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CodeNet4Repair

like 0

s857269950

p03814

u905582793

Wrong Answer

Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.

s=input() for i in range(len(s)): if s[i]=="A": pA = i for i in range(len(s)): if s[-i-1] == "Z": pZ = i print(len(s)-pA-pZ-1)

s399431856

Accepted

s=input() for i in range(len(s)): if s[i]=="A": pA = i break for i in range(len(s)): if s[-i-1] == "Z": pZ = i break print(len(s)-pA-pZ)

s078344719

p03673

u626468554

Wrong Answer

You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.

#n = int(input()) #n,k = map(int,input().split()) #x = list(map(int,input().split())) n = int(input()) a = list(input()) memo = '' for i in range(n): if i%2==0: memo = memo + a[i] else: memo = a[i] + memo print(*list(memo))

s681691079

Accepted

n = int(input()) a = list(map(int,input().split())) li1 = [] li2 = [] for i in range(n): if i%2==0: li1.append(a[i]) else: li2.append(a[i]) ans = [] if n%2==0: for i in range(n//2): ans.append(li2[(i+1)*(-1)]) for i in range(n//2): ans.append(li1[i]) else: for i in range(n//2+1): ans.append(li1[(i+1)*(-1)]) for i in range(n//2): ans.append(li2[i]) print(*ans)

s973695994

p03598

u875769753

Wrong Answer

There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.

N = int(input()) K = int(input()) lsx = list(map(int,input().split())) ans = 0 for i in range(N): ans += min(abs(lsx[i]),abs(lsx[i]-K)) print(ans)

s577576423

Accepted

N = int(input()) K = int(input()) lsx = list(map(int,input().split())) ans = 0 for i in range(N): ans += min(abs(lsx[i]),abs(lsx[i]-K)) print(2*ans)

s931646531

p03524

u464205401

Wrong Answer

Snuke has a string S consisting of three kinds of letters: `a`, `b` and `c`. He has a phobia for palindromes, and wants to permute the characters in S so that S will not contain a palindrome of length 2 or more as a substring. Determine whether this is possible.

import collections s = list(input()) n = len(s) flg = True for i in range(n-1): if len(set(s[i:i+2]))==1: flg = False for i in range(n-2): tmp = s[i:i+3] if tmp == tmp[::-1]: flg = False print('YES' if flg else 'NO')

s966569031

Accepted

import collections s = list(input()) n = len(s) cnt = [0,0,0] for i in range(n): if s[i]=="a": cnt[0]+=1 elif s[i]=="b": cnt[1]+=1 else: cnt[2]+=1 print('YES' if max(cnt)-min(cnt) <= 1 else 'NO')

s646514362

p03478

u803865203

Wrong Answer

Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).

N,A,B = map(int,input().split()) ans = 0 for i in range(int(N)+1): array = list(map(int,list(str(i)))) print(array) if(sum(array)>=A and sum(array)<=B): ans += int(i) print(ans)

s782391329

Accepted

N,A,B = map(int,input().split()) ans = 0 for i in range(1,int(N)+1): sum = 0 num = i while(True): sum += num%10 num = num//10 if(num == 0): if(sum >= A and sum <= B): ans += i break else: break print(ans)

s898955101

p02413

u382316013

Wrong Answer

Your task is to perform a simple table calculation. Write a program which reads the number of rows r, columns c and a table of r × c elements, and prints a new table, which includes the total sum for each row and column.

r, c = [int(i) for i in input().split()] data = [] sum_row = [0] * (c + 1) for ri in range(r): data.append([int(i) for i in input().split()]) data[ri].append(sum(data[ri])) print(" ".join([str(d) for d in data[ri]])) for ci in range(c + 1): sum_row[ci] += data[ri][ci] print(" ".join([str(s) for s in sum_row])) from pprint import pprint pprint(data)

s576144484

Accepted

r, c = [int(i) for i in input().split()] data = [] sum_row = [0] * (c + 1) for ri in range(r): data.append([int(i) for i in input().split()]) data[ri].append(sum(data[ri])) print(" ".join([str(d) for d in data[ri]])) for ci in range(c + 1): sum_row[ci] += data[ri][ci] print(" ".join([str(s) for s in sum_row]))

s078320145

p02646

u263660661

Wrong Answer

Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.

a, v = map(int, input().split()) b, w = map(int, input().split()) t = int(input()) if v <= w: print("No") elif t < abs(a - b)/abs(v - w): print("No") else: print("Yes")

s430673963

Accepted

a, v = map(int, input().split()) b, w = map(int, input().split()) t = int(input()) if v <= w: print("NO") elif t < abs(a - b) / abs(v - w): print("NO") else: print("YES")

s457806602

p03351

u357867755

Wrong Answer

Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.

a, b, c, d = map(int, input().split()) if abs(c - a) <= d: print('Yes') elif (abs(b - a) <= d) and (abs(c-b) <= d): print('Yes') else: print('No') print(abs(b-a))

s598136167

Accepted

a, b, c, d = map(int, input().split()) if abs(c - a) <= d: print('Yes') elif (abs(b - a) <= d) and (abs(b - c) <= d): print('Yes') else: print('No')

s239344601

p03712

u826263061

Wrong Answer

You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.

h, w = list(map(int, input().split())) a = '*'*(w+2) print(a) for i in range(h): s = input() print('*'+s+'*') print(a)

s627618317

Accepted

# -*- coding: utf-8 -*- """ Created on Sun Sep 16 16:57:33 2018 @author: maezawa """ h, w = list(map(int, input().split())) a = '#'*(w+2) b = [] for i in range(h): s = input() b.append('#'+s+'#') print(a) for s in b: print(s) print(a)

s155938177

p03251

u347452770

Wrong Answer

Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.

import sys n,m,X,Y=map(int,input().split()) x=list(map(int,input().split())) y=list(map(int,input().split())) d=min(y)-max(x) if d<=0: print("War") sys.exit() else: for i in range(d+1): Z=X+i if (Z<=X or Z>Y) or (max(x)>=Z or min(y)<Z): print("War") sys.exit() print("No War")

s833488960

Accepted

import sys n,m,X,Y=map(int,input().split()) x=list(map(int,input().split())) y=list(map(int,input().split())) d=min(y)-max(x) warFlag = False if d<=0: print("War") sys.exit() else: for i in range(1, Y - X + 1): Z=X+i if (max(x)>=Z or min(y)<Z): warFlag = True else: #max(x) < Z, max(y) >= Z warFlag = False break if warFlag: print("War") else: print("No War")

s982428385

p02406

u567380442

Wrong Answer

In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }

print(*[i for i in range(1, int(input())) if i % 3 == 0 or '3' in str(i)])

s279404443

Accepted

n = int(input()) print(' ', end='') print(*[i for i in range(1, 1 + n) if i % 3 == 0 or '3' in str(i)])

s269569376

p03494

u905582793

Wrong Answer

There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.

n=int(input()) a=list(map(int,input().split())) ans=0 for i in range(100): for j in range(n): if a[j] // 2 ==0: a[j]=a[j]//2 if j==n: ans+=1 else: break else: continue break print(ans)

s901285034

Accepted

n=int(input()) a=list(map(int,input().split())) ans=0 for i in range(100): for j in range(n): if a[j] % 2 ==0: a[j]=a[j]//2 if j==n-1: ans+=1 else: break else: continue break print(ans)

s962411356

p03024

u003501233

Wrong Answer

Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.

S=input() if S.count("x") <= 7: print("Yes") else: print("No")

s875708173

Accepted

S=input() if S.count("x") <= 7: print("YES") else: print("NO")

s003111517

p03448

u340781749

Wrong Answer

You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.

a = int(input()) b = int(input()) c = int(input()) x = int(input()) ans = 0 for p in range(a + 1): for q in range(b + 1): r = (x - p * 500 - q * 100) // 50 if r <= c: ans += 1 print(ans)

s721457267

Accepted

a = int(input()) b = int(input()) c = int(input()) x = int(input()) ans = 0 for p in range(a + 1): for q in range(b + 1): r = (x - p * 500 - q * 100) // 50 if 0 <= r <= c: ans += 1 print(ans)

s331417520

p03696

u502389123

Wrong Answer

You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.

N = int(input()) S = input() needleft = 0 needright = 0 ans = '' for i in range(N): if S[i] == ')': ans += '(' * needright ans += ')' * needright if needright == 0: needleft += 1 needright = 0 elif S[i] == '(': ans += '(' * needleft ans += ')' * needleft if needleft == 0: needright += 1 needleft = 0 if i == N-1: if needleft: ans += '(' * needleft ans += ')' * needleft else: ans += '(' * needright ans += ')' * needright print(ans)

s451669706

Accepted

N = int(input()) S = input() cnt = 0 s = '' for i in range(N): if S[i] == '(': cnt += 1 else: cnt -= 1 if cnt < 0: s += '(' cnt = 0 s += S if cnt: s += ')' * cnt print(s)

s002160519

p03050

u598720217

Wrong Answer

Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied: * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds. Find all favorite numbers and print the sum of those.

import math N = int(input()) n = int(math.sqrt(N)) count=0 for i in range(1,int(n)): d = N - i if d%i==0: count+=int(d/i) print(count)

s110142101

Accepted

import math N = int(input()) n = int(math.sqrt(N)) count=0 for i in range(1,n+1): d = N - i if d%i==0 and i < (N // i) -1: k = d//i count+=k print(count)

s856624692

p03943

u474270503

Wrong Answer

Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

A=map(int,input().split()) print('Yes' if 2*max(A)==sum(A) else 'No')

s492543887

Accepted

A=list(map(int,input().split())) print('Yes' if 2*max(A)==sum(A) else 'No')

s331709269

p03457

u740284863

Wrong Answer

AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.

n = int(input()) for i in range(n): t, x, y = map(int, input().split()) if x + y < t or (x + y + t) % 2: print("No") exit() print("Yes")

s904599365

Accepted

import sys input = sys.stdin.readline n = int(input()) t,x,y = 0,0,0 for i in range(n): t_,x_,y_ = map(int,input().split()) X = abs(x_ - x) Y = abs(y_ - y) T = t_ - t if ( X + Y > T ) or (T % 2 != ( X + Y ) % 2): print("No") exit() t = t_ x = x_ y = y_ print("Yes")

s958675938

p03637

u023229441

Wrong Answer

We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective.

n=int(input()) A=list(map(int,input().split())) for i in range(n): if A[i]%4==0: A[i]==0 elif A[i]%2==0: A[i]==2 else: A[i]==1 a=A.count(0) b=A.count(2) c=A.count(1) if c<=a : print("Yes") else: print("No")

s119816949

Accepted

n=int(input()) A=list(map(int,input().split())) for i in range(n): if A[i]%4==0: A[i]=0 elif A[i]%2==0: A[i]=2 else: A[i]=1 a=A.count(0) b=A.count(2) c=A.count(1) if c<=a : print("Yes") elif b==0 and c==a+1: print("Yes") else: print("No")

s297486840

p03958

u981931040

Wrong Answer

There are K pieces of cakes. Mr. Takahashi would like to eat one cake per day, taking K days to eat them all. There are T types of cake, and the number of the cakes of type i (1 ≤ i ≤ T) is a_i. Eating the same type of cake two days in a row would be no fun, so Mr. Takahashi would like to decide the order for eating cakes that minimizes the number of days on which he has to eat the same type of cake as the day before. Compute the minimum number of days on which the same type of cake as the previous day will be eaten.

import heapq K, T = map(int, input().split()) a = tuple(map(int, input().split())) cakes = [] heapq.heapify(cakes) for i in range(T): heapq.heappush(cakes, [-1*a[i], i]) print(cakes) now = -1 ans = 0 for _ in range(K): cake_num, cake_idx = heapq.heappop(cakes) cake_num *= -1 if cake_idx == now and len(cakes) > 1 and cakes[0][1]: next_cake_num, next_cake_idx = heapq.heappop(cakes) next_cake_num -= 1 heapq.heappush(cakes, [next_cake_num, next_cake_idx]) now = next_cake_idx else: cake_num -= 1 if now == cake_idx: ans += 1 now = cake_idx heapq.heappush(cakes, [cake_num, cake_idx]) print(ans)

s662958812

Accepted

import heapq K, T = map(int, input().split()) a = tuple(map(int, input().split())) cakes = [] heapq.heapify(cakes) for i in range(T): heapq.heappush(cakes, [-1*a[i], i]) # print(cakes) now = -1 ans = 0 for _ in range(K): cake_num, cake_idx = heapq.heappop(cakes) cake_num *= -1 if cake_idx == now and len(cakes) > 1 and cakes[0][0] != 0: next_cake_num, next_cake_idx = heapq.heappop(cakes) next_cake_num *= -1 next_cake_num -= 1 heapq.heappush(cakes, [-1 * next_cake_num, next_cake_idx]) now = next_cake_idx else: cake_num -= 1 if now == cake_idx: ans += 1 now = cake_idx heapq.heappush(cakes, [-1 * cake_num, cake_idx]) # print(cakes) print(ans)

s162238411

p03957

u838930283

Wrong Answer

This contest is `CODEFESTIVAL`, which can be shortened to the string `CF` by deleting some characters. Mr. Takahashi, full of curiosity, wondered if he could obtain `CF` from other strings in the same way. You are given a string s consisting of uppercase English letters. Determine whether the string `CF` can be obtained from the string s by deleting some characters.

s = input() result = "No" for i in range(len(s)): if s[i] == "C": for j in range(i+1, len(s)): if s[j] == "F": result = "YES" break print(result)

s043694767

Accepted

s = input() result = "No" for i in range(len(s)): if s[i] == "C": for j in range(i+1, len(s)): if s[j] == "F": result = "Yes" break print(result)

s877952174

p02260

u114315703

Wrong Answer

Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠ mini.

N = int(input()) A = [int(e) for e in input().split()] def selection_sort(A, N): steps = 0 for i in range(0, N): minj = i for j in range(i, N): if A[minj] > A[j]: minj = j steps += 1 A[minj], A[i] = A[i], A[minj] print(A) print(steps) return A selection_sort(A, N)

s982501945

Accepted

N = int(input()) A = [int(e) for e in input().split()] def selection_sort(A, N): steps = 0 for i in range(0, N): minj = i for j in range(i, N): if A[minj] > A[j]: minj = j if i != minj: steps += 1 A[minj], A[i] = A[i], A[minj] for e in A[:-1]: print(e, end=' ') print(A[-1]) print(steps) return A selection_sort(A, N)

s671747010

p03963

u774160580

Wrong Answer

There are N balls placed in a row. AtCoDeer the deer is painting each of these in one of the K colors of his paint cans. For aesthetic reasons, any two adjacent balls must be painted in different colors. Find the number of the possible ways to paint the balls.

N, K = map(int, input().split()) print(K * pow(K, N - 1))

s277792873

Accepted

N, K = map(int, input().split()) print(K * pow(K - 1, N - 1))

s493245996

p03854

u962451849

Wrong Answer

You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

s = input().replace('eraser', '').replace('erase', '') \ .replace('dreamer', '').replace('dream', '') if s: print('NO') else: print('YES') print(s)

s067356462

Accepted

s = input().replace('eraser', '').replace('erase', '') \ .replace('dreamer', '').replace('dream', '') if s: print('NO') else: print('YES')

s711061681

p00002

u945249026

Wrong Answer

Write a program which computes the digit number of sum of two integers a and b.

import sys a = [] for line in sys.stdin: a.append(line) for i in a: print(i[0] + i[1])

s995239498

Accepted

while True: try: a, b = map(int,input().split()) s = str(a + b) print(len(s)) except EOFError: break

s759778161

p02407

u639421643

Wrong Answer

Write a program which reads a sequence and prints it in the reverse order.

length = int(input()) nums = list(map(int, input().split())) print(nums[::-1])

s432077989

Accepted

length = int(input()) nums = input().split() print(" ".join(nums[::-1]))

s067283366

p03623

u240793404

Wrong Answer

Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.

a,b,c=map(int,input().split()); print("A" if abs(b-a) < abs(c-b) else "B")

s992149155

Accepted

a,b,c=map(int,input().split()); print("A" if abs(b-a) < abs(c-a) else "B")

s018796994

p03455

u014047173

Wrong Answer

AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.

a, b = map(int, input().split()) if (a % 2 == 0 ) or ( b % 2 == 0 ) : 'even' else : 'odd'

s076673238

Accepted

a, b = map(int, input().split()) if (a % 2 == 0 ) or ( b % 2 == 0 ) : print('Even') else : print('Odd')

s593620309

p02258

u826549974

Wrong Answer

You can obtain profits from foreign exchange margin transactions. For example, if you buy 1000 dollar at a rate of 100 yen per dollar, and sell them at a rate of 108 yen per dollar, you can obtain (108 - 100) × 1000 = 8000 yen. Write a program which reads values of a currency $R_t$ at a certain time $t$ ($t = 0, 1, 2, ... n-1$), and reports the maximum value of $R_j - R_i$ where $j > i$ .

n = int(input()) A = [int(input()) for i in range(n)] max = A[1]-A[0] for i in range(1,n-1): if(max < A[i+1]-A[i]): max = A[i+1]-A[i] print(max)

s772538033

Accepted

n = int(input()) A = [int(input()) for i in range(n)] max = A[1]-A[0] min = A[0] for i in range(1,n): if(max < A[i]-min): max = A[i]-min if(min > A[i]): min = A[i] print(max)

s648280006

p03455

u050428930

Wrong Answer

AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.

a,b=map(int,input().split()) t = a*b%2 if t==0: print("even") else: print("odd")

s195204363

Accepted

a,b=map(int,input().split()) t = a*b%2 if t==0: print("Even") else: print("Odd")

s275662604

p02795

u841924362

Wrong Answer

We have a grid with H rows and W columns, where all the squares are initially white. You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions: * Choose one row, then paint all the squares in that row black. * Choose one column, then paint all the squares in that column black. At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations.

import math H = int(input()) W = int(input()) N = int(input()) if(H >= W): print(math.ceil(N // H)) else: print(math.ceil(N // W))

s023709157

Accepted

import math H = int(input()) W = int(input()) N = int(input()) if(H >= W): print(math.ceil(N / H)) else: print(math.ceil(N / W))

s675913842

p02410

u369313788

Wrong Answer

Write a program which reads a $ n \times m$ matrix $A$ and a $m \times 1$ vector $b$, and prints their product $Ab$. A column vector with m elements is represented by the following equation. \\[ b = \left( \begin{array}{c} b_1 \\\ b_2 \\\ : \\\ b_m \\\ \end{array} \right) \\] A $n \times m$ matrix with $m$ column vectors, each of which consists of $n$ elements, is represented by the following equation. \\[ A = \left( \begin{array}{cccc} a_{11} & a_{12} & ... & a_{1m} \\\ a_{21} & a_{22} & ... & a_{2m} \\\ : & : & : & : \\\ a_{n1} & a_{n2} & ... & a_{nm} \\\ \end{array} \right) \\] $i$-th element of a $m \times 1$ column vector $b$ is represented by $b_i$ ($i = 1, 2, ..., m$), and the element in $i$-th row and $j$-th column of a matrix $A$ is represented by $a_{ij}$ ($i = 1, 2, ..., n,$ $j = 1, 2, ..., m$). The product of a $n \times m$ matrix $A$ and a $m \times 1$ column vector $b$ is a $n \times 1$ column vector $c$, and $c_i$ is obtained by the following formula: \\[ c_i = \sum_{j=1}^m a_{ij}b_j = a_{i1}b_1 + a_{i2}b_2 + ... + a_{im}b_m \\]

n, m = [int(i) for i in input().split()] matrix = [] for ni in range(n): matrix.append([int(x) for x in input().split()]) vector = [] for ni in range(m): vector.append(int(input())) print(matrix) print() print(vector) for ni in range(n): sum = 0 for mi in range(n): sum += matrix[ni][mi]* vector[mi] print(sum)

s536276728

Accepted

n, m = [int(i) for i in input().split()] matrix = [] for ni in range(n): matrix.append([int(a) for a in input().split()]) vector = [] for mi in range(m): vector.append(int(input())) for ni in range(n): sum = 0 for mi in range(m): sum += matrix[ni][mi] * vector[mi] print(sum)

s018978227

p03565

u226912938

Wrong Answer

E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.

import re s = str(input()) t = str(input()) NS = [] for i in range(len(s)): s_res = s[:i] s_obj = s[i:] s_res.replace('?', 'a') s_w = s_obj.replace('?', '.') s_re = re.compile(s_w) matchobj = s_re.fullmatch(t) if matchobj: start = matchobj.start() end = matchobj.end() print('-' * 30) print('i' + str(i)) print(start) print(end) print('-' * 30) new_s = s_res + s_obj[:start] + t + s_obj[end:] new_s = new_s.replace('?', 'a') NS.append(new_s) if len(NS) >= 1: NS = sorted(NS) ans = NS[0] else: ans = 'UNRESTORABLE' print(ans)

s169907538

Accepted

import re s = str(input()) t = str(input()) NS = [] for i in range(len(s)): s_res = s[:i] s_res = s_res.replace('?', 'a') s_obj = s[i:i+len(t)] s_obj = s_obj.replace('?', '.') s_comp = re.compile(s_obj) s_res_f = s[i+len(t):] matchobj = s_comp.fullmatch(t) if matchobj: start = matchobj.start() new_s = s_res + t + s_res_f new_s = new_s.replace('?', 'a') NS.append(new_s) if len(NS) == 0: ans = 'UNRESTORABLE' else: ans = sorted(NS)[0] print(ans)

s739232234

p02397

u957680575

Wrong Answer

Write a program which reads two integers x and y, and prints them in ascending order.

A=1 B=1 x=[] while True: if A==0 and B==0: break else: A, B= map(int, input().split()) if A > B: print(B,A) else: print(A,B)

s282657678

Accepted

x=[] while True: A, B= map(int, input().split()) if A==0 and B==0: break else: if A > B: print(B,A) else: print(A,B)

s311771067

p04043

u143212659

Wrong Answer

Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

#!/usr/bin/env python3 # -*- coding: utf-8 -*- def main(): ABC = list(map(int, input().split())) print('Yes' if ABC.count(5) == 2 and ABC.count(7) == 1 else 'No') if __name__ == "__main__": main()

s472115789

Accepted

#!/usr/bin/env python3 # -*- coding: utf-8 -*- def main(): ABC = list(map(int, input().split())) ABC.sort() print('YES' if [5, 5, 7] == ABC else 'NO') if __name__ == "__main__": main()

s001101512

p04030

u252964975

Wrong Answer

Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?

S=str(input()) str_ = "" for i in range(len(S)): if S[i] == "0" or "1": str_ = str_ + S[i] elif len(str_) != 0: str_ = str_.rstrip() print(str_)

s937103075

Accepted

S=str(input()) str_ = "" for i in range(len(S)): if S[i] == "0" or S[i] == "1": str_ = str_ + S[i] elif len(str_) != 0: str_ = str_[:-1] print(str_)

s621819452

p03610

u922926087

Wrong Answer

You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.

input()[::2]

s116234370

Accepted

# -*- coding: utf-8 -*- """ Created on Wed Feb 27 01:06:34 2019 @author: yuta """ def oddstring(s): print(s[::2]) oddstring(input())

s829139781

p03149

u799215419

Wrong Answer

You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".

import re S = input() pattern = r"(.+keyence|k.+eyence|ke.+yence|key.+ence|keye.+nce|keyen.+ce|keyenc.+e|keyence.+|keyence)" if re.fullmatch(pattern, S): print("YES") else: print("NO")

s938057964

Accepted

nums = list(map(int, input().split())) if 1 in nums and 9 in nums and 7 in nums and 4 in nums: print("YES") else: print("NO")

s103638490

p03643

u202634017

Wrong Answer

This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.

N = input() print("abc{}".format(N))

s959950565

Accepted

N = input() print("ABC"+N)

s316773300

p03779

u017415492

Wrong Answer

There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.

x=int(input()) #n=1 #while 1/2*(n**2+n)<x: # n+=1 #print(n) left=0 right=x while right-left>1: mid=(left+right)//2 if (mid**2+mid)/2==x: break if (mid**2+mid)/2<x: left=mid else: right=mid print(right)

s494573903

Accepted

x=int(input()) #n=1 #while 1/2*(n**2+n)<x: # n+=1 #print(n) left=0 right=x while right-left>1: mid=(left+right)//2 if (mid**2+mid)/2==x: right=mid break if (mid**2+mid)/2<x: left=mid else: right=mid print(right)

s340134302

p03861

u803647747

Wrong Answer

You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?

from decimal import Decimal a, b, x = map(Decimal, input().split()) int(Decimal(b/x) - Decimal(a/x))

s866175613

Accepted

from decimal import Decimal a, b, x = map(Decimal, input().split()) bx = (b/x) aq, amod = divmod(a, x) if a==0 or amod==0: print(int(bx-aq)+1) else: print(int(bx-aq))

s567908132

p03486

u439312138

Wrong Answer

You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.

s = input() s_asc = ''.join(sorted(s)) t = input() t_desc = ''.join(sorted(t, reverse=True)) print(s_asc) print(t_desc) if s_asc < t_desc: print('Yes') else: print('No')

s215163728

Accepted

s = input() s_asc = ''.join(sorted(s)) t = input() t_desc = ''.join(sorted(t, reverse=True)) if s_asc < t_desc: print('Yes') else: print('No')

s789789591

p03447

u871596687

Wrong Answer

You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?

X = int(input()) A = int(input()) B = int(input()) N = 0 while True: if (X-A-B*N)> 0: N +=1 else: print(X-A-B*N) break

s600967124

Accepted

X = int(input()) A = int(input()) B = int(input()) N = 0 while True: if (X-A-B*N)>= 0: N +=1 else: print(X-A-B*(N-1)) break

s178475352

p03131

u311379832

Wrong Answer

Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the maximum possible number of biscuits in Snuke's pocket after K operations.

K, A, B = map(int, input().split()) if A >= B - 1: print(K + 1)

s489322176

Accepted

K, A, B = map(int, input().split()) if A >= B - 1 or A >= K: print(K + 1) else: K -= (A + 1) ans = 0 if K >= 2: ans = B + ((B - A) * (K // 2)) ans += K % 2 elif K != 0: ans = B + 1 else: ans += B print(ans)

s586612811

p02694

u616619529

Wrong Answer

Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?

# -*- coding: utf-8 -*- x = int(input()) money = 100 year = 0 while (money <= x): money = int(money*1.01) print('{}'.format(year))

s525569827

Accepted

# -*- coding: utf-8 -*- x = int(input()) money = 100 year = 0 while (money < x): money = int(money * 1.01) year = year + 1 print('{}'.format(year))

s245919798

p03486

u257974487

Wrong Answer

You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.

s = input() t = input() p = [] q = [] m = len(s) n = len(t) ans = 0 for i in range(m): p.append(s[i]) p.sort() for j in range(n): q.append(t[j]) q.sort() if m >= n: ans = "No" else: ans = "Yes" for k in range(min(m,n)): if ord(p[k]) < ord(q[k]): ans = "Yes" break elif ord(p[k]) > ord(q[k]): ans = "No" break print(ans)

s515599923

Accepted

s = input() t = input() p = [] q = [] m = len(s) n = len(t) if m >= n: ans = "No" else: ans = "Yes" for i in range(m): p.append(s[i]) p.sort() for j in range(n): q.append(t[j]) q.sort(reverse=True) for k in range(min(m,n)): if ord(p[k]) < ord(q[k]): ans = "Yes" break elif ord(p[k]) > ord(q[k]): ans = "No" break print(ans)

s177493995

p03214

u691170040

Wrong Answer

Niwango-kun is an employee of Dwango Co., Ltd. One day, he is asked to generate a thumbnail from a video a user submitted. To generate a thumbnail, he needs to select a frame of the video according to the following procedure: * Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video. * Select t-th frame whose representation a_t is nearest to the average of all frame representations. * If there are multiple such frames, select the frame with the smallest index. Find the index t of the frame he should select to generate a thumbnail.

N = int(input()) at = input().split(" ") a = [] for i in at: a.append(int(i)) avg = sum(a) / N mdif, flag = abs(a[0]-avg), 0 for i in range(len(a)): if a[i]-avg < mdif: mdif = a[i] - avg flag = i print(flag)

s321042773

Accepted

N = int(input()) at = input().split(" ") a = [] for i in at: a.append(int(i)) avg = sum(a) / N mdif, flag = abs(a[0]-avg), 0 for i in range(len(a)): if abs(a[i]-avg) < mdif: mdif = abs(a[i]-avg) flag = i print(flag)

s651846610

p02694

u464205401

Wrong Answer

Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?

x=int(input()) a=100 cnt=0 while x>=a: a=int(a*1.01) # print(a) cnt+=1 print(cnt)

s559074683

Accepted

x=int(input()) a=100 cnt=0 while x>a: a=int(a*1.01) cnt+=1 print(cnt)

s643464567

p03377

u736729525

Wrong Answer

There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.

A, B, X = [int(x) for x in input().split()] def solve(A, B, X): if A + B < X: print("NO") elif A > X: print("NO") else: print("YES")

s468267117

Accepted

A, B, X = [int(x) for x in input().split()] if A + B < X: print("NO") elif A > X: print("NO") else: print("YES")

s109200861

p03730

u780459096

Wrong Answer

We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.

# -*- coding: utf-8 -*- a, b, c = map(int, input().split()) rest = a % b num = a while True: if num % b == c: print('Yes') exit() num+=a if num % b == rest: print('No') exit()

s342976496

Accepted

# -*- coding: utf-8 -*- a, b, c = map(int, input().split()) rest = a % b num = a while True: if num % b == c: print('YES') exit() num+=a if num % b == rest: print('NO') exit()

s032830875

p03469

u103902792

Wrong Answer

On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.

s= input() if s[4]=='2017': print('2018'+s[4:]) else: print(s)

s525784857

Accepted

s= input() print(s[:3] + '8' + s[4:])

s488084945

p03997

u263691873

Wrong Answer

You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.

a=int(input()) b=int(input()) h = int(input()) print((a+b)*h/2)

s820305714

Accepted

a=int(input()) b=int(input()) h = int(input()) print(int((a+b)*h/2))

s730377166

p02972

u845333844

Wrong Answer

There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.

n=int(input()) a=list(map(int,input().split())) b=[0]*n for i in range(n,0,-1): m=n//i total=0 for j in range(1,m+1): total+=a[i*j-1] b[i-1]=total%2 ans=sum(b) print(ans) print(*b)

s122170973

Accepted

n=int(input()) a=list(map(int,input().split())) b=[0]*n for i in range(n,0,-1): m=n//i total=0 for j in range(1,m+1): if j==1: total+=a[i-1] else: total+=b[i*j-1] b[i-1]=total%2 ans=sum(b) print(ans) if ans != 0: l=[] for i in range(n): if b[i]==1: l.append(i+1) print(*l)

s550275877

p02397

u666221014

Wrong Answer

Write a program which reads two integers x and y, and prints them in ascending order.

print( " ".join( sorted( input().split())))

s010385178

Accepted

while True : a, b = [int(i) for i in input().split()] if a == b == 0 : break elif a >= b : print(b, a) else : print(a,b)

s367973556

p02261

u150984829

Wrong Answer

Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).

n=int(input()) c=input().split();d=c[:] for i in range(n-1): for j in range(0,n-i-1): if c[j][1]>c[j+1][1]:c[j],c[j+1]=c[j+1],c[j] m=d.index(min(d[i:]),i) if d[i][1]>=d[m][1]:d[i],d[m]=d[m],d[i] print(*c);print("Stable") print(*d) b={i[1]for i in c} print(["Stable","Not stable"][len(b)<len(c)])

s101393687

Accepted

n=int(input()) c=input().split();d=c[:] for i in range(n-1): for j in range(n-i-1): if c[j][1]>c[j+1][1]:c[j:j+2]=c[j+1],c[j] m=i for j in range(i,n): if d[m][1]>d[j][1]:m=j d[i],d[m]=d[m],d[i] print(*c);print("Stable") print(*d);print(['Not s','S'][c==d]+'table')

s647900971

p03610

u544050502

Wrong Answer

You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.

print(input()[1::2])

s474094729

Accepted

print(input()[::2])

s946835948

p02290

u629780968

Wrong Answer

For given three points p1, p2, p, find the projection point x of p onto p1p2.

def dot(a, b): return a.real * b.real + a.imag * b.imag def projection(a, b): return a * dot(a, b) / (abs(a) ** 2) def solve(p0,p1,p2): a=p1-p0 b=p2-p0 pro=projection(a,b) t=p0+pro return t def main(): x0,y0,x1,y1=map(float,input().split()) p0=complex(x0,y0) p1=complex(x1,y1) q=int(input()) for i in range(q): p2=complex(*map(float,input().split())) t=solve(p0,p1,p2) print('{:.10f}{:.10f}'.format(t.real,t.imag)) main()

s273581809

Accepted

def dot(a, b): return a.real * b.real + a.imag * b.imag def projection(a, b): return a * dot(a, b) / (abs(a) ** 2) def solve(p0,p1,p2): a=p1-p0 b=p2-p0 pro=projection(a,b) t=p0+pro return t def main(): x0,y0,x1,y1=map(float,input().split()) p0=complex(x0,y0) p1=complex(x1,y1) q=int(input()) for i in range(q): p2=complex(*map(float,input().split())) t=solve(p0,p1,p2) print("{:.10f} {:.10f}".format(t.real,t.imag)) main()

s575726233

p03387

u513081876

Wrong Answer

You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.

A, B, C = map(int, input().split()) sta = max(A, B, C) ans = 0 if (B-A) >= 2: kake = (B-A)//2 A = A*kake ans += kake if (B-C) >= 2: kake = (B-C)//2 C = C*kake ans += kake if A == B == C: print(ans) elif A == C: print(ans+1) else: print(ans+2)

s590617593

Accepted

A, B, C = map(int, input().split()) ans = 0 max_ABC = max(A, B, C) if max_ABC != A: ans += (max_ABC - A) // 2 A += 2 * ((max_ABC-A) // 2) if max_ABC != B: ans += (max_ABC - B) // 2 B += 2 * ((max_ABC-B) // 2) if max_ABC != C: ans += (max_ABC - C) // 2 C += 2 * ((max_ABC-C) // 2) if 3 * max_ABC - (A+B+C) == 2: ans += 1 elif 3 * max_ABC - (A+B+C) == 1: ans += 2 print(ans)

s228884575

p03862

u281610856

Wrong Answer

There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies. Snuke can perform the following operation any number of times: * Choose a box containing at least one candy, and eat one of the candies in the chosen box. His objective is as follows: * Any two neighboring boxes contain at most x candies in total. Find the minimum number of operations required to achieve the objective.

def main(): n, x = map(int, input().split()) A = list(map(int, input().split())) ans = 0 if A[0] > x: ans += A[0] - x A[0] = x print(ans) for i in range(n - 1): j = i + 1 if j < n and A[i] + A[j] > x: diff = A[i] + A[j] - x ans += diff A[j] -= diff print(ans) if __name__ == '__main__': main()

s405338909

Accepted

def main(): n, x = map(int, input().split()) A = list(map(int, input().split())) ans = 0 if A[0] > x: ans += A[0] - x A[0] = x for i in range(n - 1): j = i + 1 if j < n and A[i] + A[j] > x: diff = A[i] + A[j] - x ans += diff A[j] -= diff print(ans) if __name__ == '__main__': main()

s631253083

p03474

u766566560

Wrong Answer

The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.

import re A, B = map(int, input().split()) S = input() result = re.match('[0-9]{A}-[0-9]{B}', S) if result: print('Yes') else: print('No')

s951446321

Accepted

import re A, B = map(int, input().split()) S = input() pattern = '[0-9]{' + str(A) + '}-[0-9]{' + str(B) + '}' result = re.match(pattern, S) if result: print('Yes') else: print('No')

s806903893

p04030

u099300051

Wrong Answer

Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?

S = input() ansS = "" ansL = 0 for i in range(len(S)): if S[i] == 'B': if len(ansS) != 0 : ansL = len(ansS) ansS = ansS[:ansL-1] else: ansS += S[i] print(int(ansS))

s554271585

Accepted

S = input() ans ='' for x in S: if x == '0': ans += x elif x == '1': ans += x else: if ans!='': ans = ans[:-1] print(ans)

s418929272

p03854

u409757418

Wrong Answer

You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

s = input() n = 0 while n == 0: if s[len(s)-5:len(s)] == "dream": s = s.rstrip("ream") s = s.rstrip("d") elif s[len(s)-5:len(s)] == "erase": s = s.rstrip("rase") s = s.rstrip("e") elif s[len(s)-7:len(s)] == "dreamer": s = s.rstrip("reamer") s = s.rstrip("d") elif s[len(s)-6:len(s)] == "eraser": s = s.rstrip("raser") s = s.rstrip("e") else: if len(s) == 0: n = 1 print("Yes") else: n = -1 print("No")

s551738598

Accepted

s = input() n = 0 while n == 0: if s[len(s)-5:len(s)] == "dream": s = s[:len(s)-5] elif s[len(s)-5:len(s)] == "erase": s = s[:len(s)-5] elif s[len(s)-7:len(s)] == "dreamer": s = s[:len(s)-7] elif s[len(s)-6:len(s)] == "eraser": s = s[:len(s)-6] else: if len(s) == 0: n = 1 print("YES") else: n = -1 print("NO")

s769124263

p03474

u908349502

Wrong Answer

The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.

a,b=map(int,input().split()) s=input() f=1 for i in range(a): if s[i]=='-': f=0 if[a]!='-': f=0 for j in range(b): if s[a+1+j]=='-': f=0 if f: print('Yes') else: print('No')

s828358781

Accepted

a,b=map(int,input().split()) s=input() f=1 for i in range(a): if s[i]=='-': f=0 if s[a]!='-': f=0 for j in range(b): if s[a+1+j]=='-': f=0 if f: print('Yes') else: print('No')

s548944614

p03158

u543954314

Wrong Answer

There are N cards. The i-th card has an integer A_i written on it. For any two cards, the integers on those cards are different. Using these cards, Takahashi and Aoki will play the following game: * Aoki chooses an integer x. * Starting from Takahashi, the two players alternately take a card. The card should be chosen in the following manner: * Takahashi should take the card with the largest integer among the remaining card. * Aoki should take the card with the integer closest to x among the remaining card. If there are multiple such cards, he should take the card with the smallest integer among those cards. * The game ends when there is no card remaining. You are given Q candidates for the value of x: X_1, X_2, ..., X_Q. For each i (1 \leq i \leq Q), find the sum of the integers written on the cards that Takahashi will take if Aoki chooses x = X_i.

n,q = map(int,input().split()) a = list(map(int,input().split())) g = [0]*(n+1) h = [0] + a[1-n%2::2] m = len(h)-1 for i in range(n): g[i+1] = g[i] + a[i] for i in range(m): h[i+1] += h[i] d = {10**10:h[-1]} for j in range(1,(n+1)//2+1): tak = g[n] - g[n-j] + h[(n-2*j)//2+n%2] key = (a[n-j]+a[n-2*j])//2 + 1 d[key] = tak l = list(d.keys()) l.sort() m = len(l) l.append(0) d[0] = g[n]-g[n-(n+1)//2] for _ in range(q): x = int(input()) ok,ng = m,-1 while ok-ng>1: mid = (ng+ok)//2 if l[mid] >= x: ok = mid else: ng = mid print(d[l[ok]])

s485233366

Accepted

import bisect n, q = map(int,input().split()) a = list(map(int,input().split())) a.reverse() cums = [0]*(n+1) cume = [0]*(n+1) for i in range(n): cums[i+1] = cums[i] + a[i] cume[i+1] = cume[i] + a[i]*(1 - i%2) bo = list() sc = list() for i in range(1,(n+1)//2): b = (a[i] + a[2*i])//2 + 1 bo.append(b) s = cums[i] + cume[-1] - cume[2*i] sc.append(s) sc.append(cums[(n+1)//2]) bo.reverse() sc.reverse() for _ in range(q): i = bisect.bisect(bo, int(input())) print(sc[i])

s462785513

p03828

u844646164

Wrong Answer

You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.

import math from collections import Counter from collections import defaultdict mod = 10**9+7 def prime(n): for i in range(2, math.ceil(math.sqrt(n))+1): while n % i == 0: n /= i prime_count[i] += 1 if n > 1: prime_count[int(n)] += 1 n = int(input()) a = 1 prime_count = defaultdict(int) for i in range(1, n+1): prime(i) print(prime_count) ans = 1 for v in prime_count.values(): ans *= (v+1) ans %= mod print(ans)

s991799615

Accepted

import math from collections import Counter from collections import defaultdict mod = 10**9+7 def prime(n): for i in range(2, math.ceil(math.sqrt(n))+1): while n % i == 0: n /= i prime_count[i] += 1 if n > 1: prime_count[int(n)] += 1 n = int(input()) a = 1 prime_count = defaultdict(int) for i in range(1, n+1): prime(i) ans = 1 for v in prime_count.values(): ans *= (v+1) ans %= mod print(ans)

s839383826

p03486

u302292660

Wrong Answer

You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.

s = input() t = input() Ls= [] Lt= [] for i in s: Ls.append(i) Ls = "".join(Ls) for i in t: Lt.append(i) Lt = "".join(Lt) L = [Ls,Lt] L = sorted(L) if L[0]==Ls: print("Yes") else: print("No")

s840286680

Accepted

s = sorted(input()) t = sorted(input())[::-1] Ls= [] Lt= [] if len(s)<len(t) and set(s) == set(s)&set(t): print("Yes") else: for i in s: Ls.append(i) Ls = "".join(Ls) for i in t: Lt.append(i) Lt = "".join(Lt) L = [Ls,Lt] L = sorted(L) if L[0]==L[1]: print("No") elif L[0]==Ls: print("Yes") else: print("No")

s204307484

p03695

u787456042

Wrong Answer

In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.

N,*A=map(int,open(0).read().split());l=[0]*9 for a in A:l[min(8,a//400)]+=1 m=sum([0<l[i]for i in range(8)]);print(min(m,1),m+l[8])

s445359273

Accepted

N,*A=map(int,open(0).read().split());l=[0]*9 for a in A:l[min(8,a//400)]+=1 m=sum([0<l[i]for i in range(8)]);print(max(m,1),m+l[8])

s219974358

p03697

u870297120

Wrong Answer

You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead.

a,b = map(int, input().split()) print(a+b if a+b >= 10 else 'error')

s283015502

Accepted

a,b = map(int, input().split()) print(a+b if a+b < 10 else 'error')

s351569527

p03227

u663710122

Wrong Answer

You are given a string S of length 2 or 3 consisting of lowercase English letters. If the length of the string is 2, print it as is; if the length is 3, print the string after reversing it.

S = input() print(S if len(S) == 2 else S[-1])

s734902329

Accepted

S = input() print(S if len(S) == 2 else S[::-1])

s040592988

p04043

u843318346

Wrong Answer

Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

arr = list(map(int,input().split())) if arr[0]==5 and arr[1]==7 and arr[2]==5: print('YES') else: print('NO')

s727935527

Accepted

arr = list(map(int,input().split())) if arr.count(5)==2 and arr.count(7)==1: print('YES') else: print('NO')

s040278991

p03472

u924374652

Time Limit Exceeded

You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?

n, h = map(int, input().split()) list_w = [] max_a = [-1, -1, 10**9] for i in range(n): a, b = list(map(int, input().split())) list_w.append([a, b]) if max_a[1] < a: max_a = [i, a, b] elif max_a[1] == a and max_a[2] > b: max_a = [i, a, b] list_w.pop(max_a[0]) list_w.sort(key=lambda x:x[1], reverse=True) count = 0 while h > 0: if h > max_a[2]: if len(list_w) > 0 and list_w[0][1] > max_a[1]: h -= list_w[0][1] list_w.pop(0) count += 1 else: if max_a[2] > max_a[1]: num = (h - max_a[2]) // max_a[1] h -= max_a[1] * num count += num else: num = h // max_a[1] h -= max_a[1] * num count += num else: h -= max_a[2] count += 1 print(count)

s783726510

Accepted

import math n, h = map(int, input().split()) list_a = [] list_b = [] for i in range(n): a, b = list(map(int, input().split())) list_a.append(a) list_b.append(b) max_a = max(list_a) use_b = [b for b in list_b if b > max_a] use_b.sort(reverse=True) count = 0 while h > 0: if len(use_b) > 0: h -= use_b.pop(0) count += 1 else: count += math.ceil(h / max_a) break print(count)

s168373250

p03380

u740284863

Wrong Answer

Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.

import math import itertools def combinations_count(n, r): return math.factorial(n) // (math.factorial(n - r) * math.factorial(r)) n = int(input()) a = list(map(int,input().split())) b = list(itertools.combinations(a,2)) ans = [0,(0,0)] #print(b) for i in range(len(b)): tmpans = combinations_count(max(b[i][0], b[i][1]),min(b[i][0], b[i][1])) if tmpans >= ans[0]: ans = [tmpans,(b[i][0], b[i][1])] print(ans[1][0],ans[1][1])

s642913715

Accepted

n = int(input()) a = list(map(int,input().split())) m = max(a) del a[a.index(m)] dis = [abs(m/2 - a[i]) for i in range(n-1)] r = a[dis.index(min(dis))] print(m,r)

s597648333

p03779

u113971909

Wrong Answer

There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.

X=int(input()) from collections import deque def dfs_2(start,goal): q = deque([]) q.append([start]) flg = 0 while flg==0: p = q.popleft() 取りだし（先頭） jp = len(p) if p[-1]-jp == goal or p[-1]+jp == goal: print(p) return jp else: p0 = [p[-1]] pp = [p[-1]+jp] pm = [p[-1]-jp] q.append(p+pp)追加 q.append(p+pm) q.append(p+p0) print(dfs_2(0,X))

s500038276

Accepted

X=int(input()) s=0 for t in range(X+1): s+=t if s>=X: print(t) break

s598667018

p02281

u939814144

Wrong Answer

Binary trees are defined recursively. A binary tree _T_ is a structure defined on a finite set of nodes that either * contains no nodes, or * is composed of three disjoint sets of nodes: \- a root node. \- a binary tree called its left subtree. \- a binary tree called its right subtree. Your task is to write a program which perform tree walks (systematically traverse all nodes in a tree) based on the following algorithms: 1. Print the root, the left subtree and right subtree (preorder). 2. Print the left subtree, the root and right subtree (inorder). 3. Print the left subtree, right subtree and the root (postorder). Here, the given binary tree consists of _n_ nodes and evey node has a unique ID from 0 to _n_ -1.

class Node: def __init__(self, node_id, left_child, right_child): self.node_id = node_id self.parent = -1 self.sibling = -1 self.degree = 0 self.depth = 0 self.height = 0 self.node_type = '' self.left_child = left_child self.right_child = right_child def binarytrees(): def input_node_list(): n = int(input()) for i in range(n): node_id, left_child, right_child = list(map(int, input().split())) node = Node(node_id, left_child, right_child) node_list.append(node) node_list.sort(key = lambda node: node.node_id) def calc_parent_sibling_degree(): if node.left_child != -1: node_list[node.left_child].parent = node.node_id node_list[node.left_child].sibling = node.right_child node.degree += 1 if node.right_child != -1: node_list[node.right_child].parent = node.node_id node_list[node.right_child].sibling = node.left_child node.degree += 1 def calc_depth(node_id): if node_list[node_id].parent == -1: return 0 else: return calc_depth(node_list[node_id].parent) + 1 def calc_height(node_id): if node_list[node_id].degree == 0: return 0 else: left_height = right_height = 0 if node_list[node_id].left_child != -1: left_height = calc_height(node_list[node_id].left_child) if node_list[node_id].right_child != -1: right_height = calc_height(node_list[node_id].right_child) return max(left_height, right_height) + 1 def calc_node_type(): if node.depth == 0: node.node_type = 'root' elif node.height == 0: node.node_type = 'leaf' else: node.node_type = 'internal node' def show_node_info(): print('node {}: parent = {}, sibling = {}, degree = {}, depth = {}, height = {}, {}' .format(node.node_id, node.parent, node.sibling, node.degree, node.depth, node.height, node.node_type)) node_list = [] input_node_list() for node in node_list: calc_parent_sibling_degree() for node in node_list: node.depth = calc_depth(node.node_id) node.height = calc_height(node.node_id) calc_node_type() show_node_info() return node_list def treewalk(): def show_preorder(node_id): current_node = node_list[node_id] print(' {}'.format(current_node.node_id), end='') if current_node.left_child != -1: show_preorder(current_node.left_child) if current_node.right_child != -1: show_preorder(current_node.right_child) def show_inorder(node_id): current_node = node_list[node_id] if current_node.left_child != -1: show_inorder(current_node.left_child) print(' {}'.format(current_node.node_id), end='') if current_node.right_child != -1: show_inorder(current_node.right_child) def show_postorder(node_id): current_node = node_list[node_id] if current_node.left_child != -1: show_postorder(current_node.left_child) if current_node.right_child != -1: show_postorder(current_node.right_child) print(' {}'.format(current_node.node_id), end='') node_list = binarytrees() print('Preorder') show_preorder(0) print('') print('Inorder') show_inorder(0) print('') print('Postorder') show_postorder(0) print('') if __name__ == '__main__': treewalk()

s363025456

Accepted

class Node: def __init__(self, node_id, left_child, right_child): self.node_id = node_id self.parent = -1 self.sibling = -1 self.degree = 0 self.depth = 0 self.height = 0 self.node_type = '' self.left_child = left_child self.right_child = right_child def binarytrees(): def input_node_list(): n = int(input()) for i in range(n): node_id, left_child, right_child = list(map(int, input().split())) node = Node(node_id, left_child, right_child) node_list.append(node) node_list.sort(key = lambda node: node.node_id) def calc_parent_sibling_degree(): if node.left_child != -1: node_list[node.left_child].parent = node.node_id node_list[node.left_child].sibling = node.right_child node.degree += 1 if node.right_child != -1: node_list[node.right_child].parent = node.node_id node_list[node.right_child].sibling = node.left_child node.degree += 1 def calc_depth(node_id): if node_list[node_id].parent == -1: return 0 else: return calc_depth(node_list[node_id].parent) + 1 def calc_height(node_id): if node_list[node_id].degree == 0: return 0 else: left_height = right_height = 0 if node_list[node_id].left_child != -1: left_height = calc_height(node_list[node_id].left_child) if node_list[node_id].right_child != -1: right_height = calc_height(node_list[node_id].right_child) return max(left_height, right_height) + 1 def calc_node_type(): if node.depth == 0: node.node_type = 'root' elif node.height == 0: node.node_type = 'leaf' else: node.node_type = 'internal node' def show_node_info(): print('node {}: parent = {}, sibling = {}, degree = {}, depth = {}, height = {}, {}' .format(node.node_id, node.parent, node.sibling, node.degree, node.depth, node.height, node.node_type)) node_list = [] input_node_list() for node in node_list: calc_parent_sibling_degree() # for node in node_list: # node.depth = calc_depth(node.node_id) # node.height = calc_height(node.node_id) # calc_node_type() # show_node_info() return node_list def treewalk(): def show_preorder(node_id): current_node = node_list[node_id] print(' {}'.format(current_node.node_id), end='') if current_node.left_child != -1: show_preorder(current_node.left_child) if current_node.right_child != -1: show_preorder(current_node.right_child) def show_inorder(node_id): current_node = node_list[node_id] if current_node.left_child != -1: show_inorder(current_node.left_child) print(' {}'.format(current_node.node_id), end='') if current_node.right_child != -1: show_inorder(current_node.right_child) def show_postorder(node_id): current_node = node_list[node_id] if current_node.left_child != -1: show_postorder(current_node.left_child) if current_node.right_child != -1: show_postorder(current_node.right_child) print(' {}'.format(current_node.node_id), end='') node_list = binarytrees() root_id = 0 for node in node_list: if node.parent == -1: root_id = node.node_id print('Preorder') show_preorder(root_id) print('') print('Inorder') show_inorder(root_id) print('') print('Postorder') show_postorder(root_id) print('') if __name__ == '__main__': treewalk()

s078574243

p04045

u013408661

Wrong Answer

Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K. She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change). However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money. Find the amount of money that she will hand to the cashier.

n,k=map(int,input().split()) d=list(map(int,input().split())) number=[] for i in range(10): number.append(i) likenumber=[i for i in number if i not in d] n=list(str(n)) n=list(map(int,n)) for i in range(len(n)-1,-1,-1): if n[i]>9: if i==len(n)-1: n.insert(0,0) n[i]-=10 n[i-1]+=1 if n[i] in d: for j in likenumber: if j<n[i]: n[i]=j continue n[i]=min(likenumber) n[i-1]+=1 print("".join(map(str,n)))

s118033173

Accepted

n,k=map(int,input().split()) d=list(map(int,input().split())) for i in range(n,10**6): flag=True for j in str(i): if int(j) in d: flag=False break if flag: print(i) exit()

s906114495

p03360

u335278042

Wrong Answer

There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?

lis = list(map(int,input().split())) K = int(input()) s=sum(lis)-max(lis)+max(lis)**K print(s)

s462732279

Accepted

lis = list(map(int,input().split())) K = int(input()) s=sum(lis)-max(lis)+max(lis)*2**K print(s)

s740833088

p02936

u818078165

Wrong Answer

Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.

n,q = map(int,input().split()) #print(n,q) edge=[] for i in range(n-1): a = list(map(int,input().split())) edge.append(a) add_list = [] for i in range(q): p = list(map(int,input().split())) add_list.append(p) #print(edge,add_list) ans = [0] * n edge_list = [[] for i in range(n)] for e in edge: edge_list[e[1]-1].append(e[0]-1) add_value = [0]*n for v in add_list: add_value[v[0]-1] += v[1] #print(edge_list) #print(add_value) dp = [-1] * n def get_value(i): if(dp[i]!=-1): return dp[i] tmp = add_value[i] for e in edge_list[i]: tmp += get_value(e) dp[i] = tmp return dp[i] for i in range(n): ans[i] = get_value(i) #print(i,ans) print(ans)

s531227673

Accepted

import sys input = sys.stdin.readline sys.setrecursionlimit(900000) n, q = map(int, input().split()) #print(n,q) edge = [] for i in range(n-1): a = list(map(int, input().split())) edge.append(a) add_list = [] for i in range(q): p = list(map(int, input().split())) add_list.append(p) #print(edge,add_list) dir_edge = [[] for i in range(n)] for e in edge: dir_edge[e[0]-1].append(e[1]-1) dir_edge[e[1]-1].append(e[0]-1) #print(dir_edge) ans = [0]*n for v in add_list: ans[v[0]-1] += v[1] #print(add_value) def dfs(index, parent): if parent != -1: ans[index] += ans[parent] for i in dir_edge[index]: if (parent != i): dfs(i,index) dfs(0, -1) print(*ans)

s776043285

p03479

u853900545

Wrong Answer

As a token of his gratitude, Takahashi has decided to give his mother an integer sequence. The sequence A needs to satisfy the conditions below: * A consists of integers between X and Y (inclusive). * For each 1\leq i \leq |A|-1, A_{i+1} is a multiple of A_i and strictly greater than A_i. Find the maximum possible length of the sequence.

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

s603880313

Accepted

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

s707821038

p04030

u016901717

Wrong Answer

Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?

t="" for i in input(): if i=="B": t=t[:-1] else: t+=i print(t)

s350277935

Accepted

t="" for i in input(): if i=="B": t=t[:-1] else: t+=i print(t)

s001981641

p03378

u109796335

Wrong Answer

There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.

input_data = input().replace("\n", " ").split(" ") n, m, x = int(input_data[0]), int(input_data[1]), int(input_data[2]) def problemB(n, m, x): masu_list = [0] * (n+1) adder = input_data[3:] for i in adder: masu_list[int(i)] = 1 front = sum(masu_list[:x]) back = sum(masu_list[x+1:]) if front <= back: return print(front) else: return print(back) problemB(n, m, x)

s167711567

Accepted

input_front = input().split(" ") input_back = input().split(" ") n, x = int(input_front[0]), int(input_front[2]) def problemB(n, x): masu_list = [0] * (n+1) adder = input_back for i in adder: masu_list[int(i)] = 1 front = sum(masu_list[:x]) back = sum(masu_list[x+1:]) if front <= back: return print(front) else: return print(back) problemB(n, x)

s398879114

p03854

u034782764

Wrong Answer

You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

s=input() a=('dream','dreamer','erase','eraser') s_2=s[::-1] a_2=(a[0][::-1],a[1][::-1],a[2][::-1],a[3][::-1]) #print(s_2) #print(a_2) while len(s_2) > 0: judge = [s_2.startswith(i) for i in a_2] #print(judge) if any(judge)==True: target=a[judge.index(True)] s_2=s_2[len(target):] if len(s_2)==0: print('Yes') else: print('NO') break

s163210971

Accepted

given = ["dreamer", "dream", "eraser", "erase"] S = input() while len(S) > 0: judge = [S.endswith(i) for i in given] if any(judge): target = given[judge.index(True)] S = S[:-len(target)] if len(S) == 0: print("YES") break else: print("NO") break

s549992674

p02663

u619551113

Wrong Answer

In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?

li = list(map(int,input().split())) h1, m1, h2, m2, k = li life_time = (h2 - h1)*60 + (m2 - m1) time = life_time - k

s620503572

Accepted

li = list(map(int,input().split())) h1, m1, h2, m2, k = li life_time = (h2 - h1)*60 + (m2 - m1) time = life_time - k print(time)

s113127252

p03544

u178432859

Wrong Answer

It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)

l = [0,1] n = int(input()) for i in range(2,n+1): l.append(l[i-1] + l[i-2]) print(l[-1])

s347692573

Accepted

l = [2,1] n = int(input()) for i in range(n-1): l.append(l[i] + l[i+1]) print(l[-1])

s263958350

p03997

u516579758

Wrong Answer

You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.

a=int(input()) b=int(input()) h=int(input()) print((a+b)*h/2)

s116840793

Accepted

a=int(input()) b=int(input()) h=int(input()) print((a+b)*h//2)

s847165915

p03163

u219417113

Wrong Answer

There are N items, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), Item i has a weight of w_i and a value of v_i. Taro has decided to choose some of the N items and carry them home in a knapsack. The capacity of the knapsack is W, which means that the sum of the weights of items taken must be at most W. Find the maximum possible sum of the values of items that Taro takes home.

import numpy as np n, W = map(int, input().split()) w, v = [0] * n, [0] * n for i in range(n): w[i], v[i] = map(int, input().split()) ndp = np.zeros(W + 1, dtype=np.int64) print(ndp) for w, v in zip(w, v): np.maximum(ndp[:-w] + v, ndp[w:], out=ndp[w:]) print(ndp[-1])

s319747855

Accepted

import numpy as np n, W = map(int, input().split()) w, v = [0] * n, [0] * n for i in range(n): w[i], v[i] = map(int, input().split()) ndp = np.zeros(W + 1, dtype=np.int64) for w, v in zip(w, v): np.maximum(ndp[:-w] + v, ndp[w:], out=ndp[w:]) print(ndp[-1])

s042699889

p02614

u253093300

Wrong Answer

We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.

H, W, K = map(int, input().split(' ')) board = [] for _ in range(H): board.append([c for c in input()]) print('1')

s199691419

Accepted

H, W, K = map(int, input().split(' ')) mat = [] for _ in range(H): mat.append(input()) ans = 0 for ib in range(1<<H): for jb in range(1<<W): count = 0 for i in range(H): for j in range(W): if ib>>i&1: continue if jb>>j&1: continue if mat[i][j] == '#': count += 1 if count == K: ans += 1 print(ans)

s145017746

p03048

u248670337

Wrong Answer

Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this?

r,g,b,n=map(int,input().split()) c=0 for i in range(n//r+1): a=r*i for j in range(a,n//g+1): a+=g*j if (n-a)%b: c+=1 print(c)

s308322989

Accepted

r,g,b,n=map(int,input().split()) a=0 for i in range(n//r+1): x=r*i for j in range((n-x)//g+1): if (n-x-g*j)%b==0: a+=1 print(a)

s127016682

p03555

u207799478

Wrong Answer

You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.

a=input() b=input() if a[0]==b[2] and a[1]==b[1] and a[2]==b[0]: print('Yes') else: print('No')

s523387918

Accepted

# coding: utf-8 # Your code here! a=input() b=input() if a[0]==b[2] and a[1]==b[1] and a[2]==b[0]: print('YES') else: print('NO')

s641772826

p03854

u811176339

Wrong Answer

You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

lt = ["dream", "dreamer", "erase", "eraser"] s = input() lt = [w[::-1] for w in lt] s = s[::-1] cond = True i = 0 while i < len(s): for t in lt: if s[i:].startswith(t): i += len(t) break else: cond = False break print("Yes" if cond else "No")

s139566402

Accepted

lt = ["dream", "dreamer", "erase", "eraser"] s = input() lt = [w[::-1] for w in lt] s = s[::-1] cond = True i = 0 while i < len(s): for t in lt: if s[i:].startswith(t): i += len(t) break else: cond = False break print("YES" if cond else "NO")

s685673583

p03761

u088552457

Wrong Answer

Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string. He will receive a headline which contains one of the strings S_1,...,S_n tomorrow. He is excited and already thinking of what string he will create. Since he does not know the string on the headline yet, he is interested in strings that can be created regardless of which string the headline contains. Find the longest string that can be created regardless of which string among S_1,...,S_n the headline contains. If there are multiple such strings, find the lexicographically smallest one among them.

import collections n = int(input()) s = [input() for i in range(n)] if n == 1: print("".join(sorted(s[0]))) else: c = [] for i in s: c.append(collections.Counter(i)) a = c[0] for i in range(1, n): a &= c[i] ans = [] a = a.most_common() print(a) for i in a: for j in range(i[1]): ans.append(i[0]) print("".join(sorted(ans)))

s823414131

Accepted

n = int(input()) ss = [] mins = 'a'*100 for _ in range(n): s = str(''.join(sorted(input()))) ss.append(s) if len(mins) > len(s): mins = s ans = '' for w in mins: ok = True for i, s in enumerate(ss[:]): if w in s: wi = s.find(w) ss[i] = s[:wi] + s[wi+1:] else: ok = False if ok: ans += w print(ans)

s894706038

p02972

u089142196

Wrong Answer

There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.

N=int(input()) A=list(map(int, input().split())) A.insert(0,0) total=sum(A) sum=0 count=0 OK_Flag=True OK_list=[] for i in range(1,N+1): sum=0 for j in range(i,N+1,i): if OK_Flag==True: #print(i,A[j-1]) sum=sum+A[j] #print(i,j,sum) #print(sum,sum%2,A[i]) if sum%2==A[i]: if A[i]==1: count=count+1 OK_list.append(A[i]) else: OK_Flag=False if OK_Flag==True: print(count) if count!=0: print(' '.join(map(str, OK_list))) else: print(-1)

s327981141

Accepted

N=int(input()) a=list(map(int, input().split())) b=[0]*N for i in range(N-1,-1,-1): summ = sum(b[2*i+1::i+1]) if summ%2==0: b[i] = a[i] else: b[i] = 1-a[i] #print(i,summ,b[i]) ans=[] print(sum(b)) for i,num in enumerate(b): if num==1: ans.append(i+1) print(*ans)

s109176971

p02613

u005569385

Wrong Answer

Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.

N = int(input()) s = [input() for i in range(N)] a = s.count("AC") b = s.count("WA") c = s.count("TLE") d = s.count("RE") print("AC * {}".format(a)) print("WA * {}".format(b)) print("TLE * {}".format(c)) print("RE * {}".format(d))

s008194399

Accepted

N = int(input()) s = [input() for i in range(N)] a = s.count("AC") b = s.count("WA") c = s.count("TLE") d = s.count("RE") print("AC x {}".format(a)) print("WA x {}".format(b)) print("TLE x {}".format(c)) print("RE x {}".format(d))

s739094443

p03470

u039192119

Wrong Answer

An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?

print(len([int(input()) for i in range(int(input()))]))

s127030418

Accepted

a=[int(input()) for i in range(int(input()))] print(len(set(a)))

s315303295

p03943

u860002137

Wrong Answer

Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

a, b, c = map(int, input().split()) print("Yes") if (a + b + c) // 2 == 0 else print("No")

s313268042

Accepted

a, b, c = map(int, input().split()) whole = a + b + c print("Yes") if (whole / 2) == max([a, b, c]) else print("No")

s282183703

p02401

u925992597

Wrong Answer

Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.

while True: try: print(eval(input())) except: break

s401345194

Accepted

while True: try: print(int(eval(input()))) except: break

s027473613

p02578

u914802579

Wrong Answer

N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.

n=int(input()) l=list(map(int,input().split())) mx=l[0] an=0 for i in range(1,n): if l[i]<mx: an+=mx-l[i] else: l[i]=mx print(an)

s538777833

Accepted

n=int(input()) l=list(map(int,input().split())) mx=l[0] an=0 for i in range(1,n): if l[i]<mx: an+=mx-l[i] else: mx=l[i] print(an)

s998533189

p03696

u039623862

Wrong Answer

You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.

n = int(input()) s = input() cnt = [0, 0] ans = '' for i in range(n): if s[i] == ')': cnt[0] += 1 else: cnt[1] += 1 print('(' * cnt[0] + s + ')'*cnt[1])

s736450027

Accepted

n = int(input()) s = input() cnt = [0, 0] add = [0, 0] ans = '' for i in range(n): if s[i] == ')': cnt[0] += 1 if cnt[0] > cnt[1]: add[1] += 1 cnt[1] += 1 else: cnt[1] += 1 print('(' * add[1] + s + ')'*(cnt[1] -cnt[0]))

s875244795

p00045

u301729341

Time Limit Exceeded

販売単価と販売数量を読み込んで、販売金額の総合計と販売数量の平均を出力するプログラムを作成してください。

P_Sum = 0 N_Sum = 0 Kazu = 0 while True: try: pri,num = map(int,input().split(",")) P_Sum += num * pri N_Sum += num Kazu += 1 except EOFError: print(P_Sum) print(round(N_Sum / Kazu))

s594649971

Accepted

P_Sum = 0 N_Sum = 0 Kazu = 0 while True: try: pri,num = map(int,input().split(",")) P_Sum += num * pri N_Sum += num Kazu += 1 except(EOFError,ValueError): print(P_Sum) print(int((N_Sum / Kazu)+ 0.5)) break

s720520842

p03048

u254050469

Wrong Answer

Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this?

n = [int(i) for i in input().split() ] rgb = n[0:3] rgb.sort(reverse=True) #rgb = list(filter(lambda x:x >= n[3],rgb)) rm = int(n[3]/rgb[0] ) + 1 c = 0 for i in range(rm): print(i) if i * rgb[0] == n[3]: c += 1 break for j in range(int((n[3] - i * rgb[0])/rgb[1]) + 1): if (n[3] - i * rgb[0] - j * rgb[1]) % rgb[2] == 0: c += 1 print (c)

s231917311

Accepted

n = [int(i) for i in input().split() ] rgb = n[0:3] rgb.sort(reverse=True) #rgb = list(filter(lambda x:x >= n[3],rgb)) rm = int(n[3]/rgb[0] ) + 1 c = 0 for i in range(rm): if i * rgb[0] == n[3]: c += 1 break for j in range(int((n[3] - i * rgb[0])/rgb[1]) + 1): if (n[3] - i * rgb[0] - j * rgb[1]) % rgb[2] == 0: c += 1 print (c)

s882700153

p03494

u248670337

Time Limit Exceeded

There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.

input() A=list(map(int,input().split())) ans=0 while all(a%2==0 for a in A): ans+=1 print(ans)

s512479952

Accepted

input() A=list(map(int,input().split())) ans=0 while all(a%2==0 for a in A): A=[a/2 for a in A] ans+=1 print(ans)