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CodeNet4Repair

like 0

s859823771

p03455

u306060982

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')

s154630834

Accepted

a,b = map(int,input().split()) if a * b % 2 == 1: print('Odd') else: print('Even')

s181881245

p03371

u974792613

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()) price1 = x*a + y*b price2 = max(x, y)*2*c if x<=y: price3 = x*2*c price3 += (y-x)*b else: price3 = y*2*c price3 += (y-x)*a print(min(price1, price2, price3))

s036769270

Accepted

a, b, c, x, y = map(int, input().split()) price1 = x * a + y * b price2 = max(x, y) * 2 * c price3 = min(x, y) * 2 * c + a * (x - min(x, y)) + b * (y - (min(x, y))) print(min(price1, price2, price3))

s129045001

p03720

u075595666

Wrong Answer

There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?

n,m = [int(i) for i in input().split()] ab = [] for i in range(m): ay = list(map(int,input().split())) ab.append(ay) ans = [] for i in range(n): ans.append(0) for i in range(m): for j in range(1,n+1): #print(i,j) if j in ab[i]: ans[j-1] += 1 print(ans)

s688317959

Accepted

n,m = [int(i) for i in input().split()] ab = [] for i in range(m): ay = list(map(int,input().split())) ab.append(ay) ans = [] for i in range(n): ans.append(0) for i in range(m): for j in range(1,n+1): #print(i,j) if j in ab[i]: ans[j-1] += 1 for i in range(n): print(ans[i])

s085672589

p03971

u137214546

Wrong Answer

There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.

def qsimulator(s, a, b): paas = 0 bpaas = 0 aflg = True bflg = True total = a + b res = [] for i in s: if i == 'c': res.append("No") elif i == 'a': if aflg and paas <= total: paas += 1 res.append("Yes") else: aflg = False res.append("No") else: if bflg and paas <= total and bpaas < b: paas += 1 bpaas += 1 res.append("Yes") else: bflg = False res.append("No") return res n, a, b = map(int, input().split()) s = input() ans = qsimulator(s, a, b) for i in ans: print(i)

s401685904

Accepted

def qsimulator(s, a, b): paas = 0 bpaas = 0 aflg = True bflg = True total = a + b res = [] for i in s: if i == 'c': res.append("No") elif i == 'a': if aflg and paas < total: paas += 1 res.append("Yes") else: aflg = False res.append("No") else: if bflg and paas < total and bpaas < b: paas += 1 bpaas += 1 res.append("Yes") else: bflg = False res.append("No") return res n, a, b = map(int, input().split()) s = input() ans = qsimulator(s, a, b) for i in ans: print(i)

s641312151

p03997

u331464808

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()) s = (a+b)*h/2 print(s)

s193582908

Accepted

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

s722561816

p02612

u642098073

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)

s237171990

Accepted

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

s760630573

p03380

u842950479

Wrong Answer

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

import math n=int(input().rstrip()) A=list(map(int, input().rstrip().split())) A.sort() work=int() bestr=int() def comb(x,y): return(math.factorial(x)/(math.factorial(x-y)*math.factorial(y))) for i in range(n-2): if comb(A[n-1],A[i-1])>work: work=comb(A[n-1],A[i-1]) bestr=A[i-1] print("{0} {1}".format(A[n-1],bestr))

s855564195

Accepted

n=int(input().rstrip()) A=list(map(int, input().rstrip().split())) A.sort() work=int() bestr=int(A[0]) #def comb(x,y): # return(math.factorial(x)/(math.factorial(x-y)*math.factorial(y))) # if comb(A[n-1],A[i-1])>work: for i in range(n-2): if abs(A[i]-A[n-1]/2)<abs(bestr-A[n-1]/2): bestr=A[i] print("{0} {1}".format(A[n-1],bestr))

s893063212

p03360

u690536347

Wrong Answer

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

l=list(map(int,input().split())) k=int(input()) m=max(l) l.pop(l.index(m)) print(sum(l)+m**k)

s887700651

Accepted

l=list(map(int,input().split())) k=int(input()) m=max(l) l.pop(l.index(m)) print(sum(l)+m*(2**k))

s329520971

p04030

u751047721

Wrong Answer

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

A = [] S = input() S=list(str(S)) for n in S: if n == "b": A.pop(-1) else: A.append(n) print("".join(A))

s432837259

Accepted

I = input() A ='' for n in I: if n == '0': A += '0' elif n == '1': A += '1' else: if len(A) > 0: A = A[:-1] print(A)

s357404829

p03486

u875028418

Wrong Answer

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

a,b = input(),input() print("Yes" if a<b else "No")

s212919300

Accepted

print("Yes" if sorted(input())<sorted(input(),reverse=1) else "No")

s543268333

p02613

u205145698

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=map(int,input().split()) S=list(map(str,input().split())) AC=S.count('AC') WA=S.count('WA') TLE=S.count('TLE') RE=S.count('RE') print("AC×", end='') print(AC) print("WA×", end='') print(WA) print("TLE×", end='') print(TLE) print("RE×", end='') print(RE)

s403731975

Accepted

N=int(input()) S=[] for i in range(0, N): val = input() S.append(val) ac=S.count('AC') wa=S.count('WA') tle=S.count('TLE') re=S.count('RE') print("AC x ", end='') print(ac) print("WA x ", end='') print(wa) print("TLE x ", end='') print(tle) print("RE x ", end='') print(re)

s023944327

p04011

u588081069

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.

String = str(input()) ## print(char.count(String) & 1) result = {} for char in String: try: result[char] += 1 except KeyError: result[char] = 1 result = list(map(lambda x: x % 2, list(result.values()))) if result == [0] * len(result): print('Yes') else: print('No')

s648354649

Accepted

N = int(input()) K = int(input()) X = int(input()) Y = int(input()) result = 0 for i in range(1, N+1): if i <= K: result += X else: result += Y print(result)

s085842488

p02389

u071333111

Wrong Answer

Write a program which calculates the area and perimeter of a given rectangle.

#!/usr/bin/env python a, b = map(int, input().split()) print(a*b)

s090454378

Accepted

#!/usr/bin/env python a, b = map(int, input().split()) print("{} {}".format(a*b, 2*(a+b)))

s110215076

p02613

u957843607

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()) str_lis = [] C0 = 0 C1 = 0 C2 = 0 C3 = 0 for i in range(N): result = input() if result == "AC": C0 += 1 elif result == "WA": C1 += 1 elif result == "TLE": C2 += 1 else: C3 += 1 print("AC × ", C0) print("WA × ", C1) print("TLE × ", C2) print("RE × ", C3)

s928887045

Accepted

N = int(input()) C0 = 0 C1 = 0 C2 = 0 C3 = 0 for i in range(N): result = input() if result == "AC": C0 += 1 elif result == "WA": C1 += 1 elif result == "TLE": C2 += 1 else: C3 += 1 print("AC x", C0) print("WA x", C1) print("TLE x", C2) print("RE x", C3)

s370181330

p03448

u198062737

Wrong Answer

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

A = int(input()) B = int(input()) C = int(input()) X = int(input()) ans = 0 for i in range(0, A + 1): for j in range(0, B + 1): for k in range(0, C + 1): if i * 500 + j * 100 + k * 50 == X: ans print(ans)

s501175206

Accepted

A = int(input()) B = int(input()) C = int(input()) X = int(input()) ans = 0 for i in range(0, A + 1): for j in range(0, B + 1): for k in range(0, C + 1): if i * 500 + j * 100 + k * 50 == X: ans += 1 print(ans)

s723646579

p03095

u787456042

Wrong Answer

You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order.

n,s=open(0);a=1 for c in set(s):a*=s.count(c)+1 print(~-a%(10**9+7))

s603508348

Accepted

n,s=open(0);a=1 for c in set(s.strip()):a*=s.count(c)+1 print(~-a%(10**9+7))

s605560671

p03351

u226912938

Wrong Answer

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

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

s514370591

Accepted

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

s996221933

p03997

u271044469

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)

s034194677

Accepted

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

s152056984

p03564

u646792990

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()) k = int(input()) ans = 1 cnt = n while ans <= k: ans *= 2 cnt -= 1 if cnt == 1: break print(ans) ans += cnt * k print(ans)

s370071994

Accepted

n = int(input()) k = int(input()) ans = 1 for _ in range(n): ans = min(2 * ans, ans + k) print(ans)

s323944514

p02613

u760760982

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.

num = int(input()) word = {"AC":0, "WA":0, "TLE":0, "RE":0} for i in range(num): judge = input() if judge in word: word[str(judge)] += 1 else: none print("AC × " + str(word["AC"])) print("WA × " + str(word["WA"])) print("TLE × " + str(word["TLE"])) print("RE × " + str(word["RE"]))

s686138300

Accepted

num = int(input()) word = {"AC":0, "WA":0, "TLE":0, "RE":0} for i in range(num): judge = input() if judge in word: word[str(judge)] += 1 else: pass print("AC x " + str(word["AC"])) print("WA x " + str(word["WA"])) print("TLE x " + str(word["TLE"])) print("RE x " + str(word["RE"]))

s413139521

p03471

u201565171

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())) flg=1 for i in range(N): for j in range(N): for k in range(N): if i+j+k==N and 10000*i+5000*j+1000*k==Y: flg=0 x=i y=j z=k if flg==1: print('-1 -1 -1') else: print(x,end=" ") print(y,end=" ") print(z,end="")

s756085021

Accepted

N,Y=list(map(int,input().split())) flg=1 for i in range(N+1): for j in range(N-i+1): k=N-(i+j) if 10000*i+5000*j+1000*k==Y: flg=0 x=i y=j z=k if flg==1: print('-1 -1 -1') else: print(x,end=" ") print(y,end=" ") print(z)

s188838560

p03457

u308588791

Wrong Answer

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

n = int(input()) for i in range(n): t, x, y = map(int, input().split()) if x + y < t or (x + y + t) % 2 == 0: print("No") exit() print("Yes")

s503937898

Accepted

N = int(input()) for _ in range(N): t, x, y = map(int, input().split()) if x + y > t or (t + x + y)%2 != 0: print("No") quit() print("Yes")

s831881453

p03494

u227210643

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.

C=int(input()) D=list(map(int,input().split())) S=0 flag=0 while(flag==0): for I in range(C): if D[I]%2==0: D[I]/=2 if I==(C-1): S+=1 else: flag=1 break print(D,S) print(S)

s328976501

Accepted

C=int(input()) D=list(map(int,input().split())) S=0 flag=0 while(flag==0): for I in range(C): if D[I]%2==0: D[I]/=2 if I==(C-1): S+=1 else: flag=1 break print(S)

s878874013

p02613

u951985579

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()) AC = 0 WA = 0 TLE = 0 RE = 0 for i in range(N): s = input() if s == 'AC': AC += 1 elif s == 'WA': WA += 1 elif s == 'TLE': TLE += 1 else: RE += 1 print('AC', '×', AC) print('WA', '×', WA) print('TLE', '×', TLE) print('RE', '×', RE)

s234310798

Accepted

ans = { 'AC' : 0, 'WA' : 0, 'TLE' : 0, 'RE' : 0 } for i in range(int(input())): ans[input()] += 1 for k, v in ans.items(): print(k, 'x', v)

s560639792

p03713

u089376182

Wrong Answer

There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle. Snuke is trying to divide the bar as evenly as possible. More specifically, he is trying to minimize S_{max} \- S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece. Find the minimum possible value of S_{max} - S_{min}.

h,w = map(int, input().split()) h1,h2 = h//3, h//3+1 w1,w2= min(w//2, w//2+1), max(w//2, w//2+1) h3,h4 = min(h//2, h//2+1), max(h//2, h//2+1) w3,w4 = w//3, w//3+1 ans = [] a,b,c = h1*w, (h-h1)*w1, (h-h1)*w2 ans.append(max([a,b,c])-min([a,b,c])) a,b,c = h2*w, (h-h2)*w1, (h-h2)*w2 ans.append(max([a,b,c])-min([a,b,c])) a,b,c = w3*h, (w-w3)*h3, (w-w3)*h4 ans.append(max([a,b,c])-min([a,b,c])) a,b,c = w4*h, (w-w4)*h3, (w-w4)*h4 ans.append(max([a,b,c])-min([a,b,c])) ans.append(h) ans.append(w)

s929755277

Accepted

h,w = map(int, input().split()) h1,h2 = h//3, h//3+1 w1,w2= min(w//2, w//2+1), max(w//2, w//2+1) h3,h4 = min(h//2, h//2+1), max(h//2, h//2+1) w3,w4 = w//3, w//3+1 ans = [] a,b,c = h1*w, (h-h1)*w1, (h-h1)*w2 ans.append(max([a,b,c])-min([a,b,c])) a,b,c = h2*w, (h-h2)*w1, (h-h2)*w2 ans.append(max([a,b,c])-min([a,b,c])) a,b,c = w3*h, (w-w3)*h3, (w-w3)*h4 ans.append(max([a,b,c])-min([a,b,c])) a,b,c = w4*h, (w-w4)*h3, (w-w4)*h4 ans.append(max([a,b,c])-min([a,b,c])) ans.append(h) ans.append(w) if w%2==0: ans.append(abs(h1*w-(h-h1)*w//2)) ans.append(abs(h2*w-(h-h2)*w//2)) if h%2==0: ans.append(abs(w3*h-(w-w3)*h//2)) ans.append(abs(w4*h-(w-w4)*h//2)) if h%3==0 or w%3==0: ans.append(0) print(min(ans))

s094839978

p02612

u415995713

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

s995652055

Accepted

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

s867981962

p03478

u655975843

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 = 0 for i in range(1, n + 1): a1 = n // 10000 a2 = n // 1000 a3 = n // 100 a4 = n // 10 k = a1 + (a2 - a1*10) + (a3 - a2*10) + (a4 - a3*10) if (k >= a) and (k <= b): sum += k print(sum)

s034536871

Accepted

n, a, b = map(str, input().split()) a = int(a) b = int(b) N = int(n) sum = 0 for i in range(1, N + 1): k = 0 num = list(str(i)) for j in range(len(num)): k = k + int(num[j]) if (k >= a) and (k <= b): sum += i print(sum)

s171272760

p03679

u217303170

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=map(int,input().split()) if a > b: print('delicious') elif a < b < x: print('safe') else: print('dangerous')

s874810131

Accepted

X,A,B=map(int,input().split()) if X<B-A: print("dangerous") elif B-A<=0: print("delicious") else: print("safe")

s764387089

p03485

u166621202

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.

import math a,b = map(int,input().split()) print(math.floor(a/b))

s766550339

Accepted

import math a,b = map(int,input().split()) print(math.ceil((a+b)/2))

s702030916

p03360

u584558499

Wrong Answer

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

numbers = list(map(int, input().split())) K = int(input()) max_index = numbers.index(max(numbers)) result = 0 print('max: ', numbers[max_index]) result = [numbers[i] * 2 ** K if i == max_index else numbers[i] for i in range(len(numbers))] print(sum(result))

s245320209

Accepted

numbers = [int(i) for i in input().split()] K = list(map(int, input().split()))[0] max_index = numbers.index(max(numbers)) result = [numbers[i] * 2 ** K if i == max_index else numbers[i] for i in range(len(numbers))] print(sum(result))

s433644014

p02928

u969062493

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()) a = list(map(int, input().split())) inv = 0 for i in range(n - 1): for j in range(i + 1, n): if a[i] > a[j]: inv += 1 print(k * (k - 1) / 2) total_inv = k * inv + k * (k - 1) / 2 * inv dst = int(total_inv % (10e+9 + 7)) print(dst)

s270422403

Accepted

n, k = map(int, input().split()) a = list(map(int, input().split())) intra_inv = 0 for i in range(n): for j in range(i + 1, n): if a[i] > a[j]: intra_inv += 1 inter_inv = 0 for i in range(n): for j in range(n): if a[i] > a[j]: inter_inv += 1 total_inv = k * intra_inv + k * (k - 1) // 2 * inter_inv dst = int(total_inv % (10**9 + 7)) print(dst)

s414475656

p03545

u170077602

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.

row = input() row = list(row) row = list(map(int, row)) for i in range(2): sum = row[0] if i == 0: op1 = "+" sum += row[1] else: op1 = "-" sum -= row[1] for j in range(1): if j == 0: op2 = "+" sum += row[2] else: op2 = "-" sum -= row[2] for k in range(1): if j == 0: op3 = "+" sum += row[3] else: op3 = "-" sum -= row[3] if sum == 7: break print(row[0],op1,row[1],op2,row[2],op3,row[3],"=7", sep="")

s366312668

Accepted

row = input() row = list(row) row = list(map(int, row)) Flag = False for i in range(2): for j in range(2): for k in range(2): sum = row[0] if i == 0: op1 = "+" sum += row[1] else: op1 = "-" sum -= row[1] if j == 0: op2 = "+" sum += row[2] else: op2 = "-" sum -= row[2] if k == 0: op3 = "+" sum += row[3] else: op3 = "-" sum -= row[3] if sum == 7: Flag = True break if Flag: break if Flag: break print(row[0],op1,row[1],op2,row[2],op3,row[3],"=7", sep="")

s115995260

p03494

u953753178

Time Limit Exceeded

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

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

s028695004

Accepted

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

s403187138

p03861

u252805217

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()) l = b - a + 1 ans = l // x + len([i for i in range(a, a+l%x) if i % x == 0]) print(x, (a, b), ans)

s390717141

Accepted

a, b, x = map(int, input().split()) def bibe(n, x): return n // x ans = bibe(b, x) - bibe(a-1, x) print(ans)

s021863815

p03693

u886712136

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,c=input().split() num=int(a+b+c) if num%4==0: print("Yes") else: print("No")

s562591761

Accepted

a,b,c=input().split() num=int(a+b+c) if num%4==0: print("YES") else: print("NO")

s399747836

p03471

u993138647

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.

s=[[int(j) for j in input().split()] for i in range (1)] N=s[0][0] Y=s[0][1] ans=list() for x in range(0,N+1): for y in range(0,N+1): for z in range(0,N+1): if 10000*x+5000*y+1000*z==Y: ans=[x,y,z] if len(ans)==0: ans=[-1,-1,-1] print(ans[0],ans[1],ans[2])

s224286320

Accepted

s=[[int(j) for j in input().split()] for i in range (1)] N=s[0][0] Y=s[0][1] ans=[-1,-1,-1] for x in range(0,N+1): for y in range(0,N+1): z=N-x-y if 10000*x+5000*y+1000*z==Y and z>=0: ans=[x,y,z] print(ans[0],ans[1],ans[2])

s013857785

p03377

u419535209

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.

def A(): A, B, X = [int(a) for a in input().split()] possible = all([ A <= X, X <= A + B ]) if possible: print('YES') else: print('NO')

s411817992

Accepted

A, B, X = [int(a) for a in input().split()] possible = all([ A <= X, X <= A + B ]) if possible: print('YES') else: print('NO')

s791765853

p03721

u798093965

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 = map(int,input().split()) count = 0 ans = 0 for i in range(n): a,b =map(int,input().split()) print(a,b) count += b if count >= k: ans = a break print(ans)

s379239944

Accepted

from operator import itemgetter n,k = map(int,input().split()) count = 0 ans = 0 a = [list(map(int,input().split())) for i in range(n)] number = sorted(a, key = lambda x: x[0]) for i in range(n): count += number[i][1] if count >= k: ans = number[i][0] break print(ans)

s476714665

p03494

u625428807

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())) count = 0 flag = False while flag == False: for ai in a: if ai % 2 == 0: ai = ai / 2 else: flag = True if flag == True: break count += 1 print(count)

s759063677

Accepted

import math n = input() a = list(map(int, input().split())) ans = float("inf") for i in a: ans = min(ans, len(bin(i)) - bin(i).rfind("1") - 1) print(round(ans))

s682088946

p02603

u346308892

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 numpy as np from functools import * import sys sys.setrecursionlimit(100000) input = sys.stdin.readline def array(size, init=0): return [[init for j in range(size[1])] for i in range(size[0])] def acinput(): return list(map(int, input().split(" "))) def II(): return int(input()) directions = np.array([[1, 0], [0, 1], [-1, 0], [0, -1]]) directions = list(map(np.array, directions)) mod = 10**9+7 def factorial(n): fact = 1 for integer in range(1, n + 1): fact *= integer return fact def permutate(x, A): n = len(x) res = [-1]*n for i in range(n): res[i] = x[A[i]] return res def generate_bin(digit): for b in range(2**digit): b = bin(b) tmp = b[2:].zfill(digit) yield list(map(int, tmp)) # yield b def search(x, count): #print("top", x, count) for d in directions: nx = d+x # print(nx) if np.all(0 <= nx) and np.all(nx < (H, W)): if field[nx[0]][nx[1]] == "E": count += 1 field[nx[0]][nx[1]] = "V" count = serch(nx, count) continue if field[nx[0]][nx[1]] == "#": field[nx[0]][nx[1]] = "V" count = serch(nx, count) return count N = II() A = acinput() A.insert(0, A[0]+1) A.append(A[-1]-1) tmp = [] for i in range(len(A)): if A[i] != A[i-1]: tmp.append(A[i]) A = tmp x = 1000 k = 0 for i in range(len(A)-1): print(i, x, k) if A[i-1] > A[i] and A[i] < A[i+1]: k += x//A[i] x = x % A[i] elif A[i-1] < A[i] and A[i] > A[i+1]: #print(i, x, k) x += A[i]*k k = 0 print(x) # print(A)

s426767576

Accepted

import numpy as np from functools import * import sys sys.setrecursionlimit(100000) input = sys.stdin.readline def array(size, init=0): return [[init for j in range(size[1])] for i in range(size[0])] def acinput(): return list(map(int, input().split(" "))) def II(): return int(input()) directions = np.array([[1, 0], [0, 1], [-1, 0], [0, -1]]) directions = list(map(np.array, directions)) mod = 10**9+7 def factorial(n): fact = 1 for integer in range(1, n + 1): fact *= integer return fact def permutate(x, A): n = len(x) res = [-1]*n for i in range(n): res[i] = x[A[i]] return res def generate_bin(digit): for b in range(2**digit): b = bin(b) tmp = b[2:].zfill(digit) yield list(map(int, tmp)) # yield b def search(x, count): #print("top", x, count) for d in directions: nx = d+x # print(nx) if np.all(0 <= nx) and np.all(nx < (H, W)): if field[nx[0]][nx[1]] == "E": count += 1 field[nx[0]][nx[1]] = "V" count = serch(nx, count) continue if field[nx[0]][nx[1]] == "#": field[nx[0]][nx[1]] = "V" count = serch(nx, count) return count N = II() A = acinput() A.insert(0, A[0]+1) A.append(A[-1]-1) tmp = [] for i in range(len(A)): if A[i] != A[i-1]: tmp.append(A[i]) A = tmp x = 1000 k = 0 for i in range(len(A)-1): #print(i, x, k) if A[i-1] > A[i] and A[i] < A[i+1]: k += x//A[i] x = x % A[i] elif A[i-1] < A[i] and A[i] > A[i+1]: #print(i, x, k) x += A[i]*k k = 0 print(x) # print(A)

s281011376

p02259

u844704750

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.

N = int(input()) A = list(map(int, input().split(" "))) def print_list(A): for a in A[:-1]: print ("%d "%a, end="") print(A[-1]) def BubbleSort(A, N): print_list(A) count = 0 flag = True while flag: flag = False for j in range(N-1, 0, -1): if A[j] < A[j-1]: v = A[j] A[j] = A[j-1] A[j-1] = v count += 1 flag = True return A, count A_sorted, c = BubbleSort(A, N) print_list(A_sorted) print(c)

s774166459

Accepted

N = int(input()) A = list(map(int, input().split(" "))) def print_list(A): for a in A[:-1]: print ("%d "%a, end="") print(A[-1]) def BubbleSort(A, N): count = 0 flag = True while flag: flag = False for j in range(N-1, 0, -1): if A[j] < A[j-1]: v = A[j] A[j] = A[j-1] A[j-1] = v count += 1 flag = True return A, count A_sorted, c = BubbleSort(A, N) print_list(A_sorted) print(c)

s294187382

p03679

u763881112

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.

import numpy as np import fractions as fra x,a,b=map(int,input().split()) if(a>=b): print("delicious") elif(x>=b): print("safe") else: print("dangerous")

s110803427

Accepted

import numpy as np import fractions as fra x,a,b=map(int,input().split()) if(a>=b): print("delicious") elif(a+x>=b): print("safe") else: print("dangerous")

s186809195

p03433

u055687574

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()) n %= 500 print(n) if n <= a: print("Yes") else: print("No")

s168487821

Accepted

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

s946725619

p03351

u626337957

Wrong Answer

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

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

s476687763

Accepted

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

s495415046

p03471

u696197059

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 = map(int,input().split()) list = {} for i in range(n): for j in range(n - i): for k in range(n - i - j): if (i*10000 + j * 5000 + k* 1000) == y: list = [i,j,k] if len(list) == 0: list = [-1,-1,-1] print(list)

s225321488

Accepted

n ,y = map(int,input().split()) list = [-1,-1,-1] for i in range(n+1): for j in range(n+1 - i): if (j *10000 + i * 5000 + (n-i-j)* 1000) == y: list = [i,j,(n-i-j)] break if (j *10000 + i * 5000 + (n-i-j)* 1000) == y: break print(list[1],list[0],list[2])

s251156906

p02850

u531599639

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.

from collections import deque n = int(input()) tree = [[] for i in range(n+1)] for _ in range(n-1): a,b = map(int,input().split()) tree[a].append(b) tree[b].append(a) m = len(max(tree,key=len)) col = [0]*(n+1) Q = deque([1]) print(tree) while Q: temp = Q.popleft() pre = col[temp] now = 1 for n in tree[temp]: if col[n]: continue if now!=pre: col[n]=now now += 1 else: col[n]=now+1 now += 2 Q.append(n) print(m,*col[2:])

s648175621

Accepted

from collections import deque n = int(input()) tree = [[] for i in range(n+1)] edge = [] for _ in range(n-1): a,b = map(int,input().split()) tree[a].append(b) tree[b].append(a) edge.append((a,b)) m = len(max(tree,key=len)) col = [0]*(n+1) col[1]='a' Q = deque([1]) #print(tree) ans = {} while Q: temp = Q.popleft() now = 1 for n in tree[temp]: if col[n]: continue if now == col[temp]: now += 1 col[n] = now ans[min(temp,n),max(temp,n)] = now now += 1 Q.append(n) print(m) for a,b in edge: print(ans[(a,b)])

s904295696

p03380

u597455618

Wrong Answer

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

from scipy.special import comb import bisect def main(): n = int(input()) a = list(map(int, input().split())) a.sort(reverse=True) ans = [0, 0] chk = 0 for i in range(n): for j in range(bisect.bisect_left(a, a[i]//2), bisect.bisect_right(a, a[i]//2+1)): tmp = max(comb(a[i], a[j], exact=True), comb(a[j], a[i], exact=True)) if chk < tmp: chk = tmp ans[0] = a[j] ans[1] = a[i] print(*ans) if __name__ == "__main__": main()

s818284323

Accepted

def main(): n = int(input()) a = list(map(int, input().split())) l = max(a) r = 0 chk = 10**9+1 for x in a: if abs(x-l//2) < chk and x != l: r = x chk = abs(x-l//2) print(l, r) if __name__ == "__main__": main()

s584473780

p03160

u896741788

Wrong Answer

There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.

n=int(input()) l=list(map(int,input().split())) dp=[float("INF")]*n dp[0]=0 for i in range(2,n): for j in range(1,3): dp[i]=min(dp[i-j]+abs(l[i]-l[i-j]),dp[i]) print(dp[-1])

s567629006

Accepted

n=int(input()) dp=[10**4*n]*n l=list(map(int,input().split())) dp[0]=0 dp[1]=abs(l[0]-l[1]) for i in range(2,n): dp[i]=min(dp[i-1]+abs(l[i]-l[i-1]),dp[i-2]+abs(l[i]-l[i-2])) print(dp[-1])

s883133401

p03644

u931889893

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.

n = int(input()) results = 0 number = 0 for i in range(n+1): if i == 0: continue count = 0 while True: if i % 2 == 0: count += 1 i = i / 2 else: if results < count: results = count number = n break print(number)

s404039257

Accepted

n = int(input()) results = 0 number = 0 for i in range(n + 1): if i == 0: continue if i == 1: number = 1 continue count = 0 v = i while v % 2 == 0: count += 1 v = v / 2 if results < count: results = count number = i continue print(number)

s113915620

p03997

u769411997

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)

s813398439

Accepted

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

s139854547

p03997

u325206354

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

s192057005

Accepted

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

s475088835

p03456

u882370611

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.

a,b=input().split() c=int(a+b) print('Yes' if c**(1/2)==c else 'No')

s606466819

Accepted

a,b=input().split() c=int(a+b) print('Yes' if (int(c**(1/2)))**2==c else 'No')

s368663524

p03657

u428132025

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.

a, b = map(int, input().split()) if a % 3 == 0 or b % 3 == 0 or (a+b) % 3 == 0: print('Yes') else: print('No')

s484320529

Accepted

a, b = map(int, input().split()) if a % 3 == 0 or b % 3 == 0 or (a+b) % 3 == 0: print('Possible') else: print('Impossible')

s677879039

p03471

u846552659

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.

# -*- coding:utf-8 -*- import sys n, y = map(int, input().split()) for a in range(n+1): for b in range(n-a+1): for c in range(n-a-b+1): if 10000*a+5000*b+1000*c == y: print(str(a)+' '+str(b)+' '+str(c)) sys.exit() else: pass print("-1 -1 -1")

s494985624

Accepted

# -*- coding:utf-8 -*- import sys n, y = map(int, input().split()) for a in range(n+1): for b in range(n+1-a): if 5000*a+9000*b == 10000*n-y and n-a-b >= 0: print(str(n-a-b)+' '+str(a)+' '+str(b)) sys.exit() else: pass ''' for a in range(n+1): for b in range(n-a+1): for c in range(n-a-b+1): if 10000*a+5000*b+1000*c == y: print(str(a)+' '+str(b)+' '+str(c)) sys.exit() else: pass ''' print("-1 -1 -1")

s557463192

p03636

u429319815

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 = (list(input())) S = s[0] + str(len(s)) + s[-1] print(S)

s599040912

Accepted

s = (list(input())) S = s[0] + str(len(s)-2) + s[-1] print(S)

s550518267

p02747

u068142202

Wrong Answer

A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.

s = input() check = "hi" if s in check: print("Yes") else: print("No")

s188539602

Accepted

import re s = input() if len(s) % 2 == 1: s += "a" li = re.split('(..)',s)[1::2] check = "hi" ans = 0 for i in li: if check not in i: ans += 1 if ans == 0: print("Yes") else: print("No")

s375183840

p03760

u027165539

Wrong Answer

Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.

o = input() e = input() s = '' for i in range(len(o)): s += o[i] if i + 1 == len(o): break s += e[i] print(s)

s648743528

Accepted

o = input() e = input() s = '' for i in range(len(o)): s += o[i] if i + 1 > len(e): break s += e[i] print(s)

s388998064

p03605

u779308281

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?

b = list(input()) num = 0 for i in b: if i == "9": num += 1 break if num == 1: print("YES") else: print("NO")

s671866835

Accepted

b = list(input()) num = 0 for i in b: if i == "9": num = 1 break if num == 1: print("Yes") else: print("No")

s832501878

p04043

u209616713

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.

w = input() l = list(w) l.sort() ans = l[0] + l[1] + l[2] if l == '557': print('YES') else: print('NO')

s966161724

Accepted

w = input() l = list(w) l.sort() ans = str(l[2]) + str(l[3]) + str(l[4]) if ans == '557': print('YES') else: print('NO')

s847519749

p03730

u646412443

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 = list(map(int, input().split())) i = 1 while(True): if (a * i) % b == 0: print('No') break elif (a * i) % b == c: print('Yes') break i += 1

s657775199

Accepted

a, b, c = list(map(int, input().split())) i = 1 while(True): if (a * i) % b == 0: print('NO') exit() elif (a * i) % b == c: print('YES') exit() i += 1

s581274823

p03644

u928784113

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.

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

s474823201

Accepted

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

s321214782

p00003

u626838615

Wrong Answer

Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so.

n = int(input()) for i in range(n): side = list(map(int,input().split())) side.sort() if(side[0]**2 + side[1]**2 == side[2]**2): print("Yes") else: print("No")

s800700164

Accepted

n = int(input()) for i in range(n): side = list(map(int,input().split())) side.sort() if(side[0]**2 + side[1]**2 == side[2]**2): print("YES") else: print("NO")

s572805811

p03470

u360061665

Wrong Answer

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

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

s276675119

Accepted

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

s574728266

p03720

u584529823

Wrong Answer

There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?

N, M = map(int, input().split()) cities = [list(map(int, input().split())) for _ in range(M)] print(cities) roads = [0] * N for city in cities: roads[city[0] - 1] += 1 roads[city[1] - 1] += 1 for road in roads: print(road)

s556226637

Accepted

N, M = map(int, input().split()) cities = [list(map(int, input().split())) for _ in range(M)] roads = [0] * N for city in cities: roads[city[0] - 1] += 1 roads[city[1] - 1] += 1 for road in roads: print(road)

s706660562

p03860

u246572032

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.

s = input() print('A' + s[1] + 'C')

s212544541

Accepted

a,b,c = input().split() print('A' + b[0] + 'C')

s022261524

p04030

u957872856

Wrong Answer

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

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

s164973702

Accepted

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

s121287427

p03150

u272557899

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() t = "keyence" p = 0 for i in range(len(s)): if p == 0: if i <= 6: if s[i] != t[i]: k = i p = 1 elif p == 1: if s[i] == t[k]: j = i p = 2 if i == len(s) - 1: print("No") exit() elif p == 2: if k + i - j <= 6: if s[i] != t[k + i - j]: print("No") exit() else: print("No") exit() print("Yes")

s683147269

Accepted

s = input() t = "keyence" if t in s: print("YES") exit() if len(s) >= 7: if "k" in s and "eyence" in s: if s[0] == "k" and s[-1] == "e" and s[-2] == "c" and s[-3] == "n" and s[-4] == "e" and s[-5] == "y" and s[-6] == "e": print("YES") exit() if "ke" in s and "yence" in s: if s[0] == "k" and s[-1] == "e" and s[-2] == "c" and s[-3] == "n" and s[-4] == "e" and s[-5] == "y" and s[1] == "e": print("YES") exit() if "key" in s and "ence" in s: if s[0] == "k" and s[-1] == "e" and s[-2] == "c" and s[-3] == "n" and s[-4] == "e" and s[2] == "y" and s[1] == "e": print("YES") exit() if "keye" in s and "nce" in s: if s[0] == "k" and s[-1] == "e" and s[-2] == "c" and s[-3] == "n" and s[3] == "e" and s[2] == "y" and s[1] == "e": print("YES") exit() if "keyen" in s and "ce" in s: if s[0] == "k" and s[-1] == "e" and s[-2] == "c" and s[4] == "n" and s[3] == "e" and s[2] == "y" and s[1] == "e": print("YES") exit() if "keyenc" in s and "e" in s: if s[0] == "k" and s[-1] == "e" and s[5] == "c" and s[4] == "n" and s[3] == "e" and s[2] == "y" and s[1] == "e": print("YES") exit() print("NO")

s162323798

p00042

u024715419

Wrong Answer

宝物がたくさん収蔵されている博物館に、泥棒が大きな風呂敷を一つだけ持って忍び込みました。盗み出したいものはたくさんありますが、風呂敷が耐えられる重さが限られており、これを超えると風呂敷が破れてしまいます。そこで泥棒は、用意した風呂敷を破らず且つ最も価値が高くなるようなお宝の組み合わせを考えなくてはなりません。 風呂敷が耐えられる重さ W、および博物館にある個々のお宝の価値と重さを読み込んで、重さの総和が W を超えない範囲で価値の総和が最大になるときの、お宝の価値総和と重さの総和を出力するプログラムを作成してください。ただし、価値の総和が最大になる組み合わせが複数あるときは、重さの総和が小さいものを出力することとします。

import itertools case = 0 while True: case += 1 w_max = int(input()) if w_max == 0: break n = int(input()) t = [] for i in range(n): t.append(list(map(int, input().split(",")))) v_best = 0 w_best = w_max p_all = itertools.product([0,1], repeat=n) for p in p_all: v = 0 w = 0 for i in range(n): v += p[i]*t[i][0] w += p[i]*t[i][1] if v > v_best and w < w_max: v_best = v w_best = w elif v == v_best and w < w_best: v_best = v w_best = w print("Case:",case) print(v_best) print(w_best)

s651325012

Accepted

case = 0 while True: case += 1 w_max = int(input()) if w_max == 0: break n = int(input()) w = [0 for i in range(n)] v = [0 for i in range(n)] for i in range(n): a, b = map(int, input().split(",")) v[i] = a w[i] = b dp = [[0 for j in range(w_max + 1)] for i in range(n + 1)] for i in range(n)[::-1]: for j in range(w_max + 1): if j < w[i]: dp[i][j] = dp[i + 1][j] else: dp[i][j] = max(dp[i + 1][j], dp[i + 1][j - w[i]] + v[i]) print("Case ",case,":",sep="") print(max(dp[0])) print(dp[0].index(max(dp[0])))

s722842247

p02260

u698762975

Wrong Answer

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

def inserrtionsort(l,n): for i in range(1,n): print(" ".join(l)) v=l[i] j=i-1 while int(v)<int(l[j]) and j>=0: l[j+1]=l[j] j-=1 l[j+1]=v print(" ".join(l)) m=int(input()) n=list(input().split()) inserrtionsort(n,m)

s724279406

Accepted

def selectionsort(l,n): c=0 for i in range(n): minj=i for j in range(i,n): if int(l[j])<int(l[minj]): minj=j l[i],l[minj]=l[minj],l[i] if i!=minj: c+=1 print(" ".join(l)) print(c) n=int(input()) l=list(input().split()) selectionsort(l,n)

s284312787

p03493

u853900545

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.

s=list(map(int,input().split())) print(s.count(1))

s344288444

Accepted

s=input() print(s.count('1'))

s026987989

p03719

u641460756

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 c>=a and c>=b: print("Yes") else: print("No")

s389688763

Accepted

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

s826275955

p03609

u305965165

Wrong Answer

We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand. How many grams of sand will the upper bulb contains after t seconds?

x, t = (int(i) for i in input().split()) print(min(x-t, 0))

s607305428

Accepted

x, t = (int(i) for i in input().split()) print(max(x-t, 0))

s131983047

p03694

u214380782

Wrong Answer

It is only six months until Christmas, and AtCoDeer the reindeer is now planning his travel to deliver gifts. There are N houses along _TopCoDeer street_. The i-th house is located at coordinate a_i. He has decided to deliver gifts to all these houses. Find the minimum distance to be traveled when AtCoDeer can start and end his travel at any positions.

x=input().split() maxX=max(x) minX=min(x) print(int(maxX)-int(minX))

s500175454

Accepted

n = int(input()) x= [int(i) for i in input().split()] maxX=max(x) minX=min(x) print(int(maxX)-int(minX))

s325209081

p02741

u812587837

Wrong Answer

Print the K-th element of the following sequence of length 32: 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51

a = list(map(int, '1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51'.split(','))) print(a) k = int(input()) print(a[k-1])

s411629132

Accepted

a = list(map(int, '1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51'.split(','))) k = int(input()) print(a[k-1])

s783094335

p03547

u642012866

Wrong Answer

In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`. When X and Y are seen as hexadecimal numbers, which is larger?

X, Y = input().split() if X > Y: print(">") elif Y < X: print("<") else: print("=")

s593113588

Accepted

X, Y = input().split() if X > Y: print(">") elif Y > X: print("<") else: print("=")

s215584786

p03486

u062691227

Wrong Answer

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

aa, bb = open(0).read().splitlines() aa = sorted(aa) bb = sorted(bb, reverse=True) for a, b in zip(aa, bb): if a == b: continue else: if a<b: print('Yes') else: print('No') break if len(aa)<len(bb): print('Yes') else: print('No')

s444783879

Accepted

aa, bb = open(0).read().splitlines() aa = sorted(aa) bb = sorted(bb, reverse=True) import sys for a, b in zip(aa, bb): if a == b: continue else: if a<b: print('Yes') sys.exit() else: print('No') sys.exit() break if len(aa)<len(bb): print('Yes') else: print('No')

s749557344

p02613

u972991614

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()) i = 1 AC = 0 WA = 0 TLE = 0 RE = 0 while i < N+1: a = str(input()) if a == "AC": AC = AC + 1 elif a == "WA": WA = WA + 1 elif a == "TLE": TLE = TLE + 1 elif a == "RE": RE = RE + 1 i = i + 1 print("AC×"+str(AC)) print("WA×" +str(WA)) print("TLE×" + str(TLE)) print("RE×"+str(RE))

s287925726

Accepted

N = int(input()) x = [] for _ in range(N): s = input() x.append(s) print('AC'+ ' '+ 'x'+ ' '+str(x.count('AC'))) print('WA'+ ' '+ 'x'+ ' '+str(x.count('WA'))) print('TLE'+ ' '+ 'x'+ ' '+str(x.count('TLE'))) print('RE'+ ' '+ 'x'+ ' '+str(x.count('RE')))

s350953204

p03486

u521866787

Wrong Answer

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

a= input() b=input() print("No" if sorted(a)>=sorted(b) else "Yes")

s051525679

Accepted

a= input() b=input() print("Yes" if sorted(a)<sorted(b, reverse=True) else "No")

s749321690

p03779

u268516119

Wrong Answer

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

import math X=int(input()) for t in range(math.ceil(math.sqrt(X))): if t*(t+1)//2>=X: print(t) break

s526661373

Accepted

import math X=int(input()) for t in range(X+1): if t*(t+1)//2>=X: print(t) break

s586482012

p02612

u153955689

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()) ans = N % 1000 print(ans)

s208130554

Accepted

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

s148568672

p03351

u143509139

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()) print('YNeos'[abs(a-c)<=d or (abs(a-b)<=d and abs(b-c)<=d)::2])

s251099525

Accepted

a,b,c,d=map(int,input().split()) print('NYoe s'[abs(a-c)<=d or (abs(a-b)<=d and abs(b-c)<=d)::2])

s351230570

p03778

u835482198

Wrong Answer

AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved.

w, a, b = map(int, input().split()) print(abs(b - a))

s045510658

Accepted

w, a, b = map(int, input().split()) if abs(a - b) <= w: print(0) else: print(abs(a - b) - w)

s137053473

p03564

u382639013

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()) K = int(input()) i = 0 ans = 1 while True: print(ans) if i == N: break if ans < K: ans *= 2 i += 1 else: ans += K i += 1 print(ans)

s324577703

Accepted

N = int(input()) K = int(input()) i = 0 ans = 1 while True: if i == N: break if ans < K: ans *= 2 i += 1 else: ans += K i += 1 print(ans)

s715866543

p03170

u884077550

Wrong Answer

There is a set A = \\{ a_1, a_2, \ldots, a_N \\} consisting of N positive integers. Taro and Jiro will play the following game against each other. Initially, we have a pile consisting of K stones. The two players perform the following operation alternately, starting from Taro: * Choose an element x in A, and remove exactly x stones from the pile. A player loses when he becomes unable to play. Assuming that both players play optimally, determine the winner.

n,k = map(int,input().split()) arr = list(map(int,input().split())) dp = [False]*(k+1) for stones in range(k+1): for x in arr: if stones >= x and dp[stones-x] == False: dp[stones] = True print(dp) if dp[k]: print("First") else: print("Second")

s963356136

Accepted

n,k = map(int,input().split()) arr = list(map(int,input().split())) dp = [False]*(k+1) for stones in range(k+1): for x in arr: if stones >= x and dp[stones-x] == False: dp[stones] = True if dp[k]: print("First") else: print("Second")

s322144240

p03494

u405328424

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())) cnt = 0 print(N) print(len(A)) while(1): exit = False for i in range(len(A)): if(A[i]%2 == 1): exit = True if(exit == True): break for i in range(len(A)): A[i] /= 2 cnt += 1 print(cnt)

s874350289

Accepted

N = int(input()) A = list(map(int,input().split())) cnt = 0 while(1): exit = False for i in range(len(A)): if(A[i]%2 == 1): exit = True if(exit == True): break for i in range(len(A)): A[i] /= 2 cnt += 1 print(cnt)

s700501518

p02275

u130834228

Wrong Answer

Counting sort can be used for sorting elements in an array which each of the n input elements is an integer in the range 0 to k. The idea of counting sort is to determine, for each input element x, the number of elements less than x as C[x]. This information can be used to place element x directly into its position in the output array B. This scheme must be modified to handle the situation in which several elements have the same value. Please see the following pseudocode for the detail: Counting-Sort(A, B, k) 1 for i = 0 to k 2 do C[i] = 0 3 for j = 1 to length[A] 4 do C[A[j]] = C[A[j]]+1 5 /* C[i] now contains the number of elements equal to i */ 6 for i = 1 to k 7 do C[i] = C[i] + C[i-1] 8 /* C[i] now contains the number of elements less than or equal to i */ 9 for j = length[A] downto 1 10 do B[C[A[j]]] = A[j] 11 C[A[j]] = C[A[j]]-1 Write a program which sorts elements of given array ascending order based on the counting sort.

def CountingSort(A, B, k): for i in range(k+1): C[i] = 0 for j in range(1, n+1): C[A[j]] += 1 for i in range(1, k+1): C[i] = C[i] + C[i-1] for j in range(1, n+1)[::-1]: B[C[A[j]]] = A[j] C[A[j]] -= 1 n = int(input()) A = list(map(int, input().split()))

s645127154

Accepted

def CountingSort(A, B, k): C = [0 for i in range(k+1)] n = len(A) for j in range(n): C[A[j]] += 1 for i in range(1, k+1): C[i] = C[i] + C[i-1] for j in range(n)[::-1]: B[C[A[j]]-1] = A[j] #print(C[A[j]]) C[A[j]] -= 1 n = int(input()) A = list(map(int, input().split())) B = [0 for i in range(n)] CountingSort(A, B, max(A)) print(*B)

s068644588

p02612

u415053349

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)

s083204242

Accepted

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

s766302750

p03679

u216631280

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 = map(int, input().split()) if a <= b: print('delicious') elif a + x <= b: print('safe') else: print('dagerous')

s953118625

Accepted

x, a, b = map(int, input().split()) if a >= b: print('delicious') elif a + x >= b: print('safe') else: print('dangerous')

s361111642

p03544

u252964975

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)

N=int(input()) L1 = 2 L2 = 1 if N==1:print(L1) if N==2:print(L2) if N>3: N=N-2 for i in range(N): L=L1+L2 L1 = L2 L2 = L print(L)

s849181806

Accepted

N=int(input()) N=N+1 L1 = 2 L2 = 1 if N==1:print(L1) if N==2:print(L2) if N>2: N=N-2 for i in range(N): L=L1+L2 L1 = L2 L2 = L print(L)

s882300214

p03612

u905582793

Wrong Answer

You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this.

n=int(input()) p=list(map(int,input().split())) p.append(0) ansls=[0] if p[0]==1: ansls[0]=1 for i in range(1,n): if p[i] ==i+1: if p[i-1] == i: ansls[-1]+=1 else: ansls.append(1) ans=0 for i in range(len(ansls)): ans+= (ansls[i]+1)//2 print(ans)

s689032542

Accepted

n=int(input()) p=list(map(int,input().split())) p.append(0) ansls=[0] if p[0]==1: ansls[0]=1 for i in range(1,n): if p[i] ==i+1: if p[i-1] == i: ansls[-1]+=1 else: ansls.append(1) ans=0 for i in range(len(ansls)): ans+= (ansls[i]+1)//2 print(ans)

s274217340

p02694

u806403461

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()) sm = 100 count = 0 while sm >= x: count += 1 sm += int(0.01*sm) print(count)

s815827431

Accepted

x = int(input()) sm = 100 count = 0 while sm < x: count += 1 sm += int(0.01*sm) print(count)

s130971922

p02659

u688203790

Wrong Answer

Compute A \times B, truncate its fractional part, and print the result as an integer.

a,b = map(float, input().split()) print(round(a * b))

s015210753

Accepted

a,b =input().split() a = int(a) b = b.replace('.','') print(a*int(b)//100)

s556308104

p03598

u652081898

Wrong Answer

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

K = int( input() ) list = map( int, input().split() ) x = 0 for num in list: a = num * 2 b = (K - num) * 2 if b <= a: x += b elif b > a: x += a print( x )

s689957546

Accepted

# -*- coding: utf-8 -*- N = int( input() ) K = int( input() ) list = map( int, input().split() ) x = 0 for num in list: a = num * 2 b = (K - num) * 2 if b <= a: x += b elif b > a: x += a print( x )

s184167798

p03957

u703890795

Wrong Answer

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

S = input() f = 0 for s in S: if f==0: if s=="C": f = 1 elif f==1: if s=="C": f = 2 if f == 2: print("Yes") else: print("No")

s015718023

Accepted

S = input() f = 0 for s in S: if f==0: if s=="C": f = 1 elif f==1: if s=="F": f = 2 if f == 2: print("Yes") else: print("No")

s673955017

p03377

u492749916

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()) if X < A: print("No") elif X - A <= B: print("Yes") else: print("No")

s002619501

Accepted

A, B, X = list(map(int, input().split())) if X < A or A + B < X: print("NO") else: print("YES")

s019660304

p04043

u897328029

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.

a, b, c = list(map(int, input().split())) s = tuple(sorted([a, b, c])) ans = 'YES' if s == (7,7,5) else 'NO' print(ans)

s090969979

Accepted

a, b, c = list(map(int, input().split())) s = tuple(sorted([a, b, c])) ans = 'YES' if s == (5,5,7) else 'NO' print(ans)

s888075775

p03485

u531599639

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+1)

s524194168

Accepted

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

s152426392

p04029

u017719431

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)

s623351652

Accepted

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

s397448006

p02742

u011276976

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()) total = h * w if w % 2: if h % 2: #both wide, high = odd ans = (total + 1)/2 print(ans) exit() ans = total/2 print(ans)

s920487902

Accepted

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

s883826439

p03434

u623735583

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()) card = list(map(int,input().split())) a = 0 b = 0 card.sort(reverse=True) #print(card) for i in range(1,n+1): print(a,b) if i % 2 == 0: b = b + card[i-1] else: a += card[i-1] #print(a,b) print(a - b)

s972596177

Accepted

n = int(input()) c = list(map(int,input().split())) a,b = 0,0 c.sort(reverse=True) for i,j in enumerate(c): if i % 2 == 0: a += j else: b += j print(a-b)