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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s667144022 | p03478 | u429541300 | 2,000 | 262,144 | Wrong Answer | 1,234 | 3,060 | 185 | Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive). | n, a, b = map(int, input().split())
ans = 0
for i in range(n+1):
m = 0
while i != 0:
m += int(n % 10)
i = i / 10
if a <= m <= b:
ans += m
print(ans)
| s851008662 | Accepted | 37 | 2,940 | 202 | n, a, b = map(int, input().split())
ans = 0
for i in range(n+1):
m = 0
i0 = i
while i != 0:
m += int(i % 10)
i = int(i / 10)
if a <= m <= b:
ans += i0
print(ans)
|
s460943702 | p03494 | u811000506 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 187 | There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform. | N = int(input())
A = list(map(int,input().split()))
A.sort()
count = 0
while(True):
if A[0] % 2 == 0:
A[0] = A[0] // 2
count += 1
else:
break
print(count) | s710393902 | Accepted | 19 | 3,064 | 280 | N = int(input())
A = list(map(int,input().split()))
i = 0
count = 0
flag = False
while (True):
for i in range(N):
if A[i] % 2 != 0:
flag = True
break
if flag == True:
break
for i in range(N):
A[i] = A[i] // 2
count += 1
print(count) |
s758116950 | p03657 | u050641473 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 128 | Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies. | A, B = map(int, input().split())
if A % 3 == 0 or B % 3 == 0 or A + B % 3 == 0:
print("Possible")
else:
print("Impossible") | s603986549 | Accepted | 18 | 2,940 | 130 | A, B = map(int, input().split())
if A % 3 == 0 or B % 3 == 0 or (A + B) % 3 == 0:
print("Possible")
else:
print("Impossible") |
s995007841 | p02396 | u796301295 | 1,000 | 131,072 | Wrong Answer | 140 | 5,608 | 125 | In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem. | a = 1
while True:
n = int(input())
if n == 0:
break
print ("case " + str(a) + ": " + str(n))
a += 1
| s332070304 | Accepted | 140 | 5,608 | 125 | a = 1
while True:
n = int(input())
if n == 0:
break
print ("Case " + str(a) + ": " + str(n))
a += 1
|
s359904804 | p02396 | u643542669 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 129 | In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem. | inpu_list = list(map(int, input().split()))
for i in range(len(inpu_list) - 1):
print("Case %d: %d" % (i + 1, inpu_list[i]))
| s110302093 | Accepted | 70 | 6,248 | 143 | array = []
x = input()
while x != "0":
array.append(x)
x = input()
for i, x in enumerate(array):
print("Case %d: %s" % (i + 1, x))
|
s451602910 | p03574 | u403355272 | 2,000 | 262,144 | Wrong Answer | 43 | 4,228 | 873 | You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process. | h,m = map(int,input().split())
s = [input() for i in range(h)]
ans = [[] for i in range(h)]
for i in range(h):
for j in range(m):
if s[i][j] == "#":
ans[i].append("#")
else:
count = 0
s1=[[i-1,j],[i+1,j],[i,j-1],[i,j+1]]
s2=[[i-1,j+1],[i+1,j+1],[i+1,j-1],[i-1,j-1]]
for temp in s1 :
i2=temp[0];j2=temp[1]
print(i2,j2)
if i2<0 or h-1<i2 or j2<0 or m-1<j2:
continue
if s[i2][j2]=="#":
count+=1
for temp in s2 :
i2=temp[0];j2=temp[1]
if i2<0 or h-1<i2 or j2 <0 or m-1<j2 :
continue
if s[i2][j2]=="#":
count+=1
ans[i].append(count)
for i in range(h):
print(*ans[i],sep="") | s365403365 | Accepted | 31 | 3,572 | 844 | h,m = map(int,input().split())
s = [input() for i in range(h)]
ans = [[] for i in range(h)]
for i in range(h):
for j in range(m):
if s[i][j] == "#":
ans[i].append("#")
else:
count = 0
s1=[[i-1,j],[i+1,j],[i,j-1],[i,j+1]]
s2=[[i-1,j+1],[i+1,j+1],[i+1,j-1],[i-1,j-1]]
for temp in s1 :
i2=temp[0];j2=temp[1]
if i2<0 or h-1<i2 or j2<0 or m-1<j2:
continue
if s[i2][j2]=="#":
count+=1
for temp in s2 :
i2=temp[0];j2=temp[1]
if i2<0 or h-1<i2 or j2 <0 or m-1<j2 :
continue
if s[i2][j2]=="#":
count+=1
ans[i].append(count)
for i in range(h):
print(*ans[i],sep="") |
s677843757 | p03434 | u000349418 | 2,000 | 262,144 | Wrong Answer | 19 | 3,060 | 166 | We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score. | n=int(input())
A=list(map(int,input().split(' ')))
A.sort()
a=0
b=0
i=0
while i < n:
if i%2==0:
a+= A[i]
else:
b += A[i]
i += 1
print(a-b) | s714942350 | Accepted | 17 | 3,060 | 171 | n=int(input())
A=list(map(int,input().split(' ')))
A.sort()
a=0
b=0
i=0
while i < n:
if i%2==0:
a+= A[i]
else:
b += A[i]
i += 1
print(abs(a-b)) |
s873635310 | p03388 | u046187684 | 2,000 | 262,144 | Wrong Answer | 19 | 3,060 | 378 | 10^{10^{10}} participants, including Takahashi, competed in two programming contests. In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th. The _score_ of a participant is the product of his/her ranks in the two contests. Process the following Q queries: * In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's. |
from math import sqrt, floor
q = int(input())
for _ in range(q):
a, b = map(int, input().split(" "))
if a == b:
print(2 * a - 2)
elif max(a, b) - min(a, b) == 1:
print(2 * min(a, b) - 2)
else:
if sqrt(a * b) == floor(sqrt(a * b)):
print(2 * floor(sqrt(a * b)) - 3)
else:
print(2 * floor(sqrt(a * b)) - 1)
| s022197937 | Accepted | 18 | 3,064 | 530 | from math import sqrt
def solve(string):
q, *ab = map(int, string.split())
ans = []
for a, b in zip(ab[::2], ab[1::2]):
c = a * b
d = int(sqrt(c))
answer = 2 * d - 2
if d * d == c:
answer -= 1
if d * (d + 1) < c:
answer += 1
if a == b:
answer += 1
ans.append(str(answer))
return "\n".join(ans)
if __name__ == '__main__':
n = int(input())
print(solve('{}\n'.format(n) + '\n'.join([input() for _ in range(n)])))
|
s307441597 | p03693 | u573670713 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 109 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4? | a,b,c = list(map(int,input().strip().split()))
if a*100 + b*10 +c %4 ==0:
print("YES")
else:
print("NO") | s273610236 | Accepted | 17 | 2,940 | 113 | a,b,c = list(map(int,input().strip().split()))
if (a *100 +b*10 +c) % 4 == 0:
print("YES")
else:
print("NO") |
s706311183 | p03475 | u187205913 | 3,000 | 262,144 | Wrong Answer | 124 | 3,064 | 384 | A railroad running from west to east in Atcoder Kingdom is now complete. There are N stations on the railroad, numbered 1 through N from west to east. Tomorrow, the opening ceremony of the railroad will take place. On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i to Station i+1 in C_i seconds. No other trains will be operated. The first train from Station i to Station i+1 will depart Station i S_i seconds after the ceremony begins. Thereafter, there will be a train that departs Station i every F_i seconds. Here, it is guaranteed that F_i divides S_i. That is, for each Time t satisfying S_i≤t and t％F_i=0, there will be a train that departs Station i t seconds after the ceremony begins and arrives at Station i+1 t+C_i seconds after the ceremony begins, where A％B denotes A modulo B, and there will be no other trains. For each i, find the earliest possible time we can reach Station N if we are at Station i when the ceremony begins, ignoring the time needed to change trains. | n = int(input())
l = [list(map(int,input().split())) for _ in range(n-1)]
ans = []
for i in range(len(l)):
m = l[i][1]
for j in range(i,len(l)):
if m<l[j][1]:
m = l[j][1]+l[j][0]
elif m==l[j][1] or (m-l[j][1])%l[j][2]==0:
m += l[j][0]
else:
m += (m-l[j][1])%l[j][2]+l[j][0]
ans.append(m)
ans.append(0)
print(ans) | s781292398 | Accepted | 127 | 3,188 | 422 | n = int(input())
l = [list(map(int,input().split())) for _ in range(n-1)]
ans = []
for i in range(len(l)):
m = l[i][1]
for j in range(i,len(l)):
if m<l[j][1]:
m = l[j][1]+l[j][0]
elif m==l[j][1] or (m-l[j][1])%l[j][2]==0:
m += l[j][0]
else:
next = l[j][2]-m%l[j][2]
m += next+l[j][0]
ans.append(m)
ans.append(0)
for i in ans:
print(i) |
s796064281 | p03418 | u692746605 | 2,000 | 262,144 | Wrong Answer | 87 | 2,940 | 142 | Takahashi had a pair of two positive integers not exceeding N, (a,b), which he has forgotten. He remembers that the remainder of a divided by b was greater than or equal to K. Find the number of possible pairs that he may have had. | N,K=map(int,input().split())
t=0
for b in range(K+1,N+1):
t+=(N//b)*(b-K)+(N%b-K+(1 if K>0 else 0) if N%b>=K else 0)if N%b else 0
print(t)
| s481324442 | Accepted | 84 | 2,940 | 121 | N,K=map(int,input().split())
t=0
for b in range(K+1,N+1):
t+=(-~N//b)*(b-K)+max(-~N%b-K,0)
print(t-(0 if K else N-K))
|
s641276527 | p03635 | u618373524 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 43 | In _K-city_ , there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city? | s = input()
print(s[0]+str(len(s)-2)+s[-1]) | s579694488 | Accepted | 17 | 2,940 | 55 | a,b = map(int,input().split())
c = (a-1)*(b-1)
print(c) |
s425279622 | p03469 | u428467389 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 129 | On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it. | S = input()
count=0
if(S[0]=="o"):
count+=1
if(S[1]=="o"):
count+=1
if(S[2]=="o"):
count+=1
print(700+100*count) | s449741183 | Accepted | 17 | 2,940 | 60 | S = input()
S = list(S)
S[3]="8"
S = "".join(S)
print(S) |
s774278114 | p03643 | u077291787 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 216 | This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer. |
def main():
X = int(input())
ans, r = X // 11 * 2, X % 11 # r: remainder
ans += 1 if 0 < r <= 6 else 2
print(ans)
if __name__ == "__main__":
main() | s887206440 | Accepted | 17 | 2,940 | 123 |
def main():
N = input().rstrip()
print("ABC", N, sep="")
if __name__ == "__main__":
main() |
s569448379 | p03478 | u829796346 | 2,000 | 262,144 | Time Limit Exceeded | 2,104 | 2,940 | 165 | Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive). | N,A,B = map(int, input().split())
ret = 0
for i in range(N):
c = 0
while i:
c,mod = divmod(i,10)
c += mod
if A <= c and c <= B:
ret += i
print(ret) | s729857710 | Accepted | 33 | 2,940 | 146 | N,A,B = map(int, input().split())
ret = 0
for i in range(N+1):
s = sum([int(c) for c in str(i)])
if A <= s and s <= B:
ret += i
print(ret) |
s269549112 | p03610 | u584870699 | 2,000 | 262,144 | Wrong Answer | 63 | 3,828 | 81 | You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1. | s=input()
for num,moji in enumerate(s):
if((num+1)%2==0):
print(moji) | s305316929 | Accepted | 17 | 3,188 | 23 | s=input()
print(s[::2]) |
s064265803 | p03644 | u697696097 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 197 | Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times. | n=int(input())
maxx=0
maxn=0
for i in range(1,n+1):
tmp=i
cnt=0
while 1:
if tmp%2==0:
tmp=tmp//2
cnt+=1
else:
break
if cnt > maxx:
maxn=tmp
print(maxn)
| s680366742 | Accepted | 17 | 3,060 | 202 | n=int(input())
maxx=0
maxn=1
for i in range(1,n+1):
tmp=i
cnt=0
while 1:
if tmp%2==0:
tmp=tmp//2
cnt+=1
else:
break
if cnt > maxx:
maxn=i
maxx=cnt
print(maxn) |
s030593627 | p04011 | u785470389 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 127 | There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee. | n = int(input())
k = int(input())
x = int(input())
y = int(input())
price = k * x
price = price + (n - k) * y
print = (price) | s572422431 | Accepted | 17 | 2,940 | 169 | n = int(input())
k = int(input())
x = int(input())
y = int(input())
if n > k:
price = k * x
price = price + (n - k) * y
elif n <= k:
price = n * x
print(price) |
s468729563 | p02256 | u227984374 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 134 | Write a program which finds the greatest common divisor of two natural numbers _a_ and _b_ | x, y = map(int, input().split())
if x > y :
pass
else :
x, y = y, x
while x % y != 0 :
y = x % y
x = y
print(y)
| s831188512 | Accepted | 30 | 5,596 | 144 | x, y = map(int, input().split())
if x > y :
pass
else :
x, y = y, x
while x % y != 0 :
d = x % y
x = y
y = d
print(y)
|
s018301110 | p03472 | u169138653 | 2,000 | 262,144 | Wrong Answer | 852 | 42,404 | 391 | You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total? | from math import ceil
n,h=map(int,input().split())
g=[list(map(int,input().split())) for i in range(n)]
l=[]
ans=0
dm=0
for i in range(n):
l.append([g[i][0],2])
l.append([g[i][1],1])
l.sort(reverse=True)
for i in range(2*n):
if l[i][1]==1:
dm+=l[i][0]
ans+=1
if dm>=h:
print(ans)
exit()
else:
h-=dm
ans+=ceil(h//l[i][0])
print(ans)
exit()
| s528282858 | Accepted | 748 | 42,404 | 390 | from math import ceil
n,h=map(int,input().split())
g=[list(map(int,input().split())) for i in range(n)]
l=[]
ans=0
dm=0
for i in range(n):
l.append([g[i][0],2])
l.append([g[i][1],1])
l.sort(reverse=True)
for i in range(2*n):
if l[i][1]==1:
dm+=l[i][0]
ans+=1
if dm>=h:
print(ans)
exit()
else:
h-=dm
ans+=ceil(h/l[i][0])
print(ans)
exit()
|
s586985377 | p03556 | u923659712 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 126 | Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer. | n=int(input())
for i in range(1,n):
if i**i>n:
print((i-1)**(i-1))
exit()
elif i**i==n:
print(i**i)
exit() | s947504727 | Accepted | 26 | 2,940 | 124 | n=int(input())
for i in range(1,n+1):
if i*i>n:
print((i-1)*(i-1))
exit()
elif i*i==n:
print(i*i)
exit() |
s411886406 | p02388 | u706217959 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 223 | Write a program which calculates the cube of a given integer x. | def main():
n = int(input())
if n < 1 or n > 100:
print("invalid answer.")
return
ans = n * 3
print("{0}".format(ans))
if __name__ == '__main__':
main() | s862266567 | Accepted | 20 | 5,596 | 196 | def main():
n = int(input())
if n <= 0 or n >= 101:
print("invalid answer.")
return
ans = n * n * n
print("{0}".format(ans))
if __name__ == '__main__':
main() |
s717222874 | p03814 | u161485415 | 2,000 | 262,144 | Wrong Answer | 39 | 9,340 | 192 | Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`. | a = input("")
a_num = a.index("A")
for i in range(len(a)-1,-1,-1):
if a[i]=="Z":
z_num = i+1
break
print("num:" + str(len(a[a_num:z_num])) + \
"text:" + a[a_num:z_num]) | s192489882 | Accepted | 38 | 9,244 | 145 | a = input("")
a_num = a.index("A")
for i in range(len(a)-1,-1,-1):
if a[i]=="Z":
z_num = i+1
break
print(len(a[a_num:z_num])) |
s678595158 | p03565 | u371942102 | 2,000 | 262,144 | Wrong Answer | 21 | 3,188 | 383 | 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 more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`. | from itertools import permutations
import re
def main():
s = input().replace('?', '.')
t = input()
for i in reversed(range(len(s) - len(t) + 1)):
print(s[i:i + len(t)])
if re.search(s[i:i + len(t)], t):
result = (s[:i] + t + s[i + len(t):]).replace('.', 'a')
print(result)
return
print('UNRESTORABLE ')
main()
| s430221685 | Accepted | 21 | 3,188 | 384 | from itertools import permutations
import re
def main():
s = input().replace('?', '.')
t = input()
for i in reversed(range(len(s) - len(t) + 1)):
# print(s[i:i + len(t)])
if re.search(s[i:i + len(t)], t):
result = (s[:i] + t + s[i + len(t):]).replace('.', 'a')
print(result)
return
print('UNRESTORABLE')
main()
|
s885564730 | p03150 | u197704813 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,064 | 636 | 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()
if S[0:7] == "keyence":
print("YES")
elif S[len(S)-8:len(S)-1] == "keyence":
print("YES")
else:
if (S[0] == "k")and(S[len(S)-7:len(S)-1] == "eyence"):
print("YES")
elif (S[0] == "ke")and(S[len(S)-6:len(S)-1] == "yence"):
print("YES")
elif (S[0] == "key")and(S[len(S)-5:len(S)-1] == "ence"):
print("YES")
elif (S[0] == "keye")and(S[len(S)-4:len(S)-1] == "nce"):
print("YES")
elif (S[0] == "keyen")and(S[len(S)-3:len(S)-1] == "ce"):
print("YES")
elif (S[0] == "keyenc")and(S[len(S)-2:len(S)-1] == "e"):
print("YES")
else:
print("NO")
| s404767190 | Accepted | 17 | 3,064 | 634 | S = input()
if S[0:7] == "keyence":
print("YES")
elif S[len(S)-7:len(S)] == "keyence":
print("YES")
else:
if (S[0:1] == "k")and(S[len(S)-6:len(S)] == "eyence"):
print("YES")
elif (S[0:2] == "ke")and(S[len(S)-5:len(S)] == "yence"):
print("YES")
elif (S[0:3] == "key")and(S[len(S)-4:len(S)] == "ence"):
print("YES")
elif (S[0:4] == "keye")and(S[len(S)-3:len(S)] == "nce"):
print("YES")
elif (S[0:5] == "keyen")and(S[len(S)-2:len(S)] == "ce"):
print("YES")
elif (S[0:6] == "keyenc")and(S[len(S)-1:len(S)] == "e"):
print("YES")
else:
print("NO")
|
s224918716 | p02748 | u581320767 | 2,000 | 1,048,576 | Wrong Answer | 599 | 40,864 | 320 | 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 tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time. You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required. | A, B, M = map(int, input().split())
ref = list(map(int, input().split()))
mic = list(map(int, input().split()))
coupons = [ list(map(int, input().split())) for i in range(M)]
ans = min(ref) + min(mic)
for i in coupons:
print(i)
price = ref[i[0]-1] + mic[i[1]-1] - i[2]
if price < ans:
ans = price
print(ans)
| s072377276 | Accepted | 456 | 38,488 | 309 | A, B, M = map(int, input().split())
ref = list(map(int, input().split()))
mic = list(map(int, input().split()))
coupons = [ list(map(int, input().split())) for i in range(M)]
ans = min(ref) + min(mic)
for i in coupons:
price = ref[i[0]-1] + mic[i[1]-1] - i[2]
if price < ans:
ans = price
print(ans)
|
s559569579 | p02664 | u607180061 | 2,000 | 1,048,576 | Wrong Answer | 88 | 10,824 | 221 | For a string S consisting of the uppercase English letters `P` and `D`, let the _doctoral and postdoctoral quotient_ of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3. We have a string T consisting of `P`, `D`, and `?`. Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient. | t = list(input())
for i in range(len(t)):
if t[i] == '?':
if i == len(t) - 1:
t[i] = 'D'
elif t[i+1] == 'D':
t[i] = 'P'
else:
t[i] = 'D'
print(''.join(t))
| s283121266 | Accepted | 82 | 10,800 | 342 | t = list(input())
tl = len(t)
for i in range(tl):
if t[i] != '?':
continue
if tl == 1:
t[i] = 'D'
elif i == tl - 1:
t[i] = 'D'
elif t[i-1] == 'P':
t[i] = 'D'
elif t[i+1] == '?':
t[i] = 'P'
elif t[i+1] == 'D':
t[i] = 'P'
else:
t[i] = 'D'
print(''.join(t))
|
s032882210 | p02613 | u631579948 | 2,000 | 1,048,576 | Wrong Answer | 153 | 16,448 | 306 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | a=int(input())
AC=[]
WA=[]
TLE=[]
RE=[]
for i in range(a):
s=(input())
if s=='AC':
AC.append(s)
if s=='TLE':
TLE.append(s)
if s=='WA':
WA.append(s)
if s=='RE':
RE.append(s)
print('AC ×',len(AC))
print('WA ×',len(WA))
print('TLE ×',len(TLE))
print('RE ×',len(RE))
| s805011114 | Accepted | 161 | 16,436 | 310 | a=int(input())
AC=[]
WA=[]
TLE=[]
RE=[]
for i in range(a):
s=(input())
if s=='AC':
AC.append(s)
if s=='TLE':
TLE.append(s)
if s=='WA':
WA.append(s)
if s=='RE':
RE.append(s)
print('AC','x',len(AC))
print('WA','x',len(WA))
print('TLE','x',len(TLE))
print('RE','x',len(RE))
|
s695357232 | p03369 | u969848070 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 41 | In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen. | a = input()
b = a.count('o')
print(700+b) | s736514061 | Accepted | 17 | 2,940 | 46 | a = input()
b = a.count('o')
print(700+b*100)
|
s217912442 | p03449 | u668726177 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 184 | 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 will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel? | n = int(input())
upper = list(map(int, input().split()))
lower = list(map(int, input().split()))
ans = 0
for i in range(n):
ans = max(ans, sum(upper[:i])+sum(lower[i:]))
print(ans) | s841915197 | Accepted | 18 | 3,060 | 186 | n = int(input())
upper = list(map(int, input().split()))
lower = list(map(int, input().split()))
ans = 0
for i in range(n):
ans = max(ans, sum(upper[:i+1])+sum(lower[i:]))
print(ans) |
s077626551 | p03457 | u065578867 | 2,000 | 262,144 | Wrong Answer | 365 | 11,816 | 552 | AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan. | n = int(input())
t = [0]
x = [0]
y = [0]
count = 0
for i in range(n):
val_list = input().split()
if val_list:
t.append(int(val_list[0]))
x.append(int(val_list[1]))
y.append(int(val_list[2]))
for i in range(n):
distance = abs(x[i + 1] - x[i]) + abs(y[i + 1] - y[i])
if distance > t[i + 1] - t[i]:
break
elif distance == t[i + 1] - t[i]:
count += 1
elif distance - (t[i + 1] - t[i]) % 2 == 0:
count += 1
else:
break
if count == n:
print('YES')
else:
print('NO')
| s990864447 | Accepted | 357 | 11,816 | 552 | n = int(input())
t = [0]
x = [0]
y = [0]
count = 0
for i in range(n):
val_list = input().split()
if val_list:
t.append(int(val_list[0]))
x.append(int(val_list[1]))
y.append(int(val_list[2]))
for i in range(n):
distance = abs(x[i + 1] - x[i]) + abs(y[i + 1] - y[i])
if distance > t[i + 1] - t[i]:
break
elif distance == t[i + 1] - t[i]:
count += 1
elif distance - (t[i + 1] - t[i]) % 2 == 0:
count += 1
else:
break
if count == n:
print('Yes')
else:
print('No')
|
s734352756 | p02399 | u072053884 | 1,000 | 131,072 | Wrong Answer | 40 | 7,676 | 113 | 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 = [int(x) for x in input().split(' ')]
d = a // b
r = a % b
f = a / b
print("{0} {1} {2}".format(d, r, f)) | s946029396 | Accepted | 20 | 7,692 | 117 | a, b = [int(x) for x in input().split(' ')]
d = a // b
r = a % b
f = a / b
print("{0} {1} {2:.5f}".format(d, r, f)) |
s285382268 | p03545 | u610042046 | 2,000 | 262,144 | Wrong Answer | 29 | 9,156 | 363 | Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted. | num = input()
ls = [0] * 3
for i in range(2 ** 3):
make7 = int(num[0])
for j in range(3):
if (i >> j) & 1:
make7 += int(num[j+1])
ls[j] = "+"
else:
make7 -= int(num[j+1])
ls[j] = "-"
if make7 == 7:
print(num[0] + ls[0] + num[1] + ls[1] + num[2] + ls[2] + num[3])
exit()
| s664576696 | Accepted | 29 | 9,148 | 370 | num = input()
ls = [0] * 3
for i in range(2 ** 3):
make7 = int(num[0])
for j in range(3):
if (i >> j) & 1:
make7 += int(num[j+1])
ls[j] = "+"
else:
make7 -= int(num[j+1])
ls[j] = "-"
if make7 == 7:
print(num[0] + ls[0] + num[1] + ls[1] + num[2] + ls[2] + num[3] + "=7")
exit()
|
s823097345 | p03471 | u985949234 | 2,000 | 262,144 | Wrong Answer | 1,191 | 3,060 | 347 | The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough. | N,Y = map(int, input().split())
f = False
for i in range(N+1):
for j in range(N+1-i):
if (Y-(10000*i + 5000*j))//1000 <=N-(i+j) and Y-(10000*i + 5000*j) >= 0:
f = True
anls = [i,j,(Y-(10000*i + 5000*j))//1000]
if f:
print(anls[0],anls[1],anls[2])
else:
print(-1,-1,-1)
| s360555598 | Accepted | 738 | 3,060 | 326 | N,Y = map(int, input().split())
f = False
for i in range(N+1):
for j in range(N+1-i):
if (Y-(10000*i + 5000*j))//1000 ==N-(i+j) and (Y-(10000*i + 5000*j))>=0:
f = True
anls = [i,j,N-(i+j)]
if f:
print(anls[0],anls[1],anls[2])
else:
print(-1,-1,-1)
|
s094766550 | p03131 | u211160392 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 183 | 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 maximum possible number of biscuits in Snuke's pocket after K operations. | K,A,B = (int(i) for i in input().split())
if A+2>=B:
print(1+K)
else:
Bis = A
K-=A-1
print(K)
Bis += int(K/2)*(B-A)
if(K)%2==1:
Bis+=1
print(Bis)
| s864822833 | Accepted | 17 | 2,940 | 170 | K,A,B = (int(i) for i in input().split())
if A+2>=B:
print(1+K)
else:
Bis = A
K-=A-1
Bis += int(K/2)*(B-A)
if(K)%2==1:
Bis+=1
print(Bis)
|
s977564221 | p03730 | u821588465 | 2,000 | 262,144 | Wrong Answer | 29 | 9,128 | 133 | We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`. | a, b, c = map(int, input().split())
if any((a * i) % b == c for i in range(1, b+1)):
print('Yes')
else:
print('No')
| s502458608 | Accepted | 29 | 9,136 | 129 | a, b, c = map(int, input().split())
if any((a * i) % b == c for i in range(1, b+1)):
print('YES')
else:
print('NO')
|
s922127524 | p03495 | u815797488 | 2,000 | 262,144 | Time Limit Exceeded | 2,104 | 26,132 | 207 | Takahashi has N balls. Initially, an integer A_i is written on the i-th ball. He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls. Find the minimum number of balls that Takahashi needs to rewrite the integers on them. | N,K = map(int, input().split())
A = [int(i) for i in input().split()]
l=[]
for i in range(N+1):
l.append(A.count(i))
ls = sorted(l,reverse=True)
j= K
ans = 0
while ls[j] != 0:
ans += ls[j]
print(ans) | s491903719 | Accepted | 104 | 32,564 | 194 | import collections as co
N,K = map(int, input().split())
A = list(map(int, input().split()))
ldic = co.Counter(A)
l = list(ldic.values())
ls = sorted(l,reverse=True)
print(sum(ls[K:])) |
s674061442 | p03563 | u962718741 | 2,000 | 262,144 | Wrong Answer | 26 | 9,088 | 38 | Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it. | print(2 * int(input()) - int(input())) | s169734813 | Accepted | 27 | 9,180 | 39 | print(-int(input()) + 2 * int(input())) |
s314547755 | p03470 | u050428930 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 159 | An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have? | N,ans,x=int(input()),0,0
s=sorted([int(input()) for i in range(N)])
print(s)
for i in range(N):
if s[i]>x:
ans+=1
x=s[i]
print(ans) | s083584223 | Accepted | 17 | 3,060 | 150 | N,ans,x=int(input()),0,0
s=sorted([int(input()) for i in range(N)])
for i in range(N):
if s[i]>x:
ans+=1
x=s[i]
print(ans) |
s844892724 | p03711 | u408791346 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 198 | Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group. | a = [1,3, 5,7,8,10,12]
b = [4,6,9,11]
c = [2]
x, y = map(int,input().split())
for i in a,b:
if x in i and y in i:
print('yes')
break
else:
print('no')
break | s901055731 | Accepted | 17 | 3,060 | 254 | a = [1,3, 5,7,8,10,12]
b = [4,6,9,11]
c = [2]
x, y = map(int,input().split())
ans = []
for i in a,b:
if x in i and y in i:
ans.append('Yes')
else:
ans.append('No')
if 'Yes' in ans:
print('Yes')
else:
print('No') |
s507409527 | p03711 | u701318346 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 133 | Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group. | x, y = input().split()
g = ['4', '6', '9', '11']
if (x == y == 2) or g.count(x) == g.count(y) == 1:
print('Yes')
else:
print('No') | s175611027 | Accepted | 17 | 3,060 | 248 | x, y = input().split()
a = ['4', '6', '9', '11']
b = ['1', '3', '5', '7', '8', '10', '12']
if (x == y == '2'):
print('Yes')
elif (a.count(x) == a.count(y) == 1):
print('Yes')
elif (b.count(x) == b.count(y) == 1):
print('Yes')
else:
print('No') |
s432774667 | p02694 | u603253967 | 2,000 | 1,048,576 | Wrong Answer | 30 | 10,120 | 530 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time? | from collections import deque, Counter, defaultdict
from itertools import chain, combinations
import json
# import numpy as np
import bisect
import sys
import math
import bisect
from functools import lru_cache
sys.setrecursionlimit(10 ** 8)
M = 10 ** 9 + 7
INF = 10 ** 17
def main():
N = int(input())
# A = [int(a) for a in input().split()]
c = 1
d = 100
while True:
d = int(d * 1.01)
if d > N:
print(c)
return
c += 1
if __name__ == "__main__":
main()
| s108865193 | Accepted | 29 | 10,120 | 531 | from collections import deque, Counter, defaultdict
from itertools import chain, combinations
import json
# import numpy as np
import bisect
import sys
import math
import bisect
from functools import lru_cache
sys.setrecursionlimit(10 ** 8)
M = 10 ** 9 + 7
INF = 10 ** 17
def main():
N = int(input())
# A = [int(a) for a in input().split()]
c = 1
d = 100
while True:
d = int(d * 1.01)
if d >= N:
print(c)
return
c += 1
if __name__ == "__main__":
main()
|
s122801027 | p03493 | u898058223 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 61 | Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble. | s=input()
ans=0
for i in range(3):
if s[i]=="1":
ans+=1 | s734063479 | Accepted | 17 | 2,940 | 72 | s=input()
ans=0
for i in range(3):
if s[i]=="1":
ans+=1
print(ans) |
s890354000 | p00017 | u567380442 | 1,000 | 131,072 | Wrong Answer | 30 | 6,720 | 399 | In cryptography, Caesar cipher is one of the simplest and most widely known encryption method. Caesar cipher is a type of substitution cipher in which each letter in the text is replaced by a letter some fixed number of positions down the alphabet. For example, with a shift of 1, 'a' would be replaced by 'b', 'b' would become 'c', 'y' would become 'z', 'z' would become 'a', and so on. In that case, a text: this is a pen is would become: uijt jt b qfo Write a program which reads a text encrypted by Caesar Chipher and prints the corresponding decoded text. The number of shift is secret and it depends on datasets, but you can assume that the decoded text includes any of the following words: "the", "this", or "that". | import sys
import string
f = sys.stdin
sentence = ''.join([line for line in f])
ceasar1 = str.maketrans(string.ascii_lowercase, string.ascii_lowercase[1:] + string.ascii_lowercase[:1])
for i in range(len(string.ascii_lowercase)):
sentence = sentence.translate(ceasar1)
for word in ['the', 'this', 'that']:
if sentence.find(word) != -1:
print(sentence)
break | s201085568 | Accepted | 30 | 6,724 | 388 | import sys
import string
f = sys.stdin
ceasar1 = str.maketrans(string.ascii_lowercase, string.ascii_lowercase[1:] + string.ascii_lowercase[:1])
for sentence in f:
for i in range(len(string.ascii_lowercase)):
sentence = sentence.translate(ceasar1)
if 'the' in sentence or 'this' in sentence or 'that' in sentence:
print(sentence, end='')
break |
s827675389 | p03068 | u840988663 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 166 | You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`. | n=input()
m=input()
l=int(input())
kotae=""
key=m[l-1]
for i in range(len(m)):
if m[i] != key:
kotae=kotae+"*"
else:
kotae=kotae+key
| s263169764 | Accepted | 17 | 3,060 | 175 | n=input()
m=input()
l=int(input())
kotae=""
key=m[l-1]
for i in range(len(m)):
if m[i] != key:
kotae=kotae+"*"
else:
kotae=kotae+key
print(kotae) |
s456161235 | p03816 | u047816928 | 2,000 | 262,144 | Wrong Answer | 91 | 17,400 | 210 | Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written. He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept. Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck. | N = int(input())
A = list(map(int, input().split()))
D = {}
for a in A:
if a not in D: D[a] = 0
D[a] += 1
n2 = 0
for k in D:
D[k] = 2 if D[k]%2==0 else 1
if D[k]==2: n2+=1
print(len(D) + n2%2) | s857062802 | Accepted | 93 | 18,272 | 210 | N = int(input())
A = list(map(int, input().split()))
D = {}
for a in A:
if a not in D: D[a] = 0
D[a] += 1
n2 = 0
for k in D:
D[k] = 2 if D[k]%2==0 else 1
if D[k]==2: n2+=1
print(len(D) - n2%2) |
s077752644 | p03679 | u071061942 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 124 | Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache. | X,A,B = map(int,input().split())
if B-A <= 0:
print("delicious")
elif B-A >= X:
print("safe")
else:
print("dangerous") | s660336828 | Accepted | 17 | 3,060 | 124 | X,A,B = map(int,input().split())
if B-A <= 0:
print("delicious")
elif B-A <= X:
print("safe")
else:
print("dangerous") |
s783945213 | p02613 | u268674745 | 2,000 | 1,048,576 | Wrong Answer | 154 | 16,316 | 357 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | n = int(input())
ac = []
wa = []
tle = []
re = []
for _ in range(n):
i = input()
if i == "WA":
wa.append(i)
elif i == "AC":
ac.append(i)
elif i == "TLE":
tle.append(i)
else:
re.append(i)
print("AC ×", str(len(ac)))
print("WA ×", str(len(wa)))
print("TLE ×", str(len(tle)))
print("RE ×", str(len(re))) | s670311216 | Accepted | 153 | 16,328 | 333 | n = int(input())
ac = []
wa = []
tle = []
re = []
for _ in range(n):
i = input()
if i == "WA":
wa.append(i)
elif i == "AC":
ac.append(i)
elif i == "TLE":
tle.append(i)
else:
re.append(i)
print("AC x", len(ac))
print("WA x", len(wa))
print("TLE x", len(tle))
print("RE x", len(re)) |
s782807743 | p03023 | u252828980 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 27 | Given an integer N not less than 3, find the sum of the interior angles of a regular polygon with N sides. Print the answer in degrees, but do not print units. | print((int(input())-3)*180) | s107413161 | Accepted | 18 | 2,940 | 27 | print((int(input())-2)*180) |
s255376550 | p02612 | u208512038 | 2,000 | 1,048,576 | Wrong Answer | 32 | 9,140 | 33 | 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. | x = int(input())
print(x % 1000) | s595269052 | Accepted | 29 | 9,156 | 81 | x = int(input())
a = x % 1000
if a == 0:
print("0")
exit()
print(1000 - a) |
s643809168 | p03493 | u041075929 | 2,000 | 262,144 | Wrong Answer | 20 | 2,940 | 255 | 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. | import sys, os
f = lambda:list(map(int,input().split()))
if 'local' in os.environ :
sys.stdin = open('./input.txt', 'r')
def solve():
s = input()
ans = 0
for i in s:
if i == '0':
ans += 1
print(ans)
solve()
| s886825699 | Accepted | 17 | 2,940 | 255 | import sys, os
f = lambda:list(map(int,input().split()))
if 'local' in os.environ :
sys.stdin = open('./input.txt', 'r')
def solve():
s = input()
ans = 0
for i in s:
if i == '1':
ans += 1
print(ans)
solve()
|
s878720571 | p02601 | u337949146 | 2,000 | 1,048,576 | Wrong Answer | 26 | 9,208 | 237 | M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful. | import math
A, B, C = list(map(int, input().split()))
K = int(input())
#A, B, C = [7, 4, 2]
#K = 3
i = math.floor(A/B)
K -= i
B = B * (2 ** i)
j = math.floor(B/C)
K -= j
C = C * (2 ** j)
if K >= 0:
print("Yes")
else:
print("No") | s758466740 | Accepted | 30 | 9,184 | 281 | import math
A, B, C = list(map(int, input().split()))
K = int(input())
#A, B, C = [7, 6, 100]
#K = 3
i = max(math.ceil(math.log2((A+1)/B)),0)
K -= i
B = B * (2 ** i)
j = max(math.ceil(math.log2((B+1)/C)),0)
K -= j
C = C * (2 ** j)
if K >= 0:
print("Yes")
else:
print("No") |
s529575552 | p03711 | u290326033 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 179 | Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group. | group = [[1, 3, 5, 7, 8, 10, 12],[4, 6, 9, 11],[2]]
x, y = list(map(int, input().split()))
for g in group:
if x in g and y in g:
print("Yes")
break
print("No") | s051952518 | Accepted | 17 | 3,060 | 199 | import sys
group = [[1, 3, 5, 7, 8, 10, 12],[4, 6, 9, 11],[2]]
x, y = list(map(int, input().split()))
for g in group:
if (x in g) and (y in g):
print("Yes")
sys.exit()
print("No") |
s567191654 | p00512 | u221679506 | 8,000 | 131,072 | Wrong Answer | 20 | 7,644 | 148 | ある工場では，各営業所から製品生産の注文を受けている． 前日の注文をまとめて，各製品の生産合計を求めたい． 入力ファイルの１行目には注文データの数 n が書いてあり， 続く n 行には製品名と注文数が空白で区切られて書いてある． 製品名は5文字以内の英大文字で書かれている． 注文データには同じ製品が含まれていることもあり，順序はバラバラである． この注文データの中に現れる同じ製品の注文数を合計し， 出力ファイルに製品名と合計を空白を区切り文字として出力しなさい． ただし，製品名に次の順序を付けて，その順で出力すること． 順序：文字の長さの小さい順に，同じ長さのときは，前から比べて 最初に異なる文字のアルファベット順とする． 入力データにおける製品数，注文数とその合計のどれも106以下である． 出力ファイルにおいては， 出力の最後の行にも改行コードを入れること． | data = {}
for i in range(int(input())):
p,n = input().split();
data[p] = data.get(p,0) + int(n)
for i, j in sorted(data.items()):
print (i, j) | s283679870 | Accepted | 30 | 7,692 | 175 | data = {}
for i in range(int(input())):
p,n = input().split();
data[p] = data.get(p,0) + int(n)
for i, j in sorted([[len(a),a] for a in data.keys()]):
print (j, data[j]) |
s644507151 | p02694 | u090406054 | 2,000 | 1,048,576 | Time Limit Exceeded | 2,206 | 8,900 | 87 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time? | x=int(input())
times=0
money=100
while money<x:
money*1.01
times+=1
print(times) | s534185034 | Accepted | 25 | 9,120 | 74 | x=int(input())
t=100
cnt=0
while t<x:
t=t+t//100
cnt+=1
print(cnt)
|
s378941049 | p02422 | u801346721 | 1,000 | 131,072 | Wrong Answer | 20 | 7,564 | 304 | Write a program which performs a sequence of commands to a given string $str$. The command is one of: * print a b: print from the a-th character to the b-th character of $str$ * reverse a b: reverse from the a-th character to the b-th character of $str$ * replace a b p: replace from the a-th character to the b-th character of $str$ with p Note that the indices of $str$ start with 0. | s = input()
n = int(input())
for i in range(n):
a = list(input().split())
if a[0] == 'replace':
temp = s[int(a[1]):int(a[2])+1]
s.replace(temp, a[3])
elif a[0] == 'reverse':
temp = s[int(a[1]):int(a[2])+1]
s.replace(temp, temp[::-1])
elif a[0] == 'print':
print(s[int(a[1]):int(a[2])+1])
| s695139508 | Accepted | 30 | 7,696 | 381 | s = input()
n = int(input())
for i in range(n):
a = list(input().split())
temp = s[int(a[1]):int(a[2])+1]
if a[0] == 'replace':
temp3 = a[3]
s = list(s)
seq = 0
for i in range(int(a[1]), int(a[2])+1):
s[i] = temp3[seq]
seq += 1
s = ''.join(s)
elif a[0] == 'reverse':
temp2 = temp[::-1]
s = s.replace(temp, temp2, 1)
elif a[0] == 'print':
print(temp)
|
s363795601 | p03563 | u048945791 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 53 | Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it. | r = int(input())
g = int(input())
print((r + g) / 2) | s888982426 | Accepted | 17 | 2,940 | 49 | r = int(input())
g = int(input())
print(2*g - r) |
s405291959 | p04029 | u068584789 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 39 | There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total? | n = int(input())
print(n * (n + 1) / 2) | s943332556 | Accepted | 20 | 3,060 | 75 | n = int(input())
if n == 1:
print(n)
else:
print(int(n * (n + 1) / 2))
|
s337204118 | p03730 | u407016024 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 142 | We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`. | A, B, C = map(int, input().split())
for i in range(A, A*B, A):
print(i)
if i % B == C:
print('Yes')
exit()
print('No') | s230829642 | Accepted | 17 | 2,940 | 129 | A, B, C = map(int, input().split())
for i in range(A, A*B, A):
if i % B == C:
print('YES')
exit()
print('NO') |
s159147184 | p00044 | u024715419 | 1,000 | 131,072 | Wrong Answer | 120 | 5,908 | 483 | 素数というのは、1 よりも大きくそれ自身か 1 でしか割りきれない整数をいいます。例えば、2 は、2 と 1 でしか割り切れないので素数ですが、12 は、12 と 1 のほかに、2, 3, 4, 6 で割りきれる数なので素数ではありません。 整数 n を入力したとき、n より小さい素数のうち最も大きいものと、n より大きい素数のうち最も小さいものを出力するプログラムを作成してください。 | n = 50000
c = [1 for i in range(n)]
c[0] = 0
i = 2
while i**2 <= n:
j = i*2
while j <= n:
c[j - 1] = 0
j += i
i += 1
while True:
try:
n = int(input())
i, a, b = 0, 0, 0
while a == 0 and b == 0:
i += 1
if n - i >= 0 and a == 0 and c[n - i] == 1:
a = n - i
if n + i < 50000 and b == 0 and c[n + i] == 1:
b = n + i
print(a,b)
except:
break
| s955989752 | Accepted | 120 | 5,908 | 498 | n = 50021
c = [1 for i in range(n)]
c[0] = 0
i = 2
while i**2 <= n:
j = i*2
while j <= n:
c[j - 1] = 0
j += i
i += 1
while True:
try:
n = int(input())
i, a, b = 0, 0, 0
while a == 0 or b == 0:
i += 1
if n - i - 1 >= 0 and a == 0 and c[n - i - 1] == 1:
a = n - i
if n + i - 1 < 50021 and b == 0 and c[n + i - 1] == 1:
b = n + i
print(a,b)
except:
break
|
s618370389 | p00079 | u032662562 | 1,000 | 131,072 | Wrong Answer | 30 | 7,528 | 583 | 凸 n 角形（すべての頂点の内角が 180 度未満である多角形、要するにへこんでいない多角形のこと）の頂点の座標を読んで、その面積を出力するプログラムを作成してください。頂点には辺のつながりの順序に従って頂点 1、頂点 2、頂点 3、‥‥頂点 n という名前がついています。 ただし、 n は 3 以上 20 以下とします。また、次の、三角形の 3 辺の長さ a, b, c から面積 S を求める公式を使ってもかまいません。 | import math
f = []
while True:
try:
st = input().strip().split(',')
print("st=",st)
x,y = list(map(float,st ))
f.append(x + y*1j)
except EOFError:
break
px = [p.real for p in f]
ox = (max(px) + min(px)) / 2.0
py = [p.imag for p in f]
oy = (max(py) + min(py)) / 2.0
fo = ox + oy * 1j
ss = 0.0
for i in range(len(f)):
if i == len(f)-1:
j = 0
else:
j = i + 1
a = abs(f[i]-fo)
b = abs(f[j]-fo)
c = abs(f[j]-f[i])
z = (a + b + c) / 2.0
s = math.sqrt(z*(z-a)*(z-b)*(z-c))
ss += s
print(ss) | s886513006 | Accepted | 30 | 7,484 | 334 | def op(u,v):
return (complex.conjugate(u)*v).imag
f = []
while True:
try:
st = input().strip().split(',')
x,y = list(map(float,st ))
f.append(x + y*1j)
except EOFError:
break
s = 0.0
fo = f[0]
for j in range(2,len(f)):
i = j - 1
s += op(f[i]-fo,f[j]-fo)/2.0
print("%.6f" % abs(s)) |
s284718611 | p03455 | u947748301 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 113 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. |
num1,num2 = map(int,"3 4".split())
if ((num1 * num2) % 2) == 0:
print("Odd")
else:
print("Even") | s234753157 | Accepted | 17 | 3,064 | 116 |
num1,num2 = map(int,input().split())
if ((num1 * num2) % 2) == 0:
print("Even")
else:
print("Odd")
|
s808373506 | p02393 | u781194524 | 1,000 | 131,072 | Wrong Answer | 20 | 5,600 | 246 | Write a program which reads three integers, and prints them in ascending order. | a,b,c=[int(x) for x in input().split()]
if a > b:
if a > c:
if b < c:
print(b,c,a,sep=" ")
else:
print(c,b,a,sep=" ")
elif b < c:
print(a,b,c,sep=" ")
else:
print(a,c,b,sep=" ")
| s264999662 | Accepted | 20 | 5,604 | 224 | a,b,c=[int(x) for x in input().split()]
if a<=b<=c:
print(a,b,c)
elif a<=c<=b:
print(a,c,b)
elif b<=a<=c:
print(b,a,c)
elif b<=c<=a:
print(b,c,a)
elif c<=a<=b:
print(c,a,b)
elif c<=b<=a:
print(c,b,a)
|
s621794027 | p02613 | u335599768 | 2,000 | 1,048,576 | Wrong Answer | 153 | 16,244 | 302 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | n = int(input());
s = [input() for i in range(n)]
ac = wa = tle = re = 0
for s_kind in s:
if s_kind == "AC":
ac += 1
elif s_kind == "WA":
wa += 1
elif s_kind == "TLE":
tle += 1
elif s_kind == "RE":
re += 1
print(ac)
print(wa)
print(tle)
print(re) | s895919293 | Accepted | 146 | 16,328 | 331 | n = int(input());
s = [input() for i in range(n)]
ac = wa = tle = re = 0
for s_kind in s:
if s_kind == "AC":
ac += 1
elif s_kind == "WA":
wa += 1
elif s_kind == "TLE":
tle += 1
elif s_kind == "RE":
re += 1
print("AC x",ac)
print("WA x",wa)
print("TLE x",tle)
print("RE x",re) |
s643323965 | p02669 | u469254913 | 2,000 | 1,048,576 | Wrong Answer | 440 | 11,300 | 1,008 | You start with the number 0 and you want to reach the number N. You can change the number, paying a certain amount of coins, with the following operations: * Multiply the number by 2, paying A coins. * Multiply the number by 3, paying B coins. * Multiply the number by 5, paying C coins. * Increase or decrease the number by 1, paying D coins. You can perform these operations in arbitrary order and an arbitrary number of times. What is the minimum number of coins you need to reach N? **You have to solve T testcases.** | # import numpy as np
# import math
# import copy
# from collections import deque
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10000000)
memo = {}
def solve():
N,A,B,C,D = map(int,input().split())
return process(N,N,A,B,C,D)
def process(M,N,A,B,C,D):
if M == 1:
return D
elif M == 0:
return 0
elif M < 0:
return D * N
global memo
if M in memo:
return memo[M]
res = process((M-M%2)//2,N,A,B,C,D) + A + (M%2) * D
res = min(res,process((M+M%2)//2,N,A,B,C,D) + A + (M%2) * D)
res = min(res,process((M-M%3)//3,N,A,B,C,D) + B + (M%3) * D)
res = min(res,process((M+3-M%3)//3,N,A,B,C,D) + B + (3-M%3) * D)
res = min(res,process((M-M%5)//5,N,A,B,C,D) + C + (M%5) * D)
res = min(res,process((M+5-M%5)//5,N,A,B,C,D) + C + (5-M%5) * D)
memo[M] = res
return res
def main():
T = int(input())
for i in range(T):
global memo
memo = {}
res = solve()
print(res)
main()
| s839963467 | Accepted | 402 | 11,204 | 970 | # import numpy as np
# import math
# import copy
# from collections import deque
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10000000)
memo = {}
def solve():
N,A,B,C,D = map(int,input().split())
global memo
memo = {0:0,1:D}
return process(N,N,A,B,C,D)
def process(M,N,A,B,C,D):
# if M < 0:
# return D * N + 1
global memo
if M in memo:
return memo[M]
res = process((M-M%2)//2,N,A,B,C,D) + A + (M%2) * D
res = min(res,process((M+M%2)//2,N,A,B,C,D) + A + (M%2) * D)
res = min(res,process((M-M%3)//3,N,A,B,C,D) + B + (M%3) * D)
res = min(res,process((M+3-M%3)//3,N,A,B,C,D) + B + (3-M%3) * D)
res = min(res,process((M-M%5)//5,N,A,B,C,D) + C + (M%5) * D)
res = min(res,process((M+5-M%5)//5,N,A,B,C,D) + C + (5-M%5) * D)
res = min(res,M*D)
memo[M] = res
return res
def main():
T = int(input())
for i in range(T):
res = solve()
print(res)
main()
|
s015700713 | p03795 | u314050667 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 87 | Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y. | n = int(input())
ans = 1
for i in range(1,n+1):
ans = (ans*i)%1000000007
print(ans) | s005989659 | Accepted | 17 | 2,940 | 81 | N = int(input())
div, mod = divmod(N, 15)
x = N * 800
y = div * 200
print(x - y) |
s058466919 | p03229 | u227082700 | 2,000 | 1,048,576 | Wrong Answer | 257 | 7,764 | 197 | You are given N integers; the i-th of them is A_i. Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like. | n=int(input())
a=[int(input())for _ in range(n)]
a.sort()
b=a[:0--n//2]
c=a[0--n//2:]
a=[]
for i in range(n//2):a+=[b[i],c[i]]
if n%2:a+=[b[-1]]
x=0
for i in range(n-1):x+=abs(a[i+1]-a[i])
print(x) | s353883304 | Accepted | 316 | 9,912 | 350 | n=int(input())
a=[int(input())for _ in range(n)]
if n==2:print(abs(a[0]-a[1]));exit()
f=[-1]
for i in range(1,n-1-(n%2==0)):
if i%2:f.append(2)
else:f.append(-2)
if n%2:f+=[-1]
else:f+=[-2,1]
b=[]
c=[]
for i in f:
b.append(i)
c.append(-i)
b.sort()
c.sort()
sb=sc=0
a.sort()
for i in range(n):
sb+=a[i]*b[i]
sc+=a[i]*c[i]
print(max(sb,sc)) |
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