wrong_submission_id stringlengths 10 10 | problem_id stringlengths 6 6 | user_id stringlengths 10 10 | time_limit float64 1k 8k | memory_limit float64 131k 1.05M | wrong_status stringclasses 2
values | wrong_cpu_time float64 10 40k | wrong_memory float64 2.94k 3.37M | wrong_code_size int64 1 15.5k | problem_description stringlengths 1 4.75k | wrong_code stringlengths 1 6.92k | acc_submission_id stringlengths 10 10 | acc_status stringclasses 1
value | acc_cpu_time float64 10 27.8k | acc_memory float64 2.94k 960k | acc_code_size int64 19 14.9k | acc_code stringlengths 19 14.9k |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s721184627 | p02842 | u547728429 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 188 | Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer).... | N = int(input())
X = int(N / 1.08)
ans = -1
while X * 1.08 <= N:
print(X)
if int(X * 1.08) == N:
ans = X
X += 1
if ans > 0:
print(ans)
else:
print(":(")
| s956850754 | Accepted | 17 | 2,940 | 180 | N = int(input())
X = int(N / 1.08)
ans = -1
while int(X * 1.08) <= N:
if int(X * 1.08) == N:
ans = X
X += 1
if ans > 0:
print(ans)
else:
print(":(")
|
s322005271 | p03486 | u970198631 | 2,000 | 262,144 | Wrong Answer | 30 | 9,076 | 150 | 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().split()
s.sort(reverse =True)
t = input().split()
t.sort()
list1 = [s,t]
list1.sort()
if list1[0] == t:
print('No')
else:
print('Yes') | s514492524 | Accepted | 27 | 8,948 | 148 | s = list(input())
s.sort()
t = list(input())
t.sort(reverse =True)
list1 = [s,t]
list1.sort()
if list1[0] == t:
print('No')
else:
print('Yes') |
s172404863 | p03998 | u697696097 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 186 | 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. * ... | import sys
d={}
d["a"]=list(input())
d["b"]=list(input())
d["c"]=list(input())
turn="a"
while 1:
turn2=d[turn].pop(0)
if len(d[turn])==0:
print(turn)
sys.exit()
turn=turn2 | s145091731 | Accepted | 43 | 5,628 | 526 | import sys
from io import StringIO
import unittest
def yn(b):
print("Yes" if b==1 else "No")
return
def resolve():
readline=sys.stdin.readline
d={"a":0,"b":1,"c":2}
win=["A","B","C"]
s=[0]*3
s[0]=list(readline().rstrip())
s[1]=list(readline().rstrip())
s[2]=list(readline().rstrip(... |
s589812209 | p03436 | u513434790 | 2,000 | 262,144 | Wrong Answer | 2,112 | 150,860 | 722 | We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j). Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player re... | from scipy.sparse.csgraph import dijkstra
H, W = map(int, input().split())
s = [input() for _ in range(H)]
graph = [[0 for _ in range(H*W)] for _ in range(H*W)]
for i in range(H):
for j in range(W-1):
if s[i][j] == s[i][j+1] == ".":
graph[j+(W*i)][j+(W*i)+1] = 1
graph[j+(W*i)+1][j... | s891184206 | Accepted | 28 | 3,700 | 610 | from collections import deque
H, W = map(int, input().split())
S = [input() for i in range(H)]
C = sum(i.count('.') for i in S)
dd = ((-1, 0), (0, -1), (1, 0), (0, 1))
que = deque([(0, 0, 0)])
used = {(0, 0)}
res = []
while que:
s, t, cost = que.popleft()
if s == H-1 and t == W-1:
res = cost
... |
s992431484 | p02743 | u929996201 | 2,000 | 1,048,576 | Wrong Answer | 27 | 9,020 | 102 | Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold? | a,b,c = map(int,input().split())
if(pow(a,2)+pow(b,2) < pow(c,2)):
print("Yes")
else:
print("No") | s879347953 | Accepted | 27 | 9,156 | 104 | a,b,c = map(int,input().split())
d = c-a-b
if(0 < d and d*d > 4*a*b):
print("Yes")
else:
print("No") |
s584358838 | p03721 | u223904637 | 2,000 | 262,144 | Wrong Answer | 512 | 29,648 | 189 | There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2... | n,k=map(int,input().split())
l=[]
for i in range(n):
l.append(list(map(int,input().split())))
l.sort()
i=n-1
ans=0
while k>0:
k=k-l[i][1]
ans=l[i][0]
i=i-1
print(ans) | s996106936 | Accepted | 498 | 27,872 | 187 | n,k=map(int,input().split())
l=[]
for i in range(n):
l.append(list(map(int,input().split())))
l.sort()
i=0
ans=0
while k>0:
k=k-l[i][1]
ans=l[i][0]
i=i+1
print(ans) |
s907125882 | p03110 | u370429695 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 171 | Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000`... | num = int(input())
total = 0
for i in range(num):
li = input().split()
if li[1] == "JPY":
total += int(li[0])
else:
total += int(float(li[0])) * 380000
print(total) | s354602476 | Accepted | 17 | 2,940 | 166 | num = int(input())
total = 0
for i in range(num):
li = input().split()
if li[1] == "JPY":
total += int(li[0])
else:
total += float(li[0]) * 380000
print(total) |
s828473344 | p03448 | u749770850 | 2,000 | 262,144 | Wrong Answer | 29 | 3,060 | 240 | 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 co... | a = int(input())
b = int(input())
c = int(input())
x = int(input())
c = 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:
c = c + 1
print(c) | s873311954 | Accepted | 50 | 3,060 | 265 | a= int(input())
b= int(input())
c= int(input())
x= int(input())
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 = ans + 1
else:
pass
print(ans) |
s317435148 | p03407 | u077019541 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 84 | An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan. | A,B,C = map(int,input().split())
if C>=A+B*2:
print("Yes")
else:
print("No") | s164438153 | Accepted | 18 | 2,940 | 81 | A,B,C = map(int,input().split())
if C>A+B:
print("No")
else:
print("Yes") |
s212036981 | p04043 | u075785512 | 2,000 | 262,144 | Wrong Answer | 37 | 3,064 | 162 | 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 ... | #coding :UTF-8
#ABC042
list=input().split(" ")
list.sort()
print(list)
if list[0]=="5" and list[1]=="5" and list[2]=="7":
print("YES")
else:
print("NO")
| s944794115 | Accepted | 39 | 3,064 | 160 | #coding :UTF-8
#ABC042
list=input().split()
list.sort()
#print(list)
if list[0]=="5" and list[1]=="5" and list[2]=="7":
print("YES")
else:
print("NO")
|
s414566189 | p02364 | u927793658 | 1,000 | 131,072 | Wrong Answer | 20 | 5,460 | 1 | Find the sum of weights of edges of the Minimum Spanning Tree for a given weighted undirected graph G = (V, E). | s889227707 | Accepted | 670 | 23,144 | 578 | from heapq import *
def find(A,x):
p = A[x]
if p == x:
return x
a = find(A,p)
A[x] = a
return a
def union(A, x, y):
if find(A,x) > find(A,y):
bx, by = find(A,y), find(A,x)
else:
bx, by = find(A,x), find(A,y)
A[y] = bx
A[by] = bx
N, M = map( int, input().spl... | |
s348011801 | p02613 | u713014053 | 2,000 | 1,048,576 | Wrong Answer | 151 | 9,224 | 389 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`,... | t = int(input())
d = dict()
d['TLE'] = 0
d['AC'] = 0
d['WA'] = 0
d['RE'] = 0
while t > 0:
s = input()
if s == 'TLE':
d['TLE'] += 1
elif s == 'WA':
d['WA'] += 1
elif s == 'RE':
d['RE'] += 1
elif s == 'AC':
d['AC'] += 1
t-=1
print('AC', 'X' ,d['AC'])
print('WA', 'X'... | s630156185 | Accepted | 156 | 9,200 | 389 | t = int(input())
d = dict()
d['TLE'] = 0
d['AC'] = 0
d['WA'] = 0
d['RE'] = 0
while t > 0:
s = input()
if s == 'TLE':
d['TLE'] += 1
elif s == 'WA':
d['WA'] += 1
elif s == 'RE':
d['RE'] += 1
elif s == 'AC':
d['AC'] += 1
t-=1
print('AC', 'x' ,d['AC'])
print('WA', 'x'... |
s249364032 | p03360 | u497883442 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 133 | There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after... | a,b,c = list(map(int, input().split()))
k = int(input())
m = max(a,b,c)
l = m
for i in range(k-1):
l *= 2
print(sum([a,b,c])+l+m) | s668367792 | Accepted | 18 | 3,060 | 131 | a,b,c = list(map(int, input().split()))
k = int(input())
m = max(a,b,c)
l = m
for i in range(k):
l *= 2
print(sum([a,b,c])-m+l) |
s426074233 | p02264 | u530663965 | 1,000 | 131,072 | Wrong Answer | 30 | 6,012 | 1,025 | _n_ _q_ _name 1 time1_ _name 2 time2_ ... _name n timen_ In the first line the number of processes _n_ and the quantum _q_ are given separated by a single space. In the following _n_ lines, names and times for the _n_ processes are given. _name i_ and _time i_ are separated by a single space. | from collections import deque
def main():
process_count, process_time, processes = get_input()
result = run_roundrobin(processes = processes, time = process_time)
for r in result:
print('{0} {1}'.format(r.name, r.time))
def run_roundrobin(processes, time, total_time = 0, result=[]):
if not p... | s159427103 | Accepted | 400 | 19,716 | 968 | from collections import deque
def main():
process_count, process_time, processes = get_input()
result = run_roundrobin(processes=processes, time=process_time)
for r in result:
print(r.name, r.time)
def run_roundrobin(processes, time, total_time=0, result=[]):
while processes:
p = pro... |
s007299087 | p03449 | u555947166 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 177 | We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You ... | N = int(input())
row1 = list(map(int, input().split()))
row2 = list(map(int, input().split()))
x = [(sum(row1[:i]) + sum(row2[i-1:N+1])) for i in range(1, N+1)]
print(min(x))
| s664467950 | Accepted | 17 | 3,060 | 176 | N = int(input())
row1 = list(map(int, input().split()))
row2 = list(map(int, input().split()))
x = [(sum(row1[:i]) + sum(row2[i-1:N+1])) for i in range(1, N+1)]
print(max(x)) |
s031098062 | p03361 | u496744988 | 2,000 | 262,144 | Wrong Answer | 24 | 3,064 | 512 | 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... | import sys
h, w = map(int, input().split())
s = [input() for _ in range(h)]
print(s)
dx = [0, 1, 0, -1]
dy = [1, 0, -1, 0]
for i in range(h):
for j in range(w):
if s[i][j] == '.':
continue
flag = 0
for k in range(4):
dh = i + dy[k]
dw = j + dx[k]
... | s398241238 | Accepted | 23 | 3,064 | 503 | import sys
h, w = map(int, input().split())
s = [input() for _ in range(h)]
dx = [0, 1, 0, -1]
dy = [1, 0, -1, 0]
for i in range(h):
for j in range(w):
if s[i][j] == '.':
continue
flag = 0
for k in range(4):
dh = i + dy[k]
dw = j + dx[k]
# pri... |
s272959393 | p03795 | u077671688 | 2,000 | 262,144 | Wrong Answer | 26 | 8,868 | 496 | 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 ... | # abc055_a
N = int(input('整数を入力してね>'))
X = 800 * N
Y = (N/15)*200
print(X - Y) | s298947749 | Accepted | 25 | 9,124 | 65 | N = int(input(''))
x = (800 * N)
y = (int(N/15)*200)
print(x - y) |
s604078447 | p03612 | u732061897 | 2,000 | 262,144 | Wrong Answer | 63 | 20,472 | 142 | You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this. | N = int(input())
P = list(map(int, input().split()))
ans = 0
for i in range(N):
if P[i] == i + 1:
ans += 1
print(max(ans - 1, 0))
| s672042508 | Accepted | 85 | 20,540 | 266 | N = int(input())
P = list(map(int, input().split()))
ans = 0
for i in range(N):
if i == N-1:
if P[i] == i+1:
ans +=1
break
a, b = P[i], P[i + 1]
if a == i + 1:
ans += 1
P[i] = b
P[i + 1] = a
print(ans)
|
s548571693 | p02401 | u179070318 | 1,000 | 131,072 | Wrong Answer | 20 | 5,612 | 273 | 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:
a,op,b = [x for x in input().split( )]
a,b = [int(y) for y in [a,b]]
if op =='+':
print(a+b)
elif op =='-':
print(a-b)
elif op =='*':
print(a*b)
elif op =='/':
print(a/b)
elif op =='?':
break
| s911807211 | Accepted | 20 | 5,604 | 274 | while True:
a,op,b = [x for x in input().split( )]
a,b = [int(y) for y in [a,b]]
if op =='+':
print(a+b)
elif op =='-':
print(a-b)
elif op =='*':
print(a*b)
elif op =='/':
print(a//b)
elif op =='?':
break
|
s119498917 | p03555 | u118147328 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 136 | You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise. | C1 = input()
C2 = input()
if C1[0] == C2[2] and\
C1[1] == C2[1] and\
C1[2] == C2[0]:
print("Yes")
else:
print("No") | s887981161 | Accepted | 17 | 2,940 | 136 | C1 = input()
C2 = input()
if C1[0] == C2[2] and\
C1[1] == C2[1] and\
C1[2] == C2[0]:
print("YES")
else:
print("NO") |
s954711612 | p03502 | u318859025 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 102 | An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number. | n=int(input())
li=[int(i) for i in str(n)]
sum=sum(li)
if n%sum==0:
print("YES")
else:
print("NO") | s774803079 | Accepted | 17 | 2,940 | 102 | n=int(input())
li=[int(i) for i in str(n)]
sum=sum(li)
if n%sum==0:
print("Yes")
else:
print("No") |
s823287216 | p03361 | u477977638 | 2,000 | 262,144 | Wrong Answer | 19 | 3,444 | 321 | 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... | h,w=map(int,input().split())
s=["."+input()+"." for i in range(h)]
s=["."*(w+2)]+s+["."*(w+2)]
flag=0
print(s)
for i in range(h):
for j in range(w):
print(s[i][j])
if s[i][j]=="#":
if s[i-1][j]=="." and s[i+1][j]=="." and s[i][j-1]=="." and s[i][j+1]==".":
flag=1
print("Yes" if flag==0 else "No"... | s467483053 | Accepted | 18 | 3,064 | 301 | h,w=map(int,input().split())
s=["."+input()+"." for i in range(h)]
s=["."*(w+2)]+s+["."*(w+2)]
flag=0
for i in range(1,h+1):
for j in range(1,w+1):
if s[i][j]=="#":
if s[i-1][j]=="." and s[i+1][j]=="." and s[i][j-1]=="." and s[i][j+1]==".":
flag=1
print("Yes" if flag==0 else "No") |
s860375296 | p00352 | u108948964 | 1,000 | 262,144 | Wrong Answer | 20 | 5,532 | 37 | Alice and Brown are brothers in a family and each receives pocket money in celebration of the coming year. They are very close and share the total amount of the money fifty-fifty. The pocket money each receives is a multiple of 1,000 yen. Write a program to calculate each one’s share given the amount of money Alice an... | inp = input().split(" ")
print(inp)
| s042933566 | Accepted | 20 | 5,580 | 54 | a, b = map(int, input().split())
print((a + b) // 2)
|
s670667389 | p03338 | u455317716 | 2,000 | 1,048,576 | Wrong Answer | 19 | 3,064 | 235 | You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when ... | N = int(input())
S = input()
result = 0
for i in range(N):
x = S[:i]
y = S[i:]
print(x,y)
x_set,y_set = set(),set()
for ii in x:
x_set.add(ii)
for ii in y:
y_set.add(ii)
result = max(len(x_set & y_set),result)
print(result) | s413991926 | Accepted | 18 | 3,060 | 266 | n = int(input()) - 1
s = input()
result_list = list()
for i in range(n):
x = s[:i+1]
y = s[i+1:]
comon_set_len = len(set(ii for ii in x) & set(iii for iii in y))
result_list.append(comon_set_len)
result_list.sort(reverse = True)
print(result_list[0]) |
s645777579 | p03719 | u444238096 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 92 | You are given three integers A, B and C. Determine whether C is not less than A and not greater than B. | A, B, C = map(int, input().split())
if C>=A and C<=B:
print("YES")
else:
print("NO") | s783125506 | Accepted | 17 | 2,940 | 92 | A, B, C = map(int, input().split())
if C>=A and C<=B:
print("Yes")
else:
print("No") |
s195713332 | p03377 | u519923151 | 2,000 | 262,144 | Wrong Answer | 29 | 9,116 | 90 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | a,b,x = map(int, input().split())
if a<= x <= a+b:
print("Yes")
else:
print("No") | s009790760 | Accepted | 28 | 9,148 | 90 | a,b,x = map(int, input().split())
if a<= x <= a+b:
print("YES")
else:
print("NO") |
s265149661 | p03759 | u053535689 | 2,000 | 262,144 | Wrong Answer | 26 | 9,100 | 103 | Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful. | # 058a
a, b, c = map(int, input().split())
if b - a == c - b:
print('Yes')
else:
print('No')
| s411796495 | Accepted | 31 | 9,132 | 103 | # 058a
a, b, c = map(int, input().split())
if b - a == c - b:
print('YES')
else:
print('NO')
|
s307323212 | p03379 | u771532493 | 2,000 | 262,144 | Wrong Answer | 2,105 | 25,220 | 133 | When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..... | N=int(input())
L=[int(i) for i in input().split()]
L.sort()
for i in range(N):
Lcopy=L[:]
del Lcopy[i]
print(Lcopy[N//2-1])
| s153841260 | Accepted | 319 | 25,228 | 153 | N=int(input())
L=[int(i) for i in input().split()]
L1=sorted(L)
a=L1[N//2]
b=L1[N//2-1]
for i in range(N):
if L[i]<a:
print(a)
else:
print(b) |
s273607289 | p02396 | u998185318 | 1,000 | 131,072 | Wrong Answer | 140 | 5,596 | 137 | In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are give... | cnt = 1
while True:
in_value = int(input())
if in_value == 0:
break
print("Case " + str(cnt) + ": " + str(in_value))
| s479474187 | Accepted | 90 | 5,908 | 295 | def input_until_zero():
rtn_list = []
while True:
in_value = int(input())
if in_value == 0:
break
rtn_list.append(in_value)
return rtn_list
rtn = input_until_zero()
for i in range(len(rtn)):
print("Case " + str(i + 1) + ": " + str(rtn[i]))
|
s625854222 | p03720 | u633548583 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 102 | There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city? | n,m=map(int,input().split())
a=list(map(int,input().split()))
print(a.count(i) for i in range (1,n+1)) | s772300761 | Accepted | 17 | 2,940 | 168 | n,m=map(int,input().split())
li=[]
for i in range(m):
a,b=map(int,input().split())
li.append(a)
li.append(b)
for j in range(1,n+1):
print(li.count(j))
|
s942772834 | p02646 | u801049006 | 2,000 | 1,048,576 | Wrong Answer | 24 | 9,176 | 161 | 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 ... | a, v = map(int, input().split())
b, w = map(int, input().split())
t = int(input())
if v - w > 0 and abs(a-b) / (v-w) > t:
print("YES")
else:
print("NO") | s675030959 | Accepted | 21 | 9,180 | 162 | a, v = map(int, input().split())
b, w = map(int, input().split())
t = int(input())
if v - w > 0 and abs(a-b) / (v-w) <= t:
print("YES")
else:
print("NO") |
s552161429 | p02692 | u296150111 | 2,000 | 1,048,576 | Wrong Answer | 142 | 16,320 | 663 | There is a game that involves three variables, denoted A, B, and C. As the game progresses, there will be N events where you are asked to make a choice. Each of these choices is represented by a string s_i. If s_i is `AB`, you must add 1 to A or B then subtract 1 from the other; if s_i is `AC`, you must add 1 to A or ... | n,a,b,c=map(int,input().split())
s=[input() for _ in range(n)]
if a+b+c==0:
print("No")
elif a+b+c==1:
for x in s:
if x=="AB":
if a==b==0:
print("No")
exit()
elif a==0:
a+=1
b-=1
else:
b+=1
a-=1
elif x=="BC":
if b==c==0:
print("No")
exit()
elif b==0:
b+=1
c-=... | s306416893 | Accepted | 186 | 17,100 | 1,666 | n,a,b,c=map(int,input().split())
s=[input() for _ in range(n)]
if a+b+c==0:
print("No")
elif a+b+c==1:
ans=[]
for x in s:
if x=="AB":
if a==b==0:
print("No")
exit()
elif a==0:
a+=1
b-=1
ans.append("A")
else:
b+=1
a-=1
ans.append("B")
elif x=="BC":
if b==c==0:
print... |
s299043070 | p03377 | u726439578 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 108 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | a,b,c =map(int,input().split())
if b>c:
print("NO")
elif a+b>=c :
print("YES")
else:
print("NO") | s371318155 | Accepted | 21 | 3,188 | 88 | a,b,c =map(int,input().split())
if a>c or a+b<c:
print("NO")
else:
print("YES") |
s060495366 | p02612 | u735091636 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,176 | 321 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | import sys
def II(): return int(input())
def MI(): return map(int,input().split())
def LI(): return list(map(int,input().split()))
def TI(): return tuple(map(int,input().split()))
def RN(N): return [input().strip() for i in range(N)]
def main():
N = II()
print(N%1000)
if __name__ == "__main__":
main(... | s207541077 | Accepted | 30 | 9,184 | 389 | import sys
def II(): return int(input())
def MI(): return map(int,input().split())
def LI(): return list(map(int,input().split()))
def TI(): return tuple(map(int,input().split()))
def RN(N): return [input().strip() for i in range(N)]
def main():
N = II()
ans = 1000-N%1000
if ans==1000:
prin... |
s898141327 | p02678 | u166340293 | 2,000 | 1,048,576 | Wrong Answer | 2,207 | 48,464 | 447 | 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 th... | N,M=map(int,input().split())
r=[]
for i in range(M):
r.append(list(map(int,input().split())))
q=[0]
d=[0]+[-1]*N
ans=[0]+[-1]*N
while q:
u=q.pop(0)
for v in range(M):
if r[v][0]-1==u:
o=r[v][1]
if d[o-1]<=0:
d[o-1]=d[u]+1
ans[o-1]=u
q.append(o-1)
elif r[v][1]-1==u:
... | s021573372 | Accepted | 1,470 | 37,268 | 399 | N,M=map(int,input().split())
r=[]
k=[]
for i in range(N):
k.append([])
for i in range(M):
a,b=map(int,input().split())
k[a-1].append(b)
k[b-1].append(a)
d=[0]+[-1]*N
ans=[0]+[-1]*N
q=[0]
while q:
u=q.pop(0)
for v in range(len(k[u])):
o=k[u][v]
if d[o-1]<=0:
d[o-1]=d[u]+1
ans[o-1]=u... |
s221075782 | p02833 | u454524105 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 133 | For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N). | n = int(input())
if n % 2 == 1: print(0)
else:
ans = 0
n /= 2
while n:
ans += n / 5
n /= 5
print(ans) | s750402218 | Accepted | 17 | 2,940 | 189 | n = int(input())
if n % 2 == 1: print(0)
else:
ans = 0
i = 1
while True:
a = 2 * 5**i
if n // a == 0: break
ans += (n // a)
i += 1
print(ans) |
s237667337 | p03693 | u217303170 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 83 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this i... | r,g,b=input().split()
x=r+g+b
y=int(x)
if y%4==0:
print('Yes')
else:
print('No')
| s140133331 | Accepted | 17 | 2,940 | 82 | r,g,b=input().split()
x=r+g+b
y=int(x)
if y%4==0:
print('YES')
else:
print('NO') |
s463954773 | p03049 | u845333844 | 2,000 | 1,048,576 | Wrong Answer | 46 | 3,064 | 351 | Snuke has N strings. The i-th string is s_i. Let us concatenate these strings into one string after arranging them in some order. Find the maximum possible number of occurrences of `AB` in the resulting string. | n=int(input())
sum=0
a=0
b=0
ab=0
for i in range(n):
x=input()
m=len(x)
if x[0]=='B' and x[m-1]=='A':
ab+=1
elif x[0]=='B':
b+=1
elif x[m-1]=='A':
a+=1
for j in range(m-1):
if x[j]=='A' and x[j+1]=='B':
sum+=1
if a==b:
print(sum+a+ab-1)
e... | s533955208 | Accepted | 45 | 3,064 | 465 | n=int(input())
sum=0
a=0
b=0
ab=0
for i in range(n):
x=input()
m=len(x)
if x[0]=='B' and x[m-1]=='A':
ab+=1
elif x[0]=='B':
b+=1
elif x[m-1]=='A':
a+=1
for j in range(m-1):
if x[j]=='A' and x[j+1]=='B':
sum+=1
if ab==0:
print(sum+min(a,b)... |
s821449514 | p03408 | u917138620 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 301 | 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 ea... | in_num = input()
in_data=[]
for i in range(0,int(in_num)):
in_data.append(input())
out_num = input()
out_data = []
for i in range(0,int(out_num)):
out_data.append(input())
max_count=0
for n in in_data:
tmp = out_data.count(n)
if tmp > max_count:
max_count = tmp
print(max_count) | s980555767 | Accepted | 17 | 3,064 | 320 | in_num = input()
in_data=[]
for i in range(0,int(in_num)):
in_data.append(input())
out_num = input()
out_data = []
for i in range(0,int(out_num)):
out_data.append(input())
max_count=0
for n in in_data:
tmp = in_data.count(n) - out_data.count(n)
if tmp > max_count:
max_count = tmp
print(max_count) |
s466295010 | p02806 | u857428111 | 2,525 | 1,048,576 | Wrong Answer | 18 | 3,064 | 559 | Niwango created a playlist of N songs. The title and the duration of the i-th song are s_i and t_i seconds, respectively. It is guaranteed that s_1,\ldots,s_N are all distinct. Niwango was doing some work while playing this playlist. (That is, all the songs were played once, in the order they appear in the playlist, w... | #!/usr/bin/env python3
import sys
input= lambda: sys.stdin.readline().rstrip()
sys.setrecursionlimit(10**9)
def pin(type=int):return map(type,input().split())
def tupin(t=int):return tuple(pin(t))
def lispin(t=int):return list(pin(t))
#%%code
def resolve():
a=sys.stdin.readlines()
#print(a)
N,X=int(a[0]),a... | s858491264 | Accepted | 18 | 3,064 | 586 | #!/usr/bin/env python3
import sys
input= lambda: sys.stdin.readline().rstrip()
sys.setrecursionlimit(10**9)
def pin(type=int):return map(type,input().split())
def tupin(t=int):return tuple(pin(t))
def lispin(t=int):return list(pin(t))
#%%code
def resolve():
a=sys.stdin.readlines()
#print(a)
N,X=int(a[0].rs... |
s664975141 | p03681 | u172569352 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 81 | Snuke has N dogs and M monkeys. He wants them to line up in a row. As a Japanese saying goes, these dogs and monkeys are on bad terms. _("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.)_ Snuke is trying to reconsile them, by arranging the animals so that there... | N, M = [int(i) for i in input().split()]
L = abs(N - M)
if L >= 2:
print(0) | s092431393 | Accepted | 703 | 5,180 | 254 | import math
N, M = [int(i) for i in input().split()]
r = pow(10, 9) + 7
L = abs(N - M)
if L >= 2:
print(0)
elif L == 0:
E = 2*math.factorial(N)*math.factorial(M)
print(E % r)
else:
E = math.factorial(N)*math.factorial(M)
print(E % r) |
s948659051 | p03696 | u703890795 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 254 | You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of... | N = int(input())
S = input()
T = []
for s in S:
if s == "(":
T.append(-1)
else:
T.append(1)
c = 0
U = ""
for i in range(N):
c += T[i]
if c > 0:
U += "("
c -= 1
if T[i] == -1:
U += "("
else:
U += ")"
print(U + ")"*(-c)) | s687014826 | Accepted | 17 | 3,060 | 265 | N = int(input())
S = input()
T = []
c = 0
m = 0
for s in S:
if s == ")":
c += 1
else:
if c > 0:
m += c
c = 0
c -= 1
if c > 0:
m += c
c = 0
print("("*m + S + ")"*(-c))
|
s502418103 | p03047 | u661439250 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 44 | Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi. How many ways are there to choose K consecutive integers? | N, K = map(int, input().split())
print(N-K) | s653472558 | Accepted | 17 | 2,940 | 46 | N, K = map(int, input().split())
print(N-K+1) |
s612380096 | p02612 | u078276601 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,096 | 47 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | N = int(input())
change = N%1000
print(change) | s661552335 | Accepted | 29 | 9,112 | 66 | N = int(input())
change = ((N+1000-1)//1000)*1000-N
print(change) |
s228354082 | p02742 | u546501940 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 137 | 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 to... | x,y = map(int,input().split())
if x ==1 or y ==1:
print(1)
exit(0)
if x%2==0 or y%2==0:
print(x*y/2)
exit(0)
print((x*y/2+1))
| s936587920 | Accepted | 17 | 2,940 | 149 | x,y = map(int,input().split())
if x ==1 or y ==1:
print(1)
exit(0)
if x%2==0 or y%2==0:
print(int(x/2*y))
exit(0)
a=int(x*y/2)+1
print(a)
|
s170046250 | p02613 | u970133396 | 2,000 | 1,048,576 | Wrong Answer | 145 | 16,484 | 183 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`,... | from collections import Counter
n = int(input().strip())
L = [input().strip() for _ in range(n)]
cnt=Counter(L)
for x in ["AC", "WA", "TLE", "RE"]:
print(x+" × "+str(cnt[x])) | s655354004 | Accepted | 146 | 16,616 | 180 | from collections import Counter
n = int(input().strip())
L = [input().strip() for _ in range(n)]
cnt=Counter(L)
for x in ["AC", "WA", "TLE", "RE"]:
print(x+" x "+str(cnt[x])) |
s811261310 | p02399 | u385991569 | 1,000 | 131,072 | Wrong Answer | 30 | 7,620 | 176 | Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number) | # -*- coding: utf-8 -*-
"""
Created on Mon Jun 12 16:30:47 2017
@author: gedyra
"""
a,b = list(map(int, input().split()))
d = int(a/b)
r = a%b
f = float(a/b)
print(d, r, f) | s913522856 | Accepted | 20 | 7,624 | 98 | a,b = list(map(int, input().split()))
d = a//b
r = a%b
f = ('{:.5f}'.format(a/b))
print(d, r, f) |
s494923422 | p03469 | u078349616 | 2,000 | 262,144 | Wrong Answer | 19 | 2,940 | 37 | 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 col... | print(input().replace("2018","2017")) | s542397398 | Accepted | 22 | 2,940 | 37 | print(input().replace("2017","2018")) |
s925170775 | p02748 | u686253980 | 2,000 | 1,048,576 | Wrong Answer | 2,105 | 36,608 | 407 | You are visiting a large electronics store to buy a refrigerator and a microwave. The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen. You have M discount tic... | A, B, M = [int(i) for i in input().split(" ")]
a = [int(i) for i in input().split(" ")]
b = [int(i) for i in input().split(" ")]
C = []
for i in range(M):
c = [int(i) for i in input().split(" ")]
C.append(c)
print(a)
m = 1000000
for i in a:
for j in b:
cost = i + j
if m >= cost:
m = cost
for c... | s778320744 | Accepted | 457 | 32,492 | 328 | A, B, M = [int(i) for i in input().split(" ")]
a = [int(i) for i in input().split(" ")]
b = [int(i) for i in input().split(" ")]
C = []
for i in range(M):
c = [int(i) for i in input().split(" ")]
C.append(c)
m = min(a) + min(b)
for c in C:
cost = a[c[0] - 1] + b[c[1] - 1] - c[2]
if m >= cost:
m = cost
... |
s050407811 | p03448 | u825528847 | 2,000 | 262,144 | Wrong Answer | 21 | 3,572 | 256 | 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 co... | A = int(input())
B = int(input())
C = int(input())
X = int(input())
ans = 0
for i in range(A+1):
for j in range(B+1):
tmp = i * 500 + j * 100
if (X - tmp) % 50 == 0 and tmp <= X:
print(i, j)
ans += 1
print(ans)
| s401871240 | Accepted | 58 | 3,316 | 222 | A = int(input())
B = int(input())
C = int(input())
X = int(input())
ans = 0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
ans += 1 * ((i * 500 + j * 100 + k * 50) == X)
print(ans)
|
s629526794 | p02612 | u229950162 | 2,000 | 1,048,576 | Wrong Answer | 32 | 9,048 | 40 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | n = int(input().rstrip())
print(n%1000) | s973228429 | Accepted | 27 | 9,056 | 47 | n = int(input().rstrip())
print((1000-n)%1000) |
s448707478 | p03359 | u859897687 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 38 | In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For ... | a=input();print(int(a[0])-(a[0]>a[1])) | s186938108 | Accepted | 17 | 2,940 | 43 | a,b=map(int,input().split())
print(a-(a>b)) |
s650523727 | p03354 | u766684188 | 2,000 | 1,048,576 | Wrong Answer | 697 | 36,100 | 886 | We have a permutation of the integers from 1 through N, p_1, p_2, .., p_N. We also have M pairs of two integers between 1 and N (inclusive), represented as (x_1,y_1), (x_2,y_2), .., (x_M,y_M). AtCoDeer the deer is going to perform the following operation on p as many times as desired so that the number of i (1 ≤ i ≤ N)... | n,m=map(int,input().split())
P=list(map(int,input().split()))
edges=[list(map(int,input().split())) for _ in range(m)]
class UnionFind:
def __init__(self,n):
self.par=[i for i in range(n+1)]
self.rank=[0]*(n+1)
def find(self,x):
if self.par[x]==x:
return x
else:
... | s975639851 | Accepted | 713 | 36,100 | 898 | #ARC097-D
n,m=map(int,input().split())
P=list(map(int,input().split()))
edges=[list(map(int,input().split())) for _ in range(m)]
class UnionFind:
def __init__(self,n):
self.par=[i for i in range(n+1)]
self.rank=[0]*(n+1)
def find(self,x):
if self.par[x]==x:
return x
... |
s123267010 | p02972 | u798818115 | 2,000 | 1,048,576 | Wrong Answer | 50 | 7,148 | 160 | 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 con... | # coding: utf-8
# Your code here!
M=int(input())
B=list(map(int,input().split()))
l=[0]*M
for i in range(1,-int(-M**0.5//1)+1):
print(i)
| s420812760 | Accepted | 787 | 14,044 | 575 | # coding: utf-8
# Your code here!
N=int(input())
A=list(map(int,input().split()))
count=0
ans=[]
hako=[0]*N
for i in range(N)[::-1]:
all=0
for j in range(i,N)[::i+1]:
if i==j:
continue
all^=hako[j]
if all==1:
if A[i]==1:
continue
else:
hak... |
s717599315 | p02608 | u347600233 | 2,000 | 1,048,576 | Wrong Answer | 2,205 | 9,008 | 257 | Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N). | n = int(input())
ans = [0] * (n + 1)
for x in range(1, n):
for y in range(1, n):
for z in range(1, n):
if x**2 + y**2 + z**2 + x*y + y*z + z*x <= n:
ans[x**2 + y**2 + z**2 + x*y + y*z + z*x] += 1
print(*ans, sep='\n') | s542137698 | Accepted | 194 | 9,588 | 325 | n = int(input())
ans = [0] * (n + 1)
a = [1, 3, 6]
for x in range(1, int(n**0.5) + 1):
for y in range(x, int(n**0.5) + 1):
for z in range(y, int(n**0.5) + 1):
v = x**2 + y**2 + z**2 + x*y + y*z + z*x
if v <= n:
ans[v] += a[len(set([x, y, z])) - 1]
print(*ans[1:], sep=... |
s265536113 | p02694 | u605327527 | 2,000 | 1,048,576 | Wrong Answer | 20 | 9,160 | 141 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or abo... | import math
n = int(input())
cash = 100
count = 0
while n >= cash:
cash = cash + math.floor(cash * 0.01)
count += 1
print(count)
| s582590841 | Accepted | 23 | 9,156 | 140 | import math
n = int(input())
cash = 100
count = 0
while n > cash:
cash = cash + math.floor(cash * 0.01)
count += 1
print(count)
|
s366353230 | p02257 | u123669391 | 1,000 | 131,072 | Wrong Answer | 20 | 5,640 | 277 | A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list. | n = int(input())
sum = 0
for _ in range(n):
a = int(input())
if a == 2:
continue
for i in range(a):
x = i + 2
if a%x == 0:
sum += 1
print(a)
break
if x > a ** 0.5:
break
print(n - sum)
| s438135678 | Accepted | 1,970 | 5,668 | 257 | n = int(input())
sum = 0
for _ in range(n):
a = int(input())
if a == 2:
continue
for i in range(a):
x = i + 2
if a%x == 0:
sum += 1
break
if x > a ** 0.5:
break
print(n - sum)
|
s401694326 | p03759 | u609987416 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 93 | Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful. | a,b,c = map(int, input().split())
if (b - a == c - b):
print("Yes")
else:
print("No") | s608919429 | Accepted | 17 | 2,940 | 93 | a,b,c = map(int, input().split())
if (b - a == c - b):
print("YES")
else:
print("NO") |
s480801960 | p03814 | u814986259 | 2,000 | 262,144 | Wrong Answer | 25 | 6,180 | 89 | 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 w... | S=list(input())
a_id=S.index("A")
z_id=list(reversed(S)).index("Z")
print(z_id - a_id +1) | s592832420 | Accepted | 25 | 6,180 | 98 | S=list(input())
l =len(S)
a_id=S.index("A")
z_id=list(reversed(S)).index("Z")
print(l-z_id - a_id) |
s750702572 | p04043 | u787059958 | 2,000 | 262,144 | Wrong Answer | 26 | 9,168 | 170 | 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 ... | ans = []
A, B, C = map(int, input().split())
ans.append(A)
ans.append(B)
ans.append(C)
if ans.count(5) == 2 and ans.count(7) == 1:
print('Yes')
else:
print('No')
| s987825104 | Accepted | 21 | 9,064 | 170 | ans = []
A, B, C = map(int, input().split())
ans.append(A)
ans.append(B)
ans.append(C)
if ans.count(5) == 2 and ans.count(7) == 1:
print('YES')
else:
print('NO')
|
s781081448 | p03829 | u873917047 | 2,000 | 262,144 | Wrong Answer | 100 | 14,252 | 221 | There are N towns on a line running east-west. The towns are numbered 1 through N, in order from west to east. Each point on the line has a one- dimensional coordinate, and a point that is farther east has a greater coordinate value. The coordinate of town i is X_i. You are now at town 1, and you want to visit all the... | #coding: UTF-8
N, A, B=map(int, input().split())
lis=list(map(int,input().split()))
#print(lis)
out=0
for i in range(0,N-1):
d=lis[i+1]-lis[i]
if d < (B-A):
out=out+d
else:
out=out+B
print(out) | s364575661 | Accepted | 101 | 14,252 | 221 | #coding: UTF-8
N, A, B=map(int, input().split())
lis=list(map(int,input().split()))
#print(lis)
out=0
for i in range(0,N-1):
d=(lis[i+1]-lis[i])*A
if d < B:
out=out+d
else:
out=out+B
print(out) |
s814129986 | p03110 | u653807637 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 171 | Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000`... | n = int(input())
bs = [input().split() for _ in range(n)]
ans = 0
for b in bs:
if b[1] == "BTC":
ans += float(b[0]) * 380000
else:
ans += int(b[0])
print(int(ans)) | s191890358 | Accepted | 17 | 2,940 | 170 | n = int(input())
bs = [input().split() for _ in range(n)]
ans = 0
for b in bs:
if b[1] == "BTC":
ans += float(b[0]) * 380000.0
else:
ans += float(b[0])
print(ans) |
s345606411 | p03478 | u441320782 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 11 | Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive). | print(2/10) | s448012045 | Accepted | 35 | 9,420 | 176 | N,A,B=map(int,input().split())
ans=[]
for i in range(1,N+1):
j=i//10000+(i%10000)//1000+(i%1000)//100+(i%100)//10+(i%10)//1
if A<=j<=B:
ans.append(i)
print(sum(ans)) |
s012242515 | p03048 | u664762434 | 2,000 | 1,048,576 | Wrong Answer | 1,300 | 3,060 | 173 | Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g... | R,G,B,N = map(int,input().split())
x = N//R
ans = 0
for i in range(x+1):
M = N - i*R
for j in range((M+1)//G):
L = M - j*G
if L%B == 0:
ans += 1
print(ans) | s158125816 | Accepted | 1,554 | 3,060 | 173 | R,G,B,N = map(int,input().split())
x = N//R
ans = 0
for i in range(x+1):
M = N - i*R
for j in range((M//G)+1):
L = M - j*G
if L%B == 0:
ans += 1
print(ans) |
s778649164 | p03861 | u183840468 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 113 | You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x? | a,b,x = [int(i) for i in input().split()]
if (b-a)%x == 0:
print(b//x - a//x)
else:
print(b//x - a//x+1) | s754901187 | Accepted | 17 | 3,064 | 89 | a,b,c = [int(i) for i in input().split()]
print(b//c - (a-1)//c if a > 0 else b//c + 1) |
s731136722 | p00020 | u518711553 | 1,000 | 131,072 | Wrong Answer | 20 | 7,288 | 27 | Write a program which replace all the lower-case letters of a given text with the corresponding captital letters. | print(input().capitalize()) | s780308590 | Accepted | 30 | 7,940 | 109 | from functools import reduce
from operator import add
print(reduce(add, map(str.capitalize, input()[:]), '')) |
s955213006 | p03730 | u760794812 | 2,000 | 262,144 | Wrong Answer | 29 | 9,148 | 112 | 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 objecti... | A,B,C = map(int,input().split())
for i in range(B):
if (i+1)*A%B== C:
print('Yes')
exit()
print('No')
| s054740555 | Accepted | 25 | 9,156 | 111 | A,B,C = map(int,input().split())
for i in range(B):
if (i+1)*A%B== C:
print('YES')
exit()
print('NO') |
s654273656 | p02409 | u780342333 | 1,000 | 131,072 | Wrong Answer | 30 | 7,708 | 394 | You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth fl... | nums = [[[0 for x in range(0, 10)] for x in range(0, 3)] for x in range(0, 4)]
input1='''1 1 3 8
3 2 2 7
4 3 8 1'''.split("\n")
print(input1)
for e in input1:
efbv = e.split(" ")
e_b, e_f, e_r, e_v = map(int, efbv)
nums[e_b - 1][e_f - 1][e_r -1] += e_v
for b in range(4):
for f in range (3):
... | s045166549 | Accepted | 30 | 5,628 | 372 | buildings = [[[0 for r in range(10)] for f in range(3)] for b in range(4)]
n = int(input())
for _ in range(n):
b, f, r, v = map(int, input().split())
buildings[b-1][f-1][r-1] += v
sep_count = len(buildings) - 1
for b in buildings:
for f in b:
print(" ",end="")
print(*f)
if sep_count !=... |
s073489618 | p02690 | u621345513 | 2,000 | 1,048,576 | Wrong Answer | 73 | 15,720 | 240 | Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X. | x = int(input())
from itertools import product
tot_comb = [el for el in product(list(range(-120, 121)), list(range(-120, 121))) if el[0]!=el[1]]
tot_comb += [(0,0)]
num_dict = {el[0]**5 - el[1]**5: el for el in tot_comb}
print(num_dict[x]) | s754326276 | Accepted | 68 | 15,724 | 264 | x = int(input())
from itertools import product
tot_comb = [el for el in product(list(range(-120, 121)), list(range(-120, 121))) if el[0]!=el[1]]
tot_comb += [(0,0)]
num_dict = {el[0]**5 - el[1]**5: el for el in tot_comb}
pair = num_dict[x]
print(pair[0], pair[1]) |
s674373108 | p03494 | u774411119 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 265 | There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform. | N=int(input())
suji=input()
I1=suji.split(" ")
I=list(map(int,I1))
wari=2
n=0
kazu=0
X=1
while n<3:
if I[n]%wari==0:
n+=1
if I[N-1]%2==0:
wari=wari*2
kazu+=1
n=0
else:
X=0
break
print(kazu) | s304924235 | Accepted | 160 | 12,508 | 301 | import math
import numpy as np
N = int(input())
A_list = list(map(int, input().split()))
np_A = np.array(A_list)
tmp_cou = 0
x = 1
while x==1:
for i in np_A:
if i % 2 == 0:
tmp_cou += 1
else:
x = 2
np_A = np_A/2
cou = math.floor(tmp_cou/N)
print(cou)
|
s036171757 | p02388 | u128236306 | 1,000 | 131,072 | Wrong Answer | 20 | 7,600 | 36 | Write a program which calculates the cube of a given integer x. | x = input()
x = int(x)
print(x ^ 3) | s719055299 | Accepted | 20 | 7,624 | 36 | x = input()
x = int(x)
print(x ** 3) |
s288752509 | p03110 | u994521204 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 158 | Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000`... | n=int(input())
c=0
for i in range(n):
x,u=input().split()
if u=='JPY':
c+=int(x)
elif u=='BTC':
c+=int(380000 * float(x))
print(c) | s782777825 | Accepted | 17 | 2,940 | 153 | n=int(input())
c=0
for i in range(n):
x,u=input().split()
if u=='JPY':
c+=int(x)
elif u=='BTC':
c+=380000 * float(x)
print(c) |
s267306023 | p03854 | u350248178 | 2,000 | 262,144 | Wrong Answer | 19 | 3,188 | 160 | 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()
s=s.replace("eraser","")
s=s.replace("erase","")
s=s.replace("dreamer","")
s=s.replace("dream","")
if s=="":
print("Yes")
else:
print("No")
| s168234642 | Accepted | 19 | 3,188 | 160 | s=input()
s=s.replace("eraser","")
s=s.replace("erase","")
s=s.replace("dreamer","")
s=s.replace("dream","")
if s=="":
print("YES")
else:
print("NO")
|
s202597438 | p03228 | u022871813 | 2,000 | 1,048,576 | Wrong Answer | 26 | 3,444 | 415 | In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other pe... | a, b, k = map(int, input().split())
for i in range(k):
if i%2 == 0:
if a%2 == 0:
a /= 2
b += 2
elif a%2 == 1:
a -= 1
a /= 2
b += a
elif i%2 == 1:
if b%2 == 0:
b /= 2
a += b
elif a%2 == 1:
... | s913490836 | Accepted | 17 | 3,064 | 390 | a, b, k = map(int, input().split())
for i in range(k):
if i%2 == 0:
if a%2 == 0:
a /= 2
b += a
elif a%2 == 1:
a -= 1
a /= 2
b += a
elif i%2 == 1:
if b%2 == 0:
b /= 2
a += b
elif b%2 == 1:
... |
s436137771 | p03720 | u369338402 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 221 | There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city? | n,m=map(int,input().split())
roads=[]
for i in range(m):
roads.append(input().split())
count=0
for j in range(1,n+1):
for k in range(m):
if roads[k][0]==j or roads[k][1]==j:
count+=1
print(count)
count=0 | s739615717 | Accepted | 19 | 3,060 | 253 | n,m=map(int,input().split())
roads=[]
for i in range(m):
roads.append(input().split())
count=0
for j in range(1,n+1):
for k in range(m):
if int(roads[k][0])==j or int(roads[k][1])==j:
count+=1
#print('road')
print(count)
count=0
|
s821848339 | p03575 | u599547273 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 486 | You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M ... | n, m = map(int, input().split(" "))
a = [list(map(int, input().split(" "))) for i in range(m)]
a_counts = [0]*n
for a_i in sum(a, []):
a_counts[a_i-1] += 1
bridge_count = 0
for i, count in enumerate(a_counts):
if count == 1:
p = i+1
while a_counts[p-1] <= 2:
for j, a_i in enumerate(a):
if a_i and p in a... | s765855814 | Accepted | 18 | 3,064 | 499 | n, m = map(int, input().split(" "))
a = [list(map(int, input().split(" "))) for i in range(m)]
a_connects = [[] for i in range(n)]
for x, y in a:
a_connects[x-1].append(y-1)
a_connects[y-1].append(x-1)
a_connect_sums = [len(connect) for connect in a_connects]
bridge_count = 0
while 1 in a_connect_sums:
x = a_conne... |
s425790771 | p03251 | u685662874 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,060 | 285 | 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 incli... | N,M,X,Y=map(int, input().split())
Xpoints=list(map(int, input().split()))
Ypoints=list(map(int, input().split()))
for z in range(-100, 101):
if X < z and z <= Y:
if max(Xpoints) < z:
if min(Ypoints) >= z:
print('No War')
else:
print('War')
| s622464221 | Accepted | 18 | 3,060 | 309 | N,M,X,Y=map(int, input().split())
Xpoints=list(map(int, input().split()))
Ypoints=list(map(int, input().split()))
isWar='War'
for z in range(-99, 101):
if X < z and z <= Y:
if max(Xpoints) < z:
if min(Ypoints) >= z:
isWar = 'No War'
break
print(isWar)
|
s685276730 | p03854 | u358957649 | 2,000 | 262,144 | Wrong Answer | 20 | 3,188 | 668 | 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`. | def main(s):
t = ""
e = ["dream","erase","eraser","dreamer"]
while len(t) < len(s) :
val = ""
c = t
for i in e:
val = t + i
if s[:len(val)] != val:
continue
else:
t = val
break
... | s737587152 | Accepted | 1,213 | 3,992 | 512 | def main(s):
e = ["dreamer","eraser","dream","erase"]
t = [""]
while len(t) > 0:
val = t.pop(0)
for i in e:
tar = val + i
if s[:len(tar)] == tar:
if s == tar:
return True
else:
... |
s071555510 | p04043 | u024422110 | 2,000 | 262,144 | Wrong Answer | 26 | 8,928 | 121 | 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 ... | a = input().split()
five = a.count(5)
seven = a.count(7)
if five == 2 and seven == 1:
print('YES')
else:
print('NO') | s980048399 | Accepted | 27 | 8,824 | 125 | a = input().split()
five = a.count('5')
seven = a.count('7')
if five == 2 and seven == 1:
print('YES')
else:
print('NO') |
s938180597 | p03645 | u085883871 | 2,000 | 262,144 | Wrong Answer | 632 | 18,912 | 554 | 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 Snuk... | import math
def solve(n, m, a, b):
islands = []
for i in range(m):
if(a[i] == 1):
islands.append(b[i])
if(b[i] == 1):
islands.append(a[i])
for i in range(m):
if(a[i] == m and b[i] in islands):
return "POSSIBLE"
if(b[i] == m and a[i] in i... | s160073796 | Accepted | 635 | 21,888 | 538 | def solve(n, m, a, b):
islands = set()
for i in range(m):
if(a[i] == 1):
islands.add(b[i])
if(b[i] == 1):
islands.add(a[i])
for i in range(m):
if(a[i] == n and b[i] in islands):
return "POSSIBLE"
if(b[i] == n and a[i] in islands):
... |
s553863673 | p03131 | u214866184 | 2,000 | 1,048,576 | Wrong Answer | 2,104 | 3,060 | 260 | Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the ... | k, a, b = map(int, input().split())
if b-a < 2 or a > k - 2:
print(k)
else:
num = a
c = k - a
if c % 2 != 0:
c -= 1
num += 1
while c > 0:
c -= 2
num += b - a
print(num) | s460169301 | Accepted | 18 | 3,060 | 222 | k, a, b = map(int, input().split())
if b - a < 2 or a > k - 1:
print(k + 1)
else:
num = a
c = k - a + 1
if c % 2 != 0:
c -= 1
num += 1
num += c//2*(b-a)
print(num) |
s687149946 | p03719 | u518556834 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 86 | You are given three integers A, B and C. Determine whether C is not less than A and not greater than B. | a,b,c = map(int,input().split())
if a <= b <= c:
print("Yes")
else:
print("No")
| s213148589 | Accepted | 17 | 2,940 | 80 | a,b,c = map(int,input().split())
if a<=c<=b:
print("Yes")
else:
print("No")
|
s358947764 | p02399 | u547492399 | 1,000 | 131,072 | Wrong Answer | 30 | 7,640 | 67 | Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number) | a, b = input().split()
a = int(a)
b = int(b)
print(a//b, a%b, a/b) | s271417202 | Accepted | 20 | 7,672 | 86 | a, b = input().split()
a = int(a)
b = int(b)
print(a//b, a%b, "{0:.5f}".format(a/b)) |
s734505881 | p03407 | u598684283 | 2,000 | 262,144 | Wrong Answer | 29 | 9,048 | 86 | An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan. | a,b,c = map(int,input().split())
if a + b >= c:
print("YES")
else:
print("No") | s793952300 | Accepted | 27 | 9,068 | 86 | a,b,c = map(int,input().split())
if a + b >= c:
print("Yes")
else:
print("No") |
s796073285 | p03067 | u161442663 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 115 | There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise. | A,B,C=map(int,input().split())
if A<B and B<C:
print("Yes")
elif C<B and B<A:
print("Yes")
else:
print("No")
| s604856314 | Accepted | 17 | 2,940 | 115 | A,C,B=map(int,input().split())
if A<B and B<C:
print("Yes")
elif C<B and B<A:
print("Yes")
else:
print("No")
|
s011514196 | p00035 | u299798926 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 628 | 平面上の異なる 4 点 $A (x_a, y_a)$, $B (x_b, y_b)$, $C (x_c, y_c)$, $D(x_d, y_d)$ の座標を読み込んで、それら 4 点を頂点とした四角形 $ABCD$ に凹みがなければ YES、凹みがあれば NO と出力するプログラムを作成してください。 凹みのある四角形とは図 1 のような四角形です。 | def judge(p1,p2,p3,p4):
t1 = (p1[0] - p3[0]) * (p2[1] - p1[1]) + (p1[1] - p3[1]) * (p1[0] - p2[0])
t2 = (p1[0] - p3[0]) * (p4[1] - p1[1]) + (p1[1] - p3[1]) * (p1[0] - p4[0])
t3 = (p2[0] - p4[0]) * (p1[1] - p2[1]) + (p2[1] - p4[1]) * (p2[0] - p1[0])
t4 = (p2[0] - p4[0]) * (p3[1] - p2[1]) + (p2[1] - p4[1]... | s820064668 | Accepted | 30 | 5,588 | 589 | def judge(p1,p2,p3,p4):
t1 = (p3[0] - p4[0]) * (p1[1] - p3[1]) + (p3[1] - p4[1]) * (p3[0] - p1[0])
t2 = (p3[0] - p4[0]) * (p2[1] - p3[1]) + (p3[1] - p4[1]) * (p3[0] - p2[0])
t3 = (p1[0] - p2[0]) * (p3[1] - p1[1]) + (p1[1] - p2[1]) * (p1[0] - p3[0])
t4 = (p1[0] - p2[0]) * (p4[1] - p1[1]) + (p1[1] - p2[1]... |
s382834850 | p03448 | u374765578 | 2,000 | 262,144 | Wrong Answer | 56 | 2,940 | 186 | 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 co... | a,b,c,x = open(0)
d=0
x = int(x)
for i in range(int(a)+1):
for j in range(int(b)+1):
for k in range(int(c)+1):
500*i + 100*j + 50*k == x
d += 1
print(d)
| s999749279 | Accepted | 51 | 2,940 | 192 | a,b,c,x = open(0)
d=0
x = int(x)
for i in range(int(a)+1):
for j in range(int(b)+1):
for k in range(int(c)+1):
if 500*i + 100*j + 50*k == x:
d += 1
print(d) |
s066983918 | p03455 | u064827461 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 114 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a, b = map(str, input().split())
n = int(a+b)
m = n**0.5
if int(m)-m == 0:
print('Yes')
else:
print('No') | s537276136 | Accepted | 17 | 2,940 | 87 | a, b = map(int, input().split())
if(a*b%2 == 0 ):
print('Even')
else:
print('Odd')
|
s107706359 | p02694 | u293523199 | 2,000 | 1,048,576 | Wrong Answer | 21 | 9,168 | 135 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or abo... | import math
X = int(input())
money = 100
year = 0
while(money <= X):
money += math.floor(money * 0.01)
year += 1
print(year) | s035309453 | Accepted | 22 | 9,164 | 134 | import math
X = int(input())
money = 100
year = 0
while(money < X):
money += math.floor(money * 0.01)
year += 1
print(year) |
s011293786 | p03486 | u553919982 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 304 | 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 = list(input())
N = len(S)
T = list(input())
M = len(T)
S.sort()
T.sort()
mmm = min(N, M)
for i in range(mmm):
if S[i] > T[i]:
print("No")
exit()
elif S[i] < T[i]:
print("Yes")
exit()
else:
continue
if N < M:
print("Yes")
else:
print("No")
| s745876744 | Accepted | 17 | 3,064 | 318 | S = list(input())
N = len(S)
T = list(input())
M = len(T)
S.sort()
T.sort(reverse = True)
mmm = min(N, M)
for i in range(mmm):
if S[i] > T[i]:
print("No")
exit()
elif S[i] < T[i]:
print("Yes")
exit()
else:
continue
if N < M:
print("Yes")
else:
print("No")
|
s357562793 | p03470 | u225482186 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 233 | 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 h... | n = int(input())
mochi = []
for i in range(n):
mochi.append(int(input()))
print (mochi)
mochi_use = sorted(mochi,reverse = True)
ans = 1
for i in range(n-1):
if mochi_use[i] > mochi_use[i+1]:
ans += 1
print(ans)
| s046515255 | Accepted | 18 | 3,060 | 218 | n = int(input())
mochi = []
for i in range(n):
mochi.append(int(input()))
mochi_use = sorted(mochi,reverse = True)
ans = 1
for i in range(n-1):
if mochi_use[i] > mochi_use[i+1]:
ans += 1
print(ans)
|
s686046560 | p03433 | u160659351 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 93 | E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins. | #88
N = int(input())
A = int(input())
if N%500 >= A:
print("Yes")
else:
print("No") | s268050713 | Accepted | 17 | 3,064 | 93 | #88
N = int(input())
A = int(input())
if N%500 <= A:
print("Yes")
else:
print("No") |
s082387053 | p03624 | u580093517 | 2,000 | 262,144 | Wrong Answer | 19 | 3,188 | 55 | You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead. | print(set(map(chr,range(97,123)))-set(input())or"None") | s308223993 | Accepted | 19 | 3,188 | 117 | letters="abcdefghijklmnopqrstuvwxyz"
ans=sorted(set(letters)^set(input()))
print("None" if len(ans) == 0 else ans[0]) |
s527514707 | p03720 | u597436499 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 300 | There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city? | n,m = map(int, input().split())
ab = []
for _ in range(m):
ab.append(list(map(int, input().split())))
ans = {}
for i in range(1, m+1):
ans.update({i : 0})
for i in range(m):
for j in range(2):
if ab[i][j] in ans:
ans[ab[i][j]] += 1
for v in ans.values():
print(v)
| s819716974 | Accepted | 17 | 3,060 | 300 | n,m = map(int, input().split())
ab = []
for _ in range(m):
ab.append(list(map(int, input().split())))
ans = {}
for i in range(1, n+1):
ans.update({i : 0})
for i in range(m):
for j in range(2):
if ab[i][j] in ans:
ans[ab[i][j]] += 1
for v in ans.values():
print(v)
|
s737404289 | p02612 | u736443076 | 2,000 | 1,048,576 | Wrong Answer | 33 | 9,140 | 43 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | n = int(input())
ans = n % 1000
print(ans)
| s148051370 | Accepted | 31 | 9,148 | 109 | n = int(input())
if n % 1000 == 0:
print(0)
elif n % 1000 != 0:
ans = 1000 - n % 1000
print(ans)
|
s698046073 | p03385 | u118211443 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 55 | You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`. | s=input()
print("YES" if len(s)==len(set(s)) else "NO") | s638829670 | Accepted | 17 | 2,940 | 55 | s=input()
print("Yes" if len(s)==len(set(s)) else "No") |
s117790207 | p00003 | u776559258 | 1,000 | 131,072 | Wrong Answer | 40 | 7,700 | 129 | Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so. | n=int(input())
for i in range(n):
a,b,c=[int(j) for j in input().split()]
if a==b and b==c:
print("YES")
else:
print("NO") | s348279267 | Accepted | 50 | 7,524 | 146 | n=int(input())
for i in range(n):
x=[int(j) for j in input().split()]
x.sort()
if x[2]**2==x[0]**2+x[1]**2:
print("YES")
else:
print("NO") |
s748326947 | p03944 | u711741418 | 2,000 | 262,144 | Wrong Answer | 24 | 3,064 | 611 | There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) po... | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
W, H, N = list(map(int, input().split()))
minX = 0
maxX = W
minY = 0
maxY = H
def calc():
print(maxX)
print(minX)
print(maxY)
print(minY)
if maxX < minX or maxY < minY:
return 0
return (maxX - minX) * (maxY - minY)
for i in range(N):
... | s050455596 | Accepted | 23 | 3,064 | 547 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
W, H, N = list(map(int, input().split()))
minX = 0
maxX = W
minY = 0
maxY = H
def calc():
if maxX < minX or maxY < minY:
return 0
return (maxX - minX) * (maxY - minY)
for i in range(N):
x, y, a = list(map(int, input().split()))
if a is 1:
... |
s152051238 | p03369 | u365375535 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 51 | 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 ... | S = input()
price = 700 + S.count("o")
print(price) | s298719225 | Accepted | 17 | 3,064 | 55 | S = input()
price = 700 + S.count("o")*100
print(price) |
s349858376 | p02742 | u150985282 | 2,000 | 1,048,576 | Wrong Answer | 19 | 2,940 | 120 | 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 to... | [H, W] = list(map(int, input().split()))
area = H*W
if area%2 == 0:
print(area/2)
if area%2 == 1:
print(area//2 + 1) | s718437440 | Accepted | 17 | 2,940 | 172 | [H, W] = list(map(int, input().split()))
area = H*W
if H == 1 or W == 1:
print(1)
else:
if area%2 == 0:
print(int(area/2))
if area%2 == 1:
print(area//2 + 1)
|
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