter is a construction that consists of one input (it can be connected to a water pipe) and *x* output pipes. When a splitter is connected to a water pipe, water flows from each output pipe. You can assume that the output pipes are ordinary pipes. For example, you can connect water supply to such pipe if there's water flowing out from it. At most one splitter can be connected to any water pipe. Vova has one splitter of each kind: with 2, 3, 4, ..., *k* outputs. Help Vova use the minimum number of splitters to build the required pipeline or otherwise state that it's impossible. Vova needs the pipeline to have exactly *n* pipes with flowing out water. Note that some of those pipes can be the output pipes of the splitters. Input Specification: The first line contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=1018, 2<=≤<=*k*<=≤<=109). Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. Output Specification: Print a single integer — the minimum number of splitters needed to build the pipeline. If it is impossible to build a pipeline with the given splitters, print -1. Demo Input: ['4 3\n', '5 5\n', '8 4\n'] Demo Output: ['2\n', '1\n', '-1\n'] Note: none

```python def main(): n, k = map(int, input().split()) print(solve(n, k)) def solve(n, k): low = 0 high = k minSplitters = -1 while low <= high: splittersTrial = low + ((high - low) // 2) # print("Splitters Trying:", splittersTrial) if is_valid(splittersTrial, n, k): minSplitters = splittersTrial high = splittersTrial - 1 # print("--Possible--") else: low = splittersTrial + 1 # print("--Not Possible--") return minSplitters def is_valid(x, totalPipes, maxSize): high = maxSize low = maxSize while x > 0: sum = ((high - low + 1) * (low + high)) // 2 pipeFlow = sum - (high - low) if pipeFlow >= totalPipes: return True low -= 1 x -= 1 return False if __name__ == "__main__": main() ```

0

525

B

Pasha and String

PROGRAMMING

1,400

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

null

Pasha got a very beautiful string *s* for his birthday, the string consists of lowercase Latin letters. The letters in the string are numbered from 1 to |*s*| from left to right, where |*s*| is the length of the given string. Pasha didn't like his present very much so he decided to change it. After his birthday Pasha spent *m* days performing the following transformations on his string — each day he chose integer *a**i* and reversed a piece of string (a segment) from position *a**i* to position |*s*|<=-<=*a**i*<=+<=1. It is guaranteed that 2·*a**i*<=≤<=|*s*|. You face the following task: determine what Pasha's string will look like after *m* days.

The first line of the input contains Pasha's string *s* of length from 2 to 2·105 characters, consisting of lowercase Latin letters. The second line contains a single integer *m* (1<=≤<=*m*<=≤<=105) — the number of days when Pasha changed his string. The third line contains *m* space-separated elements *a**i* (1<=≤<=*a**i*; 2·*a**i*<=≤<=|*s*|) — the position from which Pasha started transforming the string on the *i*-th day.

In the first line of the output print what Pasha's string *s* will look like after *m* days.

[ "abcdef\n1\n2\n", "vwxyz\n2\n2 2\n", "abcdef\n3\n1 2 3\n" ]

[ "aedcbf\n", "vwxyz\n", "fbdcea\n" ]

none

750

[ { "input": "abcdef\n1\n2", "output": "aedcbf" }, { "input": "vwxyz\n2\n2 2", "output": "vwxyz" }, { "input": "abcdef\n3\n1 2 3", "output": "fbdcea" }, { "input": "jc\n5\n1 1 1 1 1", "output": "cj" }, { "input": "wljqgdlxyc\n13\n3 4 3 3 5 4 4 2 4 4 5 3 3", "output": "wyjldgqxlc" }, { "input": "keicnqmuqinhsmtudqcilocxkbqgzhbkitmqwttdyoyvcbxincwjryzknubpacsngorexaldfurondbednowemnnlphhboycfavs\n2\n5 12", "output": "keiccyobhhphsmtudqcilocxkbqgzhbkitmqwttdyoyvcbxincwjryzknubpacsngorexaldfurondbednowemnnlniqumqnfavs" }, { "input": "xwcxggxvfqbdklewbxkjzibmufnaywuxsqvwakefxbbkfandvigasbhbatsxyqxicrosatfsfybedklsaztyyiuurfbrzmwumujy\n100\n14 43 30 13 8 19 33 7 8 14 15 35 5 18 44 1 35 1 18 7 50 47 9 49 28 29 39 37 27 17 19 12 5 24 37 42 37 23 35 31 10 26 5 38 40 34 42 47 2 40 43 34 16 25 14 45 35 38 46 48 49 27 49 38 10 49 5 7 3 3 41 25 24 34 37 33 17 50 48 11 40 43 48 10 9 50 18 39 32 13 26 40 37 16 45 50 27 3 7 31", "output": "xjcxggxvfbbruliyyxkjzikdebnfyftxsorcaxqyxbtkfhbdvigasnababsxfekiwvqsauwsayfumblsaztbweukdfqrzmwumuwy" } ]

1,521,466,804

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

6

109

15,257,600

n = input() b = list(n) a = int(input()) c = list(map(int,input().split())) d = [0]*50001 sum1 = 0 for i in range(a): d[c[i]]+=1 for i in range(1,50001): if d[i] % 2 != 0 : sum1 += 1 if sum1 % 2 != 0: b[i-1],b[len(n)-i] = b[len(n)-i],b[i-1] elif sum1 % 2 == 1 and i <= int(len(n)/2): b[i - 1], b[len(n) - i] = b[len(n) - i], b[i - 1] print(''.join(b))

Title: Pasha and String Time Limit: None seconds Memory Limit: None megabytes Problem Description: Pasha got a very beautiful string *s* for his birthday, the string consists of lowercase Latin letters. The letters in the string are numbered from 1 to |*s*| from left to right, where |*s*| is the length of the given string. Pasha didn't like his present very much so he decided to change it. After his birthday Pasha spent *m* days performing the following transformations on his string — each day he chose integer *a**i* and reversed a piece of string (a segment) from position *a**i* to position |*s*|<=-<=*a**i*<=+<=1. It is guaranteed that 2·*a**i*<=≤<=|*s*|. You face the following task: determine what Pasha's string will look like after *m* days. Input Specification: The first line of the input contains Pasha's string *s* of length from 2 to 2·105 characters, consisting of lowercase Latin letters. The second line contains a single integer *m* (1<=≤<=*m*<=≤<=105) — the number of days when Pasha changed his string. The third line contains *m* space-separated elements *a**i* (1<=≤<=*a**i*; 2·*a**i*<=≤<=|*s*|) — the position from which Pasha started transforming the string on the *i*-th day. Output Specification: In the first line of the output print what Pasha's string *s* will look like after *m* days. Demo Input: ['abcdef\n1\n2\n', 'vwxyz\n2\n2 2\n', 'abcdef\n3\n1 2 3\n'] Demo Output: ['aedcbf\n', 'vwxyz\n', 'fbdcea\n'] Note: none

```python n = input() b = list(n) a = int(input()) c = list(map(int,input().split())) d = [0]*50001 sum1 = 0 for i in range(a): d[c[i]]+=1 for i in range(1,50001): if d[i] % 2 != 0 : sum1 += 1 if sum1 % 2 != 0: b[i-1],b[len(n)-i] = b[len(n)-i],b[i-1] elif sum1 % 2 == 1 and i <= int(len(n)/2): b[i - 1], b[len(n) - i] = b[len(n) - i], b[i - 1] print(''.join(b)) ```

-1

289

B

Polo the Penguin and Matrix

PROGRAMMING

1,400

[ "brute force", "dp", "implementation", "sortings", "ternary search" ]

null

Little penguin Polo has an *n*<=×<=*m* matrix, consisting of integers. Let's index the matrix rows from 1 to *n* from top to bottom and let's index the columns from 1 to *m* from left to right. Let's represent the matrix element on the intersection of row *i* and column *j* as *a**ij*. In one move the penguin can add or subtract number *d* from some matrix element. Find the minimum number of moves needed to make all matrix elements equal. If the described plan is impossible to carry out, say so.

The first line contains three integers *n*, *m* and *d* (1<=≤<=*n*,<=*m*<=≤<=100,<=1<=≤<=*d*<=≤<=104) — the matrix sizes and the *d* parameter. Next *n* lines contain the matrix: the *j*-th integer in the *i*-th row is the matrix element *a**ij* (1<=≤<=*a**ij*<=≤<=104).

In a single line print a single integer — the minimum number of moves the penguin needs to make all matrix elements equal. If that is impossible, print "-1" (without the quotes).

[ "2 2 2\n2 4\n6 8\n", "1 2 7\n6 7\n" ]

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

none

1,000

[ { "input": "2 2 2\n2 4\n6 8", "output": "4" }, { "input": "1 2 7\n6 7", "output": "-1" }, { "input": "3 2 1\n5 7\n1 2\n5 100", "output": "104" }, { "input": "3 3 3\n5 8 5\n11 11 17\n14 5 2", "output": "12" }, { "input": "3 3 3\n5 8 5\n11 11 17\n14 5 3", "output": "-1" }, { "input": "2 2 4\n5 5\n5 5", "output": "0" }, { "input": "7 4 5\n7 7 7 12\n7 12 12 7\n7 7 7 7\n7 7 12 7\n7 7 12 12\n12 12 7 12\n7 7 7 7", "output": "9" }, { "input": "7 7 47\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91", "output": "0" }, { "input": "7 7 47\n47 47 47 47 47 47 47\n47 47 47 47 77 47 47\n47 47 47 47 47 47 47\n47 47 47 47 47 47 47\n47 47 47 47 47 47 47\n47 47 47 47 47 47 47\n47 47 47 47 47 127 47", "output": "-1" }, { "input": "7 5 47\n9583 1734 4601 5353 2110\n3802 5165 7985 6293 324\n7045 653 9160 7891 4930\n1781 3520 4178 2298 3943\n1405 2956 5447 5494 6528\n3097 1640 7750 4883 8032\n4225 8455 1875 4789 4366", "output": "1508" }, { "input": "1 1 1\n1", "output": "0" }, { "input": "1 1 8\n12", "output": "0" } ]

1,600,780,892

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

2

280

0

n , m , d = list(map(int,input().split())) a = [] b = [] odd = True for i in range(n): temp = list(map(int,input().split())) for j in temp: a.append(j) b.append(j%d) if(len(set(b))!=1): print(-1) else: a = sorted(a) mid = a[n//2] ans = 0 for i in a: ans+=abs(i-mid)//d print(ans)

Title: Polo the Penguin and Matrix Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little penguin Polo has an *n*<=×<=*m* matrix, consisting of integers. Let's index the matrix rows from 1 to *n* from top to bottom and let's index the columns from 1 to *m* from left to right. Let's represent the matrix element on the intersection of row *i* and column *j* as *a**ij*. In one move the penguin can add or subtract number *d* from some matrix element. Find the minimum number of moves needed to make all matrix elements equal. If the described plan is impossible to carry out, say so. Input Specification: The first line contains three integers *n*, *m* and *d* (1<=≤<=*n*,<=*m*<=≤<=100,<=1<=≤<=*d*<=≤<=104) — the matrix sizes and the *d* parameter. Next *n* lines contain the matrix: the *j*-th integer in the *i*-th row is the matrix element *a**ij* (1<=≤<=*a**ij*<=≤<=104). Output Specification: In a single line print a single integer — the minimum number of moves the penguin needs to make all matrix elements equal. If that is impossible, print "-1" (without the quotes). Demo Input: ['2 2 2\n2 4\n6 8\n', '1 2 7\n6 7\n'] Demo Output: ['4\n', '-1\n'] Note: none

```python n , m , d = list(map(int,input().split())) a = [] b = [] odd = True for i in range(n): temp = list(map(int,input().split())) for j in temp: a.append(j) b.append(j%d) if(len(set(b))!=1): print(-1) else: a = sorted(a) mid = a[n//2] ans = 0 for i in a: ans+=abs(i-mid)//d print(ans) ```

0

4

A

Watermelon

PROGRAMMING

800

[ "brute force", "math" ]

A. Watermelon

1

64

One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem. Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.

The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.

Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.

[ "8\n" ]

[ "YES\n" ]

For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).

0

[ { "input": "8", "output": "YES" }, { "input": "5", "output": "NO" }, { "input": "4", "output": "YES" }, { "input": "3", "output": "NO" }, { "input": "2", "output": "NO" }, { "input": "1", "output": "NO" }, { "input": "7", "output": "NO" }, { "input": "6", "output": "YES" }, { "input": "10", "output": "YES" }, { "input": "9", "output": "NO" }, { "input": "53", "output": "NO" }, { "input": "77", "output": "NO" }, { "input": "32", "output": "YES" }, { "input": "44", "output": "YES" }, { "input": "98", "output": "YES" }, { "input": "99", "output": "NO" }, { "input": "90", "output": "YES" }, { "input": "67", "output": "NO" }, { "input": "100", "output": "YES" }, { "input": "88", "output": "YES" } ]

1,689,782,969

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

30

0

n=int(input("enter number of lines")) for i in range(n): m=input("enter the words") d=len(m) print(m[0],len(m),m[-1])

Title: Watermelon Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem. Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. Input Specification: The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. Output Specification: Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. Demo Input: ['8\n'] Demo Output: ['YES\n'] Note: For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).

```python n=int(input("enter number of lines")) for i in range(n): m=input("enter the words") d=len(m) print(m[0],len(m),m[-1]) ```

-1

109

A

Lucky Sum of Digits

PROGRAMMING

1,000

[ "brute force", "implementation" ]

A. Lucky Sum of Digits

2

256

Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.

The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.

Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.

[ "11\n", "10\n" ]

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

none

500

[ { "input": "11", "output": "47" }, { "input": "10", "output": "-1" }, { "input": "64", "output": "4477777777" }, { "input": "1", "output": "-1" }, { "input": "4", "output": "4" }, { "input": "7", "output": "7" }, { "input": "12", "output": "444" }, { "input": "1000000", "output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "47", "output": "44477777" }, { "input": "100", "output": "4444777777777777" }, { "input": "700", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "485", "output": "44447777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "111", "output": "444447777777777777" }, { "input": "85", "output": "4477777777777" }, { "input": "114", "output": "444477777777777777" }, { "input": "474", "output": "444777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "74", "output": "47777777777" }, { "input": "1000", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "1024", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "4444", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "45784", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "10000", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "9854", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "186", "output": "477777777777777777777777777" }, { "input": "10416", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "10417", "output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "3840", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "100000", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "9876", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "99999", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "777777", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "854759", "output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "11000", "output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "18951", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999999", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "888887", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999998", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "40008", "output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "10691", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "5", "output": "-1" }, { "input": "6", "output": "-1" }, { "input": "9", "output": "-1" }, { "input": "8", "output": "44" }, { "input": "2", "output": "-1" }, { "input": "3", "output": "-1" }, { "input": "999997", "output": 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"4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999990", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999980", "output": 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"7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." } ]

1,683,806,012

2,147,483,647

Python 3

COMPILATION_ERROR

TESTS

0

import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int luckyFours = n / 4; int luckySevens = 0; while (luckyFours >= 0) { int remainingSum = n - luckyFours * 4; if (remainingSum % 7 == 0) { luckySevens = remainingSum / 7; break; } luckyFours--; } if (luckyFours < 0) { System.out.println("-1"); } else { for (int i = 0; i < luckyFours; i++) { System.out.print("4"); } for (int i = 0; i < luckySevens; i++) { System.out.print("7"); } } } }

Title: Lucky Sum of Digits Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task. Input Specification: The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number. Output Specification: Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1. Demo Input: ['11\n', '10\n'] Demo Output: ['47\n', '-1\n'] Note: none

```python import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int luckyFours = n / 4; int luckySevens = 0; while (luckyFours >= 0) { int remainingSum = n - luckyFours * 4; if (remainingSum % 7 == 0) { luckySevens = remainingSum / 7; break; } luckyFours--; } if (luckyFours < 0) { System.out.println("-1"); } else { for (int i = 0; i < luckyFours; i++) { System.out.print("4"); } for (int i = 0; i < luckySevens; i++) { System.out.print("7"); } } } } ```

-1

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" } ]

1,695,219,459

2,147,483,647

Python 3

OK

TESTS

25

46

2,355,200

n = int(input()) games = input() win_A = [differ for differ in games if differ == 'A'] if len(win_A) == n/2: print("Friendship") elif len(win_A) > n/2: print("Anton") else: print("Danik")

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 n = int(input()) games = input() win_A = [differ for differ in games if differ == 'A'] if len(win_A) == n/2: print("Friendship") elif len(win_A) > n/2: print("Anton") else: print("Danik") ```

3

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" } ]

1,699,030,783

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

2

2,000

12,800,000

n=int(input()) st=list(map(int,input().split())) ans=0 for i in st: if min(st) < i < max(st): ans+=1 print(ans)

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()) st=list(map(int,input().split())) ans=0 for i in st: if min(st) < i < max(st): ans+=1 print(ans) ```

0

922

A

Cloning Toys

PROGRAMMING

1,300

[ "implementation" ]

null

Imp likes his plush toy a lot. Recently, he found a machine that can clone plush toys. Imp knows that if he applies the machine to an original toy, he additionally gets one more original toy and one copy, and if he applies the machine to a copied toy, he gets two additional copies. Initially, Imp has only one original toy. He wants to know if it is possible to use machine to get exactly *x* copied toys and *y* original toys? He can't throw toys away, and he can't apply the machine to a copy if he doesn't currently have any copies.

The only line contains two integers *x* and *y* (0<=≤<=*x*,<=*y*<=≤<=109) — the number of copies and the number of original toys Imp wants to get (including the initial one).

Print "Yes", if the desired configuration is possible, and "No" otherwise. You can print each letter in arbitrary case (upper or lower).

[ "6 3\n", "4 2\n", "1000 1001\n" ]

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

In the first example, Imp has to apply the machine twice to original toys and then twice to copies.

500

[ { "input": "6 3", "output": "Yes" }, { "input": "4 2", "output": "No" }, { "input": "1000 1001", "output": "Yes" }, { "input": "1000000000 999999999", "output": "Yes" }, { "input": "81452244 81452247", "output": "No" }, { "input": "188032448 86524683", "output": "Yes" }, { "input": "365289629 223844571", "output": "No" }, { "input": "247579518 361164458", "output": "No" }, { "input": "424836699 793451637", "output": "No" }, { "input": "602093880 930771525", "output": "No" }, { "input": "779351061 773124120", "output": "Yes" }, { "input": "661640950 836815080", "output": "No" }, { "input": "543930839 974134967", "output": "No" }, { "input": "16155311 406422145", "output": "No" }, { "input": "81601559 445618240", "output": "No" }, { "input": "963891449 582938127", "output": "No" }, { "input": "141148629 351661795", "output": "No" }, { "input": "318405810 783948974", "output": "No" }, { "input": "495662991 921268861", "output": "No" }, { "input": "1 0", "output": "No" }, { "input": "0 1", "output": "Yes" }, { "input": "0 0", "output": "No" }, { "input": "453462237 167520068", "output": "Yes" }, { "input": "630719418 9872663", "output": "Yes" }, { "input": "807976599 442159843", "output": "No" }, { "input": "690266488 579479730", "output": "No" }, { "input": "771581370 589752968", "output": "No" }, { "input": "948838551 727072855", "output": "No" }, { "input": "831128440 790763814", "output": "No" }, { "input": "303352912 928083702", "output": "No" }, { "input": "185642801 65403588", "output": "Yes" }, { "input": "67932690 202723476", "output": "No" }, { "input": "540157163 340043363", "output": "No" }, { "input": "422447052 772330542", "output": "No" }, { "input": "599704233 541054210", "output": "Yes" }, { "input": "481994122 678374097", "output": "No" }, { "input": "48564714 743566477", "output": "No" }, { "input": "225821895 880886365", "output": "No" }, { "input": "403079076 313173543", "output": "Yes" }, { "input": "1000000000 1000000000", "output": "No" }, { "input": "1 1", "output": "No" }, { "input": "1 2", "output": "Yes" }, { "input": "2 1", "output": "No" }, { "input": "2 2", "output": "No" }, { "input": "2 0", "output": "No" }, { "input": "0 2", "output": "No" }, { "input": "1000000000 1", "output": "No" }, { "input": "777777 0", "output": "No" }, { "input": "10 1", "output": "No" }, { "input": "7 0", "output": "No" }, { "input": "3 0", "output": "No" }, { "input": "3 2", "output": "Yes" } ]

1,592,948,387

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

19

109

6,656,000

x,y = map(int, input().split()) if (y == 1 and x > 0) or x == y: print('No') elif y == 1 and x == 0 : print('Yes') elif x < y: if x == y - 1: print('Yes') else: print('No') elif x > y: x -= y - 1 if x % 2 == 0: print('Yes') else: print('No')

Title: Cloning Toys Time Limit: None seconds Memory Limit: None megabytes Problem Description: Imp likes his plush toy a lot. Recently, he found a machine that can clone plush toys. Imp knows that if he applies the machine to an original toy, he additionally gets one more original toy and one copy, and if he applies the machine to a copied toy, he gets two additional copies. Initially, Imp has only one original toy. He wants to know if it is possible to use machine to get exactly *x* copied toys and *y* original toys? He can't throw toys away, and he can't apply the machine to a copy if he doesn't currently have any copies. Input Specification: The only line contains two integers *x* and *y* (0<=≤<=*x*,<=*y*<=≤<=109) — the number of copies and the number of original toys Imp wants to get (including the initial one). Output Specification: Print "Yes", if the desired configuration is possible, and "No" otherwise. You can print each letter in arbitrary case (upper or lower). Demo Input: ['6 3\n', '4 2\n', '1000 1001\n'] Demo Output: ['Yes\n', 'No\n', 'Yes\n'] Note: In the first example, Imp has to apply the machine twice to original toys and then twice to copies.

```python x,y = map(int, input().split()) if (y == 1 and x > 0) or x == y: print('No') elif y == 1 and x == 0 : print('Yes') elif x < y: if x == y - 1: print('Yes') else: print('No') elif x > y: x -= y - 1 if x % 2 == 0: print('Yes') else: print('No') ```

0

248

A

Cupboards

PROGRAMMING

800

[ "implementation" ]

null

One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house. Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find. And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open. Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position. Your task is to write a program that will determine the required number of seconds *t*.

The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero. The numbers in the lines are separated by single spaces.

In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.

[ "5\n0 1\n1 0\n0 1\n1 1\n0 1\n" ]

[ "3\n" ]

none

500

[ { "input": "5\n0 1\n1 0\n0 1\n1 1\n0 1", "output": "3" }, { "input": "2\n0 0\n0 0", "output": "0" }, { "input": "3\n0 1\n1 1\n1 1", "output": "1" }, { "input": "8\n0 1\n1 0\n0 1\n1 1\n0 1\n1 0\n0 1\n1 0", "output": "7" }, { "input": "8\n1 0\n1 0\n1 0\n0 1\n0 1\n1 1\n1 1\n0 1", "output": "6" }, { "input": "15\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0", "output": "0" }, { "input": "5\n1 0\n1 0\n1 0\n0 1\n0 1", "output": "4" } ]

1,610,422,503

2,147,483,647

PyPy 3

OK

TESTS

57

436

2,867,200

t = int(input()) left = 0 right = 0 ans = 0 for i in range(t): l, r = map(int, input().split()) left += l right += r ans += min(t-left, left-0) ans += min(t-right, right) print(ans)

Title: Cupboards Time Limit: None seconds Memory Limit: None megabytes Problem Description: One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house. Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find. And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open. Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position. Your task is to write a program that will determine the required number of seconds *t*. Input Specification: The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero. The numbers in the lines are separated by single spaces. Output Specification: In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs. Demo Input: ['5\n0 1\n1 0\n0 1\n1 1\n0 1\n'] Demo Output: ['3\n'] Note: none

```python t = int(input()) left = 0 right = 0 ans = 0 for i in range(t): l, r = map(int, input().split()) left += l right += r ans += min(t-left, left-0) ans += min(t-right, right) print(ans) ```

3

344

A

Magnets

PROGRAMMING

800

[ "implementation" ]

null

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

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

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

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

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

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

500

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

1,698,099,394

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

30

0

l = int(input()) n = list(map(int,input().split())) x = 1 for i in range(l-1): if n[i] == n[i+1]: x += 0 elif n[i] != n[i+1]: x += 1 print(x)

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

```python l = int(input()) n = list(map(int,input().split())) x = 1 for i in range(l-1): if n[i] == n[i+1]: x += 0 elif n[i] != n[i+1]: x += 1 print(x) ```

-1

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" } ]

1,636,953,252

2,147,483,647

Python 3

OK

TESTS

30

92

0

s=input() up=0 low=0 for i in s: if(i.islower()): low=low+1 elif(i.isupper()): up=up+1 if(up>low): print(s.upper()) elif(low>up): print(s.lower()) elif(low==up): print(s.lower())

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

```python s=input() up=0 low=0 for i in s: if(i.islower()): low=low+1 elif(i.isupper()): up=up+1 if(up>low): print(s.upper()) elif(low>up): print(s.lower()) elif(low==up): print(s.lower()) ```

3.977

914

A

Perfect Squares

PROGRAMMING

900

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

null

Given an array *a*1,<=*a*2,<=...,<=*a**n* of *n* integers, find the largest number in the array that is not a perfect square. A number *x* is said to be a perfect square if there exists an integer *y* such that *x*<==<=*y*2.

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=106<=≤<=*a**i*<=≤<=106) — the elements of the array. It is guaranteed that at least one element of the array is not a perfect square.

Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.

[ "2\n4 2\n", "8\n1 2 4 8 16 32 64 576\n" ]

[ "2\n", "32\n" ]

In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.

500

[ { "input": "2\n4 2", "output": "2" }, { "input": "8\n1 2 4 8 16 32 64 576", "output": "32" }, { "input": "3\n-1 -4 -9", "output": "-1" }, { "input": "5\n918375 169764 598796 76602 538757", "output": "918375" }, { "input": "5\n804610 765625 2916 381050 93025", "output": "804610" }, { "input": "5\n984065 842724 127449 525625 573049", "output": "984065" }, { "input": "2\n226505 477482", "output": "477482" }, { "input": "2\n370881 659345", "output": "659345" }, { "input": "2\n4 5", "output": "5" }, { "input": "2\n3 4", "output": "3" }, { "input": "2\n999999 1000000", "output": "999999" }, { "input": "3\n-1 -2 -3", "output": "-1" }, { "input": "2\n-1000000 1000000", "output": "-1000000" }, { "input": "2\n-1 0", "output": "-1" }, { "input": "1\n2", "output": "2" }, { "input": "1\n-1", "output": "-1" }, { "input": "35\n-871271 -169147 -590893 -400197 -476793 0 -15745 -890852 -124052 -631140 -238569 -597194 -147909 -928925 -587628 -569656 -581425 -963116 -665954 -506797 -196044 -309770 -701921 -926257 -152426 -991371 -624235 -557143 -689886 -59804 -549134 -107407 -182016 -24153 -607462", "output": "-15745" }, { "input": "16\n-882343 -791322 0 -986738 -415891 -823354 -840236 -552554 -760908 -331993 -549078 -863759 -913261 -937429 -257875 -602322", "output": "-257875" }, { "input": "71\n908209 289 44521 240100 680625 274576 212521 91809 506944 499849 3844 15376 592900 58081 240100 984064 732736 257049 600625 180625 130321 580644 261121 75625 46225 853776 485809 700569 817216 268324 293764 528529 25921 399424 175561 99856 295936 20736 611524 13924 470596 574564 5329 15376 676 431649 145161 697225 41616 550564 514089 9409 227529 1681 839056 3721 552049 465124 38809 197136 659344 214369 998001 44944 3844 186624 362404 -766506 739600 10816 299209", "output": "-766506" }, { "input": "30\n192721 -950059 -734656 625 247009 -423468 318096 622521 678976 777924 1444 748303 27556 62001 795664 89401 221841 -483208 467856 477109 196 -461813 831744 772641 574564 -519370 861184 67600 -717966 -259259", "output": "748303" }, { "input": "35\n628849 962361 436921 944784 444889 29241 -514806 171396 685584 -823202 -929730 6982 198025 783225 552049 -957165 782287 -659167 -414846 695556 -336330 41616 963781 71289 119639 952576 -346713 178929 232324 121802 393266 841 649636 179555 998001", "output": "963781" }, { "input": "53\n280988 756430 -515570 -248578 170649 -21608 642677 216770 827291 589500 940901 216097 -118956 -919104 -319264 -761585 289479 499613 588276 883036 480518 -323196 -274570 -406556 -381484 -956025 702135 -445274 -783543 136593 153664 897473 352651 737974 -21123 -284944 501734 898033 604429 624138 40804 248782 -786059 -304592 -209210 -312904 419820 -328648 -47331 -919227 -280955 104827 877304", "output": "940901" }, { "input": "15\n256 -227055 427717 827239 462070 66049 987533 -175306 -552810 -867915 -408251 -693957 -972981 -245827 896904", "output": "987533" }, { "input": "3\n-1 1 0", "output": "-1" }, { "input": "2\n0 -5", "output": "-5" }, { "input": "3\n-1 -2 0", "output": "-1" }, { "input": "2\n-5 0", "output": "-5" }, { "input": "1\n-439", "output": "-439" }, { "input": "1\n-1000000", "output": "-1000000" }, { "input": "1\n-917455", "output": "-917455" }, { "input": "3\n1 1 -1", "output": "-1" }, { "input": "2\n131073 1", "output": "131073" }, { "input": "2\n99999 3", "output": "99999" }, { "input": "2\n-524272 -1000000", "output": "-524272" }, { "input": "2\n15 131073", "output": "131073" } ]

1,600,019,899

2,147,483,647

PyPy 3

RUNTIME_ERROR

TESTS

2

202

2,150,400

import math n=int(input()) a=list(map(int,input().split())) ans=0 for i in range(n): if math.sqrt(a[i])!=int(math.sqrt(a[i])): ans=max(ans,a[i]) print(ans)

Title: Perfect Squares Time Limit: None seconds Memory Limit: None megabytes Problem Description: Given an array *a*1,<=*a*2,<=...,<=*a**n* of *n* integers, find the largest number in the array that is not a perfect square. A number *x* is said to be a perfect square if there exists an integer *y* such that *x*<==<=*y*2. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=106<=≤<=*a**i*<=≤<=106) — the elements of the array. It is guaranteed that at least one element of the array is not a perfect square. Output Specification: Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists. Demo Input: ['2\n4 2\n', '8\n1 2 4 8 16 32 64 576\n'] Demo Output: ['2\n', '32\n'] Note: In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.

```python import math n=int(input()) a=list(map(int,input().split())) ans=0 for i in range(n): if math.sqrt(a[i])!=int(math.sqrt(a[i])): ans=max(ans,a[i]) print(ans) ```

-1

11

A

Increasing Sequence

PROGRAMMING

900

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

A. Increasing Sequence

1

64

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

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

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

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

[ "3\n" ]

none

0

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

1,601,195,068

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

7

218

307,200

n, d = map(int, input().split()) array = list(map(int, input().split())) res = 0 for i in range(1, len(array)): if array[i - 1] >= array[i]: res += (array[i - 1] - array[i]) // d + 1 array[i] += (array[i - 1] - array[i]) // d + 1 print(res)

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

```python n, d = map(int, input().split()) array = list(map(int, input().split())) res = 0 for i in range(1, len(array)): if array[i - 1] >= array[i]: res += (array[i - 1] - array[i]) // d + 1 array[i] += (array[i - 1] - array[i]) // d + 1 print(res) ```

0

912

A

Tricky Alchemy

PROGRAMMING

800

[ "implementation" ]

null

During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals. Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough. Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.

The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal. The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained.

Print a single integer — the minimum number of crystals that Grisha should acquire in addition.

[ "4 3\n2 1 1\n", "3 9\n1 1 3\n", "12345678 87654321\n43043751 1000000000 53798715\n" ]

[ "2\n", "1\n", "2147483648\n" ]

In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.

500

[ { "input": "4 3\n2 1 1", "output": "2" }, { "input": "3 9\n1 1 3", "output": "1" }, { "input": "12345678 87654321\n43043751 1000000000 53798715", "output": "2147483648" }, { "input": "12 12\n3 5 2", "output": "0" }, { "input": "770 1390\n170 442 311", "output": "12" }, { "input": "3555165 6693472\n1499112 556941 3075290", "output": "3089339" }, { "input": "0 0\n1000000000 1000000000 1000000000", "output": "7000000000" }, { "input": "1 1\n0 1 0", "output": "0" }, { "input": "117708228 562858833\n118004008 360437130 154015822", "output": "738362681" }, { "input": "999998118 700178721\n822106746 82987112 547955384", "output": "1753877029" }, { "input": "566568710 765371101\n60614022 80126928 809950465", "output": "1744607222" }, { "input": "448858599 829062060\n764716760 97644201 203890025", "output": "1178219122" }, { "input": "626115781 966381948\n395190569 820194184 229233367", "output": "1525971878" }, { "input": "803372962 103701834\n394260597 837711458 623172928", "output": "3426388098" }, { "input": "980630143 241021722\n24734406 928857659 312079781", "output": "1624075280" }, { "input": "862920032 378341609\n360240924 241342224 337423122", "output": "974174021" }, { "input": "40177212 515661496\n64343660 963892207 731362684", "output": "3694721078" }, { "input": "217434393 579352456\n694817470 981409480 756706026", "output": "4825785129" }, { "input": "394691574 716672343\n398920207 72555681 150645586", "output": "475704521" }, { "input": "276981463 853992230\n29394015 90072954 839552440", "output": "1754738044" }, { "input": "843552056 919184611\n341530221 423649259 101547519", "output": "263157645" }, { "input": "20809236 56504497\n972004030 441166533 495487081", "output": "4235488636" }, { "input": "198066417 825228166\n602477839 532312735 520830423", "output": "2808777834" }, { "input": "80356306 962548053\n601547868 549830008 914769984", "output": "4004161345" }, { "input": "257613487 394835231\n642087093 567347282 308709545", "output": "2692548667" }, { "input": "139903376 532155119\n641157122 289897263 629020178", "output": "3077110809" }, { "input": "612127849 669475006\n271630930 676010757 22959739", "output": "682559736" }, { "input": "0 0\n0 0 0", "output": "0" }, { "input": "1000000000 1000000000\n499999998 4 333333332", "output": "0" }, { "input": "1000000000 1000000000\n1000000000 1000000000 1000000000", "output": "5000000000" }, { "input": "4 3\n1 0 1", "output": "0" }, { "input": "4 12\n1 2 3", "output": "0" }, { "input": "4 20\n1 2 1", "output": "0" }, { "input": "100 10\n2 3 4", "output": "5" }, { "input": "6 0\n1 1 1", "output": "4" }, { "input": "25 5\n3 3 3", "output": "7" }, { "input": "48 27\n22 39 20", "output": "107" }, { "input": "4 0\n1 1 1", "output": "4" } ]

1,669,133,229

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

4

31

0

y, b = input().split() y = int(y) b = int(b) y_, g_, b_ = input().split() y_ = int(y_) g_ = int(g_) b_ = int(b_) total = 0 if (y + b) - (y_ * 2 + g_ * 1 + g_ * 1 + b_ * 3) < 0: total = abs((y + b) - (y_ * 2 + g_ * 1 + g_ * 1 + b_ * 3)) print(total)

Title: Tricky Alchemy Time Limit: None seconds Memory Limit: None megabytes Problem Description: During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals. Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough. Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls. Input Specification: The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal. The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained. Output Specification: Print a single integer — the minimum number of crystals that Grisha should acquire in addition. Demo Input: ['4 3\n2 1 1\n', '3 9\n1 1 3\n', '12345678 87654321\n43043751 1000000000 53798715\n'] Demo Output: ['2\n', '1\n', '2147483648\n'] Note: In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.

```python y, b = input().split() y = int(y) b = int(b) y_, g_, b_ = input().split() y_ = int(y_) g_ = int(g_) b_ = int(b_) total = 0 if (y + b) - (y_ * 2 + g_ * 1 + g_ * 1 + b_ * 3) < 0: total = abs((y + b) - (y_ * 2 + g_ * 1 + g_ * 1 + b_ * 3)) print(total) ```

0

21

B

Intersection

PROGRAMMING

2,000

[ "implementation", "math" ]

B. Intersection

1

256

You are given two set of points. The first set is determined by the equation *A*1*x*<=+<=*B*1*y*<=+<=*C*1<==<=0, and the second one is determined by the equation *A*2*x*<=+<=*B*2*y*<=+<=*C*2<==<=0. Write the program which finds the number of points in the intersection of two given sets.

The first line of the input contains three integer numbers *A*1,<=*B*1,<=*C*1 separated by space. The second line contains three integer numbers *A*2,<=*B*2,<=*C*2 separated by space. All the numbers are between -100 and 100, inclusive.

Print the number of points in the intersection or -1 if there are infinite number of points.

[ "1 1 0\n2 2 0\n", "1 1 0\n2 -2 0\n" ]

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

none

1,000

[ { "input": "1 1 0\n2 2 0", "output": "-1" }, { "input": "1 1 0\n2 -2 0", "output": "1" }, { "input": "0 0 0\n0 0 0", "output": "-1" }, { "input": "1 1 1\n1 1 1", "output": "-1" }, { "input": "8 3 -4\n-5 2 7", "output": "1" }, { "input": "-1 -1 0\n0 -1 -1", "output": "1" }, { "input": "-1 -1 0\n1 1 -1", "output": "0" }, { "input": "-1 -1 1\n0 -1 0", "output": "1" }, { "input": "0 0 0\n1 -1 -1", "output": "-1" }, { "input": "0 0 1\n-1 1 -1", "output": "0" }, { "input": "0 1 -1\n-1 1 -1", "output": "1" }, { "input": "1 0 -1\n0 0 -1", "output": "0" }, { "input": "0 1 1\n1 0 0", "output": "1" }, { "input": "1 0 0\n0 0 1", "output": "0" }, { "input": "1 -1 -1\n1 -1 0", "output": "0" }, { "input": "1 0 0\n0 1 1", "output": "1" }, { "input": "1 -1 1\n-1 -1 1", "output": "1" }, { "input": "1 0 0\n0 1 1", "output": "1" }, { "input": "-1 -1 1\n-1 -1 -1", "output": "0" }, { "input": "-1 -1 0\n1 1 -1", "output": "0" }, { "input": "0 0 0\n0 1 -1", "output": "-1" }, { "input": "0 1 1\n0 1 -1", "output": "0" }, { "input": "0 1 0\n1 1 0", "output": "1" }, { "input": "1 0 1\n-1 0 -1", "output": "-1" }, { "input": "1 1 0\n1 0 -1", "output": "1" }, { "input": "0 -1 -1\n-1 0 0", "output": "1" }, { "input": "1 0 0\n1 0 0", "output": "-1" }, { "input": "1 1 1\n-1 -1 0", "output": "0" }, { "input": "-1 -1 -1\n0 -1 1", "output": "1" }, { "input": "0 -1 0\n0 0 1", "output": "0" }, { "input": "0 1 1\n-1 -1 1", "output": "1" }, { "input": "0 -1 0\n0 -1 1", "output": "0" }, { "input": "0 1 1\n0 1 1", "output": "-1" }, { "input": "1 -1 0\n-1 -1 1", "output": "1" }, { "input": "0 1 1\n0 1 -1", "output": "0" }, { "input": "1 0 1\n1 0 0", "output": "0" }, { "input": "1 1 1\n0 0 0", "output": "-1" }, { "input": "1 0 1\n-1 -1 -1", "output": "1" }, { "input": "1 -1 1\n0 0 0", "output": "-1" }, { "input": "0 1 1\n-1 -1 0", "output": "1" }, { "input": "-1 0 1\n1 0 0", "output": "0" }, { "input": "0 1 -1\n0 0 1", "output": "0" }, { "input": "0 -1 0\n1 1 1", "output": "1" }, { "input": "1 0 1\n0 1 1", "output": "1" }, { "input": "0 0 0\n1 1 -1", "output": "-1" }, { "input": "1 -1 1\n1 1 1", "output": "1" }, { "input": "1 0 -1\n-1 0 1", "output": "-1" }, { "input": "1 0 1\n1 -1 1", "output": "1" }, { "input": "1 -1 -1\n-1 -1 -1", "output": "1" }, { "input": "0 -1 1\n0 0 -1", "output": "0" }, { "input": "0 0 -1\n1 -1 -1", "output": "0" }, { "input": "1 1 0\n-1 0 0", "output": "1" }, { "input": "1 0 -1\n0 -1 0", "output": "1" }, { "input": "1 -1 0\n-1 1 0", "output": "-1" }, { "input": "1 -1 1\n1 -1 0", "output": "0" }, { "input": "-1 -1 -1\n-1 1 0", "output": "1" }, { "input": "-1 0 1\n1 -1 1", "output": "1" }, { "input": "1 -1 0\n0 -1 -1", "output": "1" }, { "input": "-1 1 0\n-1 0 -1", "output": "1" }, { "input": "-1 -1 -1\n1 -1 1", "output": "1" }, { "input": "-1 -1 0\n1 1 1", "output": "0" }, { "input": "0 1 -1\n-1 0 0", "output": "1" }, { "input": "0 0 0\n0 0 0", "output": "-1" }, { "input": "0 1 1\n1 0 -1", "output": "1" }, { "input": "0 1 -1\n0 0 0", "output": "-1" }, { "input": "1 -1 0\n-1 1 0", "output": "-1" }, { "input": "0 0 0\n0 1 0", "output": "-1" }, { "input": "0 -1 1\n1 -1 1", "output": "1" }, { "input": "1 0 0\n0 1 0", "output": "1" }, { "input": "-1 1 0\n0 -1 1", "output": "1" }, { "input": "-1 0 -1\n1 1 0", "output": "1" }, { "input": "0 -1 0\n1 1 -1", "output": "1" }, { "input": "-1 -1 1\n-1 0 1", "output": "1" }, { "input": "0 1 0\n1 0 1", "output": "1" }, { "input": "1 0 0\n-1 0 -1", "output": "0" }, { "input": "-1 -1 0\n1 -1 1", "output": "1" }, { "input": "1 1 1\n-1 -1 -1", "output": "-1" }, { "input": "1 -1 0\n-1 1 0", "output": "-1" }, { "input": "-1 -1 1\n-1 1 0", "output": "1" }, { "input": "0 0 1\n1 0 -1", "output": "0" }, { "input": "0 -1 -2\n0 1 0", "output": "0" }, { "input": "0 -1 0\n2 -2 2", "output": "1" }, { "input": "1 -1 2\n-1 0 0", "output": "1" }, { "input": "-2 0 2\n0 0 2", "output": "0" }, { "input": "-1 0 -1\n1 -1 -1", "output": "1" }, { "input": "-1 2 0\n-2 1 -2", "output": "1" }, { "input": "0 2 0\n0 1 2", "output": "0" }, { "input": "2 2 2\n0 -2 0", "output": "1" }, { "input": "-2 0 -2\n2 -2 -2", "output": "1" }, { "input": "2 2 -1\n-2 1 1", "output": "1" }, { "input": "-2 -1 1\n0 -1 0", "output": "1" }, { "input": "-2 1 1\n0 0 -2", "output": "0" }, { "input": "-1 2 -2\n0 2 1", "output": "1" }, { "input": "1 2 -2\n-1 2 0", "output": "1" }, { "input": "0 0 2\n0 -1 -1", "output": "0" }, { "input": "2 1 1\n1 2 1", "output": "1" }, { "input": "-2 -1 2\n1 1 1", "output": "1" }, { "input": "0 -1 -1\n-2 -2 -1", "output": "1" }, { "input": "-1 0 -1\n0 -2 1", "output": "1" }, { "input": "1 1 2\n0 1 0", "output": "1" }, { "input": "-2 1 1\n2 1 -1", "output": "1" }, { "input": "-1 -2 1\n-1 -2 2", "output": "0" }, { "input": "0 -2 1\n-2 2 2", "output": "1" }, { "input": "0 -1 2\n-1 -1 0", "output": "1" }, { "input": "1 -1 -2\n1 2 -2", "output": "1" }, { "input": "-2 -1 0\n-2 2 2", "output": "1" }, { "input": "-1 1 0\n0 -1 0", "output": "1" }, { "input": "-1 -2 2\n-1 0 -2", "output": "1" }, { "input": "0 1 -1\n1 0 -2", "output": "1" }, { "input": "-1 -2 -2\n-2 1 0", "output": "1" }, { "input": "1 -1 2\n0 0 -2", "output": "0" }, { "input": "2 -1 2\n0 -2 1", "output": "1" }, { "input": "1 0 -1\n2 0 1", "output": "0" }, { "input": "-2 -1 0\n-2 0 -1", "output": "1" }, { "input": "-1 1 1\n0 1 1", "output": "1" }, { "input": "1 1 1\n1 1 -2", "output": "0" }, { "input": "1 2 1\n1 -1 1", "output": "1" }, { "input": "-2 -2 0\n0 -2 -1", "output": "1" }, { "input": "-1 -1 0\n-1 -2 -1", "output": "1" }, { "input": "-2 -2 -2\n1 1 -1", "output": "0" }, { "input": "0 0 0\n0 0 1", "output": "0" }, { "input": "0 0 -87\n0 0 0", "output": "0" }, { "input": "0 0 1\n0 0 1", "output": "0" }, { "input": "100 100 100\n-100 100 -100", "output": "1" }, { "input": "12 -32 89\n0 67 -23", "output": "1" }, { "input": "0 0 1\n0 0 2", "output": "0" }, { "input": "0 5 0\n0 0 5", "output": "0" }, { "input": "0 1 1\n0 -1 -1", "output": "-1" }, { "input": "1 1 0\n2 2 1", "output": "0" }, { "input": "0 0 5\n0 0 5", "output": "0" }, { "input": "0 5 0\n0 5 1", "output": "0" }, { "input": "0 1 1\n0 1 2", "output": "0" }, { "input": "0 1 1\n0 2 3", "output": "0" }, { "input": "2 2 -3\n2 2 -2", "output": "0" }, { "input": "3 3 3\n3 3 4", "output": "0" }, { "input": "0 1 2\n0 2 3", "output": "0" }, { "input": "0 0 1\n1 1 1", "output": "0" }, { "input": "5 0 1\n7 0 2", "output": "0" }, { "input": "4 6 1\n2 3 1", "output": "0" }, { "input": "0 0 0\n0 1 2", "output": "-1" } ]

1,572,137,199

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

32

109

0

import sys A1, B1, C1 = map(int, input().split()) A2, B2, C2 = map(int, input().split()) if ((A1 == 0 and B1 == 0) or (A2 == 0 and B2 == 0)): if ((A1 == 0 and B1 == 0 and C1 != 0) or (A2 == 0 and B2 == 0 and C2 != 0)): print(0) else: print(-1) elif (A1 * B2 == A2 * B1): if (A1 != 0): print(0) if (A1 * C2 != A2 * C1) else print(-1) else: print(0) if (B1 * C2 != B2 * C1) else print(0) else: print(1)

Title: Intersection Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: You are given two set of points. The first set is determined by the equation *A*1*x*<=+<=*B*1*y*<=+<=*C*1<==<=0, and the second one is determined by the equation *A*2*x*<=+<=*B*2*y*<=+<=*C*2<==<=0. Write the program which finds the number of points in the intersection of two given sets. Input Specification: The first line of the input contains three integer numbers *A*1,<=*B*1,<=*C*1 separated by space. The second line contains three integer numbers *A*2,<=*B*2,<=*C*2 separated by space. All the numbers are between -100 and 100, inclusive. Output Specification: Print the number of points in the intersection or -1 if there are infinite number of points. Demo Input: ['1 1 0\n2 2 0\n', '1 1 0\n2 -2 0\n'] Demo Output: ['-1\n', '1\n'] Note: none

```python import sys A1, B1, C1 = map(int, input().split()) A2, B2, C2 = map(int, input().split()) if ((A1 == 0 and B1 == 0) or (A2 == 0 and B2 == 0)): if ((A1 == 0 and B1 == 0 and C1 != 0) or (A2 == 0 and B2 == 0 and C2 != 0)): print(0) else: print(-1) elif (A1 * B2 == A2 * B1): if (A1 != 0): print(0) if (A1 * C2 != A2 * C1) else print(-1) else: print(0) if (B1 * C2 != B2 * C1) else print(0) else: print(1) ```

0

732

A

Buy a Shovel

PROGRAMMING

800

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

null

Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop. In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9). What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.

The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins". Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.

Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.

[ "117 3\n", "237 7\n", "15 2\n" ]

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

In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change. In the second example it is enough for Polycarp to buy one shovel. In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.

500

[ { "input": "117 3", "output": "9" }, { "input": "237 7", "output": "1" }, { "input": "15 2", "output": "2" }, { "input": "1 1", "output": "1" }, { "input": "1 9", "output": "9" }, { "input": "1000 3", "output": "1" }, { "input": "1000 1", "output": "1" }, { "input": "1000 9", "output": "1" }, { "input": "1 2", "output": "2" }, { "input": "999 9", "output": "1" }, { "input": "999 8", "output": "2" }, { "input": "105 6", "output": "2" }, { "input": "403 9", "output": "3" }, { "input": "546 4", "output": "4" }, { "input": "228 9", "output": "5" }, { "input": "57 2", "output": "6" }, { "input": "437 9", "output": "7" }, { "input": "997 6", "output": "8" }, { "input": "109 1", "output": "9" }, { "input": "998 9", "output": "5" }, { "input": "4 2", "output": "3" }, { "input": "9 3", "output": "7" }, { "input": "8 2", "output": "4" }, { "input": "1 3", "output": "3" }, { "input": "1 4", "output": "4" }, { "input": "1 5", "output": "5" }, { "input": "1 6", "output": "6" }, { "input": "1 7", "output": "7" }, { "input": "1 8", "output": "8" }, { "input": "100 3", "output": "1" }, { "input": "1000 2", "output": "1" }, { "input": "1000 4", "output": "1" }, { "input": "1000 5", "output": "1" }, { "input": "1000 6", "output": "1" }, { "input": "1000 7", "output": "1" }, { "input": "1000 8", "output": "1" }, { "input": "23 4", "output": "8" }, { "input": "33 1", "output": "7" }, { "input": "33 2", "output": "4" }, { "input": "666 5", "output": "5" }, { "input": "2 3", "output": "5" }, { "input": "5 5", "output": "1" }, { "input": "3 6", "output": "2" }, { "input": "12 4", "output": "2" }, { "input": "15 5", "output": "1" }, { "input": "2 5", "output": "5" }, { "input": "25 5", "output": "1" }, { "input": "2 9", "output": "5" }, { "input": "6 7", "output": "5" }, { "input": "8 9", "output": "5" }, { "input": "2 7", "output": "5" }, { "input": "4 7", "output": "5" }, { "input": "2 1", "output": "5" }, { "input": "261 1", "output": "1" } ]

1,699,638,683

2,147,483,647

Python 3

OK

TESTS

54

46

0

v=0 pm=0 k,r=map(int,input().split()) flag=True while(flag): if((pm%10==0 and pm!=0) or (pm%10==r)): break v=v+1 pm=pm+k print(v)

Title: Buy a Shovel Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop. In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9). What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel. Input Specification: The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins". Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels. Output Specification: Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change. Demo Input: ['117 3\n', '237 7\n', '15 2\n'] Demo Output: ['9\n', '1\n', '2\n'] Note: In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change. In the second example it is enough for Polycarp to buy one shovel. In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.

```python v=0 pm=0 k,r=map(int,input().split()) flag=True while(flag): if((pm%10==0 and pm!=0) or (pm%10==r)): break v=v+1 pm=pm+k print(v) ```

3

50

A

Domino piling

PROGRAMMING

800

[ "greedy", "math" ]

A. Domino piling

2

256

You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions.

In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).

Output one number — the maximal number of dominoes, which can be placed.

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

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

none

500

[ { "input": "2 4", "output": "4" }, { "input": "3 3", "output": "4" }, { "input": "1 5", "output": "2" }, { "input": "1 6", "output": "3" }, { "input": "1 15", "output": "7" }, { "input": "1 16", "output": "8" }, { "input": "2 5", "output": "5" }, { "input": "2 6", "output": "6" }, { "input": "2 7", "output": "7" }, { "input": "2 14", "output": "14" }, { "input": "2 15", "output": "15" }, { "input": "1 4", "output": "2" }, { "input": "2 16", "output": "16" }, { "input": "3 5", "output": "7" }, { "input": "3 6", "output": "9" }, { "input": "3 10", "output": "15" }, { "input": "3 14", "output": "21" }, { "input": "3 15", "output": "22" }, { "input": "3 16", "output": "24" }, { "input": "5 7", "output": "17" }, { "input": "16 16", "output": "128" }, { "input": "15 16", "output": "120" }, { "input": "2 3", "output": "3" }, { "input": "15 15", "output": "112" }, { "input": "14 16", "output": "112" }, { "input": "11 13", "output": "71" }, { "input": "5 16", "output": "40" }, { "input": "8 15", "output": "60" }, { "input": "2 2", "output": "2" }, { "input": "3 4", "output": "6" }, { "input": "4 4", "output": "8" }, { "input": "1 1", "output": "0" }, { "input": "1 2", "output": "1" }, { "input": "1 3", "output": "1" }, { "input": "14 15", "output": "105" } ]

1,611,633,618

2,147,483,647

Python 3

OK

TESTS

35

124

0

x = list(map(int, input().strip().split(" "))) print(int((x[0]*x[1])/2))

Title: Domino piling Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions. Input Specification: In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). Output Specification: Output one number — the maximal number of dominoes, which can be placed. Demo Input: ['2 4\n', '3 3\n'] Demo Output: ['4\n', '4\n'] Note: none

```python x = list(map(int, input().strip().split(" "))) print(int((x[0]*x[1])/2)) ```

3.969

222

A

Shooshuns and Sequence

PROGRAMMING

1,200

[ "brute force", "implementation" ]

null

One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps: 1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence. The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.

The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found.

Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.

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

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

In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one. In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.

500

[ { "input": "3 2\n3 1 1", "output": "1" }, { "input": "3 1\n3 1 1", "output": "-1" }, { "input": "1 1\n1", "output": "0" }, { "input": "2 1\n1 1", "output": "0" }, { "input": "2 1\n2 1", "output": "-1" }, { "input": "4 4\n1 2 3 4", "output": "3" }, { "input": "4 3\n1 2 3 4", "output": "-1" }, { "input": "5 3\n2 1 1 1 1", "output": "1" }, { "input": "5 3\n1 1 1 1 1", "output": "0" }, { "input": "5 5\n1 1 1 1 1", "output": "0" }, { "input": "10 1\n1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "10 1\n1 2 1 1 1 1 1 1 1 1", "output": "-1" }, { "input": "10 1\n2 1 1 1 1 1 1 1 1 1", "output": "-1" }, { "input": "10 2\n2 1 1 1 1 1 1 1 1 1", "output": "1" }, { "input": "10 2\n1 2 1 1 1 1 1 1 1 1", "output": "-1" }, { "input": "10 3\n3 2 1 1 1 1 1 1 1 1", "output": "2" }, { "input": "10 1\n1 2 3 4 5 6 7 8 9 10", "output": "-1" }, { "input": "10 10\n1 2 3 4 5 6 7 8 9 10", "output": "9" }, { "input": "20 10\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1", "output": "-1" }, { "input": "20 11\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1", "output": "10" }, { "input": "20 9\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1", "output": "-1" }, { "input": "20 10\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "20 20\n20 1 19 2 18 3 17 4 16 5 15 6 14 7 13 8 12 9 11 10", "output": "19" } ]

1,540,481,385

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

218

0

n, k = map(int, input().split()) nums = list(map(int, input().split())) ans = -1 for i in range(n): nums = nums[1:] + [nums[k - 1]] if nums.count(nums[0]) == n: ans = i + 1 break print(ans)

Title: Shooshuns and Sequence Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps: 1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence. The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same. Input Specification: The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found. Output Specification: Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1. Demo Input: ['3 2\n3 1 1\n', '3 1\n3 1 1\n'] Demo Output: ['1\n', '-1\n'] Note: In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one. In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.

```python n, k = map(int, input().split()) nums = list(map(int, input().split())) ans = -1 for i in range(n): nums = nums[1:] + [nums[k - 1]] if nums.count(nums[0]) == n: ans = i + 1 break print(ans) ```

0

none

0

[ "none" ]

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.

0

[ { "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 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"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 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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" } ]

1,689,158,578

2,147,483,647

PyPy 3-64

OK

TESTS

54

186

13,209,600

n=int(input()) if n%2==0: print(-1) else: print(*range(n)) print(*range(n)) print(*[(2*i)%n for i in range(n)])

Title: none 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==0: print(-1) else: print(*range(n)) print(*range(n)) print(*[(2*i)%n for i in range(n)]) ```

3

520

B

Two Buttons

PROGRAMMING

1,400

[ "dfs and similar", "graphs", "greedy", "implementation", "math", "shortest paths" ]

null

Vasya has found a strange device. On the front panel of a device there are: a red button, a blue button and a display showing some positive integer. After clicking the red button, device multiplies the displayed number by two. After clicking the blue button, device subtracts one from the number on the display. If at some point the number stops being positive, the device breaks down. The display can show arbitrarily large numbers. Initially, the display shows number *n*. Bob wants to get number *m* on the display. What minimum number of clicks he has to make in order to achieve this result?

The first and the only line of the input contains two distinct integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=104), separated by a space .

Print a single number — the minimum number of times one needs to push the button required to get the number *m* out of number *n*.

[ "4 6\n", "10 1\n" ]

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

In the first example you need to push the blue button once, and then push the red button once. In the second example, doubling the number is unnecessary, so we need to push the blue button nine times.

1,000

[ { "input": "4 6", "output": "2" }, { "input": "10 1", "output": "9" }, { "input": "1 2", "output": "1" }, { "input": "2 1", "output": "1" }, { "input": "1 3", "output": "3" }, { "input": "3 1", "output": "2" }, { "input": "2 10", "output": "5" }, { "input": "100 99", "output": "1" }, { "input": "99 100", "output": "50" }, { "input": "10 17", "output": "3" }, { "input": "666 6666", "output": "255" }, { "input": "6666 666", "output": "6000" }, { "input": "1 8192", "output": "13" }, { "input": "1 8193", "output": "27" }, { "input": "9999 10000", "output": "5000" }, { "input": "10000 9999", "output": "1" }, { "input": "10000 1", "output": "9999" }, { "input": "1234 5678", "output": "528" }, { "input": "9102 9103", "output": "4552" }, { "input": "8192 1", "output": "8191" }, { "input": "9912 1023", "output": "8889" }, { "input": "100 500", "output": "41" }, { "input": "9997 9999", "output": "4999" }, { "input": "5000 10000", "output": "1" }, { "input": "4000 7997", "output": "3" }, { "input": "1 10000", "output": "20" }, { "input": "1 8191", "output": "14" }, { "input": "9999 10000", "output": "5000" }, { "input": "9097 9998", "output": "4099" }, { "input": "886 9383", "output": "305" }, { "input": "1 9", "output": "7" }, { "input": "1918 10000", "output": "671" }, { "input": "1 3", "output": "3" }, { "input": "1 4", "output": "2" }, { "input": "2 10000", "output": "19" }, { "input": "3 10000", "output": "17" }, { "input": "4 10000", "output": "18" }, { "input": "9998 10000", "output": "4999" }, { "input": "5000 10000", "output": "1" }, { "input": "5001 10000", "output": "2" }, { "input": "10000 1", "output": "9999" }, { "input": "1 9999", "output": "21" }, { "input": "7777 9999", "output": "2779" }, { "input": "2 100", "output": "9" }, { "input": "10 8722", "output": "18" }, { "input": "848 4561", "output": "283" }, { "input": "9967 9973", "output": "4982" }, { "input": "5555 10000", "output": "556" }, { "input": "999 10000", "output": "378" }, { "input": "3 6", "output": "1" }, { "input": "9 38", "output": "8" } ]

1,700,122,751

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

4

31

0

n, m = map(int, input().split()) if n > m: print(n - m) exit() k1 = n - (m // 2) + 1 + m % 2 k2 = n while n < m: n *= 2 k2 += n - m print(min(k1, k2))

Title: Two Buttons Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya has found a strange device. On the front panel of a device there are: a red button, a blue button and a display showing some positive integer. After clicking the red button, device multiplies the displayed number by two. After clicking the blue button, device subtracts one from the number on the display. If at some point the number stops being positive, the device breaks down. The display can show arbitrarily large numbers. Initially, the display shows number *n*. Bob wants to get number *m* on the display. What minimum number of clicks he has to make in order to achieve this result? Input Specification: The first and the only line of the input contains two distinct integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=104), separated by a space . Output Specification: Print a single number — the minimum number of times one needs to push the button required to get the number *m* out of number *n*. Demo Input: ['4 6\n', '10 1\n'] Demo Output: ['2\n', '9\n'] Note: In the first example you need to push the blue button once, and then push the red button once. In the second example, doubling the number is unnecessary, so we need to push the blue button nine times.

```python n, m = map(int, input().split()) if n > m: print(n - m) exit() k1 = n - (m // 2) + 1 + m % 2 k2 = n while n < m: n *= 2 k2 += n - m print(min(k1, k2)) ```

0

888

D

Almost Identity Permutations

PROGRAMMING

1,600

[ "combinatorics", "dp", "math" ]

null

A permutation *p* of size *n* is an array such that every integer from 1 to *n* occurs exactly once in this array. Let's call a permutation an almost identity permutation iff there exist at least *n*<=-<=*k* indices *i* (1<=≤<=*i*<=≤<=*n*) such that *p**i*<==<=*i*. Your task is to count the number of almost identity permutations for given numbers *n* and *k*.

The first line contains two integers *n* and *k* (4<=≤<=*n*<=≤<=1000, 1<=≤<=*k*<=≤<=4).

Print the number of almost identity permutations for given *n* and *k*.

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

[ "1\n", "7\n", "31\n", "76\n" ]

none

0

[ { "input": "4 1", "output": "1" }, { "input": "4 2", "output": "7" }, { "input": "5 3", "output": "31" }, { "input": "5 4", "output": "76" }, { "input": "200 1", "output": "1" }, { "input": "200 2", "output": "19901" }, { "input": "200 3", "output": "2646701" }, { "input": "200 4", "output": "584811251" }, { "input": "400 1", "output": "1" }, { "input": "400 2", "output": "79801" }, { "input": "400 3", "output": "21253401" }, { "input": "400 4", "output": "9477912501" }, { "input": "600 1", "output": "1" }, { "input": "600 2", "output": "179701" }, { "input": "600 3", "output": "71820101" }, { "input": "600 4", "output": "48187303751" }, { "input": "800 1", "output": "1" }, { "input": "800 2", "output": "319601" }, { "input": "800 3", "output": "170346801" }, { "input": "800 4", "output": "152620985001" }, { "input": "1000 1", "output": "1" }, { "input": "1000 2", "output": "499501" }, { "input": "1000 3", "output": "332833501" }, { "input": "1000 4", "output": "373086956251" }, { "input": "4 4", "output": "24" } ]

1,663,361,763

2,147,483,647

PyPy 3-64

OK

TESTS

25

62

0

## Fast I/O import io,os,sys # Fast input. Use s = input().decode() for strings # input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline # Fast output def print(*args, sep = ' ', end = '\n'): string = sep.join(map(str, args))+end sys.stdout.write(string) def debug(*args, sep = ' ', end = '\n'): string = "Debug: " + sep.join(map(str, args)) + end sys.stderr.write(string) # from collections import Counter, defaultdict # from functools import lru_cache # from math import floor, ceil, sqrt, gcd # from sys import stderr def facto(n): if n <= 1: return 1 return n * facto(n-1) def D(n): if n == 0: return 1 elif n&1: return n * D(n - 1) - 1 return n * D(n - 1) + 1 def A(k, n): # Number of arrangments for a small k r = 1 for i in range(n, n-k, -1): r *= i return r def binom(k, n): # n choose k for a small k return A(k, n) // facto(k) def exact(k, n): return D(k) * binom(k, n) N, K = map(int, input().split()) ans = 0 for k in range(K+1): ans += exact(k, N) print(ans) # # @PoustouFlan # Code:Choke?! # [ 极客 ] #

Title: Almost Identity Permutations Time Limit: None seconds Memory Limit: None megabytes Problem Description: A permutation *p* of size *n* is an array such that every integer from 1 to *n* occurs exactly once in this array. Let's call a permutation an almost identity permutation iff there exist at least *n*<=-<=*k* indices *i* (1<=≤<=*i*<=≤<=*n*) such that *p**i*<==<=*i*. Your task is to count the number of almost identity permutations for given numbers *n* and *k*. Input Specification: The first line contains two integers *n* and *k* (4<=≤<=*n*<=≤<=1000, 1<=≤<=*k*<=≤<=4). Output Specification: Print the number of almost identity permutations for given *n* and *k*. Demo Input: ['4 1\n', '4 2\n', '5 3\n', '5 4\n'] Demo Output: ['1\n', '7\n', '31\n', '76\n'] Note: none

```python ## Fast I/O import io,os,sys # Fast input. Use s = input().decode() for strings # input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline # Fast output def print(*args, sep = ' ', end = '\n'): string = sep.join(map(str, args))+end sys.stdout.write(string) def debug(*args, sep = ' ', end = '\n'): string = "Debug: " + sep.join(map(str, args)) + end sys.stderr.write(string) # from collections import Counter, defaultdict # from functools import lru_cache # from math import floor, ceil, sqrt, gcd # from sys import stderr def facto(n): if n <= 1: return 1 return n * facto(n-1) def D(n): if n == 0: return 1 elif n&1: return n * D(n - 1) - 1 return n * D(n - 1) + 1 def A(k, n): # Number of arrangments for a small k r = 1 for i in range(n, n-k, -1): r *= i return r def binom(k, n): # n choose k for a small k return A(k, n) // facto(k) def exact(k, n): return D(k) * binom(k, n) N, K = map(int, input().split()) ans = 0 for k in range(K+1): ans += exact(k, N) print(ans) # # @PoustouFlan # Code:Choke?! # [ 极客 ] # ```

3

6

C

Alice, Bob and Chocolate

PROGRAMMING

1,200

[ "greedy", "two pointers" ]

C. Alice, Bob and Chocolate

2

64

Alice and Bob like games. And now they are ready to start a new game. They have placed *n* chocolate bars in a line. Alice starts to eat chocolate bars one by one from left to right, and Bob — from right to left. For each chocololate bar the time, needed for the player to consume it, is known (Alice and Bob eat them with equal speed). When the player consumes a chocolate bar, he immediately starts with another. It is not allowed to eat two chocolate bars at the same time, to leave the bar unfinished and to make pauses. If both players start to eat the same bar simultaneously, Bob leaves it to Alice as a true gentleman. How many bars each of the players will consume?

The first line contains one integer *n* (1<=≤<=*n*<=≤<=105) — the amount of bars on the table. The second line contains a sequence *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000), where *t**i* is the time (in seconds) needed to consume the *i*-th bar (in the order from left to right).

Print two numbers *a* and *b*, where *a* is the amount of bars consumed by Alice, and *b* is the amount of bars consumed by Bob.

[ "5\n2 9 8 2 7\n" ]

[ "2 3\n" ]

none

0

[ { "input": "5\n2 9 8 2 7", "output": "2 3" }, { "input": "1\n1", "output": "1 0" }, { "input": "2\n1 1", "output": "1 1" }, { "input": "3\n1 1 1", "output": "2 1" }, { "input": "4\n1 1 1 1", "output": "2 2" }, { "input": "1\n6", "output": "1 0" }, { "input": "2\n8 2", "output": "1 1" }, { "input": "3\n6 8 4", "output": "1 2" }, { "input": "4\n4 3 2 8", "output": "3 1" }, { "input": "5\n1 9 8 2 7", "output": "2 3" }, { "input": "6\n1 4 7 3 2 1", "output": "3 3" }, { "input": "11\n7 9 1 8 7 1 7 1 4 8 1", "output": "4 7" }, { "input": "12\n4 2 7 1 9 5 6 2 7 8 5 7", "output": "7 5" }, { "input": "13\n9 9 9 9 9 9 9 9 9 9 9 9 9", "output": "7 6" } ]

1,586,014,604

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

0

248

0

n = int(input()) c = [int(i) for i in input().split()] a = 0 b = n-1 ta = c[0] tb = c[-1] while a < b - 1: if ta <= tb: ta += c[a] a += 1 else: tb += c[b] b -= 1 print(a+1,n-b)

Title: Alice, Bob and Chocolate Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: Alice and Bob like games. And now they are ready to start a new game. They have placed *n* chocolate bars in a line. Alice starts to eat chocolate bars one by one from left to right, and Bob — from right to left. For each chocololate bar the time, needed for the player to consume it, is known (Alice and Bob eat them with equal speed). When the player consumes a chocolate bar, he immediately starts with another. It is not allowed to eat two chocolate bars at the same time, to leave the bar unfinished and to make pauses. If both players start to eat the same bar simultaneously, Bob leaves it to Alice as a true gentleman. How many bars each of the players will consume? Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=105) — the amount of bars on the table. The second line contains a sequence *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000), where *t**i* is the time (in seconds) needed to consume the *i*-th bar (in the order from left to right). Output Specification: Print two numbers *a* and *b*, where *a* is the amount of bars consumed by Alice, and *b* is the amount of bars consumed by Bob. Demo Input: ['5\n2 9 8 2 7\n'] Demo Output: ['2 3\n'] Note: none

```python n = int(input()) c = [int(i) for i in input().split()] a = 0 b = n-1 ta = c[0] tb = c[-1] while a < b - 1: if ta <= tb: ta += c[a] a += 1 else: tb += c[b] b -= 1 print(a+1,n-b) ```

0

873

B

Balanced Substring

PROGRAMMING

1,500

[ "dp", "implementation" ]

null

You are given a string *s* consisting only of characters 0 and 1. A substring [*l*,<=*r*] of *s* is a string *s**l**s**l*<=+<=1*s**l*<=+<=2... *s**r*, and its length equals to *r*<=-<=*l*<=+<=1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring. You have to determine the length of the longest balanced substring of *s*.

The first line contains *n* (1<=≤<=*n*<=≤<=100000) — the number of characters in *s*. The second line contains a string *s* consisting of exactly *n* characters. Only characters 0 and 1 can appear in *s*.

If there is no non-empty balanced substring in *s*, print 0. Otherwise, print the length of the longest balanced substring.

[ "8\n11010111\n", "3\n111\n" ]

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

In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible. In the second example it's impossible to find a non-empty balanced substring.

0

[ { "input": "8\n11010111", "output": "4" }, { "input": "3\n111", "output": "0" }, { "input": "11\n00001000100", "output": "2" }, { "input": "10\n0100000000", "output": "2" }, { "input": "13\n0001000011010", "output": "6" }, { "input": "14\n00000100101011", "output": "10" }, { "input": "14\n01111101111111", "output": "2" }, { "input": "18\n110010101101111111", "output": "10" }, { "input": "11\n00010000011", "output": "4" }, { "input": "10\n1000010110", "output": "6" }, { "input": "15\n100000100000011", "output": "4" }, { "input": "18\n011010101110111101", "output": "8" }, { "input": "10\n0011011111", "output": "6" }, { "input": "3\n011", "output": "2" }, { "input": "14\n11111111111111", "output": "0" }, { "input": "65\n11001000110001001011011110111100000000101001001010101111000100010", "output": "48" }, { "input": "10\n0011111000", "output": "10" }, { "input": "13\n1110000001110", "output": "12" }, { "input": "3\n110", "output": "2" }, { "input": "4\n1110", "output": "2" }, { "input": "9\n001011001", "output": "8" }, { "input": "2\n10", "output": "2" }, { "input": "2\n01", "output": "2" }, { "input": "12\n110000110000", "output": "8" }, { "input": "3\n001", "output": "2" }, { "input": "14\n11000011000000", "output": "8" }, { "input": "19\n0000011111111100000", "output": "18" }, { "input": "45\n011010001100001111110001011100000001101100111", "output": "44" }, { "input": "18\n000001111111100000", "output": "16" }, { "input": "4\n0101", "output": "4" }, { "input": "12\n000101011001", "output": "10" } ]

1,507,863,045

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

15

218

716,800

from sys import stdin one = "1" zero = "0" def is_ballanced(s: str): return s.count(one) == s.count(zero) cache = {} def longest_ballanced_len(l, r, s): if l == r: return 0 if r - l + 1 == 2: if s[l] == s[r]: return 0 else: return 2 if (l, r) in cache: return cache[l, r] if is_ballanced(s[l:r + 1]): length = r - l + 1 cache[l, r] = length return length l1 = longest_ballanced_len(l + 1, r, s) l2 = longest_ballanced_len(l, r - 1, s) length = max(l1, l2) cache[l, r] = length return length def main(): stdin.readline() s = stdin.readline().strip() print(longest_ballanced_len(0, len(s) - 1, s)) if __name__ == "__main__": main()

Title: Balanced Substring Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a string *s* consisting only of characters 0 and 1. A substring [*l*,<=*r*] of *s* is a string *s**l**s**l*<=+<=1*s**l*<=+<=2... *s**r*, and its length equals to *r*<=-<=*l*<=+<=1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring. You have to determine the length of the longest balanced substring of *s*. Input Specification: The first line contains *n* (1<=≤<=*n*<=≤<=100000) — the number of characters in *s*. The second line contains a string *s* consisting of exactly *n* characters. Only characters 0 and 1 can appear in *s*. Output Specification: If there is no non-empty balanced substring in *s*, print 0. Otherwise, print the length of the longest balanced substring. Demo Input: ['8\n11010111\n', '3\n111\n'] Demo Output: ['4\n', '0\n'] Note: In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible. In the second example it's impossible to find a non-empty balanced substring.

```python from sys import stdin one = "1" zero = "0" def is_ballanced(s: str): return s.count(one) == s.count(zero) cache = {} def longest_ballanced_len(l, r, s): if l == r: return 0 if r - l + 1 == 2: if s[l] == s[r]: return 0 else: return 2 if (l, r) in cache: return cache[l, r] if is_ballanced(s[l:r + 1]): length = r - l + 1 cache[l, r] = length return length l1 = longest_ballanced_len(l + 1, r, s) l2 = longest_ballanced_len(l, r - 1, s) length = max(l1, l2) cache[l, r] = length return length def main(): stdin.readline() s = stdin.readline().strip() print(longest_ballanced_len(0, len(s) - 1, s)) if __name__ == "__main__": main() ```

-1

375

A

Divisible by Seven

PROGRAMMING

1,600

[ "math", "number theory" ]

null

You have number *a*, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7. Number *a* doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.

The first line contains positive integer *a* in the decimal record. It is guaranteed that the record of number *a* contains digits: 1, 6, 8, 9. Number *a* doesn't contain any leading zeroes. The decimal representation of number *a* contains at least 4 and at most 106 characters.

Print a number in the decimal notation without leading zeroes — the result of the permutation. If it is impossible to rearrange the digits of the number *a* in the required manner, print 0.

[ "1689\n", "18906\n" ]

[ "1869\n", "18690\n" ]

none

500

[ { "input": "1689", "output": "1869" }, { "input": "18906", "output": "18690" }, { "input": "2419323689", "output": "2432391689" }, { "input": "8589157262", "output": "5857221986" }, { "input": "2717172350336955863014903670481525170997949309274087058935108848979319747543008692128164875210350026", "output": "2771723503355630149036704815251709979493092740870589351088489793197475430086921281648752103500261986" }, { "input": "9825995656040286793128006047268547610068699214477842995873286607346639816314908021369221299622234988", "output": "2599556040286793280060472685476100686992144778429958732866073466398163149080213692212996222349881968" }, { "input": "100000000689", "output": "186900000000" }, { "input": "16891", "output": "16198" }, { "input": "16892", "output": "21896" }, { "input": "16893", "output": "31689" }, { "input": "16894", "output": "41986" }, { "input": "16895", "output": "51968" }, { "input": "16896", "output": "61698" }, { "input": "16897", "output": "71869" }, { "input": "16898", "output": "86198" }, { "input": "16899", "output": "91896" }, { "input": "4048169", "output": "4041968" }, { "input": "10994168", "output": "94116890" }, { "input": "168903", "output": "316890" }, { "input": "11689", "output": "16198" }, { "input": "91111168", "output": "11111968" }, { "input": "16890000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "18690000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "1689999999999", "output": "9999999991968" }, { "input": "9883291673084", "output": "8329730841698" }, { "input": "6198", "output": "1869" } ]

1,611,899,215

2,147,483,647

PyPy 3

TIME_LIMIT_EXCEEDED

TESTS

2

1,000

119,705,600

from itertools import permutations # Main function def find_gcd_permutations_of_digit(given_number): """ Returns the gcd of all numbers that can be obtained by permuting the digits of given_number """ s = str(given_number) perms = permutations(s) num_list = [] for p in perms: if p[0] == '0': continue num_list.append(int(''.join(p))) for n in num_list: if(n%7==0): print(n) return print(0) return num = int(input()) find_gcd_permutations_of_digit(num)

Title: Divisible by Seven Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have number *a*, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7. Number *a* doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes. Input Specification: The first line contains positive integer *a* in the decimal record. It is guaranteed that the record of number *a* contains digits: 1, 6, 8, 9. Number *a* doesn't contain any leading zeroes. The decimal representation of number *a* contains at least 4 and at most 106 characters. Output Specification: Print a number in the decimal notation without leading zeroes — the result of the permutation. If it is impossible to rearrange the digits of the number *a* in the required manner, print 0. Demo Input: ['1689\n', '18906\n'] Demo Output: ['1869\n', '18690\n'] Note: none

```python from itertools import permutations # Main function def find_gcd_permutations_of_digit(given_number): """ Returns the gcd of all numbers that can be obtained by permuting the digits of given_number """ s = str(given_number) perms = permutations(s) num_list = [] for p in perms: if p[0] == '0': continue num_list.append(int(''.join(p))) for n in num_list: if(n%7==0): print(n) return print(0) return num = int(input()) find_gcd_permutations_of_digit(num) ```

0

485

A

Factory

PROGRAMMING

1,400

[ "implementation", "math", "matrices" ]

null

One industrial factory is reforming working plan. The director suggested to set a mythical detail production norm. If at the beginning of the day there were *x* details in the factory storage, then by the end of the day the factory has to produce (remainder after dividing *x* by *m*) more details. Unfortunately, no customer has ever bought any mythical detail, so all the details produced stay on the factory. The board of directors are worried that the production by the given plan may eventually stop (that means that there will be а moment when the current number of details on the factory is divisible by *m*). Given the number of details *a* on the first day and number *m* check if the production stops at some moment.

The first line contains two integers *a* and *m* (1<=≤<=*a*,<=*m*<=≤<=105).

Print "Yes" (without quotes) if the production will eventually stop, otherwise print "No".

[ "1 5\n", "3 6\n" ]

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

none

1,000

[ { "input": "1 5", "output": "No" }, { "input": "3 6", "output": "Yes" }, { "input": "1 8", "output": "Yes" }, { "input": "2 3", "output": "No" }, { "input": "3 24", "output": "Yes" }, { "input": "1 1", "output": "Yes" }, { "input": "100000 100000", "output": "Yes" }, { "input": "1 99989", "output": "No" }, { "input": "512 2", "output": "Yes" }, { "input": "100 24", "output": "No" }, { "input": "1 100000", "output": "No" }, { "input": "100000 1", "output": "Yes" }, { "input": "3 99929", "output": "No" }, { "input": "99961 99971", "output": "No" }, { "input": "1 65536", "output": "Yes" }, { "input": "4 65536", "output": "Yes" }, { "input": "3 65536", "output": "Yes" }, { "input": "32768 65536", "output": "Yes" }, { "input": "65535 65536", "output": "Yes" }, { "input": "1 65535", "output": "No" }, { "input": "98812 100000", "output": "No" }, { "input": "10 5", "output": "Yes" }, { "input": "6 8", "output": "Yes" } ]

1,596,716,784

2,147,483,647

Python 3

COMPILATION_ERROR

TESTS

0

a,m=map(int,input().split()) ans="No" if a%m==0: ans="Yes" break else: a+=a%m print(ans)

Title: Factory Time Limit: None seconds Memory Limit: None megabytes Problem Description: One industrial factory is reforming working plan. The director suggested to set a mythical detail production norm. If at the beginning of the day there were *x* details in the factory storage, then by the end of the day the factory has to produce (remainder after dividing *x* by *m*) more details. Unfortunately, no customer has ever bought any mythical detail, so all the details produced stay on the factory. The board of directors are worried that the production by the given plan may eventually stop (that means that there will be а moment when the current number of details on the factory is divisible by *m*). Given the number of details *a* on the first day and number *m* check if the production stops at some moment. Input Specification: The first line contains two integers *a* and *m* (1<=≤<=*a*,<=*m*<=≤<=105). Output Specification: Print "Yes" (without quotes) if the production will eventually stop, otherwise print "No". Demo Input: ['1 5\n', '3 6\n'] Demo Output: ['No\n', 'Yes\n'] Note: none

```python a,m=map(int,input().split()) ans="No" if a%m==0: ans="Yes" break else: a+=a%m print(ans) ```

-1

585

B

Phillip and Trains

PROGRAMMING

1,700

[ "dfs and similar", "graphs", "shortest paths" ]

null

The mobile application store has a new game called "Subway Roller". The protagonist of the game Philip is located in one end of the tunnel and wants to get out of the other one. The tunnel is a rectangular field consisting of three rows and *n* columns. At the beginning of the game the hero is in some cell of the leftmost column. Some number of trains rides towards the hero. Each train consists of two or more neighbouring cells in some row of the field. All trains are moving from right to left at a speed of two cells per second, and the hero runs from left to right at the speed of one cell per second. For simplicity, the game is implemented so that the hero and the trains move in turns. First, the hero moves one cell to the right, then one square up or down, or stays idle. Then all the trains move twice simultaneously one cell to the left. Thus, in one move, Philip definitely makes a move to the right and can move up or down. If at any point, Philip is in the same cell with a train, he loses. If the train reaches the left column, it continues to move as before, leaving the tunnel. Your task is to answer the question whether there is a sequence of movements of Philip, such that he would be able to get to the rightmost column.

Each test contains from one to ten sets of the input data. The first line of the test contains a single integer *t* (1<=≤<=*t*<=≤<=10 for pretests and tests or *t*<==<=1 for hacks; see the Notes section for details) — the number of sets. Then follows the description of *t* sets of the input data. The first line of the description of each set contains two integers *n*,<=*k* (2<=≤<=*n*<=≤<=100,<=1<=≤<=*k*<=≤<=26) — the number of columns on the field and the number of trains. Each of the following three lines contains the sequence of *n* character, representing the row of the field where the game is on. Philip's initial position is marked as 's', he is in the leftmost column. Each of the *k* trains is marked by some sequence of identical uppercase letters of the English alphabet, located in one line. Distinct trains are represented by distinct letters. Character '.' represents an empty cell, that is, the cell that doesn't contain either Philip or the trains.

For each set of the input data print on a single line word YES, if it is possible to win the game and word NO otherwise.

[ "2\n16 4\n...AAAAA........\ns.BBB......CCCCC\n........DDDDD...\n16 4\n...AAAAA........\ns.BBB....CCCCC..\n.......DDDDD....\n", "2\n10 4\ns.ZZ......\n.....AAABB\n.YYYYYY...\n10 4\ns.ZZ......\n....AAAABB\n.YYYYYY...\n" ]

[ "YES\nNO\n", "YES\nNO\n" ]

In the first set of the input of the first sample Philip must first go forward and go down to the third row of the field, then go only forward, then go forward and climb to the second row, go forward again and go up to the first row. After that way no train blocks Philip's path, so he can go straight to the end of the tunnel. Note that in this problem the challenges are restricted to tests that contain only one testset.

750

[]

1,468,024,312

2,147,483,647

Python 3

TIME_LIMIT_EXCEEDED

TESTS

2

1,000

0

t = int(input()) def dfs(x, y, n, l): #print (x, y) if y < 0 or y > 2: return 0 if x >= n-1: return 1 for i in range(4): if l[y][x - 2 + i] != '.': return 0 a = dfs(x + 3, y - 1, n, l) b = dfs(x + 3, y, n, l) c = dfs(x + 3, y + 1, n, l) return a or b or c for x in range(t): l = [] n, k = map(int, input().split()) for i in range(3): s = input() l.append(s) if dfs(3,1,n,l) or dfs(3,0,n,l) or dfs(3,2,n,l): print ('YES') else: print ('NO')

Title: Phillip and Trains Time Limit: None seconds Memory Limit: None megabytes Problem Description: The mobile application store has a new game called "Subway Roller". The protagonist of the game Philip is located in one end of the tunnel and wants to get out of the other one. The tunnel is a rectangular field consisting of three rows and *n* columns. At the beginning of the game the hero is in some cell of the leftmost column. Some number of trains rides towards the hero. Each train consists of two or more neighbouring cells in some row of the field. All trains are moving from right to left at a speed of two cells per second, and the hero runs from left to right at the speed of one cell per second. For simplicity, the game is implemented so that the hero and the trains move in turns. First, the hero moves one cell to the right, then one square up or down, or stays idle. Then all the trains move twice simultaneously one cell to the left. Thus, in one move, Philip definitely makes a move to the right and can move up or down. If at any point, Philip is in the same cell with a train, he loses. If the train reaches the left column, it continues to move as before, leaving the tunnel. Your task is to answer the question whether there is a sequence of movements of Philip, such that he would be able to get to the rightmost column. Input Specification: Each test contains from one to ten sets of the input data. The first line of the test contains a single integer *t* (1<=≤<=*t*<=≤<=10 for pretests and tests or *t*<==<=1 for hacks; see the Notes section for details) — the number of sets. Then follows the description of *t* sets of the input data. The first line of the description of each set contains two integers *n*,<=*k* (2<=≤<=*n*<=≤<=100,<=1<=≤<=*k*<=≤<=26) — the number of columns on the field and the number of trains. Each of the following three lines contains the sequence of *n* character, representing the row of the field where the game is on. Philip's initial position is marked as 's', he is in the leftmost column. Each of the *k* trains is marked by some sequence of identical uppercase letters of the English alphabet, located in one line. Distinct trains are represented by distinct letters. Character '.' represents an empty cell, that is, the cell that doesn't contain either Philip or the trains. Output Specification: For each set of the input data print on a single line word YES, if it is possible to win the game and word NO otherwise. Demo Input: ['2\n16 4\n...AAAAA........\ns.BBB......CCCCC\n........DDDDD...\n16 4\n...AAAAA........\ns.BBB....CCCCC..\n.......DDDDD....\n', '2\n10 4\ns.ZZ......\n.....AAABB\n.YYYYYY...\n10 4\ns.ZZ......\n....AAAABB\n.YYYYYY...\n'] Demo Output: ['YES\nNO\n', 'YES\nNO\n'] Note: In the first set of the input of the first sample Philip must first go forward and go down to the third row of the field, then go only forward, then go forward and climb to the second row, go forward again and go up to the first row. After that way no train blocks Philip's path, so he can go straight to the end of the tunnel. Note that in this problem the challenges are restricted to tests that contain only one testset.

```python t = int(input()) def dfs(x, y, n, l): #print (x, y) if y < 0 or y > 2: return 0 if x >= n-1: return 1 for i in range(4): if l[y][x - 2 + i] != '.': return 0 a = dfs(x + 3, y - 1, n, l) b = dfs(x + 3, y, n, l) c = dfs(x + 3, y + 1, n, l) return a or b or c for x in range(t): l = [] n, k = map(int, input().split()) for i in range(3): s = input() l.append(s) if dfs(3,1,n,l) or dfs(3,0,n,l) or dfs(3,2,n,l): print ('YES') else: print ('NO') ```

0

43

A

Football

PROGRAMMING

1,000

[ "strings" ]

A. Football

2

256

One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.

Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.

[ "1\nABC\n", "5\nA\nABA\nABA\nA\nA\n" ]

[ "ABC\n", "A\n" ]

none

500

[ { "input": "1\nABC", "output": "ABC" }, { "input": "5\nA\nABA\nABA\nA\nA", "output": "A" }, { "input": "2\nXTSJEP\nXTSJEP", "output": "XTSJEP" }, { "input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ", "output": "XZYDJAEDZ" }, { "input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD", "output": "QCCYXL" }, { "input": "3\nAZID\nEERWBC\nEERWBC", "output": "EERWBC" }, { "input": "3\nHNCGYL\nHNCGYL\nHNCGYL", "output": "HNCGYL" }, { "input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG", "output": "ZZWZTG" }, { "input": "4\nA\nA\nKUDLJMXCSE\nA", "output": "A" }, { "input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW", "output": "PHBTW" }, { "input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN", "output": "PKUZYTFYWN" }, { "input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH", "output": "HH" }, { "input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW", "output": "W" }, { "input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP", "output": "XBCP" }, { "input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS", "output": "PYPAS" }, { "input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC", "output": "NC" }, { "input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI", "output": "VOCI" }, { "input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA", "output": "HA" }, { "input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS", "output": "G" }, { "input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL", "output": "WL" } ]

1,637,454,060

2,147,483,647

Python 3

OK

TESTS

34

124

102,400

from collections import defaultdict n = int(input()) r = [] for _ in range(n): r.append(input()) dr = defaultdict(int) for i in r: dr[i] += 1 print(max(dr.items(), key=lambda a: a[1])[0])

Title: Football Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams. Output Specification: Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner. Demo Input: ['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n'] Demo Output: ['ABC\n', 'A\n'] Note: none

```python from collections import defaultdict n = int(input()) r = [] for _ in range(n): r.append(input()) dr = defaultdict(int) for i in r: dr[i] += 1 print(max(dr.items(), key=lambda a: a[1])[0]) ```

3.968809

258

A

Little Elephant and Bits

PROGRAMMING

1,100

[ "greedy", "math" ]

null

The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper. To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes). The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.

The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.

In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.

[ "101\n", "110010\n" ]

[ "11\n", "11010\n" ]

In the first sample the best strategy is to delete the second digit. That results in number 112 = 310. In the second sample the best strategy is to delete the third or fourth digits — that results in number 110102 = 2610.

500

[ { "input": "101", "output": "11" }, { "input": "110010", "output": "11010" }, { "input": "10000", "output": "1000" }, { "input": "1111111110", "output": "111111111" }, { "input": "10100101011110101", "output": "1100101011110101" }, { "input": "111010010111", "output": "11110010111" }, { "input": "11110111011100000000", "output": "1111111011100000000" }, { "input": "11110010010100001110110101110011110110100111101", "output": "1111010010100001110110101110011110110100111101" }, { "input": "1001011111010010100111111", "output": "101011111010010100111111" }, { "input": "1111111111", "output": "111111111" }, { "input": "1111111111111111111100111101001110110111111000001111110101001101001110011000001011001111111000110101", "output": "111111111111111111110111101001110110111111000001111110101001101001110011000001011001111111000110101" }, { "input": "11010110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100", "output": "1110110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100" }, { "input": "11111111111111111111111110110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011", "output": "1111111111111111111111111110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011" }, { "input": "11100010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011", "output": "1110010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011" }, { "input": "11", "output": "1" }, { "input": "111", "output": "11" }, { "input": "111111", "output": "11111" }, { "input": "11111", "output": "1111" }, { "input": "1111", "output": "111" } ]

1,609,081,593

2,147,483,647

Python 3

OK

TESTS

37

216

409,600

s=input() if "0" in s: a=s.index("0") s=s[:a]+s[a+1:] else: s=s[:-1] print(s)

Title: Little Elephant and Bits Time Limit: None seconds Memory Limit: None megabytes Problem Description: The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper. To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes). The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation. Input Specification: The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits. Output Specification: In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem. Demo Input: ['101\n', '110010\n'] Demo Output: ['11\n', '11010\n'] Note: In the first sample the best strategy is to delete the second digit. That results in number 112 = 310. In the second sample the best strategy is to delete the third or fourth digits — that results in number 110102 = 2610.

```python s=input() if "0" in s: a=s.index("0") s=s[:a]+s[a+1:] else: s=s[:-1] print(s) ```

3

384

A

Coder

PROGRAMMING

800

[ "implementation" ]

null

Iahub likes chess very much. He even invented a new chess piece named Coder. A Coder can move (and attack) one square horizontally or vertically. More precisely, if the Coder is located at position (*x*,<=*y*), he can move to (or attack) positions (*x*<=+<=1,<=*y*), (*x*–1,<=*y*), (*x*,<=*y*<=+<=1) and (*x*,<=*y*–1). Iahub wants to know how many Coders can be placed on an *n*<=×<=*n* chessboard, so that no Coder attacks any other Coder.

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

On the first line print an integer, the maximum number of Coders that can be placed on the chessboard. On each of the next *n* lines print *n* characters, describing the configuration of the Coders. For an empty cell print an '.', and for a Coder print a 'C'. If there are multiple correct answers, you can print any.

[ "2\n" ]

[ "2\nC.\n.C\n" ]

none

500

[ { "input": "2", "output": "2\nC.\n.C" }, { "input": "3", "output": "5\nC.C\n.C.\nC.C" }, { "input": "4", "output": "8\nC.C.\n.C.C\nC.C.\n.C.C" }, { "input": "10", "output": "50\nC.C.C.C.C.\n.C.C.C.C.C\nC.C.C.C.C.\n.C.C.C.C.C\nC.C.C.C.C.\n.C.C.C.C.C\nC.C.C.C.C.\n.C.C.C.C.C\nC.C.C.C.C.\n.C.C.C.C.C" }, { "input": "15", "output": "113\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C" }, { "input": "100", "output": "5000\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.\n.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.\n.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "101", "output": "5101\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C\n.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "500", "output": "125000\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.\n...." }, { "input": "501", "output": "125501\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C\n..." }, { "input": "755", "output": "285013\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "888", "output": "394272\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "998", "output": "498002\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "999", "output": "499001\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "1000", "output": "500000\nC.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C.C..." }, { "input": "1", "output": "1\nC" } ]

1,672,337,763

2,147,483,647

Python 3

COMPILATION_ERROR

TESTS

0

n = int(input()) if n%2 == 0: print(n**2//2) else: print(n**2//2+1) a = '' b = '' for x in range(1,n+1): if x%2 == 1: a+='C' b+='.' else: a+='.' b+='C' for i in range(1,n+1):

Title: Coder Time Limit: None seconds Memory Limit: None megabytes Problem Description: Iahub likes chess very much. He even invented a new chess piece named Coder. A Coder can move (and attack) one square horizontally or vertically. More precisely, if the Coder is located at position (*x*,<=*y*), he can move to (or attack) positions (*x*<=+<=1,<=*y*), (*x*–1,<=*y*), (*x*,<=*y*<=+<=1) and (*x*,<=*y*–1). Iahub wants to know how many Coders can be placed on an *n*<=×<=*n* chessboard, so that no Coder attacks any other Coder. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000). Output Specification: On the first line print an integer, the maximum number of Coders that can be placed on the chessboard. On each of the next *n* lines print *n* characters, describing the configuration of the Coders. For an empty cell print an '.', and for a Coder print a 'C'. If there are multiple correct answers, you can print any. Demo Input: ['2\n'] Demo Output: ['2\nC.\n.C\n'] Note: none

```python n = int(input()) if n%2 == 0: print(n**2//2) else: print(n**2//2+1) a = '' b = '' for x in range(1,n+1): if x%2 == 1: a+='C' b+='.' else: a+='.' b+='C' for i in range(1,n+1): ```

-1

102

B

Sum of Digits

PROGRAMMING

1,000

[ "implementation" ]

B. Sum of Digits

2

265

Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?

The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.

Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.

[ "0\n", "10\n", "991\n" ]

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

In the first sample the number already is one-digit — Herald can't cast a spell. The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once. The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.

1,000

[ { "input": "0", "output": "0" }, { "input": "10", "output": "1" }, { "input": "991", "output": "3" }, { "input": "99", "output": "2" }, { "input": "100", "output": "1" }, { "input": "123456789", "output": "2" }, { "input": "32", "output": "1" }, { "input": "86", "output": "2" }, { "input": "2", "output": "0" }, { "input": "8", "output": "0" }, { "input": "34", "output": "1" }, { "input": "13", "output": "1" }, { "input": "28", "output": "2" }, { "input": "23", "output": "1" }, { "input": "57", "output": "2" }, { "input": "29", "output": "2" }, { "input": "353869285", "output": "3" }, { "input": "549548646", "output": "2" }, { "input": "858893424", "output": "2" }, { "input": "644818852", "output": "3" }, { "input": "360322525", "output": "3" }, { "input": "122937520", "output": "2" }, { "input": "288403032", "output": "2" }, { "input": "677257481", "output": "3" }, { "input": "58059561", "output": "3" }, { "input": "211288847", "output": "2" }, { "input": "339900034079539584", "output": "2" }, { "input": "784084029310269952", "output": "3" }, { "input": "774730557264864000", "output": "3" }, { "input": "478233807148352256", "output": "3" }, { "input": "165646874056812544", "output": "3" }, { "input": "477533739511673792", "output": "3" }, { "input": "660119944377002240", "output": "3" }, { "input": "448375218506078080", "output": "3" }, { "input": "77213334785795616", "output": "3" }, { "input": "165767221702271872", "output": "3" }, { "input": "1", "output": "0" }, { "input": "5", "output": "0" }, { "input": "8", "output": "0" }, { "input": "156161456146164104103460452045416165146141414651641564105461465156445621465146514562146145465165145145614561465145614561", "output": "3" }, { "input": "9", "output": "0" } ]

1,642,680,060

2,147,483,647

PyPy 3

TIME_LIMIT_EXCEEDED

TESTS

6

2,000

74,137,600

n= int(input()) def sumDigits(n: int) -> int: return 0 if n==0 else (n%10)+sumDigits(n//10) count=0 while len(str(n))!=1: n=sumDigits(n) count+=1 print(count)

Title: Sum of Digits Time Limit: 2 seconds Memory Limit: 265 megabytes Problem Description: Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit? Input Specification: The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes. Output Specification: Print the number of times a number can be replaced by the sum of its digits until it only contains one digit. Demo Input: ['0\n', '10\n', '991\n'] Demo Output: ['0\n', '1\n', '3\n'] Note: In the first sample the number already is one-digit — Herald can't cast a spell. The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once. The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.

```python n= int(input()) def sumDigits(n: int) -> int: return 0 if n==0 else (n%10)+sumDigits(n//10) count=0 while len(str(n))!=1: n=sumDigits(n) count+=1 print(count) ```

0

864

F

Cities Excursions

PROGRAMMING

2,700

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

null

There are *n* cities in Berland. Some pairs of them are connected with *m* directed roads. One can use only these roads to move from one city to another. There are no roads that connect a city to itself. For each pair of cities (*x*,<=*y*) there is at most one road from *x* to *y*. A path from city *s* to city *t* is a sequence of cities *p*1, *p*2, ... , *p**k*, where *p*1<==<=*s*, *p**k*<==<=*t*, and there is a road from city *p**i* to city *p**i*<=+<=1 for each *i* from 1 to *k*<=-<=1. The path can pass multiple times through each city except *t*. It can't pass through *t* more than once. A path *p* from *s* to *t* is ideal if it is the lexicographically minimal such path. In other words, *p* is ideal path from *s* to *t* if for any other path *q* from *s* to *t* *p**i*<=<<=*q**i*, where *i* is the minimum integer such that *p**i*<=≠<=*q**i*. There is a tourist agency in the country that offers *q* unusual excursions: the *j*-th excursion starts at city *s**j* and ends in city *t**j*. For each pair *s**j*, *t**j* help the agency to study the ideal path from *s**j* to *t**j*. Note that it is possible that there is no ideal path from *s**j* to *t**j*. This is possible due to two reasons: - there is no path from *s**j* to *t**j*; - there are paths from *s**j* to *t**j*, but for every such path *p* there is another path *q* from *s**j* to *t**j*, such that *p**i*<=><=*q**i*, where *i* is the minimum integer for which *p**i*<=≠<=*q**i*. The agency would like to know for the ideal path from *s**j* to *t**j* the *k**j*-th city in that path (on the way from *s**j* to *t**j*). For each triple *s**j*, *t**j*, *k**j* (1<=≤<=*j*<=≤<=*q*) find if there is an ideal path from *s**j* to *t**j* and print the *k**j*-th city in that path, if there is any.

The first line contains three integers *n*, *m* and *q* (2<=≤<=*n*<=≤<=3000,0<=≤<=*m*<=≤<=3000, 1<=≤<=*q*<=≤<=4·105) — the number of cities, the number of roads and the number of excursions. Each of the next *m* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*), denoting that the *i*-th road goes from city *x**i* to city *y**i*. All roads are one-directional. There can't be more than one road in each direction between two cities. Each of the next *q* lines contains three integers *s**j*, *t**j* and *k**j* (1<=≤<=*s**j*,<=*t**j*<=≤<=*n*, *s**j*<=≠<=*t**j*, 1<=≤<=*k**j*<=≤<=3000).

In the *j*-th line print the city that is the *k**j*-th in the ideal path from *s**j* to *t**j*. If there is no ideal path from *s**j* to *t**j*, or the integer *k**j* is greater than the length of this path, print the string '-1' (without quotes) in the *j*-th line.

[ "7 7 5\n1 2\n2 3\n1 3\n3 4\n4 5\n5 3\n4 6\n1 4 2\n2 6 1\n1 7 3\n1 3 2\n1 3 5\n" ]

[ "2\n-1\n-1\n2\n-1\n" ]

none

3,000

[ { "input": "7 7 5\n1 2\n2 3\n1 3\n3 4\n4 5\n5 3\n4 6\n1 4 2\n2 6 1\n1 7 3\n1 3 2\n1 3 5", "output": "2\n-1\n-1\n2\n-1" }, { "input": "3 4 5\n1 3\n2 1\n3 1\n3 2\n1 2 1\n2 3 2\n2 3 3\n1 3 1\n3 2 1", "output": "-1\n1\n3\n1\n-1" }, { "input": "2 0 2\n2 1 2\n2 1 1", "output": "-1\n-1" }, { "input": "4 0 1\n3 1 1", "output": "-1" }, { "input": "2 1 3\n1 2\n1 2 1\n1 2 2\n1 2 1", "output": "1\n2\n1" }, { "input": "2 1 5\n1 2\n1 2 1\n1 2 1\n1 2 1\n1 2 1\n1 2 1", "output": "1\n1\n1\n1\n1" }, { "input": "2 0 1\n2 1 1", "output": "-1" }, { "input": "5 7 10\n2 5\n3 5\n5 4\n5 2\n4 1\n1 4\n2 4\n4 2 3\n5 2 1\n1 3 3\n1 3 3\n3 2 2\n4 2 2\n3 5 3\n1 5 1\n3 5 1\n5 2 2", "output": "-1\n5\n-1\n-1\n5\n-1\n-1\n-1\n3\n2" }, { "input": "10 11 10\n10 9\n5 10\n5 8\n2 5\n1 8\n3 4\n7 2\n4 6\n8 10\n6 7\n9 3\n3 2 1\n10 7 1\n7 5 2\n7 5 3\n10 5 2\n2 5 2\n10 4 2\n6 2 2\n6 2 1\n1 3 1", "output": "3\n10\n2\n5\n9\n5\n9\n7\n6\n1" }, { "input": "10 15 10\n7 8\n1 4\n5 6\n10 9\n3 4\n10 3\n8 10\n1 2\n8 9\n10 1\n10 2\n4 5\n6 7\n2 3\n9 10\n2 1 3\n1 2 2\n2 3 1\n5 4 3\n3 5 2\n8 6 2\n2 7 1\n9 8 2\n1 9 2\n8 10 3", "output": "4\n2\n2\n7\n4\n9\n2\n10\n2\n10" }, { "input": "10 16 20\n5 9\n6 3\n2 1\n9 2\n4 6\n5 10\n10 5\n6 8\n4 7\n4 8\n8 7\n7 4\n3 10\n9 10\n7 8\n1 2\n4 5 1\n9 1 1\n6 3 2\n6 3 2\n9 2 3\n3 2 6\n3 10 2\n6 10 3\n4 3 1\n6 10 3\n1 2 1\n4 3 3\n7 4 1\n7 1 7\n4 10 3\n5 9 2\n3 10 1\n3 1 1\n8 1 6\n6 10 3", "output": "4\n9\n3\n3\n-1\n-1\n10\n10\n4\n10\n1\n3\n7\n9\n3\n9\n3\n3\n10\n10" }, { "input": "15 25 20\n9 7\n3 1\n7 3\n11 13\n1 6\n6 15\n12 13\n8 12\n4 11\n4 8\n3 5\n1 14\n3 12\n15 13\n5 2\n13 5\n15 10\n13 6\n13 9\n14 9\n2 14\n10 1\n13 10\n11 12\n2 12\n7 3 2\n10 8 2\n11 12 2\n4 8 1\n13 9 1\n6 8 1\n4 8 1\n9 8 1\n14 8 1\n11 1 2\n9 1 1\n7 6 1\n7 4 2\n11 2 1\n4 8 1\n12 5 1\n13 14 1\n9 3 1\n14 8 2\n7 10 2", "output": "3\n-1\n12\n4\n-1\n-1\n4\n-1\n-1\n-1\n9\n7\n-1\n11\n4\n12\n-1\n9\n-1\n3" }, { "input": "20 20 40\n4 5\n6 7\n13 14\n10 11\n9 10\n19 20\n2 3\n1 2\n16 17\n8 9\n7 8\n15 16\n11 12\n5 6\n18 19\n17 18\n2 6\n14 15\n12 13\n3 4\n3 19 9\n3 18 13\n2 19 19\n2 18 12\n3 18 12\n1 18 8\n1 20 29\n3 19 12\n3 20 28\n1 19 27\n2 18 5\n2 18 28\n2 20 30\n1 20 26\n3 20 7\n1 20 23\n2 18 26\n3 18 21\n2 20 23\n1 20 23\n3 20 19\n2 18 27\n2 19 16\n2 18 10\n1 18 3\n3 18 25\n2 19 24\n3 18 21\n2 19 28\n2 20 23\n1 19 17\n2 19 28\n3 18 15\n3 19 29\n1 20 16\n2 19 22\n2 19 4\n1 20 22\n1 20 20\n2 19 13", "output": "11\n15\n-1\n13\n14\n8\n-1\n14\n-1\n-1\n6\n-1\n-1\n-1\n9\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n17\n11\n3\n-1\n-1\n-1\n-1\n-1\n17\n-1\n17\n-1\n16\n-1\n5\n-1\n20\n14" }, { "input": "20 20 40\n3 4\n12 13\n2 3\n12 17\n4 5\n9 10\n11 12\n10 11\n12 15\n8 9\n5 6\n12 20\n12 16\n12 18\n1 2\n6 7\n15 11\n12 19\n12 14\n7 8\n3 9 19\n3 9 5\n2 9 22\n1 6 3\n2 9 22\n2 6 21\n3 6 21\n2 9 5\n1 4 29\n3 8 13\n2 5 19\n1 3 10\n1 3 4\n1 5 19\n2 6 24\n3 9 21\n2 3 9\n2 9 24\n2 3 30\n1 9 22\n2 8 28\n2 6 27\n1 3 17\n2 5 9\n3 2 30\n2 6 16\n1 6 21\n2 8 15\n3 5 23\n3 8 17\n1 5 12\n2 5 10\n3 9 4\n1 2 29\n1 2 26\n3 5 28\n2 4 28\n2 7 19\n3 8 5\n2 4 23", "output": "-1\n7\n-1\n3\n-1\n-1\n-1\n6\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n6\n-1\n-1\n-1\n-1\n-1\n7\n-1" } ]

1,507,174,453

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

77

819,200

# -*- coding: utf-8 -*- import math import collections import bisect import heapq import time import random """ created by shhuan at 2017/10/5 11:00 """ N, M, Q = map(int, input().split()) P = collections.defaultdict(list) for i in range(M): s, t = map(int, input().split()) P[s].append(t) E = [] for i in range(Q): s, t, k = map(int, input().split()) E.append((s, t, k)) INF = False def dfs(start, end, path, visited, circle): global INF if INF: return [] if start == end: return [path] ans = [] for to in sorted(P[start]): if to not in visited: ans.extend(dfs(to, end, path+[to], visited | {to}, circle)) else: if not circle: i = path.index(to) for j in range(i+1, len(path)): if dfs(path[j-1], end, path+path[i:j], visited, True): INF = True return [] return ans for s, t, k in E: INF = False paths = dfs(s, t, [s], set(), False) if INF: print(-1) continue if not paths: print(-1) continue paths = sorted(paths) # print(s, t, k, paths) if paths[0] > paths[-1]: print(-1) else: path = paths[0] if k >= len(path): print(-1) else: print(path[k-1]) # 2 3 4 5 4 5 4 5 .... 4 6

Title: Cities Excursions Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* cities in Berland. Some pairs of them are connected with *m* directed roads. One can use only these roads to move from one city to another. There are no roads that connect a city to itself. For each pair of cities (*x*,<=*y*) there is at most one road from *x* to *y*. A path from city *s* to city *t* is a sequence of cities *p*1, *p*2, ... , *p**k*, where *p*1<==<=*s*, *p**k*<==<=*t*, and there is a road from city *p**i* to city *p**i*<=+<=1 for each *i* from 1 to *k*<=-<=1. The path can pass multiple times through each city except *t*. It can't pass through *t* more than once. A path *p* from *s* to *t* is ideal if it is the lexicographically minimal such path. In other words, *p* is ideal path from *s* to *t* if for any other path *q* from *s* to *t* *p**i*<=<<=*q**i*, where *i* is the minimum integer such that *p**i*<=≠<=*q**i*. There is a tourist agency in the country that offers *q* unusual excursions: the *j*-th excursion starts at city *s**j* and ends in city *t**j*. For each pair *s**j*, *t**j* help the agency to study the ideal path from *s**j* to *t**j*. Note that it is possible that there is no ideal path from *s**j* to *t**j*. This is possible due to two reasons: - there is no path from *s**j* to *t**j*; - there are paths from *s**j* to *t**j*, but for every such path *p* there is another path *q* from *s**j* to *t**j*, such that *p**i*<=><=*q**i*, where *i* is the minimum integer for which *p**i*<=≠<=*q**i*. The agency would like to know for the ideal path from *s**j* to *t**j* the *k**j*-th city in that path (on the way from *s**j* to *t**j*). For each triple *s**j*, *t**j*, *k**j* (1<=≤<=*j*<=≤<=*q*) find if there is an ideal path from *s**j* to *t**j* and print the *k**j*-th city in that path, if there is any. Input Specification: The first line contains three integers *n*, *m* and *q* (2<=≤<=*n*<=≤<=3000,0<=≤<=*m*<=≤<=3000, 1<=≤<=*q*<=≤<=4·105) — the number of cities, the number of roads and the number of excursions. Each of the next *m* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*), denoting that the *i*-th road goes from city *x**i* to city *y**i*. All roads are one-directional. There can't be more than one road in each direction between two cities. Each of the next *q* lines contains three integers *s**j*, *t**j* and *k**j* (1<=≤<=*s**j*,<=*t**j*<=≤<=*n*, *s**j*<=≠<=*t**j*, 1<=≤<=*k**j*<=≤<=3000). Output Specification: In the *j*-th line print the city that is the *k**j*-th in the ideal path from *s**j* to *t**j*. If there is no ideal path from *s**j* to *t**j*, or the integer *k**j* is greater than the length of this path, print the string '-1' (without quotes) in the *j*-th line. Demo Input: ['7 7 5\n1 2\n2 3\n1 3\n3 4\n4 5\n5 3\n4 6\n1 4 2\n2 6 1\n1 7 3\n1 3 2\n1 3 5\n'] Demo Output: ['2\n-1\n-1\n2\n-1\n'] Note: none

```python # -*- coding: utf-8 -*- import math import collections import bisect import heapq import time import random """ created by shhuan at 2017/10/5 11:00 """ N, M, Q = map(int, input().split()) P = collections.defaultdict(list) for i in range(M): s, t = map(int, input().split()) P[s].append(t) E = [] for i in range(Q): s, t, k = map(int, input().split()) E.append((s, t, k)) INF = False def dfs(start, end, path, visited, circle): global INF if INF: return [] if start == end: return [path] ans = [] for to in sorted(P[start]): if to not in visited: ans.extend(dfs(to, end, path+[to], visited | {to}, circle)) else: if not circle: i = path.index(to) for j in range(i+1, len(path)): if dfs(path[j-1], end, path+path[i:j], visited, True): INF = True return [] return ans for s, t, k in E: INF = False paths = dfs(s, t, [s], set(), False) if INF: print(-1) continue if not paths: print(-1) continue paths = sorted(paths) # print(s, t, k, paths) if paths[0] > paths[-1]: print(-1) else: path = paths[0] if k >= len(path): print(-1) else: print(path[k-1]) # 2 3 4 5 4 5 4 5 .... 4 6 ```

0

716

A

Crazy Computer

PROGRAMMING

800

[ "implementation" ]

null

ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear! More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen. For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen. You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything.

The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively. The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word.

Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*.

[ "6 5\n1 3 8 14 19 20\n", "6 1\n1 3 5 7 9 10\n" ]

[ "3", "2" ]

The first sample is already explained in the problem statement. For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1.

500

[ { "input": "6 5\n1 3 8 14 19 20", "output": "3" }, { "input": "6 1\n1 3 5 7 9 10", "output": "2" }, { "input": "1 1\n1000000000", "output": "1" }, { "input": "5 5\n1 7 12 13 14", "output": "4" }, { "input": "2 1000000000\n1 1000000000", "output": "2" }, { "input": "3 5\n1 10 20", "output": "1" }, { "input": "3 10\n1 2 3", "output": "3" }, { "input": "2 1\n1 100", "output": "1" }, { "input": "3 1\n1 2 10", "output": "1" }, { "input": "2 1\n1 2", "output": "2" } ]

1,591,536,606

2,147,483,647

Python 3

OK

TESTS

81

186

8,908,800

n,c = map(int,input().split()) count=1 my_list=list(map(int,input().split())) for i in range(n): if i!=0: if my_list[i]-my_list[i-1]<=c: count+=1 else: count=1 print(count)

Title: Crazy Computer Time Limit: None seconds Memory Limit: None megabytes Problem Description: ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear! More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen. For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen. You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything. Input Specification: The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively. The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word. Output Specification: Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*. Demo Input: ['6 5\n1 3 8 14 19 20\n', '6 1\n1 3 5 7 9 10\n'] Demo Output: ['3', '2'] Note: The first sample is already explained in the problem statement. For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1.

```python n,c = map(int,input().split()) count=1 my_list=list(map(int,input().split())) for i in range(n): if i!=0: if my_list[i]-my_list[i-1]<=c: count+=1 else: count=1 print(count) ```

3

126

B

Password

PROGRAMMING

1,700

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

null

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

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

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

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

[ "fix", "Just a legend" ]

none

1,000

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

1,610,462,632

2,147,483,647

PyPy 3

RUNTIME_ERROR

TESTS

1

248

2,150,400

#!/usr/bin/env python3 # created : 2020. 12. 31. 23:59 import os from sys import stdin, stdout def getTp(w): m = len(w) tp = [0 for i in range(m)] j = 0 for i in range(1,m): while j > 0 and w[i] != w[j]: j = tp[j-1] if w[i] == w[j]: j += 1 tp[i] = j return tp def kmp(s, w): # ans = [] tp = getTp(w) n, m = len(s), len(w) j = 0 for i in range(1, n-1): while j > 0 and s[i] != w[j]: j = tp[j-1] if s[i] == w[j]: if j == m-1: # ans.append(i-j) # j = fp[j] return True else: j += 1 # return ans return False def solve(tc): s = stdin.readline().strip() tp = getTp(s) p = len(s)-1 if tp[p] == 0: print("Just a legend") return while p>0: w = s[:tp[p]] if kmp(s, w): print(w) return p = tp[p-1] print("Just a legend") tcs = 1 # tcs = int(stdin.readline().strip()) tc = 1 while tc <= tcs: solve(tc) tc += 1

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

```python #!/usr/bin/env python3 # created : 2020. 12. 31. 23:59 import os from sys import stdin, stdout def getTp(w): m = len(w) tp = [0 for i in range(m)] j = 0 for i in range(1,m): while j > 0 and w[i] != w[j]: j = tp[j-1] if w[i] == w[j]: j += 1 tp[i] = j return tp def kmp(s, w): # ans = [] tp = getTp(w) n, m = len(s), len(w) j = 0 for i in range(1, n-1): while j > 0 and s[i] != w[j]: j = tp[j-1] if s[i] == w[j]: if j == m-1: # ans.append(i-j) # j = fp[j] return True else: j += 1 # return ans return False def solve(tc): s = stdin.readline().strip() tp = getTp(s) p = len(s)-1 if tp[p] == 0: print("Just a legend") return while p>0: w = s[:tp[p]] if kmp(s, w): print(w) return p = tp[p-1] print("Just a legend") tcs = 1 # tcs = int(stdin.readline().strip()) tc = 1 while tc <= tcs: solve(tc) tc += 1 ```

-1

349

A

Cinema Line

PROGRAMMING

1,100

[ "greedy", "implementation" ]

null

The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?

The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.

Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".

[ "4\n25 25 50 50\n", "2\n25 100\n", "4\n50 50 25 25\n" ]

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

none

500

[ { "input": "4\n25 25 50 50", "output": "YES" }, { "input": "2\n25 100", "output": "NO" }, { "input": "4\n50 50 25 25", "output": "NO" }, { "input": "3\n25 50 100", "output": "NO" }, { "input": "10\n25 25 25 25 25 25 25 25 25 25", "output": "YES" }, { "input": "10\n50 50 50 50 50 50 50 50 50 50", "output": "NO" }, { "input": "10\n100 100 100 100 100 100 100 100 100 100", "output": "NO" }, { "input": "7\n25 25 25 25 50 100 50", "output": "YES" }, { "input": "2\n25 25", "output": "YES" }, { "input": "3\n50 100 100", "output": "NO" }, { "input": "3\n25 50 50", "output": "NO" }, { "input": "2\n50 25", "output": "NO" }, { "input": "14\n25 25 25 25 25 25 25 50 50 50 100 100 100 100", "output": "NO" }, { "input": "5\n25 25 25 100 100", "output": "NO" }, { "input": "7\n25 50 25 50 25 50 100", "output": "NO" }, { "input": "5\n25 50 25 50 100", "output": "NO" }, { "input": "5\n25 25 50 50 50", "output": "NO" }, { "input": "5\n25 25 50 100 50", "output": "NO" }, { "input": "5\n25 25 50 50 100", "output": "NO" }, { "input": "1\n50", "output": "NO" } ]

1,621,512,463

2,147,483,647

Python 3

OK

TESTS

40

342

5,120,000

n=int(input()) seq=[int(x) for x in input().split()] count25=0 count50=0 count100=0 for k,v in enumerate(seq): val=25*count25+50*count50+100*count100 if val-(v-25)>=0: x=v-25 if x==75: if count50>=1 and count25>=1: count50-=1 count25-=1 elif count25>=3: count25-=3 else: print('NO') exit() count100+=1 elif x==25: if count25>=1: count25-=1 else: print('NO') exit() count50+=1 elif x==0: count25+=1 else: print("NO") exit() print('YES')

Title: Cinema Line Time Limit: None seconds Memory Limit: None megabytes Problem Description: The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line? Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line. Output Specification: Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO". Demo Input: ['4\n25 25 50 50\n', '2\n25 100\n', '4\n50 50 25 25\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: none

```python n=int(input()) seq=[int(x) for x in input().split()] count25=0 count50=0 count100=0 for k,v in enumerate(seq): val=25*count25+50*count50+100*count100 if val-(v-25)>=0: x=v-25 if x==75: if count50>=1 and count25>=1: count50-=1 count25-=1 elif count25>=3: count25-=3 else: print('NO') exit() count100+=1 elif x==25: if count25>=1: count25-=1 else: print('NO') exit() count50+=1 elif x==0: count25+=1 else: print("NO") exit() print('YES') ```

3

767

A

Snacktower

PROGRAMMING

1,100

[ "data structures", "implementation" ]

null

According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top. Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower. However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it. Write a program that models the behavior of Ankh-Morpork residents.

The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks. The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.

Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.

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

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

In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.

500

[ { "input": "3\n3 1 2", "output": "3 \n\n2 1 " }, { "input": "5\n4 5 1 2 3", "output": "5 4 \n\n\n3 2 1 " }, { "input": "1\n1", "output": "1 " }, { "input": "2\n1 2", "output": "2 1 " }, { "input": "10\n5 1 6 2 8 3 4 10 9 7", "output": "10 \n9 8 \n7 6 5 4 3 2 1 " }, { "input": "30\n16 10 4 29 5 28 12 21 11 30 18 6 14 3 17 22 20 15 9 1 27 19 24 26 13 25 2 23 8 7", "output": "30 29 28 \n\n\n\n\n\n\n\n\n\n\n27 \n\n\n26 \n\n25 24 \n\n23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 \n8 \n7 6 5 4 3 2 1 " }, { "input": "100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 4 80 72 39", "output": "100 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n99 98 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 \n\n\n\n\n\n\n\n\n80 79 78 77 76 75 74 73 \n72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 \n39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 " }, { "input": "2\n2 1", "output": "2 \n1 " } ]

1,661,083,549

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

4

108

0

n = int(input()) a = [int(x) for x in input().split()] wait=[] snack = n for i in a: if i == snack: if wait: wait=[i]+wait print(*wait) snack = n-len(wait) wait=[] else: print(i);snack-=1 else: print() wait.insert(0,i)

Title: Snacktower Time Limit: None seconds Memory Limit: None megabytes Problem Description: According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top. Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower. However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it. Write a program that models the behavior of Ankh-Morpork residents. Input Specification: The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks. The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*. Output Specification: Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty. Demo Input: ['3\n3 1 2\n', '5\n4 5 1 2 3\n'] Demo Output: ['3\n\xa0\n2 1', '5 4\n\xa0\n\xa0\n3 2 1\n'] Note: In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.

```python n = int(input()) a = [int(x) for x in input().split()] wait=[] snack = n for i in a: if i == snack: if wait: wait=[i]+wait print(*wait) snack = n-len(wait) wait=[] else: print(i);snack-=1 else: print() wait.insert(0,i) ```

0

61

A

Ultra-Fast Mathematician

PROGRAMMING

800

[ "implementation" ]

A. Ultra-Fast Mathematician

2

256

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

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

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

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

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

none

500

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

1,529,995,795

2,147,483,647

Python 3

OK

TESTS

102

93

0

def main(): n_ = input() m_ = input() n = int(n_, base=2) m = int(m_, base=2) print('{0:0{1}b}'.format(n ^ m, len(n_))) if __name__ == "__main__": main()

Title: Ultra-Fast Mathematician Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate. Input Specification: There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. Output Specification: Write one line — the corresponding answer. Do not omit the leading 0s. Demo Input: ['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n'] Demo Output: ['1110001\n', '111\n', '0100\n', '00010\n'] Note: none

```python def main(): n_ = input() m_ = input() n = int(n_, base=2) m = int(m_, base=2) print('{0:0{1}b}'.format(n ^ m, len(n_))) if __name__ == "__main__": main() ```

3.97675

1

A

Theatre Square

PROGRAMMING

1,000

[ "math" ]

A. Theatre Square

1

256

Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.

The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).

Write the needed number of flagstones.

[ "6 6 4\n" ]

[ "4\n" ]

none

0

[ { "input": "6 6 4", "output": "4" }, { "input": "1 1 1", "output": "1" }, { "input": "2 1 1", "output": "2" }, { "input": "1 2 1", "output": "2" }, { "input": "2 2 1", "output": "4" }, { "input": "2 1 2", "output": "1" }, { "input": "1 1 3", "output": "1" }, { "input": "2 3 4", "output": "1" }, { "input": "1000000000 1000000000 1", "output": "1000000000000000000" }, { "input": "12 13 4", "output": "12" }, { "input": "222 332 5", "output": "3015" }, { "input": "1000 1000 10", "output": "10000" }, { "input": "1001 1000 10", "output": "10100" }, { "input": "100 10001 1000000000", "output": "1" }, { "input": "1000000000 1000000000 1000000000", "output": "1" }, { "input": "1000000000 1000000000 999999999", "output": "4" }, { "input": "1000000000 1000000000 192", "output": "27126743055556" }, { "input": "1000000000 987654321 1", "output": "987654321000000000" }, { "input": "456784567 1000000000 51", "output": "175618850864484" }, { "input": "39916800 134217728 40320", "output": "3295710" } ]

1,673,622,931

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

31

0

from math import ceil n,m,a = input().split() res = ceil(int(n)/int(a))*ceil(int(n)/int(a)) print(res)

Title: Theatre Square Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. Input Specification: The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). Output Specification: Write the needed number of flagstones. Demo Input: ['6 6 4\n'] Demo Output: ['4\n'] Note: none

```python from math import ceil n,m,a = input().split() res = ceil(int(n)/int(a))*ceil(int(n)/int(a)) print(res) ```

0

939

B

Hamster Farm

PROGRAMMING

1,000

[ "implementation" ]

null

Dima has a hamsters farm. Soon *N* hamsters will grow up on it and Dima will sell them in a city nearby. Hamsters should be transported in boxes. If some box is not completely full, the hamsters in it are bored, that's why each box should be completely full with hamsters. Dima can buy boxes at a factory. The factory produces boxes of *K* kinds, boxes of the *i*-th kind can contain in themselves *a**i* hamsters. Dima can buy any amount of boxes, but he should buy boxes of only one kind to get a wholesale discount. Of course, Dima would buy boxes in such a way that each box can be completely filled with hamsters and transported to the city. If there is no place for some hamsters, Dima will leave them on the farm. Find out how many boxes and of which type should Dima buy to transport maximum number of hamsters.

The first line contains two integers *N* and *K* (0<=≤<=*N*<=≤<=1018, 1<=≤<=*K*<=≤<=105) — the number of hamsters that will grow up on Dima's farm and the number of types of boxes that the factory produces. The second line contains *K* integers *a*1, *a*2, ..., *a**K* (1<=≤<=*a**i*<=≤<=1018 for all *i*) — the capacities of boxes.

Output two integers: the type of boxes that Dima should buy and the number of boxes of that type Dima should buy. Types of boxes are numbered from 1 to *K* in the order they are given in input. If there are many correct answers, output any of them.

[ "19 3\n5 4 10\n", "28 3\n5 6 30\n" ]

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

none

1,000

[ { "input": "19 3\n5 4 10", "output": "2 4" }, { "input": "28 3\n5 6 30", "output": "1 5" }, { "input": "1 1\n1", "output": "1 1" }, { "input": "0 2\n2 3", "output": "1 0" }, { "input": "30 4\n4 5 5 4", "output": "2 6" }, { "input": "120 7\n109 92 38 38 49 38 92", "output": "3 3" }, { "input": "357 40\n12 10 12 11 12 12 12 10 10 10 12 12 12 12 12 10 12 10 10 10 11 10 12 10 12 10 12 10 10 12 12 12 12 10 10 10 12 12 12 12", "output": "4 32" }, { "input": "587 100\n92 92 76 95 61 60 64 79 64 96 63 92 60 61 95 71 60 61 65 63 84 76 98 63 90 61 61 71 63 61 95 90 79 71 77 67 63 61 63 60 100 71 98 88 67 95 60 61 79 76 70 61 64 65 64 77 96 95 84 100 67 60 84 92 70 100 63 79 61 77 92 74 60 90 84 80 76 61 88 79 64 61 79 60 61 67 98 98 92 76 61 60 80 77 77 76 63 88 99 70", "output": "19 9" }, { "input": "98765 30\n89 841 599 240 356 599 92 305 305 536 356 92 622 1000 751 522 89 149 356 598 305 518 996 92 622 536 356 91 779 770", "output": "28 1085" }, { "input": "947264836 50\n977141206 956777871 186433588 538218068 759102378 327484438 88827268 266300062 670616672 756092978 414181331 913675814 898008516 343057716 99416265 236586817 52751842 550467703 684435578 844649988 917709231 550467703 493542638 707106470 414181331 198095018 913675814 99416265 550467703 679553907 186433588 355713982 670616672 977141206 504598561 327484438 414181331 463431413 546229641 132582931 463431413 759102378 273063720 683939057 924604119 759102378 463431413 52751842 552131077 903126340", "output": "16 4" }, { "input": "600003000040000507 10\n334302557805985467 334302557805985467 681026146296527968 157006854340095780 188330644415015186 803011712275585087 638039699540420111 638039699540420111 600874219702299205 923891462598005659", "output": "5 3" }, { "input": "666 2\n1 300", "output": "1 666" }, { "input": "899999999999999991 1\n199999999999999998", "output": "1 4" }, { "input": "10 1\n11", "output": "1 0" }, { "input": "999999999999999999 1\n500000000000000000", "output": "1 1" }, { "input": "2 1\n2", "output": "1 1" }, { "input": "199999999999999999 1\n100000000000000000", "output": "1 1" }, { "input": "999999999999999999 1\n1000000000000000000", "output": "1 0" }, { "input": "1000000000000000000 1\n500000000000000001", "output": "1 1" }, { "input": "1000000000000000000 1\n2", "output": "1 500000000000000000" }, { "input": "1000000000000000000 5\n500000000000000010 500000000000000010 500000000000000010 500000000000000010 500000000000000030", "output": "5 1" }, { "input": "1000000000000000000 1\n900000000000000000", "output": "1 1" } ]

1,623,757,399

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

18

171

9,216,000

n,k= map(int,input().split()) a = list(map(int,input().split())) list1=[] for i in range(k): list1.append(n // a[i]) for i in range(k): list1[i]=a[i]*list1[i] max=0 for i in range(k): if (n == 0): index=i+1 quant=0 break if (max < list1[i]): max=list1[i] index=i+1 quant= n // a[i] print(index,quant)

Title: Hamster Farm Time Limit: None seconds Memory Limit: None megabytes Problem Description: Dima has a hamsters farm. Soon *N* hamsters will grow up on it and Dima will sell them in a city nearby. Hamsters should be transported in boxes. If some box is not completely full, the hamsters in it are bored, that's why each box should be completely full with hamsters. Dima can buy boxes at a factory. The factory produces boxes of *K* kinds, boxes of the *i*-th kind can contain in themselves *a**i* hamsters. Dima can buy any amount of boxes, but he should buy boxes of only one kind to get a wholesale discount. Of course, Dima would buy boxes in such a way that each box can be completely filled with hamsters and transported to the city. If there is no place for some hamsters, Dima will leave them on the farm. Find out how many boxes and of which type should Dima buy to transport maximum number of hamsters. Input Specification: The first line contains two integers *N* and *K* (0<=≤<=*N*<=≤<=1018, 1<=≤<=*K*<=≤<=105) — the number of hamsters that will grow up on Dima's farm and the number of types of boxes that the factory produces. The second line contains *K* integers *a*1, *a*2, ..., *a**K* (1<=≤<=*a**i*<=≤<=1018 for all *i*) — the capacities of boxes. Output Specification: Output two integers: the type of boxes that Dima should buy and the number of boxes of that type Dima should buy. Types of boxes are numbered from 1 to *K* in the order they are given in input. If there are many correct answers, output any of them. Demo Input: ['19 3\n5 4 10\n', '28 3\n5 6 30\n'] Demo Output: ['2 4\n', '1 5\n'] Note: none

```python n,k= map(int,input().split()) a = list(map(int,input().split())) list1=[] for i in range(k): list1.append(n // a[i]) for i in range(k): list1[i]=a[i]*list1[i] max=0 for i in range(k): if (n == 0): index=i+1 quant=0 break if (max < list1[i]): max=list1[i] index=i+1 quant= n // a[i] print(index,quant) ```

-1

544

A

Set of Strings

PROGRAMMING

1,100

[ "implementation", "strings" ]

null

You are given a string *q*. A sequence of *k* strings *s*1,<=*s*2,<=...,<=*s**k* is called beautiful, if the concatenation of these strings is string *q* (formally, *s*1<=+<=*s*2<=+<=...<=+<=*s**k*<==<=*q*) and the first characters of these strings are distinct. Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist.

The first line contains a positive integer *k* (1<=≤<=*k*<=≤<=26) — the number of strings that should be in a beautiful sequence. The second line contains string *q*, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive.

If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next *k* lines print the beautiful sequence of strings *s*1,<=*s*2,<=...,<=*s**k*. If there are multiple possible answers, print any of them.

[ "1\nabca\n", "2\naaacas\n", "4\nabc\n" ]

[ "YES\nabca\n", "YES\naaa\ncas\n", "NO\n" ]

In the second sample there are two possible answers: {"*aaaca*", "*s*"} and {"*aaa*", "*cas*"}.

500

[ { "input": "1\nabca", "output": "YES\nabca" }, { "input": "2\naaacas", "output": "YES\naaa\ncas" }, { "input": "4\nabc", "output": "NO" }, { "input": "3\nnddkhkhkdndknndkhrnhddkrdhrnrrnkkdnnndndrdhnknknhnrnnkrrdhrkhkrkhnkhkhhrhdnrndnknrrhdrdrkhdrkkhkrnkk", "output": "YES\nn\ndd\nkhkhkdndknndkhrnhddkrdhrnrrnkkdnnndndrdhnknknhnrnnkrrdhrkhkrkhnkhkhhrhdnrndnknrrhdrdrkhdrkkhkrnkk" }, { "input": "26\nbiibfmmfifmffbmmfmbmbmiimbmiffmffibibfbiffibibiiimbffbbfbifmiibffbmbbbfmfibmibfffibfbffmfmimbmmmfmfm", "output": "NO" }, { "input": "3\nkydoybxlfeugtrbvqnrjtzshorrsrwsxkvlwyolbaadtzpmyyfllxuciia", "output": "YES\nk\ny\ndoybxlfeugtrbvqnrjtzshorrsrwsxkvlwyolbaadtzpmyyfllxuciia" }, { "input": "3\nssussususskkskkskuusksuuussksukkskuksukukusssususuususkkuukssuksskusukkssuksskskuskusussusskskksksus", "output": "YES\nss\nussususs\nkkskkskuusksuuussksukkskuksukukusssususuususkkuukssuksskusukkssuksskskuskusussusskskksksus" }, { "input": "5\naaaaabcdef", "output": "YES\naaaaa\nb\nc\nd\nef" }, { "input": "3\niiiiiiimiriiriwmimtmwrhhxmbmhwgghhgbqhywebrblyhlxjrthoooltehrmdhqhuodjmsjwcgrfnttiitpmqvbhlafwtzyikc", "output": "YES\niiiiiii\nmi\nriiriwmimtmwrhhxmbmhwgghhgbqhywebrblyhlxjrthoooltehrmdhqhuodjmsjwcgrfnttiitpmqvbhlafwtzyikc" }, { "input": "20\ngggggllglgllltgtlglttstsgtttsslhhlssghgagtlsaghhoggtfgsaahtotdodthfltdxggxislnttlanxonhnkddtigppitdh", "output": "NO" }, { "input": "16\nkkkkkkyykkynkknkkonyokdndkyonokdywkwykdkdotknnwzkoywiooinkcyzyntcdnitnppnpziomyzdspomoqmomcyrrospppn", "output": "NO" }, { "input": "15\nwwwgggowgwwhoohwgwghwyohhggywhyyodgwydwgggkhgyydqyggkgkpokgthqghidhworprodtcogqkwgtfiodwdurcctkmrfmh", "output": "YES\nwww\nggg\nowgww\nhoohwgwghw\nyohhggywhyyo\ndgwydwggg\nkhgyyd\nqyggkgk\npokg\nthqgh\nidhwo\nrprodt\ncogqkwgt\nfiodwd\nurcctkmrfmh" }, { "input": "15\nnnnnnntnttttttqqnqqynnqqwwnnnwneenhwtyhhoqeyeqyeuthwtnhtpnphhwetjhouhwnpojvvovoswwjryrwerbwwpbvrwvjj", "output": "YES\nnnnnnn\ntntttttt\nqqnqq\nynnqq\nwwnnnwn\neen\nhwtyhh\noqeyeqye\nuthwtnht\npnphhwet\njhouhwnpoj\nvvovo\nswwj\nryrwer\nbwwpbvrwvjj" }, { "input": "15\nvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv", "output": "NO" }, { "input": "1\niiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiaaaaaiiiiaiaiiiiaaiaiiiaiiaiaaiaiiaiiiiiaiiiaiiiaiaiaai", "output": "YES\niiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiaaaaaiiiiaiaiiiiaaiaiiiaiiaiaaiaiiaiiiiiaiiiaiiiaiaiaai" }, { "input": "26\nvvvnnsnnnpsnnswwspncvshtncwphaphmwnwkhvvhuvctvnehemowkmtzissswjaxuuvphzrmfzihamdqmmyhhijbitlipgltyy", "output": "YES\nvvv\nnn\nsnnn\npsnns\nwwspn\ncvs\nh\ntncwph\naph\nmwnw\nkhvvh\nuvctvn\nehem\nowkmt\nz\nisssw\nja\nxuuvphz\nrm\nfziham\nd\nqmm\nyhhij\nbit\nlip\ngltyy" }, { "input": "26\njexzsbwaih", "output": "NO" }, { "input": "1\nk", "output": "YES\nk" }, { "input": "1\nzz", "output": "YES\nzz" }, { "input": "3\nziw", "output": "YES\nz\ni\nw" }, { "input": "26\ntjmbyqwuahlixegopkzrfndcsv", "output": "YES\nt\nj\nm\nb\ny\nq\nw\nu\na\nh\nl\ni\nx\ne\ng\no\np\nk\nz\nr\nf\nn\nd\nc\ns\nv" }, { "input": "25\nvobekscyadzqwnjxruplifmthg", "output": "YES\nv\no\nb\ne\nk\ns\nc\ny\na\nd\nz\nq\nw\nn\nj\nx\nr\nu\np\nl\ni\nf\nm\nt\nhg" }, { "input": "26\nlllplzkkzflzflffzznnnnfgflqlttlmtnkzlztskngyymitqagattkdllyutzimsrskpapcmuupjdopxqlnhqcscwvdtxbflefy", "output": "YES\nlll\npl\nz\nkkz\nflzflffzz\nnnnnf\ngfl\nql\nttl\nmtnkzlzt\nskng\nyym\nitq\nagattk\ndlly\nutzims\nrskpap\ncmuup\njd\nop\nxqln\nhqcsc\nw\nvdtx\nbfl\nefy" }, { "input": "25\nkkrrkrkrkrsrskpskbrppdsdbgbkrbllkbswdwcchgskmkhwiidicczlscsodtjglxbmeotzxnmbjmoqgkquglaoxgcykxvbhdi", "output": "YES\nkk\nrrkrkrkr\nsrsk\npsk\nbrpp\ndsdb\ngbkrb\nllkbs\nwdw\ncc\nhgsk\nmkhw\niidicc\nzlscs\nod\nt\njgl\nxbm\neotzx\nnmbjmo\nqgkq\nugl\naoxgc\nykx\nvbhdi" }, { "input": "25\nuuuuuccpucubccbupxubcbpujiliwbpqbpyiweuywaxwqasbsllwehceruytjvphytraawgbjmerfeymoayujqranlvkpkiypadr", "output": "YES\nuuuuu\ncc\npucu\nbccbup\nxubcbpu\nj\ni\nli\nwbp\nqbp\nyiw\neuyw\naxwqa\nsbsllwe\nhce\nruy\ntj\nvphytraaw\ngbj\nmer\nfeym\noayujqra\nnlv\nkpkiypa\ndr" }, { "input": "26\nxxjxodrogovufvohrodliretxxyjqnrbzmicorptkjafiwmsbwml", "output": "YES\nxx\njx\no\nd\nro\ngo\nv\nu\nfvo\nhrod\nl\nir\ne\ntxx\nyj\nq\nnr\nb\nz\nmi\ncor\npt\nkj\nafi\nwm\nsbwml" }, { "input": "26\npjhsxjbvkqntwmsdnrguecaofylzti", "output": "YES\np\nj\nh\ns\nxj\nb\nv\nk\nq\nn\nt\nw\nms\ndn\nr\ng\nu\ne\nc\na\no\nf\ny\nl\nzt\ni" }, { "input": "25\nrrrrqqwrlqrwglrlylwhrrwyvrhvzgvqahrhgsvavtggyduayivxzgeicinlnrkapoepbsfyjjrt", "output": "YES\nrrrr\nqq\nwr\nlqrw\nglrl\nylw\nhrrwy\nvrhv\nzgvq\nahrhg\nsvav\ntggy\nd\nuay\niv\nxzg\nei\nci\nnlnr\nka\np\noep\nbs\nfy\njjrt" }, { "input": "26\ncccccccaacwwaxxaacczacnnnqqwnaggzqrwagcnabxnrcvgjqjamqzgdntzanaxvjfwqlvdttuzjoxiwtkqvrydospmpeirhg", "output": "YES\nccccccc\naac\nwwa\nxxaacc\nzac\nnnn\nqqwna\nggzq\nrwagcna\nbxnrc\nvg\njqja\nmqzg\ndn\ntzanaxvj\nfwq\nlvdtt\nuzj\nox\niwt\nkqvr\nydo\ns\npmp\neir\nhg" }, { "input": "4\nsssssssssssssssssssssssssslsslslsllsslssslslssllaaslsaalsasaassllasasalrlrslarlaarlsrsaslasarlr", "output": "YES\nssssssssssssssssssssssssss\nlsslslsllsslssslslssll\naaslsaalsasaassllasasal\nrlrslarlaarlsrsaslasarlr" }, { "input": "26\na", "output": "NO" }, { "input": "26\nab", "output": "NO" }, { "input": "2\nab", "output": "YES\na\nb" } ]

1,631,008,504

2,147,483,647

PyPy 3

RUNTIME_ERROR

TESTS

0

124

22,220,800

import math def main_function(): k = int(input()) s = input() counter = 0 hash_of_a = [-1 for i in range(150)] collector = [] for i in range(len(s)): if hash_of_a[ord(s[i])] == -1: hash_of_a[ord(s[i])] = i collector.append(i) counter += 1 if counter < k: print("NO") else: print("YES") collector.append(len(s)) collector.sort() indexi = 0 internal_counter = collector[indexi] print(collector) for j in range(len(s) + 1): if j < collector[indexi + 1]: print(s[j], end="") else: k -= 1 print("") if k == 1: print(s[j:]) break else: print(s[j]) indexi += 1 main_function()

Title: Set of Strings Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a string *q*. A sequence of *k* strings *s*1,<=*s*2,<=...,<=*s**k* is called beautiful, if the concatenation of these strings is string *q* (formally, *s*1<=+<=*s*2<=+<=...<=+<=*s**k*<==<=*q*) and the first characters of these strings are distinct. Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist. Input Specification: The first line contains a positive integer *k* (1<=≤<=*k*<=≤<=26) — the number of strings that should be in a beautiful sequence. The second line contains string *q*, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive. Output Specification: If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next *k* lines print the beautiful sequence of strings *s*1,<=*s*2,<=...,<=*s**k*. If there are multiple possible answers, print any of them. Demo Input: ['1\nabca\n', '2\naaacas\n', '4\nabc\n'] Demo Output: ['YES\nabca\n', 'YES\naaa\ncas\n', 'NO\n'] Note: In the second sample there are two possible answers: {"*aaaca*", "*s*"} and {"*aaa*", "*cas*"}.

```python import math def main_function(): k = int(input()) s = input() counter = 0 hash_of_a = [-1 for i in range(150)] collector = [] for i in range(len(s)): if hash_of_a[ord(s[i])] == -1: hash_of_a[ord(s[i])] = i collector.append(i) counter += 1 if counter < k: print("NO") else: print("YES") collector.append(len(s)) collector.sort() indexi = 0 internal_counter = collector[indexi] print(collector) for j in range(len(s) + 1): if j < collector[indexi + 1]: print(s[j], end="") else: k -= 1 print("") if k == 1: print(s[j:]) break else: print(s[j]) indexi += 1 main_function() ```

-1

990

A

Commentary Boxes

PROGRAMMING

1,000

[ "implementation", "math" ]

null

Berland Football Cup starts really soon! Commentators from all over the world come to the event. Organizers have already built $n$ commentary boxes. $m$ regional delegations will come to the Cup. Every delegation should get the same number of the commentary boxes. If any box is left unoccupied then the delegations will be upset. So each box should be occupied by exactly one delegation. If $n$ is not divisible by $m$, it is impossible to distribute the boxes to the delegations at the moment. Organizers can build a new commentary box paying $a$ burles and demolish a commentary box paying $b$ burles. They can both build and demolish boxes arbitrary number of times (each time paying a corresponding fee). It is allowed to demolish all the existing boxes. What is the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by $m$)?

The only line contains four integer numbers $n$, $m$, $a$ and $b$ ($1 \le n, m \le 10^{12}$, $1 \le a, b \le 100$), where $n$ is the initial number of the commentary boxes, $m$ is the number of delegations to come, $a$ is the fee to build a box and $b$ is the fee to demolish a box.

Output the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by $m$). It is allowed that the final number of the boxes is equal to $0$.

[ "9 7 3 8\n", "2 7 3 7\n", "30 6 17 19\n" ]

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

In the first example organizers can build $5$ boxes to make the total of $14$ paying $3$ burles for the each of them. In the second example organizers can demolish $2$ boxes to make the total of $0$ paying $7$ burles for the each of them. In the third example organizers are already able to distribute all the boxes equally among the delegations, each one get $5$ boxes.

0

[ { "input": "9 7 3 8", "output": "15" }, { "input": "2 7 3 7", "output": "14" }, { "input": "30 6 17 19", "output": "0" }, { "input": "500000000001 1000000000000 100 100", "output": "49999999999900" }, { "input": "1000000000000 750000000001 10 100", "output": "5000000000020" }, { "input": "1000000000000 750000000001 100 10", "output": "2499999999990" }, { "input": "42 1 1 1", "output": "0" }, { "input": "1 1000000000000 1 100", "output": "100" }, { "input": "7 2 3 7", "output": "3" }, { "input": "999999999 2 1 1", "output": "1" }, { "input": "999999999999 10000000007 100 100", "output": "70100" }, { "input": "10000000001 2 1 1", "output": "1" }, { "input": "29 6 1 2", "output": "1" }, { "input": "99999999999 6 100 100", "output": "300" }, { "input": "1000000000000 7 3 8", "output": "8" }, { "input": "99999999999 2 1 1", "output": "1" }, { "input": "1 2 1 1", "output": "1" }, { "input": "999999999999 2 1 1", "output": "1" }, { "input": "9 2 1 1", "output": "1" }, { "input": "17 4 5 5", "output": "5" }, { "input": "100000000000 3 1 1", "output": "1" }, { "input": "100 7 1 1", "output": "2" }, { "input": "1000000000000 3 100 100", "output": "100" }, { "input": "70 3 10 10", "output": "10" }, { "input": "1 2 5 1", "output": "1" }, { "input": "1000000000000 3 1 1", "output": "1" }, { "input": "804289377 846930887 78 16", "output": "3326037780" }, { "input": "1000000000000 9 55 55", "output": "55" }, { "input": "957747787 424238336 87 93", "output": "10162213695" }, { "input": "25 6 1 2", "output": "2" }, { "input": "22 7 3 8", "output": "8" }, { "input": "10000000000 1 1 1", "output": "0" }, { "input": "999999999999 2 10 10", "output": "10" }, { "input": "999999999999 2 100 100", "output": "100" }, { "input": "100 3 3 8", "output": "6" }, { "input": "99999 2 1 1", "output": "1" }, { "input": "100 3 2 5", "output": "4" }, { "input": "1000000000000 13 10 17", "output": "17" }, { "input": "7 2 1 2", "output": "1" }, { "input": "10 3 1 2", "output": "2" }, { "input": "5 2 2 2", "output": "2" }, { "input": "100 3 5 2", "output": "2" }, { "input": "7 2 1 1", "output": "1" }, { "input": "70 4 1 1", "output": "2" }, { "input": "10 4 1 1", "output": "2" }, { "input": "6 7 41 42", "output": "41" }, { "input": "10 3 10 1", "output": "1" }, { "input": "5 5 2 3", "output": "0" }, { "input": "1000000000000 3 99 99", "output": "99" }, { "input": "7 3 100 1", "output": "1" }, { "input": "7 2 100 5", "output": "5" }, { "input": "1000000000000 1 23 33", "output": "0" }, { "input": "30 7 1 1", "output": "2" }, { "input": "100 3 1 1", "output": "1" }, { "input": "90001 300 100 1", "output": "1" }, { "input": "13 4 1 2", "output": "2" }, { "input": "1000000000000 6 1 3", "output": "2" }, { "input": "50 4 5 100", "output": "10" }, { "input": "999 2 1 1", "output": "1" }, { "input": "5 2 5 5", "output": "5" }, { "input": "20 3 3 3", "output": "3" }, { "input": "3982258181 1589052704 87 20", "output": "16083055460" }, { "input": "100 3 1 3", "output": "2" }, { "input": "7 3 1 1", "output": "1" }, { "input": "19 10 100 100", "output": "100" }, { "input": "23 3 100 1", "output": "2" }, { "input": "25 7 100 1", "output": "4" }, { "input": "100 9 1 2", "output": "2" }, { "input": "9999999999 2 1 100", "output": "1" }, { "input": "1000000000000 2 1 1", "output": "0" }, { "input": "10000 3 1 1", "output": "1" }, { "input": "22 7 1 6", "output": "6" }, { "input": "100000000000 1 1 1", "output": "0" }, { "input": "18 7 100 1", "output": "4" }, { "input": "10003 4 1 100", "output": "1" }, { "input": "3205261341 718648876 58 11", "output": "3637324207" }, { "input": "8 3 100 1", "output": "2" }, { "input": "15 7 1 1", "output": "1" }, { "input": "1000000000000 1 20 20", "output": "0" }, { "input": "16 7 3 2", "output": "4" }, { "input": "1000000000000 1 1 1", "output": "0" }, { "input": "7 3 1 100", "output": "2" }, { "input": "16 3 1 100", "output": "2" }, { "input": "13 4 1 10", "output": "3" }, { "input": "10 4 5 5", "output": "10" }, { "input": "14 3 1 100", "output": "1" }, { "input": "100 33 100 1", "output": "1" }, { "input": "22 7 1 8", "output": "6" }, { "input": "10 4 2 1", "output": "2" }, { "input": "6 4 2 2", "output": "4" }, { "input": "17 4 2 1", "output": "1" }, { "input": "7 3 100 10", "output": "10" }, { "input": "702 7 3 2", "output": "4" }, { "input": "8 3 1 5", "output": "1" }, { "input": "3 2 5 2", "output": "2" }, { "input": "99 19 1 7", "output": "15" }, { "input": "16 3 100 1", "output": "1" }, { "input": "100 34 1 100", "output": "2" }, { "input": "100 33 1 1", "output": "1" }, { "input": "2 3 4 3", "output": "4" }, { "input": "15 4 4 10", "output": "4" }, { "input": "1144108931 470211273 45 79", "output": "11993619960" }, { "input": "2 3 3 4", "output": "3" }, { "input": "29 5 4 9", "output": "4" }, { "input": "15 7 1 5", "output": "5" }, { "input": "1 1 1 1", "output": "0" }, { "input": "1 1 3 4", "output": "0" }, { "input": "10 12 2 1", "output": "4" }, { "input": "1 2 3 4", "output": "3" } ]

1,529,321,961

2,147,483,647

Python 3

OK

TESTS

109

93

0

n, m, a, b = [int(i) for i in input().split()] if n % m != 0: print(min((m - n % m) * a, n % m * b)) else: print(0)

Title: Commentary Boxes Time Limit: None seconds Memory Limit: None megabytes Problem Description: Berland Football Cup starts really soon! Commentators from all over the world come to the event. Organizers have already built $n$ commentary boxes. $m$ regional delegations will come to the Cup. Every delegation should get the same number of the commentary boxes. If any box is left unoccupied then the delegations will be upset. So each box should be occupied by exactly one delegation. If $n$ is not divisible by $m$, it is impossible to distribute the boxes to the delegations at the moment. Organizers can build a new commentary box paying $a$ burles and demolish a commentary box paying $b$ burles. They can both build and demolish boxes arbitrary number of times (each time paying a corresponding fee). It is allowed to demolish all the existing boxes. What is the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by $m$)? Input Specification: The only line contains four integer numbers $n$, $m$, $a$ and $b$ ($1 \le n, m \le 10^{12}$, $1 \le a, b \le 100$), where $n$ is the initial number of the commentary boxes, $m$ is the number of delegations to come, $a$ is the fee to build a box and $b$ is the fee to demolish a box. Output Specification: Output the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by $m$). It is allowed that the final number of the boxes is equal to $0$. Demo Input: ['9 7 3 8\n', '2 7 3 7\n', '30 6 17 19\n'] Demo Output: ['15\n', '14\n', '0\n'] Note: In the first example organizers can build $5$ boxes to make the total of $14$ paying $3$ burles for the each of them. In the second example organizers can demolish $2$ boxes to make the total of $0$ paying $7$ burles for the each of them. In the third example organizers are already able to distribute all the boxes equally among the delegations, each one get $5$ boxes.

```python n, m, a, b = [int(i) for i in input().split()] if n % m != 0: print(min((m - n % m) * a, n % m * b)) else: print(0) ```

3

1

A

Theatre Square

PROGRAMMING

1,000

[ "math" ]

A. Theatre Square

1

256

Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.

The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).

Write the needed number of flagstones.

[ "6 6 4\n" ]

[ "4\n" ]

none

0

[ { "input": "6 6 4", "output": "4" }, { "input": "1 1 1", "output": "1" }, { "input": "2 1 1", "output": "2" }, { "input": "1 2 1", "output": "2" }, { "input": "2 2 1", "output": "4" }, { "input": "2 1 2", "output": "1" }, { "input": "1 1 3", "output": "1" }, { "input": "2 3 4", "output": "1" }, { "input": "1000000000 1000000000 1", "output": "1000000000000000000" }, { "input": "12 13 4", "output": "12" }, { "input": "222 332 5", "output": "3015" }, { "input": "1000 1000 10", "output": "10000" }, { "input": "1001 1000 10", "output": "10100" }, { "input": "100 10001 1000000000", "output": "1" }, { "input": "1000000000 1000000000 1000000000", "output": "1" }, { "input": "1000000000 1000000000 999999999", "output": "4" }, { "input": "1000000000 1000000000 192", "output": "27126743055556" }, { "input": "1000000000 987654321 1", "output": "987654321000000000" }, { "input": "456784567 1000000000 51", "output": "175618850864484" }, { "input": "39916800 134217728 40320", "output": "3295710" } ]

1,689,225,805

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

31

0

from math import ceil if __name__=='__main__': inp = list(map(int, input().split())) m, n, a = inp print(ceil(m/a)*2)

Title: Theatre Square Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. Input Specification: The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). Output Specification: Write the needed number of flagstones. Demo Input: ['6 6 4\n'] Demo Output: ['4\n'] Note: none

```python from math import ceil if __name__=='__main__': inp = list(map(int, input().split())) m, n, a = inp print(ceil(m/a)*2) ```

0

526

B

Om Nom and Dark Park

PROGRAMMING

1,400

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

null

Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him. The park consists of 2*n*<=+<=1<=-<=1 squares connected by roads so that the scheme of the park is a full binary tree of depth *n*. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2*n*,<=2*n*<=+<=1,<=...,<=2*n*<=+<=1<=-<=1 and these exits lead straight to the Om Nom friends' houses. From each square *i* (2<=≤<=*i*<=<<=2*n*<=+<=1) there is a road to the square . Thus, it is possible to go from the park entrance to each of the exits by walking along exactly *n* roads. Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe. He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.

The first line contains integer *n* (1<=≤<=*n*<=≤<=10) — the number of roads on the path from the entrance to any exit. The next line contains 2*n*<=+<=1<=-<=2 numbers *a*2,<=*a*3,<=... *a*2*n*<=+<=1<=-<=1 — the initial numbers of street lights on each road of the park. Here *a**i* is the number of street lights on the road between squares *i* and . All numbers *a**i* are positive integers, not exceeding 100.

Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.

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

[ "5\n" ]

Picture for the sample test. Green color denotes the additional street lights.

500

[ { "input": "2\n1 2 3 4 5 6", "output": "5" }, { "input": "2\n1 2 3 3 2 2", "output": "0" }, { "input": "1\n39 52", "output": "13" }, { "input": "2\n59 96 34 48 8 72", "output": "139" }, { "input": "3\n87 37 91 29 58 45 51 74 70 71 47 38 91 89", "output": "210" }, { "input": "5\n39 21 95 89 73 90 9 55 85 32 30 21 68 59 82 91 20 64 52 70 6 88 53 47 30 47 34 14 11 22 42 15 28 54 37 48 29 3 14 13 18 77 90 58 54 38 94 49 45 66 13 74 11 14 64 72 95 54 73 79 41 35", "output": "974" }, { "input": "1\n49 36", "output": "13" }, { "input": "1\n77 88", "output": "11" }, { "input": "1\n1 33", "output": "32" }, { "input": "2\n72 22 81 23 14 75", "output": "175" }, { "input": "2\n100 70 27 1 68 52", "output": "53" }, { "input": "2\n24 19 89 82 22 21", "output": "80" }, { "input": "3\n86 12 92 91 3 68 57 56 76 27 33 62 71 84", "output": "286" }, { "input": "3\n14 56 53 61 57 45 40 44 31 9 73 2 61 26", "output": "236" }, { "input": "3\n35 96 7 43 10 14 16 36 95 92 16 50 59 55", "output": "173" }, { "input": "4\n1 97 18 48 96 65 24 91 17 45 36 27 74 93 78 86 39 55 53 21 26 68 31 33 79 63 80 92 1 26", "output": "511" }, { "input": "4\n25 42 71 29 50 30 99 79 77 24 76 66 68 23 97 99 65 17 75 62 66 46 48 4 40 71 98 57 21 92", "output": "603" }, { "input": "4\n49 86 17 7 3 6 86 71 36 10 27 10 58 64 12 16 88 67 93 3 15 20 58 87 97 91 11 6 34 62", "output": "470" }, { "input": "5\n16 87 36 16 81 53 87 35 63 56 47 91 81 95 80 96 91 7 58 99 25 28 47 60 7 69 49 14 51 52 29 30 83 23 21 52 100 26 91 14 23 94 72 70 40 12 50 32 54 52 18 74 5 15 62 3 48 41 24 25 56 43", "output": "1060" }, { "input": "5\n40 27 82 94 38 22 66 23 18 34 87 31 71 28 95 5 14 61 76 52 66 6 60 40 68 77 70 63 64 18 47 13 82 55 34 64 30 1 29 24 24 9 65 17 29 96 61 76 72 23 32 26 90 39 54 41 35 66 71 29 75 48", "output": "1063" }, { "input": "5\n64 72 35 68 92 95 45 15 77 16 26 74 61 65 18 22 32 19 98 97 14 84 70 23 29 1 87 28 88 89 73 79 69 88 43 60 64 64 66 39 17 27 46 71 18 83 73 20 90 77 49 70 84 63 50 72 26 87 26 37 78 65", "output": "987" }, { "input": "6\n35 61 54 77 70 50 53 70 4 66 58 47 76 100 78 5 43 50 55 93 13 93 59 92 30 74 22 23 98 70 19 56 90 92 19 7 28 53 45 77 42 91 71 56 19 83 100 53 13 93 37 13 70 60 16 13 76 3 12 22 17 26 50 6 63 7 25 41 92 29 36 80 11 4 10 14 77 75 53 82 46 24 56 46 82 36 80 75 8 45 24 22 90 34 45 76 18 38 86 43 7 49 80 56 90 53 12 51 98 47 44 58 32 4 2 6 3 60 38 72 74 46 30 86 1 98", "output": "2499" }, { "input": "6\n63 13 100 54 31 15 29 58 59 44 2 99 70 33 97 14 70 12 73 42 65 71 68 67 87 83 43 84 18 41 37 22 81 24 27 11 57 28 83 92 39 1 56 15 16 67 16 97 31 52 50 65 63 89 8 52 55 20 71 27 28 35 86 92 94 60 10 65 83 63 89 71 34 20 78 40 34 62 2 86 100 81 87 69 25 4 52 17 57 71 62 38 1 3 54 71 34 85 20 60 80 23 82 47 4 19 7 18 14 18 28 27 4 55 26 71 45 9 2 40 67 28 32 19 81 92", "output": "2465" }, { "input": "6\n87 62 58 32 81 92 12 50 23 27 38 39 64 74 16 35 84 59 91 87 14 48 90 47 44 95 64 45 31 11 67 5 80 60 36 15 91 3 21 2 40 24 37 69 5 50 23 37 49 19 68 21 49 9 100 94 45 41 22 31 31 48 25 70 25 25 95 88 82 1 37 53 49 31 57 74 94 45 55 93 43 37 13 85 59 72 15 68 3 90 96 55 100 64 63 69 43 33 66 84 57 97 87 34 23 89 97 77 39 89 8 92 68 13 50 36 95 61 71 96 73 13 30 49 57 89", "output": "2513" } ]

1,428,166,717

1,417

Python 3

OK

TESTS

38

46

0

def B(): s = 0 n = int(input()) k = (1 << (n + 1)) - 1 a = [0, 0] + list(map(int, input().split())) for i in range(k, 1, -2): u, v = a[i], a[i - 1] if u > v: u, v = v, u s += v - u a[i >> 1] += v return s print(B())

Title: Om Nom and Dark Park Time Limit: None seconds Memory Limit: None megabytes Problem Description: Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him. The park consists of 2*n*<=+<=1<=-<=1 squares connected by roads so that the scheme of the park is a full binary tree of depth *n*. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2*n*,<=2*n*<=+<=1,<=...,<=2*n*<=+<=1<=-<=1 and these exits lead straight to the Om Nom friends' houses. From each square *i* (2<=≤<=*i*<=<<=2*n*<=+<=1) there is a road to the square . Thus, it is possible to go from the park entrance to each of the exits by walking along exactly *n* roads. Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe. He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=10) — the number of roads on the path from the entrance to any exit. The next line contains 2*n*<=+<=1<=-<=2 numbers *a*2,<=*a*3,<=... *a*2*n*<=+<=1<=-<=1 — the initial numbers of street lights on each road of the park. Here *a**i* is the number of street lights on the road between squares *i* and . All numbers *a**i* are positive integers, not exceeding 100. Output Specification: Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe. Demo Input: ['2\n1 2 3 4 5 6\n'] Demo Output: ['5\n'] Note: Picture for the sample test. Green color denotes the additional street lights.

```python def B(): s = 0 n = int(input()) k = (1 << (n + 1)) - 1 a = [0, 0] + list(map(int, input().split())) for i in range(k, 1, -2): u, v = a[i], a[i - 1] if u > v: u, v = v, u s += v - u a[i >> 1] += v return s print(B()) ```

3

614

A

Link/Cut Tree

PROGRAMMING

1,500

[ "brute force", "implementation" ]

null

Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure. Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?) Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!

The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109).

Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes).

[ "1 10 2\n", "2 4 5\n" ]

[ "1 2 4 8 ", "-1" ]

Note to the first sample: numbers 20 = 1, 21 = 2, 22 = 4, 23 = 8 lie within the specified range. The number 24 = 16 is greater then 10, thus it shouldn't be printed.

500

[ { "input": "1 10 2", "output": "1 2 4 8 " }, { "input": "2 4 5", "output": "-1" }, { "input": "18102 43332383920 28554", "output": "28554 815330916 " }, { "input": "19562 31702689720 17701", "output": "313325401 " }, { "input": "11729 55221128400 313", "output": "97969 30664297 9597924961 " }, { "input": "5482 100347128000 342", "output": "116964 40001688 13680577296 " }, { "input": "3680 37745933600 10", "output": "10000 100000 1000000 10000000 100000000 1000000000 10000000000 " }, { "input": "17098 191120104800 43", "output": "79507 3418801 147008443 6321363049 " }, { "input": "10462 418807699200 2", "output": "16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 " }, { "input": "30061 641846400000 3", "output": "59049 177147 531441 1594323 4782969 14348907 43046721 129140163 387420489 1162261467 3486784401 10460353203 31381059609 94143178827 282429536481 " }, { "input": "1 1000000000000000000 2", "output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 549755813888 1099511627776 2199023255552 4398046511104 8796093022208 17592186044416 35184372088832 70368744177664 140737488355328 281474976710656 562949953421312 1125899906842624 2251799813685248 4503599627370496 900719925474099..." }, { "input": "32 2498039712000 4", "output": "64 256 1024 4096 16384 65536 262144 1048576 4194304 16777216 67108864 268435456 1073741824 4294967296 17179869184 68719476736 274877906944 1099511627776 " }, { "input": "1 2576683920000 2", "output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 549755813888 1099511627776 2199023255552 " }, { "input": "5 25 5", "output": "5 25 " }, { "input": "1 90 90", "output": "1 90 " }, { "input": "95 2200128528000 68", "output": "4624 314432 21381376 1453933568 98867482624 " }, { "input": "64 426314644000 53", "output": "2809 148877 7890481 418195493 22164361129 " }, { "input": "198765 198765 198765", "output": "198765 " }, { "input": "42 2845016496000 12", "output": "144 1728 20736 248832 2985984 35831808 429981696 5159780352 61917364224 743008370688 " }, { "input": "6 6 3", "output": "-1" }, { "input": "1 10 11", "output": "1 " }, { "input": "2 10 11", "output": "-1" }, { "input": "87 160 41", "output": "-1" }, { "input": "237171123124584251 923523399718980912 7150", "output": "-1" }, { "input": "101021572000739548 453766043506276015 8898", "output": "-1" }, { "input": "366070689449360724 928290634811046396 8230", "output": "-1" }, { "input": "438133886369772308 942612870269666780 7193", "output": "-1" }, { "input": "10 10 10", "output": "10 " }, { "input": "16 16 256", "output": "-1" }, { "input": "1 1000000000000000000 1000000000", "output": "1 1000000000 1000000000000000000 " }, { "input": "1000000000000000000 1000000000000000000 1000000000", "output": "1000000000000000000 " }, { "input": "1000000000 1000000000000000000 1000000000", "output": "1000000000 1000000000000000000 " }, { "input": "1 1 4", "output": "1 " }, { "input": "1 999999999999999999 1000000000", "output": "1 1000000000 " }, { "input": "1 1000000000000000000 999999990", "output": "1 999999990 999999980000000100 " }, { "input": "1 1000000000000000000 999999984", "output": "1 999999984 999999968000000256 " }, { "input": "1 1000000000000000000 324325", "output": "1 324325 105186705625 34114678301828125 " }, { "input": "1 1000000000000000000 999999523", "output": "1 999999523 999999046000227529 " }, { "input": "1 243 3", "output": "1 3 9 27 81 243 " }, { "input": "62769392426654367 567152589733560993 688813", "output": "326816522793383797 " }, { "input": "1 1000000000000000000 690852001", "output": "1 690852001 477276487285704001 " }, { "input": "1 1000000000000000000 918745157", "output": "1 918745157 844092663510954649 " }, { "input": "1 1000000000000000000 131299843", "output": "1 131299843 17239648771824649 " }, { "input": "2861381721051425 2861381721051425 1234", "output": "-1" } ]

1,584,862,331

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

2

155

0

#q=int(input()) q=1 for _ in range(q): l,r,k=map(int,input().split()) j=0 ans=[] while(l<= k**j and k**j<=r): ans.append(k**j) j+=1 if len(ans)==0: print(-1) else: print(*ans)

Title: Link/Cut Tree Time Limit: None seconds Memory Limit: None megabytes Problem Description: Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure. Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?) Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him! Input Specification: The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109). Output Specification: Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes). Demo Input: ['1 10 2\n', '2 4 5\n'] Demo Output: ['1 2 4 8 ', '-1'] Note: Note to the first sample: numbers 20 = 1, 21 = 2, 22 = 4, 23 = 8 lie within the specified range. The number 24 = 16 is greater then 10, thus it shouldn't be printed.

```python #q=int(input()) q=1 for _ in range(q): l,r,k=map(int,input().split()) j=0 ans=[] while(l<= k**j and k**j<=r): ans.append(k**j) j+=1 if len(ans)==0: print(-1) else: print(*ans) ```

0

227

B

Effective Approach

PROGRAMMING

1,100

[ "implementation" ]

null

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

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

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

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

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

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

1,000

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

1,666,673,436

2,147,483,647

Python 3

OK

TESTS

34

342

14,336,000

n = int(input()) a = list(map(int,input().split())) q = int(input()) res1 = 0 res2 = 0 mapp = {} for i in range(len(a)) : mapp[a[i]] = i query = list(map(int,input().split())) for i in query : res1+=(mapp[i]+1) res2+=(n-mapp[i]) print(str(res1)+" "+str(res2))

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

```python n = int(input()) a = list(map(int,input().split())) q = int(input()) res1 = 0 res2 = 0 mapp = {} for i in range(len(a)) : mapp[a[i]] = i query = list(map(int,input().split())) for i in query : res1+=(mapp[i]+1) res2+=(n-mapp[i]) print(str(res1)+" "+str(res2)) ```

3

365

A

Good Number

PROGRAMMING

1,100

[ "implementation" ]

null

Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).

The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).

Print a single integer — the number of *k*-good numbers in *a*.

[ "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n", "2 1\n1\n10\n" ]

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

none

500

[ { "input": "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560", "output": "10" }, { "input": "2 1\n1\n10", "output": "1" }, { "input": "1 0\n1000000000", "output": "1" }, { "input": "1 1\n1000000000", "output": "1" }, { "input": "6 0\n10\n102\n120\n1032\n1212103\n1999999", "output": "5" }, { "input": "1 3\n1000000000", "output": "0" }, { "input": "1 9\n1000000000", "output": "0" }, { "input": "1 0\n1", "output": "0" }, { "input": "1 1\n1", "output": "0" }, { "input": "1 3\n1", "output": "0" }, { "input": "1 9\n1", "output": "0" }, { "input": "2 8\n123456780\n123", "output": "1" }, { "input": "2 8\n12345678\n1230", "output": "0" }, { "input": "6 1\n10\n102\n120\n1032\n1212103\n1999999", "output": "5" }, { "input": "6 2\n10\n102\n120\n1032\n1212103\n1999999", "output": "4" }, { "input": "6 3\n10\n102\n120\n1032\n1212103\n1999999", "output": "2" }, { "input": "6 9\n10\n102\n120\n1032\n1212103\n1999999", "output": "0" }, { "input": "1 2\n2220", "output": "0" } ]

1,697,645,814

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

15

0

count_time=False if count_time: import time start_time = time.time() #----------------------------------------- n,k=map(int,input().split()) a=set(map(str,range(k+1))) t=0 for _ in range(n): s=set(input()) if a==s:t+=1 print(t) #------------------------------------------ if count_time: end_time = time.time() print('----------------\nRunning time: {} s' .format(end_time - start_time))

Title: Good Number Time Limit: None seconds Memory Limit: None megabytes Problem Description: Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*). Input Specification: The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109). Output Specification: Print a single integer — the number of *k*-good numbers in *a*. Demo Input: ['10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n', '2 1\n1\n10\n'] Demo Output: ['10\n', '1\n'] Note: none

```python count_time=False if count_time: import time start_time = time.time() #----------------------------------------- n,k=map(int,input().split()) a=set(map(str,range(k+1))) t=0 for _ in range(n): s=set(input()) if a==s:t+=1 print(t) #------------------------------------------ if count_time: end_time = time.time() print('----------------\nRunning time: {} s' .format(end_time - start_time)) ```

0

59

A

Word

PROGRAMMING

800

[ "implementation", "strings" ]

A. Word

2

256

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

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

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

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

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

none

500

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

1,573,452,929

2,147,483,647

Python 3

OK

TESTS

30

218

0

s=input() up=0 lo=0 for i in range(len(s)): if s[i]==s[i].upper(): up+=1 elif s[i]==s[i].lower(): lo+=1 if up>lo: print(s.upper()) else: print(s.lower())

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

```python s=input() up=0 lo=0 for i in range(len(s)): if s[i]==s[i].upper(): up+=1 elif s[i]==s[i].lower(): lo+=1 if up>lo: print(s.upper()) else: print(s.lower()) ```

3.9455

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" } ]

1,695,654,485

2,147,483,647

Python 3

OK

TESTS

62

61

0

result = "" s = input().split(" ") a = int(s[0]) b = int(s[1]) if 1 <= a and a <= b and b <= 10: amountOfYears = 0 while True: a = a * 3 b = b * 2 amountOfYears += 1 if a > b: result = amountOfYears break print(f"{result}")

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 result = "" s = input().split(" ") a = int(s[0]) b = int(s[1]) if 1 <= a and a <= b and b <= 10: amountOfYears = 0 while True: a = a * 3 b = b * 2 amountOfYears += 1 if a > b: result = amountOfYears break print(f"{result}") ```

3

227

B

Effective Approach

PROGRAMMING

1,100

[ "implementation" ]

null

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

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

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

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

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

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

1,000

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

1,689,780,227

2,147,483,647

PyPy 3

TIME_LIMIT_EXCEEDED

TESTS

6

2,000

13,619,200

def vasya_and_petya_approach(a,e): for i in range(len(a)): if a[i] == e: return i+1 , len(a)-i def effective_approach(): #counters for vasya and petya v,p = 0,0 n = int(input()) #len of list a = [int(i) for i in input().split()] input()#no need to get the number of queries we can get it from the len of the queries list queries = [int(i) for i in input().split()] for i in queries: v += vasya_and_petya_approach(a,i)[0] p += vasya_and_petya_approach(a,i)[1] #return results print(v,p) effective_approach()

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

```python def vasya_and_petya_approach(a,e): for i in range(len(a)): if a[i] == e: return i+1 , len(a)-i def effective_approach(): #counters for vasya and petya v,p = 0,0 n = int(input()) #len of list a = [int(i) for i in input().split()] input()#no need to get the number of queries we can get it from the len of the queries list queries = [int(i) for i in input().split()] for i in queries: v += vasya_and_petya_approach(a,i)[0] p += vasya_and_petya_approach(a,i)[1] #return results print(v,p) effective_approach() ```

0

131

A

cAPS lOCK

PROGRAMMING

1,000

[ "implementation", "strings" ]

null

wHAT DO WE NEED cAPS LOCK FOR? Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage. Let's consider that a word has been typed with the Caps lock key accidentally switched on, if: - either it only contains uppercase letters; - or all letters except for the first one are uppercase. In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed. Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.

The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.

Print the result of the given word's processing.

[ "cAPS\n", "Lock\n" ]

[ "Caps", "Lock\n" ]

none

500

[ { "input": "cAPS", "output": "Caps" }, { "input": "Lock", "output": "Lock" }, { "input": "cAPSlOCK", "output": "cAPSlOCK" }, { "input": "CAPs", "output": "CAPs" }, { "input": "LoCK", "output": "LoCK" }, { "input": "OOPS", "output": "oops" }, { "input": "oops", "output": "oops" }, { "input": "a", "output": "A" }, { "input": "A", "output": "a" }, { "input": "aA", "output": "Aa" }, { "input": "Zz", "output": "Zz" }, { "input": "Az", "output": "Az" }, { "input": "zA", "output": "Za" }, { "input": "AAA", "output": "aaa" }, { "input": "AAa", "output": "AAa" }, { "input": "AaR", "output": "AaR" }, { "input": "Tdr", "output": "Tdr" }, { "input": "aTF", "output": "Atf" }, { "input": "fYd", "output": "fYd" }, { "input": "dsA", "output": "dsA" }, { "input": "fru", "output": "fru" }, { "input": "hYBKF", "output": "Hybkf" }, { "input": "XweAR", "output": "XweAR" }, { "input": "mogqx", "output": "mogqx" }, { "input": "eOhEi", "output": "eOhEi" }, { "input": "nkdku", "output": "nkdku" }, { "input": "zcnko", "output": "zcnko" }, { "input": "lcccd", "output": "lcccd" }, { "input": "vwmvg", "output": "vwmvg" }, { "input": "lvchf", "output": "lvchf" }, { "input": "IUNVZCCHEWENCHQQXQYPUJCRDZLUXCLJHXPHBXEUUGNXOOOPBMOBRIBHHMIRILYJGYYGFMTMFSVURGYHUWDRLQVIBRLPEVAMJQYO", "output": "iunvzcchewenchqqxqypujcrdzluxcljhxphbxeuugnxooopbmobribhhmirilyjgyygfmtmfsvurgyhuwdrlqvibrlpevamjqyo" }, { "input": "OBHSZCAMDXEJWOZLKXQKIVXUUQJKJLMMFNBPXAEFXGVNSKQLJGXHUXHGCOTESIVKSFMVVXFVMTEKACRIWALAGGMCGFEXQKNYMRTG", "output": "obhszcamdxejwozlkxqkivxuuqjkjlmmfnbpxaefxgvnskqljgxhuxhgcotesivksfmvvxfvmtekacriwalaggmcgfexqknymrtg" }, { "input": "IKJYZIKROIYUUCTHSVSKZTETNNOCMAUBLFJCEVANCADASMZRCNLBZPQRXESHEEMOMEPCHROSRTNBIDXYMEPJSIXSZQEBTEKKUHFS", "output": "ikjyzikroiyuucthsvskztetnnocmaublfjcevancadasmzrcnlbzpqrxesheemomepchrosrtnbidxymepjsixszqebtekkuhfs" }, { "input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE", "output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype" }, { "input": "uCKJZRGZJCPPLEEYJTUNKOQSWGBMTBQEVPYFPIPEKRVYQNTDPANOIXKMPINNFUSZWCURGBDPYTEKBEKCPMVZPMWAOSHJYMGKOMBQ", "output": "Uckjzrgzjcppleeyjtunkoqswgbmtbqevpyfpipekrvyqntdpanoixkmpinnfuszwcurgbdpytekbekcpmvzpmwaoshjymgkombq" }, { "input": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR", "output": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR" }, { "input": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE", "output": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE" }, { "input": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ", "output": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ" }, { "input": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm", "output": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm" }, { "input": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm", "output": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm" }, { "input": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg", "output": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg" }, { "input": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc", "output": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc" }, { "input": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv", "output": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv" }, { "input": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect", "output": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect" }, { "input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE", "output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype" }, { "input": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu", "output": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu" }, { "input": "aBACABa", "output": "aBACABa" }, { "input": "AAAAAAAAAAAAAAAAAAAAAAAAaa", "output": "AAAAAAAAAAAAAAAAAAAAAAAAaa" }, { "input": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA" }, { "input": "dDDDDDDDDDDDDD", "output": "Dddddddddddddd" }, { "input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" }, { "input": "z", "output": "Z" }, { "input": "AZ", "output": "az" }, { "input": "Z", "output": "z" }, { "input": "aAAAA", "output": "Aaaaa" }, { "input": "F", "output": "f" } ]

1,697,447,545

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

30

0

str1=input("输入字母") if len(str1)<=100: print(str1.title())

Title: cAPS lOCK Time Limit: None seconds Memory Limit: None megabytes Problem Description: wHAT DO WE NEED cAPS LOCK FOR? Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage. Let's consider that a word has been typed with the Caps lock key accidentally switched on, if: - either it only contains uppercase letters; - or all letters except for the first one are uppercase. In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed. Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged. Input Specification: The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive. Output Specification: Print the result of the given word's processing. Demo Input: ['cAPS\n', 'Lock\n'] Demo Output: ['Caps', 'Lock\n'] Note: none

```python str1=input("输入字母") if len(str1)<=100: print(str1.title()) ```

-1

0

none

0

[ "none" ]

null

Vasya has a sequence of cubes and exactly one integer is written on each cube. Vasya exhibited all his cubes in a row. So the sequence of numbers written on the cubes in the order from the left to the right equals to *a*1,<=*a*2,<=...,<=*a**n*. While Vasya was walking, his little brother Stepan played with Vasya's cubes and changed their order, so now the sequence of numbers written on the cubes became equal to *b*1,<=*b*2,<=...,<=*b**n*. Stepan said that he swapped only cubes which where on the positions between *l* and *r*, inclusive, and did not remove or add any other cubes (i. e. he said that he reordered cubes between positions *l* and *r*, inclusive, in some way). Your task is to determine if it is possible that Stepan said the truth, or it is guaranteed that Stepan deceived his brother.

The first line contains three integers *n*, *l*, *r* (1<=≤<=*n*<=≤<=105, 1<=≤<=*l*<=≤<=*r*<=≤<=*n*) — the number of Vasya's cubes and the positions told by Stepan. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the sequence of integers written on cubes in the Vasya's order. The third line contains the sequence *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=*n*) — the sequence of integers written on cubes after Stepan rearranged their order. It is guaranteed that Stepan did not remove or add other cubes, he only rearranged Vasya's cubes.

Print "LIE" (without quotes) if it is guaranteed that Stepan deceived his brother. In the other case, print "TRUTH" (without quotes).

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

[ "TRUTH\n", "LIE\n", "TRUTH\n" ]

In the first example there is a situation when Stepan said the truth. Initially the sequence of integers on the cubes was equal to [3, 4, 2, 3, 1]. Stepan could at first swap cubes on positions 2 and 3 (after that the sequence of integers on cubes became equal to [3, 2, 4, 3, 1]), and then swap cubes in positions 3 and 4 (after that the sequence of integers on cubes became equal to [3, 2, 3, 4, 1]). In the second example it is not possible that Stepan said truth because he said that he swapped cubes only between positions 1 and 2, but we can see that it is guaranteed that he changed the position of the cube which was on the position 3 at first. So it is guaranteed that Stepan deceived his brother. In the third example for any values *l* and *r* there is a situation when Stepan said the truth.

0

[ { "input": "5 2 4\n3 4 2 3 1\n3 2 3 4 1", "output": "TRUTH" }, { "input": "3 1 2\n1 2 3\n3 1 2", "output": "LIE" }, { "input": "4 2 4\n1 1 1 1\n1 1 1 1", "output": "TRUTH" }, { "input": "5 1 3\n2 2 2 1 2\n2 2 2 1 2", "output": "TRUTH" }, { "input": "7 1 4\n2 5 5 5 4 3 4\n2 5 5 5 4 3 4", "output": "TRUTH" }, { "input": "10 1 10\n6 7 6 1 10 10 9 5 3 9\n7 10 9 6 1 5 9 3 10 6", "output": "TRUTH" }, { "input": "1 1 1\n1\n1", "output": "TRUTH" }, { "input": "4 3 4\n1 2 3 4\n2 1 3 4", "output": "LIE" }, { "input": "7 2 4\n1 2 3 4 5 7 6\n1 2 3 4 5 6 7", "output": "LIE" }, { "input": "5 1 2\n1 2 3 4 5\n1 2 3 5 4", "output": "LIE" }, { "input": "8 3 6\n5 3 1 1 1 1 3 5\n3 3 1 1 1 1 5 5", "output": "LIE" }, { "input": "4 2 2\n2 1 2 2\n1 2 2 2", "output": "LIE" } ]

1,492,200,880

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

1

77

23,142,400

size,index1,index2 = map(int, input().split()) a = list(map(int, input().split())) b = list(map(int, input().split())) i = 0 while i<index1-1: if a[i] != b[i]: print("LIE") exit() i+=1 i = index2+1 while i <size: if a[i] != b[i]: print("LIE") exit() i+=1 print("TRUTH")

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya has a sequence of cubes and exactly one integer is written on each cube. Vasya exhibited all his cubes in a row. So the sequence of numbers written on the cubes in the order from the left to the right equals to *a*1,<=*a*2,<=...,<=*a**n*. While Vasya was walking, his little brother Stepan played with Vasya's cubes and changed their order, so now the sequence of numbers written on the cubes became equal to *b*1,<=*b*2,<=...,<=*b**n*. Stepan said that he swapped only cubes which where on the positions between *l* and *r*, inclusive, and did not remove or add any other cubes (i. e. he said that he reordered cubes between positions *l* and *r*, inclusive, in some way). Your task is to determine if it is possible that Stepan said the truth, or it is guaranteed that Stepan deceived his brother. Input Specification: The first line contains three integers *n*, *l*, *r* (1<=≤<=*n*<=≤<=105, 1<=≤<=*l*<=≤<=*r*<=≤<=*n*) — the number of Vasya's cubes and the positions told by Stepan. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the sequence of integers written on cubes in the Vasya's order. The third line contains the sequence *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=*n*) — the sequence of integers written on cubes after Stepan rearranged their order. It is guaranteed that Stepan did not remove or add other cubes, he only rearranged Vasya's cubes. Output Specification: Print "LIE" (without quotes) if it is guaranteed that Stepan deceived his brother. In the other case, print "TRUTH" (without quotes). Demo Input: ['5 2 4\n3 4 2 3 1\n3 2 3 4 1\n', '3 1 2\n1 2 3\n3 1 2\n', '4 2 4\n1 1 1 1\n1 1 1 1\n'] Demo Output: ['TRUTH\n', 'LIE\n', 'TRUTH\n'] Note: In the first example there is a situation when Stepan said the truth. Initially the sequence of integers on the cubes was equal to [3, 4, 2, 3, 1]. Stepan could at first swap cubes on positions 2 and 3 (after that the sequence of integers on cubes became equal to [3, 2, 4, 3, 1]), and then swap cubes in positions 3 and 4 (after that the sequence of integers on cubes became equal to [3, 2, 3, 4, 1]). In the second example it is not possible that Stepan said truth because he said that he swapped cubes only between positions 1 and 2, but we can see that it is guaranteed that he changed the position of the cube which was on the position 3 at first. So it is guaranteed that Stepan deceived his brother. In the third example for any values *l* and *r* there is a situation when Stepan said the truth.

```python size,index1,index2 = map(int, input().split()) a = list(map(int, input().split())) b = list(map(int, input().split())) i = 0 while i<index1-1: if a[i] != b[i]: print("LIE") exit() i+=1 i = index2+1 while i <size: if a[i] != b[i]: print("LIE") exit() i+=1 print("TRUTH") ```

0

802

G

Fake News (easy)

PROGRAMMING

800

[ "implementation", "strings" ]

null

As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...

The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z).

Output YES if the string *s* contains heidi as a subsequence and NO otherwise.

[ "abcheaibcdi\n", "hiedi\n" ]

[ "YES", "NO" ]

A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*.

0

[ { "input": "abcheaibcdi", "output": "YES" }, { "input": "hiedi", "output": "NO" }, { "input": "ihied", "output": "NO" }, { "input": "diehi", "output": "NO" }, { "input": "deiih", "output": "NO" }, { "input": "iheid", "output": "NO" }, { "input": "eihdi", "output": "NO" }, { "input": "ehdii", "output": "NO" }, { "input": "edhii", "output": "NO" }, { "input": "deiih", "output": "NO" }, { "input": "ehdii", "output": "NO" }, { "input": "eufyajkssayhjhqcwxmctecaeepjwmfoscqprpcxsqfwnlgzsmmuwuoruantipholrauvxydfvftwfzhnckxswussvlidcojiciflpvkcxkkcmmvtfvxrkwcpeelwsuzqgamamdtdgzscmikvojfvqehblmjczkvtdeymgertgkwfwfukafqlfdhtedcctixhyetdypswgagrpyto", "output": "YES" }, { "input": "arfbvxgdvqzuloojjrwoyqqbxamxybaqltfimofulusfebodjkwwrgwcppkwiodtpjaraglyplgerrpqjkpoggjmfxhwtqrijpijrcyxnoodvwpyjfpvqaoazllbrpzananbrvvybboedidtuvqquklkpeflfaltukjhzjgiofombhbmqbihgtapswykfvlgdoapjqntvqsaohmbvnphvyyhvhavslamczuqifxnwknkaenqmlvetrqogqxmlptgrmqvxzdxdmwobjesmgxckpmawtioavwdngyiwkzypfnxcovwzdohshwlavwsthdssiadhiwmhpvgkrbezm", "output": "YES" }, { "input": "zcectngbqnejjjtsfrluummmqabzqbyccshjqbrjthzhlbmzjfxugvjouwhumsgrnopiyakfadjnbsesamhynsbfbfunupwbxvohfmpwlcpxhovwpfpciclatgmiufwdvtsqrsdcymvkldpnhfeisrzhyhhlkwdzthgprvkpyldeysvbmcibqkpudyrraqdlxpjecvwcvuiklcrsbgvqasmxmtxqzmawcjtozioqlfflinnxpeexbzloaeqjvglbdeufultpjqexvjjjkzemtzuzmxvawilcqdrcjzpqyhtwfphuonzwkotthsaxrmwtnlmcdylxqcfffyndqeouztluqwlhnkkvzwcfiscikv", "output": "YES" }, { "input": "plqaykgovxkvsiahdbglktdlhcqwelxxmtlyymrsyubxdskvyjkrowvcbpdofpjqspsrgpakdczletxujzlsegepzleipiyycpinzxgwjsgslnxsotouddgfcybozfpjhhocpybfjbaywsehbcfrayvancbrumdfngqytnhihyxnlvilrqyhnxeckprqafofelospffhtwguzjbbjlzbqrtiielbvzutzgpqxosiaqznndgobcluuqlhmffiowkjdlkokehtjdyjvmxsiyxureflmdomerfekxdvtitvwzmdsdzplkpbtafxqfpudnhfqpoiwvjnylanunmagoweobdvfjgepbsymfutrjarlxclhgavpytiiqwvojrptofuvlohzeguxdsrihsbucelhhuedltnnjgzxwyblbqvnoliiydfinzlogbvucwykryzcyibnniggbkdkdcdgcsbvvnavtyhtkanrblpvomvjs", "output": "YES" }, { "input": "fbldqzggeunkpwcfirxanmntbfrudijltoertsdvcvcmbwodbibsrxendzebvxwydpasaqnisrijctsuatihxxygbeovhxjdptdcppkvfytdpjspvrannxavmkmisqtygntxkdlousdypyfkrpzapysfpdbyprufwzhunlsfugojddkmxzinatiwfxdqmgyrnjnxvrclhxyuwxtshoqdjptmeecvgmrlvuwqtmnfnfeeiwcavwnqmyustawbjodzwsqmnjxhpqmgpysierlwbbdzcwprpsexyvreewcmlbvaiytjlxdqdaqftefdlmtmmjcwvfejshymhnouoshdzqcwzxpzupkbcievodzqkqvyjuuxxwepxjalvkzufnveji", "output": "YES" }, { "input": "htsyljgoelbbuipivuzrhmfpkgderqpoprlxdpasxhpmxvaztccldtmujjzjmcpdvsdghzpretlsyyiljhjznseaacruriufswuvizwwuvdioazophhyytvbiogttnnouauxllbdn", "output": "YES" }, { "input": "ikmxzqdzxqlvgeojsnhqzciujslwjyzzexnregabdqztpplosdakimjxmuqccbnwvzbajoiqgdobccwnrwmixohrbdarhoeeelzbpigiybtesybwefpcfx", "output": "YES" }, { "input": "bpvbpjvbdfiodsmahxpcubjxdykesubnypalhypantshkjffmxjmelblqnjdmtaltneuyudyevkgedkqrdmrfeemgpghwrifcwincfixokfgurhqbcfzeajrgkgpwqwsepudxulywowwxzdxkumsicsvnzfxspmjpaixgejeaoyoibegosqoyoydmphfpbutrrewyjecowjckvpcceoamtfbitdneuwqfvnagswlskmsmkhmxyfsrpqwhxzocyffiumcy", "output": "YES" }, { "input": "vllsexwrazvlfvhvrtqeohvzzresjdiuhomfpgqcxpqdevplecuaepixhlijatxzegciizpvyvxuembiplwklahlqibykfideysjygagjbgqkbhdhkatddcwlxboinfuomnpc", "output": "YES" }, { "input": "pnjdwpxmvfoqkjtbhquqcuredrkwqzzfjmdvpnbqtypzdovemhhclkvigjvtprrpzbrbcbatkucaqteuciuozytsptvsskkeplaxdaqmjkmef", "output": "NO" }, { "input": "jpwfhvlxvsdhtuozvlmnfiotrgapgjxtcsgcjnodcztupysvvvmjpzqkpommadppdrykuqkcpzojcwvlogvkddedwbggkrhuvtsvdiokehlkdlnukcufjvqxnikcdawvexxwffxtriqbdmkahxdtygodzohwtdmmuvmatdkvweqvaehaxiefpevkvqpyxsrhtmgjsdfcwzqobibeduooldrmglbinrepmunizheqzvgqvpdskhxfidxfnbisyizhepwyrcykcmjxnkyfjgrqlkixcvysa", "output": "YES" }, { "input": "aftcrvuumeqbfvaqlltscnuhkpcifrrhnutjinxdhhdbzvizlrapzjdatuaynoplgjketupgaejciosofuhcgcjdcucarfvtsofgubtphijciswsvidnvpztlaarydkeqxzwdhfbmullkimerukusbrdnnujviydldrwhdfllsjtziwfeaiqotbiprespmxjulnyunkdtcghrzvhtcychkwatqqmladxpvmvlkzscthylbzkpgwlzfjqwarqvdeyngekqvrhrftpxnkfcibbowvnqdkulcdydspcubwlgoyinpnzgidbgunparnueddzwtzdiavbprbbg", "output": "YES" }, { "input": "oagjghsidigeh", "output": "NO" }, { "input": "chdhzpfzabupskiusjoefrwmjmqkbmdgboicnszkhdrlegeqjsldurmbshijadlwsycselhlnudndpdhcnhruhhvsgbthpruiqfirxkhpqhzhqdfpyozolbionodypfcqfeqbkcgmqkizgeyyelzeoothexcoaahedgrvoemqcwccbvoeqawqeuusyjxmgjkpfwcdttfmwunzuwvsihliexlzygqcgpbdiawfvqukikhbjerjkyhpcknlndaystrgsinghlmekbvhntcpypmchcwoglsmwwdulqneuabuuuvtyrnjxfcgoothalwkzzfxakneusezgnnepkpipzromqubraiggqndliz", "output": "YES" }, { "input": "lgirxqkrkgjcutpqitmffvbujcljkqardlalyigxorscczuzikoylcxenryhskoavymexysvmhbsvhtycjlmzhijpuvcjshyfeycvvcfyzytzoyvxajpqdjtfiatnvxnyeqtfcagfftafllhhjhplbdsrfpctkqpinpdfrtlzyjllfbeffputywcckupyslkbbzpgcnxgbmhtqeqqehpdaokkjtatrhyiuusjhwgiiiikxpzdueasemosmmccoakafgvxduwiuflovhhfhffgnnjhoperhhjtvocpqytjxkmrknnknqeglffhfuplopmktykxuvcmbwpoeisrlyyhdpxfvzseucofyhziuiikihpqheqdyzwigeaqzhxzvporgisxgvhyicqyejovqloibhbunsvsunpvmdckkbuokitdzleilfwutcvuuytpupizinfjrzhxudsmjcjyfcpfgthujjowdwtgbvi", "output": "YES" }, { "input": "uuehrvufgerqbzyzksmqnewacotuimawhlbycdbsmhshrsbqwybbkwjwsrkwptvlbbwjiivqugzrxxwgidrcrhrwsmwgeoleptfamzefgaeyxouxocrpvomjrazmxrnffdwrrmblgdiabdncvfougtmjgvvazasnygdrigbsrieoonirlivfyodvulouslxosswgpdexuldmkdbpdlgutiotvxjyecbrsvbmqxrlcpcipjjncduyqtohlzybvlemmfdeubihwlwqglkgjvnwrbgydcpwklmjeewqklmqdbajqgrpnynaxfvxjzgibqerxyhnxenrmcdqaaeksbzyrcaepozqpetaurlhjuxxhwppuhgoihxdxbmxeiahyaqkbknktlzkheaarjoqqrsyeducvoygwalgarldcdlqogfvsncejssmx", "output": "YES" }, { "input": "iiopulfjxoitgiusqrhgbkiyzinphjtclodbkkydetylvuimkhdkklmyoacmekdvjpuxcrvqnjhqhhbfenlpzpwijtykqziocilvtpqhxuyrphdlamawjuzgjwiebkqyrzyqgtkcrhntjxqmcgkrqcslogjegfrivzidfedeegwbbsopvxvdoididlpypjogxaodtueebbwuwilacunqpozczcgrpaxxrtogpxgfkudtxchoravrrdtimhenwmnafxaruuojduxxglefejycfcyxllfimkszmbrhcwpnwjqgwvzysmlaaifdxfjjpgjmcksiigpanappjdloiolojmcqbnpnjjzaufdpjaknylmyvolhwypygwepmqwpiglpcnpypnudhdzpdvgzosyjthzcwtytxq", "output": "YES" } ]

1,666,073,103

2,147,483,647

Python 3

OK

TESTS

58

46

0

def news(s): h= "heidi" c= 0 m = len(s) for k in range(m): if s[k] == h[c]: c+= 1 if c == 5: break if c == 5: print("YES") else: print( "NO") s = input() news(s)

Title: Fake News (easy) Time Limit: None seconds Memory Limit: None megabytes Problem Description: As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it... Input Specification: The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z). Output Specification: Output YES if the string *s* contains heidi as a subsequence and NO otherwise. Demo Input: ['abcheaibcdi\n', 'hiedi\n'] Demo Output: ['YES', 'NO'] Note: A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*.

```python def news(s): h= "heidi" c= 0 m = len(s) for k in range(m): if s[k] == h[c]: c+= 1 if c == 5: break if c == 5: print("YES") else: print( "NO") s = input() news(s) ```

3

849

A

Odds and Ends

PROGRAMMING

1,000

[ "implementation" ]

null

Where do odds begin, and where do they end? Where does hope emerge, and will they ever break? Given an integer sequence *a*1,<=*a*2,<=...,<=*a**n* of length *n*. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers. A subsegment is a contiguous slice of the whole sequence. For example, {3,<=4,<=5} and {1} are subsegments of sequence {1,<=2,<=3,<=4,<=5,<=6}, while {1,<=2,<=4} and {7} are not.

The first line of input contains a non-negative integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains *n* space-separated non-negative integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — the elements of the sequence.

Output "Yes" if it's possible to fulfill the requirements, and "No" otherwise. You can output each letter in any case (upper or lower).

[ "3\n1 3 5\n", "5\n1 0 1 5 1\n", "3\n4 3 1\n", "4\n3 9 9 3\n" ]

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

In the first example, divide the sequence into 1 subsegment: {1, 3, 5} and the requirements will be met. In the second example, divide the sequence into 3 subsegments: {1, 0, 1}, {5}, {1}. In the third example, one of the subsegments must start with 4 which is an even number, thus the requirements cannot be met. In the fourth example, the sequence can be divided into 2 subsegments: {3, 9, 9}, {3}, but this is not a valid solution because 2 is an even number.

500

[ { "input": "3\n1 3 5", "output": "Yes" }, { "input": "5\n1 0 1 5 1", "output": "Yes" }, { "input": "3\n4 3 1", "output": "No" }, { "input": "4\n3 9 9 3", "output": "No" }, { "input": "1\n1", "output": "Yes" }, { "input": "5\n100 99 100 99 99", "output": "No" }, { "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": "No" }, { "input": "1\n0", "output": "No" }, { "input": "2\n1 1", "output": "No" }, { "input": "2\n10 10", "output": "No" }, { "input": "2\n54 21", "output": "No" }, { "input": "5\n0 0 0 0 0", "output": "No" }, { "input": "5\n67 92 0 26 43", "output": "Yes" }, { "input": "15\n45 52 35 80 68 80 93 57 47 32 69 23 63 90 43", "output": "Yes" }, { "input": "15\n81 28 0 82 71 64 63 89 87 92 38 30 76 72 36", "output": "No" }, { "input": "50\n49 32 17 59 77 98 65 50 85 10 40 84 65 34 52 25 1 31 61 45 48 24 41 14 76 12 33 76 44 86 53 33 92 58 63 93 50 24 31 79 67 50 72 93 2 38 32 14 87 99", "output": "No" }, { "input": "55\n65 69 53 66 11 100 68 44 43 17 6 66 24 2 6 6 61 72 91 53 93 61 52 96 56 42 6 8 79 49 76 36 83 58 8 43 2 90 71 49 80 21 75 13 76 54 95 61 58 82 40 33 73 61 46", "output": "No" }, { "input": "99\n73 89 51 85 42 67 22 80 75 3 90 0 52 100 90 48 7 15 41 1 54 2 23 62 86 68 2 87 57 12 45 34 68 54 36 49 27 46 22 70 95 90 57 91 90 79 48 89 67 92 28 27 25 37 73 66 13 89 7 99 62 53 48 24 73 82 62 88 26 39 21 86 50 95 26 27 60 6 56 14 27 90 55 80 97 18 37 36 70 2 28 53 36 77 39 79 82 42 69", "output": "Yes" }, { "input": "99\n99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99", "output": "Yes" }, { "input": "100\n61 63 34 45 20 91 31 28 40 27 94 1 73 5 69 10 56 94 80 23 79 99 59 58 13 56 91 59 77 78 88 72 80 72 70 71 63 60 41 41 41 27 83 10 43 14 35 48 0 78 69 29 63 33 42 67 1 74 51 46 79 41 37 61 16 29 82 28 22 14 64 49 86 92 82 55 54 24 75 58 95 31 3 34 26 23 78 91 49 6 30 57 27 69 29 54 42 0 61 83", "output": "No" }, { "input": "6\n1 2 2 2 2 1", "output": "No" }, { "input": "3\n1 2 1", "output": "Yes" }, { "input": "4\n1 3 2 3", "output": "No" }, { "input": "6\n1 1 1 1 1 1", "output": "No" }, { "input": "6\n1 1 0 0 1 1", "output": "No" }, { "input": "4\n1 4 9 3", "output": "No" }, { "input": "4\n1 0 1 1", "output": "No" }, { "input": "10\n1 0 0 1 1 1 1 1 1 1", "output": "No" }, { "input": "10\n9 2 5 7 8 3 1 9 4 9", "output": "No" }, { "input": "99\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 2", "output": "No" }, { "input": "6\n1 2 1 2 2 1", "output": "No" }, { "input": "6\n1 0 1 0 0 1", "output": "No" }, { "input": "4\n1 3 4 7", "output": "No" }, { "input": "8\n1 1 1 2 1 1 1 1", "output": "No" }, { "input": "3\n1 1 2", "output": "No" }, { "input": "5\n1 2 1 2 1", "output": "Yes" }, { "input": "5\n5 4 4 2 1", "output": "Yes" }, { "input": "6\n1 3 3 3 3 1", "output": "No" }, { "input": "7\n1 2 1 2 2 2 1", "output": "Yes" }, { "input": "4\n1 2 2 1", "output": "No" }, { "input": "6\n1 2 3 4 6 5", "output": "No" }, { "input": "5\n1 1 2 2 2", "output": "No" }, { "input": "5\n1 0 0 1 1", "output": "Yes" }, { "input": "3\n1 2 4", "output": "No" }, { "input": "3\n1 0 2", "output": "No" }, { "input": "5\n1 1 1 0 1", "output": "Yes" }, { "input": "4\n3 9 2 3", "output": "No" }, { "input": "6\n1 1 1 4 4 1", "output": "No" }, { "input": "6\n1 2 3 5 6 7", "output": "No" }, { "input": "6\n1 1 1 2 2 1", "output": "No" }, { "input": "6\n1 1 1 0 0 1", "output": "No" }, { "input": "5\n1 2 2 5 5", "output": "Yes" }, { "input": "5\n1 3 2 4 5", "output": "Yes" }, { "input": "8\n1 2 3 5 7 8 8 5", "output": "No" }, { "input": "10\n1 1 1 2 1 1 1 1 1 1", "output": "No" }, { "input": "4\n1 0 0 1", "output": "No" }, { "input": "7\n1 0 1 1 0 0 1", "output": "Yes" }, { "input": "7\n1 4 5 7 6 6 3", "output": "Yes" }, { "input": "4\n2 2 2 2", "output": "No" }, { "input": "5\n2 3 4 5 6", "output": "No" }, { "input": "4\n1 1 2 1", "output": "No" }, { "input": "3\n1 2 3", "output": "Yes" }, { "input": "6\n1 3 3 2 2 3", "output": "No" }, { "input": "4\n1 1 2 3", "output": "No" }, { "input": "4\n1 2 3 5", "output": "No" }, { "input": "5\n3 4 4 3 3", "output": "Yes" }, { "input": "4\n3 2 2 3", "output": "No" }, { "input": "6\n1 1 1 1 2 1", "output": "No" }, { "input": "6\n1 1 2 2 1 1", "output": "No" }, { "input": "10\n3 4 2 4 3 2 2 4 4 3", "output": "No" }, { "input": "7\n1 2 4 3 2 4 5", "output": "Yes" }, { "input": "28\n75 51 25 52 13 7 34 29 5 59 68 56 13 2 9 37 59 83 18 32 36 30 20 43 92 76 78 67", "output": "No" }, { "input": "79\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 18", "output": "No" }, { "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": "No" } ]

1,558,099,864

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

93

0

a=int(input('')) b = [int(z) for z in input('').split()] if b[0]%2==0: print('No') if len(b)%2==0: print('No') else: print('Yes')

Title: Odds and Ends Time Limit: None seconds Memory Limit: None megabytes Problem Description: Where do odds begin, and where do they end? Where does hope emerge, and will they ever break? Given an integer sequence *a*1,<=*a*2,<=...,<=*a**n* of length *n*. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers. A subsegment is a contiguous slice of the whole sequence. For example, {3,<=4,<=5} and {1} are subsegments of sequence {1,<=2,<=3,<=4,<=5,<=6}, while {1,<=2,<=4} and {7} are not. Input Specification: The first line of input contains a non-negative integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains *n* space-separated non-negative integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — the elements of the sequence. Output Specification: Output "Yes" if it's possible to fulfill the requirements, and "No" otherwise. You can output each letter in any case (upper or lower). Demo Input: ['3\n1 3 5\n', '5\n1 0 1 5 1\n', '3\n4 3 1\n', '4\n3 9 9 3\n'] Demo Output: ['Yes\n', 'Yes\n', 'No\n', 'No\n'] Note: In the first example, divide the sequence into 1 subsegment: {1, 3, 5} and the requirements will be met. In the second example, divide the sequence into 3 subsegments: {1, 0, 1}, {5}, {1}. In the third example, one of the subsegments must start with 4 which is an even number, thus the requirements cannot be met. In the fourth example, the sequence can be divided into 2 subsegments: {3, 9, 9}, {3}, but this is not a valid solution because 2 is an even number.

```python a=int(input('')) b = [int(z) for z in input('').split()] if b[0]%2==0: print('No') if len(b)%2==0: print('No') else: print('Yes') ```

0

1,006

B

Polycarp's Practice

PROGRAMMING

1,200

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

null

Polycarp is practicing his problem solving skill. He has a list of $n$ problems with difficulties $a_1, a_2, \dots, a_n$, respectively. His plan is to practice for exactly $k$ days. Each day he has to solve at least one problem from his list. Polycarp solves the problems in the order they are given in his list, he cannot skip any problem from his list. He has to solve all $n$ problems in exactly $k$ days. Thus, each day Polycarp solves a contiguous sequence of (consecutive) problems from the start of the list. He can't skip problems or solve them multiple times. As a result, in $k$ days he will solve all the $n$ problems. The profit of the $j$-th day of Polycarp's practice is the maximum among all the difficulties of problems Polycarp solves during the $j$-th day (i.e. if he solves problems with indices from $l$ to $r$ during a day, then the profit of the day is $\max\limits_{l \le i \le r}a_i$). The total profit of his practice is the sum of the profits over all $k$ days of his practice. You want to help Polycarp to get the maximum possible total profit over all valid ways to solve problems. Your task is to distribute all $n$ problems between $k$ days satisfying the conditions above in such a way, that the total profit is maximum. For example, if $n = 8, k = 3$ and $a = [5, 4, 2, 6, 5, 1, 9, 2]$, one of the possible distributions with maximum total profit is: $[5, 4, 2], [6, 5], [1, 9, 2]$. Here the total profit equals $5 + 6 + 9 = 20$.

The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2000$) — the number of problems and the number of days, respectively. The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2000$) — difficulties of problems in Polycarp's list, in the order they are placed in the list (i.e. in the order Polycarp will solve them).

In the first line of the output print the maximum possible total profit. In the second line print exactly $k$ positive integers $t_1, t_2, \dots, t_k$ ($t_1 + t_2 + \dots + t_k$ must equal $n$), where $t_j$ means the number of problems Polycarp will solve during the $j$-th day in order to achieve the maximum possible total profit of his practice. If there are many possible answers, you may print any of them.

[ "8 3\n5 4 2 6 5 1 9 2\n", "5 1\n1 1 1 1 1\n", "4 2\n1 2000 2000 2\n" ]

[ "20\n3 2 3", "1\n5\n", "4000\n2 2\n" ]

The first example is described in the problem statement. In the second example there is only one possible distribution. In the third example the best answer is to distribute problems in the following way: $[1, 2000], [2000, 2]$. The total profit of this distribution is $2000 + 2000 = 4000$.

0

[ { "input": "8 3\n5 4 2 6 5 1 9 2", "output": "20\n4 1 3" }, { "input": "5 1\n1 1 1 1 1", "output": "1\n5" }, { "input": "4 2\n1 2000 2000 2", "output": "4000\n2 2" }, { "input": "1 1\n2000", "output": "2000\n1" }, { "input": "1 1\n1234", "output": "1234\n1" }, { "input": "3 2\n1 1 1", "output": "2\n2 1" }, { "input": "4 2\n3 5 1 1", "output": "8\n1 3" }, { "input": "5 3\n5 5 6 7 1", "output": "18\n2 1 2" }, { "input": "6 4\n1 1 1 1 2 2", "output": "6\n3 1 1 1" }, { "input": "5 3\n5 5 6 6 4", "output": "17\n2 1 2" }, { "input": "16 15\n14 4 9 12 17 1 1 8 12 13 6 9 17 2 18 12", "output": "154\n1 1 1 1 1 2 1 1 1 1 1 1 1 1 1" }, { "input": "1 1\n1996", "output": "1996\n1" }, { "input": "5 3\n5 5 5 9 10", "output": "24\n3 1 1" }, { "input": "18 15\n18 2 13 1 18 3 2 18 18 20 9 2 20 20 4 20 9 12", "output": "204\n1 2 2 1 2 1 1 1 1 1 1 1 1 1 1" }, { "input": "5 3\n1 20 20 50 50", "output": "120\n3 1 1" }, { "input": "8 3\n15 14 11 19 17 14 14 8", "output": "51\n1 3 4" }, { "input": "5 2\n15 20 6 19 6", "output": "39\n2 3" }, { "input": "6 3\n5 5 5 5 6 9", "output": "20\n4 1 1" }, { "input": "5 3\n2 2 2 3 3", "output": "8\n3 1 1" }, { "input": "7 3\n2 2 2 2 2 3 3", "output": "8\n5 1 1" }, { "input": "6 5\n1 1 6 6 6 6", "output": "25\n2 1 1 1 1" }, { "input": "8 4\n1 2 2 2 2 3 4 5", "output": "14\n5 1 1 1" }, { "input": "6 4\n1 1 1 5 5 5", "output": "16\n3 1 1 1" }, { "input": "6 3\n1 2 2 2 4 5", "output": "11\n4 1 1" }, { "input": "18 6\n17 17 19 14 10 20 18 16 6 7 2 15 14 16 13 6 12 11", "output": "107\n1 1 1 3 1 11" }, { "input": "6 3\n1 1 2 2 3 4", "output": "9\n4 1 1" }, { "input": "8 3\n5 4 2 5 6 1 9 2", "output": "20\n4 1 3" } ]

1,615,812,881

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

46

307,200

''' Online Python Compiler. Code, Compile, Run and Debug python program online. Write your code in this editor and press "Run" button to execute it. ''' n,k=map(int,input().split()) l=list(map(int,input().split())) d={} for i in range(n): if(l[i] not in d): d[l[i]]=i l.sort(reverse=True) p=[] s=0 for i in range(k): s+=l[i] p.append(d[l[i]]) print(p) print(s) p.sort() for i in range(k): if(i==k-1): if(i==0): print(n) else: print(n-p[i-1]-1) elif(i==0): print(p[i]+1,end=" ") else: print(p[i]-p[i-1],end=" ") print()

Title: Polycarp's Practice Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp is practicing his problem solving skill. He has a list of $n$ problems with difficulties $a_1, a_2, \dots, a_n$, respectively. His plan is to practice for exactly $k$ days. Each day he has to solve at least one problem from his list. Polycarp solves the problems in the order they are given in his list, he cannot skip any problem from his list. He has to solve all $n$ problems in exactly $k$ days. Thus, each day Polycarp solves a contiguous sequence of (consecutive) problems from the start of the list. He can't skip problems or solve them multiple times. As a result, in $k$ days he will solve all the $n$ problems. The profit of the $j$-th day of Polycarp's practice is the maximum among all the difficulties of problems Polycarp solves during the $j$-th day (i.e. if he solves problems with indices from $l$ to $r$ during a day, then the profit of the day is $\max\limits_{l \le i \le r}a_i$). The total profit of his practice is the sum of the profits over all $k$ days of his practice. You want to help Polycarp to get the maximum possible total profit over all valid ways to solve problems. Your task is to distribute all $n$ problems between $k$ days satisfying the conditions above in such a way, that the total profit is maximum. For example, if $n = 8, k = 3$ and $a = [5, 4, 2, 6, 5, 1, 9, 2]$, one of the possible distributions with maximum total profit is: $[5, 4, 2], [6, 5], [1, 9, 2]$. Here the total profit equals $5 + 6 + 9 = 20$. Input Specification: The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2000$) — the number of problems and the number of days, respectively. The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2000$) — difficulties of problems in Polycarp's list, in the order they are placed in the list (i.e. in the order Polycarp will solve them). Output Specification: In the first line of the output print the maximum possible total profit. In the second line print exactly $k$ positive integers $t_1, t_2, \dots, t_k$ ($t_1 + t_2 + \dots + t_k$ must equal $n$), where $t_j$ means the number of problems Polycarp will solve during the $j$-th day in order to achieve the maximum possible total profit of his practice. If there are many possible answers, you may print any of them. Demo Input: ['8 3\n5 4 2 6 5 1 9 2\n', '5 1\n1 1 1 1 1\n', '4 2\n1 2000 2000 2\n'] Demo Output: ['20\n3 2 3', '1\n5\n', '4000\n2 2\n'] Note: The first example is described in the problem statement. In the second example there is only one possible distribution. In the third example the best answer is to distribute problems in the following way: $[1, 2000], [2000, 2]$. The total profit of this distribution is $2000 + 2000 = 4000$.

```python ''' Online Python Compiler. Code, Compile, Run and Debug python program online. Write your code in this editor and press "Run" button to execute it. ''' n,k=map(int,input().split()) l=list(map(int,input().split())) d={} for i in range(n): if(l[i] not in d): d[l[i]]=i l.sort(reverse=True) p=[] s=0 for i in range(k): s+=l[i] p.append(d[l[i]]) print(p) print(s) p.sort() for i in range(k): if(i==k-1): if(i==0): print(n) else: print(n-p[i-1]-1) elif(i==0): print(p[i]+1,end=" ") else: print(p[i]-p[i-1],end=" ") print() ```

0

270

A

Fancy Fence

PROGRAMMING

1,100

[ "geometry", "implementation", "math" ]

null

Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot. He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*. Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?

The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.

For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.

[ "3\n30\n60\n90\n" ]

[ "NO\nYES\nYES\n" ]

In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second test case, the fence is a regular triangle, and in the last test case — a square.

500

[ { "input": "3\n30\n60\n90", "output": "NO\nYES\nYES" }, { "input": "6\n1\n2\n3\n170\n179\n25", "output": "NO\nNO\nNO\nYES\nYES\nNO" } ]

1,689,600,717

2,147,483,647

PyPy 3-64

OK

TESTS

3

122

0

found=set() n=3 while 1: ang=(180*n-360)/n if ang>179: break if int(ang)*n==180*n-360: found.add(int(ang)) n+=1 t=int(input()) for _ in range(t): a=int(input()) if a in found: print('YES') else: print('NO')

Title: Fancy Fence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot. He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*. Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*? Input Specification: The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees. Output Specification: For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible. Demo Input: ['3\n30\n60\n90\n'] Demo Output: ['NO\nYES\nYES\n'] Note: In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In the second test case, the fence is a regular triangle, and in the last test case — a square.

```python found=set() n=3 while 1: ang=(180*n-360)/n if ang>179: break if int(ang)*n==180*n-360: found.add(int(ang)) n+=1 t=int(input()) for _ in range(t): a=int(input()) if a in found: print('YES') else: print('NO') ```

3

234

C

Weather

PROGRAMMING

1,300

[ "dp", "implementation" ]

null

Scientists say a lot about the problems of global warming and cooling of the Earth. Indeed, such natural phenomena strongly influence all life on our planet. Our hero Vasya is quite concerned about the problems. He decided to try a little experiment and observe how outside daily temperature changes. He hung out a thermometer on the balcony every morning and recorded the temperature. He had been measuring the temperature for the last *n* days. Thus, he got a sequence of numbers *t*1,<=*t*2,<=...,<=*t**n*, where the *i*-th number is the temperature on the *i*-th day. Vasya analyzed the temperature statistics in other cities, and came to the conclusion that the city has no environmental problems, if first the temperature outside is negative for some non-zero number of days, and then the temperature is positive for some non-zero number of days. More formally, there must be a positive integer *k* (1<=≤<=*k*<=≤<=*n*<=-<=1) such that *t*1<=<<=0,<=*t*2<=<<=0,<=...,<=*t**k*<=<<=0 and *t**k*<=+<=1<=><=0,<=*t**k*<=+<=2<=><=0,<=...,<=*t**n*<=><=0. In particular, the temperature should never be zero. If this condition is not met, Vasya decides that his city has environmental problems, and gets upset. You do not want to upset Vasya. Therefore, you want to select multiple values of temperature and modify them to satisfy Vasya's condition. You need to know what the least number of temperature values needs to be changed for that.

The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105) — the number of days for which Vasya has been measuring the temperature. The second line contains a sequence of *n* integers *t*1,<=*t*2,<=...,<=*t**n* (|*t**i*|<=≤<=109) — the sequence of temperature values. Numbers *t**i* are separated by single spaces.

Print a single integer — the answer to the given task.

[ "4\n-1 1 -2 1\n", "5\n0 -1 1 2 -5\n" ]

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

Note to the first sample: there are two ways to change exactly one number so that the sequence met Vasya's condition. You can either replace the first number 1 by any negative number or replace the number -2 by any positive number.

0

[ { "input": "4\n-1 1 -2 1", "output": "1" }, { "input": "5\n0 -1 1 2 -5", "output": "2" }, { "input": "6\n0 0 0 0 0 0", "output": "6" }, { "input": "6\n-1 -2 -3 -4 5 6", "output": "0" }, { "input": "8\n1 2 -1 0 10 2 12 13", "output": "3" }, { "input": "7\n-1 -2 -3 3 -1 3 4", "output": "1" }, { "input": "2\n3 -5", "output": "2" }, { "input": "50\n4 -8 0 -1 -3 -9 0 -2 0 1 -1 0 7 -10 9 7 0 -10 5 0 1 -6 9 -9 3 -3 3 7 4 -8 -8 3 3 -1 0 2 -6 10 7 -1 -6 -3 -4 2 3 0 -4 0 7 -9", "output": "26" }, { "input": "90\n52 -89 17 64 11 -61 92 51 42 -92 -14 -100 21 -88 73 -11 84 72 -80 -78 5 -70 -70 80 91 -89 87 -74 63 -79 -94 52 82 79 81 40 69 -15 33 -52 18 30 -39 99 84 -98 44 69 -75 0 60 -89 51 -92 83 73 16 -43 17 0 51 9 -53 86 86 -50 0 -80 3 0 86 0 -76 -45 0 -32 45 81 47 15 -62 21 4 -82 77 -67 -64 -12 0 -50", "output": "42" }, { "input": "10\n-19 -29 -21 -6 29 89 -74 -22 18 -13", "output": "3" }, { "input": "100\n-782 365 -283 769 -58 224 1000 983 7 595 -963 -267 -934 -187 -609 693 -316 431 859 -753 865 -421 861 -728 -793 621 -311 414 -101 -196 120 84 633 -362 989 94 206 19 -949 -629 489 376 -391 165 50 22 -209 735 565 61 -321 -256 890 34 343 -326 984 -268 -609 385 717 81 372 -391 271 -89 297 -510 797 -425 -276 573 510 560 165 -482 511 541 -491 60 168 -805 235 -657 -679 -617 -212 816 -98 901 380 103 608 -257 -643 333 8 355 743 -801", "output": "40" } ]

1,620,687,936

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

4

92

6,963,200

input_txt = open("input.txt", "r") n = int(input_txt.readline()) t = list(map(int, input_txt.readline().split())) res = [] def check_left(x): left = 0 for a in range(0, x): if t[a] > -1: left += 1 return left def check_right(y): right = 0 for b in range(y, n): if t[b] < 1: right += 1 return right i = 0 while i < n: left_res = check_left(i) right_res = check_right(i) total_res = left_res + right_res res.append(total_res) i += 1 output_txt = open("output.txt", "w") output_txt.write(str(min(res))) input_txt.close() output_txt.close()

Title: Weather Time Limit: None seconds Memory Limit: None megabytes Problem Description: Scientists say a lot about the problems of global warming and cooling of the Earth. Indeed, such natural phenomena strongly influence all life on our planet. Our hero Vasya is quite concerned about the problems. He decided to try a little experiment and observe how outside daily temperature changes. He hung out a thermometer on the balcony every morning and recorded the temperature. He had been measuring the temperature for the last *n* days. Thus, he got a sequence of numbers *t*1,<=*t*2,<=...,<=*t**n*, where the *i*-th number is the temperature on the *i*-th day. Vasya analyzed the temperature statistics in other cities, and came to the conclusion that the city has no environmental problems, if first the temperature outside is negative for some non-zero number of days, and then the temperature is positive for some non-zero number of days. More formally, there must be a positive integer *k* (1<=≤<=*k*<=≤<=*n*<=-<=1) such that *t*1<=<<=0,<=*t*2<=<<=0,<=...,<=*t**k*<=<<=0 and *t**k*<=+<=1<=><=0,<=*t**k*<=+<=2<=><=0,<=...,<=*t**n*<=><=0. In particular, the temperature should never be zero. If this condition is not met, Vasya decides that his city has environmental problems, and gets upset. You do not want to upset Vasya. Therefore, you want to select multiple values of temperature and modify them to satisfy Vasya's condition. You need to know what the least number of temperature values needs to be changed for that. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105) — the number of days for which Vasya has been measuring the temperature. The second line contains a sequence of *n* integers *t*1,<=*t*2,<=...,<=*t**n* (|*t**i*|<=≤<=109) — the sequence of temperature values. Numbers *t**i* are separated by single spaces. Output Specification: Print a single integer — the answer to the given task. Demo Input: ['4\n-1 1 -2 1\n', '5\n0 -1 1 2 -5\n'] Demo Output: ['1\n', '2\n'] Note: Note to the first sample: there are two ways to change exactly one number so that the sequence met Vasya's condition. You can either replace the first number 1 by any negative number or replace the number -2 by any positive number.

```python input_txt = open("input.txt", "r") n = int(input_txt.readline()) t = list(map(int, input_txt.readline().split())) res = [] def check_left(x): left = 0 for a in range(0, x): if t[a] > -1: left += 1 return left def check_right(y): right = 0 for b in range(y, n): if t[b] < 1: right += 1 return right i = 0 while i < n: left_res = check_left(i) right_res = check_right(i) total_res = left_res + right_res res.append(total_res) i += 1 output_txt = open("output.txt", "w") output_txt.write(str(min(res))) input_txt.close() output_txt.close() ```

0

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" } ]

1,694,628,756

2,147,483,647

PyPy 3-64

RUNTIME_ERROR

TESTS

0

61

2,867,200

n=int(input("")) l=[] for i in range(2*n): inp=input("") if inp!='': l.append(inp) for i in range(0,len(l)): if len(l[i])<=10: print(l[i]) else: w=l[i] print(w[0]+str(len(l[i])-2)+w[-1])

Title: Way Too Long Words Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. Output Specification: Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. Demo Input: ['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n'] Demo Output: ['word\nl10n\ni18n\np43s\n'] Note: none

```python n=int(input("")) l=[] for i in range(2*n): inp=input("") if inp!='': l.append(inp) for i in range(0,len(l)): if len(l[i])<=10: print(l[i]) else: w=l[i] print(w[0]+str(len(l[i])-2)+w[-1]) ```

-1

750

A

New Year and Hurry

PROGRAMMING

800

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

null

Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem. Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first. How many problems can Limak solve if he wants to make it to the party?

The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.

Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.

[ "3 222\n", "4 190\n", "7 1\n" ]

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

In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2. In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight. In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.

500

[ { "input": "3 222", "output": "2" }, { "input": "4 190", "output": "4" }, { "input": "7 1", "output": "7" }, { "input": "10 135", "output": "6" }, { "input": "10 136", "output": "5" }, { "input": "1 1", "output": "1" }, { "input": "1 240", "output": "0" }, { "input": "10 1", "output": "9" }, { "input": "10 240", "output": "0" }, { "input": "9 240", "output": "0" }, { "input": "9 1", "output": "9" }, { "input": "9 235", "output": "1" }, { "input": "9 236", "output": "0" }, { "input": "5 225", "output": "2" }, { "input": "5 226", "output": "1" }, { "input": "4 210", "output": "3" }, { "input": "4 211", "output": "2" }, { "input": "4 191", "output": "3" }, { "input": "10 165", "output": "5" }, { "input": "10 166", "output": "4" }, { "input": "8 100", "output": "7" }, { "input": "8 101", "output": "6" }, { "input": "8 60", "output": "8" }, { "input": "8 61", "output": "7" }, { "input": "10 15", "output": "9" }, { "input": "10 16", "output": "8" }, { "input": "4 100", "output": "4" }, { "input": "4 101", "output": "4" }, { "input": "7 167", "output": "4" }, { "input": "10 164", "output": "5" }, { "input": "9 170", "output": "4" }, { "input": "8 160", "output": "5" }, { "input": "1 100", "output": "1" }, { "input": "8 123", "output": "6" }, { "input": "2 99", "output": "2" }, { "input": "10 88", "output": "7" }, { "input": "1 235", "output": "1" }, { "input": "1 1", "output": "1" }, { "input": "4 240", "output": "0" }, { "input": "1 55", "output": "1" }, { "input": "1 240", "output": "0" }, { "input": "3 240", "output": "0" }, { "input": "10 240", "output": "0" }, { "input": "2 240", "output": "0" }, { "input": "10 1", "output": "9" }, { "input": "9 1", "output": "9" }, { "input": "2 236", "output": "0" }, { "input": "10 2", "output": "9" }, { "input": "3 239", "output": "0" }, { "input": "1 237", "output": "0" }, { "input": "7 8", "output": "7" }, { "input": "10 235", "output": "1" } ]

1,679,027,092

2,147,483,647

Python 3

OK

TESTS

52

46

0

n,k=list(map(int,input().split())) s=240-k c=0 b=0 for i in range(1,n+1): c=c+i*5 if c<=s: b=b+1 else: break print(b)

Title: New Year and Hurry Time Limit: None seconds Memory Limit: None megabytes Problem Description: Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem. Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first. How many problems can Limak solve if he wants to make it to the party? Input Specification: The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house. Output Specification: Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier. Demo Input: ['3 222\n', '4 190\n', '7 1\n'] Demo Output: ['2\n', '4\n', '7\n'] Note: In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2. In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight. In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.

```python n,k=list(map(int,input().split())) s=240-k c=0 b=0 for i in range(1,n+1): c=c+i*5 if c<=s: b=b+1 else: break print(b) ```

3

75

C

Modified GCD

PROGRAMMING

1,600

[ "binary search", "number theory" ]

C. Modified GCD

2

256

Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers. A common divisor for two positive numbers is a number which both numbers are divisible by. But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor *d* between two integers *a* and *b* that is in a given range from *low* to *high* (inclusive), i.e. *low*<=≤<=*d*<=≤<=*high*. It is possible that there is no common divisor in the given range. You will be given the two integers *a* and *b*, then *n* queries. Each query is a range from *low* to *high* and you have to answer each query.

The first line contains two integers *a* and *b*, the two integers as described above (1<=≤<=*a*,<=*b*<=≤<=109). The second line contains one integer *n*, the number of queries (1<=≤<=*n*<=≤<=104). Then *n* lines follow, each line contains one query consisting of two integers, *low* and *high* (1<=≤<=*low*<=≤<=*high*<=≤<=109).

Print *n* lines. The *i*-th of them should contain the result of the *i*-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.

[ "9 27\n3\n1 5\n10 11\n9 11\n" ]

[ "3\n-1\n9\n" ]

none

1,500

[ { "input": "9 27\n3\n1 5\n10 11\n9 11", "output": "3\n-1\n9" }, { "input": "48 72\n2\n8 29\n29 37", "output": "24\n-1" }, { "input": "90 100\n10\n51 61\n6 72\n1 84\n33 63\n37 69\n18 21\n9 54\n49 90\n14 87\n37 90", "output": "-1\n10\n10\n-1\n-1\n-1\n10\n-1\n-1\n-1" }, { "input": "84 36\n1\n18 32", "output": "-1" }, { "input": "90 36\n16\n13 15\n5 28\n11 30\n26 35\n2 8\n19 36\n3 17\n5 14\n4 26\n22 33\n16 33\n18 27\n4 17\n1 2\n29 31\n18 36", "output": "-1\n18\n18\n-1\n6\n-1\n9\n9\n18\n-1\n18\n18\n9\n2\n-1\n18" }, { "input": "84 90\n18\n10 75\n2 40\n30 56\n49 62\n19 33\n5 79\n61 83\n13 56\n73 78\n1 18\n23 35\n14 72\n22 33\n1 21\n8 38\n54 82\n6 80\n57 75", "output": "-1\n6\n-1\n-1\n-1\n6\n-1\n-1\n-1\n6\n-1\n-1\n-1\n6\n-1\n-1\n6\n-1" }, { "input": "84 100\n16\n10 64\n3 61\n19 51\n42 67\n51 68\n12 40\n10 47\n52 53\n37 67\n2 26\n23 47\n17 75\n49 52\n3 83\n63 81\n8 43", "output": "-1\n4\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n4\n-1\n-1\n-1\n4\n-1\n-1" }, { "input": "36 60\n2\n17 25\n16 20", "output": "-1\n-1" }, { "input": "90 100\n8\n55 75\n46 68\n44 60\n32 71\n43 75\n23 79\n47 86\n11 57", "output": "-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1" }, { "input": "90 36\n8\n1 19\n10 12\n14 28\n21 24\n8 8\n33 34\n10 26\n15 21", "output": "18\n-1\n18\n-1\n-1\n-1\n18\n18" }, { "input": "48 80\n19\n1 1\n16 16\n1 16\n16 48\n16 80\n16 1000000000\n1000000000 1000000000\n1 1000000000\n500000000 1000000000\n15 17\n17 17\n15 15\n8 8\n8 15\n8 16\n8 17\n7 17\n7 15\n9 15", "output": "1\n16\n16\n16\n16\n16\n-1\n16\n-1\n16\n-1\n-1\n8\n8\n16\n16\n16\n8\n-1" }, { "input": "31607 999002449\n18\n31607 31607\n31606 31608\n31607 31608\n31606 31607\n31606 31606\n31608 31608\n1 31607\n1 31606\n1 31608\n1 1000000000\n31607 1000000000\n31606 1000000000\n31608 1000000000\n1000000000 1000000000\n1 1\n2 31606\n2 31607\n2 31608", "output": "31607\n31607\n31607\n31607\n-1\n-1\n31607\n1\n31607\n31607\n31607\n31607\n-1\n-1\n1\n-1\n31607\n31607" }, { "input": "999999937 999999929\n12\n999999929 999999937\n1 1\n1 1000000000\n2 1000000000\n1 2\n999999937 999999937\n999999929 999999929\n2 2\n3 3\n1 100\n1 999999937\n1 999999929", "output": "-1\n1\n1\n-1\n1\n-1\n-1\n-1\n-1\n1\n1\n1" } ]

1,648,980,912

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

92

0

def upper_bound(arr, target, i, j): while i < j: mid = int(i + (j - i) / 2) if target > arr[mid]: i = mid + 1 else: j = mid return mid read = lambda: list(map(int, input().split())) write = lambda x: print(*x) a, b = read() from math import gcd g = gcd(a, b) factors = [0] for i in range(1, int(g**0.5)+1): if g % i == 0: factors.append(i) factors.append(g//i) if i*i == g: factors.pop() factors.sort( ) print(factors) q = int(input()) for ___ in range(q): a, b = read() pos = upper_bound(factors, b, 0, len(factors)) if factors[pos] < a: print(-1) else: print(factors[pos])

Title: Modified GCD Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers. A common divisor for two positive numbers is a number which both numbers are divisible by. But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor *d* between two integers *a* and *b* that is in a given range from *low* to *high* (inclusive), i.e. *low*<=≤<=*d*<=≤<=*high*. It is possible that there is no common divisor in the given range. You will be given the two integers *a* and *b*, then *n* queries. Each query is a range from *low* to *high* and you have to answer each query. Input Specification: The first line contains two integers *a* and *b*, the two integers as described above (1<=≤<=*a*,<=*b*<=≤<=109). The second line contains one integer *n*, the number of queries (1<=≤<=*n*<=≤<=104). Then *n* lines follow, each line contains one query consisting of two integers, *low* and *high* (1<=≤<=*low*<=≤<=*high*<=≤<=109). Output Specification: Print *n* lines. The *i*-th of them should contain the result of the *i*-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query. Demo Input: ['9 27\n3\n1 5\n10 11\n9 11\n'] Demo Output: ['3\n-1\n9\n'] Note: none

```python def upper_bound(arr, target, i, j): while i < j: mid = int(i + (j - i) / 2) if target > arr[mid]: i = mid + 1 else: j = mid return mid read = lambda: list(map(int, input().split())) write = lambda x: print(*x) a, b = read() from math import gcd g = gcd(a, b) factors = [0] for i in range(1, int(g**0.5)+1): if g % i == 0: factors.append(i) factors.append(g//i) if i*i == g: factors.pop() factors.sort( ) print(factors) q = int(input()) for ___ in range(q): a, b = read() pos = upper_bound(factors, b, 0, len(factors)) if factors[pos] < a: print(-1) else: print(factors[pos]) ```

0

61

A

Ultra-Fast Mathematician

PROGRAMMING

800

[ "implementation" ]

A. Ultra-Fast Mathematician

2

256

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

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

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

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

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

none

500

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

1,591,048,989

2,147,483,647

PyPy 3

OK

TESTS

102

155

0

#Ultra-Fast Mathematician #Problem Link : http://codeforces.com/problemset/problem/61/A n , m = input() , input() for x , y in zip(n ,m): print("0" if x == y else "1", end='')

Title: Ultra-Fast Mathematician Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate. Input Specification: There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. Output Specification: Write one line — the corresponding answer. Do not omit the leading 0s. Demo Input: ['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n'] Demo Output: ['1110001\n', '111\n', '0100\n', '00010\n'] Note: none

```python #Ultra-Fast Mathematician #Problem Link : http://codeforces.com/problemset/problem/61/A n , m = input() , input() for x , y in zip(n ,m): print("0" if x == y else "1", end='') ```

3.96125

658

A

Bear and Reverse Radewoosh

PROGRAMMING

800

[ "implementation" ]

null

Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order. There will be *n* problems. The *i*-th problem has initial score *p**i* and it takes exactly *t**i* minutes to solve it. Problems are sorted by difficulty — it's guaranteed that *p**i*<=<<=*p**i*<=+<=1 and *t**i*<=<<=*t**i*<=+<=1. A constant *c* is given too, representing the speed of loosing points. Then, submitting the *i*-th problem at time *x* (*x* minutes after the start of the contest) gives *max*(0,<= *p**i*<=-<=*c*·*x*) points. Limak is going to solve problems in order 1,<=2,<=...,<=*n* (sorted increasingly by *p**i*). Radewoosh is going to solve them in order *n*,<=*n*<=-<=1,<=...,<=1 (sorted decreasingly by *p**i*). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie. You may assume that the duration of the competition is greater or equal than the sum of all *t**i*. That means both Limak and Radewoosh will accept all *n* problems.

The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=50,<=1<=≤<=*c*<=≤<=1000) — the number of problems and the constant representing the speed of loosing points. The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=1000,<=*p**i*<=<<=*p**i*<=+<=1) — initial scores. The third line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000,<=*t**i*<=<<=*t**i*<=+<=1) where *t**i* denotes the number of minutes one needs to solve the *i*-th problem.

Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points.

[ "3 2\n50 85 250\n10 15 25\n", "3 6\n50 85 250\n10 15 25\n", "8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76\n" ]

[ "Limak\n", "Radewoosh\n", "Tie\n" ]

In the first sample, there are 3 problems. Limak solves them as follows: 1. Limak spends 10 minutes on the 1-st problem and he gets 50 - *c*·10 = 50 - 2·10 = 30 points. 1. Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points. 1. He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points. So, Limak got 30 + 35 + 150 = 215 points. Radewoosh solves problem in the reversed order: 1. Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points. 1. He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem. 1. He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets *max*(0, 50 - 2·50) = *max*(0, - 50) = 0 points. Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins. In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway. In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4.

500

[ { "input": "3 2\n50 85 250\n10 15 25", "output": "Limak" }, { "input": "3 6\n50 85 250\n10 15 25", "output": "Radewoosh" }, { "input": "8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76", "output": "Tie" }, { "input": "4 1\n3 5 6 9\n1 2 4 8", "output": "Limak" }, { "input": "4 1\n1 3 6 10\n1 5 7 8", "output": "Radewoosh" }, { "input": "4 1\n2 4 5 10\n2 3 9 10", "output": "Tie" }, { "input": "18 4\n68 97 121 132 146 277 312 395 407 431 458 461 595 634 751 855 871 994\n1 2 3 4 9 10 13 21 22 29 31 34 37 38 39 41 48 49", "output": "Radewoosh" }, { "input": "50 1\n5 14 18 73 137 187 195 197 212 226 235 251 262 278 287 304 310 322 342 379 393 420 442 444 448 472 483 485 508 515 517 523 559 585 618 627 636 646 666 682 703 707 780 853 937 951 959 989 991 992\n30 84 113 173 199 220 235 261 266 277 300 306 310 312 347 356 394 396 397 409 414 424 446 462 468 487 507 517 537 566 594 643 656 660 662 668 706 708 773 774 779 805 820 827 868 896 929 942 961 995", "output": "Tie" }, { "input": "4 1\n4 6 9 10\n2 3 4 5", "output": "Radewoosh" }, { "input": "4 1\n4 6 9 10\n3 4 5 7", "output": "Radewoosh" }, { "input": "4 1\n1 6 7 10\n2 7 8 10", "output": "Tie" }, { "input": "4 1\n4 5 7 9\n1 4 5 8", "output": "Limak" }, { "input": "50 1\n6 17 44 82 94 127 134 156 187 211 212 252 256 292 294 303 352 355 379 380 398 409 424 434 480 524 584 594 631 714 745 756 777 778 789 793 799 821 841 849 859 878 879 895 925 932 944 952 958 990\n15 16 40 42 45 71 99 100 117 120 174 181 186 204 221 268 289 332 376 394 403 409 411 444 471 487 499 539 541 551 567 589 619 623 639 669 689 722 735 776 794 822 830 840 847 907 917 927 936 988", "output": "Radewoosh" }, { "input": "50 10\n25 49 52 73 104 117 127 136 149 164 171 184 226 251 257 258 286 324 337 341 386 390 428 453 464 470 492 517 543 565 609 634 636 660 678 693 710 714 729 736 739 749 781 836 866 875 956 960 977 979\n2 4 7 10 11 22 24 26 27 28 31 35 37 38 42 44 45 46 52 53 55 56 57 59 60 61 64 66 67 68 69 71 75 76 77 78 79 81 83 85 86 87 89 90 92 93 94 98 99 100", "output": "Limak" }, { "input": "50 10\n11 15 25 71 77 83 95 108 143 150 182 183 198 203 213 223 279 280 346 348 350 355 375 376 412 413 415 432 470 545 553 562 589 595 607 633 635 637 688 719 747 767 771 799 842 883 905 924 942 944\n1 3 5 6 7 10 11 12 13 14 15 16 19 20 21 23 25 32 35 36 37 38 40 41 42 43 47 50 51 54 55 56 57 58 59 60 62 63 64 65 66 68 69 70 71 72 73 75 78 80", "output": "Radewoosh" }, { "input": "32 6\n25 77 141 148 157 159 192 196 198 244 245 255 332 392 414 457 466 524 575 603 629 700 738 782 838 841 845 847 870 945 984 985\n1 2 4 5 8 9 10 12 13 14 15 16 17 18 20 21 22 23 24 26 28 31 38 39 40 41 42 43 45 47 48 49", "output": "Radewoosh" }, { "input": "5 1\n256 275 469 671 842\n7 9 14 17 26", "output": "Limak" }, { "input": "2 1000\n1 2\n1 2", "output": "Tie" }, { "input": "3 1\n1 50 809\n2 8 800", "output": "Limak" }, { "input": "1 13\n866\n10", "output": "Tie" }, { "input": "15 1\n9 11 66 128 199 323 376 386 393 555 585 718 935 960 971\n3 11 14 19 20 21 24 26 32 38 40 42 44 47 50", "output": "Limak" }, { "input": "1 10\n546\n45", "output": "Tie" }, { "input": "50 20\n21 43 51 99 117 119 158 167 175 190 196 244 250 316 335 375 391 403 423 428 451 457 460 480 487 522 539 559 566 584 598 602 604 616 626 666 675 730 771 787 828 841 861 867 886 889 898 970 986 991\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", "output": "Limak" }, { "input": "50 21\n13 20 22 38 62 84 118 135 141 152 170 175 194 218 227 229 232 253 260 263 278 313 329 357 396 402 422 452 454 533 575 576 580 594 624 644 653 671 676 759 789 811 816 823 831 833 856 924 933 987\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", "output": "Tie" }, { "input": "1 36\n312\n42", "output": "Tie" }, { "input": "1 1000\n1\n1000", "output": "Tie" }, { "input": "1 1\n1000\n1", "output": "Tie" }, { "input": "50 35\n9 17 28 107 136 152 169 174 186 188 201 262 291 312 324 330 341 358 385 386 393 397 425 431 479 498 502 523 530 540 542 554 578 588 622 623 684 696 709 722 784 819 836 845 850 932 945 969 983 984\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", "output": "Tie" }, { "input": "50 20\n12 113 116 120 138 156 167 183 185 194 211 228 234 261 278 287 310 317 346 361 364 397 424 470 496 522 527 536 611 648 668 704 707 712 717 752 761 766 815 828 832 864 872 885 889 901 904 929 982 993\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", "output": "Limak" } ]

1,698,693,389

2,147,483,647

Python 3

OK

TESTS

29

31

0

n,c = map(int,input().split()) P = list(map(int,input().split())) T = list(map(int,input().split())) L, R = 0, 0 tl,tr = 0, 0 for i in range(n): tl += T[i] tr += T[n-1-i] L += max(0,P[i] - c*tl) R += max(0, P[n-1-i] - c*tr) if L == R: print("Tie") elif L > R: print("Limak") else: print("Radewoosh")

Title: Bear and Reverse Radewoosh Time Limit: None seconds Memory Limit: None megabytes Problem Description: Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order. There will be *n* problems. The *i*-th problem has initial score *p**i* and it takes exactly *t**i* minutes to solve it. Problems are sorted by difficulty — it's guaranteed that *p**i*<=<<=*p**i*<=+<=1 and *t**i*<=<<=*t**i*<=+<=1. A constant *c* is given too, representing the speed of loosing points. Then, submitting the *i*-th problem at time *x* (*x* minutes after the start of the contest) gives *max*(0,<= *p**i*<=-<=*c*·*x*) points. Limak is going to solve problems in order 1,<=2,<=...,<=*n* (sorted increasingly by *p**i*). Radewoosh is going to solve them in order *n*,<=*n*<=-<=1,<=...,<=1 (sorted decreasingly by *p**i*). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie. You may assume that the duration of the competition is greater or equal than the sum of all *t**i*. That means both Limak and Radewoosh will accept all *n* problems. Input Specification: The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=50,<=1<=≤<=*c*<=≤<=1000) — the number of problems and the constant representing the speed of loosing points. The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=1000,<=*p**i*<=<<=*p**i*<=+<=1) — initial scores. The third line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000,<=*t**i*<=<<=*t**i*<=+<=1) where *t**i* denotes the number of minutes one needs to solve the *i*-th problem. Output Specification: Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points. Demo Input: ['3 2\n50 85 250\n10 15 25\n', '3 6\n50 85 250\n10 15 25\n', '8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76\n'] Demo Output: ['Limak\n', 'Radewoosh\n', 'Tie\n'] Note: In the first sample, there are 3 problems. Limak solves them as follows: 1. Limak spends 10 minutes on the 1-st problem and he gets 50 - *c*·10 = 50 - 2·10 = 30 points. 1. Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points. 1. He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points. So, Limak got 30 + 35 + 150 = 215 points. Radewoosh solves problem in the reversed order: 1. Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points. 1. He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem. 1. He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets *max*(0, 50 - 2·50) = *max*(0, - 50) = 0 points. Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins. In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway. In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4.

```python n,c = map(int,input().split()) P = list(map(int,input().split())) T = list(map(int,input().split())) L, R = 0, 0 tl,tr = 0, 0 for i in range(n): tl += T[i] tr += T[n-1-i] L += max(0,P[i] - c*tl) R += max(0, P[n-1-i] - c*tr) if L == R: print("Tie") elif L > R: print("Limak") else: print("Radewoosh") ```

3

794

B

Cutting Carrot

PROGRAMMING

1,200

[ "geometry", "math" ]

null

Igor the analyst has adopted *n* little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into *n* pieces of equal area. Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to *h*. Igor wants to make *n*<=-<=1 cuts parallel to the base to cut the carrot into *n* pieces. He wants to make sure that all *n* pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

The first and only line of input contains two space-separated integers, *n* and *h* (2<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=105).

The output should contain *n*<=-<=1 real numbers *x*1,<=*x*2,<=...,<=*x**n*<=-<=1. The number *x**i* denotes that the *i*-th cut must be made *x**i* units away from the apex of the carrot. In addition, 0<=<<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=-<=1<=<<=*h* must hold. Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10<=-<=6. Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .

[ "3 2\n", "2 100000\n" ]

[ "1.154700538379 1.632993161855\n", "70710.678118654752\n" ]

Definition of isosceles triangle: [https://en.wikipedia.org/wiki/Isosceles_triangle](https://en.wikipedia.org/wiki/Isosceles_triangle).

1,000

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1,494,669,604

1,504

Python 3

OK

TESTS

31

62

307,200

import sys from math import sqrt def solve(): n, h = map(int, input().split()) ans = [0]*(n - 1) for i in range(1, n): ans[i - 1] = h * sqrt(i / n) print(*ans) if __name__ == '__main__': solve()

Title: Cutting Carrot Time Limit: None seconds Memory Limit: None megabytes Problem Description: Igor the analyst has adopted *n* little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into *n* pieces of equal area. Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to *h*. Igor wants to make *n*<=-<=1 cuts parallel to the base to cut the carrot into *n* pieces. He wants to make sure that all *n* pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area? Input Specification: The first and only line of input contains two space-separated integers, *n* and *h* (2<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=105). Output Specification: The output should contain *n*<=-<=1 real numbers *x*1,<=*x*2,<=...,<=*x**n*<=-<=1. The number *x**i* denotes that the *i*-th cut must be made *x**i* units away from the apex of the carrot. In addition, 0<=<<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=-<=1<=<<=*h* must hold. Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10<=-<=6. Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if . Demo Input: ['3 2\n', '2 100000\n'] Demo Output: ['1.154700538379 1.632993161855\n', '70710.678118654752\n'] Note: Definition of isosceles triangle: [https://en.wikipedia.org/wiki/Isosceles_triangle](https://en.wikipedia.org/wiki/Isosceles_triangle).

```python import sys from math import sqrt def solve(): n, h = map(int, input().split()) ans = [0]*(n - 1) for i in range(1, n): ans[i - 1] = h * sqrt(i / n) print(*ans) if __name__ == '__main__': solve() ```

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" } ]

1,607,950,338

2,147,483,647

Python 3

OK

TESTS

45

342

7,680,000

n=int(input()) a=list(map(int,input().split())) l,r,c=0,sum(a),0 for i in range(n-1): l+=a[i] r-=a[i] if l==r: c+=1 print(c)

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 n=int(input()) a=list(map(int,input().split())) l,r,c=0,sum(a),0 for i in range(n-1): l+=a[i] r-=a[i] if l==r: c+=1 print(c) ```

3.85728

381

A

Sereja and Dima

PROGRAMMING

800

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

null

Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins. Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move. Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.

The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.

On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.

[ "4\n4 1 2 10\n", "7\n1 2 3 4 5 6 7\n" ]

[ "12 5\n", "16 12\n" ]

In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.

500

[ { "input": "4\n4 1 2 10", "output": "12 5" }, { "input": "7\n1 2 3 4 5 6 7", "output": "16 12" }, { "input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13", "output": "613 418" }, { "input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24", "output": "644 500" }, { "input": "1\n3", "output": "3 0" }, { "input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646", "output": "6848 6568" }, { "input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727", "output": "9562 9561" }, { "input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12", "output": "315 315" }, { "input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304", "output": "3238 2222" }, { "input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325", "output": "5246 4864" }, { "input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71", "output": "8147 7807" }, { "input": "1\n1", "output": "1 0" } ]

1,697,037,118

2,147,483,647

Python 3

OK

TESTS

34

31

0

n = int(input()) a = list(map(int, input().split())) s = 0 d = 0 for i in range(n): if i % 2 == 0: if a[0] > a[-1]: s += a[0] a.remove(a[0]) else: s += a[-1] a.remove(a[-1]) else: if a[0] > a[-1]: d += a[0] a.remove(a[0]) else: d += a[-1] a.remove(a[-1]) print(s, d)

Title: Sereja and Dima Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins. Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move. Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000. Output Specification: On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game. Demo Input: ['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n'] Demo Output: ['12 5\n', '16 12\n'] Note: In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.

```python n = int(input()) a = list(map(int, input().split())) s = 0 d = 0 for i in range(n): if i % 2 == 0: if a[0] > a[-1]: s += a[0] a.remove(a[0]) else: s += a[-1] a.remove(a[-1]) else: if a[0] > a[-1]: d += a[0] a.remove(a[0]) else: d += a[-1] a.remove(a[-1]) print(s, d) ```

3

811

A

Vladik and Courtesy

PROGRAMMING

800

[ "brute force", "implementation" ]

null

At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn. More formally, the guys take turns giving each other one candy more than they received in the previous turn. This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.

Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively.

Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.

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

[ "Valera\n", "Vladik\n" ]

Illustration for first test case: <img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/> Illustration for second test case: <img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/>

500

[ { "input": "1 1", "output": "Valera" }, { "input": "7 6", "output": "Vladik" }, { "input": "25 38", "output": "Vladik" }, { "input": "8311 2468", "output": "Valera" }, { "input": "250708 857756", "output": "Vladik" }, { "input": "957985574 24997558", "output": "Valera" }, { "input": "999963734 999994456", "output": "Vladik" }, { "input": "1000000000 1000000000", "output": "Vladik" }, { "input": "946 879", "output": "Valera" }, { "input": "10819 45238", "output": "Vladik" }, { "input": "101357 236928", "output": "Vladik" }, { "input": "1033090 7376359", "output": "Vladik" }, { "input": "9754309 9525494", "output": "Valera" }, { "input": "90706344 99960537", "output": "Vladik" }, { "input": "965161805 908862070", "output": "Valera" }, { "input": "9 11", "output": "Valera" }, { "input": "3 2", "output": "Vladik" }, { "input": "6 6", "output": "Vladik" }, { "input": "4 4", "output": "Valera" }, { "input": "5 5", "output": "Valera" }, { "input": "5 4", "output": "Valera" }, { "input": "12345680 1", "output": "Valera" }, { "input": "9 10", "output": "Valera" }, { "input": "678 76687", "output": "Vladik" }, { "input": "1 678", "output": "Vladik" }, { "input": "45 1678", "output": "Vladik" }, { "input": "3 3", "output": "Vladik" }, { "input": "10 11", "output": "Valera" }, { "input": "2 1", "output": "Valera" }, { "input": "1 2", "output": "Vladik" }, { "input": "2 2", "output": "Vladik" }, { "input": "4 5", "output": "Valera" }, { "input": "9 6", "output": "Valera" }, { "input": "1 5", "output": "Vladik" }, { "input": "7 8", "output": "Vladik" }, { "input": "1000000000 999982505", "output": "Valera" }, { "input": "12 12", "output": "Vladik" }, { "input": "1000 950", "output": "Valera" }, { "input": "10 9", "output": "Valera" }, { "input": "100 9", "output": "Valera" }, { "input": "1000 996", "output": "Vladik" }, { "input": "9 5", "output": "Valera" } ]

1,634,286,765

2,147,483,647

Python 3

OK

TESTS

42

46

0

a, b = map(int, input().split()) for i in range(1, int(1e9 + 1)): if i % 2 == 0: if i > b: print("Valera") quit() else: b -= i else: if i > a: print("Vladik") quit() else: a -= i

Title: Vladik and Courtesy Time Limit: None seconds Memory Limit: None megabytes Problem Description: At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn. More formally, the guys take turns giving each other one candy more than they received in the previous turn. This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy. Input Specification: Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively. Output Specification: Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise. Demo Input: ['1 1\n', '7 6\n'] Demo Output: ['Valera\n', 'Vladik\n'] Note: Illustration for first test case: <img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/> Illustration for second test case: <img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/>

```python a, b = map(int, input().split()) for i in range(1, int(1e9 + 1)): if i % 2 == 0: if i > b: print("Valera") quit() else: b -= i else: if i > a: print("Vladik") quit() else: a -= i ```

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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 1000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 1000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ..." }, { "input": "100 999", "output": "999 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 999 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 0 999 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0..." }, { "input": "99 998", "output": "998 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 998 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 0 998 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0..." }, { "input": "100 997", "output": "997 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 997 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 " } ]

1,621,582,428

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

0

61

0

print("A")

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 print("A") ```

0

61

A

Ultra-Fast Mathematician

PROGRAMMING

800

[ "implementation" ]

A. Ultra-Fast Mathematician

2

256

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

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

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

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

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

none

500

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"0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111", "output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101" }, { "input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010", "output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101" }, { "input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000", "output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001" }, { "input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011", "output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110" }, { "input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011", "output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011" }, { "input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011", "output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011" }, { "input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010", "output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011" }, { "input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001", "output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110" }, { "input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111", "output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010" }, { "input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011", "output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011" }, { "input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100", "output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000" }, { "input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001", "output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101" }, { "input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110", "output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101" }, { "input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011", "output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110" }, { "input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111", "output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110" }, { "input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000", "output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001" }, { "input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010", "output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101" }, { "input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101", "output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011" }, { "input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011", "output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010", "output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011" }, { "input": "0\n0", "output": "0" }, { "input": "0\n1", "output": "1" } ]

1,685,610,943

2,147,483,647

PyPy 3-64

OK

TESTS

102

62

0

n1 = list(input()) n2 = list(input()) output = '' for i, k in zip(n1,n2): if i != k: output = output + '1' else: output = output + '0' print(output)

Title: Ultra-Fast Mathematician Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate. Input Specification: There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. Output Specification: Write one line — the corresponding answer. Do not omit the leading 0s. Demo Input: ['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n'] Demo Output: ['1110001\n', '111\n', '0100\n', '00010\n'] Note: none

```python n1 = list(input()) n2 = list(input()) output = '' for i, k in zip(n1,n2): if i != k: output = output + '1' else: output = output + '0' print(output) ```

3.9845

447

B

DZY Loves Strings

PROGRAMMING

1,000

[ "greedy", "implementation" ]

null

DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get?

The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103). The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103). The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000.

Print a single integer — the largest possible value of the resulting string DZY could get.

[ "abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n" ]

[ "41\n" ]

In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.

1,000

[ { "input": "abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "41" }, { "input": "mmzhr\n3\n443 497 867 471 195 670 453 413 579 466 553 881 847 642 269 996 666 702 487 209 257 741 974 133 519 453", "output": "29978" }, { "input": "ajeeseerqnpaujubmajpibxrccazaawetywxmifzehojf\n23\n359 813 772 413 733 654 33 87 890 433 395 311 801 852 376 148 914 420 636 695 583 733 664 394 407 314", "output": "1762894" }, { "input": "uahngxejpomhbsebcxvelfsojbaouynnlsogjyvktpwwtcyddkcdqcqs\n34\n530 709 150 660 947 830 487 142 208 276 885 542 138 214 76 184 273 753 30 195 722 236 82 691 572 585", "output": "2960349" }, { "input": "xnzeqmouqyzvblcidmhbkqmtusszuczadpooslqxegldanwopilmdwzbczvrwgnwaireykwpugvpnpafbxlyggkgawghysufuegvmzvpgcqyjkoadcreaguzepbendwnowsuekxxivkziibxvxfoilofxcgnxvfefyezfhevfvtetsuhwtyxdlkccdkvqjl\n282\n170 117 627 886 751 147 414 187 150 960 410 70 576 681 641 729 798 877 611 108 772 643 683 166 305 933", "output": "99140444" }, { "input": "pplkqmluhfympkjfjnfdkwrkpumgdmbkfbbldpepicbbmdgafttpopzdxsevlqbtywzkoxyviglbbxsohycbdqksrhlumsldiwzjmednbkcjishkiekfrchzuztkcxnvuykhuenqojrmzaxlaoxnljnvqgnabtmcftisaazzgbmubmpsorygyusmeonrhrgphnfhlaxrvyhuxsnnezjxmdoklpquzpvjbxgbywppmegzxknhfzyygrmejleesoqfwheulmqhonqaukyuejtwxskjldplripyihbfpookxkuehiwqthbfafyrgmykuxglpplozycgydyecqkgfjljfqvigqhuxssqqtfanwszduwbsoytnrtgc\n464\n838 95 473 955 690 84 436 19 179 437 674 626 377 365 781 4 733 776 462 203 119 256 381 668 855 686", "output": "301124161" }, { "input": "qkautnuilwlhjsldfcuwhiqtgtoihifszlyvfaygrnivzgvwthkrzzdtfjcirrjjlrmjtbjlzmjeqmuffsjorjyggzefwgvmblvotvzffnwjhqxorpowzdcnfksdibezdtfjjxfozaghieksbmowrbeehuxlesmvqjsphlvauxiijm\n98\n121 622 0 691 616 959 838 161 581 862 876 830 267 812 598 106 337 73 588 323 999 17 522 399 657 495", "output": "30125295" }, { "input": "tghyxqfmhz\n8\n191 893 426 203 780 326 148 259 182 140 847 636 778 97 167 773 219 891 758 993 695 603 223 779 368 165", "output": "136422" }, { "input": "nyawbfjxnxjiyhwkydaruozobpphgjqdpfdqzezcsoyvurnapu\n30\n65 682 543 533 990 148 815 821 315 916 632 771 332 513 472 864 12 73 548 687 660 572 507 192 226 348", "output": "2578628" }, { "input": "pylrnkrbcjgoytvdnhmlvnkknijkdgdhworlvtwuonrkhrilkewcnofodaumgvnsisxooswgrgtvdeauyxhkipfoxrrtysuepjcf\n60\n894 206 704 179 272 337 413 828 119 182 330 46 440 102 250 191 242 539 678 783 843 431 612 567 33 338", "output": "9168707" }, { "input": "vhjnkrxbyhjhnjrxvwxmhxwoxttbtqosfxtcuvhfjlkyfspeypthsdkkwnqdpxdlnxsgtzvkrgqosgfjrwetqbxgoarkjhrjbspzgblsapifltkfxbfdbxqwoohlgyzijmiwnpmveybyzvasoctxsmgjehpyysmqblwnmkappbecklqjfmxhlyceordroflnposohfplrvijxbwvqdtvzhobtrumiujnyrfbwthvciinuveoizkccelxtaveiiagryqnyvsgfnipnavrtmdqlcnldepocbpzmqnarkdvykds\n276\n364 244 798 82 582 9 309 950 286 547 892 371 569 159 705 975 740 845 655 179 130 993 255 552 882 657", "output": "144901921" }, { "input": "gsaddmezrnttfalbwlqbnedumvikplfosw\n12\n290 850 872 361 483 895 152 118 974 619 701 154 899 285 328 712 669 984 407 340 851 775 324 892 554 860", "output": "809931" }, { "input": "a\n0\n5 1 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": "5" }, { "input": "lol\n3\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", "output": "21" } ]

1,530,722,704

2,147,483,647

PyPy 3

OK

TESTS

24

140

0

string=input() k=int(input()) l=list(map(int,input().split())) sum=0 i=0 while(i<len(string)): temp=ord(string[i])-ord('a') sum+=l[temp]*(i+1) i=i+1 i=i+1 while(i<=len(string)+k): sum+=max(l)*i i=i+1 print(sum)

Title: DZY Loves Strings Time Limit: None seconds Memory Limit: None megabytes Problem Description: DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get? Input Specification: The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103). The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103). The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000. Output Specification: Print a single integer — the largest possible value of the resulting string DZY could get. Demo Input: ['abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n'] Demo Output: ['41\n'] Note: In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.

```python string=input() k=int(input()) l=list(map(int,input().split())) sum=0 i=0 while(i<len(string)): temp=ord(string[i])-ord('a') sum+=l[temp]*(i+1) i=i+1 i=i+1 while(i<=len(string)+k): sum+=max(l)*i i=i+1 print(sum) ```

3

847

M

Weather Tomorrow

PROGRAMMING

1,000

[ "implementation", "math" ]

null

Vasya came up with his own weather forecasting method. He knows the information about the average air temperature for each of the last *n* days. Assume that the average air temperature for each day is integral. Vasya believes that if the average temperatures over the last *n* days form an arithmetic progression, where the first term equals to the average temperature on the first day, the second term equals to the average temperature on the second day and so on, then the average temperature of the next (*n*<=+<=1)-th day will be equal to the next term of the arithmetic progression. Otherwise, according to Vasya's method, the temperature of the (*n*<=+<=1)-th day will be equal to the temperature of the *n*-th day. Your task is to help Vasya predict the average temperature for tomorrow, i. e. for the (*n*<=+<=1)-th day.

The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days for which the average air temperature is known. The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (<=-<=1000<=≤<=*t**i*<=≤<=1000) — where *t**i* is the average temperature in the *i*-th day.

Print the average air temperature in the (*n*<=+<=1)-th day, which Vasya predicts according to his method. Note that the absolute value of the predicted temperature can exceed 1000.

[ "5\n10 5 0 -5 -10\n", "4\n1 1 1 1\n", "3\n5 1 -5\n", "2\n900 1000\n" ]

[ "-15\n", "1\n", "-5\n", "1100\n" ]

In the first example the sequence of the average temperatures is an arithmetic progression where the first term is 10 and each following terms decreases by 5. So the predicted average temperature for the sixth day is - 10 - 5 = - 15. In the second example the sequence of the average temperatures is an arithmetic progression where the first term is 1 and each following terms equals to the previous one. So the predicted average temperature in the fifth day is 1. In the third example the average temperatures do not form an arithmetic progression, so the average temperature of the fourth day equals to the temperature of the third day and equals to - 5. In the fourth example the sequence of the average temperatures is an arithmetic progression where the first term is 900 and each the following terms increase by 100. So predicted average temperature in the third day is 1000 + 100 = 1100.

0

[ { "input": "5\n10 5 0 -5 -10", "output": "-15" }, { "input": "4\n1 1 1 1", "output": "1" }, { "input": "3\n5 1 -5", "output": "-5" }, { "input": "2\n900 1000", "output": "1100" }, { "input": "2\n1 2", "output": "3" }, { "input": "3\n2 5 8", "output": "11" }, { "input": "4\n4 1 -2 -5", "output": "-8" }, { "input": "10\n-1000 -995 -990 -985 -980 -975 -970 -965 -960 -955", "output": "-950" }, { "input": "11\n-1000 -800 -600 -400 -200 0 200 400 600 800 1000", "output": "1200" }, { "input": "31\n1000 978 956 934 912 890 868 846 824 802 780 758 736 714 692 670 648 626 604 582 560 538 516 494 472 450 428 406 384 362 340", "output": "318" }, { "input": "5\n1000 544 88 -368 -824", "output": "-1280" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "33\n456 411 366 321 276 231 186 141 96 51 6 -39 -84 -129 -174 -219 -264 -309 -354 -399 -444 -489 -534 -579 -624 -669 -714 -759 -804 -849 -894 -939 -984", "output": "-1029" }, { "input": "77\n-765 -742 -719 -696 -673 -650 -627 -604 -581 -558 -535 -512 -489 -466 -443 -420 -397 -374 -351 -328 -305 -282 -259 -236 -213 -190 -167 -144 -121 -98 -75 -52 -29 -6 17 40 63 86 109 132 155 178 201 224 247 270 293 316 339 362 385 408 431 454 477 500 523 546 569 592 615 638 661 684 707 730 753 776 799 822 845 868 891 914 937 960 983", "output": "1006" }, { "input": "3\n2 4 8", "output": "8" }, { "input": "4\n4 1 -3 -5", "output": "-5" }, { "input": "10\n-1000 -995 -990 -984 -980 -975 -970 -965 -960 -955", "output": "-955" }, { "input": "11\n-999 -800 -600 -400 -200 0 200 400 600 800 1000", "output": "1000" }, { "input": "51\n-9 10 30 50 70 90 110 130 150 170 190 210 230 250 270 290 310 330 350 370 390 410 430 450 470 490 510 530 550 570 590 610 630 650 670 690 710 730 750 770 790 810 830 850 870 890 910 930 950 970 990", "output": "990" }, { "input": "100\n10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120 122 124 126 128 130 132 134 136 138 140 142 144 146 148 150 152 154 156 158 160 162 164 166 168 170 172 174 176 178 180 182 184 186 188 190 192 194 196 198 200 202 204 206 207", "output": "207" }, { "input": "2\n1000 1000", "output": "1000" }, { "input": "2\n-1000 1000", "output": "3000" }, { "input": "2\n1000 -1000", "output": "-3000" }, { "input": "2\n-1000 -1000", "output": "-1000" }, { "input": "100\n-85 -80 -76 -72 -68 -64 -60 -56 -52 -48 -44 -40 -36 -32 -28 -24 -20 -16 -12 -8 -4 0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 100 104 108 112 116 120 124 128 132 136 140 144 148 152 156 160 164 168 172 176 180 184 188 192 196 200 204 208 212 216 220 224 228 232 236 240 244 248 252 256 260 264 268 272 276 280 284 288 292 296 300 304 308 312", "output": "312" }, { "input": "4\n1 2 4 5", "output": "5" } ]

1,508,328,624

924

Python 3

OK

TESTS

25

46

5,529,600

import sys n = int(input()) a = list(map(int, input().split())) d = a[1] - a[0] flag = False for i in range(1, n): if a[i] - a[i - 1] != d: flag = True break if not flag: print(a[n - 1] + d) else: print(a[n - 1])

Title: Weather Tomorrow Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya came up with his own weather forecasting method. He knows the information about the average air temperature for each of the last *n* days. Assume that the average air temperature for each day is integral. Vasya believes that if the average temperatures over the last *n* days form an arithmetic progression, where the first term equals to the average temperature on the first day, the second term equals to the average temperature on the second day and so on, then the average temperature of the next (*n*<=+<=1)-th day will be equal to the next term of the arithmetic progression. Otherwise, according to Vasya's method, the temperature of the (*n*<=+<=1)-th day will be equal to the temperature of the *n*-th day. Your task is to help Vasya predict the average temperature for tomorrow, i. e. for the (*n*<=+<=1)-th day. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days for which the average air temperature is known. The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (<=-<=1000<=≤<=*t**i*<=≤<=1000) — where *t**i* is the average temperature in the *i*-th day. Output Specification: Print the average air temperature in the (*n*<=+<=1)-th day, which Vasya predicts according to his method. Note that the absolute value of the predicted temperature can exceed 1000. Demo Input: ['5\n10 5 0 -5 -10\n', '4\n1 1 1 1\n', '3\n5 1 -5\n', '2\n900 1000\n'] Demo Output: ['-15\n', '1\n', '-5\n', '1100\n'] Note: In the first example the sequence of the average temperatures is an arithmetic progression where the first term is 10 and each following terms decreases by 5. So the predicted average temperature for the sixth day is - 10 - 5 = - 15. In the second example the sequence of the average temperatures is an arithmetic progression where the first term is 1 and each following terms equals to the previous one. So the predicted average temperature in the fifth day is 1. In the third example the average temperatures do not form an arithmetic progression, so the average temperature of the fourth day equals to the temperature of the third day and equals to - 5. In the fourth example the sequence of the average temperatures is an arithmetic progression where the first term is 900 and each the following terms increase by 100. So predicted average temperature in the third day is 1000 + 100 = 1100.

```python import sys n = int(input()) a = list(map(int, input().split())) d = a[1] - a[0] flag = False for i in range(1, n): if a[i] - a[i - 1] != d: flag = True break if not flag: print(a[n - 1] + d) else: print(a[n - 1]) ```

3

294

A

Shaass and Oskols

PROGRAMMING

800

[ "implementation", "math" ]

null

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

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

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

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

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

none

500

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

1,664,655,304

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

2

92

0

n_lines = int(input()) birds = [int(item) for item in input().split()] shots = int(input()) A = [] for i in range(shots): x, y = map(int, input().split()) lst = [x, y] A.append(lst) for i in range(shots): if (A[i][0]-1 +1) <= 5: birds[A[i][0]-1 +1] = birds[A[i][0]-1 +1] + birds[A[i][0]-1] - A[i][1] if (A[i][0]-1 -1) >= 0: birds[A[i][0]-1 -1] = birds[A[i][0]-1 -1] + (A[i][1]) - 1 birds[A[i][0] - 1] = 0 for i in birds: print(i)

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

```python n_lines = int(input()) birds = [int(item) for item in input().split()] shots = int(input()) A = [] for i in range(shots): x, y = map(int, input().split()) lst = [x, y] A.append(lst) for i in range(shots): if (A[i][0]-1 +1) <= 5: birds[A[i][0]-1 +1] = birds[A[i][0]-1 +1] + birds[A[i][0]-1] - A[i][1] if (A[i][0]-1 -1) >= 0: birds[A[i][0]-1 -1] = birds[A[i][0]-1 -1] + (A[i][1]) - 1 birds[A[i][0] - 1] = 0 for i in birds: print(i) ```

-1

94

A

Restoring Password

PROGRAMMING

900

[ "implementation", "strings" ]

A. Restoring Password

2

256

Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password". Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address. Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits. Help Igor K. restore his ISQ account by the encrypted password and encryption specification.

The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9.

Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists.

[ "01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n", "10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n" ]

[ "12345678\n", "30234919\n" ]

none

500

[ { "input": "01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110", "output": "12345678" }, { "input": "10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000", "output": "30234919" }, { "input": "00010101101110110101100110101100010101100010101111000101011010011010110010000011\n0101010110\n0001001101\n1001101011\n0000100011\n0010101111\n1110110101\n0001010110\n0110111000\n0000111110\n0010000011", "output": "65264629" }, { "input": "10100100010010010011011001101000100100110110011010011001101011000100110110011010\n1111110011\n1001000111\n1001000100\n1100010011\n0110011010\n0010000001\n1110101110\n0010000110\n0010010011\n1010010001", "output": "98484434" }, { "input": "00101100011111010001001000000110110000000110010011001111111010110010001011000000\n0010000001\n0110010011\n0010000010\n1011001000\n0011111110\n0110001000\n1111010001\n1011000000\n0000100110\n0010110001", "output": "96071437" }, { "input": "10001110111110000001000010001010001110110000100010100010111101101101010000100010\n0000010110\n1101010111\n1000101111\n0001011110\n0011110101\n0101100100\n0110110101\n0000100010\n1000111011\n1110000001", "output": "89787267" }, { "input": "10010100011001010001010101001101010100110100111011001010111100011001000010100000\n0011100000\n1001100100\n0001100100\n0010100000\n0101010011\n0010101110\n0010101111\n0100111011\n1001010001\n1111111110", "output": "88447623" }, { "input": "01101100111000000101011011001110000001011111111000111111100001011010001001011001\n1000000101\n0101101000\n0101110101\n1101011110\n0000101100\n1111111000\n0001001101\n0110111011\n0110110011\n1001011001", "output": "80805519" }, { "input": "11100011000100010110010011101010101010011110001100011010111110011000011010110111\n1110001100\n0110101111\n0100111010\n0101000000\n1001100001\n1010101001\n0000100010\n1010110111\n1100011100\n0100010110", "output": "09250147" }, { "input": "10000110110000010100000010001000111101110110101011110111000100001101000000100010\n0000010100\n0000110001\n0110101011\n1101110001\n1000011011\n0000110100\n0011110111\n1000110010\n0000100010\n0000011011", "output": "40862358" }, { "input": "01000000010000000110100101000110110000100100000001101100001000011111111001010001\n1011000010\n1111101010\n0111110011\n0000000110\n0000001001\n0001111111\n0110010010\n0100000001\n1011001000\n1001010001", "output": "73907059" }, { "input": "01111000111110011001110101110011110000111110010001101100110110100111101011001101\n1110010001\n1001100000\n1100001000\n1010011110\n1011001101\n0111100011\n1101011100\n1110011001\n1111000011\n0010000101", "output": "57680434" }, { "input": "01001100101000100010001011110001000101001001100010010000001001001100101001011111\n1001011111\n1110010111\n0111101011\n1000100010\n0011100101\n0100000010\n0010111100\n0100010100\n1001100010\n0100110010", "output": "93678590" }, { "input": "01110111110000111011101010110110101011010100110111000011101101110101011101001000\n0110000101\n1010101101\n1101010111\n1101011100\n0100110111\n0111011111\n1100011001\n0111010101\n0000111011\n1101001000", "output": "58114879" }, { "input": "11101001111100110101110011010100110011011110100111010110110011000111000011001101\n1100011100\n1100110101\n1011101000\n0011011110\n0011001101\n0100010001\n1110100111\n1010101100\n1110110100\n0101101100", "output": "61146904" }, { "input": "10101010001011010001001001011000100101100001011011101010101110101010001010101000\n0010110101\n1010011010\n1010101000\n1011010001\n1010101011\n0010010110\n0110100010\n1010100101\n0001011011\n0110100001", "output": "23558422" }, { "input": "11110101001100010000110100001110101011011111010100110001000001001010001001101111\n0101101100\n1001101111\n1010101101\n0100101000\n1111110000\n0101010010\n1100010000\n1111010100\n1101000011\n1011111111", "output": "76827631" }, { "input": "10001100110000110111100011001101111110110011110101000011011100001101110000110111\n0011110101\n0101100011\n1000110011\n1011011001\n0111111011\n0101111011\n0000110111\n0100001110\n1000000111\n0110110111", "output": "26240666" }, { "input": "10000100010000111101100100111101111011101000001001100001000110000010010000111101\n1001001111\n0000111101\n1000010001\n0110011101\n0110101000\n1011111001\n0111101110\n1000001001\n1101011111\n0001010100", "output": "21067271" }, { "input": "01101111000110111100011011110001101111001010001100101000110001010101100100000010\n1010001100\n0011010011\n0101010110\n1111001100\n1100011000\n0100101100\n1001100101\n0110111100\n0011001101\n0100000010", "output": "77770029" }, { "input": "10100111011010001011111000000111100000010101000011000010111101010000111010011101\n1010011101\n1010111111\n0110100110\n1111000100\n1110000001\n0000101111\n0011111000\n1000110001\n0101000011\n1010001011", "output": "09448580" }, { "input": "10000111111000011111001010101010010011111001001111000010010100100011000010001100\n1101101110\n1001001111\n0000100101\n1100111010\n0010101010\n1110000110\n1100111101\n0010001100\n1110000001\n1000011111", "output": "99411277" }, { "input": "10110110111011001111101100111100111111011011011011001111110110010011100010000111\n0111010011\n0111101100\n1001101010\n0101000101\n0010000111\n0011111101\n1011001111\n1101111000\n1011011011\n1001001110", "output": "86658594" }, { "input": "01001001100101100011110110111100000110001111001000100000110111110010000000011000\n0100100110\n1000001011\n1000111110\n0000011000\n0101100011\n1101101111\n1111001000\n1011011001\n1000001101\n0010101000", "output": "04536863" }, { "input": "10010100011101000011100100001100101111000010111100000010010000001001001101011101\n1001000011\n1101000011\n1001010001\n1101011101\n1000010110\n0011111101\n0010111100\n0000100100\n1010001000\n0101000110", "output": "21066773" }, { "input": "01111111110101111111011111111111010010000001100000101000100100111001011010001001\n0111111111\n0101111111\n0100101101\n0001100000\n0011000101\n0011100101\n1101001000\n0010111110\n1010001001\n1111000111", "output": "01063858" }, { "input": "00100011111001001010001111000011101000001110100000000100101011101000001001001010\n0010001111\n1001001010\n1010011001\n0011100111\n1000111000\n0011110000\n0000100010\n0001001010\n1111110111\n1110100000", "output": "01599791" }, { "input": "11011101000100110100110011010101100011111010011010010011010010010010100110101111\n0100110100\n1001001010\n0001111101\n1101011010\n1101110100\n1100110101\n0110101111\n0110001111\n0001101000\n1010011010", "output": "40579016" }, { "input": "10000010111101110110011000111110000011100110001111100100000111000011011000001011\n0111010100\n1010110110\n1000001110\n1110000100\n0110001111\n1101110110\n1100001101\n1000001011\n0000000101\n1001000001", "output": "75424967" }, { "input": "11101100101110111110111011111010001111111111000001001001000010001111111110110010\n0101100001\n1111010011\n1110111110\n0100110100\n1110011111\n1000111111\n0010010000\n1110110010\n0011000010\n1111000001", "output": "72259657" }, { "input": "01011110100101111010011000001001100000101001110011010111101011010000110110010101\n0100111100\n0101110011\n0101111010\n0110000010\n0101001111\n1101000011\n0110010101\n0111011010\n0001101110\n1001110011", "output": "22339256" }, { "input": "01100000100101111000100001100010000110000010100100100001100000110011101001110000\n0101111000\n1001110000\n0001000101\n0110110111\n0010100100\n1000011000\n1101110110\n0110000010\n0001011010\n0011001110", "output": "70554591" }, { "input": "11110011011000001001111100110101001000010100100000110011001110011111100100100001\n1010011000\n1111001101\n0100100001\n1111010011\n0100100000\n1001111110\n1010100111\n1000100111\n1000001001\n1100110011", "output": "18124952" }, { "input": "10001001011000100101010110011101011001110010000001010110000101000100101111101010\n0101100001\n1100001100\n1111101010\n1000100101\n0010000001\n0100010010\n0010110110\n0101100111\n0000001110\n1101001110", "output": "33774052" }, { "input": "00110010000111001001001100100010010111101011011110001011111100000101000100000001\n0100000001\n1011011110\n0010111111\n0111100111\n0100111001\n0000010100\n1001011110\n0111001001\n0100010011\n0011001000", "output": "97961250" }, { "input": "01101100001000110101101100101111101110010011010111100011010100010001101000110101\n1001101001\n1000110101\n0110110000\n0111100100\n0011010111\n1110111001\n0001000110\n0000000100\n0001101001\n1011001011", "output": "21954161" }, { "input": "10101110000011010110101011100000101101000110100000101101101101110101000011110010\n0110100000\n1011011011\n0011110010\n0001110110\n0010110100\n1100010010\n0001101011\n1010111000\n0011010110\n0111010100", "output": "78740192" }, { "input": "11000101011100100111010000010001000001001100101100000011000000001100000101011010\n1100010101\n1111101011\n0101011010\n0100000100\n1000110111\n1100100111\n1100101100\n0111001000\n0000110000\n0110011111", "output": "05336882" }, { "input": "11110100010000101110010110001000001011100101100010110011011011111110001100110110\n0101100010\n0100010001\n0000101110\n1100110110\n0101000101\n0011001011\n1111010001\n1000110010\n1111111000\n1010011111", "output": "62020383" }, { "input": "00011001111110000011101011010001010111100110100101000110011111011001100000001100\n0111001101\n0101011110\n0001100111\n1101011111\n1110000011\n0000001100\n0111010001\n1101100110\n1010110100\n0110100101", "output": "24819275" }, { "input": "10111110010011111001001111100101010111010011111001001110101000111110011001111101\n0011111001\n0101011101\n0100001010\n0001110010\n1001111101\n0011101010\n1111001001\n1100100001\n1001101000\n1011111001", "output": "90010504" }, { "input": "01111101111100101010001001011110111001110111110111011111011110110111111011011111\n1111110111\n0010000101\n0110000100\n0111111011\n1011100111\n1100101010\n1011011111\n1100010001\n0111110111\n0010010111", "output": "85948866" }, { "input": "01111100000111110000110010111001111100001001101010110010111010001000101001101010\n0100010101\n1011110101\n1010100100\n1010000001\n1001101010\n0101100110\n1000100010\n0111110000\n1100101110\n0110010110", "output": "77874864" }, { "input": "11100011010000000010011110010111001011111001000111000000001000000000100111100101\n0000000010\n1110001101\n0011010101\n0111100101\n1001000111\n1101001111\n0111010110\n1100101111\n0110000000\n1101101011", "output": "10374003" }, { "input": "01111011100111101110011001000110001111101000111110100100100001011111001011100010\n0110010100\n1100010001\n0111101110\n1001001000\n1010011011\n1000111110\n0010110101\n1011100010\n0101111100\n0110010001", "output": "22955387" }, { "input": "11011010001100000011000100110011010101000110011110110000001100111100001000011111\n0000100010\n1000011111\n1101101000\n0110011110\n0011110000\n1100000011\n0010001100\n0101101000\n0001001100\n1101010100", "output": "25893541" }, { "input": "01011001011111010010101111011001000011001100011101101111011011010011101011110110\n0100001100\n0101100101\n1111111011\n1111010010\n1111101100\n1100011101\n1011000011\n1101001110\n1011110110\n0110001010", "output": "13805878" }, { "input": "11110011011000111111001100111110001111111100000010111100110100110011111111001101\n1111001101\n1001101010\n1100110010\n0011001111\n0001011110\n1000110011\n1000111111\n0110001010\n1001011101\n1100000010", "output": "06369030" }, { "input": "01110011110010000011011001011000001000010110010110011001100001100110001100101000\n0000100001\n0110011000\n1010000010\n1110011101\n0111001111\n1100101000\n0010000011\n0110010000\n1100100101\n0110010110", "output": "46909115" }, { "input": "00001011001111110111111111011111111101110101110100010111010010100101100001010110\n1111110111\n0001010110\n0111011011\n0111000001\n1010010110\n0101110100\n0001000101\n0000111000\n0110100001\n0000101100", "output": "90005541" } ]

1,680,635,383

2,147,483,647

Python 3

OK

TESTS

50

92

0

s = input() d = {} for i in range(10): d[input()] = str(i) ans = '' for i in range(0, 80, 10): ans += d[s[i:i+10]] print(ans)

Title: Restoring Password Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password". Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address. Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits. Help Igor K. restore his ISQ account by the encrypted password and encryption specification. Input Specification: The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9. Output Specification: Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists. Demo Input: ['01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n', '10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n'] Demo Output: ['12345678\n', '30234919\n'] Note: none

```python s = input() d = {} for i in range(10): d[input()] = str(i) ans = '' for i in range(0, 80, 10): ans += d[s[i:i+10]] print(ans) ```

3.977

928

A

Login Verification

PROGRAMMING

1,200

[ "*special", "strings" ]

null

When registering in a social network, users are allowed to create their own convenient login to make it easier to share contacts, print it on business cards, etc. Login is an arbitrary sequence of lower and uppercase latin letters, digits and underline symbols («_»). However, in order to decrease the number of frauds and user-inattention related issues, it is prohibited to register a login if it is similar with an already existing login. More precisely, two logins *s* and *t* are considered similar if we can transform *s* to *t* via a sequence of operations of the following types: - transform lowercase letters to uppercase and vice versa; - change letter «O» (uppercase latin letter) to digit «0» and vice versa; - change digit «1» (one) to any letter among «l» (lowercase latin «L»), «I» (uppercase latin «i») and vice versa, or change one of these letters to other. For example, logins «Codeforces» and «codef0rces» as well as «OO0OOO00O0OOO0O00OOO0OO_lol» and «OO0OOO0O00OOO0O00OO0OOO_1oI» are considered similar whereas «Codeforces» and «Code_forces» are not. You're given a list of existing logins with no two similar amonst and a newly created user login. Check whether this new login is similar with any of the existing ones.

The first line contains a non-empty string *s* consisting of lower and uppercase latin letters, digits and underline symbols («_») with length not exceeding 50 — the login itself. The second line contains a single integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of existing logins. The next *n* lines describe the existing logins, following the same constraints as the user login (refer to the first line of the input). It's guaranteed that no two existing logins are similar.

Print «Yes» (without quotes), if user can register via this login, i.e. none of the existing logins is similar with it. Otherwise print «No» (without quotes).

[ "1_wat\n2\n2_wat\nwat_1\n", "000\n3\n00\nooA\noOo\n", "_i_\n3\n__i_\n_1_\nI\n", "La0\n3\n2a0\nLa1\n1a0\n", "abc\n1\naBc\n", "0Lil\n2\nLIL0\n0Ril\n" ]

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

In the second sample case the user wants to create a login consisting of three zeros. It's impossible due to collision with the third among the existing. In the third sample case the new login is similar with the second one.

500

[ { "input": "1_wat\n2\n2_wat\nwat_1", "output": "Yes" }, { "input": "000\n3\n00\nooA\noOo", "output": "No" }, { "input": "_i_\n3\n__i_\n_1_\nI", "output": "No" }, { "input": "La0\n3\n2a0\nLa1\n1a0", "output": "No" }, { "input": "abc\n1\naBc", "output": "No" }, { "input": "0Lil\n2\nLIL0\n0Ril", "output": "Yes" }, { "input": "iloO\n3\niIl0\noIl0\nIooO", "output": "Yes" }, { "input": "L1il0o1L1\n5\niLLoLL\noOI1Io10il\nIoLLoO\nO01ilOoI\nI10l0o", "output": "Yes" }, { "input": "ELioO1lOoOIOiLoooi1iolul1O\n7\nOoEIuOIl1ui1010uiooOoi0Oio001L0EoEolO0\nOLIoOEuoE11u1u1iLOI0oO\nuEOuO0uIOOlO01OlEI0E1Oo0IO1LI0uE0LILO0\nEOo0Il11iIOOOIiuOiIiiLOLEOOII001EE\niOoO0LOulioE0OLIIIulli01OoiuOOOoOlEiI0EiiElIIu0\nlE1LOE1Oil\n1u0EOliIiIOl1u110il0l1O0u", "output": "Yes" }, { "input": "0blo7X\n20\n1oobb6\nXIXIO2X\n2iYI2\n607XXol\n2I6io22\nOl10I\nbXX0Lo\nolOOb7X\n07LlXL\nlXY17\n12iIX2\n7lL70\nbOo11\n17Y6b62\n0O6L7\n1lX2L\n2iYl6lI\n7bXIi1o\niLIY2\n0OIo1X", "output": "Yes" }, { "input": "lkUL\n25\nIIfL\nokl\nfoo\ni0U\noko\niIoU\nUUv\nvli\nv0Uk\n0Of\niill\n1vkl\nUIf\nUfOO\nlvLO\nUUo0\nIOf1\nlovL\nIkk\noIv\nLvfU\n0UI\nkol\n1OO0\n1OOi", "output": "Yes" }, { "input": "L1lo\n3\nOOo1\nL1lo\n0lOl", "output": "No" }, { "input": "LIoooiLO\n5\nLIoooiLO\nl0o01I00\n0OOl0lLO01\nil10i0\noiloi", "output": "No" }, { "input": "1i1lQI\n7\nuLg1uLLigIiOLoggu\nLLLgIuQIQIIloiQuIIoIO0l0o000\n0u1LQu11oIuooIl0OooLg0i0IQu1O1lloI1\nQuQgIQi0LOIliLOuuuioLQou1l\nlLIO00QLi01LogOliOIggII1\no0Ll1uIOQl10IL0IILQ\n1i1lQI", "output": "No" }, { "input": "oIzz1\n20\n1TTl0O\nloF0LT\n1lLzo\noi0Ov\nFlIF1zT\nzoITzx\n0TIFlT\nl1vllil\nOviix1F\nLFvI1lL\nLIl0loz\nixz1v\n1i1vFi\nTIFTol\noIzz1\nIvTl0o\nxv1U0O\niiiioF\n1oiLUlO\nxToxv1", "output": "No" }, { "input": "00L0\n25\n0il\nIlkZ\nL0I\n00L0\nBd0\nZLd\n0d1k\nddk\nIdl\nkBd\nkBOL\nZ1lI\nkBL\nLOko\noZ0i\nZ1lO\nLiOk\niBld\nLO0d\ndIo\nZ10\n1k1i\n0o0L\nIoBd\ni0B0", "output": "No" }, { "input": "Z\n1\nz", "output": "No" }, { "input": "0\n1\no", "output": "No" }, { "input": "0\n1\nO", "output": "No" }, { "input": "o\n1\n0", "output": "No" }, { "input": "o\n1\nO", "output": "No" }, { "input": "o\n1\no", "output": "No" }, { "input": "O\n1\no", "output": "No" }, { "input": "O\n1\n0", "output": "No" }, { "input": "1\n1\nl", "output": "No" }, { "input": "1\n1\nL", "output": "No" }, { "input": "1\n1\ni", "output": "No" }, { "input": "1\n1\nI", "output": "No" }, { "input": "1\n1\no", "output": "Yes" }, { "input": "i\n1\n1", "output": "No" }, { "input": "i\n1\nL", "output": "No" }, { "input": "i\n1\nl", "output": "No" }, { "input": "I\n1\nL", "output": "No" }, { "input": "I\n1\nl", "output": "No" }, { "input": "I\n1\n1", "output": "No" }, { "input": "l\n1\n1", "output": "No" }, { "input": "l\n1\nL", "output": "No" }, { "input": "l\n1\ni", "output": "No" }, { "input": "l\n1\nI", "output": "No" }, { "input": "L\n1\nl", "output": "No" }, { "input": "L\n1\n1", "output": "No" }, { "input": "L\n1\ni", "output": "No" }, { "input": "L\n1\nI", "output": "No" } ]

1,652,754,621

2,147,483,647

PyPy 3-64

OK

TESTS

73

108

31,027,200

def f(s): return s.lower().replace('i', '1').replace('0', 'o').replace('l', '1') s=f(input()) n=int(input()) a=[f(input()) for i in range(n)] if s in a: print("No") else: print("Yes")

Title: Login Verification Time Limit: None seconds Memory Limit: None megabytes Problem Description: When registering in a social network, users are allowed to create their own convenient login to make it easier to share contacts, print it on business cards, etc. Login is an arbitrary sequence of lower and uppercase latin letters, digits and underline symbols («_»). However, in order to decrease the number of frauds and user-inattention related issues, it is prohibited to register a login if it is similar with an already existing login. More precisely, two logins *s* and *t* are considered similar if we can transform *s* to *t* via a sequence of operations of the following types: - transform lowercase letters to uppercase and vice versa; - change letter «O» (uppercase latin letter) to digit «0» and vice versa; - change digit «1» (one) to any letter among «l» (lowercase latin «L»), «I» (uppercase latin «i») and vice versa, or change one of these letters to other. For example, logins «Codeforces» and «codef0rces» as well as «OO0OOO00O0OOO0O00OOO0OO_lol» and «OO0OOO0O00OOO0O00OO0OOO_1oI» are considered similar whereas «Codeforces» and «Code_forces» are not. You're given a list of existing logins with no two similar amonst and a newly created user login. Check whether this new login is similar with any of the existing ones. Input Specification: The first line contains a non-empty string *s* consisting of lower and uppercase latin letters, digits and underline symbols («_») with length not exceeding 50 — the login itself. The second line contains a single integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of existing logins. The next *n* lines describe the existing logins, following the same constraints as the user login (refer to the first line of the input). It's guaranteed that no two existing logins are similar. Output Specification: Print «Yes» (without quotes), if user can register via this login, i.e. none of the existing logins is similar with it. Otherwise print «No» (without quotes). Demo Input: ['1_wat\n2\n2_wat\nwat_1\n', '000\n3\n00\nooA\noOo\n', '_i_\n3\n__i_\n_1_\nI\n', 'La0\n3\n2a0\nLa1\n1a0\n', 'abc\n1\naBc\n', '0Lil\n2\nLIL0\n0Ril\n'] Demo Output: ['Yes\n', 'No\n', 'No\n', 'No\n', 'No\n', 'Yes\n'] Note: In the second sample case the user wants to create a login consisting of three zeros. It's impossible due to collision with the third among the existing. In the third sample case the new login is similar with the second one.

```python def f(s): return s.lower().replace('i', '1').replace('0', 'o').replace('l', '1') s=f(input()) n=int(input()) a=[f(input()) for i in range(n)] if s in a: print("No") else: print("Yes") ```

3

459

B

Pashmak and Flowers

PROGRAMMING

1,300

[ "combinatorics", "implementation", "sortings" ]

null

Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible! Your task is to write a program which calculates two things: 1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.

The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109).

The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.

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

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

In the third sample the maximum beauty difference is 2 and there are 4 ways to do this: 1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers.

500

[ { "input": "2\n1 2", "output": "1 1" }, { "input": "3\n1 4 5", "output": "4 1" }, { "input": "5\n3 1 2 3 1", "output": "2 4" }, { "input": "2\n1 1", "output": "0 1" }, { "input": "3\n1 1 1", "output": "0 3" }, { "input": "4\n1 1 1 1", "output": "0 6" }, { "input": "5\n1 1 1 1 1", "output": "0 10" }, { "input": "5\n2 2 2 2 2", "output": "0 10" }, { "input": "10\n2 2 2 2 2 2 2 2 2 2", "output": "0 45" }, { "input": "3\n2 2 2", "output": "0 3" }, { "input": "3\n3 3 3", "output": "0 3" }, { "input": "2\n10000000 100000000", "output": "90000000 1" }, { "input": "5\n5 5 5 5 5", "output": "0 10" }, { "input": "5\n3 3 3 3 3", "output": "0 10" }, { "input": "6\n1 1 1 1 1 1", "output": "0 15" }, { "input": "2\n5 6", "output": "1 1" }, { "input": "10\n1 1 1 1 1 1 1 1 1 1", "output": "0 45" }, { "input": "10\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "0 45" }, { "input": "4\n4 4 4 4", "output": "0 6" }, { "input": "7\n1 1 1 1 1 1 1", "output": "0 21" }, { "input": "11\n1 1 1 1 1 1 1 1 1 1 1", "output": "0 55" }, { "input": "8\n8 8 8 8 8 8 8 8", "output": "0 28" }, { "input": "2\n3 2", "output": "1 1" } ]

1,663,343,396

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

5

109

22,528,000

kol=int(input()) a=list(map(int, input().split(' '))) count=0 lim=min(a) r=0 d=0 m=max(a) if max(a)-min(a)==0 and max(a)==min(a): print('0'+' '+'1') else: count=a.count(m) d=a.count(lim) r=(m-lim)*count*d print(str(m-lim)+' '+str(count*d))

Title: Pashmak and Flowers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible! Your task is to write a program which calculates two things: 1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way. Input Specification: The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109). Output Specification: The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively. Demo Input: ['2\n1 2\n', '3\n1 4 5\n', '5\n3 1 2 3 1\n'] Demo Output: ['1 1', '4 1', '2 4'] Note: In the third sample the maximum beauty difference is 2 and there are 4 ways to do this: 1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers.

```python kol=int(input()) a=list(map(int, input().split(' '))) count=0 lim=min(a) r=0 d=0 m=max(a) if max(a)-min(a)==0 and max(a)==min(a): print('0'+' '+'1') else: count=a.count(m) d=a.count(lim) r=(m-lim)*count*d print(str(m-lim)+' '+str(count*d)) ```

0

831

B

Keyboard Layouts

PROGRAMMING

800

[ "implementation", "strings" ]

null

There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet. You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order. You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout. Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.

The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout. The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout. The third line contains a non-empty string *s* consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of *s* does not exceed 1000.

Print the text if the same keys were pressed in the second layout.

[ "qwertyuiopasdfghjklzxcvbnm\nveamhjsgqocnrbfxdtwkylupzi\nTwccpQZAvb2017\n", "mnbvcxzlkjhgfdsapoiuytrewq\nasdfghjklqwertyuiopzxcvbnm\n7abaCABAABAcaba7\n" ]

[ "HelloVKCup2017\n", "7uduGUDUUDUgudu7\n" ]

none

750

[ { "input": "qwertyuiopasdfghjklzxcvbnm\nveamhjsgqocnrbfxdtwkylupzi\nTwccpQZAvb2017", "output": "HelloVKCup2017" }, { "input": "mnbvcxzlkjhgfdsapoiuytrewq\nasdfghjklqwertyuiopzxcvbnm\n7abaCABAABAcaba7", "output": "7uduGUDUUDUgudu7" }, { "input": "ayvguplhjsoiencbkxdrfwmqtz\nkhzvtbspcndierqumlojyagfwx\n3", "output": "3" }, { "input": "oaihbljgekzsxucwnqyrvfdtmp\nwznqcfvrthjibokeglmudpayxs\ntZ8WI33UZZytE8A99EvJjck228LxUQtL5A8q7O217KrmdhpmdhN7JEdVXc8CRm07TFidlIou9AKW9cCl1c4289rfU87oXoSCwHpZO7ggC2GmmDl0KGuA2IimDco2iKaBKl46H089r2tw16mhzI44d2X6g3cnoD0OU5GvA8l89nhNpzTbY9FtZ2wE3Y2a5EC7zXryudTZhXFr9EEcX8P71fp6694aa02B4T0w1pDaVml8FM3N2qB78DBrS723Vpku105sbTJEdBpZu77b1C47DujdoR7rjm5k2nsaPBqX93EfhW95Mm0sBnFtgo12gS87jegSR5u88tM5l420dkt1l1b18UjatzU7P2i9KNJA528caiEpE3JtRw4m4TJ7M1zchxO53skt3Fqvxk2C51gD8XEY7YJC2xmTUqyEUFmPX581Gow2HWq4jaP8FK87", "output": "yJ8EN33OJJmyT8Z99TdVvkh228FbOLyF5Z8l7W217HuxaqsxaqG7VTaDBk8KUx07YPnafNwo9ZHE9kKf1k4289upO87wBwIKeQsJW7rrK2RxxAf0HRoZ2NnxAkw2nHzCHf46Q089u2ye16xqjN44a2B6r3kgwA0WO5RdZ8f89gqGsjYcM9PyJ2eT3M2z5TK7jBumoaYJqBPu9TTkB8S71ps6694zz02C4Y0e1sAzDxf8PX3G2lC78ACuI723Dsho105icYVTaCsJo77c1K47AovawU7uvx5h2gizSClB93TpqE95Xx0iCgPyrw12rI87vtrIU5o88yX5f420ahy1f1c18OvzyjO7S2n9HGVZ528kznTsT3VyUe4x4YV7X1jkqbW53ihy3Pldbh2K51rA8BTM7MVK2bxYOlmTOPxSB581Rwe2QEl4vzS8PH87" }, { "input": "aymrnptzhklcbuxfdvjsgqweio\nwzsavqryltmjnfgcedxpiokbuh\nB5", "output": "N5" }, { "input": "unbclszprgiqjodxeawkymvfth\ncxfwbdvuqlotkgparmhsyinjze\nk081O", "output": "s081G" }, { "input": "evfsnczuiodgbhqmlypkjatxrw\nhvsockwjxtgreqmyanlzidpbuf\n306QMPpaqZ", "output": "306MYLldmW" }, { "input": "pbfjtvryklwmuhxnqsoceiadgz\ntaipfdvlzemhjsnkwyocqgrxbu\nTm9H66Ux59PuGe3lEG94q18u11Dda6w59q1hAAIvHR1qquKI2Xf5ZFdKAPhcEnqKT6BF6Oh16P48YvrIKWGDlRcx9BZwwEF64o0As", "output": "Fh9S66Jn59TjBq3eQB94w18j11Xxr6m59w1sRRGdSV1wwjZG2Ni5UIxZRTscQkwZF6AI6Os16T48LdvGZMBXeVcn9AUmmQI64o0Ry" }, { "input": "rtqgahmkeoldsiynjbuwpvcxfz\noxqiuwflvebnapyrmcghtkdjzs\nJqNskelr3FNjbDhfKPfPXxlqOw72p9BVBwf0tN8Ucs48Vlfjxqo9V3ruU5205UgTYi3JKFbW91NLQ1683315VJ4RSLFW7s26s6uZKs5cO2wAT4JS8rCytZVlPWXdNXaCTq06F1v1Fj2zq7DeJbBSfM5Eko6vBndR75d46mf5Pq7Ark9NARTtQ176ukljBdaqXRsYxrBYl7hda1V7sy38hfbjz59HYM9U55P9eh1CX7tUE44NFlQu7zSjSBHyS3Tte2XaXD3O470Q8U20p8W5rViIh8lsn2TvmcdFdxrF3Ye26J2ZK0BR3KShN597WSJmHJTl4ZZ88IMhzHi6vFyr7MuGYNFGebTB573e6Crwj8P18h344yd8sR2NPge36Y3QC8Y2uW577CO2w4fz", "output": "MqRalvbo3ZRmcNwzLTzTJjbqEh72t9CKChz0xR8Gda48Kbzmjqe9K3ogG5205GiXYp3MLZcH91RBQ1683315KM4OABZH7a26a6gSLa5dE2hUX4MA8oDyxSKbTHJnRJuDXq06Z1k1Zm2sq7NvMcCAzF5Vle6kCrnO75n46fz5Tq7Uol9RUOXxQ176glbmCnuqJOaYjoCYb7wnu1K7ay38wzcms59WYF9G55T9vw1DJ7xGV44RZbQg7sAmACWyA3Xxv2JuJN3E470Q8G20t8H5oKpPw8bar2XkfdnZnjoZ3Yv26M2SL0CO3LAwR597HAMfWMXb4SS88PFwsWp6kZyo7FgIYRZIvcXC573v6Dohm8T18w344yn8aO2RTiv36Y3QD8Y2gH577DE2h4zs" }, { "input": "buneohqdgxjsafrmwtzickvlpy\nzblwamjxifyuqtnrgdkchpoves\n4RZf8YivG6414X1GdDfcCbc10GA0Wz8514LI9D647XzPb66UNh7lX1rDQv0hQvJ7aqhyh1Z39yABGKn24g185Y85ER5q9UqPFaQ2JeK97wHZ78CMSuU8Zf091mePl2OX61BLe5KdmUWodt4BXPiseOZkZ4SZ27qtBM4hT499mCirjy6nB0ZqjQie4Wr3uhW2mGqBlHyEZbW7A6QnsNX9d3j5aHQN0H6GF8J0365KWuAmcroutnJD6l6HI3kSSq17Sdo2htt9y967y8sc98ZAHbutH1m9MOVT1E9Mb5UIK3qNatk9A0m2i1fQl9A65204Q4z4O4rQf374YEq0s2sfmQNW9K7E1zSbj51sGINJVr5736Gw8aW6u9Cjr0sjffXctLopJ0YQ47xD1yEP6bB3odG7slgiM8hJ9BuwfGUwN8tbAgJU8wMI2L0P446MO", "output": "4NKt8ScoI6414F1IxXthHzh10IQ0Gk8514VC9X647FkEz66BLm7vF1nXJo0mJoY7qjmsm1K39sQZIPl24i185S85WN5j9BjETqJ2YwP97gMK78HRUbB8Kt091rwEv2AF61ZVw5PxrBGaxd4ZFEcuwAKpK4UK27jdZR4mD499rHcnys6lZ0KjyJcw4Gn3bmG2rIjZvMsWKzG7Q6JluLF9x3y5qMJL0M6IT8Y0365PGbQrhnabdlYX6v6MC3pUUj17Uxa2mdd9s967s8uh98KQMzbdM1r9RAOD1W9Rz5BCP3jLqdp9Q0r2c1tJv9Q65204J4k4A4nJt374SWj0u2utrJLG9P7W1kUzy51uICLYOn5736Ig8qG6b9Hyn0uyttFhdVaeY0SJ47fX1sWE6zZ3axI7uvicR8mY9ZbgtIBgL8dzQiYB8gRC2V0E446RA" }, { "input": "qwertyuiopasdfghjklzxcvbnm\nqwertyuiopasdfghjklzxcvbnm\nqwertyuiopasdfghjklzxcvbnmPOIUYTREWQLKJHGFDSAMNBVCXZ12345678900987654321ASDFGHJKLqwertyuiopZXCVBNM", "output": "qwertyuiopasdfghjklzxcvbnmPOIUYTREWQLKJHGFDSAMNBVCXZ12345678900987654321ASDFGHJKLqwertyuiopZXCVBNM" }, { "input": "qwertyuiopasdfghjklzxcvbnm\nmnbvcxzlkjhgfdsapoiuytrewq\nasdfghjklzxcvbnmqwertyuiopASDFGHJKLQWERTYUIOPZXCVBNM12345678900987654321QWSDFGVBNxcvghjkoWQEDFGHNJMzxcfghjkl", "output": "hgfdsapoiuytrewqmnbvcxzlkjHGFDSAPOIMNBVCXZLKJUYTREWQ12345678900987654321MNGFDSREWytrsapokNMBFDSAWPQuytdsapoi" } ]

1,637,668,872

2,147,483,647

Python 3

OK

TESTS

19

61

0

str1 = input() str2 = input() str3 = input() list1 = [] for i in range(len(str3)): k = 0 temp = ord(str3[i]) if (temp>=65 and temp<= 90) or (temp>=97 and temp<=122): if temp >= 65 and temp <= 90: k = 1 str4 = str3[i].lower() else: str4 = str3[i] b = str1.find(str4) if k == 1: list1.append(str2[b].upper()) elif k == 0: list1.append(str2[b]) else: list1.append(str3[i]) print(''.join(list1))

Title: Keyboard Layouts Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet. You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order. You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout. Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters. Input Specification: The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout. The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout. The third line contains a non-empty string *s* consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of *s* does not exceed 1000. Output Specification: Print the text if the same keys were pressed in the second layout. Demo Input: ['qwertyuiopasdfghjklzxcvbnm\nveamhjsgqocnrbfxdtwkylupzi\nTwccpQZAvb2017\n', 'mnbvcxzlkjhgfdsapoiuytrewq\nasdfghjklqwertyuiopzxcvbnm\n7abaCABAABAcaba7\n'] Demo Output: ['HelloVKCup2017\n', '7uduGUDUUDUgudu7\n'] Note: none

```python str1 = input() str2 = input() str3 = input() list1 = [] for i in range(len(str3)): k = 0 temp = ord(str3[i]) if (temp>=65 and temp<= 90) or (temp>=97 and temp<=122): if temp >= 65 and temp <= 90: k = 1 str4 = str3[i].lower() else: str4 = str3[i] b = str1.find(str4) if k == 1: list1.append(str2[b].upper()) elif k == 0: list1.append(str2[b]) else: list1.append(str3[i]) print(''.join(list1)) ```

3

109

A

Lucky Sum of Digits

PROGRAMMING

1,000

[ "brute force", "implementation" ]

A. Lucky Sum of Digits

2

256

Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.

The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.

Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.

[ "11\n", "10\n" ]

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

none

500

[ { "input": "11", "output": "47" }, { "input": "10", "output": "-1" }, { "input": "64", "output": "4477777777" }, { "input": "1", "output": "-1" }, { "input": "4", "output": "4" }, { "input": "7", "output": "7" }, { "input": "12", "output": "444" }, { "input": "1000000", "output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "47", "output": "44477777" }, { "input": "100", "output": "4444777777777777" }, { "input": "700", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "485", "output": "44447777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "111", "output": "444447777777777777" }, { "input": "85", "output": "4477777777777" }, { "input": "114", "output": "444477777777777777" }, { "input": "474", "output": "444777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "74", "output": "47777777777" }, { "input": "1000", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "1024", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "4444", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "45784", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "10000", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "9854", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "186", "output": "477777777777777777777777777" }, { "input": "10416", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "10417", "output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "3840", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "100000", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "9876", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "99999", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "777777", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "854759", "output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "11000", "output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "18951", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999999", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "888887", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999998", "output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "40008", "output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "10691", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "5", "output": "-1" }, { "input": "6", "output": "-1" }, { "input": "9", "output": "-1" }, { "input": "8", "output": "44" }, { "input": "2", "output": "-1" }, { "input": "3", "output": "-1" }, { "input": "999997", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999996", "output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999990", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "999980", "output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "800000", "output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "980000", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." } ]

1,625,433,241

2,147,483,647

Python 3

OK

TESTS

51

310

2,662,400

n = int(input()) pairs = [] for b in range(10000000): if b * 7 > n: break x = (n - b * 7) if not x%4: a = x // 4 pairs.append((a, b)) if not pairs: print(-1) quit() m = (float("inf"), float("inf")) for a, b in pairs: if a + b < sum(m): m = (a, b) elif a + b == sum(m) and b < m[1]: m = (a, b) print(m[0] * "4" + m[1] * "7")

Title: Lucky Sum of Digits Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task. Input Specification: The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number. Output Specification: Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1. Demo Input: ['11\n', '10\n'] Demo Output: ['47\n', '-1\n'] Note: none

```python n = int(input()) pairs = [] for b in range(10000000): if b * 7 > n: break x = (n - b * 7) if not x%4: a = x // 4 pairs.append((a, b)) if not pairs: print(-1) quit() m = (float("inf"), float("inf")) for a, b in pairs: if a + b < sum(m): m = (a, b) elif a + b == sum(m) and b < m[1]: m = (a, b) print(m[0] * "4" + m[1] * "7") ```

3.917541

721

A

One-dimensional Japanese Crossword

PROGRAMMING

800

[ "implementation" ]

null

Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)). Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword. Help Adaltik find the numbers encrypting the row he drew.

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).

The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row. The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.

[ "3\nBBW\n", "5\nBWBWB\n", "4\nWWWW\n", "4\nBBBB\n", "13\nWBBBBWWBWBBBW\n" ]

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

The last sample case correspond to the picture in the statement.

500

[ { "input": "3\nBBW", "output": "1\n2 " }, { "input": "5\nBWBWB", "output": "3\n1 1 1 " }, { "input": "4\nWWWW", "output": "0" }, { "input": "4\nBBBB", "output": "1\n4 " }, { "input": "13\nWBBBBWWBWBBBW", "output": "3\n4 1 3 " }, { "input": "1\nB", "output": "1\n1 " }, { "input": "2\nBB", "output": "1\n2 " }, { "input": "100\nWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWB", "output": "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 " }, { "input": "1\nW", "output": "0" }, { "input": "2\nWW", "output": "0" }, { "input": "2\nWB", "output": "1\n1 " }, { "input": "2\nBW", "output": "1\n1 " }, { "input": "3\nBBB", "output": "1\n3 " }, { "input": "3\nBWB", "output": "2\n1 1 " }, { "input": "3\nWBB", "output": "1\n2 " }, { "input": "3\nWWB", "output": "1\n1 " }, { "input": "3\nWBW", "output": "1\n1 " }, { "input": "3\nBWW", "output": "1\n1 " }, { "input": "3\nWWW", "output": "0" }, { "input": "100\nBBBWWWWWWBBWWBBWWWBBWBBBBBBBBBBBWBBBWBBWWWBBWWBBBWBWWBBBWWBBBWBBBBBWWWBWWBBWWWWWWBWBBWWBWWWBWBWWWWWB", "output": "21\n3 2 2 2 11 3 2 2 3 1 3 3 5 1 2 1 2 1 1 1 1 " }, { "input": "5\nBBBWB", "output": "2\n3 1 " }, { "input": "5\nBWWWB", "output": "2\n1 1 " }, { "input": "5\nWWWWB", "output": "1\n1 " }, { "input": "5\nBWWWW", "output": "1\n1 " }, { "input": "5\nBBBWW", "output": "1\n3 " }, { "input": "5\nWWBBB", "output": "1\n3 " }, { "input": "10\nBBBBBWWBBB", "output": "2\n5 3 " }, { "input": "10\nBBBBWBBWBB", "output": "3\n4 2 2 " }, { "input": "20\nBBBBBWWBWBBWBWWBWBBB", "output": "6\n5 1 2 1 1 3 " }, { "input": "20\nBBBWWWWBBWWWBWBWWBBB", "output": "5\n3 2 1 1 3 " }, { "input": "20\nBBBBBBBBWBBBWBWBWBBB", "output": "5\n8 3 1 1 3 " }, { "input": "20\nBBBWBWBWWWBBWWWWBWBB", "output": "6\n3 1 1 2 1 2 " }, { "input": "40\nBBBBBBWWWWBWBWWWBWWWWWWWWWWWBBBBBBBBBBBB", "output": "5\n6 1 1 1 12 " }, { "input": "40\nBBBBBWBWWWBBWWWBWBWWBBBBWWWWBWBWBBBBBBBB", "output": "9\n5 1 2 1 1 4 1 1 8 " }, { "input": "50\nBBBBBBBBBBBWWWWBWBWWWWBBBBBBBBWWWWWWWBWWWWBWBBBBBB", "output": "7\n11 1 1 8 1 1 6 " }, { "input": "50\nWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW", "output": "0" }, { "input": "50\nBBBBBWWWWWBWWWBWWWWWBWWWBWWWWWWBBWBBWWWWBWWWWWWWBW", "output": "9\n5 1 1 1 1 2 2 1 1 " }, { "input": "50\nWWWWBWWBWWWWWWWWWWWWWWWWWWWWWWWWWBWBWBWWWWWWWBBBBB", "output": "6\n1 1 1 1 1 5 " }, { "input": "50\nBBBBBWBWBWWBWBWWWWWWBWBWBWWWWWWWWWWWWWBWBWWWWBWWWB", "output": "12\n5 1 1 1 1 1 1 1 1 1 1 1 " }, { "input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "1\n50 " }, { "input": "100\nBBBBBBBBBBBWBWWWWBWWBBWBBWWWWWWWWWWBWBWWBWWWWWWWWWWWBBBWWBBWWWWWBWBWWWWBWWWWWWWWWWWBWWWWWBBBBBBBBBBB", "output": "15\n11 1 1 2 2 1 1 1 3 2 1 1 1 1 11 " }, { "input": "100\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "1\n100 " }, { "input": "100\nBBBBBBBBBBBBBBBBBBBBWBWBWWWWWBWWWWWWWWWWWWWWBBWWWBWWWWBWWBWWWWWWBWWWWWWWWWWWWWBWBBBBBBBBBBBBBBBBBBBB", "output": "11\n20 1 1 1 2 1 1 1 1 1 20 " }, { "input": "100\nBBBBWWWWWWWWWWWWWWWWWWWWWWWWWBWBWWWWWBWBWWWWWWBBWWWWWWWWWWWWBWWWWBWWWWWWWWWWWWBWWWWWWWBWWWWWWWBBBBBB", "output": "11\n4 1 1 1 1 2 1 1 1 1 6 " }, { "input": "5\nBWBWB", "output": "3\n1 1 1 " }, { "input": "10\nWWBWWWBWBB", "output": "3\n1 1 2 " }, { "input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "1\n50 " }, { "input": "50\nBBBBBBBBBBBBBBBBBWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "2\n17 31 " }, { "input": "100\nBBBBBBBBBBBBBBBBBBBBBBBBWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "2\n24 42 " }, { "input": "90\nWWBWWBWBBWBBWWBWBWBBBWBWBBBWBWBWBWBWBWBWBWBBBBBWBBWWWWBWBBWBWWBBBWBWBWWBWBWBWBWWWWWWBWBBBB", "output": "30\n1 1 2 2 1 1 3 1 3 1 1 1 1 1 1 1 5 2 1 2 1 3 1 1 1 1 1 1 1 4 " }, { "input": "100\nBWWWBWBWBBBBBWBWWBWBWWWBWBWBWWBBWWBBBWBBBWWBWBWWBBBBWBWBBBWBWBBWWWWWWBWWBBBBWBWBWWBWBWWWBWBWWBWBWWWB", "output": "31\n1 1 1 5 1 1 1 1 1 1 2 3 3 1 1 4 1 3 1 2 1 4 1 1 1 1 1 1 1 1 1 " }, { "input": "90\nWBWBBBBBBWWWBBWWBWWWBBWWBWWWBWBBWBWBBWWWWBWBWBBWBBWBWWWBBWBBWWWWBWBBWWWBBBWBBWBWBBBBWWBWWB", "output": "25\n1 6 2 1 2 1 1 2 1 2 1 1 2 2 1 2 2 1 2 3 2 1 4 1 1 " }, { "input": "80\nBBWWBBBWBBWWWWBBWBWBBWWWWWBWBBWWBWBWBWBWBWWBWWBWWWBWBBWBBWBBWBBBWWBBBBBBBWBBBWBB", "output": "23\n2 3 2 2 1 2 1 2 1 1 1 1 1 1 1 1 2 2 2 3 7 3 2 " }, { "input": "65\nWWWWBWWWBBBBBWWWWWWBBBWWBBBBWWWWWWWWBBBWWWWBWBWWBBWWWWBWWWBBWBBBB", "output": "11\n1 5 3 4 3 1 1 2 1 2 4 " } ]

1,588,153,686

2,147,483,647

Python 3

OK

TESTS

54

124

0

n=int(input()) s=input() a=[] i=0 c=1 while(i<n): if(s[i]=='B'): if(i!=n-1 and s[i+1]=='B'): c+=1 else: a.append(c) c=1 i+=1 print(len(a)) if(len(a)!=0): print(*a)

Title: One-dimensional Japanese Crossword Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)). Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword. Help Adaltik find the numbers encrypting the row he drew. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew). Output Specification: The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row. The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right. Demo Input: ['3\nBBW\n', '5\nBWBWB\n', '4\nWWWW\n', '4\nBBBB\n', '13\nWBBBBWWBWBBBW\n'] Demo Output: ['1\n2 ', '3\n1 1 1 ', '0\n', '1\n4 ', '3\n4 1 3 '] Note: The last sample case correspond to the picture in the statement.

```python n=int(input()) s=input() a=[] i=0 c=1 while(i<n): if(s[i]=='B'): if(i!=n-1 and s[i+1]=='B'): c+=1 else: a.append(c) c=1 i+=1 print(len(a)) if(len(a)!=0): print(*a) ```

3

629

A

Far Relative’s Birthday Cake

PROGRAMMING

800

[ "brute force", "combinatorics", "constructive algorithms", "implementation" ]

null

Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird! The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be? Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.

In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake. Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.

Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.

[ "3\n.CC\nC..\nC.C\n", "4\nCC..\nC..C\n.CC.\n.CC.\n" ]

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

If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are: 1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3)

500

[ { "input": "3\n.CC\nC..\nC.C", "output": "4" }, { "input": "4\nCC..\nC..C\n.CC.\n.CC.", "output": "9" }, { "input": "5\n.CCCC\nCCCCC\n.CCC.\nCC...\n.CC.C", "output": "46" }, { "input": "7\n.CC..CC\nCC.C..C\nC.C..C.\nC...C.C\nCCC.CCC\n.CC...C\n.C.CCC.", "output": "84" }, { "input": "8\n..C....C\nC.CCC.CC\n.C..C.CC\nCC......\nC..C..CC\nC.C...C.\nC.C..C..\nC...C.C.", "output": "80" }, { "input": "9\n.C...CCCC\nC.CCCC...\n....C..CC\n.CC.CCC..\n.C.C..CC.\nC...C.CCC\nCCC.C...C\nCCCC....C\n..C..C..C", "output": "144" }, { "input": "10\n..C..C.C..\n..CC..C.CC\n.C.C...C.C\n..C.CC..CC\n....C..C.C\n...C..C..C\nCC.CC....C\n..CCCC.C.C\n..CC.CCC..\nCCCC..C.CC", "output": "190" }, { "input": "11\nC.CC...C.CC\nCC.C....C.C\n.....C..CCC\n....C.CC.CC\nC..C..CC...\nC...C...C..\nCC..CCC.C.C\n..C.CC.C..C\nC...C.C..CC\n.C.C..CC..C\n.C.C.CC.C..", "output": "228" }, { "input": "21\n...CCC.....CC..C..C.C\n..CCC...CC...CC.CCC.C\n....C.C.C..CCC..C.C.C\n....CCC..C..C.CC.CCC.\n...CCC.C..C.C.....CCC\n.CCC.....CCC..C...C.C\nCCCC.C...CCC.C...C.CC\nC..C...C.CCC..CC..C..\nC...CC..C.C.CC..C.CC.\nCC..CCCCCCCCC..C....C\n.C..CCCC.CCCC.CCC...C\nCCC...CCC...CCC.C..C.\n.CCCCCCCC.CCCC.CC.C..\n.C.C..C....C.CCCCCC.C\n...C...C.CCC.C.CC..C.\nCCC...CC..CC...C..C.C\n.CCCCC...C.C..C.CC.C.\n..CCC.C.C..CCC.CCC...\n..C..C.C.C.....CC.C..\n.CC.C...C.CCC.C....CC\n...C..CCCC.CCC....C..", "output": "2103" }, { "input": "20\nC.C.CCC.C....C.CCCCC\nC.CC.C..CCC....CCCC.\n.CCC.CC...CC.CCCCCC.\n.C...CCCC..C....CCC.\n.C..CCCCCCC.C.C.....\nC....C.C..CCC.C..CCC\n...C.C.CC..CC..CC...\nC...CC.C.CCCCC....CC\n.CC.C.CCC....C.CCC.C\nCC...CC...CC..CC...C\nC.C..CC.C.CCCC.C.CC.\n..CCCCC.C.CCC..CCCC.\n....C..C..C.CC...C.C\nC..CCC..CC..C.CC..CC\n...CC......C.C..C.C.\nCC.CCCCC.CC.CC...C.C\n.C.CC..CC..CCC.C.CCC\nC..C.CC....C....C...\n..CCC..CCC...CC..C.C\n.C.CCC.CCCCCCCCC..CC", "output": "2071" }, { "input": "17\nCCC..C.C....C.C.C\n.C.CC.CC...CC..C.\n.CCCC.CC.C..CCC.C\n...CCC.CC.CCC.C.C\nCCCCCCCC..C.CC.CC\n...C..C....C.CC.C\nCC....CCC...C.CC.\n.CC.C.CC..C......\n.CCCCC.C.CC.CCCCC\n..CCCC...C..CC..C\nC.CC.C.CC..C.C.C.\nC..C..C..CCC.C...\n.C..CCCC..C......\n.CC.C...C..CC.CC.\nC..C....CC...CC..\nC.CC.CC..C.C..C..\nCCCC...C.C..CCCC.", "output": "1160" }, { "input": "15\nCCCC.C..CCC....\nCCCCCC.CC.....C\n...C.CC.C.C.CC.\nCCCCCCC..C..C..\nC..CCC..C.CCCC.\n.CC..C.C.C.CC.C\n.C.C..C..C.C..C\n...C...C..CCCC.\n.....C.C..CC...\nCC.C.C..CC.C..C\n..CCCCC..CCC...\nCC.CC.C..CC.CCC\n..CCC...CC.C..C\nCC..C.C..CCC..C\n.C.C....CCC...C", "output": "789" }, { "input": "1\n.", "output": "0" }, { "input": "3\n.CC\nC..\nC.C", "output": "4" }, { "input": "13\nC.C...C.C.C..\nCC.CCCC.CC..C\n.C.CCCCC.CC..\nCCCC..C...C..\n...CC.C.C...C\n.CC.CCC...CC.\nCC.CCCCCC....\n.C...C..CC..C\nCCCC.CC...C..\n.C.CCC..C.CC.\n..C...CC..C.C\n..C.CCC..CC.C\n.C...CCC.CC.C", "output": "529" }, { "input": "16\n.C.C.C.C.C...C.C\n..C..C.CCCCCC...\n..C.C.C.C..C..C.\n.CC....C.CCC..C.\n.C.CCC..C....CCC\nCC..C.CC..C.C.CC\n...C..C..CC..CC.\n.CCC..C.CC.C.C..\n.CC.C..........C\nC...C....CC..C..\nC.CCC.C..C..C...\n.CCCCCCCCCCCC..C\n..C.C.CC.CC.CCC.\nCC..C.C....C..CC\nC.CCC..C..C.C.CC\n.C.CCC.CC..CCC.C", "output": "874" }, { "input": "2\nCC\nCC", "output": "4" }, { "input": "3\nC..\nC..\nC..", "output": "3" } ]

1,631,544,519

2,147,483,647

Python 3

OK

TESTS

48

78

6,963,200

def nC2(N): return N*(N-1)//2 N = int(input()) H = list(list(input()) for i in range(N)) happiness = sum(nC2(H[i].count('C')) for i in range(N)) V = [[H[j][i] for j in range(N)] for i in range(N)] happiness += sum(nC2(V[i].count('C')) for i in range(N)) print(happiness)

Title: Far Relative’s Birthday Cake Time Limit: None seconds Memory Limit: None megabytes Problem Description: Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird! The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be? Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column. Input Specification: In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake. Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'. Output Specification: Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column. Demo Input: ['3\n.CC\nC..\nC.C\n', '4\nCC..\nC..C\n.CC.\n.CC.\n'] Demo Output: ['4\n', '9\n'] Note: If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are: 1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3)

```python def nC2(N): return N*(N-1)//2 N = int(input()) H = list(list(input()) for i in range(N)) happiness = sum(nC2(H[i].count('C')) for i in range(N)) V = [[H[j][i] for j in range(N)] for i in range(N)] happiness += sum(nC2(V[i].count('C')) for i in range(N)) print(happiness) ```

3

825

D

Suitable Replacement

PROGRAMMING

1,500

[ "binary search", "greedy", "implementation" ]

null

You are given two strings *s* and *t* consisting of small Latin letters, string *s* can also contain '?' characters. Suitability of string *s* is calculated by following metric: Any two letters can be swapped positions, these operations can be performed arbitrary number of times over any pair of positions. Among all resulting strings *s*, you choose the one with the largest number of non-intersecting occurrences of string *t*. Suitability is this number of occurrences. You should replace all '?' characters with small Latin letters in such a way that the suitability of string *s* is maximal.

The first line contains string *s* (1<=≤<=|*s*|<=≤<=106). The second line contains string *t* (1<=≤<=|*t*|<=≤<=106).

Print string *s* with '?' replaced with small Latin letters in such a way that suitability of that string is maximal. If there are multiple strings with maximal suitability then print any of them.

[ "?aa?\nab\n", "??b?\nza\n", "abcd\nabacaba\n" ]

[ "baab\n", "azbz\n", "abcd\n" ]

In the first example string "baab" can be transformed to "abab" with swaps, this one has suitability of 2. That means that string "baab" also has suitability of 2. In the second example maximal suitability you can achieve is 1 and there are several dozens of such strings, "azbz" is just one of them. In the third example there are no '?' characters and the suitability of the string is 0.

0

[ { "input": "?aa?\nab", "output": "baab" }, { "input": "??b?\nza", "output": "azbz" }, { "input": "abcd\nabacaba", "output": "abcd" }, { "input": "mqwstphetbfrsyxuzdww\nrutseqtsbh", "output": "mqwstphetbfrsyxuzdww" }, { "input": "????????????????????\nxwkxsxlrre", "output": "eekkllrrrrsswwxxxxxx" }, { "input": "fzjqgrgzzzghrwgwhfjq\nggwfrjzzqh", "output": "fzjqgrgzzzghrwgwhfjq" }, { "input": "ng?\nvu", "output": "ngz" }, { "input": "???\nb", "output": "bbb" }, { "input": "??a?\na", "output": "aaaa" }, { "input": "c?aa???a?cdcc??\nabcadbaccb", "output": "cbaabbzazcdcczz" }, { "input": "cc?cdb?????b?cc?b?\ncabdacd", "output": "ccacdbaaaddbdcczbz" }, { "input": "ccaaa?bc?baccbccccbca?cccab??aacccbaac?bccacc?accccaccbcbbcbc\ncbaaacccc", "output": "ccaaaabcabaccbccccbcaacccabaaaacccbaaczbccacczaccccaccbcbbcbc" }, { "input": "aacaacbabc?bcca?ccca\nbc", "output": "aacaacbabcbbccabccca" } ]

1,585,479,073

3,073

PyPy 3

OK

TESTS

24

561

38,502,400

def solve(mid): tmp = 0 for i in cnt_t: if cnt_t[i] * mid <= cnt_s[i]: pass else: tmp += cnt_t[i] * mid - cnt_s[i] return tmp <= cnt_s["?"] def solve2(mid): tmp = deque([]) for i in cnt_t: if cnt_t[i] * mid <= cnt_s[i]: pass else: for _ in range(cnt_t[i] * mid - cnt_s[i]): tmp.append(i) return tmp from collections import deque import sys input = sys.stdin.readline s = input()[:-1] t = input()[:-1] cnt_s = {"?": 0} cnt_t = {} for i, char in enumerate(t): if char not in cnt_t: cnt_t[char] = 1 cnt_s[char] = 0 else: cnt_t[char] += 1 for i, char in enumerate(s): if char not in cnt_s: cnt_s[char] = 1 else: cnt_s[char] += 1 ok = -1 ng = 10 ** 9 while abs(ok - ng) > 1: mid = (ok + ng) // 2 if solve(mid): ok = mid else: ng = mid s = list(s) q = solve2(ok) for i in range(len(s)): if s[i] == "?": if q: s[i] = q.pop() else: s[i] = "a" print("".join(s))

Title: Suitable Replacement Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given two strings *s* and *t* consisting of small Latin letters, string *s* can also contain '?' characters. Suitability of string *s* is calculated by following metric: Any two letters can be swapped positions, these operations can be performed arbitrary number of times over any pair of positions. Among all resulting strings *s*, you choose the one with the largest number of non-intersecting occurrences of string *t*. Suitability is this number of occurrences. You should replace all '?' characters with small Latin letters in such a way that the suitability of string *s* is maximal. Input Specification: The first line contains string *s* (1<=≤<=|*s*|<=≤<=106). The second line contains string *t* (1<=≤<=|*t*|<=≤<=106). Output Specification: Print string *s* with '?' replaced with small Latin letters in such a way that suitability of that string is maximal. If there are multiple strings with maximal suitability then print any of them. Demo Input: ['?aa?\nab\n', '??b?\nza\n', 'abcd\nabacaba\n'] Demo Output: ['baab\n', 'azbz\n', 'abcd\n'] Note: In the first example string "baab" can be transformed to "abab" with swaps, this one has suitability of 2. That means that string "baab" also has suitability of 2. In the second example maximal suitability you can achieve is 1 and there are several dozens of such strings, "azbz" is just one of them. In the third example there are no '?' characters and the suitability of the string is 0.

```python def solve(mid): tmp = 0 for i in cnt_t: if cnt_t[i] * mid <= cnt_s[i]: pass else: tmp += cnt_t[i] * mid - cnt_s[i] return tmp <= cnt_s["?"] def solve2(mid): tmp = deque([]) for i in cnt_t: if cnt_t[i] * mid <= cnt_s[i]: pass else: for _ in range(cnt_t[i] * mid - cnt_s[i]): tmp.append(i) return tmp from collections import deque import sys input = sys.stdin.readline s = input()[:-1] t = input()[:-1] cnt_s = {"?": 0} cnt_t = {} for i, char in enumerate(t): if char not in cnt_t: cnt_t[char] = 1 cnt_s[char] = 0 else: cnt_t[char] += 1 for i, char in enumerate(s): if char not in cnt_s: cnt_s[char] = 1 else: cnt_s[char] += 1 ok = -1 ng = 10 ** 9 while abs(ok - ng) > 1: mid = (ok + ng) // 2 if solve(mid): ok = mid else: ng = mid s = list(s) q = solve2(ok) for i in range(len(s)): if s[i] == "?": if q: s[i] = q.pop() else: s[i] = "a" print("".join(s)) ```

3

400

B

Inna and New Matrix of Candies

PROGRAMMING

1,200

[ "brute force", "implementation", "schedules" ]

null

Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Candy Matrix 2: Reload". The field for the new game is a rectangle table of size *n*<=×<=*m*. Each line of the table contains one cell with a dwarf figurine, one cell with a candy, the other cells of the line are empty. The game lasts for several moves. During each move the player should choose all lines of the matrix where dwarf is not on the cell with candy and shout "Let's go!". After that, all the dwarves from the chosen lines start to simultaneously move to the right. During each second, each dwarf goes to the adjacent cell that is located to the right of its current cell. The movement continues until one of the following events occurs: - some dwarf in one of the chosen lines is located in the rightmost cell of his row; - some dwarf in the chosen lines is located in the cell with the candy. The point of the game is to transport all the dwarves to the candy cells. Inna is fabulous, as she came up with such an interesting game. But what about you? Your task is to play this game optimally well. Specifically, you should say by the given game field what minimum number of moves the player needs to reach the goal of the game.

The first line of the input contains two integers *n* and *m* (1<=≤<=*n*<=≤<=1000; 2<=≤<=*m*<=≤<=1000). Next *n* lines each contain *m* characters — the game field for the "Candy Martix 2: Reload". Character "*" represents an empty cell of the field, character "G" represents a dwarf and character "S" represents a candy. The matrix doesn't contain other characters. It is guaranteed that each line contains exactly one character "G" and one character "S".

In a single line print a single integer — either the minimum number of moves needed to achieve the aim of the game, or -1, if the aim cannot be achieved on the given game field.

[ "3 4\n*G*S\nG**S\n*G*S\n", "1 3\nS*G\n" ]

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

none

1,000

[ { "input": "3 4\n*G*S\nG**S\n*G*S", "output": "2" }, { "input": "1 3\nS*G", "output": "-1" }, { "input": "10 10\nG********S\n*G*******S\n**G******S\n***G*****S\n****G****S\n*****G***S\n******G**S\n*******G*S\n********GS\nG********S", "output": "9" }, { "input": "5 10\nG***S*****\nG****S****\n***GS*****\nG*S*******\nG***S*****", "output": "4" }, { "input": "4 8\nG*S*****\n****G*S*\nG*****S*\n**G***S*", "output": "3" }, { "input": "4 10\n***G****S*\n*****GS***\nG****S****\nG*******S*", "output": "3" }, { "input": "1 2\nSG", "output": "-1" }, { "input": "1 2\nGS", "output": "1" }, { "input": "1 4\nSG**", "output": "-1" } ]

1,620,120,436

2,147,483,647

Python 3

OK

TESTS

34

155

2,560,000

temp=list(map(int,input().split())) n=temp[0] m=temp[1] arr=[] def combo(arr,n,c): arr1=[x for x in arr if x!=0] if len(arr1)==0: return c else: t=min(arr1) if max(arr)==0: return c else: for i in range(n): if arr[i]!=0: arr[i]=arr[i]-t c+=1 return combo(arr,n,c) for i in range(n): str1=input() a=str1.find('S') b=str1.find('G') arr.append(a-b) if min(arr)<0: print(-1) else: print(combo(arr,n,0))

Title: Inna and New Matrix of Candies Time Limit: None seconds Memory Limit: None megabytes Problem Description: Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Candy Matrix 2: Reload". The field for the new game is a rectangle table of size *n*<=×<=*m*. Each line of the table contains one cell with a dwarf figurine, one cell with a candy, the other cells of the line are empty. The game lasts for several moves. During each move the player should choose all lines of the matrix where dwarf is not on the cell with candy and shout "Let's go!". After that, all the dwarves from the chosen lines start to simultaneously move to the right. During each second, each dwarf goes to the adjacent cell that is located to the right of its current cell. The movement continues until one of the following events occurs: - some dwarf in one of the chosen lines is located in the rightmost cell of his row; - some dwarf in the chosen lines is located in the cell with the candy. The point of the game is to transport all the dwarves to the candy cells. Inna is fabulous, as she came up with such an interesting game. But what about you? Your task is to play this game optimally well. Specifically, you should say by the given game field what minimum number of moves the player needs to reach the goal of the game. Input Specification: The first line of the input contains two integers *n* and *m* (1<=≤<=*n*<=≤<=1000; 2<=≤<=*m*<=≤<=1000). Next *n* lines each contain *m* characters — the game field for the "Candy Martix 2: Reload". Character "*" represents an empty cell of the field, character "G" represents a dwarf and character "S" represents a candy. The matrix doesn't contain other characters. It is guaranteed that each line contains exactly one character "G" and one character "S". Output Specification: In a single line print a single integer — either the minimum number of moves needed to achieve the aim of the game, or -1, if the aim cannot be achieved on the given game field. Demo Input: ['3 4\n*G*S\nG**S\n*G*S\n', '1 3\nS*G\n'] Demo Output: ['2\n', '-1\n'] Note: none

```python temp=list(map(int,input().split())) n=temp[0] m=temp[1] arr=[] def combo(arr,n,c): arr1=[x for x in arr if x!=0] if len(arr1)==0: return c else: t=min(arr1) if max(arr)==0: return c else: for i in range(n): if arr[i]!=0: arr[i]=arr[i]-t c+=1 return combo(arr,n,c) for i in range(n): str1=input() a=str1.find('S') b=str1.find('G') arr.append(a-b) if min(arr)<0: print(-1) else: print(combo(arr,n,0)) ```

3

129

A

Cookies

PROGRAMMING

900

[ "implementation" ]

null

Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?

The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.

Print in the only line the only number — the sought number of ways. If there are no such ways print 0.

[ "1\n1\n", "10\n1 2 2 3 4 4 4 2 2 2\n", "11\n2 2 2 2 2 2 2 2 2 2 99\n" ]

[ "1\n", "8\n", "1\n" ]

In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies. In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total. In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.

500

[ { "input": "1\n1", "output": "1" }, { "input": "10\n1 2 2 3 4 4 4 2 2 2", "output": "8" }, { "input": "11\n2 2 2 2 2 2 2 2 2 2 99", "output": "1" }, { "input": "2\n1 1", "output": "0" }, { "input": "2\n2 2", "output": "2" }, { "input": "2\n1 2", "output": "1" }, { "input": "7\n7 7 7 7 7 7 7", "output": "7" }, { "input": "8\n1 2 3 4 5 6 7 8", "output": "4" }, { "input": "100\n1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2", "output": "50" }, { "input": "99\n99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99", "output": "49" }, { "input": "82\n43 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99 54 34 59 83 33 15 51 26 40 12 66 65 19 30 29 78 92 60 25 13 19 84 71 73 12 24 54 49 16 41 11 40 57 59 34 40 39 9 71 83 1 77 79 53 94 47 78 55 77 85 29 52 80 90 53 77 97 97 27 79 28 23 83 25 26 22 49 86 63 56 3 32", "output": "51" }, { "input": "47\n61 97 76 94 91 22 2 68 62 73 90 47 16 79 44 71 98 68 43 6 53 52 40 27 68 67 43 96 14 91 60 61 96 24 97 13 32 65 85 96 81 77 34 18 23 14 80", "output": "21" }, { "input": "69\n71 1 78 74 58 89 30 6 100 90 22 61 11 59 14 74 27 25 78 61 45 19 25 33 37 4 52 43 53 38 9 100 56 67 69 38 76 91 63 60 93 52 28 61 9 98 8 14 57 63 89 64 98 51 36 66 36 86 13 82 50 91 52 64 86 78 78 83 81", "output": "37" }, { "input": "52\n38 78 36 75 19 3 56 1 39 97 24 79 84 16 93 55 96 64 12 24 1 86 80 29 12 32 36 36 73 39 76 65 53 98 30 20 28 8 86 43 70 22 75 69 62 65 81 25 53 40 71 59", "output": "28" }, { "input": "74\n81 31 67 97 26 75 69 81 11 13 13 74 77 88 52 20 52 64 66 75 72 28 41 54 26 75 41 91 75 15 18 36 13 83 63 61 14 48 53 63 19 67 35 48 23 65 73 100 44 55 92 88 99 17 73 25 83 7 31 89 12 80 98 39 42 75 14 29 81 35 77 87 33 94", "output": "47" }, { "input": "44\n46 56 31 31 37 71 94 2 14 100 45 72 36 72 80 3 38 54 42 98 50 32 31 42 62 31 45 50 95 100 18 17 64 22 18 25 52 56 70 57 43 40 81 28", "output": "15" }, { "input": "22\n28 57 40 74 51 4 45 84 99 12 95 14 92 60 47 81 84 51 31 91 59 42", "output": "11" }, { "input": "59\n73 45 94 76 41 49 65 13 74 66 36 25 47 75 40 23 92 72 11 32 32 8 81 26 68 56 41 8 76 47 96 55 70 11 84 14 83 18 70 22 30 39 28 100 48 11 92 45 78 69 86 1 54 90 98 91 13 17 35", "output": "33" }, { "input": "63\n20 18 44 94 68 57 16 43 74 55 68 24 21 95 76 84 50 50 47 86 86 12 58 55 28 72 86 18 34 45 81 88 3 72 41 9 60 90 81 93 12 6 9 6 2 41 1 7 9 29 81 14 64 80 20 36 67 54 7 5 35 81 22", "output": "37" }, { "input": "28\n49 84 48 19 44 91 11 82 96 95 88 90 71 82 87 25 31 23 18 13 98 45 26 65 35 12 31 14", "output": "15" }, { "input": "61\n34 18 28 64 28 45 9 77 77 20 63 92 79 16 16 100 86 2 91 91 57 15 31 95 10 88 84 5 82 83 53 98 59 17 97 80 76 80 81 3 91 81 87 93 61 46 10 49 6 22 21 75 63 89 21 81 30 19 67 38 77", "output": "35" }, { "input": "90\n41 90 43 1 28 75 90 50 3 70 76 64 81 63 25 69 83 82 29 91 59 66 21 61 7 55 72 49 38 69 72 20 64 58 30 81 61 29 96 14 39 5 100 20 29 98 75 29 44 78 97 45 26 77 73 59 22 99 41 6 3 96 71 20 9 18 96 18 90 62 34 78 54 5 41 6 73 33 2 54 26 21 18 6 45 57 43 73 95 75", "output": "42" }, { "input": "45\n93 69 4 27 20 14 71 48 79 3 32 26 49 30 57 88 13 56 49 61 37 32 47 41 41 70 45 68 82 18 8 6 25 20 15 13 71 99 28 6 52 34 19 59 26", "output": "23" }, { "input": "33\n29 95 48 49 91 10 83 71 47 25 66 36 51 12 34 10 54 74 41 96 89 26 89 1 42 33 1 62 9 32 49 65 78", "output": "15" }, { "input": "34\n98 24 42 36 41 82 28 58 89 34 77 70 76 44 74 54 66 100 13 79 4 88 21 1 11 45 91 29 87 100 29 54 82 78", "output": "13" }, { "input": "29\n91 84 26 84 9 63 52 9 65 56 90 2 36 7 67 33 91 14 65 38 53 36 81 83 85 14 33 95 51", "output": "17" }, { "input": "100\n2 88 92 82 87 100 78 28 84 43 78 32 43 33 97 19 15 52 29 84 57 72 54 13 99 28 82 79 40 70 34 92 91 53 9 88 27 43 14 92 72 37 26 37 20 95 19 34 49 64 33 37 34 27 80 79 9 54 99 68 25 4 68 73 46 66 24 78 3 87 26 52 50 84 4 95 23 83 39 58 86 36 33 16 98 2 84 19 53 12 69 60 10 11 78 17 79 92 77 59", "output": "45" }, { "input": "100\n2 95 45 73 9 54 20 97 57 82 88 26 18 71 25 27 75 54 31 11 58 85 69 75 72 91 76 5 25 80 45 49 4 73 8 81 81 38 5 12 53 77 7 96 90 35 28 80 73 94 19 69 96 17 94 49 69 9 32 19 5 12 46 29 26 40 59 59 6 95 82 50 72 2 45 69 12 5 72 29 39 72 23 96 81 28 28 56 68 58 37 41 30 1 90 84 15 24 96 43", "output": "53" }, { "input": "100\n27 72 35 91 13 10 35 45 24 55 83 84 63 96 29 79 34 67 63 92 48 83 18 77 28 27 49 66 29 88 55 15 6 58 14 67 94 36 77 7 7 64 61 52 71 18 36 99 76 6 50 67 16 13 41 7 89 73 61 51 78 22 78 32 76 100 3 31 89 71 63 53 15 85 77 54 89 33 68 74 3 23 57 5 43 89 75 35 9 86 90 11 31 46 48 37 74 17 77 8", "output": "40" }, { "input": "100\n69 98 69 88 11 49 55 8 25 91 17 81 47 26 15 73 96 71 18 42 42 61 48 14 92 78 35 72 4 27 62 75 83 79 17 16 46 80 96 90 82 54 37 69 85 21 67 70 96 10 46 63 21 59 56 92 54 88 77 30 75 45 44 29 86 100 51 11 65 69 66 56 82 63 27 1 51 51 13 10 3 55 26 85 34 16 87 72 13 100 81 71 90 95 86 50 83 55 55 54", "output": "53" }, { "input": "100\n34 35 99 64 2 66 78 93 20 48 12 79 19 10 87 7 42 92 60 79 5 2 24 89 57 48 63 92 74 4 16 51 7 12 90 48 87 17 18 73 51 58 97 97 25 38 15 97 96 73 67 91 6 75 14 13 87 79 75 3 15 55 35 95 71 45 10 13 20 37 82 26 2 22 13 83 97 84 39 79 43 100 54 59 98 8 61 34 7 65 75 44 24 77 73 88 34 95 44 77", "output": "55" }, { "input": "100\n15 86 3 1 51 26 74 85 37 87 64 58 10 6 57 26 30 47 85 65 24 72 50 40 12 35 91 47 91 60 47 87 95 34 80 91 26 3 36 39 14 86 28 70 51 44 28 21 72 79 57 61 16 71 100 94 57 67 36 74 24 21 89 85 25 2 97 67 76 53 76 80 97 64 35 13 8 32 21 52 62 61 67 14 74 73 66 44 55 76 24 3 43 42 99 61 36 80 38 66", "output": "52" }, { "input": "100\n45 16 54 54 80 94 74 93 75 85 58 95 79 30 81 2 84 4 57 23 92 64 78 1 50 36 13 27 56 54 10 77 87 1 5 38 85 74 94 82 30 45 72 83 82 30 81 82 82 3 69 82 7 92 39 60 94 42 41 5 3 17 67 21 79 44 79 96 28 3 53 68 79 89 63 83 1 44 4 31 84 15 73 77 19 66 54 6 73 1 67 24 91 11 86 45 96 82 20 89", "output": "51" }, { "input": "100\n84 23 50 32 90 71 92 43 58 70 6 82 7 55 85 19 70 89 12 26 29 56 74 30 2 27 4 39 63 67 91 81 11 33 75 10 82 88 39 43 43 80 68 35 55 67 53 62 73 65 86 74 43 51 14 48 42 92 83 57 22 33 24 99 5 27 78 96 7 28 11 15 8 38 85 67 5 92 24 96 57 59 14 95 91 4 9 18 45 33 74 83 64 85 14 51 51 94 29 2", "output": "53" }, { "input": "100\n77 56 56 45 73 55 32 37 39 50 30 95 79 21 44 34 51 43 86 91 39 30 85 15 35 93 100 14 57 31 80 79 38 40 88 4 91 54 7 95 76 26 62 84 17 33 67 47 6 82 69 51 17 2 59 24 11 12 31 90 12 11 55 38 72 49 30 50 42 46 5 97 9 9 30 45 86 23 19 82 40 42 5 40 35 98 35 32 60 60 5 28 84 35 21 49 68 53 68 23", "output": "48" }, { "input": "100\n78 38 79 61 45 86 83 83 86 90 74 69 2 84 73 39 2 5 20 71 24 80 54 89 58 34 77 40 39 62 2 47 28 53 97 75 88 98 94 96 33 71 44 90 47 36 19 89 87 98 90 87 5 85 34 79 82 3 42 88 89 63 35 7 89 30 40 48 12 41 56 76 83 60 80 80 39 56 77 4 72 96 30 55 57 51 7 19 11 1 66 1 91 87 11 62 95 85 79 25", "output": "48" }, { "input": "100\n5 34 23 20 76 75 19 51 17 82 60 13 83 6 65 16 20 43 66 54 87 10 87 73 50 24 16 98 33 28 80 52 54 82 26 92 14 13 84 92 94 29 61 21 60 20 48 94 24 20 75 70 58 27 68 45 86 89 29 8 67 38 83 48 18 100 11 22 46 84 52 97 70 19 50 75 3 7 52 53 72 41 18 31 1 38 49 53 11 64 99 76 9 87 48 12 100 32 44 71", "output": "58" }, { "input": "100\n76 89 68 78 24 72 73 95 98 72 58 15 2 5 56 32 9 65 50 70 94 31 29 54 89 52 31 93 43 56 26 35 72 95 51 55 78 70 11 92 17 5 54 94 81 31 78 95 73 91 95 37 59 9 53 48 65 55 84 8 45 97 64 37 96 34 36 53 66 17 72 48 99 23 27 18 92 84 44 73 60 78 53 29 68 99 19 39 61 40 69 6 77 12 47 29 15 4 8 45", "output": "53" }, { "input": "100\n82 40 31 53 8 50 85 93 3 84 54 17 96 59 51 42 18 19 35 84 79 31 17 46 54 82 72 49 35 73 26 89 61 73 3 50 12 29 25 77 88 21 58 24 22 89 96 54 82 29 96 56 77 16 1 68 90 93 20 23 57 22 31 18 92 90 51 14 50 72 31 54 12 50 66 62 2 34 17 45 68 50 87 97 23 71 1 72 17 82 42 15 20 78 4 49 66 59 10 17", "output": "54" }, { "input": "100\n32 82 82 24 39 53 48 5 29 24 9 37 91 37 91 95 1 97 84 52 12 56 93 47 22 20 14 17 40 22 79 34 24 2 69 30 69 29 3 89 21 46 60 92 39 29 18 24 49 18 40 22 60 13 77 50 39 64 50 70 99 8 66 31 90 38 20 54 7 21 5 56 41 68 69 20 54 89 69 62 9 53 43 89 81 97 15 2 52 78 89 65 16 61 59 42 56 25 32 52", "output": "49" }, { "input": "100\n72 54 23 24 97 14 99 87 15 25 7 23 17 87 72 31 71 87 34 82 51 77 74 85 62 38 24 7 84 48 98 21 29 71 70 84 25 58 67 92 18 44 32 9 81 15 53 29 63 18 86 16 7 31 38 99 70 32 89 16 23 11 66 96 69 82 97 59 6 9 49 80 85 19 6 9 52 51 85 74 53 46 73 55 31 63 78 61 34 80 77 65 87 77 92 52 89 8 52 31", "output": "44" }, { "input": "100\n56 88 8 19 7 15 11 54 35 50 19 57 63 72 51 43 50 19 57 90 40 100 8 92 11 96 30 32 59 65 93 47 62 3 50 41 30 50 72 83 61 46 83 60 20 46 33 1 5 18 83 22 34 16 41 95 63 63 7 59 55 95 91 29 64 60 64 81 45 45 10 9 88 37 69 85 21 82 41 76 42 34 47 78 51 83 65 100 13 22 59 76 63 1 26 86 36 94 99 74", "output": "46" }, { "input": "100\n27 89 67 60 62 80 43 50 28 88 72 5 94 11 63 91 18 78 99 3 71 26 12 97 74 62 23 24 22 3 100 72 98 7 94 32 12 75 61 88 42 48 10 14 45 9 48 56 73 76 70 70 79 90 35 39 96 37 81 11 19 65 99 39 23 79 34 61 35 74 90 37 73 23 46 21 94 84 73 58 11 89 13 9 10 85 42 78 73 32 53 39 49 90 43 5 28 31 97 75", "output": "53" }, { "input": "100\n33 24 97 96 1 14 99 51 13 65 67 20 46 88 42 44 20 49 5 89 98 83 15 40 74 83 58 3 10 79 34 2 69 28 37 100 55 52 14 8 44 94 97 89 6 42 11 28 30 33 55 56 20 57 52 25 75 1 87 42 62 41 37 12 54 85 95 80 42 36 94 96 28 76 54 36 4 17 26 24 62 15 17 79 84 36 92 78 74 91 96 77 54 92 81 91 62 98 37 37", "output": "43" }, { "input": "100\n86 24 61 15 11 85 1 31 47 36 23 36 59 34 3 27 16 29 82 28 58 52 52 66 71 61 98 39 60 20 67 41 67 90 73 29 92 17 70 95 58 98 58 32 21 73 46 56 87 72 80 75 40 27 94 31 59 92 93 37 14 99 96 21 97 23 81 91 52 52 96 94 92 28 38 29 52 16 57 27 17 24 91 21 79 55 96 98 95 94 23 78 79 12 77 35 32 75 100 82", "output": "51" }, { "input": "100\n88 85 41 37 69 21 7 69 36 5 92 26 64 75 22 46 67 20 70 22 62 66 38 24 47 49 68 30 90 31 67 86 86 82 9 51 43 45 48 42 73 44 31 94 45 60 54 66 20 87 11 94 34 32 87 66 56 28 75 39 37 90 72 93 55 72 31 42 30 71 87 61 4 12 12 81 23 61 56 98 71 32 30 33 96 63 92 16 8 78 47 91 47 54 49 3 81 82 41 5", "output": "47" }, { "input": "1\n2", "output": "1" }, { "input": "5\n1 1 3 2 2", "output": "3" } ]

1,695,333,442

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

62

0

n = int(input()) a = list(map(int, input().split())) counter = 0 for x in a: if x % 2 == 0: counter += 1 if counter == 0: print(0) elif counter % 2 == 0: print(counter) else: print(len(a) - counter)

Title: Cookies Time Limit: None seconds Memory Limit: None megabytes Problem Description: Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even? Input Specification: The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag. Output Specification: Print in the only line the only number — the sought number of ways. If there are no such ways print 0. Demo Input: ['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n'] Demo Output: ['1\n', '8\n', '1\n'] Note: In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies. In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total. In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.

```python n = int(input()) a = list(map(int, input().split())) counter = 0 for x in a: if x % 2 == 0: counter += 1 if counter == 0: print(0) elif counter % 2 == 0: print(counter) else: print(len(a) - counter) ```

0

766

B

Mahmoud and a Triangle

PROGRAMMING

1,000

[ "constructive algorithms", "geometry", "greedy", "math", "number theory", "sortings" ]

null

Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle. Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.

The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has.

In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.

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

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

For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.

1,000

[ { "input": "5\n1 5 3 2 4", "output": "YES" }, { "input": "3\n4 1 2", "output": "NO" }, { "input": "30\n197 75 517 39724 7906061 1153471 3 15166 168284 3019844 272293 316 16 24548 42 118 5792 5 9373 1866366 4886214 24 2206 712886 104005 1363 836 64273 440585 3576", "output": "NO" }, { "input": "30\n229017064 335281886 247217656 670601882 743442492 615491486 544941439 911270108 474843964 803323771 177115397 62179276 390270885 754889875 881720571 902691435 154083299 328505383 761264351 182674686 94104683 357622370 573909964 320060691 33548810 247029007 812823597 946798893 813659359 710111761", "output": "YES" }, { "input": "40\n740553458 532562042 138583675 75471987 487348843 476240280 972115023 103690894 546736371 915774563 35356828 819948191 138721993 24257926 761587264 767176616 608310208 78275645 386063134 227581756 672567198 177797611 87579917 941781518 274774331 843623616 981221615 630282032 118843963 749160513 354134861 132333165 405839062 522698334 29698277 541005920 856214146 167344951 398332403 68622974", "output": "YES" }, { "input": "40\n155 1470176 7384 765965701 1075 4 561554 6227772 93 16304522 1744 662 3 292572860 19335 908613 42685804 347058 20 132560 3848974 69067081 58 2819 111752888 408 81925 30 11951 4564 251 26381275 473392832 50628 180819969 2378797 10076746 9 214492 31291", "output": "NO" }, { "input": "3\n1 1000000000 1000000000", "output": "YES" }, { "input": "4\n1 1000000000 1000000000 1000000000", "output": "YES" }, { "input": "3\n1 1000000000 1", "output": "NO" }, { "input": "5\n1 2 3 5 2", "output": "YES" }, { "input": "41\n19 161 4090221 118757367 2 45361275 1562319 596751 140871 97 1844 310910829 10708344 6618115 698 1 87059 33 2527892 12703 73396090 17326460 3 368811 20550 813975131 10 53804 28034805 7847 2992 33254 1139 227930 965568 261 4846 503064297 192153458 57 431", "output": "NO" }, { "input": "42\n4317083 530966905 202811311 104 389267 35 1203 18287479 125344279 21690 859122498 65 859122508 56790 1951 148683 457 1 22 2668100 8283 2 77467028 13405 11302280 47877251 328155592 35095 29589769 240574 4 10 1019123 6985189 629846 5118 169 1648973 91891 741 282 3159", "output": "YES" }, { "input": "43\n729551585 11379 5931704 330557 1653 15529406 729551578 278663905 1 729551584 2683 40656510 29802 147 1400284 2 126260 865419 51 17 172223763 86 1 534861 450887671 32 234 25127103 9597697 48226 7034 389 204294 2265706 65783617 4343 3665990 626 78034 106440137 5 18421 1023", "output": "YES" }, { "input": "44\n719528276 2 235 444692918 24781885 169857576 18164 47558 15316043 9465834 64879816 2234575 1631 853530 8 1001 621 719528259 84 6933 31 1 3615623 719528266 40097928 274835337 1381044 11225 2642 5850203 6 527506 18 104977753 76959 29393 49 4283 141 201482 380 1 124523 326015", "output": "YES" }, { "input": "45\n28237 82 62327732 506757 691225170 5 970 4118 264024506 313192 367 14713577 73933 691225154 6660 599 691225145 3473403 51 427200630 1326718 2146678 100848386 1569 27 163176119 193562 10784 45687 819951 38520653 225 119620 1 3 691225169 691225164 17445 23807072 1 9093493 5620082 2542 139 14", "output": "YES" }, { "input": "44\n165580141 21 34 55 1 89 144 17711 2 377 610 987 2584 13 5 4181 6765 10946 1597 8 28657 3 233 75025 121393 196418 317811 9227465 832040 1346269 2178309 3524578 5702887 1 14930352 102334155 24157817 39088169 63245986 701408733 267914296 433494437 514229 46368", "output": "NO" }, { "input": "3\n1 1000000000 999999999", "output": "NO" }, { "input": "5\n1 1 1 1 1", "output": "YES" }, { "input": "10\n1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000", "output": "NO" }, { "input": "5\n2 3 4 10 20", "output": "YES" }, { "input": "6\n18 23 40 80 160 161", "output": "YES" }, { "input": "4\n5 6 7 888", "output": "YES" }, { "input": "9\n1 1 2 2 4 5 10 10 20", "output": "YES" }, { "input": "7\n3 150 900 4 500 1500 5", "output": "YES" }, { "input": "3\n2 2 3", "output": "YES" }, { "input": "7\n1 2 100 200 250 1000000 2000000", "output": "YES" }, { "input": "8\n2 3 5 5 5 6 6 13", "output": "YES" }, { "input": "3\n2 3 4", "output": "YES" }, { "input": "6\n1 1 1 4 5 100", "output": "YES" }, { "input": "13\n1 2 3 5 8 13 22 34 55 89 144 233 377", "output": "YES" }, { "input": "4\n2 3 4 8", "output": "YES" }, { "input": "3\n5 6 7", "output": "YES" }, { "input": "5\n1 4 5 6 1000000", "output": "YES" }, { "input": "4\n5 6 7 20", "output": "YES" }, { "input": "6\n1 1 1 1 1 65", "output": "YES" }, { "input": "4\n3 4 5 100", "output": "YES" }, { "input": "3\n2 4 5", "output": "YES" }, { "input": "7\n1 1 1 1 1 10 1000", "output": "YES" }, { "input": "4\n1 1 2 3", "output": "NO" }, { "input": "11\n1 2 5 6 7 8 9 17 18 19 100", "output": "YES" }, { "input": "4\n5 16 20 200", "output": "YES" }, { "input": "5\n17 6 3 3 1", "output": "YES" }, { "input": "3\n1 1 1", "output": "YES" }, { "input": "6\n1 1 1 2 3 5", "output": "YES" }, { "input": "4\n2 4 6 6", "output": "YES" }, { "input": "9\n1 2 4 4 4 4 7 8 20", "output": "YES" }, { "input": "9\n1 1 2 5 5 5 10 10 20", "output": "YES" }, { "input": "7\n3 150 600 4 1700 6000 5", "output": "YES" }, { "input": "5\n5761 20966 27841 28800 29399", "output": "YES" }, { "input": "9\n1 2 3 6 7 10 11 12 24", "output": "YES" }, { "input": "4\n1 2 1 1", "output": "YES" }, { "input": "5\n1 1 2 3 4", "output": "YES" } ]

1,689,581,719

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

31

0

n=int(input()) a=list(map(int, input().split())) count=0 for i in range(n-2): for j in range(i+1,n-1): for k in range(i+2,n): if a[i]+a[j]>a[k] and a[j]+a[k]>a[i] and a[k]+a[i]>a[j]: count+=1 if count: print("YES") else: print("NO") '''A + B > C and B + C > A and C + A > B inputCopy 5 1 5 3 2 4 outputCopy YES inputCopy 3 4 1 2 outputCopy NO'''

Title: Mahmoud and a Triangle Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle. Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area. Input Specification: The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has. Output Specification: In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise. Demo Input: ['5\n1 5 3 2 4\n', '3\n4 1 2\n'] Demo Output: ['YES\n', 'NO\n'] Note: For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.

```python n=int(input()) a=list(map(int, input().split())) count=0 for i in range(n-2): for j in range(i+1,n-1): for k in range(i+2,n): if a[i]+a[j]>a[k] and a[j]+a[k]>a[i] and a[k]+a[i]>a[j]: count+=1 if count: print("YES") else: print("NO") '''A + B > C and B + C > A and C + A > B inputCopy 5 1 5 3 2 4 outputCopy YES inputCopy 3 4 1 2 outputCopy NO''' ```

0

677

A

Vanya and Fence

PROGRAMMING

800

[ "implementation" ]

null

Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*. Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?

The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.

Print a single integer — the minimum possible valid width of the road.

[ "3 7\n4 5 14\n", "6 1\n1 1 1 1 1 1\n", "6 5\n7 6 8 9 10 5\n" ]

[ "4\n", "6\n", "11\n" ]

In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4. In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough. In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11.

500

[ { "input": "3 7\n4 5 14", "output": "4" }, { "input": "6 1\n1 1 1 1 1 1", "output": "6" }, { "input": "6 5\n7 6 8 9 10 5", "output": "11" }, { "input": "10 420\n214 614 297 675 82 740 174 23 255 15", "output": "13" }, { "input": "10 561\n657 23 1096 487 785 66 481 554 1000 821", "output": "15" }, { "input": "100 342\n478 143 359 336 162 333 385 515 117 496 310 538 469 539 258 676 466 677 1 296 150 560 26 213 627 221 255 126 617 174 279 178 24 435 70 145 619 46 669 566 300 67 576 251 58 176 441 564 569 194 24 669 73 262 457 259 619 78 400 579 222 626 269 47 80 315 160 194 455 186 315 424 197 246 683 220 68 682 83 233 290 664 273 598 362 305 674 614 321 575 362 120 14 534 62 436 294 351 485 396", "output": "144" }, { "input": "100 290\n244 49 276 77 449 261 468 458 201 424 9 131 300 88 432 394 104 77 13 289 435 259 111 453 168 394 156 412 351 576 178 530 81 271 228 564 125 328 42 372 205 61 180 471 33 360 567 331 222 318 241 117 529 169 188 484 202 202 299 268 246 343 44 364 333 494 59 236 84 485 50 8 428 8 571 227 205 310 210 9 324 472 368 490 114 84 296 305 411 351 569 393 283 120 510 171 232 151 134 366", "output": "145" }, { "input": "1 1\n1", "output": "1" }, { "input": "1 1\n2", "output": "2" }, { "input": "46 71\n30 26 56 138 123 77 60 122 73 45 79 10 130 3 14 1 38 46 128 50 82 16 32 68 28 98 62 106 2 49 131 11 114 39 139 70 40 50 45 137 33 30 35 136 135 19", "output": "63" }, { "input": "20 723\n212 602 293 591 754 91 1135 640 80 495 845 928 1399 498 926 1431 1226 869 814 1386", "output": "31" }, { "input": "48 864\n843 1020 751 1694 18 1429 1395 1174 272 1158 1628 1233 1710 441 765 561 778 748 1501 1200 563 1263 1398 1687 1518 1640 1591 839 500 466 1603 1587 1201 1209 432 868 1159 639 649 628 9 91 1036 147 896 1557 941 518", "output": "75" }, { "input": "26 708\n549 241 821 734 945 1161 566 1268 216 30 1142 730 529 1014 255 168 796 1148 89 113 1328 286 743 871 1259 1397", "output": "41" }, { "input": "75 940\n1620 1745 1599 441 64 1466 1496 1239 1716 1475 778 106 1136 1212 1261 444 781 257 1071 747 626 232 609 1544 682 1326 469 1361 1460 1450 1207 1319 922 625 1737 1057 1698 592 692 80 1016 541 1254 201 682 1007 847 206 1066 809 259 109 240 1611 219 1455 1326 1377 1827 786 42 1002 1382 1592 543 1866 1198 334 1524 1760 340 1566 955 257 1118", "output": "116" } ]

1,693,822,940

2,147,483,647

Python 3

OK

TESTS

29

46

0

f,h=map(int,input().split()) p=list(map(int,input().split())) a1=0 b1=0 for i in p: if i>h: a1=a1+2 else: b1=b1+1 print(a1+b1)

Title: Vanya and Fence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*. Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard? Input Specification: The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person. Output Specification: Print a single integer — the minimum possible valid width of the road. Demo Input: ['3 7\n4 5 14\n', '6 1\n1 1 1 1 1 1\n', '6 5\n7 6 8 9 10 5\n'] Demo Output: ['4\n', '6\n', '11\n'] Note: In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4. In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough. In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11.

```python f,h=map(int,input().split()) p=list(map(int,input().split())) a1=0 b1=0 for i in p: if i>h: a1=a1+2 else: b1=b1+1 print(a1+b1) ```

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" } ]

1,696,944,125

2,147,483,647

Python 3

OK

TESTS

36

124

1,228,800

from fractions import Fraction y, w = map(int, input().split()) Numerator = 7-max(y,w) Denominator = 6 temp = Numerator/Denominator if Numerator == 0: print("0/1") elif Numerator == Denominator: print("1/1") else: x = Fraction(Numerator, Denominator) print(x)

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 from fractions import Fraction y, w = map(int, input().split()) Numerator = 7-max(y,w) Denominator = 6 temp = Numerator/Denominator if Numerator == 0: print("0/1") elif Numerator == Denominator: print("1/1") else: x = Fraction(Numerator, Denominator) print(x) ```

3.928845

837

A

Text Volume

PROGRAMMING

800

[ "implementation" ]

null

You are given a text of single-space separated words, consisting of small and capital Latin letters. Volume of the word is number of capital letters in the word. Volume of the text is maximum volume of all words in the text. Calculate the volume of the given text.

The first line contains one integer number *n* (1<=≤<=*n*<=≤<=200) — length of the text. The second line contains text of single-space separated words *s*1,<=*s*2,<=...,<=*s**i*, consisting only of small and capital Latin letters.

Print one integer number — volume of text.

[ "7\nNonZERO\n", "24\nthis is zero answer text\n", "24\nHarbour Space University\n" ]

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

In the first example there is only one word, there are 5 capital letters in it. In the second example all of the words contain 0 capital letters.

0

[ { "input": "7\nNonZERO", "output": "5" }, { "input": "24\nthis is zero answer text", "output": "0" }, { "input": "24\nHarbour Space University", "output": "1" }, { "input": "2\nWM", "output": "2" }, { "input": "200\nLBmJKQLCKUgtTxMoDsEerwvLOXsxASSydOqWyULsRcjMYDWdDCgaDvBfATIWPVSXlbcCLHPYahhxMEYUiaxoCebghJqvmRnaNHYTKLeOiaLDnATPZAOgSNfBzaxLymTGjfzvTegbXsAthTxyDTcmBUkqyGlVGZhoazQzVSoKbTFcCRvYsgSCwjGMxBfWEwMHuagTBxkz", "output": "105" }, { "input": "199\no A r v H e J q k J k v w Q F p O R y R Z o a K R L Z E H t X y X N y y p b x B m r R S q i A x V S u i c L y M n N X c C W Z m S j e w C w T r I S X T D F l w o k f t X u n W w p Z r A k I Y E h s g", "output": "1" }, { "input": "200\nhCyIdivIiISmmYIsCLbpKcTyHaOgTUQEwnQACXnrLdHAVFLtvliTEMlzBVzTesQbhXmcqvwPDeojglBMIjOXANfyQxCSjOJyO SIqOTnRzVzseGIDDYNtrwIusScWSuEhPyEmgQIVEzXofRptjeMzzhtUQxJgcUWILUhEaaRmYRBVsjoqgmyPIKwSajdlNPccOOtWrez", "output": "50" }, { "input": "1\ne", "output": "0" }, { "input": "1\nA", "output": "1" }, { "input": "200\nABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ KLMNOPRSTU KLMNOPRSTU KLMNOPRSTU VWXYZABCDE KLMNOPRSTU KLMNOPRSTU KLMNOPRSTU KLMNOPRSTU KZ", "output": "10" }, { "input": "200\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "200" }, { "input": "200\nffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff", "output": "0" }, { "input": "24\nHarbour Space UniversitY", "output": "2" }, { "input": "5\naA AA", "output": "2" }, { "input": "10\nas AS ASDA", "output": "4" }, { "input": "10\nas AS ASDZ", "output": "4" }, { "input": "3\na A", "output": "1" }, { "input": "24\nHarbour space UniversitY", "output": "2" }, { "input": "10\nas AS ASAa", "output": "3" }, { "input": "15\naAb ABCDFGRHTJS", "output": "11" }, { "input": "53\nsdfAZEZR AZE dfdf dsdRFGSDF ZZDZSD dfsd ERBGF dsfsdfR", "output": "6" }, { "input": "10\nABC ABc AB", "output": "3" }, { "input": "10\nA c de CDE", "output": "3" }, { "input": "4\nA AB", "output": "2" }, { "input": "18\nHARbour Space UNIV", "output": "4" }, { "input": "13\na b c d e f A", "output": "1" }, { "input": "6\nA B CA", "output": "2" }, { "input": "4\naa A", "output": "1" }, { "input": "3\nA a", "output": "1" } ]

1,565,763,264

2,147,483,647

Python 3

OK

TESTS

29

124

0

n = int(input()) s = input().split() cur_min, cur_max = 0, 0 for i in range(len(s)): for k in range(len(s[i])): if s[i][k].istitle(): cur_min += 1 if cur_min > cur_max: cur_max = cur_min cur_min = 0 print(cur_max)

Title: Text Volume Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a text of single-space separated words, consisting of small and capital Latin letters. Volume of the word is number of capital letters in the word. Volume of the text is maximum volume of all words in the text. Calculate the volume of the given text. Input Specification: The first line contains one integer number *n* (1<=≤<=*n*<=≤<=200) — length of the text. The second line contains text of single-space separated words *s*1,<=*s*2,<=...,<=*s**i*, consisting only of small and capital Latin letters. Output Specification: Print one integer number — volume of text. Demo Input: ['7\nNonZERO\n', '24\nthis is zero answer text\n', '24\nHarbour Space University\n'] Demo Output: ['5\n', '0\n', '1\n'] Note: In the first example there is only one word, there are 5 capital letters in it. In the second example all of the words contain 0 capital letters.

```python n = int(input()) s = input().split() cur_min, cur_max = 0, 0 for i in range(len(s)): for k in range(len(s[i])): if s[i][k].istitle(): cur_min += 1 if cur_min > cur_max: cur_max = cur_min cur_min = 0 print(cur_max) ```

3

688

B

Lovely Palindromes

PROGRAMMING

1,000

[ "constructive algorithms", "math" ]

null

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

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

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

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

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

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

1,000

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

1,628,508,592

2,147,483,647

Python 3

OK

TESTS

40

93

7,372,800

# from sys import stdin, stdout # import re # import string # import math # import statistics # import itertools # import collections # import cmath # import calendar # import datetime # import operator #----------Profiling---------- # import cProfile # import pstats string = input() ans = '' for i in range(-1, -len(string)-1, -1): ans = ans + string[i] print(string+ans)

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

```python # from sys import stdin, stdout # import re # import string # import math # import statistics # import itertools # import collections # import cmath # import calendar # import datetime # import operator #----------Profiling---------- # import cProfile # import pstats string = input() ans = '' for i in range(-1, -len(string)-1, -1): ans = ans + string[i] print(string+ans) ```

3

915

A

Garden

PROGRAMMING

900

[ "implementation" ]

null

Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden. Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden. See the examples for better understanding.

The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively. The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour. It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.

Print one integer number — the minimum number of hours required to water the garden.

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

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

In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden. In the second test we can choose only the bucket that allows us to water the segment of length 1.

0

[ { "input": "3 6\n2 3 5", "output": "2" }, { "input": "6 7\n1 2 3 4 5 6", "output": "7" }, { "input": "5 97\n1 10 50 97 2", "output": "1" }, { "input": "5 97\n1 10 50 100 2", "output": "97" }, { "input": "100 100\n2 46 24 18 86 90 31 38 84 49 58 28 15 80 14 24 87 56 62 87 41 87 55 71 87 32 41 56 91 32 24 75 43 42 35 30 72 53 31 26 54 61 87 85 36 75 44 31 7 38 77 57 61 54 70 77 45 96 39 57 11 8 91 42 52 15 42 30 92 41 27 26 34 27 3 80 32 86 26 97 63 91 30 75 14 7 19 23 45 11 8 43 44 73 11 56 3 55 63 16", "output": "50" }, { "input": "100 91\n13 13 62 96 74 47 81 46 78 21 20 42 4 73 25 30 76 74 58 28 25 52 42 48 74 40 82 9 25 29 17 22 46 64 57 95 81 39 47 86 40 95 97 35 31 98 45 98 47 78 52 63 58 14 89 97 17 95 28 22 20 36 68 38 95 16 2 26 54 47 42 31 31 81 21 21 65 40 82 53 60 71 75 33 96 98 6 22 95 12 5 48 18 27 58 62 5 96 36 75", "output": "7" }, { "input": "8 8\n8 7 6 5 4 3 2 1", "output": "1" }, { "input": "3 8\n4 3 2", "output": "2" }, { "input": "3 8\n2 4 2", "output": "2" }, { "input": "3 6\n1 3 2", "output": "2" }, { "input": "3 6\n3 2 5", "output": "2" }, { "input": "3 8\n4 2 1", "output": "2" }, { "input": "5 6\n2 3 5 1 2", "output": "2" }, { "input": "2 6\n5 3", "output": "2" }, { "input": "4 12\n6 4 3 1", "output": "2" }, { "input": "3 18\n1 9 6", "output": "2" }, { "input": "3 9\n3 2 1", "output": "3" }, { "input": "3 6\n5 3 2", "output": "2" }, { "input": "2 10\n5 2", "output": "2" }, { "input": "2 18\n6 3", "output": "3" }, { "input": "4 12\n1 2 12 3", "output": "1" }, { "input": "3 7\n3 2 1", "output": "7" }, { "input": "3 6\n3 2 1", "output": "2" }, { "input": "5 10\n5 4 3 2 1", "output": "2" }, { "input": "5 16\n8 4 2 1 7", "output": "2" }, { "input": "6 7\n6 5 4 3 7 1", "output": "1" }, { "input": "2 6\n3 2", "output": "2" }, { "input": "2 4\n4 1", "output": "1" }, { "input": "6 8\n2 4 1 3 5 7", "output": "2" }, { "input": "6 8\n6 5 4 3 2 1", "output": "2" }, { "input": "6 15\n5 2 3 6 4 3", "output": "3" }, { "input": "4 8\n2 4 8 1", "output": "1" }, { "input": "2 5\n5 1", "output": "1" }, { "input": "4 18\n3 1 1 2", "output": "6" }, { "input": "2 1\n2 1", "output": "1" }, { "input": "3 10\n2 10 5", "output": "1" }, { "input": "5 12\n12 4 4 4 3", "output": "1" }, { "input": "3 6\n6 3 2", "output": "1" }, { "input": "2 2\n2 1", "output": "1" }, { "input": "3 18\n1 9 3", "output": "2" }, { "input": "3 8\n7 2 4", "output": "2" }, { "input": "2 100\n99 1", "output": "100" }, { "input": "4 12\n1 3 4 2", "output": "3" }, { "input": "3 6\n2 3 1", "output": "2" }, { "input": "4 6\n3 2 5 12", "output": "2" }, { "input": "4 97\n97 1 50 10", "output": "1" }, { "input": "3 12\n1 12 2", "output": "1" }, { "input": "4 12\n1 4 3 2", "output": "3" }, { "input": "1 1\n1", "output": "1" }, { "input": "3 19\n7 1 1", "output": "19" }, { "input": "5 12\n12 4 3 4 4", "output": "1" }, { "input": "3 8\n8 4 2", "output": "1" }, { "input": "3 3\n3 2 1", "output": "1" }, { "input": "5 6\n3 2 4 2 2", "output": "2" }, { "input": "2 16\n8 4", "output": "2" }, { "input": "3 6\n10 2 3", "output": "2" }, { "input": "5 3\n2 4 5 3 6", "output": "1" }, { "input": "11 99\n1 2 3 6 5 4 7 8 99 33 66", "output": "1" }, { "input": "3 12\n3 12 2", "output": "1" }, { "input": "5 25\n24 5 15 25 23", "output": "1" }, { "input": "2 4\n8 1", "output": "4" }, { "input": "4 100\n2 50 4 1", "output": "2" }, { "input": "3 28\n7 14 1", "output": "2" }, { "input": "4 8\n2 8 4 1", "output": "1" }, { "input": "4 6\n6 1 2 3", "output": "1" }, { "input": "2 12\n4 3", "output": "3" }, { "input": "4 12\n1 2 4 3", "output": "3" }, { "input": "5 12\n2 3 12 6 4", "output": "1" }, { "input": "4 4\n1 2 2 4", "output": "1" }, { "input": "3 6\n2 3 2", "output": "2" }, { "input": "4 21\n21 20 21 2", "output": "1" }, { "input": "3 8\n3 4 2", "output": "2" }, { "input": "1 25\n25", "output": "1" }, { "input": "99 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99", "output": "1" }, { "input": "98 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98", "output": "1" }, { "input": "79 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79", "output": "1" }, { "input": "4 32\n1 1 1 1", "output": "32" }, { "input": "1 100\n1", "output": "100" }, { "input": "2 100\n7 1", "output": "100" }, { "input": "7 24\n1 3 6 4 5 2 7", "output": "4" }, { "input": "6 87\n1 2 8 4 5 7", "output": "87" }, { "input": "1 88\n1", "output": "88" }, { "input": "1 89\n1", "output": "89" } ]

1,531,147,438

2,147,483,647

Python 3

OK

TESTS

83

124

0

a=[int(q) for q in input().strip().split()] b=[int(q) for q in input().strip().split()] min=200 for k in range(a[0]): if a[1]%b[k]==0: if a[1]/b[k]<min: min=a[1]/b[k] print(int(min))

Title: Garden Time Limit: None seconds Memory Limit: None megabytes Problem Description: Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden. Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden. See the examples for better understanding. Input Specification: The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively. The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour. It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket. Output Specification: Print one integer number — the minimum number of hours required to water the garden. Demo Input: ['3 6\n2 3 5\n', '6 7\n1 2 3 4 5 6\n'] Demo Output: ['2\n', '7\n'] Note: In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden. In the second test we can choose only the bucket that allows us to water the segment of length 1.

```python a=[int(q) for q in input().strip().split()] b=[int(q) for q in input().strip().split()] min=200 for k in range(a[0]): if a[1]%b[k]==0: if a[1]/b[k]<min: min=a[1]/b[k] print(int(min)) ```

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" } ]

1,533,991,624

2,147,483,647

Python 3

OK

TESTS

81

248

0

n = int(input()) ans = [0, 0, 0] for x in range(n): a = [int(y) for y in input().split()] ans[0] += a[0] ans[1] += a[1] ans[2] += a[2] if ans == [0, 0, 0]: print("YES") else: print("NO")

Title: Young Physicist Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). Output Specification: Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. Demo Input: ['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n'] Demo Output: ['NO', 'YES'] Note: none

```python n = int(input()) ans = [0, 0, 0] for x in range(n): a = [int(y) for y in input().split()] ans[0] += a[0] ans[1] += a[1] ans[2] += a[2] if ans == [0, 0, 0]: print("YES") else: print("NO") ```

3.938

714

A

Meeting of Old Friends

PROGRAMMING

1,100

[ "implementation", "math" ]

null

Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya! Sonya is an owl and she sleeps during the day and stay awake from minute *l*1 to minute *r*1 inclusive. Also, during the minute *k* she prinks and is unavailable for Filya. Filya works a lot and he plans to visit Sonya from minute *l*2 to minute *r*2 inclusive. Calculate the number of minutes they will be able to spend together.

The only line of the input contains integers *l*1, *r*1, *l*2, *r*2 and *k* (1<=≤<=*l*1,<=*r*1,<=*l*2,<=*r*2,<=*k*<=≤<=1018, *l*1<=≤<=*r*1, *l*2<=≤<=*r*2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.

Print one integer — the number of minutes Sonya and Filya will be able to spend together.

[ "1 10 9 20 1\n", "1 100 50 200 75\n" ]

[ "2\n", "50\n" ]

In the first sample, they will be together during minutes 9 and 10. In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.

500

[ { "input": "1 10 9 20 1", "output": "2" }, { "input": "1 100 50 200 75", "output": "50" }, { "input": "6 6 5 8 9", "output": "1" }, { "input": "1 1000000000 1 1000000000 1", "output": "999999999" }, { "input": "5 100 8 8 8", "output": "0" }, { "input": "1 1000000000000000000 2 99999999999999999 1000000000", "output": "99999999999999997" }, { "input": "1 1 1 1 1", "output": "0" }, { "input": "1 2 3 4 5", "output": "0" }, { "input": "1 1000000000 2 999999999 3141592", "output": "999999997" }, { "input": "24648817341102 41165114064236 88046848035 13602161452932 10000831349205", "output": "0" }, { "input": "1080184299348 34666828555290 6878390132365 39891656267344 15395310291636", "output": "27788438422925" }, { "input": "11814 27385 22309 28354 23595", "output": "5076" }, { "input": "4722316546398 36672578279675 796716437180 33840047334985 13411035401708", "output": "29117730788587" }, { "input": "14300093617438 14381698008501 6957847034861 32510754974307 66056597033082", "output": "81604391064" }, { "input": "700062402405871919 762322967106512617 297732773882447821 747309903322652819 805776739998108178", "output": "47247500916780901" }, { "input": "59861796371397621 194872039092923459 668110259718450585 841148673332698972 928360292123223779", "output": "0" }, { "input": "298248781360904821 346420922793050061 237084570581741798 726877079564549183 389611850470532358", "output": "48172141432145241" }, { "input": "420745791717606818 864206437350900994 764928840030524015 966634105370748487 793326512080703489", "output": "99277597320376979" }, { "input": "519325240668210886 776112702001665034 360568516809443669 875594219634943179 994594983925273138", "output": "256787461333454149" }, { "input": "170331212821058551 891149660635282032 125964175621755330 208256491683509799 526532153531983174", "output": "37925278862451249" }, { "input": "1 3 3 5 3", "output": "0" }, { "input": "1 5 8 10 9", "output": "0" }, { "input": "1 2 4 5 10", "output": "0" }, { "input": "1 2 2 3 5", "output": "1" }, { "input": "2 4 3 7 3", "output": "1" }, { "input": "1 2 9 10 1", "output": "0" }, { "input": "5 15 1 10 5", "output": "5" }, { "input": "1 4 9 20 25", "output": "0" }, { "input": "2 4 1 2 5", "output": "1" }, { "input": "10 1000 1 100 2", "output": "91" }, { "input": "1 3 3 8 10", "output": "1" }, { "input": "4 6 6 8 9", "output": "1" }, { "input": "2 3 1 4 3", "output": "1" }, { "input": "1 2 2 3 100", "output": "1" }, { "input": "1 2 100 120 2", "output": "0" }, { "input": "1 3 5 7 4", "output": "0" }, { "input": "1 3 5 7 5", "output": "0" }, { "input": "1 4 8 10 6", "output": "0" }, { "input": "1 2 5 6 100", "output": "0" }, { "input": "1 2 5 10 20", "output": "0" }, { "input": "1 2 5 6 7", "output": "0" }, { "input": "2 5 7 12 6", "output": "0" }, { "input": "10 20 50 100 80", "output": "0" }, { "input": "1 2 5 10 2", "output": "0" }, { "input": "1 2 5 6 4", "output": "0" }, { "input": "5 9 1 2 3", "output": "0" }, { "input": "50 100 1 20 3", "output": "0" }, { "input": "10 20 3 7 30", "output": "0" }, { "input": "1 5 10 10 100", "output": "0" }, { "input": "100 101 1 2 3", "output": "0" }, { "input": "1 5 10 20 6", "output": "0" }, { "input": "1 10 15 25 5", "output": "0" }, { "input": "1 2 5 10 3", "output": "0" }, { "input": "2 3 5 6 100", "output": "0" }, { "input": "1 2 4 5 6", "output": "0" }, { "input": "6 10 1 2 40", "output": "0" }, { "input": "20 30 1 5 1", "output": "0" }, { "input": "20 40 50 100 50", "output": "0" }, { "input": "1 1 4 9 2", "output": "0" }, { "input": "1 2 5 6 1", "output": "0" }, { "input": "1 100 400 500 450", "output": "0" }, { "input": "5 6 1 2 5", "output": "0" }, { "input": "1 10 21 30 50", "output": "0" }, { "input": "100 200 300 400 101", "output": "0" }, { "input": "2 8 12 16 9", "output": "0" }, { "input": "1 5 7 9 6", "output": "0" }, { "input": "300 400 100 200 101", "output": "0" }, { "input": "1 2 2 3 10", "output": "1" }, { "input": "1 10 100 200 5", "output": "0" }, { "input": "1 3 3 4 4", "output": "1" }, { "input": "10 20 30 40 25", "output": "0" }, { "input": "1 2 5 10 1", "output": "0" }, { "input": "2 4 8 10 1", "output": "0" }, { "input": "2 5 10 15 7", "output": "0" }, { "input": "100 200 5 10 1", "output": "0" }, { "input": "1 2 100 200 300", "output": "0" }, { "input": "30 100 10 20 25", "output": "0" }, { "input": "10 20 1 5 6", "output": "0" }, { "input": "4 5 1 2 4", "output": "0" }, { "input": "11 100 1 9 1000", "output": "0" }, { "input": "1 1 10 10 228", "output": "0" }, { "input": "5 7 10 20 15", "output": "0" }, { "input": "1 3 8 9 7", "output": "0" }, { "input": "1 10 2 8 8", "output": "6" }, { "input": "1 5 9 15 1", "output": "0" }, { "input": "1 3 5 6 12", "output": "0" }, { "input": "1 100 500 1000 3", "output": "0" }, { "input": "1 1 1 1 2", "output": "1" }, { "input": "1 1000 100 1000 200", "output": "900" }, { "input": "4 5 1 4 1", "output": "1" }, { "input": "1 5 5 7 3", "output": "1" }, { "input": "1 4 4 10 11", "output": "1" }, { "input": "1 1 3 4 100", "output": "0" }, { "input": "1 4 3 5 6", "output": "2" }, { "input": "10 100 20 30 40", "output": "11" }, { "input": "5 9 1 11 7", "output": "4" } ]

1,475,151,221

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

3

46

0

a=[int(x) for x in input().strip().split()] b=[] if a[1]>a[2] and a[0]>a[2]: keep=1 elif a[1]>a[2]: keep=a[1]-a[2]+1 if (a[4]>a[2] and a[4]<a[1]): keep-=1 print(keep)

Title: Meeting of Old Friends Time Limit: None seconds Memory Limit: None megabytes Problem Description: Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya! Sonya is an owl and she sleeps during the day and stay awake from minute *l*1 to minute *r*1 inclusive. Also, during the minute *k* she prinks and is unavailable for Filya. Filya works a lot and he plans to visit Sonya from minute *l*2 to minute *r*2 inclusive. Calculate the number of minutes they will be able to spend together. Input Specification: The only line of the input contains integers *l*1, *r*1, *l*2, *r*2 and *k* (1<=≤<=*l*1,<=*r*1,<=*l*2,<=*r*2,<=*k*<=≤<=1018, *l*1<=≤<=*r*1, *l*2<=≤<=*r*2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks. Output Specification: Print one integer — the number of minutes Sonya and Filya will be able to spend together. Demo Input: ['1 10 9 20 1\n', '1 100 50 200 75\n'] Demo Output: ['2\n', '50\n'] Note: In the first sample, they will be together during minutes 9 and 10. In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.

```python a=[int(x) for x in input().strip().split()] b=[] if a[1]>a[2] and a[0]>a[2]: keep=1 elif a[1]>a[2]: keep=a[1]-a[2]+1 if (a[4]>a[2] and a[4]<a[1]): keep-=1 print(keep) ```

0

379

A

New Year Candles

PROGRAMMING

1,000

[ "implementation" ]

null

Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles. Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle. Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.

The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).

Print a single integer — the number of hours Vasily can light up the room for.

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

[ "7\n", "8\n" ]

Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.

500

[ { "input": "4 2", "output": "7" }, { "input": "6 3", "output": "8" }, { "input": "1000 1000", "output": "1001" }, { "input": "123 5", "output": "153" }, { "input": "1000 2", "output": "1999" }, { "input": "1 2", "output": "1" }, { "input": "1 3", "output": "1" }, { "input": "1 4", "output": "1" }, { "input": "2 2", "output": "3" }, { "input": "3 2", "output": "5" }, { "input": "3 3", "output": "4" }, { "input": "999 2", "output": "1997" }, { "input": "1000 3", "output": "1499" }, { "input": "1000 4", "output": "1333" }, { "input": "1 1000", "output": "1" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "10 4", "output": "13" }, { "input": "4 3", "output": "5" }, { "input": "91 5", "output": "113" }, { "input": "777 17", "output": "825" }, { "input": "100 4", "output": "133" }, { "input": "5 3", "output": "7" }, { "input": "6 4", "output": "7" }, { "input": "26 8", "output": "29" }, { "input": "9 4", "output": "11" }, { "input": "20 3", "output": "29" }, { "input": "17 3", "output": "25" } ]

1,679,066,062

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

3

62

0

''' import sys # Fast input input = sys.stdin.readline # Fast output def print(*args, **kwargs): sep = kwargs.get('sep', ' ') end = kwargs.get('end', '') file = kwargs.get('file', sys.stdout) flush = kwargs.get('flush', False) output = sep.join(map(str, args)) + end file.write(output) if flush: file.flush() #FACTORIAL def fact(n): if n==0: return 1 else: return n*fact(n-1) #HCF def hcf(a,b): if a%b==0: return a else: return hcf(b,a%b) #EUCLID APPLICATIONS def extended_euclid(a, b): if b == 0: return (1, 0, a) else: x, y, gcd = extended_euclid(b, a % b) return (y, x - (a // b) * y, gcd) def inverse_modulo(a, m): x, y, gcd = extended_euclid(a, m) if gcd == 1: return x % m else: return None def chinese_remainder_theorem(a1, m1, a2, m2): if a1 == a2: return a1 else: inv1 = inverse_modulo(m1, m2) inv2 = inverse_modulo(m2, m1) x = (a1*inv2*m2 + a2*inv1*m1) % (m1*m2) return x #PRIME IN O(root n) def isprime(n): bool=True for i in range(2,int(n**0.5)+1): if n%i==0: bool=False break return bool #BINARY EXPO def binary(n,m,p): ret=1 while m: if m%2==1: ret=(ret*n)%p n=(n*n)%p m=m//2 return ret #TOTIENT def totient(n): totient_value = n for i in range(2, int(n**0.5)+1): if n % i == 0: totient_value //= i totient_value *= i-1 while n % i == 0: n //= i if n > 1: totient_value //= n totient_value *= n-1 return totient_value #DAILY USE for _ in range(int(input())): n,m=map(int,input().split()) l=list(map(int,input().split())) n=int(input()) ''' for _ in range(1): n,m=map(int,input().split()) sum=0 while n>=m: rem=n%m sum+=n n=n//m n+=rem print(sum+n)

Title: New Year Candles Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles. Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle. Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number. Input Specification: The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000). Output Specification: Print a single integer — the number of hours Vasily can light up the room for. Demo Input: ['4 2\n', '6 3\n'] Demo Output: ['7\n', '8\n'] Note: Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.

```python ''' import sys # Fast input input = sys.stdin.readline # Fast output def print(*args, **kwargs): sep = kwargs.get('sep', ' ') end = kwargs.get('end', '') file = kwargs.get('file', sys.stdout) flush = kwargs.get('flush', False) output = sep.join(map(str, args)) + end file.write(output) if flush: file.flush() #FACTORIAL def fact(n): if n==0: return 1 else: return n*fact(n-1) #HCF def hcf(a,b): if a%b==0: return a else: return hcf(b,a%b) #EUCLID APPLICATIONS def extended_euclid(a, b): if b == 0: return (1, 0, a) else: x, y, gcd = extended_euclid(b, a % b) return (y, x - (a // b) * y, gcd) def inverse_modulo(a, m): x, y, gcd = extended_euclid(a, m) if gcd == 1: return x % m else: return None def chinese_remainder_theorem(a1, m1, a2, m2): if a1 == a2: return a1 else: inv1 = inverse_modulo(m1, m2) inv2 = inverse_modulo(m2, m1) x = (a1*inv2*m2 + a2*inv1*m1) % (m1*m2) return x #PRIME IN O(root n) def isprime(n): bool=True for i in range(2,int(n**0.5)+1): if n%i==0: bool=False break return bool #BINARY EXPO def binary(n,m,p): ret=1 while m: if m%2==1: ret=(ret*n)%p n=(n*n)%p m=m//2 return ret #TOTIENT def totient(n): totient_value = n for i in range(2, int(n**0.5)+1): if n % i == 0: totient_value //= i totient_value *= i-1 while n % i == 0: n //= i if n > 1: totient_value //= n totient_value *= n-1 return totient_value #DAILY USE for _ in range(int(input())): n,m=map(int,input().split()) l=list(map(int,input().split())) n=int(input()) ''' for _ in range(1): n,m=map(int,input().split()) sum=0 while n>=m: rem=n%m sum+=n n=n//m n+=rem print(sum+n) ```

0

455

A

Boredom

PROGRAMMING

1,500

[ "dp" ]

null

Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it. Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player. Alex is a perfectionist, so he decided to get as many points as possible. Help him.

The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence. The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105).

Print a single integer — the maximum number of points that Alex can earn.

[ "2\n1 2\n", "3\n1 2 3\n", "9\n1 2 1 3 2 2 2 2 3\n" ]

[ "2\n", "4\n", "10\n" ]

Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.

500

[ { "input": "2\n1 2", "output": "2" }, { "input": "3\n1 2 3", "output": "4" }, { "input": "9\n1 2 1 3 2 2 2 2 3", "output": "10" }, { "input": "5\n3 3 4 5 4", "output": "11" }, { "input": "5\n5 3 5 3 4", "output": "16" }, { "input": "5\n4 2 3 2 5", "output": "9" }, { "input": "10\n10 5 8 9 5 6 8 7 2 8", "output": "46" }, { "input": "10\n1 1 1 1 1 1 2 3 4 4", "output": "14" }, { "input": "100\n6 6 8 9 7 9 6 9 5 7 7 4 5 3 9 1 10 3 4 5 8 9 6 5 6 4 10 9 1 4 1 7 1 4 9 10 8 2 9 9 10 5 8 9 5 6 8 7 2 8 7 6 2 6 10 8 6 2 5 5 3 2 8 8 5 3 6 2 1 4 7 2 7 3 7 4 10 10 7 5 4 7 5 10 7 1 1 10 7 7 7 2 3 4 2 8 4 7 4 4", "output": "296" }, { "input": "100\n6 1 5 7 10 10 2 7 3 7 2 10 7 6 3 5 5 5 3 7 2 4 2 7 7 4 2 8 2 10 4 7 9 1 1 7 9 7 1 10 10 9 5 6 10 1 7 5 8 1 1 5 3 10 2 4 3 5 2 7 4 9 5 10 1 3 7 6 6 9 3 6 6 10 1 10 6 1 10 3 4 1 7 9 2 7 8 9 3 3 2 4 6 6 1 2 9 4 1 2", "output": "313" }, { "input": "100\n7 6 3 8 8 3 10 5 3 8 6 4 6 9 6 7 3 9 10 7 5 5 9 10 7 2 3 8 9 5 4 7 9 3 6 4 9 10 7 6 8 7 6 6 10 3 7 4 5 7 7 5 1 5 4 8 7 3 3 4 7 8 5 9 2 2 3 1 6 4 6 6 6 1 7 10 7 4 5 3 9 2 4 1 5 10 9 3 9 6 8 5 2 1 10 4 8 5 10 9", "output": "298" }, { "input": "100\n2 10 9 1 2 6 7 2 2 8 9 9 9 5 6 2 5 1 1 10 7 4 5 5 8 1 9 4 10 1 9 3 1 8 4 10 8 8 2 4 6 5 1 4 2 2 1 2 8 5 3 9 4 10 10 7 8 6 1 8 2 6 7 1 6 7 3 10 10 3 7 7 6 9 6 8 8 10 4 6 4 3 3 3 2 3 10 6 8 5 5 10 3 7 3 1 1 1 5 5", "output": "312" }, { "input": "100\n4 9 7 10 4 7 2 6 1 9 1 8 7 5 5 7 6 7 9 8 10 5 3 5 7 10 3 2 1 3 8 9 4 10 4 7 6 4 9 6 7 1 9 4 3 5 8 9 2 7 10 5 7 5 3 8 10 3 8 9 3 4 3 10 6 5 1 8 3 2 5 8 4 7 5 3 3 2 6 9 9 8 2 7 6 3 2 2 8 8 4 5 6 9 2 3 2 2 5 2", "output": "287" }, { "input": "100\n4 8 10 1 8 8 8 1 10 3 1 8 6 8 6 1 10 3 3 3 3 7 2 1 1 6 10 1 7 9 8 10 3 8 6 2 1 6 5 6 10 8 9 7 4 3 10 5 3 9 10 5 10 8 8 5 7 8 9 5 3 9 9 2 7 8 1 10 4 9 2 8 10 10 5 8 5 1 7 3 4 5 2 5 9 3 2 5 6 2 3 10 1 5 9 6 10 4 10 8", "output": "380" }, { "input": "100\n4 8 10 1 8 8 8 1 10 3 1 8 6 8 6 1 10 3 3 3 3 7 2 1 1 6 10 1 7 9 8 10 3 8 6 2 1 6 5 6 10 8 9 7 4 3 10 5 3 9 10 5 10 8 8 5 7 8 9 5 3 9 9 2 7 8 1 10 4 9 2 8 10 10 5 8 5 1 7 3 4 5 2 5 9 3 2 5 6 2 3 10 1 5 9 6 10 4 10 8", "output": "380" }, { "input": "100\n10 5 8 4 4 4 1 4 5 8 3 10 2 4 1 10 8 1 1 6 8 4 2 9 1 3 1 7 7 9 3 5 5 8 6 9 9 4 8 1 3 3 2 6 1 5 4 5 3 5 5 6 7 5 7 9 3 5 4 9 2 6 8 1 1 7 7 3 8 9 8 7 3 2 4 1 6 1 3 9 4 2 2 8 5 10 1 8 8 5 1 5 6 9 4 5 6 5 10 2", "output": "265" }, { "input": "100\n7 5 1 8 5 6 6 2 6 2 7 7 3 6 2 4 4 2 10 2 2 2 10 6 6 1 5 10 9 1 5 9 8 9 4 1 10 5 7 5 7 6 4 8 8 1 7 8 3 8 2 1 8 4 10 3 5 6 6 10 9 6 5 1 10 7 6 9 9 2 10 10 9 1 2 1 7 7 4 10 1 10 5 5 3 8 9 8 1 4 10 2 4 5 4 4 1 6 2 9", "output": "328" }, { "input": "100\n5 6 10 7 1 7 10 1 9 1 5 1 4 1 3 3 7 9 1 6 1 6 5 7 1 6 3 1 3 6 3 8 2 4 1 5 2 10 7 3 10 4 10 1 5 4 2 9 7 9 5 7 10 4 1 4 8 9 3 1 3 7 7 4 3 7 7 10 6 9 5 5 6 5 3 9 8 8 5 5 4 10 9 4 10 4 1 8 3 5 4 10 9 3 10 4 10 7 10 9", "output": "324" }, { "input": "10\n7 4 5 3 9 1 10 3 4 5", "output": "34" }, { "input": "10\n8 9 6 5 6 4 10 9 1 4", "output": "39" }, { "input": "10\n1 7 1 4 9 10 8 2 9 9", "output": "40" }, { "input": "1\n100000", "output": "100000" } ]

1,698,019,671

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

5

124

2,764,800

from typing import Counter n=int(input()) lst=list(map(int,input().split())) dp=[0]*(max(lst)+1) c=Counter(lst) for i in range(max(lst)+1): dp[i]=i*c[i] if i-2>=0: dp[i]+=dp[i-2] print(max(dp))

Title: Boredom Time Limit: None seconds Memory Limit: None megabytes Problem Description: Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it. Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player. Alex is a perfectionist, so he decided to get as many points as possible. Help him. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence. The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105). Output Specification: Print a single integer — the maximum number of points that Alex can earn. Demo Input: ['2\n1 2\n', '3\n1 2 3\n', '9\n1 2 1 3 2 2 2 2 3\n'] Demo Output: ['2\n', '4\n', '10\n'] Note: Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.

```python from typing import Counter n=int(input()) lst=list(map(int,input().split())) dp=[0]*(max(lst)+1) c=Counter(lst) for i in range(max(lst)+1): dp[i]=i*c[i] if i-2>=0: dp[i]+=dp[i-2] print(max(dp)) ```

0

208

A

Dubstep

PROGRAMMING

900

[ "strings" ]

null

Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them. Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club. For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX". Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.

The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.

Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.

[ "WUBWUBABCWUB\n", "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n" ]

[ "ABC ", "WE ARE THE CHAMPIONS MY FRIEND " ]

In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya. In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".

500

[ { "input": "WUBWUBABCWUB", "output": "ABC " }, { "input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB", "output": "WE ARE THE CHAMPIONS MY FRIEND " }, { "input": "WUBWUBWUBSR", "output": "SR " }, { "input": "RWUBWUBWUBLWUB", "output": "R L " }, { "input": "ZJWUBWUBWUBJWUBWUBWUBL", "output": "ZJ J L " }, { "input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB", "output": "C B E Q " }, { "input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB", "output": "JKD WBIRAQKF YE WV " }, { "input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB", "output": "KSDHEMIXUJ R S H " }, { "input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB", "output": "OG X I KO " }, { "input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH", "output": "Q QQ I WW JOPJPBRH " }, { "input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB", "output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C " }, { "input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV", "output": "E IQMJNIQ GZZBQZAUHYP PMR DCV " }, { "input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB", "output": "FV BPS RXNETCJ JDMBH B V B " }, { "input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL", "output": "FBQ IDFSY CTWDM SXO QI L " }, { "input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL", "output": "I QLHD YIIKZDFQ CX U K NL " }, { "input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE", "output": "K UPDYXGOKU AGOAH IZD IY V P E " }, { "input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB", "output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ " }, { "input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB", "output": "PAMJGY XGPQM TKGSXUY E N H E " }, { "input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB", "output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB " }, { "input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM", "output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M " }, { "input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW", "output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W " }, { "input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG", "output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G " }, { "input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN", "output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N " }, { "input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG", "output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG " }, { "input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB", "output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L " }, { "input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB", "output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U " }, { "input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB", "output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ " }, { "input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB", "output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J " }, { "input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO", "output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O " }, { "input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR", "output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R " }, { "input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB", "output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM " }, { "input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB", "output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H " }, { "input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB", "output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S " }, { "input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB", "output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J " }, { "input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF", "output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F " }, { "input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY", "output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y " }, { "input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB", "output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO " }, { "input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW", "output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W " }, { "input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD", "output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D " }, { "input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB", "output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A " }, { "input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB", "output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y " }, { "input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB", "output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R " }, { "input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY", "output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y " }, { "input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE", "output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E " }, { "input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS", "output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S " }, { "input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP", "output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP " }, { "input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB", "output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y " }, { "input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL", "output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL " }, { "input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO", "output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O " }, { "input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU", "output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU " }, { "input": "A", "output": "A " }, { "input": "WUBA", "output": "A " }, { "input": "WUBWUBA", "output": "A " }, { "input": "AWUBWUBWUB", "output": "A " }, { "input": "AWUBBWUBCWUBD", "output": "A B C D " }, { "input": "WUBWWUBWUBWUBUWUBWUBBWUB", "output": "W U B " }, { "input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA " }, { "input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA", "output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA " }, { "input": "WUWUBBWWUBUB", "output": "WU BW UB " }, { "input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB", "output": "WUAB " }, { "input": "ZWUB", "output": "Z " }, { "input": "WU", "output": "WU " }, { "input": "UB", "output": "UB " }, { "input": "U", "output": "U " }, { "input": "WUBW", "output": "W " }, { "input": "WUBWU", "output": "WU " }, { "input": "WUWUB", "output": "WU " }, { "input": "UBWUB", "output": "UB " }, { "input": "WUWUBUBWUBUWUB", "output": "WU UB U " }, { "input": "WUBWWUBAWUB", "output": "W A " }, { "input": "WUUUUU", "output": "WUUUUU " } ]

1,676,064,133

2,147,483,647

Python 3

OK

TESTS

71

92

0

def restore_song(dubstep_remix): original_song = dubstep_remix.replace("WUB", " ") original_song = original_song.strip() return original_song dubstep_remix = input() print(restore_song(dubstep_remix))

Title: Dubstep Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them. Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club. For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX". Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song. Input Specification: The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word. Output Specification: Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space. Demo Input: ['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n'] Demo Output: ['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND '] Note: In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya. In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".

```python def restore_song(dubstep_remix): original_song = dubstep_remix.replace("WUB", " ") original_song = original_song.strip() return original_song dubstep_remix = input() print(restore_song(dubstep_remix)) ```

3

712

A

Memory and Crow

PROGRAMMING

800

[ "implementation", "math" ]

null

There are *n* integers *b*1,<=*b*2,<=...,<=*b**n* written in a row. For all *i* from 1 to *n*, values *a**i* are defined by the crows performing the following procedure: - The crow sets *a**i* initially 0. - The crow then adds *b**i* to *a**i*, subtracts *b**i*<=+<=1, adds the *b**i*<=+<=2 number, and so on until the *n*'th number. Thus, *a**i*<==<=*b**i*<=-<=*b**i*<=+<=1<=+<=*b**i*<=+<=2<=-<=*b**i*<=+<=3.... Memory gives you the values *a*1,<=*a*2,<=...,<=*a**n*, and he now wants you to find the initial numbers *b*1,<=*b*2,<=...,<=*b**n* written in the row? Can you do it?

The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of integers written in the row. The next line contains *n*, the *i*'th of which is *a**i* (<=-<=109<=≤<=*a**i*<=≤<=109) — the value of the *i*'th number.

Print *n* integers corresponding to the sequence *b*1,<=*b*2,<=...,<=*b**n*. It's guaranteed that the answer is unique and fits in 32-bit integer type.

[ "5\n6 -4 8 -2 3\n", "5\n3 -2 -1 5 6\n" ]

[ "2 4 6 1 3 \n", "1 -3 4 11 6 \n" ]

In the first sample test, the crows report the numbers 6, - 4, 8, - 2, and 3 when he starts at indices 1, 2, 3, 4 and 5 respectively. It is easy to check that the sequence 2 4 6 1 3 satisfies the reports. For example, 6 = 2 - 4 + 6 - 1 + 3, and - 4 = 4 - 6 + 1 - 3. In the second sample test, the sequence 1, - 3, 4, 11, 6 satisfies the reports. For example, 5 = 11 - 6 and 6 = 6.

500

[ { "input": "5\n6 -4 8 -2 3", "output": "2 4 6 1 3 " }, { "input": "5\n3 -2 -1 5 6", "output": "1 -3 4 11 6 " }, { "input": "10\n13 -2 532 -63 -23 -63 -64 -23 12 10", "output": "11 530 469 -86 -86 -127 -87 -11 22 10 " }, { "input": "10\n0 0 0 0 0 0 0 0 0 0", "output": "0 0 0 0 0 0 0 0 0 0 " }, { "input": "10\n1 -1 1 -1 1 -1 1 -1 1 -1", "output": "0 0 0 0 0 0 0 0 0 -1 " }, { "input": "10\n-1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000", "output": "0 0 0 0 0 0 0 0 0 1000000000 " }, { "input": "10\n124 532 -642 6482 -124 952 -235 7594 34 -12", "output": "656 -110 5840 6358 828 717 7359 7628 22 -12 " }, { "input": "10\n1294 35852 32749 537295 12048 53729 29357 58320 64739 1240", "output": "37146 68601 570044 549343 65777 83086 87677 123059 65979 1240 " } ]

1,567,518,461

1,361

Python 3

OK

TESTS

49

327

8,499,200

n = int(input()) a = [int(x) for x in input().split()] b = [0]*(n) b[n-1] = a[n-1] for i in range(n-2,-1,-1): b[i] = a[i] + a[i+1] print(*b)

Title: Memory and Crow Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* integers *b*1,<=*b*2,<=...,<=*b**n* written in a row. For all *i* from 1 to *n*, values *a**i* are defined by the crows performing the following procedure: - The crow sets *a**i* initially 0. - The crow then adds *b**i* to *a**i*, subtracts *b**i*<=+<=1, adds the *b**i*<=+<=2 number, and so on until the *n*'th number. Thus, *a**i*<==<=*b**i*<=-<=*b**i*<=+<=1<=+<=*b**i*<=+<=2<=-<=*b**i*<=+<=3.... Memory gives you the values *a*1,<=*a*2,<=...,<=*a**n*, and he now wants you to find the initial numbers *b*1,<=*b*2,<=...,<=*b**n* written in the row? Can you do it? Input Specification: The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of integers written in the row. The next line contains *n*, the *i*'th of which is *a**i* (<=-<=109<=≤<=*a**i*<=≤<=109) — the value of the *i*'th number. Output Specification: Print *n* integers corresponding to the sequence *b*1,<=*b*2,<=...,<=*b**n*. It's guaranteed that the answer is unique and fits in 32-bit integer type. Demo Input: ['5\n6 -4 8 -2 3\n', '5\n3 -2 -1 5 6\n'] Demo Output: ['2 4 6 1 3 \n', '1 -3 4 11 6 \n'] Note: In the first sample test, the crows report the numbers 6, - 4, 8, - 2, and 3 when he starts at indices 1, 2, 3, 4 and 5 respectively. It is easy to check that the sequence 2 4 6 1 3 satisfies the reports. For example, 6 = 2 - 4 + 6 - 1 + 3, and - 4 = 4 - 6 + 1 - 3. In the second sample test, the sequence 1, - 3, 4, 11, 6 satisfies the reports. For example, 5 = 11 - 6 and 6 = 6.

```python n = int(input()) a = [int(x) for x in input().split()] b = [0]*(n) b[n-1] = a[n-1] for i in range(n-2,-1,-1): b[i] = a[i] + a[i+1] print(*b) ```

3

978

A

Remove Duplicates

PROGRAMMING

800

[ "implementation" ]

null

Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements. Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.

The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array. The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array.

In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates. In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left.

[ "6\n1 5 5 1 6 1\n", "5\n2 4 2 4 4\n", "5\n6 6 6 6 6\n" ]

[ "3\n5 6 1 \n", "2\n2 4 \n", "1\n6 \n" ]

In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$. In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$. In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$.

0

[ { "input": "6\n1 5 5 1 6 1", "output": "3\n5 6 1 " }, { "input": "5\n2 4 2 4 4", "output": "2\n2 4 " }, { "input": "5\n6 6 6 6 6", "output": "1\n6 " }, { "input": "7\n1 2 3 4 2 2 3", "output": "4\n1 4 2 3 " }, { "input": "9\n100 100 100 99 99 99 100 100 100", "output": "2\n99 100 " }, { "input": "27\n489 489 487 488 750 230 43 645 42 42 489 42 973 42 973 750 645 355 868 112 868 489 750 489 887 489 868", "output": "13\n487 488 230 43 42 973 645 355 112 750 887 489 868 " }, { "input": "40\n151 421 421 909 117 222 909 954 227 421 227 954 954 222 421 227 421 421 421 151 421 227 222 222 222 222 421 183 421 227 421 954 222 421 954 421 222 421 909 421", "output": "8\n117 151 183 227 954 222 909 421 " }, { "input": "48\n2 2 2 903 903 2 726 2 2 2 2 2 2 2 2 2 2 726 2 2 2 2 2 2 2 726 2 2 2 2 62 2 2 2 2 2 2 2 2 726 62 726 2 2 2 903 903 2", "output": "4\n62 726 903 2 " }, { "input": "1\n1", "output": "1\n1 " }, { "input": "13\n5 37 375 5 37 33 37 375 37 2 3 3 2", "output": "6\n5 33 375 37 3 2 " }, { "input": "50\n1 2 3 4 5 4 3 2 1 2 3 2 1 4 5 5 4 3 2 1 1 2 3 4 5 4 3 2 1 2 3 2 1 4 5 5 4 3 2 1 4 3 2 5 1 6 6 6 6 6", "output": "6\n4 3 2 5 1 6 " }, { "input": "47\n233 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "2\n233 1 " }, { "input": "47\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", "output": "1\n1 " }, { "input": "2\n964 964", "output": "1\n964 " }, { "input": "2\n1000 1000", "output": "1\n1000 " }, { "input": "1\n1000", "output": "1\n1000 " }, { "input": "45\n991 991 996 996 992 992 999 1000 998 1000 992 999 996 999 991 991 999 993 992 999 1000 997 992 999 996 991 994 996 991 999 1000 993 999 997 999 992 991 997 991 998 998 995 998 994 993", "output": "10\n996 1000 999 992 997 991 995 998 994 993 " }, { "input": "6\n994 993 1000 998 991 994", "output": "5\n993 1000 998 991 994 " }, { "input": "48\n992 995 992 991 994 992 995 999 996 993 999 995 993 992 1000 992 997 996 991 993 992 998 998 998 999 995 992 992 993 992 992 995 996 995 997 991 997 991 999 994 994 997 1000 998 1000 992 1000 999", "output": "10\n993 996 995 991 994 997 998 992 1000 999 " }, { "input": "3\n6 6 3", "output": "2\n6 3 " }, { "input": "36\n999 1000 993 993 1000 999 996 997 998 995 995 997 999 995 1000 998 998 994 993 998 994 999 1000 995 996 994 991 991 999 996 993 999 996 998 991 997", "output": "9\n1000 995 994 993 999 996 998 991 997 " }, { "input": "49\n991 997 995 991 995 998 994 996 997 991 998 993 992 999 996 995 991 991 999 997 995 993 997 1000 997 993 993 994 999 994 992 991 1000 996 994 996 996 999 992 992 1000 991 997 993 991 994 997 991 996", "output": "10\n998 995 999 992 1000 993 994 997 991 996 " }, { "input": "2\n1000 999", "output": "2\n1000 999 " }, { "input": "28\n997 994 991 994 994 995 1000 992 995 994 994 995 991 996 991 996 991 999 999 993 994 997 995 992 991 992 998 1000", "output": "10\n996 999 993 994 997 995 991 992 998 1000 " }, { "input": "15\n991 995 995 1000 991 993 997 998 1000 994 1000 992 991 995 992", "output": "8\n993 997 998 994 1000 991 995 992 " }, { "input": "6\n1000 1 1000 2 1000 1", "output": "3\n2 1000 1 " }, { "input": "6\n1000 1 1000 2 1 1000", "output": "3\n2 1 1000 " }, { "input": "7\n1 1000 1 999 1000 1 2", "output": "4\n999 1000 1 2 " } ]

1,691,779,824

2,147,483,647

Python 3

OK

TESTS

28

46

0

n = int(input()) a = list(map(int, input().split())) ans = [] for i in range(n): if(a[i] not in a[i+1:]): ans.append(a[i]) print(len(ans)) print(*ans)

Title: Remove Duplicates Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements. Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed. Input Specification: The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array. The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array. Output Specification: In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates. In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left. Demo Input: ['6\n1 5 5 1 6 1\n', '5\n2 4 2 4 4\n', '5\n6 6 6 6 6\n'] Demo Output: ['3\n5 6 1 \n', '2\n2 4 \n', '1\n6 \n'] Note: In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$. In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$. In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$.

```python n = int(input()) a = list(map(int, input().split())) ans = [] for i in range(n): if(a[i] not in a[i+1:]): ans.append(a[i]) print(len(ans)) print(*ans) ```

3

918

B

Radio Station

PROGRAMMING

900

[ "implementation", "strings" ]

null

As the guys fried the radio station facilities, the school principal gave them tasks as a punishment. Dustin's task was to add comments to nginx configuration for school's website. The school has *n* servers. Each server has a name and an ip (names aren't necessarily unique, but ips are). Dustin knows the ip and name of each server. For simplicity, we'll assume that an nginx command is of form "command ip;" where command is a string consisting of English lowercase letter only, and ip is the ip of one of school servers. Each ip is of form "a.b.c.d" where *a*, *b*, *c* and *d* are non-negative integers less than or equal to 255 (with no leading zeros). The nginx configuration file Dustin has to add comments to has *m* commands. Nobody ever memorizes the ips of servers, so to understand the configuration better, Dustin has to comment the name of server that the ip belongs to at the end of each line (after each command). More formally, if a line is "command ip;" Dustin has to replace it with "command ip; #name" where name is the name of the server with ip equal to ip. Dustin doesn't know anything about nginx, so he panicked again and his friends asked you to do his task for him.

The first line of input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000). The next *n* lines contain the names and ips of the servers. Each line contains a string name, name of the server and a string ip, ip of the server, separated by space (1<=≤<=|*name*|<=≤<=10, *name* only consists of English lowercase letters). It is guaranteed that all ip are distinct. The next *m* lines contain the commands in the configuration file. Each line is of form "command ip;" (1<=≤<=|*command*|<=≤<=10, command only consists of English lowercase letters). It is guaranteed that ip belongs to one of the *n* school servers.

Print *m* lines, the commands in the configuration file after Dustin did his task.

[ "2 2\nmain 192.168.0.2\nreplica 192.168.0.1\nblock 192.168.0.1;\nproxy 192.168.0.2;\n", "3 5\ngoogle 8.8.8.8\ncodeforces 212.193.33.27\nserver 138.197.64.57\nredirect 138.197.64.57;\nblock 8.8.8.8;\ncf 212.193.33.27;\nunblock 8.8.8.8;\ncheck 138.197.64.57;\n" ]

[ "block 192.168.0.1; #replica\nproxy 192.168.0.2; #main\n", "redirect 138.197.64.57; #server\nblock 8.8.8.8; #google\ncf 212.193.33.27; #codeforces\nunblock 8.8.8.8; #google\ncheck 138.197.64.57; #server\n" ]

none

1,000

[ { "input": "2 2\nmain 192.168.0.2\nreplica 192.168.0.1\nblock 192.168.0.1;\nproxy 192.168.0.2;", "output": "block 192.168.0.1; #replica\nproxy 192.168.0.2; #main" }, { "input": "3 5\ngoogle 8.8.8.8\ncodeforces 212.193.33.27\nserver 138.197.64.57\nredirect 138.197.64.57;\nblock 8.8.8.8;\ncf 212.193.33.27;\nunblock 8.8.8.8;\ncheck 138.197.64.57;", "output": "redirect 138.197.64.57; #server\nblock 8.8.8.8; #google\ncf 212.193.33.27; #codeforces\nunblock 8.8.8.8; #google\ncheck 138.197.64.57; #server" }, { "input": "10 10\nittmcs 112.147.123.173\njkt 228.40.73.178\nfwckqtz 88.28.31.198\nkal 224.226.34.213\nnacuyokm 49.57.13.44\nfouynv 243.18.250.17\ns 45.248.83.247\ne 75.69.23.169\nauwoqlch 100.44.219.187\nlkldjq 46.123.169.140\ngjcylatwzi 46.123.169.140;\ndxfi 88.28.31.198;\ngv 46.123.169.140;\nety 88.28.31.198;\notbmgcrn 46.123.169.140;\nw 112.147.123.173;\np 75.69.23.169;\nvdsnigk 46.123.169.140;\nmmc 46.123.169.140;\ngtc 49.57.13.44;", "output": "gjcylatwzi 46.123.169.140; #lkldjq\ndxfi 88.28.31.198; #fwckqtz\ngv 46.123.169.140; #lkldjq\nety 88.28.31.198; #fwckqtz\notbmgcrn 46.123.169.140; #lkldjq\nw 112.147.123.173; #ittmcs\np 75.69.23.169; #e\nvdsnigk 46.123.169.140; #lkldjq\nmmc 46.123.169.140; #lkldjq\ngtc 49.57.13.44; #nacuyokm" }, { "input": "1 1\nervbfot 185.32.99.2\nzygoumbmx 185.32.99.2;", "output": "zygoumbmx 185.32.99.2; #ervbfot" }, { "input": "1 2\ny 245.182.246.189\nlllq 245.182.246.189;\nxds 245.182.246.189;", "output": "lllq 245.182.246.189; #y\nxds 245.182.246.189; #y" }, { "input": "2 1\ntdwmshz 203.115.124.110\neksckjya 201.80.191.212\nzbtjzzue 203.115.124.110;", "output": "zbtjzzue 203.115.124.110; #tdwmshz" }, { "input": "8 5\nfhgkq 5.19.189.178\nphftablcr 75.18.177.178\nxnpcg 158.231.167.176\ncfahrkq 26.165.124.191\nfkgtnqtfoh 230.13.13.129\nt 101.24.94.85\nvjoirslx 59.6.179.72\ntwktmskb 38.194.117.184\nrvzzlygosc 26.165.124.191;\ndcsgxrkgv 101.24.94.85;\nyvmyppn 59.6.179.72;\ngpdjjuq 75.18.177.178;\nvdviz 101.24.94.85;", "output": "rvzzlygosc 26.165.124.191; #cfahrkq\ndcsgxrkgv 101.24.94.85; #t\nyvmyppn 59.6.179.72; #vjoirslx\ngpdjjuq 75.18.177.178; #phftablcr\nvdviz 101.24.94.85; #t" }, { "input": "2 1\nmain 1.1.1.1\nget 11.1.1.1\ncommand 11.1.1.1;", "output": "command 11.1.1.1; #get" }, { "input": "2 2\nmain 0.0.0.255\nmainn 0.0.2.55\nblock 0.0.0.255;\nblock 0.0.2.55;", "output": "block 0.0.0.255; #main\nblock 0.0.2.55; #mainn" }, { "input": "2 2\nmain 0.0.63.0\nreplica 0.38.0.56\nblock 0.0.63.0;\nproxy 0.38.0.56;", "output": "block 0.0.63.0; #main\nproxy 0.38.0.56; #replica" }, { "input": "2 2\nmain 192.168.10.12\nreplica 192.167.10.12\nblock 192.168.10.12;\nproxy 192.167.10.12;", "output": "block 192.168.10.12; #main\nproxy 192.167.10.12; #replica" }, { "input": "2 1\nneserver 185.218.47.91\nserver 255.255.255.255\nblock 255.255.255.255;", "output": "block 255.255.255.255; #server" }, { "input": "2 1\nalp 22.222.30.10\nbet 222.22.30.10\nblock 22.222.30.10;", "output": "block 22.222.30.10; #alp" } ]

1,699,352,995

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

30

0

n, m = map(int, input().split()) ip_to_name = {} for i in range(n): name, ip = input().split() ip_to_name[ip] = name # Process each command and add comments for i in range(m): command, ip = input().split() name = ip_to_name[ip[:-1]] # Remove the trailing semicolon new_command = f"{command} {ip}; #{name}" print(new_command)

Title: Radio Station Time Limit: None seconds Memory Limit: None megabytes Problem Description: As the guys fried the radio station facilities, the school principal gave them tasks as a punishment. Dustin's task was to add comments to nginx configuration for school's website. The school has *n* servers. Each server has a name and an ip (names aren't necessarily unique, but ips are). Dustin knows the ip and name of each server. For simplicity, we'll assume that an nginx command is of form "command ip;" where command is a string consisting of English lowercase letter only, and ip is the ip of one of school servers. Each ip is of form "a.b.c.d" where *a*, *b*, *c* and *d* are non-negative integers less than or equal to 255 (with no leading zeros). The nginx configuration file Dustin has to add comments to has *m* commands. Nobody ever memorizes the ips of servers, so to understand the configuration better, Dustin has to comment the name of server that the ip belongs to at the end of each line (after each command). More formally, if a line is "command ip;" Dustin has to replace it with "command ip; #name" where name is the name of the server with ip equal to ip. Dustin doesn't know anything about nginx, so he panicked again and his friends asked you to do his task for him. Input Specification: The first line of input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000). The next *n* lines contain the names and ips of the servers. Each line contains a string name, name of the server and a string ip, ip of the server, separated by space (1<=≤<=|*name*|<=≤<=10, *name* only consists of English lowercase letters). It is guaranteed that all ip are distinct. The next *m* lines contain the commands in the configuration file. Each line is of form "command ip;" (1<=≤<=|*command*|<=≤<=10, command only consists of English lowercase letters). It is guaranteed that ip belongs to one of the *n* school servers. Output Specification: Print *m* lines, the commands in the configuration file after Dustin did his task. Demo Input: ['2 2\nmain 192.168.0.2\nreplica 192.168.0.1\nblock 192.168.0.1;\nproxy 192.168.0.2;\n', '3 5\ngoogle 8.8.8.8\ncodeforces 212.193.33.27\nserver 138.197.64.57\nredirect 138.197.64.57;\nblock 8.8.8.8;\ncf 212.193.33.27;\nunblock 8.8.8.8;\ncheck 138.197.64.57;\n'] Demo Output: ['block 192.168.0.1; #replica\nproxy 192.168.0.2; #main\n', 'redirect 138.197.64.57; #server\nblock 8.8.8.8; #google\ncf 212.193.33.27; #codeforces\nunblock 8.8.8.8; #google\ncheck 138.197.64.57; #server\n'] Note: none

```python n, m = map(int, input().split()) ip_to_name = {} for i in range(n): name, ip = input().split() ip_to_name[ip] = name # Process each command and add comments for i in range(m): command, ip = input().split() name = ip_to_name[ip[:-1]] # Remove the trailing semicolon new_command = f"{command} {ip}; #{name}" print(new_command) ```

0

771

A

Bear and Friendship Condition

PROGRAMMING

1,500

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

null

Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures). There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves. Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z. For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well. Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.

The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends. The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input.

If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).

[ "4 3\n1 3\n3 4\n1 4\n", "4 4\n3 1\n2 3\n3 4\n1 2\n", "10 4\n4 3\n5 10\n8 9\n1 2\n", "3 2\n1 2\n2 3\n" ]

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

The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not.

250

[ { "input": "4 3\n1 3\n3 4\n1 4", "output": "YES" }, { "input": "4 4\n3 1\n2 3\n3 4\n1 2", "output": "NO" }, { "input": "10 4\n4 3\n5 10\n8 9\n1 2", "output": "YES" }, { "input": "3 2\n1 2\n2 3", "output": "NO" }, { "input": "3 0", "output": "YES" }, { "input": "15 42\n8 1\n3 14\n7 14\n12 3\n7 9\n6 7\n6 12\n14 12\n3 10\n10 14\n6 3\n3 13\n13 10\n7 12\n7 2\n6 10\n11 4\n9 3\n8 4\n7 3\n2 3\n2 10\n9 13\n2 14\n6 14\n13 2\n1 4\n13 6\n7 10\n13 14\n12 10\n13 7\n12 2\n9 10\n13 12\n2 6\n9 14\n6 9\n12 9\n11 1\n2 9\n11 8", "output": "YES" }, { "input": "20 80\n17 4\n10 1\n11 10\n17 7\n15 10\n14 15\n13 1\n18 13\n3 13\n12 7\n9 13\n10 12\n14 12\n18 11\n4 7\n10 13\n11 3\n19 8\n14 7\n10 17\n14 3\n7 11\n11 14\n19 5\n10 14\n15 17\n3 1\n9 10\n11 1\n4 1\n11 4\n9 1\n12 3\n13 7\n1 14\n11 12\n7 1\n9 12\n18 15\n17 3\n7 15\n4 10\n7 18\n7 9\n12 17\n14 18\n3 18\n18 17\n9 15\n14 4\n14 9\n9 18\n12 4\n7 10\n15 4\n4 18\n15 13\n1 12\n7 3\n13 11\n4 13\n5 8\n12 18\n12 15\n17 9\n11 15\n3 10\n18 10\n4 3\n15 3\n13 12\n9 4\n9 11\n14 17\n13 17\n3 9\n13 14\n1 17\n15 1\n17 11", "output": "NO" }, { "input": "99 26\n64 17\n48 70\n71 50\n3 50\n9 60\n61 64\n53 50\n25 12\n3 71\n71 53\n3 53\n65 70\n9 25\n9 12\n59 56\n39 60\n64 69\n65 94\n70 94\n25 60\n60 12\n94 48\n17 69\n61 17\n65 48\n61 69", "output": "NO" }, { "input": "3 1\n1 2", "output": "YES" }, { "input": "3 2\n3 2\n1 3", "output": "NO" }, { "input": "3 3\n2 3\n1 2\n1 3", "output": "YES" }, { "input": "4 2\n4 1\n2 1", "output": "NO" }, { "input": "4 3\n3 1\n2 1\n3 2", "output": "YES" }, { "input": "5 9\n1 2\n5 1\n3 1\n1 4\n2 4\n5 3\n5 4\n2 3\n5 2", "output": "NO" }, { "input": "10 5\n9 5\n1 2\n6 8\n6 3\n10 6", "output": "NO" }, { "input": "10 8\n10 7\n9 7\n5 7\n6 8\n3 5\n8 10\n3 4\n7 8", "output": "NO" }, { "input": "10 20\n8 2\n8 3\n1 8\n9 5\n2 4\n10 1\n10 5\n7 5\n7 8\n10 7\n6 5\n3 7\n1 9\n9 8\n7 2\n2 10\n2 1\n6 4\n9 7\n4 3", "output": "NO" }, { "input": "150000 10\n62562 50190\n48849 60549\n139470 18456\n21436 25159\n66845 120884\n99972 114453\n11631 99153\n62951 134848\n78114 146050\n136760 131762", "output": "YES" }, { "input": "150000 0", "output": "YES" }, { "input": "4 4\n1 2\n2 3\n3 4\n1 4", "output": "NO" }, { "input": "30 73\n25 2\n2 16\n20 12\n16 20\n7 18\n11 15\n13 11\n30 29\n16 12\n12 25\n2 1\n18 14\n9 8\n28 16\n2 9\n22 21\n1 25\n12 28\n14 7\n4 9\n26 7\n14 27\n12 2\n29 22\n1 9\n13 15\n3 10\n1 12\n8 20\n30 24\n25 20\n4 1\n4 12\n20 1\n8 4\n2 28\n25 16\n16 8\n20 4\n9 12\n21 30\n23 11\n19 6\n28 4\n29 21\n9 28\n30 10\n22 24\n25 8\n27 26\n25 4\n28 20\n9 25\n24 29\n20 9\n18 26\n1 28\n30 22\n23 15\n28 27\n8 2\n23 13\n12 8\n14 26\n16 4\n28 25\n8 1\n4 2\n9 16\n20 2\n18 27\n28 8\n27 7", "output": "NO" }, { "input": "5 4\n1 2\n2 5\n3 4\n4 5", "output": "NO" }, { "input": "4 4\n1 2\n2 3\n3 4\n4 1", "output": "NO" }, { "input": "6 6\n1 2\n2 4\n4 3\n1 5\n5 6\n6 3", "output": "NO" }, { "input": "3 2\n1 2\n1 3", "output": "NO" }, { "input": "6 6\n1 2\n2 3\n3 4\n4 5\n5 6\n1 6", "output": "NO" }, { "input": "4 4\n1 2\n1 3\n2 4\n3 4", "output": "NO" }, { "input": "6 9\n1 4\n1 5\n1 6\n2 4\n2 5\n2 6\n3 4\n3 5\n3 6", "output": "NO" }, { "input": "4 3\n1 2\n1 3\n3 4", "output": "NO" }, { "input": "4 3\n1 2\n1 3\n2 4", "output": "NO" }, { "input": "6 6\n1 2\n2 3\n3 4\n4 5\n5 6\n6 1", "output": "NO" }, { "input": "4 3\n1 2\n1 3\n1 4", "output": "NO" }, { "input": "150000 3\n150000 149999\n149998 149999\n149998 150000", "output": "YES" }, { "input": "10 15\n1 2\n1 3\n2 4\n2 5\n3 4\n3 5\n4 5\n1 6\n6 7\n6 8\n7 9\n7 10\n8 9\n8 10\n9 10", "output": "NO" }, { "input": "5 5\n1 2\n2 3\n3 4\n4 5\n5 1", "output": "NO" }, { "input": "5 4\n1 2\n1 3\n1 4\n1 5", "output": "NO" }, { "input": "9 9\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 1", "output": "NO" }, { "input": "4 5\n1 2\n1 3\n1 4\n2 3\n3 4", "output": "NO" }, { "input": "4 4\n1 2\n1 3\n4 2\n4 3", "output": "NO" }, { "input": "5 5\n1 4\n5 3\n5 1\n2 3\n4 5", "output": "NO" }, { "input": "4 5\n1 2\n1 3\n2 3\n1 4\n2 4", "output": "NO" }, { "input": "4 5\n1 2\n1 3\n1 4\n2 3\n2 4", "output": "NO" }, { "input": "6 12\n1 2\n1 4\n1 5\n1 6\n2 3\n2 5\n2 6\n3 4\n3 5\n3 6\n4 5\n4 6", "output": "NO" }, { "input": "1000 4\n100 132\n100 164\n196 132\n196 164", "output": "NO" } ]

1,687,868,517

2,147,483,647

Python 3

OK

TESTS

96

608

83,763,200

n, m = [int(x) for x in input().split()] d = {} aux = set() for x in range(m): a, b = input().split() d.setdefault(a, {a}).add(b) d.setdefault(b, {b}).add(a) res = "YES" for x, y in d.items(): if x not in aux: if all([d[u] == y for u in y]): aux.update(y) else: res = "NO" break print(res)

Title: Bear and Friendship Condition Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures). There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves. Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z. For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well. Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes. Input Specification: The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends. The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input. Output Specification: If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes). Demo Input: ['4 3\n1 3\n3 4\n1 4\n', '4 4\n3 1\n2 3\n3 4\n1 2\n', '10 4\n4 3\n5 10\n8 9\n1 2\n', '3 2\n1 2\n2 3\n'] Demo Output: ['YES\n', 'NO\n', 'YES\n', 'NO\n'] Note: The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not.

```python n, m = [int(x) for x in input().split()] d = {} aux = set() for x in range(m): a, b = input().split() d.setdefault(a, {a}).add(b) d.setdefault(b, {b}).add(a) res = "YES" for x, y in d.items(): if x not in aux: if all([d[u] == y for u in y]): aux.update(y) else: res = "NO" break print(res) ```

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" } ]

1,626,092,673

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

124

6,656,000

n=int(input()) a=input() arr=list(map(int,a.split())) for i in range(n): for j in range(n): bft=1

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=int(input()) a=input() arr=list(map(int,a.split())) for i in range(n): for j in range(n): bft=1 ```

0

544

B

Sea and Islands

PROGRAMMING

1,400

[ "constructive algorithms", "implementation" ]

null

A map of some object is a rectangular field consisting of *n* rows and *n* columns. Each cell is initially occupied by the sea but you can cover some some cells of the map with sand so that exactly *k* islands appear on the map. We will call a set of sand cells to be island if it is possible to get from each of them to each of them by moving only through sand cells and by moving from a cell only to a side-adjacent cell. The cells are called to be side-adjacent if they share a vertical or horizontal side. It is easy to see that islands do not share cells (otherwise they together form a bigger island). Find a way to cover some cells with sand so that exactly *k* islands appear on the *n*<=×<=*n* map, or determine that no such way exists.

The single line contains two positive integers *n*, *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=*n*2) — the size of the map and the number of islands you should form.

If the answer doesn't exist, print "NO" (without the quotes) in a single line. Otherwise, print "YES" in the first line. In the next *n* lines print the description of the map. Each of the lines of the description must consist only of characters 'S' and 'L', where 'S' is a cell that is occupied by the sea and 'L' is the cell covered with sand. The length of each line of the description must equal *n*. If there are multiple answers, you may print any of them. You should not maximize the sizes of islands.

[ "5 2\n", "5 25\n" ]

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

none

1,000

[ { "input": "5 2", "output": "YES\nSSSSS\nLLLLL\nSSSSS\nLLLLL\nSSSSS" }, { "input": "5 25", "output": "NO" }, { "input": "82 6047", "output": "NO" }, { "input": "6 5", "output": "YES\nLSLSLS\nSLSLSS\nSSSSSS\nSSSSSS\nSSSSSS\nSSSSSS" }, { "input": "10 80", "output": "NO" }, { "input": "48 1279", "output": "NO" }, { "input": "40 1092", "output": "NO" }, { "input": "9 12", "output": "YES\nLSLSLSLSL\nSLSLSLSLS\nLSLSLSSSS\nSSSSSSSSS\nSSSSSSSSS\nSSSSSSSSS\nSSSSSSSSS\nSSSSSSSSS\nSSSSSSSSS" }, { "input": "43 146", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSS..." }, { "input": "100 5000", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS..." }, { "input": "100 4999", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS..." }, { "input": "100 5001", "output": "NO" }, { "input": "99 4901", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nS..." }, { "input": "99 4900", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nS..." }, { "input": "99 4902", "output": "NO" }, { "input": "99 9801", "output": "NO" }, { "input": "99 10", "output": "YES\nLSLSLSLSLSLSLSLSLSLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nS..." }, { "input": "99 1", "output": "YES\nLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nS..." }, { "input": "100 10000", "output": "NO" }, { "input": "100 10", "output": "YES\nLSLSLSLSLSLSLSLSLSLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS..." }, { "input": "50 1200", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL..." }, { "input": "50 1438", "output": "NO" }, { "input": "50 2447", "output": "NO" }, { "input": "49 1719", "output": "NO" }, { "input": "51 1996", "output": "NO" }, { "input": "51 1981", "output": "NO" }, { "input": "34 1060", "output": "NO" }, { "input": "74 3901", "output": "NO" }, { "input": "65 617", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS..." }, { "input": "89 497", "output": "YES\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS\nLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSL\nSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLSLS..." }, { "input": "34 621", "output": "NO" }, { "input": "1 0", "output": "YES\nS" }, { "input": "10 0", "output": "YES\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS" }, { "input": "11 0", "output": "YES\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS\nSSSSSSSSSSS" }, { "input": "99 0", "output": "YES\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nS..." }, { "input": "100 0", "output": "YES\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS\nSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS..." }, { "input": "1 1", "output": "YES\nL" }, { "input": "2 1", "output": "YES\nLS\nSS" }, { "input": "2 0", "output": "YES\nSS\nSS" }, { "input": "2 2", "output": "YES\nLS\nSL" }, { "input": "2 3", "output": "NO" }, { "input": "2 4", "output": "NO" } ]

1,623,298,987

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

3

61

0

n,k = list(map(int,input().split())) ans = 0 for i in range(1,n+1): ans+=i if i%2==0: ans-=1 if ans<k:print("NO");exit() print("YES") f = 0 for i in range(n): for j in range(n): if f==0 and j%2 and k>0: print('L',end="");k-=1 elif f==1 and j%2==0 and k>0 : print('L',end="");k-=1 else: print('S',end="") f^=0 print()

Title: Sea and Islands Time Limit: None seconds Memory Limit: None megabytes Problem Description: A map of some object is a rectangular field consisting of *n* rows and *n* columns. Each cell is initially occupied by the sea but you can cover some some cells of the map with sand so that exactly *k* islands appear on the map. We will call a set of sand cells to be island if it is possible to get from each of them to each of them by moving only through sand cells and by moving from a cell only to a side-adjacent cell. The cells are called to be side-adjacent if they share a vertical or horizontal side. It is easy to see that islands do not share cells (otherwise they together form a bigger island). Find a way to cover some cells with sand so that exactly *k* islands appear on the *n*<=×<=*n* map, or determine that no such way exists. Input Specification: The single line contains two positive integers *n*, *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=*n*2) — the size of the map and the number of islands you should form. Output Specification: If the answer doesn't exist, print "NO" (without the quotes) in a single line. Otherwise, print "YES" in the first line. In the next *n* lines print the description of the map. Each of the lines of the description must consist only of characters 'S' and 'L', where 'S' is a cell that is occupied by the sea and 'L' is the cell covered with sand. The length of each line of the description must equal *n*. If there are multiple answers, you may print any of them. You should not maximize the sizes of islands. Demo Input: ['5 2\n', '5 25\n'] Demo Output: ['YES\nSSSSS\nLLLLL\nSSSSS\nLLLLL\nSSSSS\n', 'NO\n'] Note: none

```python n,k = list(map(int,input().split())) ans = 0 for i in range(1,n+1): ans+=i if i%2==0: ans-=1 if ans<k:print("NO");exit() print("YES") f = 0 for i in range(n): for j in range(n): if f==0 and j%2 and k>0: print('L',end="");k-=1 elif f==1 and j%2==0 and k>0 : print('L',end="");k-=1 else: print('S',end="") f^=0 print() ```

0

387

B

George and Round

PROGRAMMING

1,200

[ "brute force", "greedy", "two pointers" ]

null

George decided to prepare a Codesecrof round, so he has prepared *m* problems for the round. Let's number the problems with integers 1 through *m*. George estimates the *i*-th problem's complexity by integer *b**i*. To make the round good, he needs to put at least *n* problems there. Besides, he needs to have at least one problem with complexity exactly *a*1, at least one with complexity exactly *a*2, ..., and at least one with complexity exactly *a**n*. Of course, the round can also have problems with other complexities. George has a poor imagination. It's easier for him to make some already prepared problem simpler than to come up with a new one and prepare it. George is magnificent at simplifying problems. He can simplify any already prepared problem with complexity *c* to any positive integer complexity *d* (*c*<=≥<=*d*), by changing limits on the input data. However, nothing is so simple. George understood that even if he simplifies some problems, he can run out of problems for a good round. That's why he decided to find out the minimum number of problems he needs to come up with in addition to the *m* he's prepared in order to make a good round. Note that George can come up with a new problem of any complexity.

The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=3000) — the minimal number of problems in a good round and the number of problems George's prepared. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*<=≤<=106) — the requirements for the complexity of the problems in a good round. The third line contains space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b*1<=≤<=*b*2...<=≤<=*b**m*<=≤<=106) — the complexities of the problems prepared by George.

Print a single integer — the answer to the problem.

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

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

In the first sample the set of the prepared problems meets the requirements for a good round. In the second sample, it is enough to come up with and prepare two problems with complexities 2 and 3 to get a good round. In the third sample it is very easy to get a good round if come up with and prepare extra problems with complexities: 2, 3, 4.

1,000

[ { "input": "3 5\n1 2 3\n1 2 2 3 3", "output": "0" }, { "input": "3 5\n1 2 3\n1 1 1 1 1", "output": "2" }, { "input": "3 1\n2 3 4\n1", "output": "3" }, { "input": "29 100\n20 32 41 67 72 155 331 382 399 412 465 470 484 511 515 529 616 637 679 715 733 763 826 843 862 903 925 979 989\n15 15 15 17 18 19 19 20 21 21 22 24 25 26 26 27 28 31 32 32 37 38 38 39 39 40 41 42 43 43 45 45 46 47 49 49 50 50 50 51 52 53 53 55 56 57 59 59 59 60 60 62 62 63 63 64 64 64 66 67 69 69 70 70 72 72 73 74 75 76 77 78 80 80 81 81 83 83 83 84 86 86 86 86 87 88 89 91 91 91 92 93 94 94 96 97 97 97 98 98", "output": "24" } ]

1,537,861,342

2,147,483,647

Python 3

OK

TESTS

41

124

614,400

n, m = map(int, input().split()) good = list(map(int, input().split())) prepared = list(map(int, input().split())) i = 0 j = 0 while i < n and j < m: if prepared[j] >= good[i]: i += 1 j += 1 else: j += 1 print(n - i)

Title: George and Round Time Limit: None seconds Memory Limit: None megabytes Problem Description: George decided to prepare a Codesecrof round, so he has prepared *m* problems for the round. Let's number the problems with integers 1 through *m*. George estimates the *i*-th problem's complexity by integer *b**i*. To make the round good, he needs to put at least *n* problems there. Besides, he needs to have at least one problem with complexity exactly *a*1, at least one with complexity exactly *a*2, ..., and at least one with complexity exactly *a**n*. Of course, the round can also have problems with other complexities. George has a poor imagination. It's easier for him to make some already prepared problem simpler than to come up with a new one and prepare it. George is magnificent at simplifying problems. He can simplify any already prepared problem with complexity *c* to any positive integer complexity *d* (*c*<=≥<=*d*), by changing limits on the input data. However, nothing is so simple. George understood that even if he simplifies some problems, he can run out of problems for a good round. That's why he decided to find out the minimum number of problems he needs to come up with in addition to the *m* he's prepared in order to make a good round. Note that George can come up with a new problem of any complexity. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=3000) — the minimal number of problems in a good round and the number of problems George's prepared. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*<=≤<=106) — the requirements for the complexity of the problems in a good round. The third line contains space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b*1<=≤<=*b*2...<=≤<=*b**m*<=≤<=106) — the complexities of the problems prepared by George. Output Specification: Print a single integer — the answer to the problem. Demo Input: ['3 5\n1 2 3\n1 2 2 3 3\n', '3 5\n1 2 3\n1 1 1 1 1\n', '3 1\n2 3 4\n1\n'] Demo Output: ['0\n', '2\n', '3\n'] Note: In the first sample the set of the prepared problems meets the requirements for a good round. In the second sample, it is enough to come up with and prepare two problems with complexities 2 and 3 to get a good round. In the third sample it is very easy to get a good round if come up with and prepare extra problems with complexities: 2, 3, 4.

```python n, m = map(int, input().split()) good = list(map(int, input().split())) prepared = list(map(int, input().split())) i = 0 j = 0 while i < n and j < m: if prepared[j] >= good[i]: i += 1 j += 1 else: j += 1 print(n - i) ```

3

0

none

0

[ "none" ]

null

Поликарп мечтает стать программистом и фанатеет от степеней двойки. Среди двух чисел ему больше нравится то, которое делится на большую степень числа 2. По заданной последовательности целых положительных чисел *a*1,<=*a*2,<=...,<=*a**n* требуется найти *r* — максимальную степень числа 2, на которую делится хотя бы одно из чисел последовательности. Кроме того, требуется вывести количество чисел *a**i*, которые делятся на *r*.

В первой строке записано целое число *n* (1<=≤<=*n*<=≤<=100) — длина последовательности *a*. Во второй строке записана последовательность целых чисел *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).

Выведите два числа: - *r* — максимальную степень двойки, на которую делится хотя бы одно из чисел заданной последовательности, - количество элементов последовательности, которые делятся на *r*.

[ "5\n80 7 16 4 48\n", "4\n21 5 3 33\n" ]

[ "16 3\n", "1 4\n" ]

В первом тестовом примере максимальная степень двойки, на которую делится хотя бы одно число, равна 16 = 24, на неё делятся числа 80, 16 и 48. Во втором тестовом примере все четыре числа нечётные, поэтому делятся только на 1 = 20. Это и будет максимальной степенью двойки для данного примера.

0

[ { "input": "5\n80 7 16 4 48", "output": "16 3" }, { "input": "4\n21 5 3 33", "output": "1 4" }, { "input": "10\n8 112 52 86 93 102 24 24 100 826791168", "output": "256 1" }, { "input": "3\n458297759 18 104", "output": "8 1" }, { "input": "7\n12 14 40 8 74 104 11", "output": "8 3" }, { "input": "11\n35 16 664311776 46 48 52 63 82 84 80 23", "output": "32 1" }, { "input": "7\n67 68 58 24 96 73 72", "output": "32 1" }, { "input": "8\n48 112 40 8 112 14 80 36", "output": "16 4" }, { "input": "10\n14 6 68 8 84 949689614 91 26 80 56", "output": "16 1" }, { "input": "4\n39 56 939117699 56", "output": "8 2" }, { "input": "5\n90 18 56 64 32", "output": "64 1" }, { "input": "9\n64 95 32 64 96 80 100 96 66", "output": "64 2" }, { "input": "20\n105407881 735510073 587127085 111067442 126807503 250859170 778634763 919694130 592496831 462347734 532487590 475786023 951527598 183635985 612791353 447723541 409812454 900700354 801564406 532793851", "output": "2 9" }, { "input": "20\n850632510 530517796 700510265 454664263 131947796 418444926 921278498 251889644 705327498 892480283 884422799 479219117 399278535 80826412 496934492 448261193 39033930 49426174 621130971 808191947", "output": "4 5" }, { "input": "20\n780355354 620754888 193377552 463211662 46248927 312489308 472238901 823707535 138518748 267363170 19751630 193171944 411443343 858525221 458019868 490268043 7864848 218005780 744553112 83590041", "output": "16 2" }, { "input": "20\n29023024 579267278 217400978 121454376 235087976 154574217 708760940 84623652 195299056 329204104 527952531 822521791 513319036 285749488 292843688 389260660 498981613 835987320 444201058 251639011", "output": "16 3" }, { "input": "20\n267784376 576420580 392773522 296581728 508523192 812838532 920098710 624114448 194991560 850559568 29915376 785467756 490019770 524237000 871021232 970867040 769417893 210139479 445850586 333230268", "output": "32 2" }, { "input": "20\n860654784 630481952 430211228 13468621 33985780 279050728 782571295 83521731 818343376 508318323 550168944 763113524 152970477 502262855 934672824 712697136 451447464 732781790 71573907 50381000", "output": "64 1" }, { "input": "20\n673865536 152236510 957204496 401364096 969402746 287701920 768559538 642049008 736330680 179648832 480094052 225156558 957671104 726304328 612058916 257008256 173639040 673864512 431405191 454360662", "output": "128 3" }, { "input": "20\n706678380 597303020 176804438 146220776 485004772 799346560 692789954 737954674 398118372 231976240 957701828 556811840 74342144 966291136 893909760 745234360 44276827 878935416 975182148 322390872", "output": "256 2" }, { "input": "20\n442107036 883530112 852749824 997931232 902004480 838557324 186049792 827163136 3843737 603467472 383038751 548720704 843680384 906511492 591629504 41722624 79778650 839163077 880599104 456148480", "output": "512 3" }, { "input": "20\n667815852 318176276 693849088 724201296 188710200 39249152 929966576 651876056 580647856 575425536 367972188 647585808 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1,458,979,048

3,448

Python 3

WRONG_ANSWER

TESTS

8

62

4,608,000

n = int(input()) a = list(map(int, input().split())) d = dict() r = 0 top = [] for el in a: while el % 2**r == 0: if el % 2 ** r == 0: if el in d: d[el] += 1 else: d[el] = 0 r += 1 r = 0 b = sorted(list(d.items()), key = lambda x: -x[1]) value = b[0][1] for el in b: if el[1] == value: top.append(el) print(2 ** top[0][1], len(top))

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Поликарп мечтает стать программистом и фанатеет от степеней двойки. Среди двух чисел ему больше нравится то, которое делится на большую степень числа 2. По заданной последовательности целых положительных чисел *a*1,<=*a*2,<=...,<=*a**n* требуется найти *r* — максимальную степень числа 2, на которую делится хотя бы одно из чисел последовательности. Кроме того, требуется вывести количество чисел *a**i*, которые делятся на *r*. Input Specification: В первой строке записано целое число *n* (1<=≤<=*n*<=≤<=100) — длина последовательности *a*. Во второй строке записана последовательность целых чисел *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). Output Specification: Выведите два числа: - *r* — максимальную степень двойки, на которую делится хотя бы одно из чисел заданной последовательности, - количество элементов последовательности, которые делятся на *r*. Demo Input: ['5\n80 7 16 4 48\n', '4\n21 5 3 33\n'] Demo Output: ['16 3\n', '1 4\n'] Note: В первом тестовом примере максимальная степень двойки, на которую делится хотя бы одно число, равна 16 = 24, на неё делятся числа 80, 16 и 48. Во втором тестовом примере все четыре числа нечётные, поэтому делятся только на 1 = 20. Это и будет максимальной степенью двойки для данного примера.

```python n = int(input()) a = list(map(int, input().split())) d = dict() r = 0 top = [] for el in a: while el % 2**r == 0: if el % 2 ** r == 0: if el in d: d[el] += 1 else: d[el] = 0 r += 1 r = 0 b = sorted(list(d.items()), key = lambda x: -x[1]) value = b[0][1] for el in b: if el[1] == value: top.append(el) print(2 ** top[0][1], len(top)) ```

0