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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s916121210
|
p02612
|
u366974168
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,084 | 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())
x=N%1000
if x==1000:
print(0)
else:
print(x)
|
s529247983
|
Accepted
| 31 | 9,104 | 67 |
N=int(input())
x=N%1000
if x==0:
print(0)
else:
print(1000-x)
|
s947499475
|
p03386
|
u782733831
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 99 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,k=map(int,input().split())
r=range(a,b+1)
for i in sorted(set(r[:3])|set(r[-3:])):
print(i)
|
s070246919
|
Accepted
| 18 | 3,060 | 99 |
a,b,k=map(int,input().split())
r=range(a,b+1)
for i in sorted(set(r[:k])|set(r[-k:])):
print(i)
|
s728502495
|
p02612
|
u114099505
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,144 | 62 |
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%1000==0:
print(0)
print(1000-1000%N)
|
s928032882
|
Accepted
| 29 | 9,096 | 103 |
N = int(input())
if N%1000==0:
print(0)
elif N<1000:
print(1000-N)
else:
print(1000-N%1000)
|
s628082603
|
p03228
|
u319805917
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,064 | 241 |
In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.
|
A,B,K=map(int,input().split())
count=1
while count!=K:
if A%2!=0:
A-=1
A,B=A/2,A/2+B
count+=1
A,B=B,A
if count%2==0:
print(int(A),int(B))
else:
print(int(B),int(A))
|
s586518110
|
Accepted
| 17 | 3,060 | 241 |
A,B,K=map(int,input().split())
count=0
while count!=K:
if A%2!=0:
A-=1
A,B=A/2,A/2+B
count+=1
A,B=B,A
if count%2==0:
print(int(A),int(B))
else:
print(int(B),int(A))
|
s991991105
|
p03963
|
u050622763
| 2,000 | 262,144 |
Wrong Answer
| 119 | 27,236 | 179 |
There are N balls placed in a row. AtCoDeer the deer is painting each of these in one of the K colors of his paint cans. For aesthetic reasons, any two adjacent balls must be painted in different colors. Find the number of the possible ways to paint the balls.
|
def ABC046_B():
import numpy as np
n,k = map(int,input().split())
print(n,k)
for i in range(n):
ans = k*np.power(k-1,n-1)
print(ans)
ABC046_B()
|
s356194952
|
Accepted
| 117 | 27,004 | 165 |
def ABC046_B():
import numpy as np
n,k = map(int,input().split())
for i in range(n):
ans = k*np.power(k-1,n-1)
print(ans)
ABC046_B()
|
s630626849
|
p03486
|
u418826171
| 2,000 | 262,144 |
Wrong Answer
| 31 | 9,156 | 357 |
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s = sorted(input())
t = sorted(input())
flag1 = True
flag2 = False
if len(s) < len(t):
for i in range(len(s)):
if s[i] != t[i]:
flag1 = False
break
else:
flag1 = False
for i in range(min(len(s), len(t))):
if s[i] < t[i]:
flag2 = True
break
if flag1 or flag2:
print('Yes')
else:
print('No')
|
s148632727
|
Accepted
| 31 | 9,096 | 103 |
s = sorted(input())
t = sorted(input(),reverse = True)
if s < t:
print('Yes')
else:
print('No')
|
s425621796
|
p02694
|
u301679431
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 9,080 | 68 |
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())
Y=100
n=0
while Y<=X:
Y=int(Y*1.01)
n+=1
print(n)
|
s489427437
|
Accepted
| 22 | 9,000 | 67 |
X=int(input())
Y=100
n=0
while Y<X:
Y=int(Y*1.01)
n+=1
print(n)
|
s730197311
|
p03485
|
u102960641
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 65 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
import math
a,b = map(int, input().split())
print(math.ceil(a+b))
|
s605939302
|
Accepted
| 18 | 2,940 | 70 |
import math
a,b = map(int, input().split())
print(math.ceil((a+b)/2))
|
s885396544
|
p03997
|
u334712262
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 66 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)//2)
|
s092977576
|
Accepted
| 17 | 2,940 | 68 |
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h//2)
|
s214851406
|
p03069
|
u166306121
| 2,000 | 1,048,576 |
Wrong Answer
| 202 | 16,692 | 1,211 |
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 that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
import sys
# gcd
from fractions import gcd
# from math import ceil, floor
# from copy import deepcopy
from itertools import accumulate
# l = ['a', 'b', 'b', 'c', 'b', 'a', 'c', 'c', 'b', 'c', 'b', 'a']
# print(S.most_common(2)) # [('b', 5), ('c', 4)]
# from collections import Counter
#
input = sys.stdin.readline
def ii(): return int(input())
def mi(): return map(int, input().rstrip().split())
def lmi(): return list(map(int, input().rstrip().split()))
def li(): return list(input().rstrip())
# template
N = ii()
S = li()
intS = [0] * N
for i in range(N):
if S[i] == '#':
intS[i] = 1
# print(intS)
acuS = list(accumulate(intS))
# print(acuS)
minS = 10 ** 9
for i in range(N):
minS = min(minS, acuS[i] + (N-1-i) - (acuS[N - 1] - acuS[i]))
# print(minS)
print(minS)
|
s457549277
|
Accepted
| 199 | 14,368 | 1,208 |
import sys
# gcd
# from fractions import gcd
# from math import ceil, floor
# from copy import deepcopy
from itertools import accumulate
# l = ['a', 'b', 'b', 'c', 'b', 'a', 'c', 'c', 'b', 'c', 'b', 'a']
# print(S.most_common(2)) # [('b', 5), ('c', 4)]
# from collections import Counter
input = sys.stdin.readline
def ii(): return int(input())
def mi(): return map(int, input().rstrip().split())
def lmi(): return list(map(int, input().rstrip().split()))
def li(): return list(input().rstrip())
# template
N = ii()
S = li()
intS = [0] * N
for i in range(N):
if S[i] == '#':
intS[i] = 1
# print(intS)
acuS = list(accumulate(intS))
# print(acuS)
minS = 10 ** 9
for i in range(N):
minS = min(minS, acuS[i] + (N-1-i) - (acuS[N - 1] - acuS[i]))
# print(minS)
print(min(minS, N - acuS[N-1]))
|
s174748603
|
p02975
|
u704563784
| 2,000 | 1,048,576 |
Wrong Answer
| 59 | 14,836 | 557 |
Snuke has N hats. The i-th hat has an integer a_i written on it. There are N camels standing in a circle. Snuke will put one of his hats on each of these camels. If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print `Yes`; otherwise, print `No`. * The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself. What is XOR? The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non- negative integers x_1, x_2, \ldots, x_n is defined as follows: - When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even. For example, 3 \oplus 5 = 6.
|
def count(l):
res = {}
for d in l:
if d in res:
res[d] += 1
else:
res[d] = 1
return res
n = float(input())
data = list(map(int, input().split()))
t = count(data)
if len(t) == 3:
if 0 not in t and all(i == n/3 for i in t.values()):
print('YES')
else:
print('NO')
elif len(t) == 2:
if 0 in t and t[0] == n/3:
print('YES')
else:
print('NO')
elif len(t) == 1:
if 0 in t:
print('YES')
else:
print('NO')
else:
print('NO')
|
s520956231
|
Accepted
| 58 | 14,116 | 606 |
def count(l):
res = {}
for d in l:
if d in res:
res[d] += 1
else:
res[d] = 1
return res
n = float(input())
data = list(map(int, input().split()))
t = count(data)
if len(t) == 3:
ks = list(t.keys())
if 0 not in t and all(i == n/3 for i in t.values()) and ks[0]^ks[1] == ks[2]:
print('Yes')
else:
print('No')
elif len(t) == 2:
if 0 in t and t[0] == n/3:
print('Yes')
else:
print('No')
elif len(t) == 1:
if 0 in t:
print('Yes')
else:
print('No')
else:
print('No')
|
s641092122
|
p03693
|
u588526762
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 316 |
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?
|
rgb_list = list(map(int, input().split()))
print(rgb_list)
rgb_list = map(str, rgb_list)
rgb = ''.join(rgb_list)
if int(rgb)%4 == 0:
print('YES')
else:
print('NO')
|
s492215392
|
Accepted
| 17 | 2,940 | 96 |
r,g,b = map(int,input().split())
if((10*g + b) % 4 == 0):
print('YES')
else:
print('NO')
|
s827505984
|
p03502
|
u732870425
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 141 |
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
N = int(input())
t = N + 0
c = 0
while 0 < t:
c += t % 10
t = t / 10
print("Yes" if N % c == 0 else "No")
|
s528970147
|
Accepted
| 18 | 2,940 | 142 |
N = int(input())
t = N + 0
c = 0
while 0 < t:
c += t % 10
t = t // 10
print("Yes" if N % c == 0 else "No")
|
s113561954
|
p03672
|
u539517139
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 150 |
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 deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
s=input()
if len(s)%2==0:
s=s[:len(s)-3]
else:
s=s[:len(s)-2]
while True:
l=len(s)
if s[:l//2-1]==s[l//2:]:
print(l)
break
s=s[:l-2]
|
s100986035
|
Accepted
| 17 | 3,060 | 148 |
s=input()
if len(s)%2==0:
s=s[:len(s)-2]
else:
s=s[:len(s)-1]
while True:
l=len(s)
if s[:l//2]==s[l//2:]:
print(l)
break
s=s[:l-2]
|
s771953645
|
p02936
|
u379142263
| 2,000 | 1,048,576 |
Wrong Answer
| 1,626 | 54,260 | 418 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
import sys
if __name__ == "__main__":
n,q = map(int,input().split())
edge = [[] for i in range(n)]
for i in range(n-1):
a,b = map(int,input().split())
a-=1
b-=1
edge[a].append(b)
edge[b].append(a)
cnt = [0]*n
for i in range(q):
p,x = map(int,input().split())
cnt[p-1] += x
for i in range(1,n):
cnt[i] += cnt[i-1]
print(*cnt)
|
s973188222
|
Accepted
| 1,164 | 55,200 | 730 |
import sys
import collections
INF = 10**18
def dfs(s,cnt):
visited = [False]*n
visited[s] = True
q = collections.deque([s])
while q:
p = q.pop()
visited[p] = True
for v in edge[p]:
if visited[v]:
continue
cnt[v] += cnt[p]
q.append(v)
return cnt
if __name__ == "__main__":
n,q = map(int,input().split())
edge = [[] for i in range(n)]
for i in range(n-1):
a,b = map(int,input().split())
a-=1
b-=1
edge[a].append(b)
edge[b].append(a)
cnt = [0]*n
for i in range(q):
v,x = map(int,input().split())
v -= 1
cnt[v] += x
ans = dfs(0,cnt)
print(*ans)
|
s284949634
|
p03470
|
u322826875
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 212 |
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
n = int(input ())
print(n)
d = [int(input()) for i in range(n)]
#print(d)
d2 = sorted(d)
#print(d2)
count=1
for i in range(n-1):
if( d2[i] < d2[i+1] ):
count += 1
print(count)
|
s715564226
|
Accepted
| 17 | 3,060 | 213 |
n = int(input ())
#print(n)
d = [int(input()) for i in range(n)]
#print(d)
d2 = sorted(d)
#print(d2)
count=1
for i in range(n-1):
if( d2[i] < d2[i+1] ):
count += 1
print(count)
|
s945097922
|
p00008
|
u273843182
| 1,000 | 131,072 |
Wrong Answer
| 30 | 5,584 | 167 |
Write a program which reads an integer n and identifies the number of combinations of a, b, c and d (0 ≤ a, b, c, d ≤ 9) which meet the following equality: a + b + c + d = n For example, for n = 35, we have 4 different combinations of (a, b, c, d): (8, 9, 9, 9), (9, 8, 9, 9), (9, 9, 8, 9), and (9, 9, 9, 8).
|
num = int(input())
count = 0
for a in range(10):
for b in range(10):
for c in range(10):
for d in range(10):
if a+b+c+d==num:
count+=1
print(int(count))
|
s106893394
|
Accepted
| 180 | 5,592 | 220 |
while True:
try:
num = int(input())
count = 0
for a in range(10):
for b in range(10):
for c in range(10):
for d in range(10):
if a+b+c+d==num:
count+=1
print(int(count))
except:
break
|
s832086175
|
p03024
|
u063073794
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 64 |
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
s=input()
if s.count("x")>=8:
print("YES")
else:
print("NO")
|
s602175391
|
Accepted
| 17 | 2,940 | 65 |
s=input()
if s.count("x")>=8:
print("NO")
else:
print("YES")
|
s755353339
|
p03251
|
u902468164
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 363 |
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
str1 = input();
str2 = str1.split(" ")
n = int(str2[0])
m = int(str2[1])
xcap = int(str2[2])
ycap = int(str2[3])
strx = input()
stry = input()
strx2 = strx.split(" ")
stry2 = stry.split(" ")
for i in range(n):
strx2[i] = int(strx2[i])
for i in range(m):
stry2[i] = int(stry2[i])
if max(strx2) >= min(stry2)-1:
print("War")
else:
print("No War")
|
s319005991
|
Accepted
| 17 | 3,064 | 381 |
str1 = input();
str2 = str1.split(" ")
n = int(str2[0])
m = int(str2[1])
xcap = int(str2[2])
ycap = int(str2[3])
strx = input()
stry = input()
strx2 = strx.split(" ")
stry2 = stry.split(" ")
for i in range(n):
strx2[i] = int(strx2[i])
for i in range(m):
stry2[i] = int(stry2[i])
if max(max(strx2),xcap) < min(min(stry2), ycap):
print("No War")
else:
print("War")
|
s094675261
|
p02612
|
u072284094
| 2,000 | 1,048,576 |
Wrong Answer
| 34 | 9,124 | 26 |
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.
|
print(int(input()) % 1000)
|
s538690814
|
Accepted
| 28 | 9,148 | 75 |
n = int(input()) % 1000
if n > 0:
print(1000 - n)
else:
print(0)
|
s415229050
|
p03502
|
u915214301
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 88 |
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
x = input()
fx = sum([int(num) for num in x])
print("YES" if int(x)%fx == 0 else "NO")
|
s571847933
|
Accepted
| 17 | 2,940 | 90 |
x = input()
fx = sum([int(num) for num in x])
print("Yes" if int(x)%fx == 0 else "No")
|
s054298874
|
p02936
|
u921811474
| 2,000 | 1,048,576 |
Wrong Answer
| 1,958 | 137,748 | 446 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
N, Q = map(int,input().split())
l = [list(map(int,input().split())) for _ in range(N-1)]
m = [list(map(int,input().split())) for _ in range(Q)]
ll = [0]*N
bb = [[i] for i in range(1,N+1)]
for i in l:
bb[i[0]-1].append(i[1])
for i in m:
ll[i[0]-1] += int(i[1])
ans = [0]*N
# print("bb=",bb)
# print("ll=",ll)
for i in range(N):
for k in bb[i]:
if k == i+1:
ans[i] += ll[i]
else:
ll[k-1] += ll[i]
|
s574592063
|
Accepted
| 1,268 | 36,896 | 341 |
n, q = map(int, input().split())
ki = [0] * (n + 1)
for i in range(n - 1):
a, b = map(int, input().split())
ki[b] = a
# print("ki=",ki)
c = [0] * (n + 1)
for i in range(q):
p, x = map(int, input().split())
c[p] += x
# print("c=",c)
for i in range(1, n + 1):
c[i] += c[ki[i]]
# print(c)
print(" ".join(map(str, c[1:])))
|
s913691444
|
p03359
|
u163320134
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 66 |
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a,b=map(int,input().split())
if a<b:
print(a)
else:
print(a-1)
|
s835177409
|
Accepted
| 19 | 3,060 | 67 |
a,b=map(int,input().split())
if a<=b:
print(a)
else:
print(a-1)
|
s995609846
|
p03416
|
u617440820
| 2,000 | 262,144 |
Wrong Answer
| 103 | 3,188 | 219 |
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
input_one = input().split()
i = int(input_one[0])
two = input_one[1]
#i = int(one)
while i <= int(two):
i = str(i)
if i[0] == i[4]:
if i[1] == i[3]:
print(int(i))
i = int(i) + 1
|
s662920184
|
Accepted
| 96 | 3,060 | 238 |
input_one = input().split()
i = int(input_one[0])
two = input_one[1]
count = 0
#i = int(one)
while i <= int(two):
i = str(i)
if i[0] == i[4]:
if i[1] == i[3]:
count += 1
i = int(i) + 1
print(count)
|
s002306271
|
p03545
|
u786020649
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,200 | 207 |
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.
|
tn=list(map(int,list(input())))
for i in range(2,16,2):
if sum(x*[1,-1][i>>k&1] for k,x in enumerate(tn))==7:
print(str(tn[0])+''.join(list(['+','-'][i>>k&1]+str(tn[k]) for k in range(1,4))))
break
|
s975911449
|
Accepted
| 29 | 9,136 | 221 |
tn=list(map(int,list(input())))
for i in range(0,16,2):
if sum(x*[1,-1][i>>k&1] for k,x in enumerate(tn))==7:
s=''.join(list(['+','-'][i>>k&1]+str(tn[k]) for k in range(1,4)))
print(str(tn[0])+s+'=7')
break
|
s636328417
|
p03131
|
u169138653
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 128 |
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=map(int,input().split())
ans=0
cnt=0
if b<=a+1 or a>k:
print(k+1)
else:
print(a+(k-a+1)*(b-a)//2+(k-a+1)*(b-a)%2)
|
s927749143
|
Accepted
| 17 | 2,940 | 121 |
k,a,b=map(int,input().split())
ans=0
cnt=0
if b<=a+1 or a>k:
print(k+1)
else:
print(a+((k-a+1)//2)*(b-a)+((k-a+1)%2))
|
s333838001
|
p03095
|
u405256066
| 2,000 | 1,048,576 |
Wrong Answer
| 293 | 22,640 | 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 concatenation of **one or more** characters from the string without changing the order.
|
import numpy as np
from sys import stdin
N=int(stdin.readline().rstrip())
S=(stdin.readline().rstrip())
S=list(S)
ans=N**2-2**(((N-len(set(S))))+1)
print(ans%(10**9+7))
|
s355429193
|
Accepted
| 60 | 3,956 | 172 |
from sys import stdin
N=int(stdin.readline().rstrip())
S=(stdin.readline().rstrip())
S=list(S)
ans=1
for i in set(S):
ans=ans*(S.count(i)+1)
ans-=1
print(ans%(10**9+7))
|
s419001724
|
p03352
|
u331036636
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 4 |
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
s228724897
|
Accepted
| 86 | 2,940 | 160 |
x = int(input())
list_all = [1]
for i in range(2,x):
if i**2 >x:break
list_all.append(max([i**j for j in range(2,x) if x >= i**j]))
print(max(list_all))
|
|
s057707326
|
p03644
|
u204800924
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 262 |
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())
result = 0
for i in range(N+1):
count = 0
flag = True
while flag == True:
if i%2 == 0:
count += 1
i = i/2
else:
flag = False
result = max([result,count])
print(2^(result))
|
s068615223
|
Accepted
| 17 | 2,940 | 88 |
N = int(input())
count = 0
while N != 1:
count += 1
N = N//2
print(2**count)
|
s269677604
|
p00735
|
u328199937
| 3,000 | 131,072 |
Wrong Answer
| 12,220 | 10,048 | 948 |
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not. A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7 ** _N_** +{1,6} number. But as it is hard to pronounce, we shall call it a _Monday-Saturday number_. For Monday-Saturday numbers _a_ and _b_ , we say _a_ is a Monday-Saturday divisor of _b_ if there exists a Monday-Saturday number _x_ such that _a_ _x_ = _b_. It is easy to show that for any Monday-Saturday numbers _a_ and _b_ , it holds that _a_ is a Monday-Saturday divisor of _b_ if and only if _a_ is a divisor of _b_ in the usual sense. We call a Monday-Saturday number a _Monday-Saturday prime_ if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday- Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number _a_ a _Monday-Saturday prime factor_ of _a_. For example, 27 is one of the Monday- Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds. Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds. Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
|
this_prime_num = [0 for i in range(300001)]
for i in range(300001):
if i % 7 == 1 or i % 7 == 6:
this_prime_num[i] = 1
for i in range(2, 300001):
j = i * 2
while this_prime_num[i] == 1 and j < len(this_prime_num):
#print(j)
this_prime_num[j] = 0
j += i
this_primenum_list = []
for i in range(2, len(this_prime_num)):
if this_prime_num[i]:
this_primenum_list.append(i)
print(this_primenum_list[:30])
ans_list = []
while True:
n = int(input())
if n == 1:
break
i = 0
ans = [n]
while this_primenum_list[i] <= n:
#print(i)
if n % this_primenum_list[i] == 0:
ans.append(this_primenum_list[i])
i += 1
if i >= len(this_primenum_list):
break
ans_list.append(ans)
for i in ans_list:
print(str(i[0]) + ":", end = "")
for j in i[1:]:
print("", end = " ")
print(j, end = "")
print()
|
s944881441
|
Accepted
| 12,200 | 10,052 | 949 |
this_prime_num = [0 for i in range(300001)]
for i in range(300001):
if i % 7 == 1 or i % 7 == 6:
this_prime_num[i] = 1
for i in range(2, 300001):
j = i * 2
while this_prime_num[i] == 1 and j < len(this_prime_num):
#print(j)
this_prime_num[j] = 0
j += i
this_primenum_list = []
for i in range(2, len(this_prime_num)):
if this_prime_num[i]:
this_primenum_list.append(i)
ans_list = []
while True:
n = int(input())
if n == 1:
break
i = 0
ans = [n]
while this_primenum_list[i] <= n:
#print(i)
if n % this_primenum_list[i] == 0:
ans.append(this_primenum_list[i])
i += 1
if i >= len(this_primenum_list):
break
ans_list.append(ans)
for i in ans_list:
print(str(i[0]) + ":", end = "")
for j in i[1:]:
print("", end = " ")
print(j, end = "")
print()
|
s889073609
|
p03569
|
u785205215
| 2,000 | 262,144 |
Wrong Answer
| 1,497 | 4,212 | 989 |
We have a string s consisting of lowercase English letters. Snuke can perform the following operation repeatedly: * Insert a letter `x` to any position in s of his choice, including the beginning and end of s. Snuke's objective is to turn s into a palindrome. Determine whether the objective is achievable. If it is achievable, find the minimum number of operations required.
|
from sys import stdin, stdout
import re
def main():
s = [i for i in stdin.readline()]
left = 0
right = 0
if len(s) == 1:
print(0)
else:
f = True
while f:
if len(s) == 1:
break
else:
while s[0]=='x' or s[-1]=='x' and len(s)>0:
if s[0] == 'x':
del s[0]
left += 1
if s[-1] == 'x':
del s[-1]
right += 1
if s[0] != s[-1]:
f = False
else:
if len(s)>1:
del s[0]
del s[-1]
if f:
if left-right == 0:
print(left + right)
elif left-right != 0:
print(abs(left-right))
else:
print(-1)
if __name__ == "__main__":
main()
|
s912816722
|
Accepted
| 69 | 3,316 | 204 |
s = input()
ans = 0
p = 0
q = len(s) - 1
while p < q:
if s[p] == s[q]:
p += 1
q -= 1
elif s[p] == "x":
p += 1
ans += 1
elif s[q] == "x":
q -= 1
ans += 1
else:
ans = -1
break
print(ans)
|
s093585549
|
p03485
|
u845937249
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 112 |
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())
import math
print(math.ceil(a/b))
|
s278181024
|
Accepted
| 17 | 2,940 | 116 |
a,b = map(int,input().split())
import math
print(math.ceil((a+b)/2))
|
s534892136
|
p03370
|
u422552722
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 243 |
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
N, X = map(int, input().split(" "))
i = 0
dough = []
count = N
while i < N:
dough.append(int(input()))
i += 1
# make at least one doughnut for each
excess = X - sum(dough)
while X <= 0:
X -= min(dough)
count += 1
print(count)
|
s394618562
|
Accepted
| 197 | 3,060 | 258 |
N, X = map(int, input().split(" "))
i = 0
dough = []
count = N
while i < N:
dough.append(int(input()))
i += 1
# make at least one doughnut for each
excess = X - sum(dough)
while excess >= 0:
excess -= min(dough)
count += 1
print(count - 1)
|
s788267359
|
p03387
|
u853869447
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 214 |
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 proved that we can always make A, B and C all equal by repeatedly performing these operations.
|
A = sorted(map(int, input().split()))
print(A)
x = 0
while A[0] < A[2] - 1:
A[0] += 2
x += 1
while A[1] < A[2] - 1:
A[1] += 2
x += 1
if A[0] == A[1]:
x = x + 1
else:
x = x + 1
print(x)
|
s137233364
|
Accepted
| 17 | 3,060 | 243 |
A = sorted(map(int, input().split()))
x = 0
while A[0] < A[2] - 1:
A[0] += 2
x += 1
while A[1] < A[2] - 1:
A[1] += 2
x += 1
if A[0] == A[1] == A[2]:
x = x
elif A[0] == A[1]:
x = x + 1
else:
x = x + 2
print(x)
|
s236870999
|
p03417
|
u932868243
| 2,000 | 262,144 |
Wrong Answer
| 23 | 9,112 | 101 |
There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region. The front and back sides of these cards can be distinguished, and initially every card faces up. We will perform the following operation once for each square contains a card: * For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square. It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed. Find the number of cards that face down after all the operations.
|
n,m=map(int,input().split())
if n>=3 and m>=3:
print(n*m-(n-2)*(m-2))
elif n==1 or m==1:
print(2)
|
s045588720
|
Accepted
| 24 | 9,172 | 150 |
n,m=map(int,input().split())
if n==m==1:
print(1)
exit()
if n>=3 and m>=3:
print((n-2)*(m-2))
elif n==1 or m==1:
print(n*m-2)
else:
print(0)
|
s029961465
|
p04043
|
u317493066
| 2,000 | 262,144 |
Wrong Answer
| 24 | 3,064 | 439 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
# -*- coding:utf-8 -*-
def check(data):
five_count = 0
seven_count = 0
for i in data:
if i == 5:
five_count += 1
elif i == 7:
seven_count += 1
if five_count == 2 and seven_count == 1:
return True
else:
return False
if __name__ == "__main__":
A, B, C = map(int, input().split())
if check([A, B, C]):
print("Yes")
else:
print("No")
|
s137316102
|
Accepted
| 23 | 3,064 | 391 |
# -*- coding:utf-8 -*-
def check(data):
five_count = 0
seven_count = 0
for i in data:
if i == 5:
five_count += 1
elif i == 7:
seven_count += 1
if five_count == 2 and seven_count == 1:
return "YES"
else:
return "NO"
if __name__ == "__main__":
A, B, C = map(int, input().split())
print(check([A, B, C]))
|
s729105655
|
p00293
|
u621997536
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,720 | 232 |
バスマニアの健次郎君は、Ａ市内のバスをよく利用しています。ある日ふと、健次郎君の家の前のバス停から出発するすべてのバスを写真に収めることを思い立ちました。このバス停には飯盛山行きと鶴ケ城行きの２つのバス路線が通ります。各路線の時刻表は手に入れましたが、１つの時刻表としてまとめた方がバス停で写真が撮りやすくなります。 健次郎君を助けるために、２つの路線の時刻表を、０時０分を基準として出発時刻が早い順に１つの時刻表としてまとめるプログラムを作成してください。
|
v = []
for i in range(2):
x = list(map(int, input().split()))
N, x = x[0], x[1:]
for j in range(N):
v.append(x[j * 2] * 60 + x[j * 2 + 1])
for i in sorted(set(v)):
print("{0}:{1:02d}".format(i // 60, i % 60))
|
s202397204
|
Accepted
| 30 | 6,724 | 239 |
v = []
for i in range(2):
x = list(map(int, input().split()))
N, x = x[0], x[1:]
for j in range(N):
v.append(x[j * 2] * 60 + x[j * 2 + 1])
print(' '.join(['{0}:{1:02d}'.format(i // 60, i % 60) for i in sorted(set(v))]))
|
s927797782
|
p03958
|
u373047809
| 1,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 78 |
There are K pieces of cakes. Mr. Takahashi would like to eat one cake per day, taking K days to eat them all. There are T types of cake, and the number of the cakes of type i (1 ≤ i ≤ T) is a_i. Eating the same type of cake two days in a row would be no fun, so Mr. Takahashi would like to decide the order for eating cakes that minimizes the number of days on which he has to eat the same type of cake as the day before. Compute the minimum number of days on which the same type of cake as the previous day will be eaten.
|
k, _, *a = map(int, open(0).read().split())
print(max(b*2 - k - 1 for b in a))
|
s602418361
|
Accepted
| 17 | 2,940 | 75 |
k, _, *a = map(int, open(0).read().split())
print(max(max(a)*2 - k - 1, 0))
|
s208287736
|
p02842
|
u736524428
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 330 |
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). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
N = int(input())
X_low = 1
X_high = N
def can_buy(price):
return price * 1.08 >= N
while X_high - X_low > 1:
x = int(X_high + X_low) // 2
if can_buy(x):
X_low = x
else:
X_high = x
if int(X_low * 1.08) == N:
print(1)
else:
print(0)
|
s949037332
|
Accepted
| 17 | 2,940 | 346 |
N = int(input())
X_low = 1
X_high = 50001
def can_buy(price):
return int(price * 1.08) <= N
while X_high - X_low > 1:
x = int(X_high + X_low) // 2
if can_buy(x):
X_low = x
else:
X_high = x
if int(X_low * 1.08) == N:
print(X_low)
else:
print(":(")
|
s899874619
|
p04030
|
u242684850
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 97 |
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
s = input()
t = ''
for x in s:
if x == 'B' and len(t) > 0:
t = t[:-1]
else:
t = t + x
|
s809230887
|
Accepted
| 19 | 2,940 | 91 |
s = input()
t = ''
for x in s:
if x == 'B':
t = t[:-1]
else:
t = t + x
print(t)
|
s436010773
|
p03997
|
u609814378
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 68 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a*b*2)/2)
|
s952572193
|
Accepted
| 17 | 2,940 | 75 |
a = int(input())
b = int(input())
h = int(input())
print(int(((a+b)*h)/2))
|
s512885728
|
p03408
|
u580404776
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,316 | 225 |
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
from collections import Counter
N=int(input())
A=[input() for _ in range(N)]
M=int(input())
B=[input() for _ in range(M)]
a=Counter(A)
b=Counter(B)
ans=0
for p in A:
print(p)
ans=max(ans,a[p]-b[p])
print(ans)
|
s901635014
|
Accepted
| 20 | 3,316 | 212 |
from collections import Counter
N=int(input())
A=[input() for _ in range(N)]
M=int(input())
B=[input() for _ in range(M)]
a=Counter(A)
b=Counter(B)
ans=0
for p in A:
ans=max(ans,a[p]-b[p])
print(ans)
|
s234866554
|
p02928
|
u023958502
| 2,000 | 1,048,576 |
Wrong Answer
| 411 | 3,188 | 554 |
We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.
|
N,K = map(int,input().split())
A = list(map(int,input().split()))
ans = 0
time_ans = 0
A_sort = sorted(A)
before = -1
b = -1
for i in range(N):
a_sorti = A_sort[i]
if a_sorti == before:
time_ans += b
continue
b = A_sort.index(a_sorti)
time_ans += b
before = a_sorti
for i in range(N):
for j in range(i + 1,N):
if A[i] > A[j]:
ans += 1
ans = int(time_ans * K * (K + 1) / 2) - K * (time_ans - ans)
print(ans)
print(ans % (10 ** 9 + 7))
|
s671856278
|
Accepted
| 421 | 3,188 | 564 |
# N = int(input())
N,K = map(int,input().split())
A = list(map(int,input().split()))
ans = 0
time_ans = 0
A_sort = sorted(A)
before = -1
b = -1
for i in range(N):
a_sorti = A_sort[i]
if a_sorti == before:
time_ans += b
continue
b = A_sort.index(a_sorti)
time_ans += b
before = a_sorti
for i in range(N - 1):
for j in range(i + 1,N):
if A[i] > A[j]:
ans += 1
ans1 = K * (K + 1) * time_ans // 2 - (time_ans - ans) * K
print(ans1 % (10 ** 9 + 7))
|
s013917893
|
p02612
|
u450288159
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,148 | 90 |
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.
|
def main():
n = int(input())
print(n%1000)
if __name__ == '__main__':
main()
|
s084570148
|
Accepted
| 27 | 9,092 | 150 |
def main():
n = int(input())
if n%1000 == 0:
print(0)
else:
print(1000 - (n%1000))
if __name__ == '__main__':
main()
|
s141305364
|
p03943
|
u572561929
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 123 |
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. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b, c = (int(x) for x in input().split())
if a == b + c or b == a + c or c == a + b:
print('YES')
else:
print('NO')
|
s254966722
|
Accepted
| 17 | 2,940 | 123 |
a, b, c = (int(x) for x in input().split())
if a == b + c or b == a + c or c == a + b:
print('Yes')
else:
print('No')
|
s843658472
|
p03605
|
u126747509
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 135 |
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
def main():
n = input()
if '9' in n:
print("YES")
else:
print("NO")
if __name__ == "__main__":
main()
|
s250164495
|
Accepted
| 17 | 2,940 | 109 |
def main():
if '9' in input(): print("Yes")
else: print("No")
if __name__ == "__main__":
main()
|
s029234312
|
p02409
|
u609407244
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,720 | 305 |
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
rooms = [[[0] * 10 for b in range(3)] for a in range(4)]
n = int(input())
for _ in range(n):
b, f, r, v = map(int, input().split())
rooms[b - 1][f - 1][r - 1] += v
for i in range(4):
for j in range(3):
print(' ' + ' '.join(map(str, rooms[i][j])))
if i < 4:
print('#' * 20)
|
s661375345
|
Accepted
| 30 | 6,724 | 305 |
rooms = [[[0] * 10 for b in range(3)] for a in range(4)]
n = int(input())
for _ in range(n):
b, f, r, v = map(int, input().split())
rooms[b - 1][f - 1][r - 1] += v
for i in range(4):
for j in range(3):
print(' ' + ' '.join(map(str, rooms[i][j])))
if i < 3:
print('#' * 20)
|
s764012140
|
p03415
|
u345094945
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 65 |
We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}. Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.
|
a=input()
b=input()
c=input()
print(a[0])
print(b[1])
print(c[2])
|
s695926024
|
Accepted
| 17 | 2,940 | 51 |
a=input()
b=input()
c=input()
print(a[0]+b[1]+c[2])
|
s319178386
|
p03160
|
u180908266
| 2,000 | 1,048,576 |
Wrong Answer
| 225 | 13,980 | 252 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
h_s = list(map(int, input().split()))
h_s = [abs(h_s[0]-h) for h in h_s]+[0]*10
# print(h_s)
dp = [float("inf")]*(N+10)
dp[0] = 0
for i in range(N):
for j in range(1, 3):
dp[i+j] = min(dp[i+j], dp[i]+abs(h_s[i]-h_s[i+j]))
|
s457575442
|
Accepted
| 210 | 13,980 | 232 |
N = int(input())
h_s = list(map(int, input().split()))+[0]*10
# print(h_s)
dp = [float("inf")]*(N+10)
dp[0] = 0
for i in range(N):
for j in range(1, 3):
dp[i+j] = min(dp[i+j], dp[i]+abs(h_s[i]-h_s[i+j]))
print(dp[N-1])
|
s689953710
|
p03563
|
u101627912
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 63 |
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.
|
# coding: utf-8
R=int(input())
G=int(input())
print((R+G)/2)
|
s973841546
|
Accepted
| 17 | 2,940 | 62 |
# coding: utf-8
R=int(input())
G=int(input())
print((G-R)+G)
|
s246871564
|
p03399
|
u844789719
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 76 |
You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses.
|
a = input()
b = input()
c = input()
d = input()
print(min(a, b) + min(c, d))
|
s700216349
|
Accepted
| 17 | 2,940 | 74 |
a, b, c, d = [int(input()) for _ in range(4)]
print(min(a, b) + min(c, d))
|
s326844128
|
p03485
|
u724154852
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 121 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
import sys
import math
input = sys.stdin.readline
x,y = map(int,input().split())
a = float(x + y)
print(math.ceil(a))
|
s621898315
|
Accepted
| 19 | 3,060 | 207 |
import sys
import math
input = sys.stdin.readline
x,y = map(int,input().split())
a = float((x + y) / 2)
print(math.ceil(a))
|
s945515163
|
p03502
|
u375616706
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 207 |
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
N = int(input())
dig = 0
while N:
dig += N % 10
if N % dig == 0:
print("Yes")
else:
print("No")
|
s615630393
|
Accepted
| 17 | 2,940 | 215 |
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
N = input()[:-1]
dig = 0
for c in N:
dig += int(c)
if int(N) % dig == 0:
print("Yes")
else:
print("No")
|
s279299063
|
p02615
|
u075303794
| 2,000 | 1,048,576 |
Wrong Answer
| 226 | 50,144 | 233 |
Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?
|
import numpy as np
N=int(input())
A=np.array(list(map(int,input().split())))
ans=A[0]
if N%2==0:
for i in range(1,(N-2)//2+1):
ans+=A[i]*2
else:
for i in range(1,(N-2)//2):
ans+=A[i]*2
else:
ans+=A[i+1]
print(ans)
|
s903299279
|
Accepted
| 353 | 49,980 | 259 |
import numpy as np
N=int(input())
A=np.array(list(map(int,input().split())))
A=sorted(A,reverse=True)
ans=A[0]
if N%2==0:
for i in range(1,(N-2)//2+1):
ans+=A[i]*2
else:
for i in range(1,(N-2)//2+1):
ans+=A[i]*2
else:
ans+=A[i+1]
print(ans)
|
s118273090
|
p02646
|
u927860986
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,184 | 209 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
A,V=map(int,input().split())
B,W=map(int,input().split())
T=int(input())
try:
print("Yes" if abs(A - B) // (V - W) <= T and abs(A - B) // (V - W) >= 0 else "No")
except ZeroDivisionError:
print("No")
|
s142651153
|
Accepted
| 23 | 9,184 | 192 |
A,V=map(int,input().split())
B,W=map(int,input().split())
T=int(input())
if V == W or V - W <= 0:
print("NO")
elif abs(A - B) / (V - W) <= T:
print("YES")
else:
print("NO")
|
s258676923
|
p03494
|
u294385082
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 166 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n = int(input())
a = list(map(int,input().split()))
count = 0
for i in range(10**10):
for j in a:
if j%2 != 0:
print(count)
exit()
count += 1
|
s183076552
|
Accepted
| 18 | 2,940 | 206 |
n = int(input())
a = list(map(int,input().split()))
count = 0
for i in range(10**10):
for j in range(n):
if a[j]%2 != 0:
print(count)
exit()
else:
a[j] = a[j]//2
count += 1
|
s742404895
|
p03494
|
u411278350
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 3,060 | 205 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
a = list(map(int, input().split()))
cnt = 0
while True:
for i in range(N):
if a[i] % 2 == 1:
break
else:
a[i] = a[i]/2
cnt += 1
print(cnt)
|
s649801659
|
Accepted
| 18 | 3,060 | 257 |
N = int(input())
a = list(map(int, input().split()))
cnt = 0
t_f = True
while t_f == True:
for i in range(N):
if a[i] % 2 == 1:
t_f = False
break
else:
a[i] = a[i]//2
cnt += 1
print(cnt - 1)
|
s689115153
|
p02606
|
u509150616
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,088 | 157 |
How many multiples of d are there among the integers between L and R (inclusive)?
|
def solve(L,R,d):
count = 0
for i in range(L,R+1):
if i % d == 0:
count += 1
return count
L, R, D = map(int, input().split())
|
s743146381
|
Accepted
| 23 | 9,088 | 177 |
def solve(L,R,d):
count = 0
for i in range(L,R+1):
if i % d == 0:
count += 1
return count
L, R, d = map(int, input().split())
print(solve(L,R,d))
|
s264811394
|
p02665
|
u672542358
| 2,000 | 1,048,576 |
Wrong Answer
| 365 | 20,144 | 270 |
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.
|
n=int(input())
l=list(map(int,input().split()))
xl=list(reversed(l))
g=[]
a=1
k=0
f=0
for i in xl:
k+=i
g.append(k)
g.reverse()
for i in range(len(g)):
if g[i]>=a:
g[i]=a
a-=l[i]
a*=2
if a<=0:
f+=1
break
if f==0:
print(sum(g))
else:
print(-1)
|
s810312355
|
Accepted
| 387 | 20,024 | 400 |
n=int(input())
l=list(map(int,input().split()))
xl=list(reversed(l))
g=[]
a=1
k=0
f=0
for i in xl:
k+=i
g.append(k)
g.reverse()
for i in range(len(g)):
if g[i]>=a:
g[i]=a
a-=l[i]
a*=2
if a<=0 and i!=len(g)-1:
f+=1
break
if a<0 and i==len(g)-1:
f+=1
break
if f==0 and n!=0:
print(sum(g))
elif n==0:
if l[0]==1:
print(1)
else:
print(-1)
else:
print(-1)
|
s913359932
|
p02601
|
u433697974
| 2,000 | 1,048,576 |
Wrong Answer
| 32 | 9,200 | 350 |
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 sys
def hantei(A, B, C, K):
i = 0
while A >= B:
if i >= K:
return False
sys.exit()
B *= 2
i += 1
while B >= C:
if i >= K:
return False
sys.exit()
C *= 2
i += 1
return True
A, B, C = map(int, input().split())
K = int(input())
if(hantei(A, B, C, K)):
print('YES')
else:
print('NO')
|
s147691808
|
Accepted
| 35 | 9,196 | 350 |
import sys
def hantei(A, B, C, K):
i = 0
while A >= B:
if i >= K:
return False
sys.exit()
B *= 2
i += 1
while B >= C:
if i >= K:
return False
sys.exit()
C *= 2
i += 1
return True
A, B, C = map(int, input().split())
K = int(input())
if(hantei(A, B, C, K)):
print('Yes')
else:
print('No')
|
s241758808
|
p03997
|
u679817762
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,020 | 90 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a + b) * h / 2)
|
s339431272
|
Accepted
| 28 | 9,084 | 97 |
a = int(input())
b = int(input())
h = int(input())
print(int(((a + b) * h) / 2))
|
s884900085
|
p03549
|
u867848444
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,188 | 269 |
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
|
import math
n,m=map(int,input().split())
ac=(n-m)*100
tle=m*1900
time=0
cnt=1
time_ac=(ac+tle)*(1/2)**m
fault=(1-(1/2)**m)
while True:
temp=cnt*time_ac*(fault**(cnt-1))
if time==time+temp:
print(math.ceil(time))
break
time+=temp
cnt+=1
|
s480047181
|
Accepted
| 20 | 3,188 | 253 |
n,m=map(int,input().split())
ac=(n-m)*100
tle=m*1900
time=0
cnt=1
time_ac=(ac+tle)*(1/2)**m
fault=(1-(1/2)**m)
while True:
temp=cnt*time_ac*(fault**(cnt-1))
if time==time+temp:
print(round(time))
break
time+=temp
cnt+=1
|
s962826682
|
p03400
|
u574053975
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 174 |
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
n=int(input())
x=input()
x=x.split()
d=int(x[0])
x=int(x[1])
o=0
for i in range(n):
a=int(input())
b=0
c=0
while b<d:
o+=1
b=c*a+1
c+=1
o+=x
print(str(o))
|
s265649746
|
Accepted
| 17 | 3,064 | 174 |
n=int(input())
x=input()
x=x.split()
d=int(x[0])
x=int(x[1])
o=0
for i in range(n):
a=int(input())
b=0
c=0
while (c*a+1)<=d:
o+=1
c+=1
o+=x
print(str(o))
|
s418579569
|
p02409
|
u067975558
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,788 | 835 |
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
floor = [
[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]],
[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]],
[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]],
[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
]
count = int(input())
for c in range(count):
(t,f,r,i) = [int(x) for x in input().split()]
floor[t-1][f-1][r-1] = i
for z in range(4):
for i in range(len(floor[z])):
for v in floor[z][i]:
print(v, end=' ')
print()
for h in range(20):
print('#', end='')
print()
|
s007836128
|
Accepted
| 40 | 6,732 | 315 |
floor = [[[0] * 10 for x in range(3)] for y in range(4)]
for c in range(int(input())):
(b,f,r,v) = [int(x) for x in input().split()]
floor[b-1][f-1][r-1] += v
for x in range(3 * 4):
if x % 3 == 0 and not x == 0:
print('#' * 20)
print('',' '.join(str(y) for y in floor[int(x / 3)][x % 3]))
|
s109088049
|
p03095
|
u856234158
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 5,492 | 341 |
You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order.
|
def counter(flag,st):
count = 0
if st == '':
return count
for i in range(len(st)):
if not st[i] in flag:
flag[st[i]] = 1
count += 1
count += counter(flag,st[i:])
del flag[st[i]]
return count
N = int(input())
S = input()
cnt = 0
for i in range(N):
flag = {}
cnt += counter(flag,S[i:])
print(cnt)
|
s132345323
|
Accepted
| 229 | 3,188 | 356 |
N = int(input())
S = input()
charCnt = []
flag = {}
for c in S:
if not c in flag:
flag[c] = 1
charCnt.append([c,1])
else :
flag[c] += 1
for i in range(len(charCnt)):
if charCnt[i][0] == c:
charCnt[i][1] += 1
break
cnt = 1
for i in range(len(charCnt)):
cnt *= (charCnt[i][1]+1)
cnt -= 1
cnt %= 10**9+7
print(cnt)
|
s388198892
|
p00015
|
u582608581
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,720 | 692 |
A country has a budget of more than 81 trillion yen. We want to process such data, but conventional integer type which uses signed 32 bit can represent up to 2,147,483,647. Your task is to write a program which reads two integers (more than or equal to zero), and prints a sum of these integers. If given integers or the sum have more than 80 digits, print "overflow".
|
def Add(numa, numb):
longer = (numa if len(numa) >= len(numb) else numb)
shorter = (numa if len(numa) < len(numb) else numb)
carry = 0
result = ''
for s in range(-1,-len(shorter) - 1, -1):
ans = int(longer[s]) + int(shorter[s]) + carry
if ans >= 10:
result = str(ans % 10) + result
carry = 1
else:
result = str(ans) + result
carry = 0
for l in range(-len(shorter) - 1, -len(longer) - 1, -1):
ans = int(longer[l]) + carry
if ans >= 10:
result = str(ans % 10) + result
carry = 1
else:
result = str(ans) + result
carry = 0
return (result if carry == 0 else '1' + result)
N = eval(input())
for _ in range(N):
a = input()
b = input()
print(Add(a, b))
|
s162504173
|
Accepted
| 20 | 7,684 | 737 |
def Add(numa, numb):
longer = (numa if len(numa) >= len(numb) else numb)
shorter = (numa if len(numa) < len(numb) else numb)
carry = 0
result = ''
for s in range(-1, -len(shorter) - 1, -1):
ans = int(longer[s]) + int(shorter[s]) + carry
if ans >= 10:
result = str(ans % 10) + result
carry = 1
else:
result = str(ans) + result
carry = 0
for l in range(-len(shorter) - 1, -len(longer) - 1, -1):
ans = int(longer[l]) + carry
if ans >= 10:
result = str(ans % 10) + result
carry = 1
else:
result = str(ans) + result
carry = 0
return (result if carry == 0 else '1' + result)
N = eval(input())
for _ in range(N):
a = input()
b = input()
ans = Add(a, b)
print('overflow' if len(ans) > 80 else ans)
|
s329059266
|
p03063
|
u591287669
| 2,000 | 1,048,576 |
Wrong Answer
| 237 | 29,480 | 335 |
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 that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
n=int(input())
s=input()
w=[0]
b=[0]
for i in range(len(s)):
if s[i]=='#':
w.append(w[-1]+1)
else:
w.append(w[-1])
if s[-i-1]=='.':
b.append(b[-1]+1)
else:
b.append(b[-1])
print(w)
print(b)
total=[]
for i in range(len(s)+1):
total.append(w[i]+b[-i-1])
print(total)
print(min(total))
|
s775762215
|
Accepted
| 190 | 20,932 | 338 |
n=int(input())
s=input()
w=[0]
b=[0]
for i in range(len(s)):
if s[i]=='#':
w.append(w[-1]+1)
else:
w.append(w[-1])
if s[-i-1]=='.':
b.append(b[-1]+1)
else:
b.append(b[-1])
#print(w)
#print(b)
total=[]
for i in range(len(s)+1):
total.append(w[i]+b[-i-1])
#print(total)
print(min(total))
|
s411738659
|
p03919
|
u397531548
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 222 |
There is a grid with H rows and W columns. The square at the i-th row and j-th column contains a string S_{i,j} of length 5. The rows are labeled with the numbers from 1 through H, and the columns are labeled with the uppercase English letters from `A` through the W-th letter of the alphabet. Exactly one of the squares in the grid contains the string `snuke`. Find this square and report its location. For example, the square at the 6-th row and 8-th column should be reported as `H6`.
|
H,M=map(int,input().split())
c=0
for i in range(H):
S=input().split()
for j in range(0,M):
if S[j]=="snuke":
a=i
b=j
else:
c+=1
print(chr(ord('A')+a) + str(b+1))
|
s914693204
|
Accepted
| 18 | 3,060 | 224 |
H,M=map(int,input().split())
c=0
for i in range(1,H+1):
S=input().split()
for j in range(0,M):
if S[j]=="snuke":
a=i
b=j
else:
c+=1
print(chr(ord('A')+b) + str(a))
|
s317441811
|
p03007
|
u983918956
| 2,000 | 1,048,576 |
Wrong Answer
| 288 | 20,456 | 700 |
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 final integer on the blackboard and a sequence of operations that maximizes the final integer.
|
N = int(input())
A = sorted(list(map(int,input().split())))
ans_list = []
p = []
n = []
p.append(A[-1])
n.append(A[0])
for a in A[1:-1]:
if a >= 0:
p.append(a)
elif a < 0:
n.append(a)
while len(p) != 1:
x = n.pop()
y = p.pop()
n.append(x-y)
ans_list.append((x, y))
while len(n) != 0:
x = p.pop()
y = n.pop()
p.append(x-y)
ans_list.append((x,y))
M = p[0]
print(M)
for ans in ans_list:
print(ans)
|
s865659779
|
Accepted
| 249 | 20,120 | 315 |
N = int(input())
A = sorted(list(map(int,input().split())))
p = A[-1]
n = A[0]
ans_list = []
for a in A[1:-1]:
if a >= 0:
ans_list.append((n, a))
n -= a
else:
ans_list.append((p, a))
p -= a
M = p - n
ans_list.append((p, n))
print(M)
for x, y in ans_list:
print(x, y)
|
s091817275
|
p03679
|
u516242950
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 149 |
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())
over = b - a
if over < x:
print("delicious")
elif x <= over < x + 1:
print("safe")
else:
print("dangerous")
|
s063271690
|
Accepted
| 17 | 2,940 | 143 |
x, a, b = map(int, input().split())
over = b - a
if over <= 0:
print("delicious")
elif over <= x:
print("safe")
else:
print("dangerous")
|
s773849852
|
p03457
|
u316733945
| 2,000 | 262,144 |
Wrong Answer
| 396 | 27,300 | 296 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
n = int(input())
txy = [list(map(int, input().split())) for _ in range(n)]
print(txy[0])
for i in range(n):
if txy[i][1] + txy[i][2] >txy[i][0]:
status = "No"
break
elif (txy[i][1]+txy[i][2])%2 != txy[i][0]%2:
status = "No"
break
else:
status = "Yes"
print(status)
|
s095697353
|
Accepted
| 390 | 3,064 | 250 |
n = int(input())
x0, y0, t0 = 0, 0, 0
for _ in range(n):
t, x, y = map(int, input().split())
if abs(x-x0) + abs(y-y0) > t-t0 or (x+y)%2 != t%2:
status = "No"
break
else:
status = "Yes"
t0, x0, y0 = t, x, y
print(status)
|
s336520560
|
p03693
|
u841568901
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,024 | 56 |
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?
|
print("No" if int(input().replace(" ",""))%4 else "Yes")
|
s649923849
|
Accepted
| 25 | 9,148 | 56 |
print("NO" if int(input().replace(" ",""))%4 else "YES")
|
s106112213
|
p04043
|
u408375121
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 200 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
l = list(map(int, input().split()))
count_5 = 0
count_7 = 0
for r in l:
if r == 5:
count_5 += 1
if r == 7:
count_7 += 1
if count_5 == 2 and count_7 == 1:
print('Yes')
else:
print('No')
|
s981623239
|
Accepted
| 17 | 2,940 | 201 |
l = list(map(int, input().split()))
count_5 = 0
count_7 = 0
for r in l:
if r == 5:
count_5 += 1
if r == 7:
count_7 += 1
if count_5 == 2 and count_7 == 1:
print('YES')
else:
print('NO')
|
s261096389
|
p04045
|
u924845460
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 159 |
Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K. She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change). However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money. Find the amount of money that she will hand to the cashier.
|
n,k=map(int, input().split())
ds=list(map(str, input().split()))
while True:
for i in ds:
if i in str(n):
n+=1
continue
break
print(str(n))
|
s352300790
|
Accepted
| 102 | 2,940 | 126 |
fn=str
n,k=map(int, input().split())
ds=list(map(fn, input().split()))
while any(i in fn(n) for i in ds):
n+=1
print(fn(n))
|
s464457813
|
p03779
|
u252828980
| 2,000 | 262,144 |
Wrong Answer
| 28 | 2,940 | 64 |
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
n = int(input())
t = 0
while t*(t+1) <=2*n:
t += 1
print(t)
|
s368225311
|
Accepted
| 40 | 9,072 | 88 |
d = int(input())
t = 0
while True:
if t*(t+1)//2 >=d:
print(t)
exit()
t +=1
|
s720172359
|
p02646
|
u746206084
| 2,000 | 1,048,576 |
Wrong Answer
| 34 | 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 her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a,v = map(int,input().split())
b,w = map(int,input().split())
t=int(input())
if abs(a-b)>(v-w)*t:
print("No")
else:
print("Yes")
|
s865717879
|
Accepted
| 22 | 9,180 | 136 |
a,v = map(int,input().split())
b,w = map(int,input().split())
t=int(input())
if abs(a-b)>(v-w)*t:
print("NO")
else:
print("YES")
|
s935921211
|
p03625
|
u950708010
| 2,000 | 262,144 |
Wrong Answer
| 112 | 14,252 | 251 |
We have N sticks with negligible thickness. The length of the i-th stick is A_i. Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides. Find the maximum possible area of the rectangle.
|
n = int(input())
a = sorted(list(int(i) for i in input().split()),reverse=True)
ans = []
for i in range (n-1):
if a[i+1] == a[i]:
ans.append(a[i])
a[i+1]= 0
print(a[i])
if len(ans)>=2:
cand = ans[0]*ans[1]
break
else:
cand = 0
print(cand)
|
s650526502
|
Accepted
| 111 | 14,252 | 238 |
n = int(input())
a = sorted(list(int(i) for i in input().split()),reverse=True)
ans = []
for i in range (n-1):
if a[i+1] == a[i]:
ans.append(a[i])
a[i+1]= 0
if len(ans)>=2:
cand = ans[0]*ans[1]
break
else:
cand = 0
print(cand)
|
s817497040
|
p03545
|
u492532572
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 475 |
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.
|
A, B, C, D = map(int, input())
for i in range(8):
sum = 0
op1 = "+" if i // 4 % 2 == 0 else "-"
op2 = "+" if i // 2 % 2 == 0 else "-"
op3 = "+" if i % 2 == 0 else "-"
if op1 == "+":
sum = A + B
else:
sum = A - B
if op2 == "+":
sum += C
else:
sum -= C
if op3 == "+":
sum += D
else:
sum -= D
if sum == 7:
print(str(A) + op1 + str(B) + op2 + str(C) + op3 + str(D))
break
|
s323740657
|
Accepted
| 18 | 3,064 | 482 |
A, B, C, D = map(int, input())
for i in range(8):
sum = 0
op1 = "+" if i // 4 % 2 == 0 else "-"
op2 = "+" if i // 2 % 2 == 0 else "-"
op3 = "+" if i % 2 == 0 else "-"
if op1 == "+":
sum = A + B
else:
sum = A - B
if op2 == "+":
sum += C
else:
sum -= C
if op3 == "+":
sum += D
else:
sum -= D
if sum == 7:
print(str(A) + op1 + str(B) + op2 + str(C) + op3 + str(D) + "=7")
break
|
s204973340
|
p03407
|
u118019047
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 85 |
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
a,b,c = map(int,input().split())
if a + b < c:
print("yes")
else:
print("No")
|
s680670474
|
Accepted
| 17 | 2,940 | 83 |
a,b,c = map(int,input().split())
if a+b>=c:
print("Yes")
else:
print("No")
|
s072044489
|
p03814
|
u361826811
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,512 | 292 |
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`.
|
import sys
readline = sys.stdin.readline
readlines = sys.stdin.readlines
sys.setrecursionlimit(10 ** 7)
S = readline().rstrip()
print(S.index('A'))
print(len(S)-(S[::-1].index('Z')+1))
print(abs(S.index('A')-(len(S)-(S[::-1].index('Z')+1)))+1)
|
s480766692
|
Accepted
| 18 | 3,516 | 275 |
import sys
import itertools
# import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
S = readline().decode('utf8')
print(-S.find('A') + S.rfind('Z') + 1)
|
s172715952
|
p02742
|
u086063386
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 90 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
h,w = map(int,input().split())
if h%2 == 1 and w%2 == 1: print(h*w/2+1)
else: print(h*w/2)
|
s888160127
|
Accepted
| 17 | 2,940 | 120 |
h,w = map(int,input().split())
ans = (h*w)//2
if h%2 == 1 and w%2 == 1: ans += 1
if h == 1 or w == 1: ans = 1
print(ans)
|
s232977715
|
p02607
|
u972658925
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,080 | 133 |
We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i. How many squares i satisfy both of the following conditions? * The assigned number, i, is odd. * The written integer is odd.
|
n = int(input())
a = list(map(int,input().split()))
cnt=0
for i in range(len(a)):
if (i+1)&1==a[i]==1:
cnt+=1
print(cnt)
|
s607589737
|
Accepted
| 25 | 9,052 | 135 |
n = int(input())
a = list(map(int,input().split()))
cnt=0
for i in range(len(a)):
if (i+1)&1==a[i]&1==1:
cnt+=1
print(cnt)
|
s222474422
|
p02742
|
u893661063
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,176 | 155 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
H, W = map(int, input().split())
if H == 1 or W == 1:
ans = 1
elif (H * W) % 2 == 0:
ans = (H * W) / 2
else:
ans = (H * W) // 2 + 1
print (ans)
|
s898435760
|
Accepted
| 26 | 8,748 | 156 |
H, W = map(int, input().split())
if H == 1 or W == 1:
ans = 1
elif (H * W) % 2 == 0:
ans = (H * W) // 2
else:
ans = (H * W) // 2 + 1
print (ans)
|
s662100719
|
p03478
|
u739360929
| 2,000 | 262,144 |
Wrong Answer
| 29 | 3,408 | 208 |
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())
someSums = 0
for i in range(1, N+1):
if A <= (i // 1000 % 10) + (i // 100 % 10) + (i // 10 % 10) + (i % 10) <= B:
someSums += i
print(i)
print(someSums)
|
s901094034
|
Accepted
| 23 | 2,940 | 212 |
N, A, B = map(int, input().split())
someSums = 0
for i in range(1, N+1):
if A <= (i // 10000) % 10 + (i // 1000) % 10 + (i // 100) % 10 + (i // 10) % 10 + (i % 10) <= B:
someSums += i
print(someSums)
|
s448071178
|
p03456
|
u189575640
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 251 |
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
import sys
# A,B,C = [int(n) for n in input().split()]
# N = int(input())
# S = str(input())
# T = str(input())
a,b = [str(n) for n in input().split()]
ab = int(a+b)
print(ab)
table = [n*n for n in range(1,101)]
print("Yes" if ab in table else "No")
|
s943594841
|
Accepted
| 18 | 3,192 | 2,169 |
import sys
a, b = input().split()
ab = int(str(a) + str(b))
s = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481, 3600, 3721, 3844, 3969, 4096, 4225, 4356, 4489, 4624, 4761, 4900, 5041, 5184, 5329, 5476, 5625, 5776, 5929, 6084, 6241, 6400, 6561, 6724, 6889, 7056, 7225, 7396, 7569, 7744, 7921, 8100, 8281, 8464, 8649, 8836, 9025, 9216, 9409, 9604, 9801, 10000, 10201, 10404, 10609, 10816, 11025, 11236, 11449, 11664, 11881, 12100, 12321, 12544, 12769, 12996, 13225, 13456, 13689, 13924, 14161, 14400, 14641, 14884, 15129, 15376, 15625, 15876, 16129, 16384, 16641, 16900, 17161, 17424, 17689, 17956, 18225, 18496, 18769, 19044, 19321, 19600, 19881, 20164, 20449, 20736, 21025, 21316, 21609, 21904, 22201, 22500, 22801, 23104, 23409, 23716, 24025, 24336, 24649, 24964, 25281, 25600, 25921, 26244, 26569, 26896, 27225, 27556, 27889, 28224, 28561, 28900, 29241, 29584, 29929, 30276, 30625, 30976, 31329, 31684, 32041, 32400, 32761, 33124, 33489, 33856, 34225, 34596, 34969, 35344, 35721, 36100, 36481, 36864, 37249, 37636, 38025, 38416, 38809, 39204, 39601, 40000, 40401, 40804, 41209, 41616, 42025, 42436,
42849, 43264, 43681, 44100, 44521, 44944, 45369, 45796, 46225, 46656, 47089, 47524, 47961, 48400, 48841, 49284, 49729, 50176, 50625, 51076, 51529, 51984, 52441, 52900, 53361, 53824, 54289, 54756, 55225, 55696, 56169, 56644, 57121, 57600, 58081, 58564, 59049,
59536, 60025, 60516, 61009, 61504, 62001, 62500, 63001, 63504, 64009, 64516, 65025, 65536, 66049, 66564, 67081, 67600, 68121, 68644, 69169, 69696, 70225, 70756, 71289, 71824, 72361, 72900, 73441, 73984, 74529, 75076, 75625, 76176, 76729, 77284, 77841, 78400, 78961, 79524, 80089, 80656, 81225, 81796, 82369, 82944, 83521, 84100, 84681, 85264, 85849, 86436, 87025, 87616, 88209, 88804, 89401, 90000, 90601, 91204, 91809, 92416, 93025, 93636, 94249, 94864, 95481, 96100, 96721, 97344, 97969, 98596, 99225, 99856]
print("Yes" if ab in s else "No")
|
s125608388
|
p02694
|
u294542073
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,104 | 248 |
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())
t = 100
result = 0
while t >= X:
t += t // 100
result += 1
print(result)
|
s623851985
|
Accepted
| 24 | 9,016 | 210 |
X = int(input())
n = 100
count = 0
while n < X:
count += 1
n = int(n * 1.01)
print(count)
|
s161222176
|
p02601
|
u763410402
| 2,000 | 1,048,576 |
Wrong Answer
| 2,205 | 9,016 | 272 |
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.
|
def main():
r,g,b=map(int,input().split())
k=int(input())
while(k>0):
if(g<=r):
g<<1
k-=1
elif(b<=g):
b<<1
k-=1
if(r<g and g<b):
print('Yes')
else:
print('No')
main()
|
s432458300
|
Accepted
| 28 | 9,068 | 309 |
def main():
r,g,b=map(int,input().split())
k=int(input())
while(k>0):
if(g<=r):
g=g<<1
k-=1
elif(b<=g):
b=b<<1
k-=1
else:
break;
if(r<g and g<b):
print('Yes')
else:
print('No')
main()
|
s839157302
|
p02398
|
u999594662
| 1,000 | 131,072 |
Time Limit Exceeded
| 40,000 | 8,220 | 132 |
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
|
cnt=0
a,b,c=map(int,input().split())
while True:
if b-a<=0:break
for i in range(a,b):
if (c%a==0):
cnt+=1
a+=1
print(cnt)
|
s427861599
|
Accepted
| 20 | 7,636 | 94 |
cnt=0
a,b,c=map(int,input().split())
for i in range(a,b+1):
if (c%i==0):
cnt+=1
print(cnt)
|
s038308692
|
p02842
|
u411237324
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 83 |
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). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
n = int(input())
x = n // 1.08
if x * 1.08 == n:
print(x)
else:
print(':(')
|
s683243494
|
Accepted
| 17 | 2,940 | 151 |
n = int(input())
x = int(n // 1.08)
if int(x * 1.08) == n:
print(x)
exit()
x = x + 1
if int(x * 1.08) == n:
print(x)
else:
print(':(')
|
s273905332
|
p03971
|
u252828980
| 2,000 | 262,144 |
Wrong Answer
| 76 | 9,204 | 458 |
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
n,a,b = map(int,input().split())
a_cnt,b_cnt,cnt = 0,0,0
s = input()
cnt = a+b
for i in range(n):
if s[i] == "a":
if a_cnt + b_cnt <= cnt:
#print(a_cnt + b_cnt)
print("Yes")
else:
print("No")
a_cnt +=1
elif s[i] == "b":
if a_cnt + b_cnt < cnt and b_cnt < b:
print("Yes")
else:
print("No")
b_cnt +=1
elif s[i] == "c":
print("No")
|
s635246156
|
Accepted
| 74 | 9,268 | 470 |
n,a,b = map(int,input().split())
a_cnt,b_cnt,cnt = 0,0,0
s = input()
cnt = a+b
for i in range(n):
if s[i] == "a":
if a_cnt + b_cnt < cnt:
a_cnt +=1#print(a_cnt + b_cnt)
print("Yes")
else:
print("No")
elif s[i] == "b":
if a_cnt + b_cnt < cnt and b_cnt < b:
b_cnt +=1
print("Yes")
else:
print("No")
elif s[i] == "c":
print("No")
|
s630215204
|
p03160
|
u821432765
| 2,000 | 1,048,576 |
Wrong Answer
| 151 | 13,928 | 221 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
h = [int(i) for i in input().split()]
dp = [float("inf")]*N
dp[0] = 0
for i in range(1, N):
dp[i] = min(dp[i], abs(h[i] - h[i-1]))
if i > 1: dp[i] = min(dp[i], abs(h[i] - h[i-2]))
print(dp[-1])
|
s569244362
|
Accepted
| 179 | 13,928 | 245 |
INF = 10**9
N = int(input())
h = [int(i) for i in input().split()] + [0, 0]
dp = [INF] * 100010
dp[0] = 0
for i in range(N):
dp[i+1] = min(dp[i+1], dp[i]+abs(h[i+1]-h[i]))
dp[i+2] = min(dp[i+2], dp[i]+abs(h[i+2]-h[i]))
print(dp[N-1])
|
s301758692
|
p03852
|
u361826811
| 2,000 | 262,144 |
Wrong Answer
| 147 | 12,400 | 296 |
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
import sys
import itertools
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
S = readline().decode('utf8')
vowel = 'aiueo'
print('vowel' if S in vowel else 'consonant')
|
s926145844
|
Accepted
| 269 | 19,748 | 305 |
import sys
import itertools
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
S = readline().decode('utf8').rstrip()
vowel = 'aiueo'
print('vowel' if S in vowel else 'consonant')
|
s203056789
|
p03679
|
u144980750
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 78 |
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.
|
a,b,c=map(int,input().split())
print("delicious" if (c-b)<=a else "dangerous")
|
s596060575
|
Accepted
| 17 | 2,940 | 98 |
a,b,c=map(int,input().split())
print("delicious" if b>=c else "safe" if (c-b)<=a else "dangerous")
|
s550033760
|
p03079
|
u940102677
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 55 |
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
print("YNEOS"[len(set(map(int,input().split())))>1::2])
|
s113747574
|
Accepted
| 17 | 2,940 | 55 |
print("YNeos"[len(set(map(int,input().split())))>1::2])
|
s115083548
|
p03740
|
u780962115
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 99 |
Alice and Brown loves games. Today, they will play the following game. In this game, there are two piles initially consisting of X and Y stones, respectively. Alice and Bob alternately perform the following operation, starting from Alice: * Take 2i stones from one of the piles. Then, throw away i of them, and put the remaining i in the other pile. Here, the integer i (1≤i) can be freely chosen as long as there is a sufficient number of stones in the pile. The player who becomes unable to perform the operation, loses the game. Given X and Y, determine the winner of the game, assuming that both players play optimally.
|
x,y=map(int,input().split())
if abs(x-y)<=1:
print("Blown")
else:
print("Alice")
|
s904457018
|
Accepted
| 17 | 2,940 | 100 |
x,y=map(int,input().split())
if abs(x-y)<=1:
print("Brown")
else:
print("Alice")
|
s674292885
|
p02613
|
u733866054
| 2,000 | 1,048,576 |
Wrong Answer
| 150 | 16,308 | 207 |
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=[str(input()) for i in range(N)]
AC=S.count("AC")
WA=S.count("WA")
TLE=S.count("TLE")
RE=S.count("RE")
print(f"AC × {AC}")
print(f"WA × {WA}")
print(f"TLE × {TLE}")
print(f"RE × {RE}")
|
s215156724
|
Accepted
| 149 | 16,192 | 203 |
N=int(input())
S=[str(input()) for i in range(N)]
AC=S.count("AC")
WA=S.count("WA")
TLE=S.count("TLE")
RE=S.count("RE")
print(f"AC x {AC}")
print(f"WA x {WA}")
print(f"TLE x {TLE}")
print(f"RE x {RE}")
|
s872795566
|
p02412
|
u471400255
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,520 | 581 |
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
s = []
while True:
cnt = 0
n, x = list(map(int, input().split()))
if n == 0 and x == 0:
break
else:
for i in range(1, n + 1):
for j in range(1, n + 1):
if i == j:
continue
else:
for k in range(1, n + 1):
if i == k or j == k:
continue
else:
if i + j + k == x:
cnt += 1
s.append(cnt)
for l in range(len(s)):
print(s[l])
|
s851546294
|
Accepted
| 540 | 7,552 | 363 |
s = []
while True:
cnt = 0
n, x = list(map(int, input().split()))
if n == 0 and x == 0:
break
else:
for i in range(1, n + 1):
for j in range(1, i):
for k in range(1, j):
if i + j + k == x:
cnt += 1
s.append(cnt)
for l in range(len(s)):
print(s[l])
|
s262331569
|
p03997
|
u668352391
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 76 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print(((a + b) * h) / 2)
|
s797110918
|
Accepted
| 17 | 2,940 | 73 |
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2))
|
s207628629
|
p03469
|
u843318346
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 34 |
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()
print('2018/'+s[6:11])
|
s934131777
|
Accepted
| 17 | 2,940 | 37 |
s = input()
print('2018/01/'+s[-2:])
|
s853170174
|
p02364
|
u855775311
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,628 | 1,046 |
Find the sum of weights of edges of the Minimum Spanning Tree for a given weighted undirected graph G = (V, E).
|
import sys
class UnionFind():
def __init__(self, n):
self.n = n
self.p = [i for i in range(n + 5)]
self.rank = [0 for i in range(n + 5)]
def find_set(self, x):
if self.p[x] != x:
self.p[x] = self.find_set(self.p[x])
return self.p[x]
def unite(self, x, y):
px, py = self.find_set(x), self.find_set(y)
if px == py:
return
if self.rank[px] > self.rank[py]:
self.p[y] = px
else:
self.p[x] = py
if self.rank[px] == self.rank[py]:
self.rank[py] += 1
def main():
sys.setrecursionlimit(int(1e5))
nvertices, nedges = map(int, input().split())
E = []
uf = UnionFind(nvertices)
for i in range(nedges):
s, t, w = map(int, input().split())
E.append((w, s, t))
E.sort()
for e in E:
print(e)
ans = 0
for w, s, t in E:
if uf.find_set(s) != uf.find_set(t):
ans += w
uf.unite(s, t)
print(ans)
main()
|
s402304955
|
Accepted
| 600 | 35,788 | 758 |
import heapq
def main():
nvertices, nedges = map(int, input().split())
Adj = [[] for i in range(nvertices)]
for i in range(nedges):
u, v, w = map(int, input().split())
Adj[u].append((v, w))
Adj[v].append((u, w))
INF = float('inf')
Q = [(0, 0)]
for i in range(1, nvertices):
Q.append((INF, i))
heapq.heapify(Q)
key = [INF] * nvertices
key[0] = 0
done = [False] * nvertices
ans = 0
while Q:
val, u = heapq.heappop(Q)
if val > key[u]:
continue
done[u] = True
ans += val
for v, w in Adj[u]:
if not done[v] and w < key[v]:
key[v] = w
heapq.heappush(Q, (w, v))
print(ans)
main()
|
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