Datasets:

TnT

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CodeNet4Repair

like 0

s577531731

p03999

u280978334

Wrong Answer

You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion. All strings that can be obtained in this way can be evaluated as formulas. Evaluate all possible formulas, and print the sum of the results.

S = input() ans = 0 digit = len(S)-1 for i in range(2**digit): bit = format(i,"0"+str(digit)+"b") temp = [] num = S[0] for j in range(digit): if j == digit - 1: if bit[j] == "1": temp.append(int(num)) temp.append(int(S[j+1])) else: temp.append(int(num+S[j+1:])) else: if bit[j] == "1": temp.append(int(num)) num = S[j+1] else: num += S[j+1] print(temp) ans += sum(temp) print(ans)

s109484778

Accepted

S = input() ans = 0 digit = len(S)-1 for i in range(2**digit): bit = format(i,"0"+str(digit)+"b") temp = [] num = S[0] for j in range(digit): if j == digit - 1: if bit[j] == "1": temp.append(int(num)) temp.append(int(S[j+1])) else: temp.append(int(num+S[j+1:])) else: if bit[j] == "1": temp.append(int(num)) num = S[j+1] else: num += S[j+1] ans += sum(temp) print(int(S) if digit == 0 else ans)

s571501408

p04029

u194894739

Wrong Answer

There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?

N = int(input()) print(N*(N-1)//2)

s668270747

Accepted

N = int(input()) print(N*(N+1)//2)

s391249543

p02281

u255317651

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.

# -*- coding: utf-8 -*- """ Created on Sat May 19 19:34:44 2018 ALDS1_7_B @author: maezawa """ n = int(input()) parent = [-1 for _ in range(n)] left = [-1 for _ in range(n)] right = [-1 for _ in range(n)] height = [None for _ in range(n)] depth = [None for _ in range(n)] def set_parent(): global parent for i in range(n): if left[i] != -1: parent[left[i]] = i if right[i] != -1: parent[right[i]] = i def set_depth(i, p): global depth if depth[i] != None: return depth[i] = p if right[i] != None: set_depth(right[i], p) if left[i] != None: set_depth(left[i], p+1) def get_depth(i): global depth if depth[i] != None: return depth[i] d = 0 u = i while parent[u] != -1: u = parent[u] d += 1 depth[i] = d return d def set_height(u): global height h1 = 0 h2 = 0 if right[u] != -1: h1 = set_height(right[u])+1 if left[u] != -1: h2 = set_height(left[u])+1 height[u] = max(h1, h2) return height[u] def get_sib(i): if right[parent[i]] == i: return left[parent[i]] else: return right[parent[i]] def get_deg(i): deg = 0 if right[i] != -1: deg += 1 if left[i] != -1: deg += 1 return deg def get_height(i): return height[i] def preparse(i): if i == -1: return print('{} '.format(i), end='') preparse(left[i]) preparse(right[i]) def inparse(i): if i == -1: return inparse(left[i]) print('{} '.format(i), end='') preparse(right[i]) def postparse(i): if i == -1: return postparse(left[i]) postparse(right[i]) print('{} '.format(i), end='') for i in range(n): line = list(map(int, input().split())) left[line[0]] = line[1] right[line[0]] = line[2] set_parent() root = parent.index(-1) print('Preorder') preparse(root) print() print('Inorder') inparse(root) print() print('Postorder') postparse(root) print()

s529787161

Accepted

# -*- coding: utf-8 -*- """ Created on Sat May 19 19:34:44 2018 ALDS1_7_B @author: maezawa """ n = int(input()) parent = [-1 for _ in range(n)] left = [-1 for _ in range(n)] right = [-1 for _ in range(n)] height = [None for _ in range(n)] depth = [None for _ in range(n)] def set_parent(): global parent for i in range(n): if left[i] != -1: parent[left[i]] = i if right[i] != -1: parent[right[i]] = i def set_depth(i, p): global depth if depth[i] != None: return depth[i] = p if right[i] != None: set_depth(right[i], p) if left[i] != None: set_depth(left[i], p+1) def get_depth(i): global depth if depth[i] != None: return depth[i] d = 0 u = i while parent[u] != -1: u = parent[u] d += 1 depth[i] = d return d def set_height(u): global height h1 = 0 h2 = 0 if right[u] != -1: h1 = set_height(right[u])+1 if left[u] != -1: h2 = set_height(left[u])+1 height[u] = max(h1, h2) return height[u] def get_sib(i): if right[parent[i]] == i: return left[parent[i]] else: return right[parent[i]] def get_deg(i): deg = 0 if right[i] != -1: deg += 1 if left[i] != -1: deg += 1 return deg def get_height(i): return height[i] def preparse(i): if i == -1: return print(' {}'.format(i), end='') preparse(left[i]) preparse(right[i]) def inparse(i): if i == -1: return inparse(left[i]) print(' {}'.format(i), end='') inparse(right[i]) def postparse(i): if i == -1: return postparse(left[i]) postparse(right[i]) print(' {}'.format(i), end='') for i in range(n): line = list(map(int, input().split())) left[line[0]] = line[1] right[line[0]] = line[2] set_parent() root = parent.index(-1) print('Preorder') preparse(root) print() print('Inorder') inparse(root) print() print('Postorder') postparse(root) print()

s605883446

p02578

u550416338

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()) A = [int(i) for i in input().split()] for i in range(N-1): if A[i] > A[i+1]: S = A[i] - A[i+1] K = A[i+1] + S A[i+1] = K print(A)

s544193680

Accepted

N = int(input()) A = [int(i) for i in input().split()] K = 0 for i in range(N-1): if A[i] > A[i+1]: S = A[i] - A[i+1] A[i+1] += S K += S else: pass print(K)

s410965421

p03860

u978494963

Wrong Answer

Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.

r = "" map(lambda x: r+x, str(input()).split(" ")) print(r)

s227033441

Accepted

r = "" for s in str(input()).split(" "): r+=s[0] print(r)

s831040694

p03730

u044632922

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`.

a, b, c = map(int, input().split()) x = a % b for i in range(1, b+1): if x*i % b == c: print('Yes') exit() print('No')

s837800106

Accepted

a, b, c = map(int, input().split()) x = a % b for i in range(1, b+1): if x*i % b == c: print('YES') exit() print('NO')

s988664175

p02694

u865108308

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()) money = 100 year = 0 while money == x: money += money / 100 year += 1 print(year)

s799125575

Accepted

x = int(input()) money = 100 year = 0 while money < x: money += money // 100 year += 1 print(year)

s401393968

p02406

u698693989

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; }

x=int(input()) print(" ",end="") for i in range(x): if (i+1)%3==0: print(i+1,end=" ")

s190182814

Accepted

n=int(input()) for i in range(1,n+1): if i % 3 == 0 or "3" in str(i): print(" {0}".format(i),end="") print("")

s725781198

p03386

u691018832

Wrong Answer

Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.

a, b, k = map(int, input().split()) match = 2*(b-a) if k >= match: for i in range(a, b): print(i) else: for i in range(a, a+k): print(i) for j in range(b-k, b): print(j)

s257113861

Accepted

a, b, k = map(int, input().split()) if 2*k >= b-a+1: for i in range(a, b+1): print(i) else: for i in range(a, a+k): print(i) for j in range(b-k+1, b+1): print(j)

s901581913

p03069

u217940964

Wrong Answer

There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.

import numpy as np N = int(input()) S = input() L = [0] * N L1 = [0] * N L2 = [0] * N L3 = [0] * N for i in range(N): L1[i] = 0 L2[i] = 1 L3[i] = 0 if S[i] == ".": # white L[i] = 0 else: L[i] = 1 L3[N-1] = 1 L = np.array(L) L1 = np.array(L1) L2 = np.array(L2) L3 = np.array(L3) a = [((L - L1)**2).sum(), ((L - L2)**2).sum(), ((L - L3)**2).sum()] print(a) ans = min(a) print(ans)

s945394481

Accepted

N = int(input()) S = input() L = [0] * N for i in range(N): if S[i] == "#": L[i] = 1 else: L[i] = 0 leftblack = 0 rightwhite = len([x for x in L if x == 0]) ans = leftblack + rightwhite for i in range(N): if S[i] == "#": leftblack += 1 else: rightwhite -=1 ans = min(ans, leftblack + rightwhite) print(ans)

s175746288

p03544

u393512980

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)

a,b=2,1 n=int(input()) for _ in range(n-2): b=a+b a=b print(b)

s487687190

Accepted

a,b=2,1 n=int(input()) for _ in range(n-1): t=b b=a+b a=t print(b)

s869409664

p02601

u206890818

Wrong Answer

M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.

import math import collections import bisect from collections import deque from copy import copy, deepcopy def main(): A, B, C = map(int, input().split()) K = int(input()) ans = 0 tmp1 = math.ceil(math.log2((A + 0.1) / B)) if tmp1 > 0: B = B * (2**tmp1) ans += tmp1 tmp2 = math.ceil(math.log2((B + 0.1) / C)) if tmp2 > 0: ans += tmp2 #print(tmp1, tmp2) if ans <= K: print("YES") else: print("NO") if __name__ == '__main__': main()

s015898968

Accepted

import math import collections import bisect from collections import deque from copy import copy, deepcopy def main(): A, B, C = map(int, input().split()) K = int(input()) ans = 0 tmp1 = math.ceil(math.log2((A + 0.1) / B)) if tmp1 > 0: B = B * (2**tmp1) ans += tmp1 tmp2 = math.ceil(math.log2((B + 0.1) / C)) if tmp2 > 0: ans += tmp2 #print(tmp1, tmp2) if ans <= K: print("Yes") else: print("No") if __name__ == '__main__': main()

s127728933

p02233

u255317651

Wrong Answer

Write a program which prints $n$-th fibonacci number for a given integer $n$. The $n$-th fibonacci number is defined by the following recursive formula: \begin{equation*} fib(n)= \left \\{ \begin{array}{ll} 1 & (n = 0) \\\ 1 & (n = 1) \\\ fib(n - 1) + fib(n - 2) & \\\ \end{array} \right. \end{equation*}

# -*- coding: utf-8 -*- """ Created on Fri May 4 20:13:10 2018 ALDS1_10_A @author: maezawa """ fib_arr = [None for _ in range(45)] fib_arr[1] = 1 fib_arr[2] = 1 def fib(n): if fib_arr[n] != None: return fib_arr[n] fib_arr[n] = fib(n-1)+fib(n-2) return fib_arr[n] n = int(input()) print(fib(n))

s945086174

Accepted

# -*- coding: utf-8 -*- """ Created on Fri May 4 20:13:10 2018 ALDS1_10_A @author: maezawa """ fib_arr = [None for _ in range(45)] fib_arr[0] = 1 fib_arr[1] = 1 def fib(n): if fib_arr[n] != None: return fib_arr[n] fib_arr[n] = fib(n-1)+fib(n-2) return fib_arr[n] n = int(input()) print(fib(n))

s588056435

p03194

u096100666

Wrong Answer

There are N integers a_1, a_2, ..., a_N not less than 1. The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P. Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.

import math N=4 P=972439611840 a=math.sqrt(P) res=1 for i in range(1,int(a)): if P % i**N: continue res = i print(res)

s251245658

Accepted

n,p = list(map(int,input().split())) i = a = o = 1 m = round(p**(1/n)) for i in range(m,0,-1): a = i**n if p%a==0: print(i) exit()

s598025841

p03448

u989074104

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()) count=0 a_max=min(x//500,a) for i in range(a_max+1): b_max=min((x-500*i)//100,b) print("100円玉最大数",b_max) for j in range(b_max+1): # print("i,j",i,j) if (x-500*i-100*j)//50<=c: count+=1 print(count)

s566402550

Accepted

a=int(input()) b=int(input()) c=int(input()) x=int(input()) count=0 a_max=min(x//500,a) for i in range(a_max+1): b_max=min((x-500*i)//100,b) for j in range(b_max+1): # print("i,j",i,j) if (x-500*i-100*j)//50<=c: count+=1 print(count)

s076610841

p02612

u236441175

Wrong Answer

We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.

N = int(input()) otsuri = N % 1000 print(otsuri)

s088200015

Accepted

N = int(input()) otsuri = N % 1000 if otsuri == 0: print(0) else: print(1000-otsuri)

s460224671

p02645

u250703962

Wrong Answer

When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him.

x = input() print(x[:4])

s081687002

Accepted

s = input() print(s[:3].lower())

s281397869

p02608

u986190948

Wrong Answer

Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).

n=int(input()) ans=[0]*n x=1 y=1 z=1 c=0 for i in range(1,n+1): if 6*i*i>n: d=i break; print(d) for i in range(1,n+1): if 1+2*i*i+i+i+i*i>n: e=i break; print(e) for i in range(d): x=i+1 y=x z=x for j in range(i,e): y=j+1 z=y for k in range(j,n): z=k+1 print(x,y,z) b=x*x+y*y+z*z+x*y+y*z+z*x if b>n: c=1 break; if x!=y and y!=z and z!=x: ans[b-1]+=6 elif (x==y and x!=z) or (y==z and x!=y): ans[b-1]+=3 elif x==y and y==z: ans[b-1]+=1 #if c==1: #break; #if c==1: #break; for i in range(n): print(ans[i])

s250887025

Accepted

n=int(input()) ans=[0]*n x=1 y=1 z=1 c=0 for i in range(1,n+1): if 6*i*i>n: d=i break; #print(d) for i in range(1,n+1): if 1+2*i*i+i+i+i*i>n: e=i break; #print(e) for i in range(d): x=i+1 y=x z=x for j in range(i,e): y=j+1 z=y for k in range(j,n): z=k+1 #print(x,y,z) b=x*x+y*y+z*z+x*y+y*z+z*x if b>n: c=1 break; if x!=y and y!=z and z!=x: ans[b-1]+=6 elif (x==y and x!=z) or (y==z and x!=y): ans[b-1]+=3 elif x==y and y==z: ans[b-1]+=1 #if c==1: #break; #if c==1: #break; for i in range(n): print(ans[i])

s803262635

p02795

u474925961

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 sys if sys.platform =='ios': sys.stdin=open('input_file.txt') N=int(input()) A=list(map(int,input().split())) B=list(map(int,input().split())) def flip(i): k=A[i] A[i]=B[i+1] B[i+1]=k l=B[i] B[i]=A[i+1] A[i+1]=B[i] if A==sorted(A): print("0") else: print("-1")

s007527091

Accepted

import sys if sys.platform =='ios': sys.stdin=open('input_file.txt') H=int(input()) W=int(input()) N=int(input()) p=max(H,W) print(int((N-1/2)//p+1))

s122387489

p04031

u842689614

Wrong Answer

Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.

N=int(input()) a=list(map(int,input().split())) a.sort() d=[[0]*N for i in range(N)] for i in range(N): for j in range(1,N): if j>i and a[j]!=a[j-1]: d[i][j]=(a[i]-a[j])**2 else: d[i][j]=d[j][i] print(min([sum(d[i]) for i in range(N)]))

s623635450

Accepted

N=int(input()) a=list(map(int,input().split())) a.sort() d=[[0]*N for i in range(max(a)-min(a)+1)] for i in range(max(a)-min(a)+1): for j in range(N): d[i][j]=(a[j]-(min(a)+i))**2 print(min([sum(dd) for dd in d]))

s228697350

p03080

u785578220

Wrong Answer

There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat.

a=int(input()) b = input() if b.count("R")< b.count("B"):print("Yes") else:print("No")

s685299257

Accepted

a=int(input()) b = input() if b.count("R")> b.count("B"):print("Yes") else:print("No")

s642284460

p03605

u484856305

Wrong Answer

It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?

n=input() if n in "9": print("Yes") else: print("No")

s438340095

Accepted

n=input() if "9" in n: print("Yes") else: print("No")

s367420615

p03644

u976162616

Wrong Answer

Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.

if __name__ == "__main__": N = int(input()) if N <= 1: print (0) exit() res_cnt = 0 res = 0 for x in range(N + 1): cnt = 0 y = x while (y > 1): if y % 2 == 0: y //= 2 cnt += 1 else: break if cnt > res_cnt: res = x print (res)

s359442829

Accepted

if __name__ == "__main__": N = int(input()) if N <= 1: print (1) exit() res_cnt = 0 res = 0 for x in range(N + 1): cnt = 0 y = x while (y > 1): if y % 2 == 0: y //= 2 cnt += 1 else: break if cnt > res_cnt: res_cnt = cnt res = x print (res)

s543925837

p04029

u757274384

Wrong Answer

There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?

# -*- coding: utf-8-*- N = int(input()) print(N*(N+1)/2)

s461436624

Accepted

# -*- coding: utf-8-*- N = int(input()) print(int(N*(N+1)/2))

s950467870

p02603

u561862393

Wrong Answer

To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?

import sys input = sys.stdin.readline import numpy as np n = int(input()) AA = [int(a) for a in input().split()] c = 0 updwn=[None]*(n-1) for a in AA: if c ==0: c +=1 base = a continue if a - base >= 0: updwn[c-1] = True base = a else: updwn[c-1] = False base = a c += 1 updwn+=[False, False] base = False total = 1000 stock = 0 if n ==2: if updwn[0]==True: total += (total//AA[0])*(AA[1]-AA[0]) else: for i in range(n): if updwn[i] == True and base == False: tdcost= AA[i] stock = total // tdcost base_cost = tdcost elif updwn[i] == True and updwn[i+1]==False: tdcost=AA[i+1] total += (tdcost - base_cost)*stock stock = 0 base = updwn[i] else: pass print(total)

s042271124

Accepted

import sys input = sys.stdin.readline import numpy as np n = int(input()) AA = [int(a) for a in input().split()] c = 0 updwn=[None]*(n-1) for a in AA: if c ==0: c +=1 base = a continue if a - base >= 0: updwn[c-1] = True base = a else: updwn[c-1] = False base = a c += 1 updwn+=[False, False] base = False total = 1000 stock = 0 for i in range(n): if updwn[i] == True and base == False: tdcost= AA[i] stock = total // tdcost base_cost = tdcost elif updwn[i-1] == True and updwn[i]==False: tdcost=AA[i] total += (tdcost - base_cost)*stock stock = 0 else: pass base = updwn[i] print(total)

s761691887

p03919

u767664985

Wrong Answer

There is a grid with H rows and W columns. The square at the i-th row and j-th column contains a string S_{i,j} of length 5. The rows are labeled with the numbers from 1 through H, and the columns are labeled with the uppercase English letters from `A` through the W-th letter of the alphabet. Exactly one of the squares in the grid contains the string `snuke`. Find this square and report its location. For example, the square at the 6-th row and 8-th column should be reported as `H6`.

H, W = map(int, input().split()) S = [list(input().split()) for _ in range(H)] for i in range(H): for j in range(W): if S[i][j] == "snuke": print(chr(97 + i).upper() + str(j + 1)) exit()

s891508547

Accepted

H, W = map(int, input().split()) S = [list(input().split()) for _ in range(H)] for i in range(H): for j in range(W): if S[i][j] == "snuke": print(chr(97 + j).upper() + str(i + 1)) exit()

s220505680

p03795

u543005033

Wrong Answer

Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.

N = int(input()) p = 1 for num in range(1,N+1): p = p * num temp = 10**9 + 7 ans = p % temp print(ans)

s365962028

Accepted

N = int(input()) # x = N * 800 temp = N // 15 y = temp * 200 ans = x-y print(ans)

s360129672

p03997

u191667127

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("a? ")) b = int(input("b? ")) h = int(input("h? ")) # Since h is even, area will always be an integer! area = int(((a+b)*h)/2) print(area)

s689220522

Accepted

a = int(input()) b = int(input()) h = int(input()) # Since h is even, area will always be an integer! area = int(((a+b)*h)/2) print(area)

s893564238

p02850

u185948224

Wrong Answer

Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different. Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.

N = int(input()) tree_dict = {} tree_lst = [] for i in range(N-1): a, b = map(int, input().split()) tree_lst.append([a, b]) if a in tree_dict: tree_dict[a].append(b) else: tree_dict[a] = [b] if b in tree_dict: tree_dict[b].append(a) else: tree_dict[b] = [a] colors = -1 for val in tree_dict.values(): colors = max(colors, len(val)) bond = [] [bond.append([0]*len(tree_dict[i])) for i in range(1,N + 1) if i in tree_dict] for a,b in tree_lst: for i in range(1, colors + 1): if i in bond[a-1]:continue elif len(bond[a-1]) < i:break else: for ni, bi in enumerate(tree_dict[a]): if bi == b:bond[a-1][ni] = i for nj, bj in enumerate(tree_dict[b]): if bj == a:bond[b-1][nj] = i print(i, bond) print(colors) for a, b in tree_lst: for ni, bi in enumerate(tree_dict[a]): if bi == b:print(bond[a-1][ni])

s380312698

Accepted

N = int(input()) ab = [list(map(int, input().split())) for _ in range(N-1)] graph = [[] for _ in range(N+1)] graph[0] = [0] edct = {} par = [-1] * (N+1) par[1] = 0 for a, b in ab: graph[a].append(b) graph[b].append(a) edct[(a, b)] = 0 nmax = 0 for i, g in enumerate(graph[1:], 1): nmax = max(nmax, len(g)) clr = 1 for j in g: if par[j] != -1: continue if clr == par[i]:clr += 1 edct[(min(i, j), max(i, j))] = clr par[j] = clr clr += 1 print(nmax) for a, b in ab: print(edct[(a, b)])

s609468206

p02409

u981238682

Wrong Answer

You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.

A = [[[0 for i in range(10)] for j in range(3)] for k in range(4)] n = int(input()) for i in range(n): b,f,r,v = map(int,input().split()) A[b-1][f-1][r-1] = v for b in range(4): for f in range(3): for r in range(10): print('',A[b][f][r],end='') print() if i != 3: print('#'*20)

s691283265

Accepted

A = [[[0 for i in range(10)] for j in range(3)] for k in range(4)] n = int(input()) for i in range(n): b,f,r,v = map(int,input().split()) A[b-1][f-1][r-1] += v for b in range(4): for f in range(3): for r in range(10): print('',A[b][f][r],end='') print() if b != 3: print('#'*20)

s286366540

p03548

u845427284

Wrong Answer

We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?

x, y, z = map(int,input().split()) a = x / (y + z) b = x % (y + z) print(a if b >= z else a - 1)

s141342936

Accepted

x, y, z = map(int,input().split()) a = x / (y + z) b = x % (y + z) print(int(a) if b >= z else int(a) - 1)

s889633121

p03494

u202406075

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.

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

s581497573

Accepted

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

s150142321

p03408

u459283268

Wrong Answer

Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.

n = int(input()) arr1 = [input() for _ in range(n)] m = int(input()) arr2 = [input() for _ in range(m)] # print(arr1, arr2) for i, j in zip(arr1, arr2): try: arr1.remove(j) except: pass # print(arr1, arr2) d = {} for i in set(arr1): d[i] = arr1.count(i) try: print(max(d.items(), key=lambda x:x[1])[0]) except: print("0")

s033984485

Accepted

n = int(input()) arr1 = [input() for _ in range(n)] m = int(input()) arr2 = [input() for _ in range(m)] r = 0 for i in set(arr1): r = max(arr1.count(i) - arr2.count(i), r) print(r)

s238811115

p03992

u282721391

Wrong Answer

This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s.

s = input('input') print(s[0:4], end=' ') print(s[4:], end='')

s095952372

Accepted

s = input() print(s[0:4], end=' ') print(s[4:])

s189167078

p03377

u317440328

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,C = map(int,input().split()) if(A+B>C): print("Yes") else: print("No")

s034003725

Accepted

A,B,C = map(int,input().split()) print("YES" if (A+B>=C)and(A<=C) else "NO")

s097909709

p03371

u196697332

Wrong Answer

"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.

A, B, C, X, Y = map(int, input().split()) if Y - X > 0: ans_1 = min(X, Y) * C * 2 + (Y - X) * B else: ans_1 = min(X, Y) * C * 2 + (X - Y) * B ans_2 = max(Y, X) * C * 2 print(min(ans_1, ans_2))

s335282485

Accepted

A, B, C, X, Y = map(int, input().split()) if Y - X > 0: ans_1 = min(X, Y) * C * 2 + (Y - X) * B else: ans_1 = min(X, Y) * C * 2 + (X - Y) * A ans_2 = max(Y, X) * C * 2 ans_3 = X * A + Y * B print(min(ans_1, ans_2, ans_3))

s092748756

p02972

u973240207

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.

from collections import defaultdict, deque import sys, heapq, bisect, math, itertools, string, queue, copy, time from fractions import gcd import numpy as np sys.setrecursionlimit(10**8) INF = float('inf') MOD = 10**9+7 EPS = 10**-7 def inp(): return int(sys.stdin.readline()) def inpl(): return list(map(int, sys.stdin.readline().split())) def inpl_str(): return list(sys.stdin.readline().split()) N = int(input()) alist = list(map(int, input().split())) ans_list = [] for i in range(len(alist)): bef = 0 for j in range(2, len(alist)): if (N-i)*j > N: break else: bef += alist[(N-i)*j-1] if alist[N-i-1] != bef%2: ans_list.append(N-i) print(len(ans_list)) print(*ans_list)

s942234118

Accepted

n = int(input()) a = list(map(int, input().split())) b = [0] * n for i in range(n, 0, -1): if sum(b[i-1::i]) % 2 == a[i-1]: continue else: b[i-1] = 1 c = [] for i in range(n): if b[i] == 1: c.append(i+1) print(sum(b)) if len(c) > 0: print(*c)

s350984809

p03476

u267325300

Time Limit Exceeded

We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime. You are given Q queries. In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.

import math N = 10**5 Q = int(input()) def eratosthenes(n): if not isinstance(n, int): raise TypeError('n is int type.') if n < 2: raise ValueError('n is more than 2') prime = [] limit = math.sqrt(n) data = [i + 1 for i in range(1, n)] while True: p = data[0] if limit <= p: return prime + data prime.append(p) data = [e for e in data if e % p != 0] primes = eratosthenes(N) like_2017 = [p for p in primes if (p + 1) // 2 in primes] c = 0 C = [0] for i in range(1, 10**5 + 1): if i in like_2017: c += 1 C.append(c) for _ in range(Q): left, right = map(int, input().split()) print(C[right] - C[left - 1])

s872397102

Accepted

import math N = 10**5 Q = int(input()) def eratosthenes(n): if not isinstance(n, int): raise TypeError('n is int type.') if n < 2: raise ValueError('n is more than 2') prime = [] limit = math.sqrt(n) data = [i + 1 for i in range(1, n)] while True: p = data[0] if limit <= p: return prime + data prime.append(p) data = [e for e in data if e % p != 0] _primes = eratosthenes(N) primes = [False] * (10**5 + 1) for p in _primes: primes[p] = True like_2017 = [False] * (10**5 + 1) for p in _primes: if primes[(p + 1) // 2]: like_2017[p] = True c = 0 C = [0] for i in range(1, 10**5 + 1): if like_2017[i]: c += 1 C.append(c) for _ in range(Q): left, right = map(int, input().split()) print(C[right] - C[left - 1])

s992674906

p02694

u258355189

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?

import math num_x = int(input()) taka = 100 count = 0 while num_x >= taka : taka = taka*0.01 + taka taka = math.floor(taka) count = count + 1 print(count)

s004268087

Accepted

import math num_x = int(input()) taka = 100 count = 0 while num_x > taka : taka = taka*0.01 + taka taka = math.floor(taka) count = count + 1 print(count)

s337969409

p02613

u440613652

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.

from collections import defaultdict d=defaultdict(int) for i in range(int(input())): s=input() d[s]+=1 print("AC x",d["AC"]) print("WA x",d["WA"]) print("TLE x",d["TLE"]) print("TLE x",d["TLE"])

s846316353

Accepted

from collections import defaultdict d=defaultdict(int) for i in range(int(input())): s=input() d[s]+=1 print("AC x",d["AC"]) print("WA x",d["WA"]) print("TLE x",d["TLE"]) print("RE x",d["RE"])

s058160171

p02259

u022407960

Wrong Answer

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

#!/usr/bin/env python # encoding: utf-8 class Solution: @staticmethod def bubble_sort(): # write your code here array_length = int(input()) unsorted_array = [int(x) for x in input().split()] flag = 1 count = 0 while flag: flag = 0 for j in range(array_length - 1, 0, -1): if unsorted_array[j] < unsorted_array[j - 1]: unsorted_array[j], unsorted_array[j - 1] = unsorted_array[j - 1], unsorted_array[j] flag = 1 count += 1 print(unsorted_array) print(count) if __name__ == '__main__': solution = Solution() solution.bubble_sort()

s952893846

Accepted

#!/usr/bin/env python # encoding: utf-8 class Solution: @staticmethod def bubble_sort(): # write your code here array_length = int(input()) array = [int(x) for x in input().split()] flag, count, cursor = 1, 0, 0 while flag: flag = 0 for j in range(array_length - 1, cursor, -1): if array[j] < array[j - 1]: array[j], array[j - 1] = array[j - 1], array[j] flag = 1 count += 1 cursor += 1 print(" ".join(map(str, array))) print(str(count)) if __name__ == '__main__': solution = Solution() solution.bubble_sort()

s916749636

p03659

u103902792

Wrong Answer

Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.

n = int(input()) A = list(map(int,input().split())) from itertools import accumulate A = list(accumulate(A)) ans = float('inf') for i in range(n-1): ans = min(ans, abs(A[-1] - A[i])) print(ans)

s203813434

Accepted

n = int(input()) A = list(map(int,input().split())) from itertools import accumulate A = list(accumulate(A)) ans = float('inf') for i in range(n-1): ans = min(ans, abs((A[-1] - A[i])-A[i])) print(ans)

s710783092

p03545

u167908302

Wrong Answer

Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

# coding:utf-8 number = input() n = 3 for bit in range(2 ** n): ope = ['-'] * n for i in range(n): if ((bit >> i) & 1): ope[n - 1 - i] = '+' # print(ope) ans = '' for num, op in zip(number, ope): ans += num + op ans += number[-1] if eval(ans) == 7: print(ans + '==7') exit()

s993779680

Accepted

# coding:utf-8 number = input() n = 3 for bit in range(2 ** n): ope = ['-'] * n for i in range(n): if ((bit >> i) & 1): ope[n - 1 - i] = '+' # print(ope) ans = '' for num, op in zip(number, ope): ans += num + op ans += number[-1] if eval(ans) == 7: print(ans + '=7') exit()

s844101378

p03826

u214434454

Wrong Answer

There are two rectangles. The lengths of the vertical sides of the first rectangle are A, and the lengths of the horizontal sides of the first rectangle are B. The lengths of the vertical sides of the second rectangle are C, and the lengths of the horizontal sides of the second rectangle are D. Print the area of the rectangle with the larger area. If the two rectangles have equal areas, print that area.

a, b, c, d = map(int, input().split()) print(max(a+b, c*d))

s261830428

Accepted

a, b, c, d = map(int, input().split()) print(max(a*b, c*d))

s943433075

p02407

u742505495

Wrong Answer

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

n = int(input()) a = list(map(int,input().split())) a.reverse() for i in range(len(a)): if i!=len(a)-1: print(a[i]) else: print(a[i],end=' ')

s454510528

Accepted

n = int(input()) a = list(map(int,input().split())) a.reverse() for i in range(len(a)): if i!=len(a)-1: print(a[i],end=' ') else: print(a[i])

s524507073

p00008

u744114948

Wrong Answer

Write a program which reads an integer n and identifies the number of combinations of a, b, c and d (0 ≤ a, b, c, d ≤ 9) which meet the following equality: a + b + c + d = n For example, for n = 35, we have 4 different combinations of (a, b, c, d): (8, 9, 9, 9), (9, 8, 9, 9), (9, 9, 8, 9), and (9, 9, 9, 8).

#! /usr/bin/python3 s=int(input()) n=0 for i in range(10): for j in range(10): for k in range(10): for l in range(10): if s == i+j+k+l: n+=1 print(n)

s192744320

Accepted

#! /usr/bin/python3 while True: try: s=int(input()) n=0 for i in range(10): for j in range(10): for k in range(10): for l in range(10): if s == i+j+k+l: n+=1 print(n) except: break

s806910502

p03606

u021916304

Wrong Answer

Joisino is working as a receptionist at a theater. The theater has 100000 seats, numbered from 1 to 100000. According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive). How many people are sitting at the theater now?

def ii():return int(input()) def iim():return map(int,input().split()) def iil():return list(map(int,input().split())) def ism():return map(str,input().split()) def isl():return list(map(str,input().split())) n = ii() ans = 0 for i in range(n): l,r = iim() ans += l-r+1 print(ans)

s460774048

Accepted

def ii():return int(input()) def iim():return map(int,input().split()) def iil():return list(map(int,input().split())) def ism():return map(str,input().split()) def isl():return list(map(str,input().split())) n = ii() ans = 0 for i in range(n): l,r = iim() ans += r-l+1 print(ans)

s228091918

p03485

u004025573

Wrong Answer

You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.

A,B=map(int,input().split()) if A%B==0: print(A//B) else: print(A//B + 1)

s707518897

Accepted

A,B=map(int,input().split()) if (A+B)%2==0: print((A+B)//2) else: print((A+B+1)//2)

s376032446

p03386

u519939795

Wrong Answer

Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.

a,b,k=map(int,input().split()) l=[int(i) for i in range(a,b+1)] l1=sorted(l,reverse=True) ans=[] if k>len(l): for a in l: print(a) else: for i in range(0,k): ans.append(l[i]) for j in range(0,k): ans.append(l1[j]) ans=set(sorted(ans)) for g in ans: print(g)

s592779095

Accepted

# coding: utf-8 A, B, K = map(int, input().split()) a = set(i for i in range(A, min(A + K, B + 1))) b = set(j for j in range(max(B - K + 1, A), B + 1)) c = list(a | b) c.sort() for k in c: print(k)

s773098813

p03455

u821432765

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()) ab=int(str(a)+str(b)) j=0 for i in range(318): if i**2==ab: print("Yes") j=1 break else: pass if j==1: pass else: print("No")

s421123880

Accepted

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

s222359080

p03387

u540762794

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.

# -*- coding: utf-8 -*- A,B,C = map(int, input().split()) s = A + B + C M = max(A,B,C) if M % 2 == 0 and s % 2 != 0: M += 1 elif M % 2 != 0 and s % 2 == 0: M += 1 ans = (3 * M - s) / 2 print(ans)

s564038482

Accepted

# -*- coding: utf-8 -*- A,B,C = map(int, input().split()) s = A + B + C M = max(A,B,C) if M % 2 == 0 and s % 2 != 0: M += 1 elif M % 2 != 0 and s % 2 == 0: M += 1 ans = (3 * M - s) / 2 print(int(ans))

s443138501

p03796

u836737505

Wrong Answer

Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.

import math N = int(input()) d = math.factorial(N)

s160918220

Accepted

import math print(math.factorial(int(input()))%1000000007)

s448323528

p03069

u096736378

Wrong Answer

There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.

n = int(input()) s = input() def inverse(num, index, first): offset = 0 if first == '*': offset = 1 count = 0 for n, i in enumerate(num): if n <= index and n % 2 == offset: count += i elif n > index and n % 2 != offset: count += i return count num = [] first = s[0] end = s[-1] pre = first count = 0 for c in s: if c == pre: count += 1 else: num.append(count) count = 1 pre = c num.append(count) print(num) if len(num) == 1: print(0) else: minlist = [] for i in range(len(num)): minlist.append(inverse(num, i, first)) print(min(minlist))

s132198020

Accepted

n = int(input()) s = input() num = [] first = s[0] end = s[-1] pre = first count = 0 for c in s: if c == pre: count += 1 else: num.append(count) count = 1 pre = c num.append(count) if len(num) == 1: print(0) else: b = 0 w = n offset = 0 if first == '#': offset = 1 w = sum(num[offset::2]) minlist = [b + w] for i in range(len(num)): if i % 2 == offset: w -= num[i] else: b += num[i] minlist.append(b + w) print(min(minlist))

s955553210

p03943

u785578220

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.

l= list(map(int, input().split())) l.sort() if l[0] == l[1] + l[2]: print("Yes") else:print("No")

s854459017

Accepted

l= list(map(int, input().split())) l.sort() if l[2] == l[1] + l[0]: print("Yes") else:print("No")

s433702919

p02601

u908763441

Wrong Answer

M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.

a, b, c = map(int, input().split()) k = int(input()) while k > 0: if (b <= a): b = b * 2 k -= 1 else: if (c <= b): c = c * 2 k -= 1 else: print('Yes') exit() print('No')

s617382057

Accepted

a, b, c = map(int, input().split()) k = int(input()) while k > 0: if (b <= a): b = b * 2 k -= 1 else: if (c <= b): c = c * 2 k -= 1 else: print('Yes') exit() if (b > a and c > b): print('Yes') else: print('No')

s930480878

p00007

u742505495

Wrong Answer

Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week. Write a program which computes the amount of the debt in n weeks.

import sys import math n = int(input()) debt = 100000 for i in range(1,n+1): debt = math.floor((debt*1.05)/1000)*1000 print(debt)

s878616667

Accepted

import sys import math n = int(input()) debt = 100000 for i in range(1,n+1): debt = math.ceil((debt*1.05)/1000)*1000 print(debt)

s430163965

p03623

u556477263

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()) if abs(a-b) > abs(a-c): print('A') else: print('B')

s178879749

Accepted

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

s028037660

p03150

u742729271

Wrong Answer

A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.

S = input() s = "keyence" flag = False ans = "No" for i in range(len(S)): if s[:i] in S: s1 = s[:i] s2 = s[i:] flag = True if flag: for i in range(len(S)-len(s1)): if s1 == S[i:i+len(s1)]: if s2 in S[i+len(s1):]: ans = "Yes" break print(ans)

s181375634

Accepted

S = input() s = "keyence" ans = "NO" for i in range(len(S)): if s == S[:i] + S[len(S)-len(s)+i:]: ans = "YES" print(ans)

s052501145

p03485

u563588763

Wrong Answer

You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.

a,b = map(int,input().split()) if int((a + b)//2) == 0: print(int((a + b)//2)) else: print(int((a + b) // 2 + 1))

s966395451

Accepted

a,b = map(int,input().split()) if int((a + b)%2) == 0: print(int((a + b)//2)) else: print(int((a + b) // 2 + 1))

s069275107

p03369

u252964975

Wrong Answer

In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.

S=str(input()) k = 0 for i in range(3): if S[i] == "○": k = k + 100 print(k)

s543550619

Accepted

S=str(input()) k = 700 for i in range(3): if S[i] == 'o': k = k + 100 print(k)

s409467098

p02612

u040642458

Wrong Answer

We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.

n = int(input()) print(1000-n)

s876416093

Accepted

n = int(input()) a = n%1000 b = 1000-a if b==1000: print(0) else: print(b)

s143065254

p03455

u079022116

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*b%2==0: print("Odd") else: print("Even")

s424154643

Accepted

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

s485698983

p03351

u780111164

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()) ans="YES" if a-c>d or c-a>d: if a-b>d or b-a>d or b-c>d or c-b>d: ans="NO" print(ans)

s174019078

Accepted

a,b,c,d=map(int,input().split()) ans="Yes" if (a-c)*(a-c) >d*d: ans="No" if (a-b)*(a-b)<=d*d and (b-c)*(b-c)<=d*d: ans="Yes" print(ans)

s669798928

p02842

u651058930

Wrong Answer

Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.

N = int(input()) est_X = int(N/1.08)*1.08 if est_X == N: print(est_X) elif est_X + 1 == N: print(est_X + 1) else: print(':(')

s207759550

Accepted

N = int(input()) est_X = int(N/1.08) if int(est_X*1.08) == N: print(est_X) elif int((est_X + 1)*1.08) == N: print(est_X + 1) else: print(':(')

s875207944

p03434

u244737745

Wrong Answer

We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.

# 2020-08-30, Sun n = int(input()) lis = sorted(list(map(int, input().split()))) a = 0 b = 0 for i in lis[::2]: a += i for i in lis[1::2]: b += i print(a - b)

s516823967

Accepted

# 2020-08-30, Sun n = int(input()) lis = sorted(list(map(int, input().split())), reverse=True) a = 0 b = 0 for i in lis[::2]: a += i for i in lis[1::2]: b += i print(a - b)

s504573234

p02578

u621596556

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()) a = list(map(int,input().split())) ans = 0 for i in range(1,n): if(a[i-1] > a[i]): while(a[i-1] == a[i]): a[i] += 1 ans += 1 print(ans)

s122885838

Accepted

n = int(input()) a = list(map(int,input().split())) ans = 0 diff = 0 for i in range(1,n): if(a[i-1] > a[i]): diff = a[i-1] - a[i] a[i] += diff ans += diff print(ans)

s527677115

p02831

u018872912

Wrong Answer

Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.

n,m =map(int,input().split()) n2=max(n,m) m2=min(n,m) while m2!=0: tmp=n2 n2=m2 m2=tmp%m2 print(n2,m2) print(int(n*m/n2))

s961999609

Accepted

n,m =map(int,input().split()) n2=max(n,m) m2=min(n,m) while m2!=0: tmp=n2 n2=m2 m2=tmp%m2 print(int(n*m/n2))

s799546742

p03493

u106184985

Wrong Answer

Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.

ss = map(int, input().split()) ans = 0 for s in ss: if s == 1: ans+=1 print(ans)

s174046686

Accepted

print(sum(list(map(int, list(input())))))

s330722982

p03433

u921773161

Wrong Answer

E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.

N = int(input()) A = int(input()) if N >= 500: while N<500: N = N-500 if N<=A: print('Yes') else: print('No') else: if N<=A: print('Yes') else: print('No')

s239181480

Accepted

N = int(input()) A = int(input()) N_500 = N%500 if A >= N_500 : print('Yes') else : print('No')

s831303081

p03433

u321354941

Wrong Answer

E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.

n = int(input()) a = int(input()) x = n % 500 if x > a: print("no") else: print("yes")

s848051369

Accepted

n = int(input()) a = int(input()) x = n % 500 if a >= x: print("Yes") else: print("No")

s381453754

p03479

u970348538

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.

import math a,b = [list(map(int, s.split())) for s in open(0)][0] math.floor(math.log2(b) - math.log2(a))+1

s624693226

Accepted

import math a,b = [list(map(int, s.split())) for s in open(0)][0] ans = 0 while a <= b: a *= 2 ans += 1 print(ans)

s983546527

p02393

u113501470

Wrong Answer

Write a program which reads three integers, and prints them in ascending order.

abc = input().split() for n in abc: n = int(n) a,b,c = abc[0],abc[1], abc[2] print(a,b,c) low = min(abc) high = max(abc) abc.remove(low) abc.remove(high) print (low, abc.pop(), high)

s757005404

Accepted

abc = input().split() for n in abc: n = int(n) a,b,c = abc[0],abc[1], abc[2] low = min(abc) high = max(abc) abc.remove(low) abc.remove(high) print (low, abc.pop(), high)

s109741388

p03998

u412481017

Wrong Answer

Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.

dic={"a":input(),"b":input(),"c":input()} print(dic) temp="a" while 1: temp2=dic[temp][:1] #print(temp) dic[temp]=dic[temp][1:] temp=temp2 #print(dic) if dic[temp]=="": if temp=="a": print("A") elif temp=="b": print("B") elif temp=="c": print("C") break

s320900289

Accepted

dic={"a":input(),"b":input(),"c":input()} #print(dic) temp="a" while 1: temp2=dic[temp][:1] #print(temp) dic[temp]=dic[temp][1:] temp=temp2 #print(dic) if dic[temp]=="": if temp=="a": print("A") elif temp=="b": print("B") elif temp=="c": print("C") break

s604073440

p03605

u360515075

Wrong Answer

It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?

N = int(input()) print ("YES" if N //10 == 9 or N % 10 == 9 else "NO")

s318381885

Accepted

N = int(input()) print ("Yes" if N //10 == 9 or N % 10 == 9 else "No")

s193308177

p03485

u239342230

Wrong Answer

You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.

x=eval(input().replace(' ','+')) print(x/2 if x%2==0 else x//2+1)

s292802732

Accepted

print(-~eval(input().replace(' ','+'))//2)

s697220470

p03943

u444856278

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.

[print(len(list(i.groupby(input())))-1)for i in[__import__('itertools')]]

s707055602

Accepted

print("Yes"if[x[0]+x[1]==x[2]for x in[sorted([int(i)for i in input().split()])]][0]else"No")

s779119986

p03853

u654558363

Wrong Answer

There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).

if __name__ == "__main__": h, w = map(int, input().split()) c = [] for i in range(h): c.append(input()) newC = [] for i in range(2*h - 1): newC.append(c[(i) // 2]) newC.append(c[len(c) - 1]) print("===") for i in newC: print(i)

s460483090

Accepted

if __name__ == "__main__": h, w = map(int, input().split()) c = [] for i in range(h): c.append(input()) newC = [] for i in range(2*h - 1): newC.append(c[(i) // 2]) newC.append(c[len(c) - 1]) for i in newC: print(i)

s122939556

p02615

u229518917

Wrong Answer

Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?

N=int(input()) A=list(map(int,input().split())) A.sort(reverse=True) print(A) print(sum(A[0:-1]))

s468864549

Accepted

N=int(input()) A=list(map(int,input().split())) A.sort(reverse=True) AA=[] AA.append(A[0]) for i in range(1,N): AA.append(A[i]) AA.append(A[i]) ans=0 for j in range(N-1): ans+=AA[j] print(ans)

s247535607

p03545

u892166393

Wrong Answer

Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

abcd=str(input()) for i in range(2**4): list=[int(abcd[0])] symbol=[] for j in range(1,4): if (i>>j)&1==True: list.append(int(abcd[j])) symbol.append('+') else: list.append(int(abcd[j])*(-1)) symbol.append('') if sum(list)==7: print(list[0],symbol[0],list[1],symbol[1],list[2],symbol[2],list[3],"=7") break

s643452826

Accepted

import sys abcd=str(input()) n=len(abcd) ans=int(abcd[0]) sign=[] for i in range(2**(n-1)): for j in range(n-1): if (i>>j)&1==True: ans+=int(abcd[j+1]) sign.append("+") else: ans-=int(abcd[j+1]) sign.append("-") if ans==7: print(abcd[0]+sign[0]+abcd[1]+sign[1]+abcd[2]+sign[2]+abcd[3]+"=7") sys.exit() else: ans=int(abcd[0]) sign=[]

s548708502

p03854

u431624930

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() flag = 0 while(len(s)>0): if (s[:6] == 'eraser'): s = s[6:] elif (s[:5] == 'erase'): s = s[5:] elif (s[:5] == 'dream'): if (s[5:8] == 'ere' or s[5:8] == 'erd'): s = s[7:] else: s = s[5:] else: flag = 1 break if (flag == 0): print('Yes') else: print('No')

s413184692

Accepted

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

s558977849

p03693

u175590965

Wrong Answer

AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?

x,a,b = map(int,input().split()) if ((100*x)+(10*a)+(b)% 4)== 0: print("Yes") else: print("No")

s844517483

Accepted

x,a,b = map(int,input().split()) if (((100*x)+(10*a)+(b))% 4) == 0: print("YES") else: print("NO")

s879936891

p03545

u297651868

Wrong Answer

Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

s=input() op=["+","-"] for i in range(2): for j in range(2): for k in range(2): formula=s[0]+op[i]+s[1]+op[j]+s[2]+op[k]+s[3] if eval(formula)==7: print(formula+"==7") exit(0)

s453484401

Accepted

a,b,c,d=list(map(str,input())) def solve(m,n): if eval(m)==7: print(m+"=7") exit(0) elif n<7: solve(m,n+2) solve(m[:n]+"-"+m[n+1:],n+2) solve(a+"+"+b+"+"+c+"+"+d,1)

s867477100

p02646

u339873473

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 = input() if (int(A) < int(B) and int(V) < int(W)): print("False") a = int(A) + (int(V) * int(T)) b = int(B) + (int(W) * int(T)) if a >= b: print("True")

s328511201

Accepted

a, v = map(int, input().split()) b, w = map(int, input().split()) t = int(input()) if abs(a-b) <= t* (v-w): print('YES') else: print('NO')

s559519170

p03643

u224873361

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.

K = int(input()) if K % 2 == 0: print(2) print("{0} {1}".format((K)//2 + 1, K//2 + 1)) else: print(2) print("{0} {1}".format((K)//2 - 1, (K)//2))

s029013558

Accepted

print("ABC" + input())

s989459499

p00005

u002010345

Wrong Answer

Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b.

def main(): a,b=(int(x) for x in input().split()) a1=a b1=b while b!=0: c=a%b a=b b=c print(a," ",a1*b1/a) main()

s872021498

Accepted

def main(): while True: try: a,b=(int(x) for x in input().split()) except: break a1=a b1=b while b!=0: c=a%b a=b b=c print(a, a1*b1//a) main()

s173932067

p03657

u101627912

Wrong Answer

Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.

# -*- coding:utf-8 -*- a,b=map(int,input().split()) if a%3==0 or b%3==0 or a+b%3==0: print("Possible") else: print("Impossible")

s289171753

Accepted

# -*- coding:utf-8 -*- a,b=map(int,input().split()) if a%3==0 or b%3==0 or (a+b)%3==0: print("Possible") else: print("Impossible")

s232556728

p02261

u408260374

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).

def bubbleSort(l, n): for i in range(n): for j in range(i, n-1): if l[j][0] > l[j+1][0]: l[j], l[j+1] = l[j+1], l[j] return l[:] def selectionSort(l, n): for i in range(n): minj = i for j in range(i+1, n): if l[j][0] < l[minj][0]: minj = j l[i], l[minj] = l[minj], l[i] return l[:] n = int(input()) l = [] for c in input().split(): l.append((int(c[1]), c[0])) bl = ' '.join([k + str(n) for n, k in bubbleSort(l, n)]) sl = ' '.join([k + str(n) for n, k in selectionSort(l, n)]) print(bl) print('Stable') print(sl) print('Stable' if sl == bl else 'Not stable')

s672776898

Accepted

def bubbleSort(l, n): for i in range(n): for j in range(n-1-i): if l[j][0] > l[j+1][0]: l[j], l[j+1] = l[j+1], l[j] return l[:] def selectionSort(l, n): for i in range(n): minj = i for j in range(i+1, n): if l[j][0] < l[minj][0]: minj = j l[i], l[minj] = l[minj], l[i] return l[:] n = int(input()) l = [] for c in input().split(): l.append((int(c[1]), c[0])) bl = ' '.join([k + str(n) for n, k in bubbleSort(l[:], n)]) sl = ' '.join([k + str(n) for n, k in selectionSort(l[:], n)]) print(bl) print('Stable') print(sl) print('Stable' if sl == bl else 'Not stable')

s862708605

p04043

u099212858

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.

mylist = list(map(int,input().split())) x = mylist.count(5) y = mylist.count(7) if x == 2 and y == 1: print("Yes") else: print("NO")

s167326061

Accepted

mylist = list(map(int,input().split())) x = mylist.count(5) y = mylist.count(7) if x == 2 and y == 1: print("YES") else: print("NO")

s156291335

p02399

u580737984

Wrong Answer

Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)

i = j = 0 n = '' m = '' line = input() while line[i] != ' ': n = n + line[i] i += 1 while i < len(line): m = m + line[i] i += 1 n = int(n) m = int(m) d = n // m r = n % m f = n / m print(d,end=' ') print(r,end=' ') print(f)

s788652451

Accepted

i = j = 0 n = '' m = '' line = input() while line[i] != ' ': n = n + line[i] i += 1 while i < len(line): m = m + line[i] i += 1 n = int(n) m = int(m) d = n // m r = n % m f = n / m print(d,end=' ') print(r,end=' ') print("{0:.5f}".format(f))

s834784503

p03672

u038052425

Wrong Answer

We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.

letters = input() print(letters) while True: not_equal = 0 length = len(letters) letters = letters[:length-1] print(letters) if len(letters) % 2 != 0: continue if int(len(letters)) % 2 == 0: half = int(len(letters)/2 -1) else: half = int((len(letters)-1)/2) i = 0 while i <= half: if letters[i] != letters[i + half + 1]: not_equal = 1 break i += 1 if not_equal == 0: print(len(letters)) break

s669548020

Accepted

letters = input() while True: not_equal = 0 length = len(letters) letters = letters[:length-1] if len(letters) % 2 != 0: continue if int(len(letters)) % 2 == 0: half = int(len(letters)/2 -1) else: half = int((len(letters)-1)/2) i = 0 while i <= half: if letters[i] != letters[i + half + 1]: not_equal = 1 break i += 1 if not_equal == 0: print(len(letters)) break

s335485446

p03579

u968404618

Wrong Answer

Rng has a connected undirected graph with N vertices. Currently, there are M edges in the graph, and the i-th edge connects Vertices A_i and B_i. Rng will add new edges to the graph by repeating the following operation: * Operation: Choose u and v (u \neq v) such that Vertex v can be reached by traversing exactly three edges from Vertex u, and add an edge connecting Vertices u and v. It is not allowed to add an edge if there is already an edge connecting Vertices u and v. Find the maximum possible number of edges that can be added.

import sys sys.setrecursionlimit(10**7) def dfs(v, color): colors[v] = color for to in g[v]: if colors[to] == color: return False if colors[to] == 0 and not dfs(to, -color): return False return True n, m = map(int, input().split()) g = [[] for _ in range(n)] for _ in range(m): a, b = map(int, input().split()) a -= 1 b -= 1 g[a].append(b) g[b].append(a) colors = [0]*n ans = -m if dfs(0,1): b = sum(colors)//2 w = n-b ans += b*w else: ans += n*(n-1)//2 print(ans)

s926981161

Accepted

import sys sys.setrecursionlimit(10**7) def dfs(v, color): colors[v] = color for to in g[v]: if colors[to] == color: return False if colors[to] == 0 and not dfs(to, -color): return False return True n, m = map(int, input().split()) g = [[] for _ in range(n)] for _ in range(m): a, b = map(int, input().split()) a -= 1 b -= 1 g[a].append(b) g[b].append(a) colors = [0]*n ans = -m if dfs(0,1): b = (sum(colors)+n)//2 w = n-b ans += b*w else: ans += n*(n-1)//2 print(ans)

s355720038

p03997

u579875569

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.

#!/usr/bin/python3 a = int(input()) b = int(input()) h = int(input()) print((a+b)*h/2)

s790216752

Accepted

#!/usr/bin/python3 a = int(input()) b = int(input()) h = int(input()) print(int((a+b)*h/2))

s421274354

p03711

u328755070

Wrong Answer

Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.

x, y = list(map(int, input().split())) a = [1,3,5,7,8,10,12] b = [4,6,9,11] if x in a and y in b: print('Yes') elif x in b and y in b: print('Yes') elif x == 2 and y == 2: print('Yes') else: print('No')

s334496224

Accepted

x, y = list(map(int, input().split())) a = [1,3,5,7,8,10,12] b = [4,6,9,11] if x in a and y in a: print('Yes') elif x in b and y in b: print('Yes') elif x == 2 and y == 2: print('Yes') else: print('No')

s762545198

p02612

u767821815

Wrong Answer

We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.

N = int(input()) print(N%1000)

s703998939

Accepted

N = int(input()) if N%1000 == 0: print(0) else: print(1000-N%1000)

s519358312

p02257

u591232952

Wrong Answer

A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list.

def prime(n): for i in range(2, int(n**0.5) + 1): if n % i: return False return True n = int(input()) c = 0 for _ in range(n): if prime(int(input())): c+=1 print(c)

s543004214

Accepted

def prime(n): for i in range(2, int(n**0.5) + 1): if n % i == 0: return False return True n = int(input()) c = 0 for _ in range(n): s = int(input()) if prime(s): c+=1 print(c)

s537590024

p03385

u702786238

Wrong Answer

You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.

s = input() if sorted(s) == "abc": print("Yes") else: print("No")

s011059777

Accepted

s = input() if "".join(sorted(s)) == "abc": print("Yes") else: print("No")

s613914435

p02833

u038408819

Wrong Answer

For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).

N = int(input()) if N % 2 == 1: print(0) quit() N /= 2 res = 0 while N: print(N) res += int(N / 5) N = int(N / 5) print(res)

s266164579

Accepted

N = int(input()) if N % 2 == 1: print(0) quit() N //= 2 s = 1 ans = 0 while N >= 5 ** s: ans += N // (5 ** s) s += 1 print(ans)

s711345659

p03456

u674722380

Wrong Answer

AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.

import math import numpy as np a, b = map(int, input().split()) c = (int(str(a) + str(b))) d = math.sqrt(c) z = np.rint(d) if z**2== c: print("Yes") print(c) else: print("No") print(c)

s756528366

Accepted

import math import numpy as np a, b = map(int, input().split()) c = (int(str(a) + str(b))) d = math.sqrt(c) z = np.rint(d) if z**2== c: print("Yes") else: print("No")

s919697081

p03456

u158703648

Wrong Answer

AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.

import math A,B = map(int, input().split()) C=str(A)+str(B) C2=int(C) C3=math.sqrt(C2) if C3.is_integer(): print("yes") else: print("no")

s309688514

Accepted

import math A,B = map(str,input().split()) if math.sqrt(int(A+B)) % 1==0: print("Yes") else: print("No")

s098460831

p03997

u842388336

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)

s879981341

Accepted

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