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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s917410590 | p03129 | u343523393 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 141 | Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1. | N, K = map(int, input().split())
count = 1
for i in range(1,N,2):
count+=1
if(count >= K):
print("Yes")
else:
print("No")
| s619430907 | Accepted | 17 | 2,940 | 209 | N, K = map(int, input().split())
if(N%2==1):
if(N//2 + 1 >= K):
print("YES")
else:
print("NO")
else:
if(N//2 >= K):
print("YES")
else:
print("NO")
# print(N//2) |
s381095769 | p03478 | u856169020 | 2,000 | 262,144 | Wrong Answer | 34 | 3,060 | 223 | 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). | def num_sum(L):
ret = 0
for l in L:
ret += int(l)
return ret
N, A, B = map(int, input().split())
ans = 0
for i in range(1, N+1):
num = num_sum(list(str(i)))
if A <= num and num <= B:
ans += num
print(ans) | s129711388 | Accepted | 33 | 3,064 | 221 | def num_sum(L):
ret = 0
for l in L:
ret += int(l)
return ret
N, A, B = map(int, input().split())
ans = 0
for i in range(1, N+1):
num = num_sum(list(str(i)))
if A <= num and num <= B:
ans += i
print(ans) |
s377934291 | p03485 | u617659131 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 55 | You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer. | a,b = map(int, input().split())
print(int(((a+b)/2)+1)) | s709664761 | Accepted | 17 | 2,940 | 55 | a,b = map(int, input().split())
print(int(((a+b+1)/2))) |
s909539046 | p02406 | u340500592 | 1,000 | 131,072 | Wrong Answer | 30 | 7,560 | 128 | In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement ... | n = int(input())
for i in range(1, n+1):
x = i
if x % 3 == 0:
print(" ")
while x:
if x % 10 == 3:
print(" ")
x /= 10 | s258055856 | Accepted | 20 | 7,676 | 192 | n = int(input())
cout = ""
for i in range(1, n+1):
x = i
if x % 3 == 0:
cout += " " + str(i)
else:
while x:
if x % 10 == 3:
cout += " " + str(i)
break
x //= 10
print(cout) |
s678222340 | p00002 | u560214129 | 1,000 | 131,072 | Wrong Answer | 20 | 7,544 | 118 | Write a program which computes the digit number of sum of two integers a and b. | a, b = map(int, input().split())
c=int(a+b)
count=0
while (c/10 != 0 ):
c=int(c/10)
count=count+1
print(count) | s104826804 | Accepted | 30 | 7,664 | 188 | import sys
for line in sys.stdin.readlines():
a, b = map(int,line.split())
c=int(a+b)
count=0
while (c/10 != 0 ):
c=int(c/10)
count=count+1
print(count) |
s360838436 | p02646 | u999750647 | 2,000 | 1,048,576 | Wrong Answer | 23 | 9,096 | 236 | 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 ... | import sys
a = list(map(int,input().split()))
b = list(map(int,input().split()))
t = int(input())
if a[1] <= b[1]:
print('NO')
sys.exit()
elif abs(a[0]-b[0])//a[1]-b[1] <= t:
print('Yes')
sys.exit()
else:
print('NO') | s003191402 | Accepted | 21 | 9,116 | 240 | import sys
a = list(map(int,input().split()))
b = list(map(int,input().split()))
t = int(input())
if a[1] <= b[1]:
print('NO')
sys.exit()
elif abs(a[0]-b[0])/abs(a[1]-b[1]) <= t:
print('YES')
sys.exit()
else:
print('NO') |
s874772592 | p04044 | u989345508 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 91 | Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically sm... | n,l=input().split()
n,l=int(n),int(l)
s=sorted([input() for i in range(n)])
print(list(s))
| s273941778 | Accepted | 17 | 3,060 | 86 | n,l=map(int,input().split())
s=[input() for i in range(n)]
s.sort()
print("".join(s))
|
s053620920 | p03861 | u604412462 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 76 | 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? | abx = list(map(int, input().split()))
print(abx[1]//abx[2] - abx[0]//abx[2]) | s772417182 | Accepted | 18 | 3,060 | 154 | abx = list(map(int, input().split()))
if abx[0]%abx[2]==0:
print(abx[1]//abx[2] - abx[0]//abx[2] + 1)
else:
print(abx[1]//abx[2] - abx[0]//abx[2]) |
s594665494 | p03457 | u643301782 | 2,000 | 262,144 | Wrong Answer | 351 | 3,064 | 306 | 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... | n = int(input())
t1 = 0
x1 = 0
y1 = 0
count = 0
for _ in range(n):
t,x,y = map(int,input().split())
t = t - t1
x = x - x1
y = y - y1
if x + y <= t and x+y % 2 == t % 2:
t1 = t
x1 = x
y1 = y
count += 1
if count == n:
print("YES")
else:
print("NO") | s611045063 | Accepted | 388 | 3,060 | 332 | n = int(input())
t1 = 0
x1 = 0
y1 = 0
count = 0
for _ in range(n):
t,x,y = map(int,input().split())
t = abs(t - t1)
x = abs(x - x1)
y = abs(y - y1)
if x + y <= t and (x+y) % 2 == t % 2:
t1 = t
x1 = x
y1 = y
count += 1
if count == n:
print("Yes")
else:
print("N... |
s205652251 | p03671 | u006425112 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 104 | 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 m... | import sys
a, b, c = map(int, sys.stdin.readline().split())
m = max([a,b,c])
print(a + b + c - 2 * m) | s262291740 | Accepted | 17 | 2,940 | 100 | import sys
a, b, c = map(int, sys.stdin.readline().split())
m = max([a,b,c])
print(a + b + c - m) |
s015614518 | p03473 | u828277092 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 31 | How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December? | m = int(input())
print(24 + m) | s483063083 | Accepted | 17 | 2,940 | 38 | m = int(input())
print(24 + (24 - m)) |
s438317623 | p03007 | u310233294 | 2,000 | 1,048,576 | Wrong Answer | 1,732 | 19,972 | 170 | There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the... | N = int(input())
A = list(map(int, input().split()))
A.sort()
xy = []
while(len(A)>1):
x = A.pop(-1)
y = A.pop(0)
xy.append([x, y])
A.append(x-y)
print(A) | s341550140 | Accepted | 1,798 | 21,760 | 836 | N = int(input())
A = list(map(int, input().split()))
A.sort(reverse=True)
AM = A.pop(0)
xy = []
if AM > 0 and A[-1] >= 0:
p = 1
elif AM <= 0 and A[-1] < 0:
p = 2
else:
p = 3
if p == 1:
while(len(A)>1):
x = A.pop(-1)
y = A.pop(0)
xy.append([x, y])
A.append(x-y)
xy.appe... |
s716154188 | p03434 | u209620426 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 167 | 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 c... | n = int(input())
l = list(map(int, input().split()))
a = 0
b = 0
l.sort()
for i in range(n):
if i%2 == 0:
a += l[i]
else:
b += l[i]
print(a-b) | s593782719 | Accepted | 18 | 3,060 | 186 | n = int(input())
l = list(map(int, input().split()))
a = 0
b = 0
l = sorted(l, reverse=True)
for i in range(n):
if i%2 == 0:
a += l[i]
else:
b += l[i]
print(a-b) |
s513849561 | p03457 | u545644875 | 2,000 | 262,144 | Wrong Answer | 395 | 32,172 | 444 | 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... | N = int(input())
A = [input().split() for _ in range(N)]
if N ==1:
n = int(A[0][0])
move = int(A[0][1]) + int(A[0][2])
if move > n or move%2 != n%2:
print("NO")
else:
print("YES")
for i in range(1,N):
H = int(A[i][1]) - int(A[i -1][1])
V = int(A[i][2]) - int(A[i -1][2])
move = H + V
n = int... | s078675974 | Accepted | 413 | 32,148 | 454 | N = int(input())
A = [input().split() for _ in range(N)]
if N ==1:
n = int(A[0][0])
move = int(A[0][1]) + int(A[0][2])
if move > n or move%2 != n%2:
print("No")
else:
print("Yes")
for i in range(1,N):
H = abs(int(A[i][1]) - int(A[i -1][1]))
V = abs(int(A[i][2]) - int(A[i -1][2]))
move = H + V... |
s473328256 | p03556 | u816631826 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 56 | 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
n = int(input())
print(math.floor(n**0.5)) | s167972693 | Accepted | 17 | 2,940 | 64 | import math
n=int(input())
m=math.floor(math.sqrt(n))
print(m*m) |
s379064355 | p02415 | u343251190 | 1,000 | 131,072 | Wrong Answer | 30 | 7,448 | 69 | Write a program which converts uppercase/lowercase letters to lowercase/uppercase for a given string. | words = input().split()
for word in words:
print(word.swapcase()) | s760542451 | Accepted | 40 | 7,520 | 39 | words = input()
print(words.swapcase()) |
s720586667 | p03599 | u243699903 | 3,000 | 262,144 | Wrong Answer | 17 | 2,940 | 60 | Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the bea... | a, b, c, d, e, f = map(int, input().split())
print(100,100) | s593501991 | Accepted | 36 | 3,064 | 1,308 | a, b, c, d, e, f = map(int, input().split())
ans = (0, 0, 0)
for i in range(0, f // (100 * a) + 1):
for j in range(0, f // (100 * b) + 1):
if i == 0 and j == 0:
continue
waterV = 100 * a * i + 100 * b * j
if waterV > f:
break
leftVolume... |
s640627773 | p03693 | u278868910 | 2,000 | 262,144 | Wrong Answer | 22 | 3,572 | 122 | 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... | from functools import reduce
num = int(reduce(lambda a,b:a+b, input().split(), ""))
print('Yes' if num%4 == 0 else 'No')
| s410380260 | Accepted | 22 | 3,572 | 122 | from functools import reduce
num = int(reduce(lambda a,b:a+b, input().split(), ""))
print('YES' if num%4 == 0 else 'NO')
|
s447589456 | p03565 | u518064858 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 428 | E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One mo... | s=input()
t=input()
l=len(t)
a=[]
for i in range(l-1,len(s)):
x=s[len(s)-1-i:len(s)-1-i+l]
cnt=0
for j in range(l):
if x[j]=="?" or x[j]==t[j]:
cnt+=1
else:
break
if cnt==l:
u=s[:len(s)-1-i].replace("?","a")
v=s[len(s)-1-i+l:].replace("?","a")
... | s747537808 | Accepted | 17 | 3,064 | 431 | s=input()
t=input()
l=len(t)
a=[]
for i in range(l-1,len(s)):
x=s[len(s)-1-i:len(s)-1-i+l]
cnt=0
for j in range(l):
if x[j]=="?" or x[j]==t[j]:
cnt+=1
else:
break
if cnt==l:
u=s[:len(s)-1-i].replace("?","a")
v=s[len(s)-1-i+l:].replace("?","a")
... |
s044755878 | p02602 | u946648678 | 2,000 | 1,048,576 | Wrong Answer | 2,206 | 31,604 | 307 | 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 m... | n, k = [int(x) for x in input().split()]
arr = [int(x) for x in input().split()]
p = 1
for i in range(k):
p *= arr[i]
grades = [p]
for i in range(k, n):
p /= arr[i-k]
p *= arr[k]
grades.append(p)
for i in range(1, len(grades)):
if grades[i] > grades[i-1]:
print("Yes")
else:
print("No")
| s812764061 | Accepted | 161 | 31,616 | 179 | n, k = [int(x) for x in input().split()]
arr = [int(x) for x in input().split()]
for i in range(k, n):
if arr[i] > arr[i-k]:
print("Yes")
else:
print("No") |
s220543460 | p03493 | u957799665 | 2,000 | 262,144 | Wrong Answer | 27 | 9,076 | 98 | Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble. | n_1 = 0
a = list(str(input()))
print(a)
for i in a:
if i == "1":
n_1 += 1
print(n_1) | s261116788 | Accepted | 31 | 8,996 | 89 | n_1 = 0
a = list(str(input()))
for i in a:
if i == "1":
n_1 += 1
print(n_1) |
s132308071 | p03007 | u327466606 | 2,000 | 1,048,576 | Wrong Answer | 262 | 14,484 | 1,396 | There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the... | import bisect
N = int(input())
A = sorted(map(int,input().split()))
i = bisect.bisect_left(A, 0)
j = bisect.bisect_right(A, 0)
if i == j:
if i == 0:
negatives = A[:1]
positives = A[1:]
elif i == N:
negatives = A[:-1]
positives = A[-1:]
else:
negatives = A[:i]
... | s215014572 | Accepted | 132 | 32,916 | 857 | from bisect import bisect
def solve():
N = int(input())
A = sorted(map(int,input().split()))
if N == 2:
res = [(max(A),min(A))]
s = max(A)-min(A)
return s, res
if A[0] >= 0:
res = []
x = A[0]
for a in A[1:-1]:
res.append((x,a))
x ... |
s279527555 | p02613 | u046466256 | 2,000 | 1,048,576 | Wrong Answer | 148 | 9,104 | 315 | 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`,... | AC = 0
WA = 0
TLE = 0
RE = 0
N = int(input())
for i in range(N):
S = input()
if(S == "AC"):
AC+=1
elif(S == "WA"):
WA+=1
elif(S == "TLE"):
TLE+=1
elif(S == "RE"):
RE+=1
print("AC × " + str(AC) + "\nWA ×" + str(WA) + "\nTLE ×" + str(TLE) + "\nRE ×" + str(RE)) | s268044459 | Accepted | 137 | 16,188 | 130 | N = int(input())
s = [input() for i in range(N)]
for v in ['AC', 'WA', 'TLE', 'RE']:
print('{0} x {1}'.format(v, s.count(v)))
|
s552421071 | p02659 | u566574814 | 2,000 | 1,048,576 | Wrong Answer | 27 | 10,068 | 78 | Compute A \times B, truncate its fractional part, and print the result as an integer. | from decimal import Decimal
a,b=map(Decimal,input().split())
print(round(a*b)) | s802185654 | Accepted | 26 | 10,068 | 95 | from decimal import Decimal
import math
a,b=map(Decimal,input().split())
print(math.floor(a*b)) |
s101468903 | p03471 | u310956674 | 2,000 | 262,144 | Wrong Answer | 1,991 | 21,244 | 236 | 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 situat... | N, Y = map(int, input().split())
for x in range(N+1):
for y in range(N+1-x):
z = N - x - y
if 10000*x + 5000*y + 1000*z == Y:
print(x, y, z)
exit()
else:
print(-1, -1, -1) | s613165151 | Accepted | 603 | 9,056 | 275 | N, Y = map(int, input().split())
ansX = -1
ansY = -1
ansZ = -1
for x in range(N+1):
for y in range(N+1-x):
z = N - x - y
if 10000*x + 5000*y + 1000*z == Y:
ansX = x
ansY = y
ansZ = z
print(ansX, ansY, ansZ) |
s461907545 | p03697 | u583326945 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 239 | You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead. | if __name__ == "__main__":
S = input()
result = "yes"
# for each character in S
for c in S:
# there is a same character in S
if S.count(c) > 1:
result = "no"
break
print(result) | s489659262 | Accepted | 17 | 2,940 | 138 | if __name__ == "__main__":
A, B = map(int, input().split())
if A + B >= 10:
print("error")
else:
print(A + B) |
s725027028 | p03150 | u942697937 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,060 | 182 | A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string. | S = input()
V = "keyence"
for i in range(len(S) - len(V) + 1):
print(i, S[:i], S[i-len(V):])
if S[:i] + S[i-len(V):] == V:
print("YES")
exit(0)
print("NO")
| s344976994 | Accepted | 17 | 2,940 | 158 | S = input()
V = "keyence"
t = len(S) - len(V)
for i in range(len(S) - t + 1):
if S[:i] + S[i+t:] == V:
print("YES")
exit(0)
print("NO")
|
s803792446 | p02397 | u587193722 | 1,000 | 131,072 | Wrong Answer | 20 | 7,620 | 110 | Write a program which reads two integers x and y, and prints them in ascending order. | x ,y = [int(i) for i in input().split()]
if x <= y:
print(x,' ',y,sep='')
else:
print(y,' ',x,sep=' ') | s494155042 | Accepted | 60 | 7,660 | 186 | while True:
x, y = [int(i) for i in input().split()]
if x > y:
print(y ,x)
elif x ==0 and y == 0:
break
else:
print(x,y)
|
s266119066 | p03486 | u547608423 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 491 | 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. | base=["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
s=input()
t=input()
S=[]
T=[]
answer="No"
for i in s:
S.append(base.index(i))
for j in t:
T.append(base.index(j))
S=sorted(S)
T=sorted(T,reverse=True)
for k in range(min(len(S),len(T))):
if S[... | s268710545 | Accepted | 17 | 3,064 | 559 | base=["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
s=input()
t=input()
S=[]
T=[]
answer="NO"
for i in s:
S.append(base.index(i))
for j in t:
T.append(base.index(j))
S=sorted(S)
T=sorted(T,reverse=True)
#print(S)
#print(T)
for k in range(min(len(S)... |
s281509352 | p03555 | u138781768 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 129 | 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. | s_a = input()
s_b = input()
if s_a[0] == s_b[2] and s_a[2] == s_a[0] and s_a[1] == s_b[1]:
print("YES")
else:
print("NO") | s915833497 | Accepted | 17 | 2,940 | 129 | s_a = input()
s_b = input()
if s_a[0] == s_b[2] and s_a[2] == s_b[0] and s_a[1] == s_b[1]:
print("YES")
else:
print("NO") |
s359147613 | p02600 | u031722966 | 2,000 | 1,048,576 | Wrong Answer | 30 | 9,164 | 124 | M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * Fr... | X = int(input())
A = 2000 - X
if A % 200 == 0:
print(str(A // 200) + "級")
else:
print(str(A // 200 + 1) + "級") | s770124343 | Accepted | 30 | 9,152 | 108 | X = int(input())
A = 2000 - X
if A % 200 == 0:
print(str(A // 200))
else:
print(str(A // 200 + 1)) |
s474012816 | p02612 | u162565714 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,148 | 63 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | n=int(input())
if n==0:
print('0')
else:
print(1000-n) | s933593685 | Accepted | 28 | 9,076 | 71 | n=int(input())
n%=1000
if n==0:
print('0')
else:
print(1000-n) |
s367239962 | p02846 | u619819312 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,064 | 347 | Takahashi and Aoki are training for long-distance races in an infinitely long straight course running from west to east. They start simultaneously at the same point and moves as follows **towards the east** : * Takahashi runs A_1 meters per minute for the first T_1 minutes, then runs at A_2 meters per minute for th... | a,b=map(int,input().split())
c,d=map(int,input().split())
e,f=map(int,input().split())
if (c>e and d>f) or (c<e and d<f):
print(0)
else:
if c<e:
c,d,e,f=e,f,c,d
s=(c-e)*a
t=(f-d)*b
if s>t:
print(0)
elif s==t:
print("infinity")
else:
print(int((s+t)/(t-s))-1-(1... | s211334731 | Accepted | 17 | 3,064 | 350 | a,b=map(int,input().split())
c,d=map(int,input().split())
e,f=map(int,input().split())
if (c>e and d>f) or (c<e and d<f):
print(0)
else:
if c<e:
c,d,e,f=e,f,c,d
s=(c-e)*a
t=(f-d)*b
if s>t:
print(0)
elif s==t:
print("infinity")
else:
print(int(s/(t-s))+int(t/(t... |
s497144130 | p02613 | u775278230 | 2,000 | 1,048,576 | Wrong Answer | 151 | 9,116 | 293 | 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`,... |
n = int(input())
j = [0] * 4
for i in range(n):
s=input()
if s == "AC":
j[0] += 1
elif s=="WA":
j[1] += 1
elif s=="TLE":
j[2] += 1
elif s == "RE":
j[3] += 1
for k in range(4):
print(f"AC × {j[k]}")
| s333288077 | Accepted | 148 | 9,124 | 332 |
n = int(input())
j = [0] * 4
moji = ["AC", "WA", "TLE", "RE"]
for i in range(n):
s=input()
if s == "AC":
j[0] += 1
elif s=="WA":
j[1] += 1
elif s=="TLE":
j[2] += 1
elif s == "RE":
j[3] += 1
for k in range(4):
print(f"{moji[k]} x {j[k]}")
|
s802152301 | p03455 | u432853936 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 97 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a,b=map(int, input().split())
if a*b%2 == 0:
print("even")
else:
print("odd")
| s101656470 | Accepted | 17 | 2,940 | 97 | a,b=map(int, input().split())
if a*b%2 == 0:
print("Even")
else:
print("Odd")
|
s972291933 | p03543 | u998835868 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 170 | We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**? | a = map(str,input())
a = list(a)
c = False
for x in range(2):
if a[x] == a[x+1]:
if a[x+1] == a[x+2]:
print("YES")
c=True
if c == False:
print("NO") | s306128756 | Accepted | 16 | 2,940 | 143 | a = map(int,input())
a = list(a)
for x in range(2):
if a[x] == a[x+1]:
if a[x+1] == a[x+2]:
print("Yes")
exit()
print("No") |
s316534519 | p03455 | u375870553 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 168 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | def main():
a, b = [int(i) for i in input().split()]
if a*b%2 == 0:
print("Odd")
else:
print("Even")
if __name__ == '__main__':
main() | s000890465 | Accepted | 17 | 2,940 | 168 | def main():
a, b = [int(i) for i in input().split()]
if a*b%2 == 0:
print("Even")
else:
print("Odd")
if __name__ == '__main__':
main() |
s534017024 | p03303 | u014139588 | 2,000 | 1,048,576 | Wrong Answer | 30 | 9,164 | 105 | You are given a string S consisting of lowercase English letters. We will write down this string, starting a new line after every w letters. Print the string obtained by concatenating the letters at the beginnings of these lines from top to bottom. | s = input()
w = int(input())
ans = []
for i in range(len(s)//w):
ans.append(s[w*i])
print("".join(ans)) | s019750772 | Accepted | 30 | 9,124 | 187 | s = input()
w = int(input())
ans = []
if len(s)%w == 0:
for i in range(len(s)//w):
ans.append(s[w*i])
else:
for i in range(len(s)//w+1):
ans.append(s[w*i])
print("".join(ans)) |
s085442845 | p02255 | u405478089 | 1,000 | 131,072 | Wrong Answer | 20 | 5,600 | 310 | 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 ... | def insertionSort(A,N):
for i in range(1,len(A)):
j = i
while (j > 0) and (A[j-1] > A[j]) :
tmp = A[j-1]
A[j-1] = A[j]
A[j] = tmp
j -= 1
print(' '.join(map(str, A)))
N = input()
N = int(N)
A = input()
A = list(map(int, A.split()))
insertionSort(A,N)
| s947593802 | Accepted | 20 | 5,600 | 306 | def insertionSort(A,N):
for i in range(N):
j = i
while (j > 0) and (A[j-1] > A[j]) :
tmp = A[j-1]
A[j-1] = A[j]
A[j] = tmp
j -= 1
print(' '.join(map(str, A)))
N = input()
N = int(N)
A = input()
A = list(map(int, A.split()))
insertionSort(A,N)
|
s371257976 | p03339 | u543954314 | 2,000 | 1,048,576 | Wrong Answer | 55 | 3,700 | 96 | There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of the... | n = int(input())
s = input()
d = {"E":0, "W":0}
for i in s:
d[i] += 1
print(min(d.values())) | s532845886 | Accepted | 126 | 3,676 | 132 | n = int(input())
s = input()
c = s.count("E")
d = c
for i in s:
if i == "E":
d -= 1
else:
d += 1
c = min(c,d)
print(c) |
s408013806 | p03408 | u702208001 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 250 | 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... | n = int(input())
blue = [input() for _ in range(n)]
n = int(input())
red = [input() for _ in range(n)]
blue_set = list(set(blue))
count = 0
for i in blue_set:
if blue.count(i) >= red.count(i):
count += blue.count(i) - red.count(i)
print(count)
| s010483167 | Accepted | 17 | 3,060 | 166 | n=[input() for _ in range(int(input()))]
m=[input() for _ in range(int(input()))]
l=list(set(n))
print(max(0,max(n.count(l[i])-m.count(l[i]) for i in range(len(l))))) |
s207358364 | p02842 | u234189749 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,064 | 273 | 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())
MIN = int(N/1.08)
MAX = int((N+1)/1.08)
if (MIN == MAX) and (int(MIN*1.08) == N):
print(MIN)
elif MIN == MAX:
print(":(")
for i in range(MIN,MAX+1):
if int(i* 1.08) == N:
print(i)
break
elif i == MAX:
print(":(") | s572058263 | Accepted | 17 | 3,064 | 139 | N = int(input())
if N%1.08 ==0:
print(int(N//1.08))
elif int((N//1.08+1)*1.08) == N:
print(int(N//1.08) +1)
else:
print(":(") |
s678677642 | p03597 | u256464928 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 47 | We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black? | N = int(input())
A = int(input())
print(N**N-A) | s077743202 | Accepted | 17 | 2,940 | 49 | N = int(input())
A = int(input())
print((N**2)-A) |
s419243963 | p03992 | u266874640 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 108 | This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the fi... | s = input()
t = 0
for i in s:
print(i, end = '')
t += 1
if t == 4:
print(' ', end = '')
| s677976435 | Accepted | 17 | 2,940 | 118 | s = input()
t = 0
for i in s:
print(i, end = '')
t += 1
if t == 4:
print(' ', end = '')
print('')
|
s849102212 | p03563 | u108784591 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 109 | 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 cont... | N=int(input())
K=int(input())
ans=1
c=0
while c<N:
if ans<K:
ans=ans*2
else:
ans=ans+K
c+=1
print(ans) | s490260976 | Accepted | 17 | 2,940 | 46 | R=int(input())
G=int(input())
a=2*G-R
print(a) |
s483583296 | p03853 | u488934106 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 137 | There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, p... | H , W = map(int, input().split())
Clist = [input().split() for i in range(H)]
for a in range(H):
print(Clist[a])
print(Clist[a]) | s560926961 | Accepted | 18 | 3,060 | 140 | H , W = map(int, input().split())
Clist = [input().split() for i in range(H)]
for a in range(H):
print(*Clist[a])
print(*Clist[a])
|
s182530771 | p02659 | u023229441 | 2,000 | 1,048,576 | Wrong Answer | 26 | 8,996 | 64 | Compute A \times B, truncate its fractional part, and print the result as an integer. | a,b=map(float,input().split())
import math
print(math.ceil(a*b)) | s436402459 | Accepted | 33 | 9,876 | 83 | from decimal import Decimal as D
a,b=map(str,input().split())
print(int(D(a)*D(b))) |
s315566947 | p02411 | u767368698 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 466 | Write a program which reads a list of student test scores and evaluates the performance for each student. The test scores for a student include scores of the midterm examination m (out of 50), the final examination f (out of 50) and the makeup examination r (out of 100). If the student does not take the examination, t... | def evl_score(m,f,r):
if m + f >= 80:
ret = 'A'
elif m + f >= 65 and m + f < 80:
ret = 'B'
elif m + f >= 50 and m + f < 65:
ret = 'C'
elif m + f >= 30 and m + f < 50:
if r >= 50:
ret = 'C'
else:
ret = 'D'
else:
ret = 'F'
re... | s541817434 | Accepted | 20 | 5,600 | 532 | def evl_score(m,f,r):
if m == -1 or f == -1:
ret = 'F'
elif m + f >= 80:
ret = 'A'
elif m + f >= 65 and m + f < 80:
ret = 'B'
elif m + f >= 50 and m + f < 65:
ret = 'C'
elif m + f >= 30 and m + f < 50:
if r >= 50:
ret = 'C'
else:
... |
s674072939 | p03623 | u051496905 | 2,000 | 262,144 | Wrong Answer | 25 | 9,036 | 127 | 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... | x , a ,b = map(int,input().split())
X = abs(a - x)
Y = abs(b - x)
Z = max(Y,X)
if Z == X:
print("A")
else:
print("B")
| s471736298 | Accepted | 26 | 9,164 | 126 | x , a ,b = map(int,input().split())
X = abs(a - x)
Y = abs(b - x)
Z = min(Y,X)
if Z == X:
print("A")
else:
print("B") |
s343476581 | p03435 | u297103202 | 2,000 | 262,144 | Wrong Answer | 20 | 3,064 | 363 | We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) ... | c1 = list(map(int, input().split()))
c2 = list(map(int, input().split()))
c3 = list(map(int, input().split()))
s = [c1,c2,c3]
print(s)
s1 = s[0][0] - s[0][1] - s[1][0] + s[1][1]
s2 = s[1][1] - s[1][2] - s[2][1] + s[2][2]
if s1== 0 and s2==0:
if s[0][0] + s[2][2] == s[0][2] + s[2][2]:
print('Yes')
else:... | s997154050 | Accepted | 18 | 3,060 | 253 | s = []
for i in range(3):
s.append(list(map(int,input().split())))
s1 = s[0][0] - s[0][1] - s[1][0] + s[1][1]
s2 = s[1][1] - s[1][2] - s[2][1] + s[2][2]
s3 = s[0][0] - s[0][2] - s[2][0] + s[2][2]
print('Yes' if s1== 0 and s2==0 and s3==0 else 'No')
|
s357498194 | p02927 | u036531287 | 2,000 | 1,048,576 | Wrong Answer | 26 | 3,064 | 337 | Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq... | k = 0
m,d = (int(i) for i in input().split())
listm = range(1,m+1)
listd = range(1,d)
for a in listm:
for b in listd:
dt = str(b)
da = dt[0]
da = int(da)
if len(dt) == 2:
db = dt[1]
db = int(db)
if db >= 2:
if da >= 2:
if a == da * db:
if a >= 2:
k = k + 1
print(a,"月",b,... | s214287457 | Accepted | 33 | 3,060 | 343 | k = 0
m,d = (int(i) for i in input().split())
listm = range(1,m + 1)
listd = range(1,d + 1)
for a in listm:
for b in listd:
if len(str(b)) >= 2:
if int(str(b)[0]) >= 2:
if int(str(b)[1]) >= 2:
#print(a,"y",b,"d")
if a == int(str(b)[0]) * int(str(b)[1]):
#print(a,"y",b,"d")
k ... |
s720221649 | p02747 | u225845681 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 190 | A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string. | S = input()
ans = 'No'
if len(S)%2 == 0:
for i in range(len(S)//2):
if S[i]+S[i+1] != 'hi':
break
elif i == (len(S)//2)-1:
ans = 'Yes'
print(ans) | s061517085 | Accepted | 17 | 2,940 | 197 | S = input()
ans = 'No'
if len(S)%2 == 0:
for i in range(len(S)//2):
if S[2*i] + S[2*i+1] != 'hi':
break
elif i == (len(S)//2)-1:
ans = 'Yes'
print(ans) |
s646619116 | p03095 | u368796742 | 2,000 | 1,048,576 | Wrong Answer | 57 | 9,156 | 168 | You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a conca... | n = int(input())
s = input()
count = [0]*26
for i in s:
count[ord(i)-ord("a")] += 1
mod = 10**9+7
ans = 1
for i in count:
ans *= (i+1)
ans %= mod
print(ans) | s896173283 | Accepted | 52 | 9,288 | 170 | n = int(input())
s = input()
count = [0]*26
for i in s:
count[ord(i)-ord("a")] += 1
mod = 10**9+7
ans = 1
for i in count:
ans *= (i+1)
ans %= mod
print(ans-1) |
s658871052 | p03943 | u383508661 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 104 | Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Not... | a,b,c=map(int,input().split())
Lis=[a,b,c]
sorted(Lis)
if a==b+c:
print("Yes")
else:
print("No") | s146414915 | Accepted | 18 | 2,940 | 89 | a,b,c=sorted(map(int,input().split()))
if a+b==c:
print("Yes")
else:
print("No") |
s738991841 | p02238 | u112247126 | 1,000 | 131,072 | Wrong Answer | 30 | 7,936 | 1,030 | Depth-first search (DFS) follows the strategy to search ”deeper” in the graph whenever possible. In DFS, edges are recursively explored out of the most recently discovered vertex $v$ that still has unexplored edges leaving it. When all of $v$'s edges have been explored, the search ”backtracks” to explore edges leaving ... | from collections import deque
def dfs(root):
time = 1
stack.append(root)
color[root] = 'gray'
arrive[root] = time
time += 1
while len(stack) > 0:
node = stack[-1]
for next in range(n):
if adjMat[node][next] == 1:
if color[next] == 'white':
... | s969423393 | Accepted | 30 | 7,784 | 711 | def dfs_recursive(u):
global time
color[u] = 'gray'
arrive[u] = time
time += 1
for v in range(n):
if adjMat[u][v] == 1 and color[v] == 'white':
dfs_recursive(v)
color[u] = 'black'
finish[u] = time
time += 1
n = int(input())
adjMat = [[0] * n for _ in range(n)]
colo... |
s012926196 | p03438 | u125205981 | 2,000 | 262,144 | Wrong Answer | 25 | 4,600 | 269 | You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N. Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal. Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two ac... | def main():
N = int(input())
A = tuple(map(int, input().split()))
B = tuple(map(int, input().split()))
asum = sum(A)
bsum = sum(B)
if A == B:
print(0)
elif asum > bsum:
print(-1)
else:
print(bsum - asum)
main()
| s351963765 | Accepted | 23 | 4,600 | 334 | def main():
N = int(input())
A = tuple(map(int, input().split()))
B = tuple(map(int, input().split()))
la = 0
lb = 0
for a, b in zip(A, B):
if a > b:
la += a - b
elif a < b:
lb += (b - a) // 2
if la > lb:
print('No')
else:
print('... |
s272934150 | p03457 | u615850856 | 2,000 | 262,144 | Wrong Answer | 405 | 3,064 | 1,429 | 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... | N = int(input())
T = 0
X = Y = 0
Flag = 'Even'
for n in range(N):
Ti,Xi,Yi = [int(x) for x in input().split()]
if abs(Xi-X) + abs(Yi-Y) > (Ti-T):
print('NO')
break
else:
if Flag == 'Even':
if (Ti-T) % 2 == 0:#Even and Even = Even
if (Xi + Yi) % 2 == 0:
... | s985676088 | Accepted | 411 | 3,064 | 1,429 | N = int(input())
T = 0
X = Y = 0
Flag = 'Even'
for n in range(N):
Ti,Xi,Yi = [int(x) for x in input().split()]
if abs(Xi-X) + abs(Yi-Y) > (Ti-T):
print('No')
break
else:
if Flag == 'Even':
if (Ti-T) % 2 == 0:#Even and Even = Even
if (Xi + Yi) % 2 == 0:
... |
s536620996 | p03761 | u859897687 | 2,000 | 262,144 | Wrong Answer | 19 | 3,064 | 380 | Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string. He will receive a headline which contains one of the strings S_1,...,S_n tomorrow. He is excited and already thinking of what string he will create. Since he does not know the string on the headlin... | n=int(input())
d=dict()
for i in input():
if i not in d:
d[i]=1
else:
d[i]+=1
for _ in range(n-1):
dd=dict()
for i in input():
if i not in dd:
dd[i]=1
else:
dd[i]+=1
for i in d.keys():
if i in dd:
d[i]=min(d[i],dd[i])
ans=[]
for i in d.keys():
for _ in range(d[i]):
... | s348003212 | Accepted | 19 | 3,064 | 403 | n=int(input())
d=dict()
for i in input():
if i not in d:
d[i]=1
else:
d[i]+=1
for _ in range(n-1):
dd=dict()
for i in input():
if i not in dd:
dd[i]=1
else:
dd[i]+=1
for i in d.keys():
if i in dd:
d[i]=min(d[i],dd[i])
else:
d[i]=0
ans=[]
for i in d.keys():
for... |
s211568124 | p02646 | u470392120 | 2,000 | 1,048,576 | Wrong Answer | 21 | 9,204 | 136 | 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 ... | #B
a,v=map(int,input().split())
b,w=map(int,input().split())
t=int(input())
if (v-w)*t >= (b-a):
print('Yes')
else:
print('No') | s358376240 | Accepted | 24 | 9,184 | 155 |
a,v=map(int,input().split())
b,w=map(int,input().split())
t=int(input())
if (v-w)*t >= abs(b-a):
ans='YES'
else:
ans='NO'
print(ans) |
s428903791 | p03523 | u247211039 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 289 | You are given a string S. Takahashi can insert the character `A` at any position in this string any number of times. Can he change S into `AKIHABARA`? | S = input()
S1 = "KIHABARA"
S2 = "AKIHBARA"
S3 = "AKIHABRA"
S4 = "AKIHABAR"
if S.find("KIHABARA") != -1:
print("YES")
elif S.find("AKIHBARA") != -1:
print("YES")
elif S.find("AKIHABRA") != -1:
print("YES")
elif S.find("AKIHABAR") != -1:
print("YES")
else:
print("No") | s521980538 | Accepted | 17 | 3,060 | 360 | s = input()
if s == "KIHBR" or s == "KIHABARA" or s == "AKIHBARA" or s == "AKIHABRA" or s == "AKIHABAR" or \
s == "AKIHABARA" or s == "AKIHBR" or s == "KIHABR" or s == "KIHBAR" or s == "KIHBRA" or \
s == "AKIHABR" or s == "AKIHBAR" or s == "AKIHBRA" or s == "KIHABAR" or s == "KIHABRA" or s == "KIHBARA":
... |
s437392211 | p03567 | u029169777 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 109 | Snuke built an online judge to hold a programming contest. When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.) Determine whether the judge can return the ... | S=input()
for i in range(len(S)-1):
if S[i]=='A' and S[i+1]=='C':
print('Yes')
else:
print('No') | s270225457 | Accepted | 17 | 2,940 | 109 | S=input()
for i in range(len(S)-1):
if S[i]=='A' and S[i+1]=='C':
print('Yes')
exit()
print('No') |
s525317183 | p03795 | u551109821 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 49 | 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 ... | n = int(input())
b = int(n/15)
print(n*800-n*200) | s602136951 | Accepted | 18 | 2,940 | 49 | n = int(input())
b = int(n/15)
print(n*800-b*200) |
s065295578 | p02665 | u785578220 | 2,000 | 1,048,576 | Wrong Answer | 769 | 670,228 | 703 | Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1. |
a = int(input())
A = list(map(int,input().split()))
ch = [1]
cH = [1]
if A[0]>0:
print(-1)
exit()
for e,i in enumerate(A[1:]):
cH.append(cH[-1]*2-i)
# ch.append((cH[-1]*2)//(2**i))
# print(ch)
# print(cH)
# ch = ch[:-1]
# ch = ch[::-1]
# ch[0]*=2
# print(ch)
cH = cH[:-1]
cH.append(cH[-1]*2)
cH = cH[:... | s632623311 | Accepted | 793 | 670,020 | 794 |
a = int(input())
A = list(map(int,input().split()))
if a == 0 and A[0] == 1:
print(1)
exit()
ch = [1]
cH = [1]
if A[0]>0:
print(-1)
exit()
for e,i in enumerate(A[1:]):
cH.append(cH[-1]*2-i)
# ch.append((cH[-1]*2)//(2**i))
# print(ch)
# print(cH)
# ch = ch[:-1]
# ch = ch[::-1]
# ch[0]*=2
# p... |
s832912004 | p03836 | u462626125 | 2,000 | 262,144 | Wrong Answer | 19 | 3,188 | 538 | Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement ... | nums = [int(x) for x in input().split()]
sx,sy,tx,ty = nums[0],nums[1],nums[2],nums[3]
dx = tx - sx
dy = ty - sy
ans = []
# 1th
for i in range(dy):
ans.append("U")
for i in range(dx):
ans.append("R")
# 2nd
for i in range(dy):
ans.append("D")
for i in range(dx):
ans.append("L")
# 3rd
ans.append("L")
for... | s562225117 | Accepted | 19 | 3,192 | 560 | nums = [int(x) for x in input().split()]
sx,sy,tx,ty = nums[0],nums[1],nums[2],nums[3]
dx = tx - sx
dy = ty - sy
ans = []
# 1th
for i in range(dy):
ans.append("U")
for i in range(dx):
ans.append("R")
# 2nd
for i in range(dy):
ans.append("D")
for i in range(dx):
ans.append("L")
# 3rd
ans.append("L")
for... |
s206561042 | p03387 | u569742427 | 2,000 | 262,144 | Wrong Answer | 20 | 3,188 | 267 | You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be ... | data=[int(_) for _ in input().split()]
data.sort()
A=data[2]
B=data[1]
C=data[0]
count=0
print(A,B,C)
while A!=B:
B+=1
C+=1
count+=1
print(A,B,C)
while C<B:
C+=2
count+=1
print(A,B,C)
if B==C:
print(count)
else:
print(count+1)
| s284244583 | Accepted | 18 | 3,060 | 223 | data=[int(_) for _ in input().split()]
data.sort()
A=data[2]
B=data[1]
C=data[0]
count=0
while A!=B:
B+=1
C+=1
count+=1
while C<B:
C+=2
count+=1
if B==C:
print(count)
else:
print(count+1)
|
s474714264 | p03455 | u293198424 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 120 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | num = [int(i) for i in input().split()]
number = num[0]* num[1]
if number%2==0:
print("Odd")
else:
print("Even") | s401922780 | Accepted | 17 | 2,940 | 120 | num = [int(i) for i in input().split()]
number = num[0]* num[1]
if number%2==0:
print("Even")
else:
print("Odd") |
s978712479 | p03573 | u391328897 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 63 | You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers. | abc = list(map(int, input().split()))
print(list(set(abc))[0])
| s432785894 | Accepted | 17 | 2,940 | 112 | a, b, c = map(int, input().split())
if a == b:
print(c)
elif a == c:
print(b)
elif b == c:
print(a)
|
s588907660 | p03047 | u839953865 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 43 | 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? | [a,b]=map(int,input().split())
print(b-a+1) | s335289852 | Accepted | 17 | 2,940 | 43 | [a,b]=map(int,input().split())
print(a-b+1) |
s557359453 | p03369 | u776871252 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 87 | 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 ... | a = list(input())
ans = 700
for i in a:
if i == "+":
ans += 100
print(ans)
| s084194137 | Accepted | 17 | 2,940 | 87 | a = list(input())
ans = 700
for i in a:
if i == "o":
ans += 100
print(ans)
|
s988664899 | p03730 | u480568292 | 2,000 | 262,144 | Wrong Answer | 19 | 2,940 | 165 | 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())
count = 0
for i in range(1000):
s = a*i
if s%b == c:
count+=1
break
else:
pass
print("Yes" if count>0 else "No") | s573429929 | Accepted | 18 | 2,940 | 165 | a,b,c = map(int,input().split())
count = 0
for i in range(1000):
s = a*i
if s%b == c:
count+=1
break
else:
pass
print("YES" if count>0 else "NO") |
s988690938 | p03455 | u552290152 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 89 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a,b = map(int, input().split())
if a%2==0 or b%2==0:
print("even")
else:
print("odd") | s127885823 | Accepted | 20 | 3,060 | 89 | a,b = map(int, input().split())
if a%2==0 or b%2==0:
print("Even")
else:
print("Odd") |
s395466050 | p02694 | u599669731 | 2,000 | 1,048,576 | Wrong Answer | 20 | 9,160 | 117 | 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())
a=100
for i in range(3761):
a=int(1.01*a)
if a>x:
print(i+1)
break | s111458938 | Accepted | 21 | 9,164 | 118 | import math
x=int(input())
a=100
for i in range(1,3761):
a=int(1.01*a)
if a>=x:
print(i)
break |
s971152687 | p00011 | u187646742 | 1,000 | 131,072 | Wrong Answer | 20 | 7,708 | 246 | Let's play Amidakuji. In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines. In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step,... | stick = [i for i in range(1, int(input()) + 1)]
net = int(input())
for _ in range(net):
a, b = list(map(int, input().split(",")))
stick[a - 1], stick[b - 1] = stick[b - 1], stick[a - 1]
print(stick)
print("\n".join(map(str, stick))) | s074781986 | Accepted | 30 | 7,688 | 224 | stick = [i for i in range(1, int(input()) + 1)]
net = int(input())
for _ in range(net):
a, b = list(map(int, input().split(",")))
stick[a - 1], stick[b - 1] = stick[b - 1], stick[a - 1]
for v in stick:
print(v) |
s961843593 | p02613 | u810356688 | 2,000 | 1,048,576 | Wrong Answer | 60 | 9,228 | 503 | 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`,... | import sys
def input(): return sys.stdin.readline().rstrip()
def main():
n = int(input())
ac, wa, tle, re =0, 0, 0, 0
for i in range(n):
s = input()
if s == "AC":
ac += 1
elif s == "WA":
wa += 1
elif s == "TLE":
tle += 1
else:
... | s602329882 | Accepted | 60 | 9,220 | 503 | import sys
def input(): return sys.stdin.readline().rstrip()
def main():
n = int(input())
ac, wa, tle, re =0, 0, 0, 0
for i in range(n):
s = input()
if s == "AC":
ac += 1
elif s == "WA":
wa += 1
elif s == "TLE":
tle += 1
else:
... |
s790793758 | p03693 | u643840641 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 72 | 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 = map(int, input().split())
print("YES" if (g+b)%4==0 else "NO") | s018841819 | Accepted | 18 | 2,940 | 75 | r, g, b = map(int, input().split())
print("YES" if (10*g+b)%4==0 else "NO") |
s444353941 | p03681 | u863442865 | 2,000 | 262,144 | Wrong Answer | 51 | 3,188 | 320 | 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 = list(map(int, input().split()))
ans = 1
a = lambda x, y: x*y % (10**9+7)
if n==m:
for i in range(1, n+1):
a(ans, i)
print((ans**2*2)%(10**9+7))
elif abs(n-m)==1:
for i in range(1, n+1):
a(ans, i)
for i in range(1, m+1):
a(ans, i)
print(ans%(10**9+7))
else:
print(0) | s763595633 | Accepted | 68 | 3,064 | 339 | n, m = list(map(int, input().split()))
ans = 1
mod = 10**9+7
def a(x, y):
return x*y%mod
if n==m:
for i in range(1, n+1):
ans = a(ans, i)
print((ans**2*2)%(mod))
elif abs(n-m)==1:
for i in range(1, n+1):
ans = a(ans, i)
for i in range(1, m+1):
ans = a(ans, i)
print(ans)
... |
s998573205 | p03155 | u021548497 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 73 | It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board. The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns. How many ways are there to choose where ... | n = int(input())
h = int(input())
w = int(input())
print((w-h+1)*(n-h+1)) | s106687537 | Accepted | 17 | 2,940 | 73 | n = int(input())
h = int(input())
w = int(input())
print((n-w+1)*(n-h+1)) |
s235159147 | p03712 | u289792007 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 158 | You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1. | h, w = map(int, input().split())
img = ['*' * (w+2)]
for i in range(0, h):
img.append('*' + input() + '*')
img.append('*' * (w+2))
[print(i) for i in img] | s571923339 | Accepted | 18 | 3,060 | 158 | h, w = map(int, input().split())
img = ['#' * (w+2)]
for i in range(0, h):
img.append('#' + input() + '#')
img.append('#' * (w+2))
[print(i) for i in img] |
s754843637 | p04043 | u578722729 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 148 | 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 ... | ABC=input().split()
five_count=ABC.count("5")
seven_count=ABC.count("7")
if five_count == 2 and seven_count==1:
print("Yes")
else:
print("NO") | s605170123 | Accepted | 17 | 2,940 | 150 | ABC=input().split()
five_count=ABC.count("5")
seven_count=ABC.count("7")
if five_count == 2 and seven_count == 1:
print("YES")
else:
print("NO") |
s065870373 | p03251 | u374929478 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 228 | 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 =(int(x) for x in input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
z = []
z = range(X+1,Y+1)
if min(z) > max(x) and max(z) <= min(y):
print('No War')
else:
print('War')
| s467862057 | Accepted | 18 | 3,060 | 252 | n, m, X, Y =(int(x) for x in input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
for z in range(X+1,Y+1):
if z > max(x) and z <= min(y):
print('No War')
break
if z == Y:
print('War') |
s300817733 | p03162 | u106166456 | 2,000 | 1,048,576 | Wrong Answer | 679 | 33,792 | 480 | Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain... | def chmax_a(l,B,C,i):
return max(B+l[i][1],C+l[i][2])
def chmax_b(l,A,C,i):
return max(A+l[i][0],C+l[i][2])
def chmax_c(l,A,B,i):
return max(B+l[i][1],A+l[i][0])
N=int(input())
happiness=[]
for i in range(N):
cost=list(map(int,input().split()))
happiness.append(cost)
answers=[]
A=0
B=0
C=0
for i in range(N-1,-1... | s454284225 | Accepted | 1,060 | 47,328 | 412 | if __name__=='__main__':
N=int(input())
happiness=[]
for i in range(N):
cost=list(map(int,input().split()))
happiness.append(cost)
dp=[[0 for j in range(3)] for i in range(N+1)]
#j:0 is from A, 1 is from B, 2 is from C
for i in range(N):
for j in range(3):
for k in range(3):
if j != k:
dp[i+1][k]... |
s786644980 | p03473 | u069744971 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 30 | How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December? | a = int(input())
print(a-48) | s535462927 | Accepted | 17 | 2,940 | 30 | a = int(input())
print(48-a) |
s084435723 | p03437 | u981931040 | 2,000 | 262,144 | Wrong Answer | 29 | 9,092 | 61 | You are given positive integers X and Y. If there exists a positive integer not greater than 10^{18} that is a multiple of X but not a multiple of Y, choose one such integer and print it. If it does not exist, print -1. | a, b = map(int, input().split())
print(a if a != b else '=1') | s034812045 | Accepted | 25 | 9,036 | 112 | a, b = map(int, input().split())
ans = -1
for i in range(1, 4):
if a * i % b != 0:
ans = a * i
print(ans)
|
s819927157 | p03069 | u457901067 | 2,000 | 1,048,576 | Wrong Answer | 214 | 15,616 | 352 | There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so t... | N = int(input())
S = input()
cumdots = [0] * (N+1)
for i in range(N):
cumdots[i+1] = cumdots[i] + (1 if S[i]=="." else 0)
print(cumdots)
hands = N+1
for i in range(N):
leftops = i - cumdots[i]
rightops = cumdots[N] - cumdots[i+1]
hands = min(hands, leftops + rightops)
print(hands) | s081789682 | Accepted | 208 | 11,144 | 353 | N = int(input())
S = input()
cumdots = [0] * (N+1)
for i in range(N):
cumdots[i+1] = cumdots[i] + (1 if S[i]=="." else 0)
#print(cumdots)
hands = N+1
for i in range(N):
leftops = i - cumdots[i]
rightops = cumdots[N] - cumdots[i+1]
hands = min(hands, leftops + rightops)
print(hands) |
s974708009 | p02264 | u591875281 | 1,000 | 131,072 | Wrong Answer | 30 | 7,964 | 700 | _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 round_robin_scheduling(N, Q, A):
time = 0
finished_process = []
while A:
process = A.popleft()
if process[1] < Q:
time += process[1]
finished_process.append([process[0],time])
else:
time += Q
A.append(... | s498035753 | Accepted | 280 | 16,208 | 517 | # encoding: utf-8
from collections import deque
def round_robin_scheduling(N, Q, A):
t = 0
while A:
process = A.popleft()
if process[1] <= Q:
t += process[1]
print(process[0],t)
else:
t += Q
A.append([process[0],process[1]-Q])
if __name_... |
s245370929 | p03470 | u818655004 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 299 | 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())
A = sorted(list(map(int, input().split())), reverse=True)
A_cards = []
B_cards = []
flag = 'A'
for card in A:
if flag == 'A':
flag = 'B'
A_cards.append(card)
elif flag == 'B':
flag = 'A'
B_cards.append(card)
print(sum(A_cards) - sum(B_cards)) | s192255136 | Accepted | 17 | 2,940 | 160 | N = int(input())
d_list = [int(input()) for i in range(N)]
result = {}
for i in d_list:
result.setdefault(i, 0)
result[i] += 1
print(len(result.keys())) |
s416037580 | p03351 | u838350081 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,064 | 576 | 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 ... |
class Position():
def __init__(self, posi, dist):
self.pos = posi
self.d = dist
def is_talk(self, tar_pos):
if abs(self.pos - tar_pos) <= self.d:
return True
else:
return False
if __name__ == '__main__':
a, b, c, d = [int(_) for _ in input... | s664767216 | Accepted | 18 | 3,064 | 554 |
class Position():
def __init__(self, posi, dist):
self.pos = posi
self.d = dist
def is_talk(self, tar_pos):
if abs(self.pos - tar_pos) <= self.d:
return True
else:
return False
if __name__ == '__main__':
a, b, c, d = [int(_) for _ in input... |
s052787082 | p03574 | u853900545 | 2,000 | 262,144 | Wrong Answer | 26 | 3,064 | 314 | 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 con... | h,w = map(int,input().split())
s = [input() for _ in range(h)]
for i in range(h):
l = ''
for j in range(w):
if s[i][j] == '#':
l += s[i][j]
else:
l += str(sum([t[max(0,j-1):min(j+2,w)].count('#')\
for t in s[max(0,i):min(h,i+2)]]))
print(l) | s819547840 | Accepted | 28 | 3,064 | 316 | h,w = map(int,input().split())
s = [input() for _ in range(h)]
for i in range(h):
l = ''
for j in range(w):
if s[i][j] == '#':
l += s[i][j]
else:
l += str(sum([t[max(0,j-1):min(j+2,w)].count('#')\
for t in s[max(0,i-1):min(h,i+2)]]))
print(l) |
s104319541 | p03433 | u260036763 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 91 | E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins. | N = int(input())
A = int(input())
if (N - A) % 500 == 0:
print('Yes')
else:
print('No') | s434397224 | Accepted | 17 | 2,940 | 87 | n = int(input())
a = int(input())
if (n % 500) <= a:
print('Yes')
else:
print('No') |
s245783518 | p01554 | u741801763 | 2,000 | 131,072 | Wrong Answer | 20 | 7,652 | 420 | ある部屋ではICカードを用いて鍵を開け閉めする電子錠システムを用いている。 このシステムは以下のように動作する。 各ユーザーが持つICカードを扉にかざすと、そのICカードのIDがシステムに渡される。 システムはIDが登録されている時、施錠されているなら開錠し、そうでないのなら施錠し、それぞれメッセージが出力される。 IDが登録されていない場合は、登録されていないというメッセージを出力し、開錠及び施錠はおこなわれない。 さて、現在システムにはN個のID(U1, U2, ……, UN)が登録されており、施錠されている。 M回ICカードが扉にかざされ、そのIDはそれぞれ順番にT1, T2, ……, TMであるとする。 この時のシステム... | if __name__ == '__main__':
n = int(input())
reg = []
for _ in range(n):
reg.append(input())
m = int(input())
state = 0
for _ in range(m):
p = input()
if p in reg and not state:
print("Opened by %s" % (p))
state = 1
elif p in reg and state:... | s261242373 | Accepted | 70 | 7,732 | 421 | if __name__ == '__main__':
n = int(input())
reg = []
for _ in range(n):
reg.append(input())
m = int(input())
state = 0
for _ in range(m):
p = input()
if p in reg and not state:
print("Opened by %s" % (p))
state = 1
elif p in reg and state:... |
s650880247 | p03574 | u036340997 | 2,000 | 262,144 | Wrong Answer | 37 | 3,700 | 537 | 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 con... | hw = input().split()
h = int(hw[0])
w = int(hw[1])
field = []
for i in range(h):
field.append(input())
ans_field = []
for i in range(h):
ans_field.append('')
for j in range(w):
s = ''
peris = [[i-1, j-1], [i-1, j], [i-1, j+1], [i, j-1], [i, j+1], [i+1, j-1], [i+1, j], [i+1, j+1]]
for peri in peris:
... | s334436342 | Accepted | 32 | 3,064 | 750 | hw = input().split()
h = int(hw[0])
w = int(hw[1])
field = []
for i in range(h):
field.append(input())
ans_field = []
for i in range(h):
ans_field.append('')
for j in range(w):
if field[i][j]=='.':
s = ''
peris = [[i-1, j-1], [i-1, j], [i-1, j+1], [i, j-1], [i, j+1], [i+1, j-1], [i+1, j],... |
s873497793 | p03719 | u332385682 | 2,000 | 262,144 | Wrong Answer | 25 | 3,444 | 229 | You are given three integers A, B and C. Determine whether C is not less than A and not greater than B. | import sys
from collections import Counter
sys.setrecursionlimit(10**7)
inf = 1<<100
def solve():
a, b, c = map(int, input().split())
print('YES' if a <= c <= b else 'NO')
if __name__ == '__main__':
solve() | s407081395 | Accepted | 17 | 2,940 | 152 | import sys
def solve():
a, b, c = map(int, input().split())
print('Yes' if a <= c <= b else 'No')
if __name__ == '__main__':
solve() |
s898368533 | p02742 | u193927973 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 261 | 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=map(int, input().split())
if h%2==0:
print(h*w/2)
else:
set1=h
if w%2==0:
print(set1*w/2)
else:
rm=w//2
print(set1*rm+h//2+1)
| s207063093 | Accepted | 18 | 2,940 | 134 | h, w=map(int, input().split())
if h==1 or w==1:
print(1)
else:
ans=(h*w)//2
if h*w%2==1:
print(ans+1)
else:
print(ans) |
s660500625 | p02398 | u936401118 | 1,000 | 131,072 | Wrong Answer | 30 | 7,556 | 125 | Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b. | a, b, c = map (int, input().split())
count = 0
for i in range(a+1, b+1):
if c % i == 0:
count += 1
print (count) | s740135072 | Accepted | 60 | 7,700 | 123 | a, b, c = map (int, input().split())
count = 0
for i in range(a, b+1):
if c % i == 0:
count += 1
print (count) |
s816923190 | p03993 | u414809621 | 2,000 | 262,144 | Wrong Answer | 91 | 14,008 | 193 | There are N rabbits, numbered 1 through N. The i-th (1≤i≤N) rabbit likes rabbit a_i. Note that no rabbit can like itself, that is, a_i≠i. For a pair of rabbits i and j (i＜j), we call the pair (i，j) a _friendly pair_ if the following condition is met. * Rabbit i likes rabbit j and rabbit j likes rabbit i. Calculat... |
N = int(input())
a = list(map(int,input().split()))
print(a)
ans=0
for i, v in enumerate(a):
if v != None:
if a[v-1] == i+1:
ans+=1
a[v-1] = None
print(ans) | s649647831 | Accepted | 88 | 14,008 | 184 |
N = int(input())
a = list(map(int,input().split()))
ans=0
for i, v in enumerate(a):
if v != None:
if a[v-1] == i+1:
ans+=1
a[v-1] = None
print(ans) |
s047935830 | p03455 | u743852057 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 103 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a,b = input().split()
result = int(a)*int(b)
if result%2 == 0:
print("even")
else:
print("odd") | s610628239 | Accepted | 17 | 2,940 | 103 | a,b = input().split()
result = int(a)*int(b)
if result%2 == 0:
print("Even")
else:
print("Odd") |
s736024577 | p02842 | u497952650 | 2,000 | 1,048,576 | Wrong Answer | 33 | 2,940 | 105 | 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())
for i in range(0,50001):
if int(i*1.08)==N:
print(N)
exit()
print(":(") | s666705063 | Accepted | 31 | 2,940 | 104 | N = int(input())
for i in range(0,N+1):
if int(i*1.08)==N:
print(i)
exit()
print(":(") |
s308161314 | p04035 | u445624660 | 2,000 | 262,144 | Wrong Answer | 98 | 19,824 | 560 | We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i. At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly: * Choose a (connected) rope with... |
n, l = map(int, input().split())
a = list(map(int, input().split()))
idx = 0
for i in range(1, n):
if a[i - 1] + a[i] >= l:
idx = i
break
if idx == 0:
print("impossibe")
exit()
print("possible")
ans = [idx]
for i in range(idx - 1, 0, -1):
ans.append(i)
for i in range(idx + 1, n):
... | s566401464 | Accepted | 97 | 19,996 | 561 |
n, l = map(int, input().split())
a = list(map(int, input().split()))
idx = 0
for i in range(1, n):
if a[i - 1] + a[i] >= l:
idx = i
break
if idx == 0:
print("Impossible")
exit()
print("Possible")
ans = [idx]
for i in range(idx - 1, 0, -1):
ans.append(i)
for i in range(idx + 1, n):
... |
s903061691 | p03471 | u100873497 | 2,000 | 262,144 | Wrong Answer | 807 | 3,060 | 173 | 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 situat... | n,y = map(int,input().split())
for a in range(n+1):
for b in range(n+1-a):
c=n-a-b
if y==a*10000+b*5000+c*1000 and n==a+b+c:
print(a,b,c)
break
| s124669672 | Accepted | 834 | 3,060 | 247 | n,y = map(int,input().split())
l=0
for a in range(n+1):
for b in range(n+1-a):
c=n-a-b
if y==a*10000+b*5000+c*1000 and n==a+b+c:
i=a
j=b
k=c
l=1
break
if l==1:
print(i,j,k)
else:
print(-1,-1,-1)
|
s648687137 | p03672 | u411858517 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 240 | We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by del... | S = list(input())
for i in range(len(S)):
if len(S)%2 != 0:
S.pop()
else:
if S[:len(S)//2] == S[len(S)//2:]:
res = len(S)
break
else:
S.pop()
S.pop()
print(res) | s551907630 | Accepted | 17 | 3,060 | 234 | S = list(input())
if len(S)%2 != 0:
S.pop()
else:
S.pop()
S.pop()
for i in range(len(S)):
if S[:len(S)//2] == S[len(S)//2:]:
res = len(S)
break
else:
S.pop()
S.pop()
print(res) |
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