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CodeNet4Repair

like 0

s375127162

p02853

u170296094

Wrong Answer

We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned.

x, y = input().split() s = 0 if x == 1: s += 300000 elif x == 2: s += 200000 elif x == 3: s += 100000 else: s += 0 if y == 1: s += 300000 elif y == 2: s += 200000 elif y == 3: s += 100000 else: s += 0 if x == 1 and y == 1: s += 400000 print(s)

s542227202

Accepted

x, y = input().split() s = 0 if int(x) == 1: s += 300000 elif int(x) == 2: s += 200000 elif int(x) == 3: s += 100000 else: s += 0 if int(y) == 1: s += 300000 elif int(y) == 2: s += 200000 elif int(y) == 3: s += 100000 else: s += 0 if int(x) == 1 and int(y) == 1: s += 400000 print(s)

s817518603

p00015

u723913470

Wrong Answer

A country has a budget of more than 81 trillion yen. We want to process such data, but conventional integer type which uses signed 32 bit can represent up to 2,147,483,647. Your task is to write a program which reads two integers (more than or equal to zero), and prints a sum of these integers. If given integers or the sum have more than 80 digits, print "overflow".

import sys N=int(input()) N=int(N/2) for i in range(N): a=input() b=input() len_a=len(a) len_b=len(b) c=[] kuriage=0 if(len_a>=len_b): for i in range(len_b): c.append((kuriage+int(a[len_a-1-i])+int(b[len_b-1-i]))%10) kuriage=(kuriage+int(a[len_a-1-i])+int(b[len_b-1-i]))//10 for i in range(len_b,len_a): c.append((kuriage+int(a[len_a-1-i]))%10) kuriage=(kuriage+int(a[len_a-1-i]))//10 else : for i in range(len_a): c.append((kuriage+int(a[len_a-1-i])+int(b[len_b-1-i]))%10) kuriage=(kuriage+int(a[len_a-1-i])+int(b[len_b-1-i]))//10 for i in range(len_a,len_b): c.append((kuriage+int(b[len_b-1-i]))%10) kuriage=(kuriage+int(b[len_b-1-i]))//10 if(kuriage==1): c.append(1) if(len(c)>=80): print('overflow') else: for i in c[::-1]: print(i,end='') print('')

s991448816

Accepted

import sys N=int(input()) for i in range(N): a=input() b=input() len_a=len(a) len_b=len(b) c=[] kuriage=0 if(len_a>=len_b): for i in range(len_b): c.append((kuriage+int(a[len_a-1-i])+int(b[len_b-1-i]))%10) kuriage=(kuriage+int(a[len_a-1-i])+int(b[len_b-1-i]))//10 for i in range(len_b,len_a): c.append((kuriage+int(a[len_a-1-i]))%10) kuriage=(kuriage+int(a[len_a-1-i]))//10 else : for i in range(len_a): c.append((kuriage+int(a[len_a-1-i])+int(b[len_b-1-i]))%10) kuriage=(kuriage+int(a[len_a-1-i])+int(b[len_b-1-i]))//10 for i in range(len_a,len_b): c.append((kuriage+int(b[len_b-1-i]))%10) kuriage=(kuriage+int(b[len_b-1-i]))//10 if(kuriage==1): c.append(1) if(len(c)>=81): print('overflow') else: for i in c[::-1]: print(i,end='') print('')

s130255396

p03846

u405660020

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())) from collections import Counter c=Counter(a) cnt_lst=[0]*n for key in c: cnt_lst[key]+=c[key] if n%2==1: if cnt_lst[0]==1: cnt_lst.pop(0) print(cnt_lst) if all([cnt_lst[i]==2 for i in range(n-1) if i%2==1]): print(2**(n//2)) else: print(0) else: print(0) else: if all([cnt_lst[i]==2 for i in range(n-1) if i%2==1]): print(2**(n//2)) else: print(0)

s857579236

Accepted

n=int(input()) a=list(map(int, input().split())) mod=10**9+7 from collections import Counter c=Counter(a) cnt_lst=[0]*n for key in c: cnt_lst[key]+=c[key] flag=True if n%2==1: if cnt_lst[0]==1: cnt_lst.pop(0) else: flag=False for i in range(n-1): if (i%2==1 and cnt_lst[i]!=2) or (i%2==0 and cnt_lst[i]!=0): flag=False else: for i in range(n): if (i%2==1 and cnt_lst[i]!=2) or (i%2==0 and cnt_lst[i]!=0): flag=False print(2**(n//2)%mod if flag else 0)

s528275076

p03645

u291988695

Wrong Answer

In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.

n,m=map(int,input().split()) l=[[0,0]]*m t=[0]*m tc=0 a=False for k in range(m): l[k]=list(map(int,input().split())) l.sort() for k in l: if k[0]==1: t[tc]=k[1] tc+=1 elif k[1]==n: if k[1] in t: a=True break if a: print("POSSIBLE") else: print("IMPOSSIBLE")

s192923037

Accepted

n,m=map(int,input().split()) l=[[0,0]]*m at=[0]*m bt=[0]*m atc=0 btc=0 a=False f=0 for k in range(m): ln=list(map(int,input().split())) if ln[0]==1: at[atc]=ln[1] atc+=1 elif ln[1]==n: bt[btc]=ln[0] btc+=1 del at[atc:] del bt[btc:] at.sort() bt.sort() c=list(set(at)^set(bt)) if len(c)!=(atc+btc): print("POSSIBLE") else: print("IMPOSSIBLE")

s696391933

p02659

u818318325

Wrong Answer

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

import math from decimal import * getcontext().prec = 28 a,b=map(str, input().split()) #a_ = Decimal(a) #b_ = Decimal(b) a = int(a) b = float(b)*100 #print(round(Decimal(Decimal(a)*Decimal(b)))) print((a*b)//100) #print(math.floor(Decimal(a)*Decimal(b)/Decimal(100.0)))

s509121276

Accepted

#import math #from decimal import * #getcontext().prec = 28 a,b=input().split() #a_ = Decimal(a) #b_ = Decimal(b) a = int(a) b = int(float(b)*1000) #print(round(Decimal(Decimal(a)*Decimal(b)))) ans = a*b print(ans//1000) #print(math.floor(Decimal(a)*Decimal(b)/Decimal(100.0)))

s547256902

p02843

u799916419

Wrong Answer

AtCoder Mart sells 1000000 of each of the six items below: * Riceballs, priced at 100 yen (the currency of Japan) each * Sandwiches, priced at 101 yen each * Cookies, priced at 102 yen each * Cakes, priced at 103 yen each * Candies, priced at 104 yen each * Computers, priced at 105 yen each Takahashi wants to buy some of them that cost exactly X yen in total. Determine whether this is possible. (Ignore consumption tax.)

def bag(x): max_num = x // 100 print(int(max_num * 5 <= x % 100)) if __name__ == "__main__": import sys in_str = '' for line in sys.stdin: in_str += line n = [int(_) for _ in in_str.split()][0] bag(n)

s510350447

Accepted

def bag(x): max_num = x // 100 print(int(max_num * 5 >= x % 100)) if __name__ == "__main__": import sys in_str = '' for line in sys.stdin: in_str += line n = [int(_) for _ in in_str.split()][0] bag(n)

s442407829

p03729

u594762426

Wrong Answer

You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.

a, b , c= map(str, input().split()) if a[len(a)-1] == b[0] and b[len(b)-1] == c[0]: print("Yes") else: print("No")

s324135070

Accepted

a, b, c = map(str, input().split()) if a[len(a)-1] == b[0] and b[len(b)-1] == c[0]: print("YES") else: print("NO")

s922237321

p02608

u571199625

Wrong Answer

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

n = int(input()) dic = {} for x in range(1,int(n**0.5)+1): for y in range(1,int(n**0.5)+1): for z in range(1,int(n**0.5)+1): ans = x**2 + y **2 + z**2 +x*y+y*z+z*x if ans in dic: dic[ans] +=1 else: dic[ans] = 1 print(dic) for i in range(n): if i+1 in dic: print(dic[i+1]) else: print(0)

s721431059

Accepted

n = int(input()) dic = {} for x in range(1,int(n**0.5)+1): for y in range(1,int(n**0.5)+1): for z in range(1,int(n**0.5)+1): ans = x**2 + y **2 + z**2 +x*y+y*z+z*x if ans in dic: dic[ans] +=1 else: dic[ans] = 1 for i in range(n): if i+1 in dic: print(dic[i+1]) else: print(0)

s681208456

p04029

u178536051

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) ret = 0 for i in range(N): ret += i+1 print(ret)

s825180723

Accepted

N = int(input()) ret = 0 for i in range(N): ret += i+1 print(ret)

s364037241

p00254

u221679506

Wrong Answer

0 から 9 の数字からなる四桁の数 N に対して以下の操作を行う。 1. N の桁それぞれの数値を大きい順に並べた結果得た数を L とする 2. N の桁それぞれの数値を小さい順に並べた結果得た数を S とする 3. 差 L-S を新しい N とする(1回分の操作終了) 4. 新しい N に対して 1. から繰り返す このとき、全桁が同じ数字(0000, 1111など)である場合を除き、あらゆる四桁の数はいつかは 6174になることが知られている。例えば N = 2012の場合 1回目 (N = 2012): L = 2210, S = 0122, L-S = 2088 2回目 (N = 2088): L = 8820, S = 0288, L-S = 8532 3回目 (N = 8532): L = 8532, S = 2358, L-S = 6174 となり、3 回の操作で 6174 に到達する。 0 から 9 の数字からなる四桁の数が与えられたとき、何回の操作で 6174 に到達するか計算するプログラムを作成して下さい。

from functools import reduce while True: n = input() if n == "0000": break n = n if len(n) >= 4 else n.zfill(4) if reduce(lambda x,y:x and y,[n[0] == i for i in n]): print("NA") else: cnt = 0 while n != "6174": s = ''.join(sorted(n)) l = ''.join(sorted(n,reverse = True)) n = str(int(l)-int(s)) n = n if len(n) >= 4 else n.zfill(4) print(n) cnt += 1 print(cnt)

s236905724

Accepted

from functools import reduce while True: n = input() if n == "0000": break n = n if len(n) >= 4 else n.zfill(4) if reduce(lambda x,y:x and y,[n[0] == i for i in n]): print("NA") else: cnt = 0 while n != "6174": s = ''.join(sorted(n)) l = ''.join(sorted(n,reverse = True)) n = str(int(l)-int(s)) n = n if len(n) >= 4 else n.zfill(4) cnt += 1 print(cnt)

s795564829

p03556

u102461423

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.

N = int(input()) print(int(N**.5))

s359242247

Accepted

N = int(input()) print(int(N**.5)**2)

s149404780

p03494

u215643129

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

s180561607

Accepted

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

s892936345

p03644

u432333240

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()) count = 0 ans = 1 for i in range(1,N): tmp_count = 0 tmp_i = i while(tmp_i%2==0): tmp_count+=1 tmp_i /= 2 print(i, tmp_count) if count < tmp_count: count = tmp_count ans = i print(ans)

s841923203

Accepted

N = int(input()) count = 0 ans = 1 for i in range(1,N+1): tmp_count = 0 tmp_i = i while(tmp_i%2==0): tmp_count+=1 tmp_i /= 2 if count < tmp_count: count = tmp_count ans = i print(ans)

s830432579

p02401

u315329386

Wrong Answer

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

while True: inst = input() if inst.find("?") > 0: break print(eval(inst))

s265916845

Accepted

while True: inst = input() if inst.find("?") > 0: break print(eval(inst.replace("/", "//")))

s154849624

p02388

u027634846

Wrong Answer

Write a program which calculates the cube of a given integer x.

x = int(input()) print(x * 3)

s481091292

Accepted

x = int(input()) print(x ** 3)

s184724796

p03556

u837852400

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.

import math print(str(int(math.sqrt(int(input())))))

s463109589

Accepted

import math print(str(int(math.pow(int(math.sqrt(int(input()))),2))))

s502789365

p03671

u165988235

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,b,c = map(int,input().split()) list = [] list.append(a) list.append(b) list.append(c) print(list) list.sort() print(list[0]+list[1])

s298203544

Accepted

a,b,c = map(int,input().split()) list = [] list.append(a) list.append(b) list.append(c) list.sort() print(list[0]+list[1])

s840827624

p04029

u349856770

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("人数を入力してね >")) ame = 0 for i in range(1, n+1): ame += i print(ame)

s512298961

Accepted

n = int(input()) ame = 0 for i in range(1, n+1): ame += i print(ame)

s762447558

p02678

u531599639

Wrong Answer

There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.

from collections import deque N,M = map(int,input().split()) cnct = [[] for i in range(N+1)] for _ in range(M): A,B = map(int,input().split()) cnct[A].append(B) cnct[B].append(A) Q = deque() Q.append(1) ans = [0]*(N+1) vstd = [0]*(N+1) while Q: temp = Q.popleft() for a in cnct[temp]: if vstd[a] == 0: ans[a] = temp vstd[a] = 1 Q.append(a) for a in ans[2:]: print(a)

s551839070

Accepted

from collections import deque N,M = map(int,input().split()) cnct = [[] for i in range(N+1)] for _ in range(M): A,B = map(int,input().split()) cnct[A].append(B) cnct[B].append(A) Q = deque() Q.append(1) ans = [0]*(N+1) vstd = [0]*(N+1) while Q: temp = Q.popleft() for a in cnct[temp]: if vstd[a] == 0: ans[a] = temp vstd[a] = 1 Q.append(a) print('Yes') for a in ans[2:]: print(a)

s585289023

p02261

u956645355

Wrong Answer

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

def main(): N = int(input()) CARD = tuple(input().split()) sort_ = [] for i in range(1, 10): for s in CARD: if str(i) in s: sort_.append(s) sort_ = tuple(sort_) a = list(CARD) bsort = tuple(bubbleSort(a, N)) ssort = tuple(selectionSort(a, N)) for s in (bsort, ssort): print(*s, sep=' ') if s == sort_: print('Stable') else: print('Not stable') def bubbleSort(c, n): flag = 1 while flag: flag = 0 for i in range(1, n): if c[i][1] < c[i-1][1]: c[i], c[i-1] = c[i-1], c[i] flag = 1 return c def selectionSort(c, n): for i in range(n): minj = i for j in range(i, n): if c[j][1] < c[minj][1]: minj = j c[i], c[minj] = c[minj], c[i] return c main()

s344388055

Accepted

def main(): N = int(input()) CARD = tuple(input().split()) sort_ = [] for i in range(1, 10): for s in CARD: if str(i) in s: sort_.append(s) sort_ = tuple(sort_) bsort = tuple(bubbleSort(list(CARD), N)) ssort = tuple(selectionSort(list(CARD), N)) for s in (bsort, ssort): print(*s, sep=' ') if s == sort_: print('Stable') else: print('Not stable') def bubbleSort(c, n): flag = 1 while flag: flag = 0 for i in range(1, n): if c[i][1] < c[i-1][1]: c[i], c[i-1] = c[i-1], c[i] flag = 1 return c def selectionSort(c, n): for i in range(n): minj = i for j in range(i, n): if c[j][1] < c[minj][1]: minj = j c[i], c[minj] = c[minj], c[i] return c main()

s094830308

p03448

u526459074

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, b, c, x = map(int, [input() for i in range(4)]) ans = 0 for i in range(a): for j in range (b): for k in range (c): if 500*i + 100*j + 50*k == x: ans += 1 print(ans)

s984359729

Accepted

a, b, c, x = map(int, [input() for i in range(4)]) ans = 0 for i in range(a+1): for j in range (b+1): for k in range (c+1): if 500*i + 100*j + 50*k == x: ans += 1 print(ans)

s708047267

p03861

u113255362

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,c=map(int,input().split()) k = 0 m = a % c if m == 0: k = a else: k = c-m + a res = 0 for i in range(k,b,c): res += 1 print(res)

s551095132

Accepted

a,b,c=map(int,input().split()) mini = 0 m = a % c if m == 0: mini = a/c else: mini = (c-m + a)/c maxi = 0 M = b % c if M == 0: maxi = b//c else: maxi = (b-M)//c res = int(maxi) - int(mini) + 1 print(res)

s548793966

p03605

u485566817

Wrong Answer

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

n = input() if n[0] == 9 or n[1] == 9: print("Yes") else: print("No")

s551187305

Accepted

n = input() if n[0] == "9" or n[1] == "9": print("Yes") else: print("No")

s062464529

p03998

u583507988

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 = list(map(str, input().split())) b = list(map(str, input().split())) c = list(map(str, input().split())) n = 'a' while True: if n == 'a': if len(a) == 0: print('A') break else: n = a.pop(0) elif n == b: if len(b) == 0: print('B') break else: n = b.pop(0) else: if len(c) == 0: print('C') break else: n = c.pop(0)

s267653041

Accepted

sa=input()+'a' sb=input()+'b' sc=input()+'c' st='a' a=0 b=0 c=0 while a<len(sa) and b<len(sb) and c<len(sc): if st=='a': st=sa[a] a+=1 elif st=='b': st=sb[b] b+=1 else: st=sc[c] c+=1 if a==len(sa): print('A') elif b==len(sb): print('B') else: print('C')

s718071245

p03471

u341855122

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 = list(map(int,input().split())) sen=0 gosen=0 man = 0 man = N[0] while(N[0]==man+gosen+sen and man > -1 and sen > -1 and gosen > -1): while(N[0]==man+gosen+sen and man >-1 and sen > -1 and gosen > -1): print(10000*man+5000*gosen+1000*sen) if(N[1]==10000*man+5000*gosen+1000*sen): print(str(man)+" "+str(gosen)+" "+str(sen)) exit() gosen -= 1 sen += 1 man -= 1 gosen += 1 print("-1 -1 -1")

s389211544

Accepted

N = list(map(int,input().split())) sen=0 gosen=0 man = N[0] while(N[0]+1>gosen): sen = 0 while(N[0]+1>sen): man = N[0] - sen - gosen if(man < 0): sen += 1 continue if(N[1]==(10000*man+5000*gosen+1000*sen)): print(str(man)+" "+str(gosen)+" "+str(sen)) exit() sen += 1 gosen += 1 print("-1 -1 -1")

s098372129

p03054

u102126195

Wrong Answer

We have a rectangular grid of squares with H horizontal rows and W vertical columns. Let (i,j) denote the square at the i-th row from the top and the j-th column from the left. On this grid, there is a piece, which is initially placed at square (s_r,s_c). Takahashi and Aoki will play a game, where each player has a string of length N. Takahashi's string is S, and Aoki's string is T. S and T both consist of four kinds of letters: `L`, `R`, `U` and `D`. The game consists of N steps. The i-th step proceeds as follows: * First, Takahashi performs a move. He either moves the piece in the direction of S_i, or does not move the piece. * Second, Aoki performs a move. He either moves the piece in the direction of T_i, or does not move the piece. Here, to move the piece in the direction of `L`, `R`, `U` and `D`, is to move the piece from square (r,c) to square (r,c-1), (r,c+1), (r-1,c) and (r+1,c), respectively. If the destination square does not exist, the piece is removed from the grid, and the game ends, even if less than N steps are done. Takahashi wants to remove the piece from the grid in one of the N steps. Aoki, on the other hand, wants to finish the N steps with the piece remaining on the grid. Determine if the piece will remain on the grid at the end of the game when both players play optimally.

h, w, n = map(int, input().split()) sr, sc = map(int, input().split()) S = input() T = input() win = 0 SR = sr SC = sc sr = SR sc = SC for i in range(n): if S[i] == "L": sc -= 1 if sc <= 0: break if T[i] == "R" and sc < w: sc += 1 if sc <= 0: win = 1 sr = SR sc = SC for i in range(n): if S[i] == "R": sc += 1 if sc > w: break if T[i] == "L" and sc > 1: sc -= 1 if sc > w: win = 1 sr = SR sc = SC for i in range(n): if S[i] == "D": sr += 1 if sr > h: break if T[i] == "U" and sr > 1: sr -= 1 if sr > h: win = 1 sr = SR sc = SC for i in range(n): if S[i] == "U": sr -= 1 if sr <= 0: break if T[i] == "D" and sr < h: sr += 1 if sr <= 0: win = 1 if win: print("No") else: print("Yes")

s210095864

Accepted

h, w, n = map(int, input().split()) sr, sc = map(int, input().split()) S = input() T = input() win = 0 SR = sr SC = sc sr = SR sc = SC for i in range(n): if S[i] == "L": sc -= 1 if sc <= 0: break if T[i] == "R" and sc < w: sc += 1 if sc <= 0: win = 1 sr = SR sc = SC for i in range(n): if S[i] == "R": sc += 1 if sc > w: break if T[i] == "L" and sc > 1: sc -= 1 if sc > w: win = 1 sr = SR sc = SC for i in range(n): if S[i] == "D": sr += 1 if sr > h: break if T[i] == "U" and sr > 1: sr -= 1 if sr > h: win = 1 sr = SR sc = SC for i in range(n): if S[i] == "U": sr -= 1 if sr <= 0: break if T[i] == "D" and sr < h: sr += 1 if sr <= 0: win = 1 if win: print("NO") else: print("YES")

s620045014

p02602

u019685451

Wrong Answer

M-kun is a student in Aoki High School, where a year is divided into N terms. There is an exam at the end of each term. According to the scores in those exams, a student is given a grade for each term, as follows: * For the first through (K-1)-th terms: not given. * For each of the K-th through N-th terms: the multiplication of the scores in the last K exams, including the exam in the graded term. M-kun scored A_i in the exam at the end of the i-th term. For each i such that K+1 \leq i \leq N, determine whether his grade for the i-th term is **strictly** greater than the grade for the (i-1)-th term.

def prod(X): val = 1 for x in X: val = val * x return val N, K = map(int, input().split()) A = list(map(int, input().split())) G = [prod(A[:K])] for i in range(K, N): val = G[-1] // A[i - K] * A[i] G.append(val) print(G) for i in range(1, len(G)): print('Yes' if G[i] > G[i - 1] else 'No')

s111576023

Accepted

N, K = map(int, input().split()) A = list(map(int, input().split())) for i in range(K, N): print('Yes' if A[i] > A[i - K] else 'No')

s681101729

p04011

u235376569

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.

a=int(input()) b=int(input()) c=int(input()) d=int(input()) ans=0 for i in range(a): if 0<a: ans+=c else: ans+=d print(d)

s614230489

Accepted

a=int(input()) b=int(input()) c=int(input()) d=int(input()) ans=0 for i in range(a): if 0<b: ans+=c else: ans+=d b=b-1 print(ans)

s915249331

p03457

u348868667

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()) txy = [] for i in range(N): txy.append(list(map(int,input().split()))) ans = "YES" t = 0 x = 0 y = 0 for i in range(N): deft = txy[i][0] - t defx = abs(txy[i][1] - x) defy = abs(txy[i][2] - y) t = txy[i][0] x = txy[i][1] y = txy[i][2] if (defx + defy) > deft: ans = "NO" break elif (deft - defx - defy)%2 != 0: ans = "NO" break else: continue print(ans)

s591169361

Accepted

N = int(input()) txy = [] for i in range(N): txy.append(list(map(int,input().split()))) ans = "Yes" t = 0 x = 0 y = 0 for i in range(N): deft = txy[i][0] - t defx = abs(txy[i][1] - x) defy = abs(txy[i][2] - y) t = txy[i][0] x = txy[i][1] y = txy[i][2] if (defx + defy) > deft: ans = "No" break elif (deft - defx - defy)%2 != 0: ans = "No" break print(ans)

s091937166

p02742

u151079949

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*W%2==0: print(H*W/2) else: print(H*W/2+1)

s025116049

Accepted

H,W=map(int,input().split()) if H!=1 and W!=1: if H*W%2==0: print(int(H*W/2)) else: print(int(H*W/2+0.5)) else: print(1)

s729344267

p03827

u980503157

Wrong Answer

You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).

firstline = int(input()) sec = input() maxN = 0 init = 0 for i in sec: if i == "I": init -= 1 else: init += 1 if init > maxN: maxN = init print(maxN)

s700011489

Accepted

firstline = int(input()) sec = input() maxN = 0 init = 0 for i in sec: if i == "I": init += 1 else: init -= 1 if init > maxN: maxN = init print(maxN)

s191912593

p03814

u246820565

Wrong Answer

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

s = input() a = s.index('A') z = s.index('Z') print(s[a:z+1])

s068712067

Accepted

s = input() a = s.index('A') for i in range(len(s)): if s[i] == 'Z': z = i+1 print(len(s[a:z]))

s201927766

p03386

u803617136

Wrong Answer

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

a, b, k = map(int, input().split()) upper = min(a + k, b + 1) bottom = min(max(b - k, upper), a + k, ) for i in range(a, upper): print(i) for j in range(bottom, b + 1): print(j)

s863193011

Accepted

a, b, k = map(int, input().split()) upper = min(a + k, b) bottom = max(b - k + 1, upper) for i in range(a, upper): print(i) for j in range(bottom, b + 1): print(j)

s058578418

p03090

u608088992

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()) S = 0 for i in range(2, N): S += i Ans = [] for i in range(2, N): Ans.append((1, i)) Ans.append((N, i)) Total = [N+1] * N Total[0], Total[N-1] = S, S for i in range(1, N-1): j = i+1 while Total[i] < S: Ans.append((i+1, j+1)) Total[i] += j+1 Total[j] += i+1 j += 1 print(len(Ans)) for a in Ans: print(" ".join(map(str, a)))

s218776046

Accepted

import sys def solve(): input = sys.stdin.readline N = int(input()) Ans = [] AnsNum = 0 if N % 2 == 0: for i in range(1, N): for j in range(i + 1, N + 1): if i + j != N + 1: Ans.append((i, j)) AnsNum += 1 else: for i in range(1, N): for j in range(i + 1, N + 1): if i + j != N: Ans.append((i, j)) AnsNum += 1 print(AnsNum) for i in range(AnsNum): print(" ".join(map(str, Ans[i]))) return 0 if __name__ == "__main__": solve()

s964118737

p02614

u609561564

Wrong Answer

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

h, w, k = map(int, input().split()) c = [] ans = 0 c = [list(input()) for _ in range(h)] print(c) ans = 0 for i in range(2 ** h): for j in range(2 ** w): cnt = 0 for m in range(h): for n in range(w): if ((i >> m) & 1 == 0) and ((j >> n) & 1 == 0): if c[m][n] == '#': cnt += 1 if cnt == k: ans += 1 print(ans)

s132113735

Accepted

h, w, k = map(int, input().split()) c = [] ans = 0 c = [list(input()) for _ in range(h)] # print(c) ans = 0 for i in range(2 ** h): for j in range(2 ** w): cnt = 0 for m in range(h): for n in range(w): if ((i >> m) & 1 == 0) and ((j >> n) & 1 == 0): if c[m][n] == '#': cnt += 1 if cnt == k: ans += 1 print(ans)

s881629399

p03433

u372345564

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.

def main(): N = int(input() ) A = int(input() ) surplus = N % 500 print(surplus) if(surplus <= A): print("Yes") else: print("No") if __name__=="__main__": main()

s589209075

Accepted

def main(): N = int(input() ) A = int(input() ) surplus = N % 500 if(surplus <= A): print("Yes") else: print("No") if __name__=="__main__": main()

s520036261

p03854

u558961961

Wrong Answer

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

S = input() result = "NO" while True: if S[-5:-1] == "dream": if len(S) == 5: result = "YES" break S.rstrip("dream") elif S[-5:-1] == "erase": if len(S) == 5: result = "YES" break S.rstrip("erase") elif S[-6:-1] == "eraser": if len(S) == 6: result = "YES" break S.rstrip("eraser") elif S[-7:-1] == "dreamer": if len(S) == 7: result = "YES" break S.rstrip("dreamer") else: break print(result)

s918451931

Accepted

S = input() result = "NO" while True: if S[-5:] == "dream": if len(S) == 5: result = "YES" break S = S[:-5] elif S[-5:] == "erase": if len(S) == 5: result = "YES" break S = S[:-5] elif S[-6:] == "eraser": if len(S) == 6: result = "YES" break S = S[:-6] elif S[-7:] == "dreamer": if len(S) == 7: result = "YES" break S = S[:-7] else: break print(result)

s227141789

p03730

u385167811

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`.

temp = input().rstrip().split(" ") A = int(temp[0]) B = int(temp[1]) C = int(temp[2]) if A == B and C > 0: print("No") elif A == 1: print("Yes") elif A + C == B: print("Yes") else: print("No")

s533278918

Accepted

temp = input().rstrip().split(" ") A = int(temp[0]) B = int(temp[1]) C = int(temp[2]) flag = 0 if A == B and C > 0: print("NO") elif A == 1: print("YES") elif A + C == B: print("YES") else: for i in range(1,10000): if (A * i) % B == C: print("YES") flag = 1 break if flag == 0: print("NO")

s848367415

p03361

u136346222

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.

# -*- coding: utf-8 -*- from sys import stdin import sys import numpy as np xy = stdin.readline().rstrip().split() x = int(xy[0]) y = int(xy[1]) data = [stdin.readline().rstrip().split() for _ in range(x)] print(*data, sep='\n') print('----------------') for i in range(x): data[i] = list(data[i][0]) arr = np.array(data) print(arr) print('----------------') OK = True for i in range(x): for j in range(y): if arr[i][j] == '#': can_paint = False if i != 0: if arr[i-1][j] == '#': can_paint =True if j != 0: if arr[i][j-1] == '#': can_paint = True if i != x-1: if arr[i+1][j] == '#': can_paint = True if j != y-1: if arr[i][j+1] == '#': can_paint = True if can_paint != True: OK = False if OK == True: print('Yes') else: print('No')

s092085925

Accepted

# -*- coding: utf-8 -*- from sys import stdin import sys import numpy as np xy = stdin.readline().rstrip().split() x = int(xy[0]) y = int(xy[1]) data = [stdin.readline().rstrip().split() for _ in range(x)] #print(*data, sep='\n') #print('----------------') for i in range(x): data[i] = list(data[i][0]) arr = np.array(data) #print(arr) #print('----------------') OK = True for i in range(x): for j in range(y): if arr[i][j] == '#': can_paint = False if i != 0: if arr[i-1][j] == '#': can_paint =True if j != 0: if arr[i][j-1] == '#': can_paint = True if i != x-1: if arr[i+1][j] == '#': can_paint = True if j != y-1: if arr[i][j+1] == '#': can_paint = True if can_paint != True: OK = False if OK == True: print('Yes') else: print('No')

s468790976

p03574

u002459665

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.

def main(): h, w = map(int, input().split()) b = [] for _ in range(h): b.append(input()) # b[1][1] -> b[0][0], b[0][1], b[0][2], b[1][0], b[1][2], b[2][0], b[2][1], b[2][2] ans = [] for i in range(h): e = [] for j in range(w): e.append(0) ans.append(e) for i in range(h): for j in range(w): count = 0 v = b[i][j] if v == "#": ans[i][j] = '#' continue for k in range(-1, 2): for l in range(-1, 2): if i + k < 0 or i + k >= h: pass elif j + l < 0 or j + l >= w: pass else: if b[i + k][j + l] == "#": count += 1 ans[i][j] = count print(ans) if __name__ == '__main__': main()

s367922107

Accepted

def main(): h, w = map(int, input().split()) b = [] for _ in range(h): b.append(input()) # b[1][1] -> b[0][0], b[0][1], b[0][2], b[1][0], b[1][2], b[2][0], b[2][1], b[2][2] ans = [] for i in range(h): e = [] for j in range(w): e.append(0) ans.append(e) for i in range(h): for j in range(w): count = 0 v = b[i][j] if v == "#": ans[i][j] = '#' continue for k in range(-1, 2): for l in range(-1, 2): if i + k < 0 or i + k >= h: pass elif j + l < 0 or j + l >= w: pass else: if b[i + k][j + l] == "#": count += 1 ans[i][j] = count for l1 in ans: for l2 in l1: print(l2, end="") print("") if __name__ == '__main__': main()

s335748843

p04030

u303119576

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?

0

s490498701

Accepted

s=input() stk=[] for c in s: if c in '01': stk.append(c) elif stk: stk.pop() print(''.join(stk))

s608735469

p03998

u131881594

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,b,c=input(),input(),input() string=[a,b,c] s=[0,0,0] ban=0 print(string) while s[0]!=len(a) and s[1]!=len(b) and s[2]!=len(c): temp=s[ban] s[ban]+=1 if string[ban][temp]=="a": ban=0 elif string[ban][temp]=="b": ban=1 else: ban=2 if s[0]==len(a): print("A") if s[1]==len(b): print("B") if s[2]==len(c): print("C")

s182160235

Accepted

a,b,c=input(),input(),input() string=[a,b,c] ans=["A","B","C"] s=[0,0,0] ban=0 while s[ban]!=len(string[ban]): temp=s[ban] s[ban]+=1 if string[ban][temp]=="a": ban=0 elif string[ban][temp]=="b": ban=1 else: ban=2 print(ans[ban])

s251970320

p03563

u331464808

Wrong Answer

Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it.

r = int(input()) g = int(input()) print(float(2*g-r))

s392594945

Accepted

r = int(input()) g = int(input()) print(2*g-r)

s614586112

p03024

u163320134

Wrong Answer

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

s=input() cnt=s.count('o') l=len(s) if 15-l+cnt>=8: print('Yes') else: print('No')

s783971964

Accepted

s=input() cnt=s.count('o') l=len(s) if 15-l+cnt>=8: print('YES') else: print('NO')

s973971336

p03251

u503228842

Wrong Answer

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

N,M,X,Y = map(int,input().split()) x = list(map(int,input().split())) y = list(map(int,input().split())) max_X = max(X,max(x)) min_Y = min(Y,min(y)) if max_X<min_Y: print("War") else:print("No war")

s212180956

Accepted

N,M,X,Y = map(int,input().split()) x = list(map(int,input().split())) y = list(map(int,input().split())) max_X = max(X,max(x)) min_Y = min(Y,min(y)) if max_X<min_Y: print("No War") else:print("War")

s320710067

p03657

u384579799

Wrong Answer

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

# -*- coding: utf-8 -*- import sys A, B = map(int, input().split()) C = A + B if C%3 == 0: print('possible') else: print('impossible')

s360140565

Accepted

# -*- coding: utf-8 -*- import sys A, B = map(int, input().split()) C = A + B if A%3 == 0: print('Possible') elif B%3 == 0: print('Possible') elif C%3 == 0: print('Possible') else: print('Impossible')

s548399403

p03380

u921773161

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.

a = list(map(int, input().split())) x = max(a) b1 = x//2 b2 = (x+1)//2 y = x z = x for i in range(len(a)): if a[i] != x: tmp = min(abs(a[i]-b1), abs(a[i]-b2)) if tmp < y: z = a[i] y = min(tmp, y) ans = x*z print(x, z)

s625928800

Accepted

n = int(input()) a = list(map(int, input().split())) x = max(a) b1 = x//2 b2 = (x+1)//2 y = x z = x for i in range(len(a)): if a[i] != x: tmp = min(abs(a[i]-b1), abs(a[i]-b2)) if tmp < y: z = a[i] y = min(tmp, y) ans = x*z print(str(x) + ' '+ str(z))

s379622134

p03637

u617515020

Wrong Answer

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

N=int(input()) a=list(map(int,input().split())) o = 0 f = 0 t = 0 for i in a: if i % 2 == 1: o += 1 if i % 4 == 0: f += 1 if i % 4 == 2: t += 1 if f >= o-1 and t % 2 == 0: print('Yes') else: print('No')

s512793528

Accepted

N=int(input()) a=list(map(int,input().split())) o = 0 f = 0 t = 0 for i in a: if i % 2 == 1: o += 1 if i % 4 == 0: f += 1 if i % 4 == 2: t += 1 if f >o-1: print('Yes') elif f == o-1 and t==0: print('Yes') else: print('No')

s814245220

p03167

u216928054

Wrong Answer

There is a grid with H horizontal rows and W vertical columns. Let (i, j) denote the square at the i-th row from the top and the j-th column from the left. For each i and j (1 \leq i \leq H, 1 \leq j \leq W), Square (i, j) is described by a character a_{i, j}. If a_{i, j} is `.`, Square (i, j) is an empty square; if a_{i, j} is `#`, Square (i, j) is a wall square. It is guaranteed that Squares (1, 1) and (H, W) are empty squares. Taro will start from Square (1, 1) and reach (H, W) by repeatedly moving right or down to an adjacent empty square. Find the number of Taro's paths from Square (1, 1) to (H, W). As the answer can be extremely large, find the count modulo 10^9 + 7.

#!/usr/bin/env python3 from collections import defaultdict from heapq import heappush, heappop import sys sys.setrecursionlimit(10**6) input = sys.stdin.buffer.readline INF = 10 ** 9 + 1 # sys.maxsize # float("inf") MOD = 10 ** 9 + 7 def debug(*x): print(*x) def solve(H, W, data): "void()" score = [[0] * (W + 1) for i in range(H + 1)] score[0][1] = 1 for y in range(1, H + 1): for x in range(1, W + 1): if data[y - 1][x - 1] == "#": score[y][x] = 0 else: score[y][x] = (score[y - 1][x] + score[y][x - 1]) % MOD # print(score) return score[H][W] def main(): H, W = map(int, input().split()) data = [input() for i in range(H)] print(solve(H, W, data)) T1 = """ 3 4 ...# .#.. .... """ T2 = """ 5 5 ..#.. ..... #...# ..... ..#.. """ T3 = """ 20 20 .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... """ def test_T3(): """ >>> as_input(T3) >>> main() 345263555 """ def test_T2(): """ >>> as_input(T2) >>> main() 24 """ def _test(): """ >>> as_input(T1) >>> main() 3 """ import doctest doctest.testmod() def as_input(s): "use in test, use given string as input file" import io global read, input f = io.StringIO(s.strip()) input = f.readline read = f.read USE_NUMBA = False if (USE_NUMBA and sys.argv[-1] == 'ONLINE_JUDGE') or sys.argv[-1] == '-c': print("compiling") from numba.pycc import CC cc = CC('my_module') cc.export('solve', solve.__doc__.strip().split()[0])(solve) cc.compile() exit() else: input = sys.stdin.buffer.readline read = sys.stdin.buffer.read if (USE_NUMBA and sys.argv[-1] != '-p') or sys.argv[-1] == "--numba": # -p: pure python mode # if not -p, import compiled module from my_module import solve # pylint: disable=all elif sys.argv[-1] == "-t": _test() sys.exit() elif sys.argv[-1] != '-p' and len(sys.argv) == 2: # input given as file input_as_file = open(sys.argv[1]) input = input_as_file.buffer.readline read = input_as_file.buffer.read main()

s058671880

Accepted

#!/usr/bin/env python3 from collections import defaultdict from heapq import heappush, heappop import sys sys.setrecursionlimit(10**6) input = sys.stdin.buffer.readline INF = 10 ** 9 + 1 # sys.maxsize # float("inf") MOD = 10 ** 9 + 7 def debug(*x): print(*x) def solve(H, W, data): "void()" score = [[0] * (W + 1) for i in range(H + 1)] score[0][1] = 1 for y in range(1, H + 1): for x in range(1, W + 1): if data[y - 1][x - 1] == ord("#"): score[y][x] = 0 else: score[y][x] = (score[y - 1][x] + score[y][x - 1]) % MOD # print(score) return score[H][W] def main(): H, W = map(int, input().split()) data = [input() for i in range(H)] print(solve(H, W, data)) T1 = """ 3 4 ...# .#.. .... """ T2 = """ 5 5 ..#.. ..... #...# ..... ..#.. """ T3 = """ 20 20 .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... .................... """ T4 = """ 5 2 .. #. .. .# .. """ def test_T4(): """ >>> as_input(T4) >>> main() 0 """ def test_T3(): """ >>> as_input(T3) >>> main() 345263555 """ def test_T2(): """ >>> as_input(T2) >>> main() 24 """ def _test(): """ >>> as_input(T1) >>> main() 3 """ import doctest doctest.testmod() def as_input(s): "use in test, use given string as input file" import io global read, input f = io.StringIO(s.strip()) def input(): return bytes(f.readline(), "ascii") def read(): return bytes(f.read(), "ascii") USE_NUMBA = False if (USE_NUMBA and sys.argv[-1] == 'ONLINE_JUDGE') or sys.argv[-1] == '-c': print("compiling") from numba.pycc import CC cc = CC('my_module') cc.export('solve', solve.__doc__.strip().split()[0])(solve) cc.compile() exit() else: input = sys.stdin.buffer.readline read = sys.stdin.buffer.read if (USE_NUMBA and sys.argv[-1] != '-p') or sys.argv[-1] == "--numba": # -p: pure python mode # if not -p, import compiled module from my_module import solve # pylint: disable=all elif sys.argv[-1] == "-t": _test() sys.exit() elif sys.argv[-1] != '-p' and len(sys.argv) == 2: # input given as file input_as_file = open(sys.argv[1]) input = input_as_file.buffer.readline read = input_as_file.buffer.read main()

s948602375

p03644

u368270116

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()) for i in range (6): if n<2**i: print(i) quit()

s108698497

Accepted

n=int(input()) for i in reversed(range(7)): if n>=2**i: print(2**i) quit()

s961054221

p02601

u760961723

Wrong Answer

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

A,B,C = map(int,input().split()) K = int(input()) import math x = math.ceil(math.log2(A/B)) if x < 0: x = 0 y = math.ceil(math.log2((B*(2**x))/C)) if y < 0: y = 0 if 2*x + y <= K: print("Yes") else: print("No")

s747339391

Accepted

A,B,C = map(int,input().split()) K = int(input()) import math x,y = 0,0 if A > B: if math.modf(math.log2(A/B))[0] != 0: x = math.ceil(math.log2(A/B)) else: x = math.ceil(math.log2(A/B))+1 elif A == B: x = 1 else: pass if B*(2**x) > C: if math.modf(math.log2((B*(2**x))/C))[0] != 0: y = math.ceil(math.log2((B*(2**x))/C)) else: y = math.ceil(math.log2((B*(2**x))/C))+1 elif B*(2**x) == C: y = 1 else: pass if x + y <= K: print("Yes") else: print("No")

s923318805

p03369

u041641933

Wrong Answer

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

s = input() price = 500 if s[0] == "o": price += 100 if s[1] == "o": price += 100 if s[2] == "o": price += 100 print(price)

s957804575

Accepted

s = input() price = 700 if s[0] == "o": price += 100 if s[1] == "o": price += 100 if s[2] == "o": price += 100 print(price)

s228802016

p03795

u532099150

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()) a = n%15 print(800*n - a)

s132523322

Accepted

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

s520911228

p03470

u247043226

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?

def solve(): N = int(input()) d = [int(input()) for _ in range(N)] d.sort() number_of_steps = 0 previous_diameter = 0 for i in range(N): if d[i] > previous_diameter: number_of_steps += 1 previous_diameter = d[i] print(number_of_steps) # solve()

s797309646

Accepted

def solve(): N = int(input()) d = [int(input()) for _ in range(N)] d.sort() number_of_steps = 0 previous_diameter = 0 for i in range(N): if d[i] > previous_diameter: number_of_steps += 1 previous_diameter = d[i] print(number_of_steps) solve()

s275244223

p03774

u303059352

Wrong Answer

There are N students and M checkpoints on the xy-plane. The coordinates of the i-th student (1 \leq i \leq N) is (a_i,b_i), and the coordinates of the checkpoint numbered j (1 \leq j \leq M) is (c_j,d_j). When the teacher gives a signal, each student has to go to the nearest checkpoint measured in _Manhattan distance_. The Manhattan distance between two points (x_1,y_1) and (x_2,y_2) is |x_1-x_2|+|y_1-y_2|. Here, |x| denotes the absolute value of x. If there are multiple nearest checkpoints for a student, he/she will select the checkpoint with the smallest index. Which checkpoint will each student go to?

while(True): try: n, m = map(int, input().split()) except EOFError: exit() a, b = [[0 for _ in range(n)] for i in range(2)] c, d = [[0 for _ in range(m)] for i in range(2)] for i in range(n): a[i], b[i] = map(int, input().split()) for i in range(m): c[i], d[i] = map(int, input().split()) for i in range(n): dist = 1 << 9 ans = 0 for j in range(m)[::-1]: ans = j + 1 if min(dist, abs(a[i] - c[j]) + abs(b[i] - d[j])) == abs(a[i] - c[j]) + abs(b[i] - d[j]) else ans dist = min(dist, abs(a[i] - c[j]) + abs(b[i] - d[j])) print(ans)

s026138326

Accepted

while(True): try: n, m = map(int, input().split()) ab = [list(map(int, input().split())) for _ in range(n)] cd = [list(map(int, input().split())) for _ in range(m)] except EOFError: exit() for a, b in ab: dist = [abs(a - c) + abs(b - d) for c, d in cd] print(dist.index(min(dist)) + 1)

s335812839

p03845

u278260569

Wrong Answer

Joisino is about to compete in the final round of a certain programming competition. In this contest, there are N problems, numbered 1 through N. Joisino knows that it takes her T_i seconds to solve problem i(1≦i≦N). Also, there are M kinds of drinks offered to the contestants, numbered 1 through M. If Joisino takes drink i(1≦i≦M), her brain will be stimulated and the time it takes for her to solve problem P_i will become X_i seconds. It does not affect the time to solve the other problems. A contestant is allowed to take exactly one of the drinks before the start of the contest. For each drink, Joisino wants to know how many seconds it takes her to solve all the problems if she takes that drink. Here, assume that the time it takes her to solve all the problems is equal to the sum of the time it takes for her to solve individual problems. Your task is to write a program to calculate it instead of her.

N = int(input()) T = list(map(int,input().split())) M = int(input()) P_X = [] for i in range(M): P_X.append(list(map(int,input().split()))) temp_sum = sum(T) print(temp_sum) for i in range(M): print(temp_sum - T[P_X[i][0] - 1] + P_X[i][1])

s851011499

Accepted

N = int(input()) T = list(map(int,input().split())) M = int(input()) P_X = [] for i in range(M): P_X.append(list(map(int,input().split()))) temp_sum = sum(T) for i in range(M): print(temp_sum - T[P_X[i][0] - 1] + P_X[i][1])

s470909782

p02608

u040660107

Wrong Answer

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

import time import itertools N = int(input()) def function(x,y,z,n): n_tmp = pow(x+y+z,2) - (x*y) - (y*z) -(x*z) if n == n_tmp: return True else: return False max = pow(10,4) count = 0 start = time.time() p = list(itertools.product(range(1,15), repeat=3)) for n in range(1,N): count = 0 for p_in in p: if function(p_in[0],p_in[1],p_in[2],n) == True: count = count + 1 print(count) elapsed_time = time.time() - start # print("elapsed_time:{0}".format(elapsed_time) + "[sec]")

s810182294

Accepted

import time import itertools import collections N = int(input()) start = time.time() def function(x,y,z): return pow(x,2) + pow(y,2) + pow(z,2) + (x * y) + (x * z ) + (y * z) p = list(itertools.product(range(1,100),repeat=3)) a = [] for p_in in p: a.append(function(p_in[0],p_in[1],p_in[2])) dict = collections.Counter(a) for n in range(1,N + 1): print(dict[n])

s940425482

p03408

u257449466

Wrong Answer

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

from collections import Counter arr_b = [] N = int(input()) for i in range(N): arr_b.append(str(input())) arr_r = [] N = int(input()) for i in range(N): arr_r.append(str(input())) c_b = Counter(arr_b) c_r = Counter(arr_r) result = max(c_b[x] for x in c_b) - max(c_r[x] for x in c_r) if result > 0: print(result) else: print(0)

s991978593

Accepted

from collections import Counter arr_b = [] N = int(input()) for i in range(N): arr_b.append(str(input())) arr_r = [] N = int(input()) for i in range(N): arr_r.append(str(input())) c_b = Counter(arr_b) c_r = Counter(arr_r) res = 0 for k_b in c_b.keys(): if k_b in c_r : if res < c_b[k_b] - c_r[k_b]: res = c_b[k_b] - c_r[k_b] else: if res < c_b[k_b]: res = c_b[k_b] print(res)

s587568448

p00045

u567380442

Wrong Answer

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

import sys f = sys.stdin sum_price = sum_amount = kind = 0 for line in f: price, amount = map(int, line.split(',')) sum_price += price * amount sum_amount += amount kind += 1 print(sum_price, round(sum_amount / kind))

s589771312

Accepted

import sys f = sys.stdin sum_price = sum_amount = kind = 0 for line in f: price, amount = map(int, line.split(',')) sum_price += price * amount sum_amount += amount kind += 1 print(sum_price) print(int(sum_amount / kind + 0.5))

s299986799

p03440

u368796742

Wrong Answer

You are given a forest with N vertices and M edges. The vertices are numbered 0 through N-1. The edges are given in the format (x_i,y_i), which means that Vertex x_i and y_i are connected by an edge. Each vertex i has a value a_i. You want to add edges in the given forest so that the forest becomes connected. To add an edge, you choose two different vertices i and j, then span an edge between i and j. This operation costs a_i + a_j dollars, and afterward neither Vertex i nor j can be selected again. Find the minimum total cost required to make the forest connected, or print `Impossible` if it is impossible.

n,m = map(int,input().split()) if n == 1: print(0) exit() if m*2 < n: print("Impossible") exit()

s154437857

Accepted

n,m = map(int,input().split()) if m == n-1: print(0) exit() if 2*(n-m-1) > n: print("Impossible") exit() class Unionfind: def __init__(self,n): self.uf = [-1]*n def find(self,x): if self.uf[x] < 0: return x else: self.uf[x] = self.find(self.uf[x]) return self.uf[x] def same(self,x,y): return self.find(x) == self.find(y) def union(self,x,y): x = self.find(x) y = self.find(y) if x == y: return False if self.uf[x] > self.uf[y]: x,y = y,x self.uf[x] += self.uf[y] self.uf[y] = x return True def size(self,x): x = self.find(x) return -self.uf[x] u = Unionfind(n) a = list(map(int,input().split())) for i in range(m): x,y = map(int,input().split()) u.union(x,y) count = 2*(n-m-1) l = [[] for i in range(n)] for i in range(n): x = u.find(i) if x < 0: l[i].append(a[i]) else: l[x].append(a[i]) ans = 0 h = [] for i in range(n): if l[i]: l[i].sort() ans += l[i][0] count -= 1 for j in l[i][1:]: h.append(j) h.sort() ans += sum(h[:count]) print(ans)

s939795018

p03214

u503111914

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()) A = list(map(int,input().split())) ave = sum(A) / N ans = 0 a = 10**10 for i in range(N): if abs(A[i] - ave) < a: ans = i print(ans)

s087964155

Accepted

N = int(input()) A = list(map(int,input().split())) ave = sum(A) / N ans = 0 a = 10**10 for i in range(N): if abs(A[i] - ave) < a: a = abs(A[i] - ave) ans = i print(ans)

s811425360

p03352

u322185540

Time Limit Exceeded

You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.

n = int(input()) ans = 0 for i in range(1,1001): for j in range(1,1001): if ans < i**j < n: ans = i**j print(ans)

s601130783

Accepted

n = int(input()) ans = 1 for i in range(1,1001): for j in range(2,1001): if i**j > n: break if ans < i**j <= n: ans = i**j print(ans)

s606865078

p02972

u237362582

Wrong Answer

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

import sys input = sys.stdin.readline def main(): N = int(input()) a = list(map(int, input().strip().split())) ans = [0] * N for i, a_r in enumerate(a[::-1]): j = N - i - 1 count = 0 now = j+1 while now <= N: count += ans[j-1] now += j+1 if a_r + count % 2 == 1: ans[j] = 1 output = "" count = 0 for i, b in enumerate(ans): if b == 1: output += str(a[i]) + " " count += 1 print(count) if count != 0: print(output) if __name__ == "__main__": main()

s142234299

Accepted

import sys input = sys.stdin.readline def main(): N = int(input()) a = list(map(int, input().strip().split())) ans = [0] * N for i, a_r in enumerate(a[::-1]): j = N - i - 1 count = 0 now = j+1 while now <= N: count += ans[now-1] now += j+1 if a_r == 0: if count % 2 == 1: ans[j] = 1 else: if count % 2 == 0: ans[j] = 1 output = "" count = 0 for i, b in enumerate(ans): if b == 1: output += str(i+1) + " " count += 1 print(count) if count != 0: print(output[:-1]) if __name__ == "__main__": main()

s009633858

p02410

u606989659

Wrong Answer

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

n,m = map(int,input().split()) a_list = [] b_list = [] c_list = [] for a in range(n): a_list.append(list(map(int,input().split()))) for b in range(n): b_list.append(int(input())) for c in range(n): c_list.append([]) for d in range(m): c_list[c].append((a_list[c][d] * b_list[c])) c_list = list(map(sum,c_list)) for c in c_list: print(c)

s707862866

Accepted

n,m = map(int,input().split()) a_list = [] b_list = [] c_list = [] for a in range(n): a_list.append(list(map(int,input().split()))) for b in range(m): b_list.append(int(input())) for c in range(n): c_list.append([]) for d in range(m): c_list[c].append((a_list[c][d] * b_list[d])) c_list = list(map(sum,c_list)) for c in c_list: print(c)

s195326842

p03795

u163320134

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-int(60/15)*200)

s252104483

Accepted

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

s360525758

p03623

u972892985

Wrong Answer

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

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

s441895897

Accepted

x, a, b = map(int,input().split()) if abs(x - a) > abs(x - b): print("B") else: print("A")

s173779012

p03494

u325119213

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.

def actual(N, A): count = 0 while True: is_all_even = all([a % 2 == 0 for a in A]) if is_all_even: A = [a // 2 for a in A] count += 1 else: break return count N = int(input()) A = map(int, input().split()) print(actual(N, A))

s232604317

Accepted

def actual(N, A): count = 0 while True: is_all_even = all([a % 2 == 0 for a in A]) if is_all_even: A = [a // 2 for a in A] count += 1 else: return count N = int(input()) A = list(map(int, input().split())) print(actual(N, A))

s815801105

p04043

u118642796

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.

if sorted(list(map(int,input().split()))) == [5,5,7]: print("Yes") else: print("No")

s587401032

Accepted

if sorted(list(map(int,input().split()))) == [5,5,7]: print("YES") else: print("NO")

s366241887

p03469

u482157295

Wrong Answer

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

s = list(input()) s[3] = "8" print(s)

s772000055

Accepted

s = list(input()) s[3] = "8" print("".join(s))

s550447847

p03351

u059684735

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('Yes' if (d > abs(a-b) and d > abs(b-c)) or d > abs(a-c) else 'No')

s316779234

Accepted

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

s437394693

p02397

u629874472

Wrong Answer

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

while True: a,b = input().split() if a =='0' and b =='0': break else: a,b =b,a print(b,a)

s588643243

Accepted

while True: a,b = map(int,input().split()) if a ==0 and b ==0: break else: if a<=b: print(a,b) else: a,b =b,a print(a,b)

s454832706

p03945

u556610039

Wrong Answer

Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa. In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white. Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

s = list(input()) ans = 0 prestr = s[0] for id in range(len(s) - 1): if prestr == s[id - 1]: continue else: ans += 1 prestr = s[id - 1] print(ans)

s612600442

Accepted

s = list(input()) ans = 0 prestr = s[0] for id in range(len(s) - 1): if prestr == s[id + 1]: continue else: ans += 1 prestr = s[id + 1] print(ans)

s047666583

p02646

u954153335

Wrong Answer

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

import sys a,v=map(int,input().split()) b,w=map(int,input().split()) t=int(input()) dist=b-a speed=v-w if speed<=0: print("No") sys.exit() if dist/speed<=t: print("Yes") else: print("No")

s652682478

Accepted

import sys a,v=map(int,input().split()) b,w=map(int,input().split()) t=int(input()) dist=abs(b-a) speed=v-w if speed<=0: print("NO") sys.exit() if dist/speed<=t: print("YES") else: print("NO")

s231589844

p03488

u784022244

Wrong Answer

A robot is put at the origin in a two-dimensional plane. Initially, the robot is facing in the positive x-axis direction. This robot will be given an instruction sequence s. s consists of the following two kinds of letters, and will be executed in order from front to back. * `F` : Move in the current direction by distance 1. * `T` : Turn 90 degrees, either clockwise or counterclockwise. The objective of the robot is to be at coordinates (x, y) after all the instructions are executed. Determine whether this objective is achievable.

S=input() x,y=map(int, input().split()) N=len(S) now=0 xs=[] ys=[] count=0 for i in range(N): if S[i]=="F": now+=1 else: if count%2==0: xs.append(now) else: ys.append(now) count+=1 now=0 if S[i]=="F" and i==N-1: if count%2==0: xs.append(now) else: ys.append(now) print(xs,ys) from collections import defaultdict D=defaultdict(int) xnow=xs.pop(0) D[xnow]=1 for i in range(len(xs)): dx=xs[i] ds=D.keys() D=defaultdict(int) for v in ds: D[v+dx]=1 D[v-dx]=1 D=dict(D) #print(D) xok=x in D D=defaultdict(int) D[0]=1 for i in range(len(ys)): dy=ys[i] ds=D.keys() D=defaultdict(int) for v in ds: D[v+dy]=1 D[v-dy]=1 D=dict(D) #print(D) yok=y in D if xok and yok: print("Yes") else: print("No")

s153287062

Accepted

S=input() x,y=map(int, input().split()) N=len(S) now=0 xs=[] ys=[] count=0 for i in range(N): if S[i]=="F": now+=1 else: if count%2==0: xs.append(now) else: ys.append(now) count+=1 now=0 if S[i]=="F" and i==N-1: if count%2==0: xs.append(now) else: ys.append(now) #print(xs,ys) from collections import defaultdict D=defaultdict(int) xnow=xs.pop(0) D[xnow]=1 for i in range(len(xs)): dx=xs[i] ds=D.keys() D=defaultdict(int) for v in ds: D[v+dx]=1 D[v-dx]=1 D=dict(D) #print(D) xok=x in D D=defaultdict(int) D[0]=1 for i in range(len(ys)): dy=ys[i] ds=D.keys() D=defaultdict(int) for v in ds: D[v+dy]=1 D[v-dy]=1 D=dict(D) #print(D) yok=y in D if xok and yok: print("Yes") else: print("No")

s684989815

p04012

u075595666

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.

w = list(input()) for i in range(ord('a'),ord('z')): if int(w.count("chr(i)"))%2 == 1: print("No") break if int(w.count("chr(i)"))%2 == 0: continue print("Yes")

s240426484

Accepted

w = list(input()) ans = "Yes" for i in w: if int(w.count(i))%2 == 1: ans = 'No' break if int(w.count(i))%2 == 0: continue print(ans)

s146678668

p02237

u733945366

Wrong Answer

There are two standard ways to represent a graph $G = (V, E)$, where $V$ is a set of vertices and $E$ is a set of edges; Adjacency list representation and Adjacency matrix representation. An adjacency-list representation consists of an array $Adj[|V|]$ of $|V|$ lists, one for each vertex in $V$. For each $u \in V$, the adjacency list $Adj[u]$ contains all vertices $v$ such that there is an edge $(u, v) \in E$. That is, $Adj[u]$ consists of all vertices adjacent to $u$ in $G$. An adjacency-matrix representation consists of $|V| \times |V|$ matrix $A = a_{ij}$ such that $a_{ij} = 1$ if $(i, j) \in E$, $a_{ij} = 0$ otherwise. Write a program which reads a directed graph $G$ represented by the adjacency list, and prints its adjacency-matrix representation. $G$ consists of $n\; (=|V|)$ vertices identified by their IDs $1, 2,.., n$ respectively.

n=int(input()); G=[[0 for i in range(n)]for j in range(n)] #print(G) for i in range(n): list=[int(m) for m in input().split()] u=list[0] k=list[1] for j in range(2,len(list)): v=list[j] G[u-1][v-1]=1 print(G)

s320210913

Accepted

n=int(input()); G=[[0 for i in range(n)]for j in range(n)] for i in range(n): list=[int(m) for m in input().split()] u=list[0] k=list[1] for j in range(2,len(list)): v=list[j] G[u-1][v-1]=1 for i in range(n): for j in range(n): if j!=n-1: print(G[i][j],end=" ") else: print(G[i][j],end="\n")

s608903762

p03545

u021763820

Wrong Answer

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

# -*- coding: utf-8 -*- n = input() m = len(n)-1 for i in range(2 ** m): ops = [] N = [int(n[0]), int(n[1]), int(n[2]), int(n[3])] for j in range(m): if (i >> j) & 1: ops.append("+") else: N[j + 1] *= -1 ops.append("-") if sum(N) == 7: print(N) print(n[0]+ops[0]+n[1]+ops[1]+n[2]+ops[2]+n[3]+"=7") break

s279627840

Accepted

# -*- coding: utf-8 -*- s = input() l = [int(i) for i in s] for i in range(2**3): ans = l[0] siki = s[0] for j in range(3): if i>>j & 1: ans += l[j+1] siki += "+"+s[j+1] else: ans -= l[j+1] siki += "-"+s[j+1] if ans == 7: print(siki+"=7") exit()

s312291709

p02390

u804558166

Wrong Answer

Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.

x=int(input()) print(x/3600, ':', (x-x/3600)/60, ':', x-(x-x/3600)/60, sep='')

s092538283

Accepted

x=int(input()) a=x//3600 b=(x-a*3600)//60 c=x-(a*3600+b*60) print(a, ':', b, ':', c, sep='')

s446388522

p03386

u404057606

Wrong Answer

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

a,b,k = map(int,input().split()) l=[] for i in range(k+1): if a+i not in l and a+i<=b: l.append(a+i) for i in range(k+1): if b-k+i not in l and b-k+i>=a: l.append(b-k+i) for i in range(len(l)): print(l[i])

s992360046

Accepted

a,b,k = map(int,input().split()) l=[] for i in range(k): if a+i not in l and a+i<=b: l.append(a+i) for i in range(k): if b-k+1+i not in l and b-k+1+i>=a: l.append(b-k+1+i) m=sorted(l) for i in range(len(m)): print(m[i])

s589112229

p02646

u560222605

Wrong Answer

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

a,v=map(int,input().split()) b,w=map(int,input().split()) t=int(input()) x=a+v*t y=b+w*t if x>=y: print('Yes') else: print('No')

s998968580

Accepted

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

s666589666

p03557

u160414758

Wrong Answer

The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower. He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i. To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar. How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.

import sys,collections sys.setrecursionlimit(10**7) def Is(): return [int(x) for x in sys.stdin.readline().split()] def Ss(): return sys.stdin.readline().split() def I(): return int(sys.stdin.readline()) def S(): return input() N = I() A,B,C = sorted(Is()),sorted(Is()),sorted(Is()) Bs = [0]*N for i in range(N): b = 0 for j in range(N): if B[i] < C[-j-1]: b = j+1 else: Bs[i] = b break sum = 0 for i in range(N): for j in range(N): if A[i] < B[-j-1]: sum += Bs[-j-1] else: break print(sum)

s227832624

Accepted

import sys,collections sys.setrecursionlimit(10**7) def Is(): return [int(x) for x in sys.stdin.readline().split()] def Ss(): return sys.stdin.readline().split() def I(): return int(sys.stdin.readline()) def S(): return input() N = I() A,B,C = sorted(Is()),sorted(Is()),sorted(Is()) Bs = [0]*N b,j = 0,0 for i in range(N): while j < N: if B[i] >= C[j]: b += 1 else: Bs[i] = N - b break if j == N-1: Bs[i] = N - b j += 1 all = sum(Bs) ans,sum,j = 0,0,0 for i in range(N): while j < N: if A[i] >= B[j]: sum += Bs[j] else: ans += all - sum break if j == N-1: ans += all - sum j += 1 print(ans)

s400857955

p03416

u044964932

Wrong Answer

Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.

def main(): a, b = map(int, input().split()) print(b-a) ans = (b-a)//100 if int(str(a)[-2:]) <= int(str(a)[:1]) and int(str(b)[-2:]) <= int(str(b)[1:]): ans += 1 print(ans) if __name__ == "__main__": main()

s820243464

Accepted

def main(): a, b = map(int, input().split()) ans = 0 for i in range(a, b+1): if str(i) == str(i)[::-1]: ans += 1 print(ans) if __name__ == "__main__": main()

s098643860

p03695

u853900545

Wrong Answer

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

n = int(input()) a = list(map(int,input().split())) for i in range(n): if a[i] < 400: a[i] = 0 elif a[i] < 800: a[i] = 1 elif a[i] < 1200: a[i] = 2 elif a[i] < 1600: a[i] = 3 elif a[i] < 2000: a[i] = 4 elif a[i] < 2400: a[i] = 5 elif a[i] < 2800: a[i] = 6 elif a[i] < 3200: a[i] = 7 else: a[i] = 8 if a.count(8) > 0: print(len(set(a)),len(set(a))+2) else: print([len(set(a)),len(set(a))])

s145708012

Accepted

n = int(input()) a = list(map(int,input().split())) for i in range(n): if a[i] < 400: a[i] = 0 elif a[i] < 800: a[i] = 1 elif a[i] < 1200: a[i] = 2 elif a[i] < 1600: a[i] = 3 elif a[i] < 2000: a[i] = 4 elif a[i] < 2400: a[i] = 5 elif a[i] < 2800: a[i] = 6 elif a[i] < 3200: a[i] = 7 else: a[i] = 8 if a.count(8) > 0: print(max(len(set(a))-1,1),len(set(a)) -1 +a.count(8)) else: print(len(set(a)),len(set(a)))

s501222705

p03486

u076764813

Wrong Answer

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

s=input() t=input() s_=sorted(s) t_=sorted(t, reverse=True) print(s_) print(t_) if s_ < t_: print("Yes") else: print("No")

s063887114

Accepted

s=input() t=input() s_=sorted(s) t_=sorted(t, reverse=True) if s_ < t_: print("Yes") else: print("No")

s241693776

p03854

u101490607

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`.

st = input() sample = ['dream', 'dreamer', 'erase', 'eraser'] sample_edit = ['dream', 'eraser','erase', 'dreamer'] sample_er = ['er'] sample_dr = ['dream'] sample_ase = ['ase', 'aser'] ret = 'No' reg_len = 0 fd_word = '' er_flg = False while( len(st) != 0 ): if reg_len == len(st): break er_flg = False reg_len = len(st) for i in range(len(sample_edit)): fd = st.startswith(sample_edit[i]) #print(st,i) if fd > 0: st = ''.join(st[len(sample_edit[i]):len(st)]) if len(st) == 0: break #if sample[i] == 'er': # break er_flg = True if er_flg == True: break if len(st) == 0: ret = 'Yes' #print(st) #print(len(st)) print( ret )

s285742056

Accepted

st = input() sample = ['dream', 'dreamer', 'erase', 'eraser'] sample_edit = ['dream', 'eraser','erase', 'er'] sample_er = ['er'] sample_dr = ['dream'] sample_ase = ['ase', 'aser'] ret = 'NO' reg_len = 0 fd_word = '' dream_flg = False end_flg = False word = '' length = len(sample_edit) #while( len(st) != 0 ): while( True ): if reg_len == len(st): break reg_len = len(st) for i in range(length): word = sample_edit[i] fd = st.startswith(word) if fd > 0: #print(st,i,dream_flg) #st = ''.join(st[len(word):len(st)]) st = st[len(word):len(st)] if word == 'dream': dream_flg = True elif word == 'er': if dream_flg == False: end_flg = True else: dream_flg = False break if end_flg == True: break if len(st) == 0: ret = 'YES' break #print(st) #print(len(st)) print( ret )

s527906073

p03563

u634079249

Wrong Answer

Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it.

import sys import os def main(): if os.getenv("LOCAL"): sys.stdin = open("input.txt", "r") R = float(sys.stdin.readline().rstrip()) G = float(sys.stdin.readline().rstrip()) print(R+(G-R)*2) if __name__ == '__main__': main()

s703202622

Accepted

import sys import os def main(): if os.getenv("LOCAL"): sys.stdin = open("input.txt", "r") R = int(sys.stdin.readline().rstrip()) G = int(sys.stdin.readline().rstrip()) print(2*G-R) if __name__ == '__main__': main()

s391785066

p03852

u544050502

Wrong Answer

Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.

c=input() print("vowel" if ["a","e","i","u","o",c].index(c)==5 else "consonant")

s965344969

Accepted

print("vowel" if input() in "aiueo" else "consonant")

s715532452

p02402

u829107115

Wrong Answer

Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.

n = int(input()) l = input().split() mini = min(l) maxi = max(l) sums = 0 for i in range(0,n): sums = sums + int(l[i]) print(f"{mini} {maxi} {sums}")

s300346472

Accepted

n = int(input()) l = list(map(int,input().split())) mini = min(l) maxi = max(l) sums = 0 for i in range(0,n): sums = sums + int(l[i]) print(f"{mini} {maxi} {sums}")

s133796230

p03473

u609255576

Wrong Answer

How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?

print(int(input()) + 24)

s851463917

Accepted

print(24 - int(input()) + 24)

s096851101

p02408

u085472528

Wrong Answer

Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.

data = ["{0} {1}" .format(s, r) for s in ('S','H','C','D') for r in range(1, 13 + 1) ] print(data)

s126970021

Accepted

data = [ "{0} {1}" .format(s, r) for s in ('S','H','C','D') for r in range(1, 13 + 1) ] count = int(input()) for c in range(count): card = input() data.remove(card) for c in data: print(c)

s646227710

p02402

u192145025

Wrong Answer

Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.

n = int(input()) s = input().split() max = s[0] min = s[0] for i in range(n): if max < s[i]: max = s[i] if min > s[i]: min = s[i] sum = min + max print(min, end =" ") print(max, end =" ") print(sum)

s619266272

Accepted

n = int(input()) s = list(map(int, input().strip().split(' '))) max = s[0] min = s[0] sum = 0 for i in range(n): sum = sum + s[i] if max < s[i]: max = s[i] if min > s[i]: min = s[i] print(min, end =" ") print(max, end =" ") print(sum)

s754530761

p00452

u797673668

Time Limit Exceeded

あなたは以下のルールでダーツゲームをすることになった. あなたは,矢を的(まと)に向かって 4 本まで投げることができる.必ずしも 4 本全てを投げる必要はなく,1 本も投げなくてもかまわない.的は N 個の部分に区切られていて,各々の部分に点数 P1,..., PN が書か れている.矢が刺さった場所の点数の合計 S があなたの得点の基礎となる.S があらかじめ決められたある点数 M 以下の場合は S がそのままあなたの得点となる.しかし,S が M を超えた場合はあなたの得点は 0 点となる. 的に書かれている点数と M の値が与えられたとき,あなたが得ることのできる点数の最大値を求めるプログラムを作成せよ.

while True: n, m = map(int, input().split()) if not n: break ps = {int(input()) for _ in range(n)} ss = [False] * (m + 1) min_p, max_p = 1e8, 0 for p in ps: ss[p] = True if min_p > p: min_p = p if max_p < p: max_p = p max_s = 0 for t in range(1, 4): new_ss = [False] * (m + 1) for i in range(min(m, max_p * t), min_p * t - 1, -1): if ss[i]: for p in ps: if i + p <= m: new_ss[i + p] = True ss = new_ss p = m while p: if ss[p]: if max_s < p: max_s = p break print(max_s)

s484812936

Accepted

from bisect import bisect_right while True: n, m = map(int, input().split()) if not n: break ps = [0] + sorted(int(input()) for _ in range(n)) p2 = set() for i, pi in enumerate(ps): for pj in ps[i:]: if pi + pj > m: break p2.add(pi + pj) p2 = sorted(p2) print(max(pi + p2[bisect_right(p2, m - pi) - 1] for pi in p2))

s491479728

p00208

u755162050

Time Limit Exceeded

ウォーターデブンに住む建築家のデブンキーさんのもとに、古い大病院を改装する仕事の依頼が舞い込んできました。 国によっては忌み数(いみかず)として嫌われる数字を部屋番号に用いたくない人がいます(日本では 4 と 9 が有名です)。しかし、この病院の部屋番号は忌み数に関係なく、1 から順番に付けられていました。 それが気になったデブンキーさんは、機材やベッドの入れ替えが全て終わる前にウォーターデブンの忌み数である「4」と「6」を除いた数字で部屋番号を付けなおしてしまいました。しかし、入れ替え作業は旧部屋番号で計画していたので、残りの作業を確実に行うには旧部屋番号を新部屋番号に変換する必要があります。計算が苦手なデブンキーさんはこのことに気づいて愕然としています。 そんなデブンキーさんのために、旧部屋番号を入力とし対応する新部屋番号を出力するプログラムを作成してください。 15 番目までの部屋番号の対応表は以下のようになります。 旧部屋番号| 1| 2| 3| 4| 5| 6| 7| 8 | 9 | 10 | 11| 12| 13| 14| 15 ---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|--- 新部屋番号| 1| 2| 3| 5 | 7| 8| 9| 10| 11| 12| 13| 15| 17| 18 | 19

while True: num = int(input()) index = 1 for n in range(1, num): index += 1 while True: str_index = str(index) if '4' in str_index or '6' in str_index: index += 1 else: break print(index)

s447663136

Accepted

""" Created by Jieyi on 9/20/16. """ def main(): while True: num = int(input()) if num == 0: break num = str(oct(num)[2:]) ans = num.translate(str.maketrans('4567', '5789')) print(ans) if __name__ == '__main__': main()

s968530070

p03827

u693048766

Wrong Answer

You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).

state = [0] for c in list(input()): if c == "D": a = state[-1] a -= 1 else: a = state[-1] a += 1 state.append(a) print(max(state))

s391676844

Accepted

state = [0] n = int(input()) chars = list(input()) for c in chars: if c == "D": a = state[-1] a -= 1 else: a = state[-1] a += 1 state.append(a) print(max(state))

s263730943

p03679

u637175065

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 math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools sys.setrecursionlimit(10**7) inf = 10**20 mod = 10**9 + 7 def LI(): return [int(x) for x in sys.stdin.readline().split()] def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()] def LF(): return [float(x) for x in sys.stdin.readline().split()] def LS(): return sys.stdin.readline().split() def I(): return int(sys.stdin.readline()) def F(): return float(sys.stdin.readline()) def S(): return input() def main(): x,a,b = LI() if a <= b: return 'delicious' if a+x <= b: return 'safe' return 'dangerous' print(main())

s651910134

Accepted

import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools sys.setrecursionlimit(10**7) inf = 10**20 mod = 10**9 + 7 def LI(): return [int(x) for x in sys.stdin.readline().split()] def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()] def LF(): return [float(x) for x in sys.stdin.readline().split()] def LS(): return sys.stdin.readline().split() def I(): return int(sys.stdin.readline()) def F(): return float(sys.stdin.readline()) def S(): return input() def main(): x,a,b = LI() if a >= b: return 'delicious' if a+x >= b: return 'safe' return 'dangerous' print(main())

s552746005

p02255

u520845247

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.

input() xs = list(map(int, input().split())) def insertion_sort(xs): for i in range(1, len(xs)): v = xs[i] j = i - 1 while j >= 0 and xs[j] > v: xs[j + 1] = xs[j] j -= 1 xs[j + 1] = v print(*xs) insertion_sort(xs)

s033328658

Accepted

input() xs = list(map(int, input().split())) def insertion_sort(xs): for i in range(1, len(xs)): v = xs[i] j = i - 1 while j >= 0 and xs[j] > v: xs[j + 1] = xs[j] j -= 1 xs[j + 1] = v print(*xs) print(*xs) insertion_sort(xs)

s745113931

p03473

u667024514

Wrong Answer

How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?

a = int(input()) print(a + 24)

s768702164

Accepted

a = int(input()) print((24-a)+24)

s131406511

p02605

u667024514

Wrong Answer

M-kun is a brilliant air traffic controller. On the display of his radar, there are N airplanes numbered 1, 2, ..., N, all flying at the same altitude. Each of the airplanes flies at a constant speed of 0.1 per second in a constant direction. The current coordinates of the airplane numbered i are (X_i, Y_i), and the direction of the airplane is as follows: * if U_i is `U`, it flies in the positive y direction; * if U_i is `R`, it flies in the positive x direction; * if U_i is `D`, it flies in the negative y direction; * if U_i is `L`, it flies in the negative x direction. To help M-kun in his work, determine whether there is a pair of airplanes that will collide with each other if they keep flying as they are now. If there is such a pair, find the number of seconds after which the first collision will happen. We assume that the airplanes are negligibly small so that two airplanes only collide when they reach the same coordinates simultaneously.

n = int(input()) ans = 10 ** 9 upanddown = [[] for i in range(200001)] leftandright = [[] for i in range(200001)] lis = [] for i in range(n): x,y,r = map(str,input().split()) x,y = int(x),int(y) lis.append([x+y,x-y,x,y,r]) if r == "U": upanddown[x].append([y,0]) elif r == "D": upanddown[x].append([y,1]) elif r == "R": leftandright[y].append([x,0]) else: leftandright[y].append([x,1]) # print(upanddown[11]) for li in upanddown: if len(li) >= 2: li.sort(key=lambda z:z[1]) li.sort(key=lambda z:z[0]) key = 0 keynum = -1 for num,r in li: if r == 0: keynum = num else: if keynum != -1: ans = min(ans,num-keynum) for li in leftandright: if len(li) >= 2: li.sort(key=lambda z:z[1]) li.sort(key=lambda z:z[0]) key = 0 keynum = -1 for num,r in li: if r == 0: keynum = num else: if keynum != -1: ans = min(ans,(num-keynum)*5) # print(ans) lis.sort(key = lambda z:z[2]) lis.sort(key = lambda z:z[0]) for i in range(n-1): if lis[i][0] == lis[i+1][0]: if (lis[i][4] == "R" and lis[i+1][4] == "U") or (lis[i][4] == "D" or lis[i+1][4] == "L"): ans = min(ans,(lis[i+1][2]-lis[i][2])*10) lis.sort(key = lambda z:z[2]) lis.sort(key = lambda z:z[1]) for i in range(n-1): if lis[i][1] == lis[i+1][1]: if (lis[i][4] == "U" and lis[i+1][4] == "L") or (lis[i][4] == "R" and lis[i+1][4] == "D"): ans = min(ans,(lis[i+1][2]-lis[i][2])*10) if ans == 10 ** 9: print("SAFE") else: print(ans)

s953170968

Accepted

n = int(input()) ans = 10 ** 10 upanddown = [[] for i in range(200001)] leftandright = [[] for i in range(200001)] lis = [] for i in range(n): x,y,r = map(str,input().split()) x,y = int(x),int(y) lis.append([x+y,x-y,x,y,r]) if r == "U": upanddown[x].append([y,0]) elif r == "D": upanddown[x].append([y,1]) elif r == "R": leftandright[y].append([x,0]) else: leftandright[y].append([x,1]) # print(upanddown[11]) for li in upanddown: if len(li) >= 2: li.sort(key=lambda z:z[1]) li.sort(key=lambda z:z[0]) key = 0 keynum = -1 for num,r in li: if r == 0: keynum = num else: if keynum != -1: ans = min(ans,(num-keynum)*5) for li in leftandright: if len(li) >= 2: li.sort(key=lambda z:z[1]) li.sort(key=lambda z:z[0]) key = 0 keynum = -1 for num,r in li: if r == 0: keynum = num else: if keynum != -1: ans = min(ans,(num-keynum)*5) # print(ans) lis.sort(key = lambda z:z[2]) lis.sort(key = lambda z:z[0]) for i in range(n-1): if lis[i][0] == lis[i+1][0]: if (lis[i][4] == "R" and lis[i+1][4] == "U") or (lis[i][4] == "D" and lis[i+1][4] == "L"): ans = min(ans,(lis[i+1][2]-lis[i][2])*10) lis.sort(key = lambda z:z[2]) lis.sort(key = lambda z:z[1]) for i in range(n-1): if lis[i][1] == lis[i+1][1]: if (lis[i][4] == "U" and lis[i+1][4] == "L") or (lis[i][4] == "R" and lis[i+1][4] == "D"): ans = min(ans,(lis[i+1][2]-lis[i][2])*10) if ans == 10 ** 10: print("SAFE") else: print(ans)

s230609501

p03434

u919633157

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()) s=list(map(int,input().split())) s.sort() a=[int(i) for i in s[::-2]] b=[int(j) for j in s[1::-2]] print(sum(a)-sum(b))

s942540935

Accepted

n=int(input()) a=sorted(list(map(int,input().split()))) print(sum(a[-1::-2])-sum(a[-2::-2]))

s507099804

p00725

u420446254

Wrong Answer

On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from ours. The game is played on an ice game board on which a square mesh is marked. They use only a single stone. The purpose of the game is to lead the stone from the start to the goal with the minimum number of moves. Fig. D-1 shows an example of a game board. Some squares may be occupied with blocks. There are two special squares namely the start and the goal, which are not occupied with blocks. (These two squares are distinct.) Once the stone begins to move, it will proceed until it hits a block. In order to bring the stone to the goal, you may have to stop the stone by hitting it against a block, and throw again. Fig. D-1: Example of board (S: start, G: goal) The movement of the stone obeys the following rules: * At the beginning, the stone stands still at the start square. * The movements of the stone are restricted to x and y directions. Diagonal moves are prohibited. * When the stone stands still, you can make it moving by throwing it. You may throw it to any direction unless it is blocked immediately(Fig. D-2(a)). * Once thrown, the stone keeps moving to the same direction until one of the following occurs: * The stone hits a block (Fig. D-2(b), (c)). * The stone stops at the square next to the block it hit. * The block disappears. * The stone gets out of the board. * The game ends in failure. * The stone reaches the goal square. * The stone stops there and the game ends in success. * You cannot throw the stone more than 10 times in a game. If the stone does not reach the goal in 10 moves, the game ends in failure. Fig. D-2: Stone movements Under the rules, we would like to know whether the stone at the start can reach the goal and, if yes, the minimum number of moves required. With the initial configuration shown in Fig. D-1, 4 moves are required to bring the stone from the start to the goal. The route is shown in Fig. D-3(a). Notice when the stone reaches the goal, the board configuration has changed as in Fig. D-3(b). Fig. D-3: The solution for Fig. D-1 and the final board configuration

while(): w, h = map(int, input().split()) if w == 0 or h == 0: break field = [] for y in range(h): field.append(map(int, input().split())) print(-1)

s282896739

Accepted

dx = [0, 1, 0, -1] dy = [1, 0, -1, 0] N_MOVE = 4 EMPTY = 0 ROCK = 1 START = 2 GOAL = 3 INF = 100000 def in_field(field, x, y): return y >=0 and y < len(field) and x >= 0 and x< len(field[0]) def move_to_rock(field, x, y, direction): while(True): x += dx[direction] y += dy[direction] if not in_field(field, x, y): return None elif field[y][x] == ROCK: x -= dx[direction] y -= dy[direction] break elif field[y][x] == GOAL: break return x, y def dfs(depth, field, x, y): if depth > 10: return None cost = INF for r in range(N_MOVE): if cost <= depth: return cost nx, ny = x + dx[r], y + dy[r] if not in_field(field, nx, ny): continue if field[ny][nx] == ROCK: continue next_pos = move_to_rock(field, x, y, r) if next_pos is None: continue nx, ny = next_pos if field[ny][nx] == GOAL: return depth rock_pos_x, rock_pos_y = nx+dx[r], ny+dy[r] assert field[rock_pos_y][rock_pos_x] == ROCK field[rock_pos_y][rock_pos_x] = EMPTY result = dfs(depth+1, field, nx, ny) if result is not None: cost = min(cost, result) field[rock_pos_y][rock_pos_x] = ROCK return cost def find_start_pos(field): h = len(field) w = len(field[0]) for y in range(h): for x in range(w): if field[y][x] == START: return x, y return None # ? def solve(): w, h = map(int, input().split()) if w == 0 or h == 0: return None field = [] for y in range(h): field.append(list(map(int, input().split()))) x, y = find_start_pos(field) res = dfs(1, field, x, y) return res if __name__ == '__main__': while(True): answer = solve() if answer is None: break elif answer == INF: print(-1) else: print(answer)