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CodeNet4Repair

like 0

s195468850

p03048

u801512570

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()) tmp=0 for i in range(1,N+1): for j in range(1,N+1): if (R*i+G*j)<=N and (N-(R*i+G*j))%B==0: tmp+=1 print(tmp)

s967396956

Accepted

R,G,B,N=map(int,input().split()) tmp=0 for i in range(N//R+1): for j in range((N-R*i)//G+1): if (N-(R*i+G*j))%B==0: tmp+=1 print(tmp)

s664084223

p02612

u231484984

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(str((N)%1000))

s043439771

Accepted

import math N = int(input()) print(str((math.ceil((N)/1000))*1000 - (N)))

s034018046

p02694

u615590527

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()) i = 100 cnt = 0 while i<X: i =i*101//1 cnt += 1 print(cnt)

s461022710

Accepted

X = int(input()) i = 100 cnt = 0 while i<X: i =i*1.01//1 cnt += 1 print(cnt)

s555570654

p03574

u779170803

Wrong Answer

You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.

INT = lambda: int(input()) INTM = lambda: map(int,input().split()) STRM = lambda: map(str,input().split()) STR = lambda: str(input()) LIST = lambda: list(map(int,input().split())) LISTS = lambda: list(map(str,input().split())) def do(): dxs=[-1,0,1] dys=[-1,0,1] h,w=INTM() s=['.'*(w+2)] ans=[[0]*(w+2) for i in range(h+2)] for i in range(h): st='.'+STR()+'.' s.append(st) s.append('.'*(w+2)) #print(s) #print(ans) for i in range(1,h+1): for j in range(1,w+1): if s[i][j]=='#': ans[i][j]='#' for dx in dxs: for dy in dys: if s[i+dx][j+dy]=='.': #print(ans) ans[i+dx][j+dy]=int(ans[i+dx][j+dy]+1) for i in range(1,h+1): print(*ans[i][1:w],sep='') if __name__ == '__main__': do()

s466240599

Accepted

INT = lambda: int(input()) INTM = lambda: map(int,input().split()) STRM = lambda: map(str,input().split()) STR = lambda: str(input()) LIST = lambda: list(map(int,input().split())) LISTS = lambda: list(map(str,input().split())) def do(): dxs=[-1,0,1] dys=[-1,0,1] h,w=INTM() s=['.'*(w+2)] ans=[[0]*(w+2) for i in range(h+2)] for i in range(h): st='.'+STR()+'.' s.append(st) s.append('.'*(w+2)) #print(s) #print(ans) for i in range(1,h+1): for j in range(1,w+1): if s[i][j]=='#': ans[i][j]='#' for dx in dxs: for dy in dys: if s[i+dx][j+dy]=='.': #print(ans) ans[i+dx][j+dy]=int(ans[i+dx][j+dy]+1) for i in range(1,h+1): print(*ans[i][1:w+1],sep='') if __name__ == '__main__': do()

s726151558

p03698

u594762426

Wrong Answer

You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.

s = str(input()) if s == set(s): print("yes") else: print("no")

s499090713

Accepted

s = str(input()) if len(s) == len(set(s)): print("yes") else: print("no")

s124951879

p03377

u464912173

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 = map(int, input().split()) print('YES' if a+b <= x else 'NO')

s135391933

Accepted

a, b, x = map(int, input().split()) print('YES' if a <= x and b >= x-a else 'NO')

s856570218

p03730

u802963389

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()) flg = False for i in range(a, a*b+1, a): if i % b == c: flg = True break if flg == True: print("Yes") else: print("No")

s199441360

Accepted

a, b, c = map(int,input().split()) flg = False for i in range(a, a*b+1, a): if i % b == c: flg = True break if flg == True: print("YES") else: print("NO")

s205903661

p02612

u546853743

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()) if n%1000==0: print('0') else: k = n k += 1000 k %= 1000 cha = k*1000-n print(cha)

s777248384

Accepted

n=int(input()) if n%1000==0: print('0') else: k = n k += 1000 k //= 1000 cha = k*1000-n print(cha)

s792704608

p03693

u791664126

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?

_,a,b=map(int,input().split()) print('YNEOS'[(a*10+b)%4>1])

s487683596

Accepted

_,a,b=map(int,input().split()) print('YNEOS'[(a*10+b)%4>0::2])

s155505803

p03854

u505830998

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`.

import sys #+++++ def main(): s = input() ww=['dream','dreamer','erase','eraser'] s=s[::-1] wwr=[w[::-1] for w in ww] #print(wwrs) while len(s)>0: pa(s) is_ok=False for w in wwr: ll=len(w) ss=s[:ll] pa((2,ss,w)) if ss == w: is_ok=True pa((1,ss,ll)) s=s[ll:] if not is_ok: return 'No' print('Yes') #+++++ isTest=False def pa(v): if isTest: print(v) if __name__ == "__main__": if sys.platform =='ios': sys.stdin=open('inputFile.txt') isTest=True else: pass #input = sys.stdin.readline ret = main() if ret is not None: print(ret)

s874245056

Accepted

import sys #+++++ def main(): s = input() ww=['dream','dreamer','erase','eraser'] s=s[::-1] wwr=[w[::-1] for w in ww] #print(wwrs) while len(s)>0: #pa(s) is_ok=False for w in wwr: ll=len(w) ss=s[:ll] #pa((2,ss,w)) if ss == w: is_ok=True #pa((1,ss,ll)) s=s[ll:] if not is_ok: print('NO') return print('YES') #+++++ isTest=False def pa(v): if isTest: print(v) if __name__ == "__main__": if sys.platform =='ios': sys.stdin=open('inputFile.txt') isTest=True else: pass #input = sys.stdin.readline ret = main() if ret is not None: print(ret)

s833839624

p03556

u539826121

Wrong Answer

Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.

x = int(input()) print(int(x**0.25))

s548317205

Accepted

n = int(input()) x = 0 while (x+1) * (x+1)<= n: x += 1 print(x*x)

s244618086

p03471

u994521204

Wrong Answer

The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.

N, Y=list(map(int, input().split())) ans=(-1, -1, -1) flag=False for x in range(Y//10000, -1, -1): for y in range((Y-10000*x)//5000, -1, -1): for z in range((Y-10000*x-5000*y)//1000,-1,-1): print(x, y, z) if N==x+y+z: ans=(x,y,z) flag=True break if flag==True: break if flag==True: break if flag==True: break print(ans)

s442539446

Accepted

N, Y=list(map(int, input().split())) ans=[-1, -1, -1] flag=False for x in range(N, -1, -1): for y in range(N-x, -1, -1): if 10000*x+5000*y+1000*(N-x-y)==Y: ans=[x,y,N-x-y] flag=True break if flag==True: break if flag==True: break print(ans[0],ans[1],ans[2])

s340560919

p02846

u738898077

Wrong Answer

Takahashi and Aoki are training for long-distance races in an infinitely long straight course running from west to east. They start simultaneously at the same point and moves as follows **towards the east** : * Takahashi runs A_1 meters per minute for the first T_1 minutes, then runs at A_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever. * Aoki runs B_1 meters per minute for the first T_1 minutes, then runs at B_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever. How many times will Takahashi and Aoki meet each other, that is, come to the same point? We do not count the start of the run. If they meet infinitely many times, report that fact.

t1,t2 = map(int,input().split()) c1,c2 = map(int,input().split()) d1,d2 = map(int,input().split()) hosei = 1 if t1*c1>t1*d1: a1=c1 a2=c2 b1=d1 b2=d2 else: a1,a2,b1,b2=d1,d2,c1,c2 print(a1,a2,b1,b2) if t1*a1+t2*a2 > t1*b1+t2*b2: exit() elif t1*a1+t2*a2 == t1*b1+t2*b2: print("infinity") else: # print((t1*(a1-b1))//(t2*(b2-a2)) + 1,t1*(a1-b1),t2*(b2-a2)) # if t2*(b2-a2) % (t2*(b2-a2)-t1*(a1-b1)) == 0: print((t2*(b2-a2)) // (t2*(b2-a2)-t1*(a1-b1))*2-hosei) # print((t2*(a2-b2)))

s190287775

Accepted

t1,t2 = map(int,input().split()) c1,c2 = map(int,input().split()) d1,d2 = map(int,input().split()) hosei = 0 if t1*c1>t1*d1: a1=c1 a2=c2 b1=d1 b2=d2 else: a1,a2,b1,b2=d1,d2,c1,c2 if t1*a1+t2*a2 > t1*b1+t2*b2: print(0) exit() elif t1*a1+t2*a2 == t1*b1+t2*b2: print("infinity") else: # print((t1*(a1-b1))//(t2*(b2-a2)) + 1,t1*(a1-b1),t2*(b2-a2)) if t2*(b2-a2) % (t2*(b2-a2)-t1*(a1-b1)) == 0: hosei = 1 print((t2*(b2-a2)) // (t2*(b2-a2)-t1*(a1-b1))*2-1-hosei) # print((t2*(a2-b2)))

s287546285

p03534

u796366385

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.

s = input() a = [s.count("a"), s.count("b"), s.count("c")] a.sort(reverse=True) print(a) if a[0] <= a[2] + 1: print("YES") else: print("NO")

s547073178

Accepted

s = input() a = [s.count("a"), s.count("b"), s.count("c")] a.sort(reverse=True) if a[0] <= a[2] + 1: print("YES") else: print("NO")

s443976112

p03636

u226912938

Wrong Answer

The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.

s = str(input()) ans = s[0] + str(len(s[1:-2])) + s[-1] print(ans)

s043328645

Accepted

s = str(input()) ans = s[0] + str(len(s[1:-1])) + s[-1] print(ans)

s338428414

p03361

u071416928

Wrong Answer

We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j). Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`. However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black. Determine if square1001 can achieve his objective.

h, w = map(int,input().split()) s = [["."]*(w+2)] + \ [list("." + input() + ".") for _ in range(h)] + \ [["."]*(w+2)] flg = 0 for i in range(1,h+1): for j in range(1,w+1): if s[i][j] == "#": if s[i-1][j] == "#" or s[i+1][j] == "#" or s[i][-1] == "#" or s[i][j+1] == "#" : flg = 1 if flg == 1 : print("Yse") else : print("No")

s993527792

Accepted

h, w = map(int,input().split()) s = [["."]*(w+2)] + \ [list("." + input() + ".") for _ in range(h)] + \ [["."]*(w+2)] for i in range(1,h+1): for j in range(1,w+1): if s[i][j] == "#": if s[i-1][j] == "." and s[i+1][j] == "." and s[i][j-1] == "." and s[i][j+1] == "." : print("No") exit() print("Yes")

s443050861

p03545

u404676457

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.

(a, b, c, d) = map(int, list(input())) print(a) if a+b+c+d == 7: print(str(a) + '+' + str(b) + '+' + str(c) + '+' + str(d) + '=7') elif a+b+c-d == 7: print(str(a) + '+' + str(b) + '+' + str(c) + '-' + str(d) + '=7') elif a+b-c+d == 7: print(str(a) + '+' + str(b) + '-' + str(c) + '+' + str(d) + '=7') elif a+b-c-d == 7: print(str(a) + '+' + str(b) + '-' + str(c) + '-' + str(d) + '=7') elif a-b+c-d == 7: print(str(a) + '-' + str(b) + '+' + str(c) + '-' + str(d) + '=7') elif a-b-c+d == 7: print(str(a) + '-' + str(b) + '-' + str(c) + '+' + str(d) + '=7') elif a-b-c-d == 7: print(str(a) + '-' + str(b) + '-' + str(c) + '-' + str(d) + '=7')

s661678380

Accepted

(a, b, c, d) = map(int, list(input())) if a+b+c+d == 7: print(str(a) + '+' + str(b) + '+' + str(c) + '+' + str(d) + '=7') elif a+b+c-d == 7: print(str(a) + '+' + str(b) + '+' + str(c) + '-' + str(d) + '=7') elif a+b-c+d == 7: print(str(a) + '+' + str(b) + '-' + str(c) + '+' + str(d) + '=7') elif a-b+c+d == 7: print(str(a) + '-' + str(b) + '+' + str(c) + '+' + str(d) + '=7') elif a+b-c-d == 7: print(str(a) + '+' + str(b) + '-' + str(c) + '-' + str(d) + '=7') elif a-b+c-d == 7: print(str(a) + '-' + str(b) + '+' + str(c) + '-' + str(d) + '=7') elif a-b-c+d == 7: print(str(a) + '-' + str(b) + '-' + str(c) + '+' + str(d) + '=7') elif a-b-c-d == 7: print(str(a) + '-' + str(b) + '-' + str(c) + '-' + str(d) + '=7')

s776494487

p02268

u926657458

Wrong Answer

You are given a sequence of _n_ integers S and a sequence of different _q_ integers T. Write a program which outputs C, the number of integers in T which are also in the set S.

import sys def binsearch(x, s): l = 0 r = len(s) - 1 while (l < r): m = int((l + r) / 2) if s[m] == x: return True else: if s[m] < x: l = m + 1 else: r = m return False ns = sys.stdin.readline() s = sys.stdin.readline().split() nt = sys.stdin.readline() t = sys.stdin.readline().split() out = "" for x in t: if (binsearch(x, s)): out += str(x) + " " print(out.strip())

s441196654

Accepted

n = int(input()) S = list(map(int,input().split())) q = int(input()) T = list(map(int,input().split())) def search(a, X): l = 0 r = len(X) while l < r: m = (l + r ) // 2 if a == X[m]: return m elif a > X[m]: l = m + 1 elif a < X[m]: r = m return None s = 0 for t in T: if search(t,S) is not None: s += 1 print(s)

s349148600

p03385

u697658632

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`.

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

s786341025

Accepted

s = input() t = [] for c in s: t.append(c) if sorted(t) == ['a', 'b', 'c']: print('Yes') else: print('No')

s817430660

p04025

u562015767

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.

import sys n = int(input()) a = list(map(int,input().split())) s = ((sum(a))//n)+1 t = sum(a)//n u = ((sum(a))//n)-1 l = [] l.append(s) l.append(t) l.append(u) ans = [] ans2 = [] aa = set(a) if len(aa) == 1: print(0) sys.exit() for j in range(len(l)): ans = [] for i in range(n): if a[i] == l[j]: continue else: b = (a[i]-l[j])**2 ans.append(b) ans2.append(sum(ans)) print(ans2) print(min(ans2))

s389444556

Accepted

import sys n = int(input()) a = list(map(int,input().split())) s = ((sum(a))//n)+1 t = sum(a)//n u = ((sum(a))//n)-1 l = [] l.append(s) l.append(t) l.append(u) ans = [] ans2 = [] aa = set(a) if len(aa) == 1: print(0) sys.exit() for j in range(len(l)): ans = [] for i in range(n): if a[i] == l[j]: continue else: b = (a[i]-l[j])**2 ans.append(b) ans2.append(sum(ans)) print(min(ans2))

s289752748

p04011

u981206782

Wrong Answer

There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.

n=int(input()) k=int(input()) x=int(input()) y=int(input()) ans=0 if n<k: ans=x*n else: ans=x*n+y*(n-k) print(ans)

s050277921

Accepted

n=int(input()) k=int(input()) x=int(input()) y=int(input()) ans=0 if n<k: ans=x*n else: ans=x*k+y*(n-k) print(ans)

s239727153

p00102

u755162050

Wrong Answer

Your task is to develop a tiny little part of spreadsheet software. Write a program which adds up columns and rows of given table as shown in the following figure:

from sys import stdin if __name__ == '__main__': for n in stdin: if int(n) == 0: break bottom_total = [] right_total = [] for _ in range(int(n)): line = input().split(' ') total = 0 for i, v in enumerate(line): print(repr(int(v)).rjust(5), end='') total += int(v) if len(bottom_total) < _ + 1: bottom_total.append(int(v)) else: bottom_total[_] += int(v) right_total.append(total) print(repr(right_total[_]).rjust(5)) for num in bottom_total: print(repr(num).rjust(5), end='') print(repr(sum(right_total)).rjust(5))

s470592631

Accepted

def output_res(row): print(''.join(map(lambda num: str(num).rjust(5), row)), str(sum(row)).rjust(5), sep='') def main(): while True: n = int(input()) if n == 0: break bottom_total = [0 for _ in range(n)] for _ in range(n): each_line = list(map(int, input().split(' '))) output_res(each_line) bottom_total = [x + y for x, y in zip(bottom_total, each_line)] output_res(bottom_total) if __name__ == '__main__': main()

s734315443

p03997

u568789901

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(int(a*b*h/2))

s586858278

Accepted

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

s828564322

p02928

u538632589

Wrong Answer

We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.

N, K = map(int, input().split()) As = list(map(int, input().split())) MOD = 10**9 + 7 K_1 = K*(1+K) % MOD K_1 /= 2 K_2 = (K_1 - K) % MOD res_next = [0 for i in range(N)] for i in range(N): a = As[i] for j in range(i+1, N): if a > As[j]: res_next[i] += 1 res_prev = [0 for i in range(N)] for i in range(N): a = As[i] for j in range(i): if a > As[j]: res_prev[i] += 1 ans = 0 for i in range(N): ans += (res_next[i] * K_1) % MOD ans += (res_prev[i] * K_2) % MOD ans %= MOD print(ans)

s743363206

Accepted

from decimal import * N, K = map(int, input().split()) As = list(map(int, input().split())) MOD = 10**9 + 7 K_1 = Decimal(K)*Decimal(1+K) / Decimal(2) K_2 = (K_1 - Decimal(K)) % Decimal(MOD) K_1 %= Decimal(MOD) res_next = [0 for i in range(N)] for i in range(N): a = As[i] for j in range(i+1, N): if a > As[j]: res_next[i] += 1 res_prev = [0 for i in range(N)] for i in range(N): a = As[i] for j in range(i): if a > As[j]: res_prev[i] += 1 ans = Decimal(0) for i in range(N): ans += (Decimal(res_next[i]) * K_1) % Decimal(MOD) ans %= Decimal(MOD) ans += (Decimal(res_prev[i]) * K_2) % Decimal(MOD) ans %= Decimal(MOD) print(int(ans))

s011675947

p03998

u792512290

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.

s_a = list(input())[::-1] s_b = list(input())[::-1] s_c = list(input())[::-1] person = "a" try: while True: print("s_a", s_a) print("s_a", s_b) print("s_a", s_c) print("---------") if person == "a": person = s_a.pop() elif person == "b": person = s_b.pop() else: person = s_c.pop() except IndexError: print(person.upper())

s155015702

Accepted

s_a = list(input())[::-1] s_b = list(input())[::-1] s_c = list(input())[::-1] person = "a" try: while True: if person == "a": person = s_a.pop() elif person == "b": person = s_b.pop() else: person = s_c.pop() except IndexError: print(person.upper())

s671425405

p03407

u380772254

Wrong Answer

An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.

a,b,c=map(int,input().split()) print('Yes' if a == c or b==c or a+b==c else 'No')

s517528273

Accepted

a,b,c=map(int,input().split()) print('Yes' if c<=a+b else 'No')

s783448267

p03214

u859897687

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()) l=list(map(int,input().split())) a=sum(l)/n ans=0 b=100 for i in range(n): if b>abs(l[i]-a): b=abs(l[i]-a) ans=i print(i)

s439452767

Accepted

n=int(input()) l=list(map(int,input().split())) a=sum(l)/n ans=0 b=100 for i in range(n): if b>abs(l[i]-a): b=abs(l[i]-a) ans=i print(ans)

s105136523

p03605

u636267225

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

s231533218

Accepted

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

s280089624

p04011

u307418002

Wrong Answer

There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.

n = int(input()) k = int(input()) x = int(input()) y = int(input()) sum = 0 for i in range( n ): if i ==1 : sum += x else : sum += y print(sum)

s393860664

Accepted

n = int(input()) k = int(input()) x = int(input()) y = int(input()) sum = 0 for i in range( n ): if i < k : sum += x else : sum += y print(sum)

s759573224

p03861

u114648678

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?

a,b,x=map(int,input().split()) print(b//x-a//x)

s371736488

Accepted

a,b,x=map(int,input().split()) print(b//x-(a-1)//x)

s325203455

p03387

u227550284

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.

import sys input = sys.stdin.readline def main(): li = list(map(int, input().split())) max_n = max(li) max_i = li.index(max_n) diff = 0 for i in range(3): if i == max_i: continue diff += max_n - li[i] print(diff) if diff % 2 == 0: print(diff//2) else: print((diff+1)//2 + 1) if __name__ == '__main__': main()

s299595652

Accepted

li = list(map(int, input().split())) sum_li = sum(li) max_li = max(li) j = 3 * max_li - sum_li if j % 2 == 0: ans = j // 2 else: ans = (j + 3) // 2 print(ans)

s160320224

p03721

u188745744

Wrong Answer

There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3.

N,K = list(map(int,input().split())) listA = [] for i in range(N): listA.append(list(map(int,input().split()))) now = 0 listA.sort() print(listA) for i in range(N): if now <= K <= now+listA[i][1]: print(listA[i][0]) exit() else: now += listA[i][1]

s721143928

Accepted

N,K = list(map(int,input().split())) listA = [] for i in range(N): listA.append(list(map(int,input().split()))) now = 0 listA.sort() for i in range(N): if now <= K <= now+listA[i][1]: print(listA[i][0]) exit() else: now += listA[i][1]

s032919091

p03564

u319690708

Wrong Answer

Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.

N = int(input()) R = int(input()) max = 0 maxcounter = 0 j = 1 for i in range(N): # counter = 0 # while j > 1: if j * 2 <= j + R: j = j * 2 # counter += 1 else: j = j + R

s067761360

Accepted

N = int(input()) R = int(input()) max = 0 maxcounter = 0 j = 1 for i in range(N): # counter = 0 # while j > 1: if j * 2 <= j + R: j = j * 2 # counter += 1 else: j = j + R print(j)

s380450651

p00228

u350508326

Wrong Answer

電卓などでよく目にするデジタル数字を表示している画面は、デジタル数字が 7 つの部分(セグメン ト)で構成されることから、「7セグメントディスプレイ」と呼ばれています。 ワカマツ社で新しく売り出す予定の製品は、7セグメントディスプレイを製品に組み込むことになり、社員であるあなたは、与えられた数字を 7 セグメントディスプレイに表示するプログラムを作成することになりました。 この7セグメントディスプレイは、次の切り替えの指示が送られてくるまで表示状態は変わりません。7 ビットからなるシグナルを送ることで、それぞれ対応したセグメントの表示情報を切り替える事ができます。ビットとは 1 か 0 の値を持つもので、ここでは 1 が「切り替え」、0 が「そのまま」を表します。 各ビットとセグメントの対応関係は下の図のようになっています。シグナルは 7 つのビットを"gfedcba"の順番に送ります。例えば、非表示の状態から「0」を表示するためには"0111111"をシグナルとしてディスプレイに送らなければなりません。「0」から「5」に変更する場合には"1010010"を送ります。続けて「5」を「1」に変更する場合には"1101011"を送ります。 表示したい n (1 ≤ n ≤ 100) 個の数字を入力とし、それらの数字 di (0 ≤ di ≤ 9) を順に 7 セグメントディスプレイに正しく表示するために必要なシグナル列を出力するプログラムを作成してください。なお、7 セグメントディスプレイの初期状態は全て非表示の状態であるものとします。

#!/usr/bin/env python3 seven_seg = { 0: '0111111', 1: '0000110', 2: '1011011', 3: '1001111', 4: '1100110', 5: '1101101', 6: '1111101', 7: '0100111', 8: '1111111', 9: '1101111' } while True: seven_seg_state = ['0'] * 7 n = int(input()) if n == -1: break for _ in range(n): in_data = int(input()) for i, s in enumerate(seven_seg_state): if seven_seg[in_data][i] == s: print("0", end="") else: print("1", end="") s = seven_seg[in_data][i] print()

s731045893

Accepted

#!/usr/bin/env python3 seven_seg = { 0: '0111111', 1: '0000110', 2: '1011011', 3: '1001111', 4: '1100110', 5: '1101101', 6: '1111101', 7: '0100111', 8: '1111111', 9: '1101111' } while True: seven_seg_state = ['0'] * 7 n = int(input()) if n == -1: break for _ in range(n): in_data = int(input()) for i in range(7): if seven_seg[in_data][i] == seven_seg_state[i]: print("0", end="") else: print("1", end="") seven_seg_state[i] = seven_seg[in_data][i] print()

s660834003

p01052

u283783177

Wrong Answer

太郎君は夏休みの間、毎日１つの映画を近所の映画館で見ることにしました。 （太郎君の夏休みは8月1日から8月31日までの31日間あります。） その映画館では、夏休みの間にn つの映画が上映されることになっています。 それぞれの映画には 1 から n までの番号が割り当てられており、i 番目の映画は8月 ai 日から8月 bi 日の間だけ上映されます。 太郎君は映画を見た時、それが初めて見る映画だった場合は 100 の幸福度を得ることができます。 しかし、過去に 1 度でも見たことのある映画だった場合は 50 の幸福度を得ます。 太郎君は上映される映画の予定表をもとに、夏休みの計画を立てることにしました。 太郎君が得られる幸福度の合計値が最大になるように映画を見たときの合計値を求めてください。 どの日も必ず1つ以上の映画が上映されていることが保証されます。

#!/usr/bin/env python3 # -*- coding: utf-8 -*- N = int(input()) As = [] Bs = [] movies = [[] for x in range(31)] for n in range(N): a, b = map(int, input().split()) a -= 1 b -= 1 As.append(a) Bs.append(b) for day in range(a, b+1): movies[day].append(n) print(movies) picked = [False] * N ans = 0 for day in range(31): # ???????????????????????? unseen_movies = [m for m in movies[day] if not picked[m]] # ??????????????????????????? if len(unseen_movies) == 0: ans += 50 else: lastday = 1000 should_see = -1 for m in unseen_movies: if Bs[m] < lastday: lastday = Bs[m] should_see = m picked[should_see] = True ans += 100 print(ans)

s393189371

Accepted

#!/usr/bin/env python3 # -*- coding: utf-8 -*- N = int(input()) As = [] Bs = [] movies = [[] for x in range(31)] for n in range(N): a, b = map(int, input().split()) a -= 1 b -= 1 As.append(a) Bs.append(b) for day in range(a, b+1): movies[day].append(n) picked = [False] * N ans = 0 for day in range(31): # ???????????????????????? unseen_movies = [m for m in movies[day] if not picked[m]] # ??????????????????????????? if len(unseen_movies) == 0: ans += 50 else: lastday = 1000 should_see = -1 for m in unseen_movies: if Bs[m] < lastday: lastday = Bs[m] should_see = m picked[should_see] = True ans += 100 print(ans)

s429625143

p02613

u761168538

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()) l=[0,0,0,0] for _ in range(n): s=input() if(s=='AC'): l[0]+=1 elif(s=='WA'): l[1]+=1 elif(s=='TLE'): l[2]+=1 elif(s=='RE'): l[3]+=1 print("AC X",l[0]) print("WA X",l[1]) print("TLE X",l[2]) print("RE X",l[3])

s191922704

Accepted

n=int(input()) l=[0,0,0,0] for _ in range(n): s=input() if(s=='AC'): l[0]+=1 elif(s=='WA'): l[1]+=1 elif(s=='TLE'): l[2]+=1 elif(s=='RE'): l[3]+=1 print("AC x",l[0]) print("WA x",l[1]) print("TLE x",l[2]) print("RE x",l[3])

s045512275

p03545

u904331908

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() p = list(map(int,s)) for x in range(2**3): ans = p[0] for y in range(3): if (x>>y)&1: ans += p[y+1] else: ans -= p[y+1] if ans == 7: kotae = format(x,"#05b") kotae = kotae[2:] print(kotae) for z in range(3): print(str(p[z]),end="") if kotae[-z-1] == str(1): print("+",end="") else: print("-",end="") print(str(p[3]) + "=7" )

s051547987

Accepted

A = input() n = len(A)-1 for i in range(2**n): pn = ['-']*n for j in range(n): if ((i>>j)&1): pn[n-1-j] = '+' ans = A[0]+pn[0]+A[1]+pn[1]+A[2]+pn[2]+A[3] if eval(ans) == 7: print(ans+'=7') break

s968323832

p03574

u928784113

Wrong Answer

You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.

H,W = map(int,input().split()) S = [list(input()) for i in range(H)] print(S) dx = [1,1,1,0,0,-1,-1,-1] dy = [1,0,-1,1,-1,1,0,-1] for j in range(H): for i in range(W): cnt = 0 for k in range(8): if 0<=i+dx[k]<W and 0<=j+dy[k]<H: if S[j+dy[k]][i+dx[k]]=="#": cnt += 1 if S[j][i] == "#": continue else: S[j][i] = cnt for i in range(H): for j in range(W): S[i][j] = str(S[i][j]) for i in range(H): print("".join(S[i]))

s104738032

Accepted

H,W = map(int,input().split()) S = [list(input()) for i in range(H)] dx = [1,1,1,0,0,-1,-1,-1] dy = [1,0,-1,1,-1,1,0,-1] for j in range(H): for i in range(W): cnt = 0 for k in range(8): if 0<=i+dx[k]<W and 0<=j+dy[k]<H: if S[j+dy[k]][i+dx[k]]=="#": cnt += 1 if S[j][i] == "#": continue else: S[j][i] = cnt for i in range(H): for j in range(W): S[i][j] = str(S[i][j]) for i in range(H): print("".join(S[i]))

s088259733

p03814

u479638406

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 = list(input()) A = 0 Z = 0 for i in range(len(s)): if s[i] == 'A': A = i break for i in range(len(s)): if s[i] == 'Z': Z = i print(Z - A)

s656120358

Accepted

s = list(input()) A = 0 Z = 0 for i in range(len(s)): if s[i] == 'A': A = i break for i in range(len(s)): if s[i] == 'Z': Z = i print(Z - A + 1)

s658841697

p02613

u642267889

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.

# Judge Status Summary N = int(input()) stringList = [input() for i in range(N)] c0 = 0 c1 = 0 c2 = 0 c3 = 0 for j in range(N): if stringList[j] == 'AC': c0 = c0 + 1 if stringList[j] == 'WA': c1 = c1 + 1 if stringList[j] == 'TLE': c2 = c2 + 1 if stringList[j] == 'RE': c3 = c3 + 1 print('AC×'+ str(c0)) print('WA×'+ str(c1)) print('TLE×'+str(c2)) print('RE×'+ str(c3))

s808636321

Accepted

# Judge Status Summary N = int(input()) stringList = [input() for i in range(N)] c0 = 0 c1 = 0 c2 = 0 c3 = 0 for j in range(N): if stringList[j] == 'AC': c0 = c0 + 1 if stringList[j] == 'WA': c1 = c1 + 1 if stringList[j] == 'TLE': c2 = c2 + 1 if stringList[j] == 'RE': c3 = c3 + 1 print('AC x '+ str(c0)) print('WA x '+ str(c1)) print('TLE x '+str(c2)) print('RE x '+ str(c3))

s408140715

p02842

u361381049

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()) a = [] for i in range(n): if int(i * 1.08) == n: print(int(i * 1.08)) exit() print(':(')

s759587519

Accepted

n = int(input()) import math for i in range(50000): if math.floor(i*1.08) == n: print(i) exit() print(':(')

s321410732

p02902

u893063840

Wrong Answer

Given is a directed graph G with N vertices and M edges. The vertices are numbered 1 to N, and the i-th edge is directed from Vertex A_i to Vertex B_i. It is guaranteed that the graph contains no self-loops or multiple edges. Determine whether there exists an induced subgraph (see Notes) of G such that the in-degree and out-degree of every vertex are both 1. If the answer is yes, show one such subgraph. Here the null graph is not considered as a subgraph.

from queue import Queue INF = 10 * 9 n = 0 g = [] def bfs(s): dist = [INF for _ in range(n)] pre = [-1 for _ in range(n)] q = Queue() dist[s] = 0 q.put(s) while not q.empty(): f = q.get() for i in g[f]: if dist[i] != INF: continue pre[i] = f dist[i] = dist[f] + 1 q.put(i) best = INF, -1 for i in range(n): if i == s: continue for j in g[i]: if j == s: best = min(best, (dist[i], i)) if best[0] == INF: return [INF for _ in range(n + 1)] t = best[1] res = [] while t != -1: res.append(t) t = pre[t] return res def main(): global n, g 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) ans = [INF for _ in range(n + 1)] for i in range(n): now = bfs(i) if len(now) < len(ans): ans = now if len(ans) == n + 1: print(-1) return print(len(ans)) for e in ans: print(e + 1) if __name__ == "__main__": main()

s293156444

Accepted

from collections import deque n, m = map(int, input().split()) ab = [list(map(int, input().split())) for _ in range(m)] INF = 10 ** 5 adj = [[] for _ in range(n)] for a, b in ab: a -= 1 b -= 1 adj[a].append(b) mn = INF for s in range(n): dq = deque([s]) d = [-1] * n p = [-1] * n d[s] = 0 last = [] while dq: u = dq.popleft() for v in adj[u]: if d[v] == -1: d[v] = d[u] + 1 p[v] = u dq.append(v) if v == s: last.append(u) for v in last: route = [v] while p[v] != s: v = p[v] route.append(v) route.append(s) size = len(route) if size < mn: mn = size ans = route[::-1] if mn != INF: print(mn) for e in ans: print(e + 1) else: print(-1)

s194037286

p02393

u506468108

Wrong Answer

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

a, b, c= map(int, input().split()) if a >= b: big = a mid = b else: big = b mid = a if c >= big: small = big big = c else: small = c if small >= mid: small_1 = mid mid = small small = small_1 print("{0} {1} {2}".format(big, mid, small))

s853338683

Accepted

a, b, c= map(int, input().split()) if a >= b: big = a mid = b else: big = b mid = a if c >= big: small = big big = c else: small = c if small >= mid: small_1 = mid mid = small small = small_1 print("{0} {1} {2}".format(small, mid, big))

s810082464

p03129

u496821919

Wrong Answer

Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.

n,k = map(int,input().split()) if 2*k-1 <= n: print("Yes") else: print("No")

s229571486

Accepted

n,k = map(int,input().split()) if 2*k-1 <= n: print("YES") else: print("NO")

s676673215

p00256

u849555829

Wrong Answer

真也君はテレビで「マヤの大予言！2012年で世界が終る？」という番組を見ました。結局、世界が終るかどうかはよくわかりませんでしたが、番組で紹介されていたマヤの「長期暦」という暦に興味を持ちました。その番組では以下のような説明をしていました。 マヤ長期暦は、右の表のような単位からなる、全部で13バクトゥン(187万2000日)で構成される非常に長い暦である。ある計算法では、この暦は紀元前3114年8月11日に始まり2012年12月21日に終わると考えられていて、このため今年の12月21日で世界が終るという説が唱えられている。しかし、13バクトゥンで１サイクルとなり、今の暦が終わったら新しいサイクルに入るだけという考えもある。 | １キン＝１日 １ウィナル＝２０キン １トゥン＝１８ウィナル １カトゥン＝２０トゥン １バクトゥン＝２０カトゥン | ---|---|--- 「ぼくの二十歳の誕生日はマヤ暦だと何日になるのかな？」真也君はいろいろな日をマヤ長期暦で表してみたくなりました。 では、真也君の代わりに、西暦とマヤ長期暦とを相互変換するプログラムを作成してください。

from datetime import date, timedelta st = date(2012, 12, 21) while True: s = input() if s == '#': break d = list(map(int, s.split('.'))) if len(d) == 3: r = 0 while d[0] > 3000: d[0] -= 400 r += 365*400+97 ed = date(d[0], d[1], d[2]) r += (ed-st).days ki = r%20 r//=20 w = r%18 r//=18 t = r%20 r//=20 ka = r%20 r//=20 b = r%13 print(str(b)+'.'+str(ka)+'.'+str(t)+'.'+str(w)+'.'+str(ki)) else: r = d[0] r *= 20 r += d[1] r *= 20 r += d[2] r *= 18 r += d[3] r *= 20 r += d[4] print((st+timedelta(days=r)).strftime("%Y.%m.%d"))

s021849729

Accepted

from datetime import date, timedelta st = date(2012, 12, 21) while True: s = input() if s == '#': break d = list(map(int, s.split('.'))) if len(d) == 3: r = 0 while d[0] > 3000: d[0] -= 400 r += 365*400+97 ed = date(d[0], d[1], d[2]) r += (ed-st).days ki = r%20 r//=20 w = r%18 r//=18 t = r%20 r//=20 ka = r%20 r//=20 b = r%13 print(str(b)+'.'+str(ka)+'.'+str(t)+'.'+str(w)+'.'+str(ki)) else: r = d[0] r *= 20 r += d[1] r *= 20 r += d[2] r *= 18 r += d[3] r *= 20 r += d[4] ed = st+timedelta(days=r) print(str(ed.year)+'.'+str(ed.month)+'.'+str(ed.day))

s802556535

p04012

u653792857

Wrong Answer

Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.

n = input() char_set = { x for x in n} char_freq = { x : n.count(x) for x in char_set } judge = 'YES' for v in char_freq.values(): if(v % 2 != 0): judge = 'NO' print(judge)

s116501254

Accepted

l=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'] w=input() for x in l: if w.count(x)%2!=0: print('No') exit() print('Yes')

s932075151

p02833

u185502200

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()) p = 5 tmp = 0 n //= 2 while n > 1: tmp += n // p n = n // p if n % 2 == 0: print(tmp) else: print(0)

s732761199

Accepted

n = int(input()) p = 5 tmp = 0 o = n n //= 2 while n > 1: tmp += n // p n = n // p if o % 2 == 0: print(tmp) else: print(0)

s326868801

p02694

u955865184

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()) t, ans = 100, 0 while t < x: t += int(t * (1.01)) ans += 1 print(ans)

s403879227

Accepted

x = int(input()) t, ans = 100, 0 while t < x: t = int(t * (1.01)) ans += 1 print(ans)

s046238387

p03737

u331464808

Wrong Answer

You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.

a,b,c = map(str,input().split()) print('a[0],b[0],c[0]'.upper())

s134543755

Accepted

a,b,c = map(str,input().split()) print((a[0]+b[0]+c[0]).upper())

s857737920

p03719

u570018655

Wrong Answer

You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.

a, b, c = map(int, input().split()) if a <= c <= b: ans = "YES" else: ans = "NO" print(ans)

s796107864

Accepted

a, b, c = map(int, input().split()) if a <= c <= b: ans = "Yes" else: ans = "No" print(ans)

s507029412

p03795

u723654028

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()) print(800*N-200//N)

s033702727

Accepted

N=int(input()) print(800*N-N//15*200)

s197283966

p03574

u870297120

Wrong Answer

You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.

h,w = map(int, input().split()) x = [list(map(int, input().replace('.', '0').replace('#', '1'))) for _ in range(h)] x1 = [[0]*(w+2)]*2 for i in range(h): x1.insert(i+1, [0]+x[i]+[0]) x2 = [['#']*w]*h for i in range(h): for j in range(w): if x[i][j] == 1: continue else: x2[i][j] = sum(x1[i][j-1:j+2])+sum(x1[i+1][j-1:j+2])+sum(x1[i+2][j-1:j+2]) print('hoge') for i in x2: print(''.join(map(str, i)).replace('0', '#'))

s579770799

Accepted

h,w = map(int, input().split()) x = [list(input()) for _ in range(h)] y = [[0]*(w+2) for _ in range(h+2)] for i in range(h): for j in range(w): if x[i][j] == '.': y[i+1][j+1] = 0 else: y[i+1][j+1] = 1 for i in range(h): for j in range(w): if x[i][j] == '.': x[i][j] = str(sum(y[i][j:j+3])+sum(y[i+1][j:j+3])+sum(y[i+2][j:j+3])) print(''.join(x[i]))

s857487045

p03970

u612721349

Wrong Answer

CODE FESTIVAL 2016 is going to be held. For the occasion, Mr. Takahashi decided to make a signboard. He intended to write `CODEFESTIVAL2016` on it, but he mistakenly wrote a different string S. Fortunately, the string he wrote was the correct length. So Mr. Takahashi decided to perform an operation that replaces a certain character with another in the minimum number of iterations, changing the string to `CODEFESTIVAL2016`. Find the minimum number of iterations for the rewrite operation.

s="C0DEFESTIVAL2O16" t=input() a=0 for c, d in zip(s, t): if c != d: a += 1 print(a)

s690532994

Accepted

s="CODEFESTIVAL2016" t=input() a=0 for c, d in zip(s, t): if c != d: a += 1 print(a)

s222538764

p03672

u762008592

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.

s = input() while(True): s1, s2 = s[0:len(s)//2], s[len(s)//2:len(s)] if s1 == s2: print(len(s1)*2) break s = s[:-2]

s757467047

Accepted

s = input() s= s[:-2] while(True): s1, s2 = s[0:len(s)//2], s[len(s)//2:len(s)+1] if s1 == s2: print(len(s1)*2) break s = s[:-2]

s721744362

p03387

u814781830

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.

ABC = list(map(int, input().split())) tmp = 0 max_num = max(ABC) abs_list = [abs(max_num-i) for i in ABC] if all(i % 2 == 0 for i in abs_list): res = int(sum(abs_list) / 2) else: res = int((sum(abs_list)) / 2) + 2 print(res)

s878994761

Accepted

ABC = list(map(int, input().split())) max_num = max(ABC) abs_list = [abs(max_num-i) for i in ABC] if sum(abs_list) % 2 == 0: res = int(sum(abs_list) / 2) else: res = int((sum(abs_list)) / 2) + 2 print(res)

s095308552

p03494

u968404618

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())) cont = 0 cont_list =[0] for a in A: if a % 2 != 0: break else: while a % 2==0: a //= 2 cont += 1 cont_list.append(cont) cont=0 print(min(cont_list))

s521570989

Accepted

n = int(input()) A = list(map(int, input().split())) cont = 0 cont_list =[] for a in A: if a % 2 != 0: cont_list.append(cont) break else: while a % 2==0: a //= 2 cont += 1 cont_list.append(cont) cont=0 print(min(cont_list))

s877115473

p03478

u934868410

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(1, n+1): if a <= sum(map(int, list(str(i)))) <= b: ans += 1 print(ans)

s823857158

Accepted

n,a,b = map(int,input().split()) ans = 0 for i in range(1, n+1): if a <= sum(map(int, list(str(i)))) <= b: ans += i print(ans)

s304377135

p04044

u825429469

Wrong Answer

Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.

l,n = map(int, input().split()) s = [str(input()) for _ in range(l)] s.sort() for i in s: print(i)

s664374674

Accepted

l,n = map(int, input().split()) s = [str(input()) for _ in range(l)] s.sort() print(''.join(s))

s120967048

p02417

u506705885

Wrong Answer

Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.

number_of_letter={'a':0,'b':0,'c':0,'d':0,'e':0,'f':0,'g':0,'h':0,'i':0,'j':0,'k':0,'l':0,'m':0,'n':0,'o':0,'p':0 \ ,'q':0,'r':0,'s':0,'t':0,'u':0,'v':0,'w':0,'x':0,'y':0,'z':0} letters=input() number_of_letter['a']=letters.count('a') number_of_letter['b']=letters.count('b') number_of_letter['c']=letters.count('c') number_of_letter['d']=letters.count('d') number_of_letter['e']=letters.count('e') number_of_letter['f']=letters.count('f') number_of_letter['g']=letters.count('g') number_of_letter['h']=letters.count('h') number_of_letter['i']=letters.count('i') number_of_letter['j']=letters.count('j') number_of_letter['k']=letters.count('k') number_of_letter['l']=letters.count('l') number_of_letter['m']=letters.count('m') number_of_letter['n']=letters.count('n') number_of_letter['o']=letters.count('o') number_of_letter['p']=letters.count('p') number_of_letter['q']=letters.count('q') number_of_letter['r']=letters.count('r') number_of_letter['s']=letters.count('s') number_of_letter['t']=letters.count('t') number_of_letter['u']=letters.count('u') number_of_letter['v']=letters.count('v') number_of_letter['w']=letters.count('w') number_of_letter['x']=letters.count('x') number_of_letter['y']=letters.count('y') number_of_letter['z']=letters.count('z') for i in range(97,123): print(chr(i),':',number_of_letter[chr(i)])

s950869472

Accepted

sente="" while True: try: t=input().lower() sente = sente+t except: break for i in range(97,123): print(chr(i),":",sente.count(chr(i)))

s905808442

p02606

u492749916

Wrong Answer

How many multiples of d are there among the integers between L and R (inclusive)?

L, R, d = list(map(int, input().split())) if L%d == 0 or R%d == 0: print(1+ R//d - L//d) else: print(R//d - L//d)

s479598904

Accepted

L, R, d = list(map(int, input().split())) if L%d == 0 and R%d == 0: print(1 + R//d - L//d) else: print(R//d - L//d)

s847226760

p03090

u211160392

Wrong Answer

You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem.

N = int(input()) ans = [[j for j in range(i+1,N)]for i in range(N)] for i in range(N): if len(ans[i]) >= N%2+i+1: ans[i][-(N%2+i+1)] = -1 for i in range(N): for j in ans[i]: if j >= 0: print(i+1,j+1)

s802196823

Accepted

N = int(input()) ans = [[j for j in range(i+1,N)]for i in range(N)] for i in range(N): if len(ans[i]) >= N%2+i+1: ans[i][-(N%2+i+1)] = -1 out = [] for i in range(N): for j in ans[i]: if j >= 0: out.append([i+1,j+1]) print(len(out)) for i,j in out: print(i,j)

s185250878

p02841

u657913472

Wrong Answer

In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month.

a,b,c,d=map(int,open(0).read().split()) print(d<2)

s514571002

Accepted

a,b,c,d=map(int,open(0).read().split()) print(c-a)

s527351320

p03485

u979078704

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) % 2 == 0: print((a + b) / 2) else: print((a + b) / 2 - 0.5)

s682706744

Accepted

a, b = map(int,input().split()) if (a + b) % 2 == 0: print(int((a + b) / 2)) else: print(int((a + b) / 2 + 0.5))

s452587006

p04029

u174849391

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*(1+N)/2)

s017238854

Accepted

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

s409479691

p03140

u128927017

Wrong Answer

You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective?

N = int(input()) A = input() B = input() C = input() count = 0 for i in range(N): if A[i] != B[i]: count+=1 if A[i] != C[i]: count+=1 elif A[i] != C[i]: count+=1 print(count)

s922419141

Accepted

N = int(input()) A = input() B = input() C = input() count = 0 for i in range(N): if A[i] != B[i]: count+=1 if A[i] != C[i] and B[i] != C[i]: count+=1 elif A[i] != C[i]: count+=1 print(count)

s501569917

p02420

u605879293

Wrong Answer

Your task is to shuffle a deck of n cards, each of which is marked by a alphabetical letter. A single shuffle action takes out h cards from the bottom of the deck and moves them to the top of the deck. The deck of cards is represented by a string as follows. abcdeefab The first character and the last character correspond to the card located at the bottom of the deck and the card on the top of the deck respectively. For example, a shuffle with h = 4 to the above deck, moves the first 4 characters "abcd" to the end of the remaining characters "eefab", and generates the following deck: eefababcd You can repeat such shuffle operations. Write a program which reads a deck (a string) and a sequence of h, and prints the final state (a string).

while True: s = input() if s == "-": break m = int(input()) for n in range(m): h = int(input()) s = s[0:h] + s[h:] print(s)

s531975191

Accepted

while True: s = input() if s == "-": break m = int(input()) for n in range(m): h = int(input()) s = s[h:] + s[0: h] print(s)

s563417069

p03407

u977349332

Wrong Answer

An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.

a,b,c = map(int, input().split()) if a+b > c: 'Yes' else: 'No'

s494898428

Accepted

a,b,c = map(int, input().split()) if a+b >= c: print('Yes') else: print('No')

s516702835

p03478

u492532572

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(" ")) sum_total = 0 for n in range(1, N + 1): sum = 0 for s in str(n): sum += int(s) if A <= sum and sum <= B: sum_total += sum print(sum_total)

s381599103

Accepted

N, A, B = map(int, input().split(" ")) sum_total = 0 for n in range(1, N + 1): sum = 0 for s in str(n): sum += int(s) if A <= sum and sum <= B: sum_total += n print(sum_total)

s614491778

p03597

u383508661

Wrong Answer

We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black?

n=int(input()) a=int(input()) print(n**-a)

s354975025

Accepted

n=int(input()) a=int(input()) print(n**2-a)

s462047247

p03680

u490489966

Wrong Answer

Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.

#B import sys n = int(input()) a = [int(input()) for _ in range(n)] lamp = 0 ans=0 for i in range(n): if lamp == i: lamp = a[i] - 1 ans += 1 # print(i,lamp) if lamp == 1: # print(ans) sys.exit() print(-1)

s932108009

Accepted

#B import sys n = int(input()) a = [int(input()) for _ in range(n)] lamp = 1 ans=0 for i in range(n): if lamp - 1 == 1: print(i) sys.exit() lamp = a[lamp - 1] print(-1)

s532386757

p03846

u620157187

Wrong Answer

There are N people, conveniently numbered 1 through N. They were standing in a row yesterday, but now they are unsure of the order in which they were standing. However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person. According to their reports, the difference above for person i is A_i. Based on these reports, find the number of the possible orders in which they were standing. Since it can be extremely large, print the answer modulo 10^9+7. Note that the reports may be incorrect and thus there may be no consistent order. In such a case, print 0.

N = int(input()) A = list(map(int, input().split())) if len(A)%2 == 0: result = 2**(len(A)//2) for i in range(len(A)//2): if A.count(1+2*i) == 2: pass else: result = 0 break else: if A.count(0) == 1: result = 2**((len(A)//2)-1) for i in range(len(A)//2): if A.count(2*i) == 2: pass else: result = 0 break else: result = 0 print(result)

s119958633

Accepted

import numpy as np N = int(input()) A = sorted(list(map(int, input().split()))) if len(A)%2 == 0: result = (2**(len(A)//2))%(10**9+7) a = np.arange(1, len(A), 2).tolist() a = sorted(a+a) if a == A: pass else: result = 0 else: result = 2**(len(A)//2)%(10**9+7) a = np.arange(2, len(A), 2).tolist() a = sorted([0]+a+a) if a == A: pass else: result = 0 print(result)

s048042603

p03476

u694665829

Wrong Answer

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.

dp = [1]*(10**5+1) dp[0],dp[1] = 0, 0 for i in range(2, 10**5+1): if dp[i] == 0: continue j = i while j + i <= 10**5: j += i if dp[j] == 1: dp[j] = 0 ni = [0]*(10**5+1) for i in range(2, 10**5+1): if dp[i] == 1 and dp[(i+1)//2] == 1: ni[i] = 1 ru = [0]*(10**5+1) for i in range(1, 10**5+1): ru[i] = ru[i-1] + ni[i] q = int(input()) for i in range(q): l, r = map(int,input().split()) print(ru[l] - ru[r-1])

s944238063

Accepted

def f(): dp = [1]*(10**5+1) dp[0],dp[1] = 0, 0 for i in range(2, 10**5+1): if dp[i] == 0: continue j = i while j + i <= 10**5: j += i if dp[j] == 1: dp[j] = 0 ni = [0]*(10**5+1) for i in range(2, 10**5+1): if dp[i] == 1 and dp[(i+1)//2] == 1: ni[i] = 1 ru = [0]*(10**5+1) for i in range(1, 10**5+1): ru[i] = ru[i-1] + ni[i] q = int(input()) for i in range(q): l, r = map(int,input().split()) print(ru[r] - ru[l-1]) if __name__ == '__main__': f()

s135352382

p03610

u252964975

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.

S=str(input()) s = "" for i in range(round(len(S)/2)): s = s + S[2*i-1] print(s)

s686457163

Accepted

import math S=str(input()) s = "" for i in range(math.ceil(len(S)/2)): s = s + S[2*i] print(s)

s938154096

p03698

u478417863

Wrong Answer

You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.

a= input().strip() b=set(a) if len(a)==len(b): print("yes") else: print("no") print(len(a),len(b))

s603087465

Accepted

a= input().strip() b=set(a) if len(a)==len(b): print("yes") else: print("no")

s102924576

p03719

u846652026

Wrong Answer

You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.

a,b,c = map(int, input().split()) print("YES" if a < c < b else "No")

s496003779

Accepted

a,b,c = map(int, input().split()) print("Yes" if a <= c <= b else "No")

s165792647

p03795

u871596687

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()) x = 800 * N y = int((N/15)*200) print(x-y)

s812470142

Accepted

n = int(input()) s = int(800 * n - 200 * (n//15)) print(s)

s626263812

p02255

u782850499

Wrong Answer

Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.

size = int(input()) element = list(map(int,input().split())) for i in range(1,len(element)): v = element[i] j = i-1 while j >=0 and element[j] > v: element[j+1] = element[j] j -=1 element[j+1] = v print(" ".join(list(map(str,element))))

s178834024

Accepted

size = int(input()) element = list(map(int,input().split())) print(" ".join(map(str,element))) for i in range(1,len(element)): v = element[i] j = i-1 while j >=0 and element[j] > v: element[j+1] = element[j] j -=1 element[j+1] = v print(" ".join(map(str,element)))

s394038872

p03836

u014333473

Wrong Answer

Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.

sx,sy,tx,ty=map(int,input().split()) dx,dy=tx-sx,ty-sy print('L'+dx*'U'+'U'+dy*'R'+'R'+'D'+dy*'L'+dx*'D'+'D'+dy*'R'+'R'+dy*'U'+'U'+'L'+'D'+dy*'D'+dx*'L')

s519785820

Accepted

sx,sy,tx,ty=map(int,input().split()) dx,dy=tx-sx,ty-sy print(dy*'U'+dx*'R'+dy*'D'+(dx+1)*'L'+(dy+1)*'U'+(dx+1)*'R'+'D'+'R'+(dy+1)*'D'+(dx+1)*'L'+'U')

s009334332

p03477

u163874353

Wrong Answer

A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.

a, b, c, d = map(int, input().split()) l = a + b r = c + d if l > r: print("Left") if l == r: print("Balanced") else: print("Right")

s235242422

Accepted

a, b, c, d = map(int, input().split()) l = a + b r = c + d if l > r: print("Left") elif l == r: print("Balanced") else: print("Right")

s024010837

p03494

u159655606

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 = input().split() A = [int(i) for i in A] check1 = 0 count1 = 0 while check1 == 0: print("this is") for i in range(len(A)): if(A[i]%2 == 0): A[i] = A[i]/2 else: check1 = 1 break count1 += 1 count1 -= 1

s456802044

Accepted

N = int(input()) A = input().split() A = [int(i) for i in A] check1 = 0 count1 = 0 while check1 == 0: for i in range(len(A)): if(A[i]%2 == 0): A[i] = A[i]/2 else: check1 = 1 break count1 += 1 count1 -= 1 print(count1)

s638004110

p02742

u967835038

Wrong Answer

We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:

h,w=map(int,input().split()) if h==1 and w==1: print(1) else: if (h*w)%2==1: print(int(h*w//2)+1) else: print(h*w/2)

s408449676

Accepted

h,w=map(int,input().split()) if h==1 or w==1: print(1) else: if (h*w)%2==1: print(int(int(h*w//2)+1)) else: print(int(h*w/2))

s476746629

p03457

u097026338

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 (t+x+y)%2==0: print("yes") exit() else: print("NO")

s909664622

Accepted

n=int(input()) ta=0 xa=0 ya=0 count=0 for i in range(n): t,x,y=map(int,input().split()) if (x-xa+y-ya)<=t-ta and (t-ta+x-xa+y-ya)%2==0: ta=t xa=x ya=y count+=1 print("Yes" if count==n else "No")

s423536661

p03861

u760961723

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?

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

s195189057

Accepted

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

s184830344

p03719

u371763408

Wrong Answer

You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.

a,b,c = map(int,input().split()) print('YES' if a <= c <= b else 'NO')

s538707817

Accepted

a,b,c = map(int,input().split()) print('Yes' if a <= c <= b else 'No')

s855012616

p03998

u588785393

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.

a = input() b = input() c = input() ac = 0 bc = 0 cc = 0 curr = 'a' while ac < len(a) and bc < len(b) and cc < len(c): if curr == 'a': curr = a[ac] ac += 1 elif curr == 'b': curr = b[bc] bc += 1 elif curr == 'c': curr = c[cc] cc += 1 if ac == 0: print("A") elif bc == 0: print("B") else: print("C")

s797000535

Accepted

a = input() b = input() c = input() ac = 0 bc = 0 cc = 0 curr = 'a' while ac <= len(a) and bc <= len(b) and cc <= len(c): if curr == 'a': if ac == len(a): print("A") break curr = a[ac] ac += 1 elif curr == 'b': if bc == len(b): print("B") break curr = b[bc] bc += 1 elif curr == 'c': if cc == len(c): print("C") break curr = c[cc] cc += 1

s604914489

p03543

u164229553

Wrong Answer

We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?

N = input() N_a = N[:3] N_b = N[1:] N_a_1 = [int(i) for i in N_a] N_b_1 = [int(i) for i in N_b] if str(int(sum(N_a_1)/3)) == N_a or str(int(sum(N_b_1)/3)) == N_b: print("Yes") else: print("No")

s679283720

Accepted

N = input() N_a = N[:3] N_b = N[1:] N_a_1 = int(N_a) N_b_1 = int(N_b) if N_a_1%111 == 0 or N_b_1%111 == 0: print("Yes") else: print("No")

s424836524

p04035

u026788530

Wrong Answer

We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i. At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly: * Choose a (connected) rope with a total length of at least L, then untie one of its knots. Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.

n,l=[int(x) for x in input().split()] a=[int(x) for x in input().split()] j=-1 for i in range(n-1): if a[i]+a[i-1]>=l: j=i break if j==-1: print("Impossible") else: print("Possible") for i in range(j): print(i+1) for i in range(n-1,j,-1): print(i)

s450150456

Accepted

n,l=[int(x) for x in input().split()] a=[int(x) for x in input().split()] j=-1 for i in range(n-1): if a[i]+a[i+1]>=l: j=i break if j==-1: print("Impossible") else: #print(j) print("Possible") for i in range(j): print(i+1) for i in range(n-1,j,-1): print(i)

s014786883

p03671

u612975321

Wrong Answer

Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.

a = list(map(int,input().split())) a.sort() print(a[:2])

s340799288

Accepted

a = list(map(int,input().split())) a.sort() print(sum(a[:2]))

s774590446

p03795

u051496905

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()) # L = 800 / N B = 0 for i in range(N): if i % 15 == 0: B += 1 L = 800 * N C = B * 200 print(L - C)

s008214476

Accepted

N = int(input()) # L = 800 / N B = 0 for i in range(1,N+1): if i % 15 == 0: B += 1 L = 800 * N C = B * 200 print(L - C)

s988512553

p03493

u795733769

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.

li = list(input()) print(li) cnt = 0 101 if (li[0] == "1"): cnt += 1 if (li[1] == "1"): cnt += 1 if (li[2] == "1"): cnt += 1 print(cnt)

s960685904

Accepted

li = list(input()) cnt = 0 101 if (li[0] == "1"): cnt += 1 if (li[1] == "1"): cnt += 1 if (li[2] == "1"): cnt += 1 print(cnt)

s052540362

p03434

u526459074

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.

n = input() a = list(map(int, input().split())) a= sorted(a, reverse=True) print(a) ans =0 for i in a: if i%2 == 0: ans += i elif i%2 ==1: ans -= i print(ans)

s097527976

Accepted

n = input() a = list(map(int, input().split())) a= sorted(a, reverse=True) ans =0 for i in range(len(a)): if i%2 == 0: ans += a[i] elif i%2 ==1: ans -= a[i] print(ans)

s820858754

p02417

u106285852

Wrong Answer

Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.

import sys # inputData inputData = input() cnt = [0]*26 #for str in inputData: for i in range(len(inputData)): if "a" == inputData[i] or inputData[i] =="A": cnt[0] += 1 elif "b" == inputData[i] or inputData[i] =="B": cnt[1] += 1 elif "c" == inputData[i] or inputData[i] =="C": cnt[2] += 1 elif "d" == inputData[i] or inputData[i] =="D": cnt[3] += 1 elif "e" == inputData[i] or inputData[i] =="E": cnt[4] += 1 elif "f" == inputData[i] or inputData[i] =="F": cnt[5] += 1 elif "g" == inputData[i] or inputData[i] =="G": cnt[6] += 1 elif "h" == inputData[i] or inputData[i] =="H": cnt[7] += 1 elif "i" == inputData[i] or inputData[i] =="I": cnt[8] += 1 elif "j" == inputData[i] or inputData[i] =="J": cnt[9] += 1 elif "k" == inputData[i] or inputData[i] =="K": cnt[10] += 1 elif "l" == inputData[i] or inputData[i] =="L": cnt[11] += 1 elif "n" == inputData[i] or inputData[i] =="B": cnt[12] += 1 elif "m" == inputData[i] or inputData[i] =="M": cnt[13] += 1 elif "o" == inputData[i] or inputData[i] =="O": cnt[14] += 1 elif "p" == inputData[i] or inputData[i] =="P": cnt[15] += 1 elif "q" == inputData[i] or inputData[i] =="Q": cnt[16] += 1 elif "r" == inputData[i] or inputData[i] =="R": cnt[17] += 1 elif "s" == inputData[i] or inputData[i] =="S": cnt[18] += 1 elif "t" == inputData[i] or inputData[i] =="T": cnt[19] += 1 elif "u" == inputData[i] or inputData[i] =="U": cnt[20] += 1 elif "v" == inputData[i] or inputData[i] =="V": cnt[21] += 1 elif "w" == inputData[i] or inputData[i] =="W": cnt[22] += 1 elif "x" == inputData[i] or inputData[i] =="X": cnt[23] += 1 elif "y" == inputData[i] or inputData[i] =="Y": cnt[24] += 1 elif "z" == inputData[i] or inputData[i] =="Z": cnt[25] += 1 # ???????????? #print("a:"+a_cnt) print ('a : %d' % (cnt[0])) print ('b : %d' % (cnt[1])) print ('c : %d' % (cnt[2])) print ('d : %d' % (cnt[3])) print ('e : %d' % (cnt[4])) print ('f : %d' % (cnt[5])) print ('g : %d' % (cnt[6])) print ('h : %d' % (cnt[7])) print ('i : %d' % (cnt[8])) print ('j : %d' % (cnt[9])) print ('k : %d' % (cnt[10])) print ('l : %d' % (cnt[11])) print ('n : %d' % (cnt[12])) print ('m : %d' % (cnt[13])) print ('o : %d' % (cnt[14])) print ('p : %d' % (cnt[15])) print ('q : %d' % (cnt[16])) print ('r : %d' % (cnt[17])) print ('s : %d' % (cnt[18])) print ('t : %d' % (cnt[19])) print ('u : %d' % (cnt[20])) print ('v : %d' % (cnt[21])) print ('W : %d' % (cnt[22])) print ('x : %d' % (cnt[23])) print ('y : %d' % (cnt[24])) print ('z : %d' % (cnt[25]))

s975607029

Accepted

# inputData # inputData = input() # cnt = [0]*26 # # if "a" == inputData[i] or inputData[i] =="A": # cnt[0] += 1 # elif "b" == inputData[i] or inputData[i] =="B": # cnt[1] += 1 # elif "c" == inputData[i] or inputData[i] =="C": # cnt[2] += 1 # elif "d" == inputData[i] or inputData[i] =="D": # cnt[3] += 1 # elif "e" == inputData[i] or inputData[i] =="E": # cnt[4] += 1 # elif "f" == inputData[i] or inputData[i] =="F": # cnt[5] += 1 # elif "g" == inputData[i] or inputData[i] =="G": # cnt[6] += 1 # elif "h" == inputData[i] or inputData[i] =="H": # cnt[7] += 1 # elif "i" == inputData[i] or inputData[i] =="I": # cnt[8] += 1 # elif "j" == inputData[i] or inputData[i] =="J": # cnt[9] += 1 # elif "k" == inputData[i] or inputData[i] =="K": # cnt[10] += 1 # elif "l" == inputData[i] or inputData[i] =="L": # cnt[11] += 1 # elif "m" == inputData[i] or inputData[i] =="M": # cnt[12] += 1 # elif "n" == inputData[i] or inputData[i] =="N": # cnt[13] += 1 # elif "o" == inputData[i] or inputData[i] =="O": # cnt[14] += 1 # elif "p" == inputData[i] or inputData[i] =="P": # cnt[15] += 1 # elif "q" == inputData[i] or inputData[i] =="Q": # cnt[16] += 1 # elif "r" == inputData[i] or inputData[i] =="R": # cnt[17] += 1 # elif "s" == inputData[i] or inputData[i] =="S": # cnt[18] += 1 # elif "t" == inputData[i] or inputData[i] =="T": # cnt[19] += 1 # elif "u" == inputData[i] or inputData[i] =="U": # cnt[20] += 1 # elif "v" == inputData[i] or inputData[i] =="V": # cnt[21] += 1 # elif "w" == inputData[i] or inputData[i] =="W": # cnt[22] += 1 # elif "x" == inputData[i] or inputData[i] =="X": # cnt[23] += 1 # elif "y" == inputData[i] or inputData[i] =="Y": # cnt[24] += 1 # elif "z" == inputData[i] or inputData[i] =="Z": # cnt[25] += 1 # # print ('a : %d' % (cnt[0])) # print ('b : %d' % (cnt[1])) # print ('c : %d' % (cnt[2])) # print ('d : %d' % (cnt[3])) # print ('e : %d' % (cnt[4])) # print ('f : %d' % (cnt[5])) # print ('g : %d' % (cnt[6])) # print ('h : %d' % (cnt[7])) # print ('i : %d' % (cnt[8])) # print ('j : %d' % (cnt[9])) # print ('k : %d' % (cnt[10])) # print ('l : %d' % (cnt[11])) # print ('m : %d' % (cnt[12])) # print ('n : %d' % (cnt[13])) # print ('o : %d' % (cnt[14])) # print ('p : %d' % (cnt[15])) # print ('r : %d' % (cnt[17])) # print ('s : %d' % (cnt[18])) # print ('t : %d' % (cnt[19])) # print ('u : %d' % (cnt[20])) # print ('v : %d' % (cnt[21])) # print ('w : %d' % (cnt[22])) # print ('x : %d' % (cnt[23])) # print ('y : %d' % (cnt[24])) import sys string = [] for line in sys.stdin: string += line.lower() A = 'abcdefghijklmnopqrstuvwxyz' for cnt0 in A: print(cnt0+' : {}'.format(string.count(cnt0)))

s574420403

p03672

u634079249

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.

import sys import os ii = lambda: int(sys.stdin.buffer.readline().rstrip()) il = lambda: list(map(int, sys.stdin.buffer.readline().split())) fl = lambda: list(map(float, sys.stdin.buffer.readline().split())) iln = lambda n: [int(sys.stdin.buffer.readline().rstrip()) for _ in range(n)] iss = lambda: sys.stdin.buffer.readline().decode().rstrip() sl = lambda: list(map(str, sys.stdin.buffer.readline().decode().split())) isn = lambda n: [sys.stdin.buffer.readline().decode().rstrip() for _ in range(n)] MOD = 10 ** 9 + 7 def main(): if os.getenv("LOCAL"): sys.stdin = open("input.txt", "r") S = list(iss()) S = S[:-1] for _ in range(len(S)): l = len(S) if l % 2 == 0 and S[:l // 2] == S[l // 2:]: print("".join(S)) exit() else: S = S[:-1] if __name__ == '__main__': main()

s029719823

Accepted

import sys import os ii = lambda: int(sys.stdin.buffer.readline().rstrip()) il = lambda: list(map(int, sys.stdin.buffer.readline().split())) fl = lambda: list(map(float, sys.stdin.buffer.readline().split())) iln = lambda n: [int(sys.stdin.buffer.readline().rstrip()) for _ in range(n)] iss = lambda: sys.stdin.buffer.readline().decode().rstrip() sl = lambda: list(map(str, sys.stdin.buffer.readline().decode().split())) isn = lambda n: [sys.stdin.buffer.readline().decode().rstrip() for _ in range(n)] MOD = 10 ** 9 + 7 def main(): if os.getenv("LOCAL"): sys.stdin = open("input.txt", "r") S = list(iss()) S = S[:-1] for _ in range(len(S)): l = len(S) if l % 2 == 0 and S[:l // 2] == S[l // 2:]: print(len(S)) exit() else: S = S[:-1] if __name__ == '__main__': main()

s445793181

p03679

u897328029

Wrong Answer

Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.

x, a, b = list(map(int, input().split())) d = b - a if d <= 0: ans = 'delicious' elif d <= a: ans = 'safe' else: ans = 'dangerous'

s050951396

Accepted

x, a, b = list(map(int, input().split())) d = b - a if d <= 0: ans = 'delicious' elif d <= x: ans = 'safe' else: ans = 'dangerous' print(ans)

s831866373

p03434

u587213169

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.

N=int(input()) cards = list(map(int, input().split())) cards.reverse() Alice=sum(cards[0:N:2]) Bob=sum(cards[1:N:2]) print(Alice-Bob)

s767676603

Accepted

N=int(input()) cards = list(map(int, input().split())) cards.sort(reverse=True) Alice=0 Bob=0 for i in range (len(cards)): if i%2==0: Alice+=cards[i] else: Bob+=cards[i] print(Alice-Bob)

s799229830

p02831

u995163736

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.

from math import gcd a, b = map(int,input().split()) print(a * b / gcd(a, b))

s508265347

Accepted

from math import gcd a, b = map(int,input().split()) print(int(a * b / gcd(a, b)))

s967436231

p04045

u085334230

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 = set(map(int, input().split())) while True: if set(str(N)) == set(str(N)) - D: print(N) break else: N += 1

s598066926

Accepted

N, K = map(int, input().split()) D = set(map(str, input().split())) f = True while f: if set(list(str(N))) == set(list(str(N))) - D: print(N) f = False else: N += 1

s250684978

p03338

u697658632

Wrong Answer

You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.

n = int(input()) s = input() v = [0] * n for c in range(ord('a'), ord('z') + 1): p = -1 for i in range(n): if s[i] == c: if p >= 0: v[i] -= 1 v[p] += 1 p = i ans = 0 vsum = 0 for i in v: vsum += i ans = max(ans, vsum) print(ans)

s770305232

Accepted

n = int(input()) s = input() v = [0] * n for c in range(ord('a'), ord('z') + 1): p = -1 for i in range(n): if ord(s[i]) == c: if p >= 0: v[i] -= 1 v[p] += 1 p = i ans = 0 vsum = 0 for i in v: vsum += i ans = max(ans, vsum) print(ans)

s306001840

p02396

u400765446

Wrong Answer

In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.

def main(): i = 1 while True: x = input() if x == '0': break else: print("case "+str(i)+": "+x) if __name__ == "__main__": main()

s776866347

Accepted

i = 1 while True: x = int(input()) if x == 0: break else: print("Case {}: {}".format(i,x)) i += 1

s769751683

p02694

u767821815

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?

F = 100 K = int(input()) count = 0 while F <= K: F *= 1.01 F = int(F) count += 1 print(count)

s124533295

Accepted

import math F = 100 K = int(input()) count = 0 while F < K: F = F + math.floor(F*0.01) count += 1 print(count)