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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s789321218
|
p03494
|
u270057938
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,108
| 14,180
| 616
|
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.
|
import numpy
import re
import math
def calc_max_num(int_list):
count = 0
even_flag = False
while True:
for i in int_list:
if i % 2 == 1:
even_flag = True
if even_flag is True:
break
else:
for i in int_list:
i /= 2
count += 1
return count
def main():
line = input()
int_num = int(line)
line = input()
line_list = re.split(" +", line)
int_list = [int(s) for s in line_list]
print(calc_max_num(int_list))
if __name__ == '__main__':
main()
|
s462206574
|
Accepted
| 150
| 12,508
| 540
|
import numpy
import re
import math
def main():
n = input()
a = list(map(int, input().split()))
ans = float("inf")
for i in a:
ans = min(ans, len(bin(i)) - bin(i).rfind("1") - 1)
print(round(ans))
if __name__ == '__main__':
main()
|
s954191352
|
p03160
|
u668785999
| 2,000
| 1,048,576
|
Wrong Answer
| 102
| 13,924
| 256
|
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.
|
def Frog1():
N = int(input())
Hs = list(map(int,input().split()))
dp = [0 for i in range(N)]
dp[0] = 0
dp[1] = abs(Hs[1]-Hs[0])
for i in range(2,N):
dp[i] = min(abs(Hs[i-1]-Hs[i])+dp[i-1],abs(Hs[i-2]-Hs[i])+dp[i-2])
Frog1()
|
s767807129
|
Accepted
| 104
| 13,924
| 275
|
def Frog1():
N = int(input())
Hs = list(map(int,input().split()))
dp = [0 for i in range(N)]
dp[0] = 0
dp[1] = abs(Hs[1]-Hs[0])
for i in range(2,N):
dp[i] = min(abs(Hs[i-1]-Hs[i])+dp[i-1],abs(Hs[i-2]-Hs[i])+dp[i-2])
print(dp[N-1])
Frog1()
|
s944074419
|
p03919
|
u102461423
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 190
|
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,W = map(int,input().split())
answer = ''
for row in range(H):
for col,s in enumerate(input().split()):
if s == 'snuke':
answer = chr(ord('A')+col) + str(row)
print(answer)
|
s568845311
|
Accepted
| 17
| 2,940
| 192
|
H,W = map(int,input().split())
answer = ''
for row in range(H):
for col,s in enumerate(input().split()):
if s == 'snuke':
answer = chr(ord('A')+col) + str(row+1)
print(answer)
|
s261547998
|
p02842
|
u336093806
| 2,000
| 1,048,576
|
Wrong Answer
| 38
| 5,016
| 142
|
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())
s = []
for i in range(50000):
s.append(int((i+1)*1.08))
if N in s:
print(int(1001/1.08+1/1.08))
else:
print(":)")
|
s280220980
|
Accepted
| 38
| 5,144
| 144
|
N = int(input())
s = []
for i in range(50000):
s.append(int((i+1)*1.08))
if N in s:
print(int(N/1.08+1/1.08))
else:
print(":(")
|
s626813581
|
p03599
|
u778700306
| 3,000
| 262,144
|
Wrong Answer
| 119
| 3,064
| 742
|
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
a,b,c,d,e,f = map(int, input().split())
waters = []
for i in range(f // (100 * a) + 1):
for j in range(f // (100 * b) + 1):
amount = 100 * (i * a + j * b)
if amount <= e:
waters.append(100 * i * a + 100 * j * b)
sugers = []
for i in range(f // c + 1):
for j in range(f // d + 1):
amount = i * c + j * d
if amount <= e:
sugers.append(i * c + j * d)
maxp = 0.0
res_w = 0
res_s = 0
for w in waters:
for s in sugers:
if w + s > f or w + s == 0:
continue
capa = w // 100 * e
if s > capa:
continue
p = 100 * s / (w + s)
if maxp < p:
maxp, res_w, res_s = p, w + s, s
print("%d %d" % (res_w, res_s))
|
s075393970
|
Accepted
| 180
| 12,440
| 760
|
a,b,c,d,e,f = map(int, input().split())
waters = []
for i in range(f // (100 * a) + 1):
for j in range(f // (100 * b) + 1):
amount = 100 * (i * a + j * b)
if amount <= f:
waters.append(amount)
sugers = []
for i in range(f // c + 1):
for j in range(f // d + 1):
amount = i * c + j * d
if amount <= f:
sugers.append(amount)
waters = set(waters)
sugers = set(sugers)
maxp = -1.0
res_w = 0
res_s = 0
for w in waters:
for s in sugers:
if w + s > f or w + s == 0:
continue
capa = w // 100 * e
if s > capa:
continue
p = 100 * s / (w + s)
if maxp < p:
maxp, res_w, res_s = p, w + s, s
print("%d %d" % (res_w, res_s))
|
s926616254
|
p03844
|
u089230684
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 77
|
Joisino wants to evaluate the formula "A op B". Here, A and B are integers, and the binary operator op is either `+` or `-`. Your task is to evaluate the formula instead of her.
|
# Python
user_input = input("Input in format A op B")
print(eval(user_input))
|
s589105476
|
Accepted
| 18
| 2,940
| 98
|
n,x,m=input().split()
if x=='+':
print(int(n)+int(m))
if x=='-':
print(int(n)-int(m))
|
s750371696
|
p03352
|
u845620905
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 76
|
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.
|
x = int(input())
ans = 1
while(ans**2 <= x):
ans += 1
print(ans**2)
|
s333919969
|
Accepted
| 17
| 2,940
| 131
|
x = int(input())
ans = 1
for i in range(2, x+1):
n = i*i
while(n <= x):
ans = max(ans, n)
n *= i
print(ans)
|
s176815104
|
p03556
|
u787679173
| 2,000
| 262,144
|
Wrong Answer
| 21
| 2,940
| 68
|
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
N = int(input())
i = 1
while i * i <= N:
i = i + 1
print(i * i)
|
s285037306
|
Accepted
| 24
| 2,940
| 78
|
N = int(input())
i = 1
while (i + 1) * (i + 1) <= N:
i = i + 1
print(i * i)
|
s171575284
|
p02612
|
u840958781
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,096
| 28
|
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())
print(n%1000)
|
s885621406
|
Accepted
| 28
| 9,012
| 64
|
n=int(input())
if n%1000==0:
print(0)
else:
print(1000-n%1000)
|
s512314897
|
p02612
|
u474122160
| 2,000
| 1,048,576
|
Wrong Answer
| 33
| 9,076
| 30
|
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())
print(n-n%1000)
|
s621226174
|
Accepted
| 29
| 9,088
| 69
|
n=int(input())
if(n%1000==0):
print("0")
else:
print(1000-n%1000)
|
s795023056
|
p03698
|
u633450100
| 2,000
| 262,144
|
Wrong Answer
| 31
| 8,996
| 127
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s = input()
list = []
for i in s:
list.append(i)
for j in list:
if j in s:
print('no')
break
print('yes')
|
s617562371
|
Accepted
| 33
| 8,988
| 212
|
import sys
k = 'abcdefghijklmnopqrstuvwxyz'
alp = {}
for i in k:
alp[i] = 0
s = input()
for i in s:
if alp[i] == 0:
alp[i] = 1
else:
print('no')
sys.exit()
print('yes')
|
s035024428
|
p02409
|
u539753516
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,600
| 289
|
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.
|
n=int(input())
a=[[[0 for _ in range(10)]for _ in range(3)]for _ in range(4)]
for i in range(n):
b,f,r,v=map(int, input().split())
a[b-1][f-1][r-1]+=v
for bb in range(b):
for ff in range(f):
print(*a[bb][ff])
if bb==2:
break
else:
print("#"*20)
|
s352556503
|
Accepted
| 30
| 5,624
| 294
|
n=int(input())
a=[[[0 for _ in range(10)]for _ in range(3)]for _ in range(4)]
for i in range(n):
b,f,r,v=map(int, input().split())
a[b-1][f-1][r-1]+=v
for bb in range(4):
for ff in range(3):
print(*[""]+a[bb][ff])
if bb==3:
break
else:
print("#"*20)
|
s785786933
|
p03711
|
u616040357
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 236
|
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
def solve():
x, y = map(int, input().split())
a = [0 , 1 , 3 , 1 , 2 , 1 , 2 , 1 , 1 , 2 , 1 , 2 , 1]
if a[x] == a[y]:
ans = "YES"
else:
ans = "NO"
print(ans)
if __name__ == '__main__':
solve()
|
s432919607
|
Accepted
| 17
| 2,940
| 236
|
def solve():
x, y = map(int, input().split())
a = [0 , 1 , 3 , 1 , 2 , 1 , 2 , 1 , 1 , 2 , 1 , 2 , 1]
if a[x] == a[y]:
ans = "Yes"
else:
ans = "No"
print(ans)
if __name__ == '__main__':
solve()
|
s100149927
|
p02741
|
u448650754
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 139
|
Print the K-th element of the following sequence of length 32: 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51
|
K=int(input())
L=[1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51]
L.sort()
print(L[K-1])
|
s883389772
|
Accepted
| 17
| 2,940
| 132
|
k=int(input())
l=[1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51]
print(l[k-1])
|
s295671597
|
p00004
|
u028347703
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,628
| 466
|
Write a program which solve a simultaneous equation: ax + by = c dx + ey = f The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
|
import sys
def det(mat):
t = 1 / (mat[0][0] * mat[1][1] - mat[0][1] * mat[1][0])
a = t * mat[1][1]
b = t * -1 * mat[0][1]
c = t * -1 * mat[1][0]
d = t * mat[0][0]
return [[a, b], [c, d]]
for line in sys.stdin:
try:
a, b, c, d, e, f = [int(i) for i in line.split()]
A = [[a, b], [d, e]]
P = [c, f]
detA = det(A)
print("%lf %lf" % (detA[0][0] * P[0] + detA[0][1] * P[1], detA[1][0] * P[0] + detA[1][1] * P[1]))
except:
break
|
s694194888
|
Accepted
| 20
| 5,624
| 470
|
import sys
def det(mat):
t = 1 / (mat[0][0] * mat[1][1] - mat[0][1] * mat[1][0])
a = t * mat[1][1]
b = t * -1 * mat[0][1]
c = t * -1 * mat[1][0]
d = t * mat[0][0]
return [[a, b], [c, d]]
for line in sys.stdin:
try:
a, b, c, d, e, f = [int(i) for i in line.split()]
A = [[a, b], [d, e]]
P = [c, f]
detA = det(A)
print("%.3lf %.3lf" % (detA[0][0] * P[0] + detA[0][1] * P[1], detA[1][0] * P[0] + detA[1][1] * P[1]))
except:
break
|
s523121821
|
p02692
|
u619819312
| 2,000
| 1,048,576
|
Wrong Answer
| 186
| 16,808
| 2,723
|
There is a game that involves three variables, denoted A, B, and C. As the game progresses, there will be N events where you are asked to make a choice. Each of these choices is represented by a string s_i. If s_i is `AB`, you must add 1 to A or B then subtract 1 from the other; if s_i is `AC`, you must add 1 to A or C then subtract 1 from the other; if s_i is `BC`, you must add 1 to B or C then subtract 1 from the other. After each choice, none of A, B, and C should be negative. Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.
|
n,a,b,c=map(int,input().split())
if a+b+c==0:
print("No")
elif a+b+c==1:
d=[]
for i in range(n):
s=input()
if s=="AB":
if a+b==0:
print("No")
exit()
else:
if a==0:
d.append("A")
else:
d.append("B")
a,b=b,a
if s=="BC":
if c+b==0:
print("No")
exit()
else:
if b==0:
d.append("B")
else:
d.append("C")
c,b=b,c
if s=="AC":
if a+c==0:
print("No")
exit()
else:
if a==0:
d.append("A")
else:
d.append("C")
a,c=c,a
print("Yes")
for i in d:
print(i)
else:
d=[]
s=[input()for i in range(n)]+["AB"]
for i in range(n):
if s[i]=="AB":
if a+b==0:
print("No")
exit()
if a+b+c==2 and a==b==1 and s[i+1]!=s[i]:
if s[i]=="AC":
d.append("A")
a+=1
b-=1
else:
d.append("B")
b+=1
a-=1
elif a==0:
d.append("A")
a+=1
b-=1
else:
d.append("B")
b+=1
a-=1
if s=="BC":
if c+b==0:
print("No")
exit()
if a+b+c==2 and c==b==1 and s[i+1]!=s[i]:
if s[i]=="AB":
d.append("B")
c-=1
b+=1
else:
d.append("C")
c+=1
b-=1
elif b==0:
d.append("B")
b+=1
c-=1
else:
d.append("C")
c+=1
b-=1
if s=="AC":
if c+a==0:
print("No")
exit()
if a+b+c==2 and c==a==1 and s[i+1]!=s[i]:
if s[i]=="AB":
d.append("A")
a+=1
c-=1
else:
d.append("C")
c+=1
a-=1
elif a==0:
d.append("A")
a+=1
c-=1
else:
d.append("C")
c+=1
a-=1
print("Yes")
for i in d:
print(i)
|
s517107684
|
Accepted
| 210
| 17,048
| 2,754
|
n,a,b,c=map(int,input().split())
if a+b+c==0:
print("No")
elif a+b+c==1:
d=[]
for i in range(n):
s=input()
if s=="AB":
if a+b==0:
print("No")
exit()
else:
if a==0:
d.append("A")
else:
d.append("B")
a,b=b,a
if s=="BC":
if c+b==0:
print("No")
exit()
else:
if b==0:
d.append("B")
else:
d.append("C")
c,b=b,c
if s=="AC":
if a+c==0:
print("No")
exit()
else:
if a==0:
d.append("A")
else:
d.append("C")
a,c=c,a
print("Yes")
for i in d:
print(i)
else:
d=[]
s=[input()for i in range(n)]+["AB"]
for i in range(n):
if s[i]=="AB":
if a+b==0:
print("No")
exit()
if a+b+c==2 and a==1 and b==1 and s[i+1]!=s[i]:
if s[i+1]=="AC":
d.append("A")
a+=1
b-=1
else:
d.append("B")
b+=1
a-=1
elif a<b:
d.append("A")
a+=1
b-=1
else:
d.append("B")
b+=1
a-=1
elif s[i]=="BC":
if c+b==0:
print("No")
exit()
if a+b+c==2 and c==1 and b==1 and s[i+1]!=s[i]:
if s[i+1]=="AB":
d.append("B")
c-=1
b+=1
else:
d.append("C")
c+=1
b-=1
elif b<c:
d.append("B")
b+=1
c-=1
else:
d.append("C")
c+=1
b-=1
elif s[i]=="AC":
if c+a==0:
print("No")
exit()
if a+b+c==2 and c==1and a==1 and s[i+1]!=s[i]:
if s[i+1]=="AB":
d.append("A")
a+=1
c-=1
else:
d.append("C")
c+=1
a-=1
elif a<c:
d.append("A")
a+=1
c-=1
else:
d.append("C")
c+=1
a-=1
print("Yes")
for i in d:
print(i)
|
s643314282
|
p04000
|
u589969467
| 3,000
| 262,144
|
Wrong Answer
| 3,331
| 939,036
| 463
|
We have a grid with H rows and W columns. At first, all cells were painted white. Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column. Compute the following: * For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?
|
h,w,n = map(int,input().split())
s = [[0 for j in range(w+2)] for i in range(h+2)]
#print(s)
for i in range(0,n):
x,y = map(int,input().split())
s[x-1][y-1] += 1
s[x-1][y] += 1
s[x-1][y+1] += 1
s[x][y-1] += 1
s[x][y] += 1
s[x][y+1] += 1
s[x+1][y-1] += 1
s[x+1][y] += 1
s[x+1][y+1] += 1
print(s)
ans = [0] * 10
for i in range(2,h):
for j in range(2,w):
ans[s[i][j]] += 1
for i in range(0,10):
print(ans[i])
|
s698643258
|
Accepted
| 2,679
| 135,604
| 840
|
def add_memo(tate,yoko):
global memo
if 2 <= tate <= h-1:
if 2 <= yoko <= w-1:
memo.append([tate,yoko])
h,w,n = map(int,input().split())
#print(s)
memo = []
for i in range(0,n):
x,y = map(int,input().split())
add_memo(x-1,y-1)
add_memo(x-1,y)
add_memo(x-1,y+1)
add_memo(x,y-1)
add_memo(x,y)
add_memo(x,y+1)
add_memo(x+1,y-1)
add_memo(x+1,y)
add_memo(x+1,y+1)
memo.sort()
ans = [0] * 10
ans[0] = (h-2)*(w-2)
tmp_x = 0
tmp_y = 0
count = 0
for i in range(0,len(memo)):
if tmp_x == memo[i][0] and tmp_y == memo[i][1]:
ans[count] -= 1
count += 1
ans[count] += 1
else:
ans[0] -= 1
count = 1
ans[count] += 1
tmp_x = memo[i][0]
tmp_y = memo[i][1]
# print(memo[i],count)
for i in range(0,10):
print(ans[i])
|
s658088014
|
p04029
|
u076363290
| 2,000
| 262,144
|
Wrong Answer
| 24
| 9,044
| 39
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N = input()
print(int(N)*(int(N)+1)/2)
|
s882521334
|
Accepted
| 29
| 9,044
| 73
|
N = input()
s = 0
i = 1
while i <= int(N):
s = s + i
i = i + 1
print(s)
|
s826453935
|
p03069
|
u510097799
| 2,000
| 1,048,576
|
Wrong Answer
| 120
| 12,512
| 175
|
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 = list(input())
b = s.count("*")
b_l = 0
lst =[n - b]
for i in range(len(s)):
b_l += (s[i] == "*")
lst.append(2 * b_l + (n - i - 1) - b)
print(min(lst))
|
s099449698
|
Accepted
| 115
| 4,840
| 155
|
n = int(input())
s = list(input())
b = 0
w = s.count(".")
m = w
for i in range(n):
if s[i] == "#":
b += 1
else:
w -= 1
m = min(m, b + w)
print(m)
|
s929439965
|
p03407
|
u926678805
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 63
|
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())
print('Yes' if a+b==c else 'No')
|
s067198969
|
Accepted
| 17
| 2,940
| 63
|
a,b,c=map(int,input().split())
print('Yes' if a+b>=c else 'No')
|
s158395656
|
p03361
|
u369212307
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,064
| 604
|
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j). Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`. However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black. Determine if square1001 can achieve his objective.
|
tate, yoko = map(int, input().split())
s = [input() for i in range(tate)]
dxy = [(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0), (1, 1)]
for i in range(tate):
for j in range(yoko):
if s[i][j] == "#":
save = []
for dx, dy in dxy:
if i + dy < 0 or i + dy > tate - 1 or j + dx < 0 or j + dx > yoko - 1: continue
if s[i + dy][j + dx] == "#":
save.append("#")
break
if "#" in save: continue
else:
print("NO")
exit()
print("YES")
|
s365371941
|
Accepted
| 21
| 3,064
| 569
|
tate, yoko = map(int, input().split())
s = [input() for i in range(tate)]
dxy = [(-1, 0), (0, -1), (0, 1), (1, 0),]
for i in range(tate):
for j in range(yoko):
if s[i][j] == "#":
save = []
for dx, dy in dxy:
if i + dy < 0 or i + dy > tate - 1 or j + dx < 0 or j + dx > yoko - 1: continue
if s[i + dy][j + dx] == "#":
save.append("#")
break
if "#" in save: continue
else:
print("No")
exit()
print("Yes")
|
s000030547
|
p03377
|
u006425112
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 83
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
s = input()
sum = 700
for i in s:
if i == "o":
sum += 100
print(sum)
|
s646676230
|
Accepted
| 17
| 2,940
| 120
|
import sys
a, b, x = map(int, sys.stdin.readline().split())
if a <= x <= a + b:
print("YES")
else:
print("NO")
|
s289177716
|
p03557
|
u873904588
| 2,000
| 262,144
|
Wrong Answer
| 429
| 28,432
| 822
|
The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower. He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i. To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar. How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.
|
import bisect
if __name__ == "__main__":
N = int(input())
A = sorted(list(map(int, input().split())))
B = sorted(list(map(int, input().split())))
C = sorted(list(map(int, input().split())))
BB = []
CC = []
print(A)
print(B)
print(C)
print('-------')
cnt = 0
for i in range(N):
b = B[i]
a_index = bisect.bisect_left(A, b)
if i == 0:
BB.append(a_index)
else:
BB.append(BB[i-1] + a_index)
for i in range(N):
c = C[i]
b_index = bisect.bisect_left(B, c)
if b_index == 0:
CC.append(0)
continue
if i == 0:
CC.append(BB[b_index - 1])
else:
CC.append(CC[i-1] + BB[b_index - 1])
# print(CC)
print(CC[-1])
|
s599384164
|
Accepted
| 411
| 26,312
| 830
|
import bisect
if __name__ == "__main__":
N = int(input())
A = sorted(list(map(int, input().split())))
B = sorted(list(map(int, input().split())))
C = sorted(list(map(int, input().split())))
BB = []
CC = []
# print(A)
# print(C)
# print('-------')
cnt = 0
for i in range(N):
b = B[i]
a_index = bisect.bisect_left(A, b)
if i == 0:
BB.append(a_index)
else:
BB.append(BB[i-1] + a_index)
for i in range(N):
c = C[i]
b_index = bisect.bisect_left(B, c)
if b_index == 0:
CC.append(0)
continue
if i == 0:
CC.append(BB[b_index - 1])
else:
CC.append(CC[i-1] + BB[b_index - 1])
# print(CC)
print(CC[-1])
|
s053133271
|
p03623
|
u858136677
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 80
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
l = input().split()
x = int(l[0])
t = int(l[1])
y = x - t
print('{0}'.format(y))
|
s983367728
|
Accepted
| 18
| 3,060
| 172
|
def abs(x):
if x >= 0:
return x
else:
return -1 * x
ls = list(map(int,input().split()))
if abs(ls[0] - ls[1]) < abs(ls[0] - ls[2]):
print ('A')
else:
print ('B')
|
s144825881
|
p03545
|
u708255304
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 239
|
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.
|
ticket = list(str(input()))
for state in range(1 << (len(ticket)-1)):
fomula = ticket[0] + ''.join(('+' if state >> i & 1 else '-') + x for i, x in enumerate(ticket[1:]))
if eval(fomula) == 7:
print(fomula)
exit()
|
s191572217
|
Accepted
| 17
| 3,060
| 467
|
numbers = list(input())
for i in range(1 << (len(numbers)-1)):
tmp = numbers[0]
for j in range(len(numbers)-1):
if (i >> j) & 1:
tmp += "+"
else:
tmp += "-"
tmp += numbers[j+1]
if eval(tmp) == 7:
print(tmp+"=7")
break
|
s593817061
|
p02402
|
u017435045
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,588
| 130
|
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
ns=input().split()
for i in range(len(ns)):
ns[i]=int(ns[i])
asum=sum(ns)
amin=min(ns)
amax=max(ns)
print(amin, amax,asum)
|
s860398078
|
Accepted
| 20
| 6,296
| 139
|
n=input()
ns=input().split()
for i in range(len(ns)):
ns[i]=int(ns[i])
asum=sum(ns)
amin=min(ns)
amax=max(ns)
print(amin, amax,asum)
|
s515456375
|
p03860
|
u683117905
| 2,000
| 262,144
|
Wrong Answer
| 26
| 8,828
| 33
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
s=input()
print("A" + s[0] + "C")
|
s807008378
|
Accepted
| 31
| 8,972
| 30
|
s=input()
print("A"+s[8]+"C")
|
s918941344
|
p04043
|
u241190800
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 173
|
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.
|
data = [int(i) for i in input().split(" ")]
print(data)
total = 0
for i in data:
if i==7 or i==5:
total += i
if total == 17:
print("YES")
else:
print("NO")
|
s003626298
|
Accepted
| 17
| 2,940
| 117
|
tupni = [int(idx) for idx in input().split(" ")]
tupni.sort()
if tupni == [5,5,7]:
print("YES")
else:
print("NO")
|
s158656415
|
p02396
|
u104171359
| 1,000
| 131,072
|
Wrong Answer
| 120
| 7,612
| 375
|
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
#!usr/bin/env python3
def main():
count = 1
while True:
x = int(input())
print('Case ' + str(count) + ': ' + str(x))
count += 1
if x == 0:
break
if __name__ == '__main__':
main()
# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ITP1_3_B
|
s083503689
|
Accepted
| 130
| 7,676
| 227
|
#!usr/bin/env python3
def main():
count = 1
while True:
x = int(input())
if x == 0:
break
print('Case %d: %d' % (count, x))
count += 1
if __name__ == '__main__':
main()
|
s753073472
|
p03407
|
u776871252
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 90
|
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("No")
else:
print("Yes")
|
s989035182
|
Accepted
| 17
| 2,940
| 91
|
A, B, C = map(int, input().split())
if A + B >= C:
print("Yes")
else:
print("No")
|
s916157824
|
p00008
|
u728901930
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,640
| 248
|
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).
|
import sys
import math as mas
n=int(input())
out=0
for i in range(10):
for j in range(10):
for k in range(10):
for l in range(10):
if(i+j+k+l==n):out+=1
print(out)
# a,b=map(int,i.split())
# print(gcd(a,b),lcm(a,b))
|
s587318634
|
Accepted
| 30
| 7,640
| 234
|
import sys
import math as mas
x=[0]*60
for i in range(10):
for j in range(10):
for k in range(10):
for l in range(10):
x[i+j+k+l]+=1
for i in sys.stdin:
print(x[int(i)])
# a,b=map(int,i.split())
# print(gcd(a,b),lcm(a,b))
|
s569797133
|
p03997
|
u912862653
| 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)*h/2)
|
s742509987
|
Accepted
| 18
| 2,940
| 73
|
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2))
|
s945183192
|
p03729
|
u534308356
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,016
| 97
|
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a, b, c = map(str, input().split())
if a == b and b == c:
print("YES")
else:
print("NO")
|
s745097500
|
Accepted
| 30
| 9,032
| 111
|
a, b, c = map(str, input().split())
if a[-1] == b[0] and b[-1] == c[0]:
print("YES")
else:
print("NO")
|
s583196906
|
p02282
|
u805716376
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 282
|
Write a program which reads two sequences of nodes obtained by the preorder tree walk and the inorder tree walk on a binary tree respectively, and prints a sequence of the nodes obtained by the postorder tree walk on the binary tree.
|
n = int(input())
pre = list(map(int, input().split()))
ino = list(map(int, input().split()))
root = pre[1]
pre = iter(pre).__next__
def dfs(l, r):
if l >= r:
return
c = pre()
m = ino.index(c)
dfs(l, m)
dfs(m+1, r)
print(c)
dfs(0, len(ino))
|
s668501857
|
Accepted
| 20
| 5,600
| 310
|
n = int(input())
pre = list(map(int, input().split()))
ino = list(map(int, input().split()))
d = []
root = pre[0]
pre = iter(pre).__next__
def dfs(l, r):
global d
if l >= r:
return
c = pre()
m = ino.index(c)
dfs(l, m)
dfs(m+1, r)
d += [c]
dfs(0, len(ino))
print(*d)
|
s766855782
|
p03836
|
u481333386
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,188
| 644
|
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
def main(sx, sy, tx, ty):
answer = ''
d_x = tx - sx
d_y = ty - sy
one_right = 'R' * d_x
one_up = 'U' * d_y
answer += one_right + one_up
one_down = 'D' * d_y
one_left = 'L' * (d_x + 1)
answer += one_down + one_left + 'L'
two_up = 'U' * (d_y + 1)
two_right = 'R' * (d_x + 1)
answer += two_up + two_right + 'DR'
two_down = 'D' * (d_y + 1)
two_left = 'L' * (d_x + 1)
answer += two_down + two_left + 'U'
return answer
# print(main(
# 0, 0, 1, 2
# ))
if __name__ == '__main__':
sx, sy, tx, ty = [int(e) for e in input().split()]
print(main(sx, sy, tx, ty))
|
s804043975
|
Accepted
| 17
| 3,064
| 638
|
def main(sx, sy, tx, ty):
answer = ''
d_x = tx - sx
d_y = ty - sy
one_right = 'R' * d_x
one_up = 'U' * d_y
answer += one_up + one_right
one_down = 'D' * d_y
one_left = 'L' * (d_x + 1)
answer += one_down + one_left
two_up = 'U' * (d_y + 1)
two_right = 'R' * (d_x + 1)
answer += two_up + two_right + 'DR'
two_down = 'D' * (d_y + 1)
two_left = 'L' * (d_x + 1)
answer += two_down + two_left + 'U'
return answer
# print(main(
# 0, 0, 1, 2
# ))
if __name__ == '__main__':
sx, sy, tx, ty = [int(e) for e in input().split()]
print(main(sx, sy, tx, ty))
|
s196465906
|
p03525
|
u747602774
| 2,000
| 262,144
|
Wrong Answer
| 36
| 9,592
| 1,396
|
In CODE FESTIVAL XXXX, there are N+1 participants from all over the world, including Takahashi. Takahashi checked and found that the _time gap_ (defined below) between the local times in his city and the i-th person's city was D_i hours. The time gap between two cities is defined as follows. For two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is defined to be min(d,24-d) hours. Here, we are using 24-hour notation. That is, the local time in the i-th person's city is either d o'clock or 24-d o'clock at the moment when the local time in Takahashi's city is 0 o'clock, for example. Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours. Find the maximum possible value of s.
|
import sys
readline = sys.stdin.readline
sys.setrecursionlimit(10**8)
mod = 10**9+7
#mod = 998244353
INF = 10**18
eps = 10**-7
N = int(readline())
D = list(map(int,readline().split()))
c = [0]*13
c[0] = 1
for d in D:
c[d] += 1
if c[0] >= 2 or c[12] >= 2:
print(0)
exit()
for i in range(1,12):
if c[i] >= 3:
print(0)
exit()
bitsearchlist = []
notbitsearch = [False]*24
notbitsearch[0] = True
if c[12]:
notbitsearch[12] = True
for i in range(1,12):
if c[i] == 1:
bitsearchlist.append(i)
elif c[i] == 2:
notbitsearch[i] = True
notbitsearch[24-i] = True
print(bitsearchlist)
print(notbitsearch)
def count(List):
anss = INF
ret = INF
for b in notbitsearch:
if b:
anss = min(anss,ret)
ret = 0
else:
ret += 1
anss = min(anss,ret)
anss += 1
return anss
L = len(bitsearchlist)
if L == 0:
print(count(notbitsearch))
exit()
ans = 0
for i in range(2**L):
for j in range(L):
if ((i >> j) & 1):
notbitsearch[bitsearchlist[j]] = True
else:
notbitsearch[24-bitsearchlist[j]] = True
ans = max(ans,count(notbitsearch))
for j in range(L):
if ((i >> j) & 1):
notbitsearch[bitsearchlist[j]] = False
else:
notbitsearch[24-bitsearchlist[j]] = False
print(ans)
|
s125171958
|
Accepted
| 35
| 9,480
| 864
|
import sys
readline = sys.stdin.readline
sys.setrecursionlimit(10**8)
mod = 10**9+7
#mod = 998244353
INF = 10**18
eps = 10**-7
N = int(readline())
D = list(map(int,readline().split()))
cnt = [0]*13
for d in D:
cnt[min(24-d,d)] += 1
cnt[0] += 1
for i in range(1,12):
if cnt[i] >= 3:
print(0)
exit()
if cnt[0] >= 2 or cnt[12] >= 2:
print(0)
exit()
where = [0]*24
where[0] = 1
b = 0
if cnt[12]:
where[12] = 1
for i in range(1,12):
c = cnt[i]
if c == 0:
continue
if c == 2:
where[i] = 1
where[24-i] = 1
b = 0
else:
if b:
where[i] = 1
b = 0
else:
where[24-i] = 1
b = 1
ans = INF
a = 1
for i in range(1,24):
if where[i]:
ans = min(ans,a)
a = 1
else:
a += 1
ans = min(ans,a)
print(ans)
|
s959356247
|
p00003
|
u058433718
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,568
| 273
|
Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so.
|
import sys
num = int(sys.stdin.readline().strip())
for i in range(num):
data = sys.stdin.readline().strip().split(' ')
a = int(data[0])
b = int(data[1])
c = int(data[2])
if a == b and b == c:
print('YES')
else:
print('NO')
|
s701001259
|
Accepted
| 30
| 7,776
| 409
|
import sys
def is_right(edges):
max_len = max(edges)
edges.remove(max_len)
return max_len ** 2 == edges[0] ** 2 + edges[1] ** 2
num = int(sys.stdin.readline().strip())
for i in range(num):
data = sys.stdin.readline().strip().split(' ')
a = int(data[0])
b = int(data[1])
c = int(data[2])
if is_right([a, b, c]):
print('YES')
else:
print('NO')
|
s478614568
|
p03080
|
u896741788
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 69
|
There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat.
|
s=input()
r=s.count("R")
b=s.count("B")
print("Yes" if r>b else "No")
|
s707557395
|
Accepted
| 17
| 2,940
| 79
|
n=input()
s=input()
r=s.count("R")
b=s.count("B")
print("Yes" if r>b else "No")
|
s129947546
|
p02846
|
u363315718
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,064
| 753
|
Takahashi and Aoki are training for long-distance races in an infinitely long straight course running from west to east. They start simultaneously at the same point and moves as follows **towards the east** : * Takahashi runs A_1 meters per minute for the first T_1 minutes, then runs at A_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever. * Aoki runs B_1 meters per minute for the first T_1 minutes, then runs at B_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever. How many times will Takahashi and Aoki meet each other, that is, come to the same point? We do not count the start of the run. If they meet infinitely many times, report that fact.
|
import sys
sys.setrecursionlimit(10**7)
INF = 10 ** 18
MOD = 10 ** 9 + 7
def LI(): return list(map(int, sys.stdin.readline().split()))
def II(): return int(sys.stdin.readline())
def LS(): return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def main():
t1, t2 = LI()
a1, a2 = LI()
b1, b2 = LI()
if a1 < b1:
tmp = b1
b1 = a1
a1 = tmp
tmp = b2
b2 = a2
a2 = tmp
l1 = (a1 - b1) * t1
l2 = (a2 - b2) * t2
l3 = l1 + l2
print(l1, l2)
if l3 == 0:
print("infinity")
return
if l3 > 0:
print(0)
return
ans = 1
print(ans + 2 * (l1 // (-l3)))
if __name__ == '__main__':
main()
|
s231964325
|
Accepted
| 17
| 3,064
| 818
|
import sys
sys.setrecursionlimit(10**7)
INF = 10 ** 18
MOD = 10 ** 9 + 7
def LI(): return list(map(int, sys.stdin.readline().split()))
def II(): return int(sys.stdin.readline())
def LS(): return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def main():
t1, t2 = LI()
a1, a2 = LI()
b1, b2 = LI()
if a1 < b1:
tmp = b1
b1 = a1
a1 = tmp
tmp = b2
b2 = a2
a2 = tmp
l1 = (a1 - b1) * t1
l2 = (a2 - b2) * t2
l3 = l1 + l2
if l3 == 0:
print("infinity")
return
if l3 > 0:
print(0)
return
ans = 1
if l1 % (-l3) == 0:
print(ans + 2 * (l1 // (-l3)) - 1)
return
print(ans + 2 * (l1 // (-l3)))
if __name__ == '__main__':
main()
|
s302171126
|
p03303
|
u363407574
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 114
|
You are given a string S consisting of lowercase English letters. We will write down this string, starting a new line after every w letters. Print the string obtained by concatenating the letters at the beginnings of these lines from top to bottom.
|
S = input()
w = int(input())
stack = []
for x in range(len(S)//w):
stack.append(S[x*w])
print("".join(stack))
|
s300290006
|
Accepted
| 18
| 3,064
| 136
|
import math
S = input()
w = int(input())
stack = []
for x in range(math.ceil(len(S)/w)):
stack.append(S[x*w])
print("".join(stack))
|
s575655644
|
p03660
|
u346812984
| 2,000
| 262,144
|
Wrong Answer
| 412
| 23,740
| 912
|
Fennec and Snuke are playing a board game. On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree. Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored. Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell. More specifically, each player performs the following action in her/his turn: * Fennec: selects an uncolored cell that is adjacent to a **black** cell, and paints it **black**. * Snuke: selects an uncolored cell that is adjacent to a **white** cell, and paints it **white**. A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.
|
import sys
from collections import deque
sys.setrecursionlimit(10 ** 6)
INF = float("inf")
MOD = 10 ** 9 + 7
def input():
return sys.stdin.readline().strip()
def bfs(edges, n_nodes):
visited = [None] * n_nodes
visited[0] = "B"
visited[-1] = "W"
q = deque([0, n_nodes - 1])
while q:
i = q.popleft()
for j in edges[i]:
if visited[j] is None:
print(i, j)
visited[j] = visited[i]
q.append(j)
return visited
def main():
N = int(input())
edges = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(int, input().split())
a -= 1
b -= 1
edges[a].append(b)
edges[b].append(a)
visited = bfs(edges, N)
if visited.count("B") > visited.count("W"):
print("Fennec")
else:
print("Snuke")
if __name__ == "__main__":
main()
|
s695151865
|
Accepted
| 280
| 22,268
| 884
|
import sys
from collections import deque
sys.setrecursionlimit(10 ** 6)
INF = float("inf")
MOD = 10 ** 9 + 7
def input():
return sys.stdin.readline().strip()
def bfs(edges, n_nodes):
visited = [None] * n_nodes
visited[0] = "B"
visited[-1] = "W"
q = deque([0, n_nodes - 1])
while q:
i = q.popleft()
for j in edges[i]:
if visited[j] is None:
visited[j] = visited[i]
q.append(j)
return visited
def main():
N = int(input())
edges = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(int, input().split())
a -= 1
b -= 1
edges[a].append(b)
edges[b].append(a)
visited = bfs(edges, N)
if visited.count("B") > visited.count("W"):
print("Fennec")
else:
print("Snuke")
if __name__ == "__main__":
main()
|
s398057391
|
p00001
|
u558004042
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,720
| 232
|
There is a data which provides heights (in meter) of mountains. The data is only for ten mountains. Write a program which prints heights of the top three mountains in descending order.
|
#!/usr/bin/env python3
#coding: utf-8
# Volume0 - 0001 (http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0001)
li = []
for x in range(10):
y = input()
li.append(y)
li.reverse()
for y in range(3):
print(li[y])
|
s459822627
|
Accepted
| 30
| 6,720
| 257
|
#!/usr/bin/env python3
#coding: utf-8
# Volume0 - 0001 (http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0001)
li = []
for x in range(10):
s = input()
s = int(s)
li.append(s)
li.sort()
li.reverse()
for y in range(3):
print(li[y])
|
s167446142
|
p03860
|
u920438243
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 46
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
line = input().split()
print("A"+line[1]+"C")
|
s108860210
|
Accepted
| 18
| 2,940
| 49
|
line = input().split()
print("A"+line[1][0]+"C")
|
s798389386
|
p03796
|
u723654028
| 2,000
| 262,144
|
Wrong Answer
| 29
| 2,940
| 87
|
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
N=int(input())
p = 1
a = 10 ** 9 + 7
for i in range(1,N+1):
p = p * i // a
print(p)
|
s809993390
|
Accepted
| 36
| 2,940
| 86
|
N=int(input())
p = 1
a = 10 ** 9 + 7
for i in range(1,N+1):
p = p * i % a
print(p)
|
s117407186
|
p03577
|
u252828980
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 18
|
Rng is going to a festival. The name of the festival is given to you as a string S, which ends with `FESTIVAL`, from input. Answer the question: "Rng is going to a festival of what?" Output the answer. Here, assume that the name of "a festival of s" is a string obtained by appending `FESTIVAL` to the end of s. For example, `CODEFESTIVAL` is a festival of `CODE`.
|
print(input()[:8])
|
s951138342
|
Accepted
| 17
| 2,940
| 19
|
print(input()[:-8])
|
s253988158
|
p03543
|
u461636820
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 165
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
# -*- coding: utf-8 -*-
def main():
N = input()
print('YES' if len(set(N[:3])) == 1 or len(set(N[1:])) == 1 else 'NO')
if __name__ == '__main__':
main()
|
s334633407
|
Accepted
| 17
| 2,940
| 165
|
# -*- coding: utf-8 -*-
def main():
N = input()
print('Yes' if len(set(N[:3])) == 1 or len(set(N[1:])) == 1 else 'No')
if __name__ == '__main__':
main()
|
s588273388
|
p03370
|
u925406312
| 2,000
| 262,144
|
Wrong Answer
| 169
| 4,468
| 303
|
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.
|
a,b = map(int,input().split())
c=[int(input()) for i in range(a)]
minbox = min(c)
flag = 0
for i in c:
b = b - i
flag += 1
while True:
print(b)
if b < minbox:
break
else:
b = b - minbox
flag += 1
print(flag)
continue
break
print(flag)
|
s417471440
|
Accepted
| 35
| 3,060
| 270
|
a,b = map(int,input().split())
c=[int(input()) for i in range(a)]
minbox = min(c)
flag = 0
for i in c:
b = b - i
flag += 1
while True:
if b < minbox:
break
else:
b = b - minbox
flag += 1
continue
break
print(flag)
|
s194441285
|
p02612
|
u952708174
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,144
| 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)
|
s484393189
|
Accepted
| 27
| 9,072
| 36
|
print((1000-int(input())%1000)%1000)
|
s595831901
|
p03494
|
u733738237
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 2,940
| 192
|
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.
|
import sys
input = sys.stdin.readline
list_n = [int(i) for i in input().split()]
x=0
while True:
if [i for i in list_n if i % 2 == 1]:
break
nums = [i//2 for i in list_n]
x += 1
print(x)
|
s487924731
|
Accepted
| 19
| 3,060
| 181
|
N = int(input())
a = list(map(int, input().split()))
point = 0
while True:
A =[x % 2 for x in a]
if A.count(0) != len(a):
break
a = [x // 2 for x in a]
point += 1
print(point)
|
s111533782
|
p02612
|
u215268883
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,004
| 34
|
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.
|
a=input()
a=int(a)
print(a % 1000)
|
s075145147
|
Accepted
| 27
| 9,056
| 85
|
N = input()
N = int(N)
r = N % 1000
if r == 0:
print(0)
else:
print(1000 - r)
|
s339156611
|
p03545
|
u169678167
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 364
|
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.
|
s = input()
for bit in range(1 << (3)):
f = s[0]
for i in range(3):
if bit & (1 << i):
f += '+'
f += s[i + 1]
if sum(list(map(int, f.split('+')))) == 7:
print(f)
exit()
|
s551576797
|
Accepted
| 18
| 3,060
| 394
|
s = input()
for bit in range(1 << (3)):
f = s[0]
for i in range(3):
if bit & (1 << i):
f += '+'
else:
f += '-'
f += s[i+1]
if eval(f) == 7:
f += "=7"
print(f)
exit()
|
s252539887
|
p03493
|
u256833330
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 38
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
l=input().split()
print(l.count('1'))
|
s932378475
|
Accepted
| 20
| 2,940
| 30
|
l=input()
print(l.count('1'))
|
s628581866
|
p03377
|
u869519920
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,128
| 83
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A,B,X = map(int,input().split())
s = "Yes"
if A > X or A+B < X:
s = "No"
print(s)
|
s911002387
|
Accepted
| 29
| 9,152
| 83
|
A,B,X = map(int,input().split())
s = "YES"
if A > X or A+B < X:
s = "NO"
print(s)
|
s498735050
|
p03360
|
u366644013
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 91
|
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?
|
a, b, c = map(int, input().split())
k = int(input())
d = max(a, b, c)
print(a+b+c+(d**k-d))
|
s167759547
|
Accepted
| 17
| 2,940
| 90
|
a = list(map(int, input().split()))
k = int(input())
d = max(a)
print(d*2**k + sum(a) - d)
|
s128391513
|
p03141
|
u325227960
| 2,000
| 1,048,576
|
Wrong Answer
| 617
| 47,808
| 253
|
There are N dishes of cuisine placed in front of Takahashi and Aoki. For convenience, we call these dishes Dish 1, Dish 2, ..., Dish N. When Takahashi eats Dish i, he earns A_i points of _happiness_ ; when Aoki eats Dish i, she earns B_i points of happiness. Starting from Takahashi, they alternately choose one dish and eat it, until there is no more dish to eat. Here, both of them choose dishes so that the following value is maximized: "the sum of the happiness he/she will earn in the end" minus "the sum of the happiness the other person will earn in the end". Find the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".
|
n=int(input())
A=[]
for i in range(n):
A.append(list(map(int,input().split())))
B=[]
for i in range(n):
B.append([i,A[i][0]-A[i][1]])
B.sort(key=lambda x:x[1])
print(B)
S=[0,0]
for i in range(n):
S[i%2]+=A[B[i][0]][i%2]
print(S[0]-S[1])
|
s756073788
|
Accepted
| 755
| 43,048
| 304
|
n=int(input())
A=[]
for i in range(n):
A.append(list(map(int,input().split())))
A.sort(reverse=1)
B=[]
for i in range(n):
B.append([i,A[i][1]+A[i][0]])
B.sort(reverse=1,key=lambda x:x[1])
#print(C)
S=[0,0]
for i in range(n):
S[i%2]+=A[B[i][0]][i%2]
print(S[0]-S[1])
|
s746631090
|
p03693
|
u517774558
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 124
|
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?
|
array = input().split()
score = int(array[1]) * 10 + int(array[2])
if score % 4 == 0:
print("Yes")
else:
print("No")
|
s513283377
|
Accepted
| 17
| 2,940
| 123
|
array = input().split()
score = int(array[1]) * 10 + int(array[2])
if score % 4 == 0:
print("YES")
else:
print("NO")
|
s759656396
|
p03605
|
u847033024
| 2,000
| 262,144
|
Wrong Answer
| 25
| 8,900
| 73
|
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?
|
a = input()
if a[0] == 9 or a[1] == 9:
print('Yes')
else:
print('No')
|
s206764285
|
Accepted
| 28
| 8,952
| 60
|
a = input()
if '9' in a:
print('Yes')
else:
print('No')
|
s144441840
|
p03433
|
u684084063
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 184
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
#-*-cording:utf-8-*-
N = int(input())
A = input().split()
A.sort()
A.reverse()
a=list(map(int,A))
Alice = a[1::2]
Bob = a[0::2]
allA=sum(Alice)
allB=sum(Bob)
Ans=allA-allB
print(-Ans)
|
s373735029
|
Accepted
| 17
| 2,940
| 133
|
#-*-cording:utf-8-*-
N = int(input())
A = int(input())
num=N//500
NUM=N-500*num
if NUM <= A:
print("Yes")
else:
print("No")
|
s905318904
|
p01223
|
u073685357
| 3,000
| 131,072
|
Wrong Answer
| 30
| 7,756
| 6,923
|
ある国で「サイゾウ」というテレビ番組が流行している． この番組は参加者がフィールドアスレチックに挑戦し， 見事攻略すれば賞金をもらえるというものである． フィールドアスレチックは高さが異なるブロックを一列に並べて作られていて， 攻略のためにはいかにして段差を登り降りするかが重要になる （図1）． この番組に参加することになったあなたの友人は， フィールドアスレチックの構造が与えられたときに， 登らなければならない最大の段差と降りなければならない最大の段差を 計算するプログラムを書いてほしいと， 凄腕プログラマーであるあなたに依頼してきた． --- 図1: アスレチックの構造の例（入力例の最初のデータセット）．
|
#coding:UTF-8
def Calculate(rows) :
up, down = 0, 0
for x in range(0, len(rows) - 1) :
tmp = rows[x + 1] - rows[x]
if tmp < 0 :
tmp = abs(tmp)
if down < tmp :
down = tmp
elif tmp > 0 :
if up < tmp :
up = tmp
else :
pass
print("{0} {1}".format(up, down))
return(up, down)
def Main() :
fields = int(input())
up, down = [], []
for x in range(fields) :
blocks_q = int(input())
blocks = input().split(' ')
blocks = [int(x) for x in blocks]
if blocks_q != len(blocks) :
pass
tmp_up, tmp_down = Calculate(blocks)
up.append(tmp_up)
down.append(tmp_down)
for x in range(len(up)) :
print("{0} {1}".format(up[x], down[x]))
if __name__ == "__main__" :
Main()
|
s513575560
|
Accepted
| 30
| 7,660
| 6,711
|
#coding:UTF-8
def Calculate(rows) :
up, down = 0, 0
for x in range(0, len(rows) - 1) :
tmp = rows[x + 1] - rows[x]
if tmp < 0 :
tmp = abs(tmp)
if down < tmp :
down = tmp
elif tmp > 0 :
if up < tmp :
up = tmp
else :
pass
return(up, down)
def Main() :
fields = int(input())
up, down = [], []
for x in range(fields) :
blocks_q = int(input())
blocks = input().split(' ')
blocks = [int(x) for x in blocks]
if blocks_q != len(blocks) :
pass
tmp_up, tmp_down = Calculate(blocks)
up.append(tmp_up)
down.append(tmp_down)
for x in range(len(up)) :
print("{0} {1}".format(up[x], down[x]))
if __name__ == "__main__" :
Main()
|
s300794018
|
p04029
|
u672898046
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 61
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n = int(input())
c = 0
for i in range(n+1):
c += i
print(i)
|
s084485006
|
Accepted
| 18
| 2,940
| 61
|
n = int(input())
c = 0
for i in range(n+1):
c += i
print(c)
|
s120462591
|
p02603
|
u514969825
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,120
| 605
|
To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?
|
N = int(input())
s_list = list(input().split())
money = 1000
stock = 0
for i in range(N):
if (i == N-1):
money += stock * int(s_list[i])
stock = 0
break
# Stock price tomorrow is higher than today
if (int(s_list[i+1]) > int(s_list[i])):
stock = stock + int(money / int(s_list[i]))
money = money - (int(s_list[i]) * int(money / int(s_list[i])))
continue
# Stock price tomorrow is lower than or equal to today
if (int(s_list[i+1]) <= int(s_list[i])):
money = money + (int(s_list[i]) * stock)
stock = 0
continue
print("money: {}".format(money))
|
s357871406
|
Accepted
| 36
| 8,832
| 612
|
N = int(input())
s_list = list(input().split())
money = 1000
stock = 0
for i in range(N):
#print("i:{}".format(i))
if (i == N-1):
money += stock * int(s_list[i])
stock = 0
break
# Stock price tomorrow is higher than today
if (int(s_list[i+1]) > int(s_list[i])):
stock = stock + int(money / int(s_list[i]))
money = money - (int(s_list[i]) * int(money / int(s_list[i])))
continue
# Stock price tomorrow is lower than or equal to today
if (int(s_list[i+1]) <= int(s_list[i])):
money = money + (int(s_list[i]) * stock)
stock = 0
continue
print(money)
|
s650304710
|
p03657
|
u301823349
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 115
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a,b=map(int,input().split())
if a%3==0 or b%3==0 or (a+b)%3==0:
print("POSSIBLE")
else:
print("IMPOSSIBLE")
|
s824244963
|
Accepted
| 17
| 2,940
| 115
|
a,b=map(int,input().split())
if a%3==0 or b%3==0 or (a+b)%3==0:
print("Possible")
else:
print("Impossible")
|
s658808140
|
p03798
|
u711539583
| 2,000
| 262,144
|
Wrong Answer
| 23
| 4,092
| 648
|
Snuke, who loves animals, built a zoo. There are N animals in this zoo. They are conveniently numbered 1 through N, and arranged in a circle. The animal numbered i (2≤i≤N-1) is adjacent to the animals numbered i-1 and i+1. Also, the animal numbered 1 is adjacent to the animals numbered 2 and N, and the animal numbered N is adjacent to the animals numbered N-1 and 1. There are two kinds of animals in this zoo: honest sheep that only speak the truth, and lying wolves that only tell lies. Snuke cannot tell the difference between these two species, and asked each animal the following question: "Are your neighbors of the same species?" The animal numbered i answered s_i. Here, if s_i is `o`, the animal said that the two neighboring animals are of the same species, and if s_i is `x`, the animal said that the two neighboring animals are of different species. More formally, a sheep answered `o` if the two neighboring animals are both sheep or both wolves, and answered `x` otherwise. Similarly, a wolf answered `x` if the two neighboring animals are both sheep or both wolves, and answered `o` otherwise. Snuke is wondering whether there is a valid assignment of species to the animals that is consistent with these responses. If there is such an assignment, show one such assignment. Otherwise, print `-1`.
|
n = int(input())
s = [c == 'o' for c in input()]
def next_s(s, flag):
d = {'S' : 'W', 'W' : 'S'}
if flag:
return s
else:
return d[s]
def sim(s0):
ans = ['.' for _ in range(n)]
ans[0] = s0[0]
ans[1] = s0[1]
for i in range(2, n):
if ans[i-1] == 'S':
ans[i] = next_s(ans[i-2], s[i-1])
else:
ans[i] = next_s(ans[i-2], not s[i-1])
res = ans[1] == ans[-1]
if s0[0] == 'S':
if res == s[0]:
return ans
else:
if res != s[0]:
return ans
return False
def solve():
for item in ["SS", "SW", "WS", "WW"]:
ans = sim(item)
if ans:
return print(''.join(ans))
return print(-1)
|
s032067181
|
Accepted
| 228
| 5,516
| 716
|
n = int(input())
s = [c == 'o' for c in input()]
def next_s(s, flag):
d = {'S' : 'W', 'W' : 'S'}
if flag:
return s
else:
return d[s]
def sim(s0):
ans = ['.' for _ in range(n+1)]
ans[0] = s0[0]
ans[1] = s0[1]
for i in range(2, n+1):
if ans[i-1] == 'S':
ans[i] = next_s(ans[i-2], s[i-1])
else:
ans[i] = next_s(ans[i-2], not s[i-1])
res = ans[1] == ans[-2]
if s0[0] == 'S':
if res == s[0] and ans[-1] == ans[0]:
return ans[:-1]
else:
if res != s[0] and ans[-1] == ans[0]:
return ans[:-1]
return False
def solve():
for item in ["SS", "SW", "WS", "WW"]:
ans = sim(item)
if ans:
return print(''.join(ans))
return print(-1)
solve()
|
s731655966
|
p03110
|
u189950446
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 215
|
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
N = int(input())
ans = 0
float(ans)
for i in range(N):
x,u = input().split()
if str(u) == 'JPY':
ans = float(ans) + float(x)
if str(u) == 'BTC':
ans = float(ans) + float(x) * 380000
print(u)
print(ans)
|
s915536464
|
Accepted
| 17
| 3,060
| 204
|
N = int(input())
ans = 0
float(ans)
for i in range(N):
x,u = input().split()
if str(u) == 'JPY':
ans = float(ans) + float(x)
if str(u) == 'BTC':
ans = float(ans) + float(x) * 380000
print(ans)
|
s231465431
|
p02613
|
u516927307
| 2,000
| 1,048,576
|
Wrong Answer
| 141
| 16,428
| 196
|
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.
|
import collections
N = int(input())
S = [input() for i in range(N)]
c = collections.Counter(S)
print("AC","×",c['AC'])
print("WA","×",c['WA'])
print("TLE","×",c['TLE'])
print("RE","×",c['RE'])
|
s060108113
|
Accepted
| 146
| 16,500
| 192
|
import collections
N = int(input())
S = [input() for i in range(N)]
c = collections.Counter(S)
print("AC","x",c['AC'])
print("WA","x",c['WA'])
print("TLE","x",c['TLE'])
print("RE","x",c['RE'])
|
s631975772
|
p03672
|
u551109821
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 147
|
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 = list(input())
n = len(S)
for i in range(n,1,-1):
if i%2==0:
if S[:i//2] == S[i//2:]:
print(i//2)
exit()
|
s748010382
|
Accepted
| 18
| 2,940
| 147
|
S = list(input())
n = len(S)
for i in range(n-1,1,-1):
if i%2==0:
if S[:i//2] == S[i//2:i]:
print(i)
exit()
|
s833578275
|
p03738
|
u474423089
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 105
|
You are given two positive integers A and B. Compare the magnitudes of these numbers.
|
a=int(input())
b=int(input())
if a<b:
print('GREATER')
elif a>b:
print('LESS')
else:
print('EQUAL')
|
s084584567
|
Accepted
| 18
| 2,940
| 201
|
def main():
a = int(input())
b = int(input())
if a > b:
print('GREATER')
elif a < b:
print('LESS')
else:
print('EQUAL')
if __name__ == '__main__':
main()
|
s764299574
|
p02613
|
u554850896
| 2,000
| 1,048,576
|
Wrong Answer
| 156
| 9,180
| 187
|
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())
d = {}
for i in range(N):
x = input()
if x in d.keys():
d[x]+=1
else:
d[x] = 1
for j in d.keys():
print(j + " " + "x" + " " + str(d[j]))
print("\n")
|
s960774965
|
Accepted
| 158
| 9,220
| 381
|
N = int(input())
ac = 0
wa = 0
tle = 0
re = 0
while (N) :
x = input()
if x == 'AC':
ac+=1
if x == 'WA':
wa +=1
if x == 'TLE':
tle += 1
if x == 'RE':
re += 1
N-=1
print('AC' + " " + "x" + " " + str(ac))
print('WA' + " " + "x" + " " + str(wa))
print('TLE' + " " + "x" + " " + str(tle))
print('RE' + " " + "x" + " " + str(re))
|
s023222390
|
p03796
|
u210827208
| 2,000
| 262,144
|
Wrong Answer
| 26
| 2,940
| 76
|
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
n=int(input())
ans=1
for i in range(n+1):
ans*=1
print(ans%(10**9+7))
|
s877089430
|
Accepted
| 36
| 2,940
| 83
|
n=int(input())
ans=1
for i in range(1,n+1):
ans=(ans*i)%(10**9+7)
print(ans)
|
s619432209
|
p02612
|
u017624958
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,144
| 50
|
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())
answer = N / 1000
print(answer)
|
s863120929
|
Accepted
| 31
| 9,080
| 101
|
import math
N = int(input())
answer = math.ceil(N / 1000)
answer = answer * 1000 - N
print(answer)
|
s158234892
|
p04044
|
u634873566
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 81
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
n, l = map(int, input().split())
d = {input() for _ in range(n)}
print(sorted(d))
|
s154147528
|
Accepted
| 17
| 3,060
| 90
|
n, l = map(int, input().split())
d = sorted([input() for _ in range(n)])
print(''.join(d))
|
s038121153
|
p03814
|
u607074939
| 2,000
| 262,144
|
Wrong Answer
| 43
| 3,516
| 198
|
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
s = str(input())
Acount = 0
Zcount = 0
for i in range (len(s)):
if s[i] == 'A':
Acount = i
break
for j in range (len(s)):
if s[j] == 'Z':
Zcount = j
print(j - i +1)
|
s007512944
|
Accepted
| 73
| 3,516
| 234
|
s = str(input())
Acount = 3000000
Zcount = 0
for i in range (len(s)):
if (s[i] == 'A' and i < Acount):
Acount = i
for j in range (len(s)):
if (s[j] == 'Z' and j > Zcount):
Zcount = j
print(Zcount - Acount +1)
|
s771173258
|
p02612
|
u974918235
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,120
| 91
|
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.
|
a = int(input())
if a > 10000000:
print(f"{a-10000000}")
else:
print(f"{a % 1000}")
|
s205137144
|
Accepted
| 28
| 9,144
| 38
|
N = int(input())
print(f"{-N % 1000}")
|
s456253133
|
p03455
|
u287097467
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
c = a*b %2
if c:
print('odd')
else:
print('even')
|
s956590725
|
Accepted
| 17
| 2,940
| 88
|
a, b = map(int, input().split())
c = a*b %2
if c:
print('Odd')
else:
print('Even')
|
s125423427
|
p03385
|
u263824932
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 221
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
S=input()
t=[a for a in S]
number=0
for a in range(2):
if 'a' in t:
number+=1
elif 'b' in t:
number+=1
elif 'c' in t:
number+=1
if number==3:
print('Yes')
else:
print('No')
|
s884497349
|
Accepted
| 18
| 2,940
| 124
|
S=input()
t=[a for a in S]
u=sorted(t)
if u[0]=='a' and u[1]=='b' and u[2]=='c':
print('Yes')
else:
print('No')
|
s745883323
|
p02406
|
u843169619
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,584
| 117
|
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
num = int(input().rstrip())
for i in range(1,num+1):
if i % 3 == 0 or '3' in str(i):
print(i)
|
s699590660
|
Accepted
| 20
| 6,124
| 123
|
n=int(input())
d=[]
for i in range(3,n+1):
if i%3==0 or "3" in str(i):
d.append(i)
print(' ',end="")
print(*d)
|
s319894887
|
p02612
|
u261036477
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,136
| 38
|
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())
A = 1000 - N
print(A)
|
s298308029
|
Accepted
| 36
| 9,196
| 275
|
N = input()
if len(N) >= 3:
A = int(N[-3])*100 + int(N[-2])*10 + int(N[-1])
if A == 0:
print(0)
else:
print(1000 - A)
elif len(N) == 2:
A = int(N[-2])*10 + int(N[-1])
print(1000 - A)
elif len(N) == 1:
A = int(N[-1])
print(1000 - A)
|
s493595106
|
p02255
|
u004192101
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,580
| 237
|
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
n = input()
A = [int(i) for i in input().split(' ')]
for i in range(1, len(A)):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j+1] = A[j]
j -= 1
A[j+1] = v
[print(i, end=' ') for i in A]
print()
|
s808644653
|
Accepted
| 40
| 8,452
| 636
|
n = input()
A = [int(i) for i in input().split(' ')]
def trace(A):
for index, v in enumerate(A):
print(v, end='')
if index != len(A) - 1:
print(' ', end='')
print()
trace(A)
# algorithm
# while j >= 0 and A[j] > v:
# A[j+1] = A[j]
# j -= 1
# A[j+1] = v
# trace(A)
for index, value in enumerate(A):
if index == 0:
continue
prev_idx = index - 1
while prev_idx >= 0 and A[prev_idx] > value:
A[prev_idx + 1] = A[prev_idx]
prev_idx -= 1
A[prev_idx + 1] = value
trace(A)
|
s554184494
|
p03149
|
u978167553
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,084
| 102
|
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
n = input().split()
print('Yes' if ('1' in n) and ('9' in n) and ('7' in n) and ('4' in n) else 'No')
|
s312433314
|
Accepted
| 35
| 8,940
| 102
|
n = input().split()
print('YES' if ('1' in n) and ('9' in n) and ('7' in n) and ('4' in n) else 'NO')
|
s369860159
|
p03377
|
u399973890
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 100
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A, B, X = map(int, input().split())
if A <= X and A+B >= X:
print('Yes')
else:
print(('No'))
|
s914389269
|
Accepted
| 18
| 3,064
| 98
|
A, B, X = map(int, input().split())
if A <= X and A+B >= X:
print('YES')
else:
print('NO')
|
s142338640
|
p03795
|
u434630332
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,112
| 157
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n = int(input())
if n >= n % 15 == 0:
x = n * 800
y = 200 * n / 15
answer = x - y
print(answer)
else:
answer = n * 800
print(answer)
|
s324942297
|
Accepted
| 27
| 9,020
| 93
|
n = int(input())
if n >= 15:
print(800 * n - ( n // 15) * 200)
else:
print(800 * n )
|
s018192509
|
p03377
|
u381416158
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 103
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A, B, X = [int(i) for i in input().split()]
if A <= X <= A + B:
print("Yes")
else:
print("No")
|
s865298552
|
Accepted
| 17
| 2,940
| 103
|
A, B, X = [int(i) for i in input().split()]
if A <= X <= A + B:
print("YES")
else:
print("NO")
|
s166205077
|
p03623
|
u302292660
| 2,000
| 262,144
|
Wrong Answer
| 16
| 2,940
| 62
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b = map(int,input().split())
print(min(abs(x-a),abs(x-b)))
|
s385144890
|
Accepted
| 17
| 2,940
| 73
|
x,a,b = map(int,input().split())
print("A" if abs(x-a)<abs(x-b) else "B")
|
s329636276
|
p02936
|
u738898077
| 2,000
| 1,048,576
|
Wrong Answer
| 1,876
| 59,120
| 410
|
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())
v = [[0,0] for i in range(n+1)]
for i in range(n-1):
a,b = map(int,input().split())
v[a].append(b)
for i in range(q):
p,x = map(int,input().split())
v[p][1] += x
q = [1]
while q:
now = q.pop()
# print(now)
for i in v[now][2:]:
v[i][1] += v[now][1]
q += v[now][2:]
# print(q)
print(v)
s = []
for i in v:
s.append(i[1])
print(*s[1:])
|
s836825357
|
Accepted
| 1,853
| 58,420
| 566
|
from collections import deque
n,q = map(int,input().split())
v = [[0,0] for i in range(n+1)]
# print(v)
for i in range(n-1):
a,b = map(int,input().split())
v[a].append(b)
v[b].append(a)
for i in range(q):
p,x = map(int,input().split())
v[p][1] += x
que = deque([1])
# print(v)
v[1][0] = 1
while que:
temp = que.popleft()
for i in v[temp][2:]:
if v[i][0] == 0:
v[i][0] = 1
que.append(i)
v[i][1] += v[temp][1]
# print(v)
ans = []
for i in v[1:]:
ans.append(i[1])
# print(i[1])
print(*ans)
|
s740166077
|
p02694
|
u113255362
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 8,908
| 125
|
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())
k = 0
res = 100
for i in range (1, 4200):
res = int(res * 1.01)
if res > X:
k = i
break
print(k)
|
s344182912
|
Accepted
| 26
| 9,096
| 122
|
X = int(input())
k = 0
res = 100
for i in range (1, 4200):
res += res // 100
if res >= X:
k = i
break
print(k)
|
s076422135
|
p03359
|
u026788530
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 100
|
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 = input().split()
a = int(a)
b = int(b)
print(a,b)
if b > a:
print(a+1)
else:
print(a)
|
s649996381
|
Accepted
| 17
| 2,940
| 90
|
a , b = input().split()
a = int(a)
b = int(b)
if b >= a:
print(a)
else:
print(a-1)
|
s535998564
|
p02806
|
u209627628
| 2,525
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 95
|
Niwango created a playlist of N songs. The title and the duration of the i-th song are s_i and t_i seconds, respectively. It is guaranteed that s_1,\ldots,s_N are all distinct. Niwango was doing some work while playing this playlist. (That is, all the songs were played once, in the order they appear in the playlist, without any pause in between.) However, he fell asleep during his work, and he woke up after all the songs were played. According to his record, it turned out that he fell asleep at the very end of the song titled X. Find the duration of time when some song was played while Niwango was asleep.
|
N = int(input())
i = 0
songs = []
while N-i:
songs.append(input().split())
i+=1
X = input()
|
s789657073
|
Accepted
| 17
| 3,060
| 279
|
N = int(input())
songs = []
result = 0
for i in range(N):
songs.append(input().split())
X = input()
point = len(songs)
for j in range(N):
if songs[j][0] == X:
point = j+1
break
for i in range(point, N):
result += int(songs[i][1])
print(result)
|
s798789804
|
p02612
|
u704157847
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,140
| 40
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print(1000 - 1000 % N)
|
s079264429
|
Accepted
| 27
| 9,152
| 63
|
n = int(input())
print(1000 - n % 1000 if n % 1000 > 0 else 0)
|
s084112473
|
p03854
|
u873134970
| 2,000
| 262,144
|
Wrong Answer
| 37
| 3,188
| 302
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
S = input()
S = S[::-1]
_ = ["dream", "dreamer", "erase", "eraser"]
_ = [_[i][::-1] for i in range(4)]
i = 0
while i < len(S):
can = False
for k in _:
if S[i:i + len(k)] == k:
can = True
i += len(k)
if not can:
print("No")
exit()
print("Yes")
|
s618881319
|
Accepted
| 38
| 3,188
| 302
|
S = input()
S = S[::-1]
_ = ["dream", "dreamer", "erase", "eraser"]
_ = [_[i][::-1] for i in range(4)]
i = 0
while i < len(S):
can = False
for k in _:
if S[i:i + len(k)] == k:
can = True
i += len(k)
if not can:
print("NO")
exit()
print("YES")
|
s764880295
|
p03545
|
u341855122
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,188
| 794
|
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 = input()
c = map(int,list(a))
b = []
for i in c:
b.append(i)
if(b[0] + b[1] + b[2] + b[3]==7):
print(a[0]+"+"+a[1]+"+"+a[2]+"+"+a[3]+"=7",end="")
elif(b[0] + b[1] + b[2] - b[3]==7):
print(a[0]+"+"+a[1]+"+"+a[2],"-"+a[3]+"=7",end="")
elif(b[0] + b[1] - b[2] + b[3]==7):
print(a[0]+"+"+a[1]+"-"+a[2],"+"+a[3]+"=7",end="")
elif(b[0] + b[1] - b[2] - b[3]==7):
print(a[0]+"+"+a[1]+"-"+a[2],"-"+a[3]+"=7",end="")
elif(b[0] - b[1] + b[2] + b[3]==7):
print(a[0]+"-"+a[1]+"+"+a[2],"+"+a[3]+"=7",end="")
elif(b[0] - b[1] + b[2] - b[3]==7):
print(a[0]+"-"+a[1]+"+"+a[2],"-"+a[3]+"=7",end="")
elif(b[0] - b[1] - b[2] + b[3]==7):
print(a[0]+"-"+a[1]+"-"+a[2],"+"+a[3]+"=7",end="")
elif(b[0] - b[1] - b[2] - b[3]==7):
print(a[0]+"-"+a[1]+"-"+a[2],"-"+a[3]+"=7",end="")
|
s576851972
|
Accepted
| 18
| 3,188
| 737
|
a = input()
c = map(int,list(a))
b = []
for i in c:
b.append(i)
if(b[0] + b[1] + b[2] + b[3]==7):
print(a[0]+"+"+a[1]+"+"+a[2]+"+"+a[3]+"=7")
elif(b[0] + b[1] + b[2] - b[3]==7):
print(a[0]+"+"+a[1]+"+"+a[2]+"-"+a[3]+"=7")
elif(b[0] + b[1] - b[2] + b[3]==7):
print(a[0]+"+"+a[1]+"-"+a[2]+"+"+a[3]+"=7")
elif(b[0] + b[1] - b[2] - b[3]==7):
print(a[0]+"+"+a[1]+"-"+a[2]+"-"+a[3]+"=7")
elif(b[0] - b[1] + b[2] + b[3]==7):
print(a[0]+"-"+a[1]+"+"+a[2]+"+"+a[3]+"=7")
elif(b[0] - b[1] + b[2] - b[3]==7):
print(a[0]+"-"+a[1]+"+"+a[2]+"-"+a[3]+"=7")
elif(b[0] - b[1] - b[2] + b[3]==7):
print(a[0]+"-"+a[1]+"-"+a[2]+"+"+a[3]+"=7")
elif(b[0] - b[1] - b[2] - b[3]==7):
print(a[0]+"-"+a[1]+"-"+a[2]+"-"+a[3]+"=7")
|
s501869964
|
p02255
|
u177808190
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,600
| 362
|
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
def insertionSort(A, N):
for i in range(1, N):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j+1] = A[j]
j -= 1
A[j+1] = v
print (' '.join([str(x) for x in A]))
if __name__ == '__main__':
length = int(input())
hoge = [int(x) for x in input().split()]
insertionSort(hoge, length)
|
s572070207
|
Accepted
| 20
| 5,612
| 407
|
def insertionSort(A, N):
for i in range(1, N):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j+1] = A[j]
j -= 1
A[j+1] = v
print (' '.join([str(x) for x in A]))
if __name__ == '__main__':
length = int(input())
hoge = [int(x) for x in input().split()]
print (' '.join([str(x) for x in hoge]))
insertionSort(hoge, length)
|
s092147216
|
p03502
|
u952708174
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 63
|
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 = input()
print('Yes' if int(N) % sum(map(int, N)) else 'No')
|
s879758671
|
Accepted
| 18
| 2,940
| 68
|
N = input()
print('Yes' if int(N) % sum(map(int, N)) == 0 else 'No')
|
s450340679
|
p03852
|
u790812284
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 112
|
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`.
|
list = ["a","b","c","d","e"]
c =input()
if c in list==True:
print("vowel")
else:
print("consonant")
|
s709548635
|
Accepted
| 17
| 2,940
| 101
|
list = ["a","i","u","e","o"]
c =input()
if c in list:
print("vowel")
else:
print("consonant")
|
s378881434
|
p03564
|
u966000628
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 112
|
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.
|
N = int(input())
K = int(input())
i = 1
for num in range(N):
if i < K:
i *= 2
else:
i += K
print (K)
|
s467153827
|
Accepted
| 17
| 2,940
| 112
|
N = int(input())
K = int(input())
i = 1
for num in range(N):
if i < K:
i *= 2
else:
i += K
print (i)
|
s797126486
|
p02841
|
u031324264
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 111
|
In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month.
|
m1, d1 = map(int, input().split())
m2, d2 = map(int, input().split())
if m1 != m2:
print(0)
else:
print(1)
|
s642105702
|
Accepted
| 17
| 2,940
| 111
|
m1, d1 = map(int, input().split())
m2, d2 = map(int, input().split())
if m1 != m2:
print(1)
else:
print(0)
|
s380686184
|
p03455
|
u845937249
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 169
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
moda = a % 2
modb = b % 2
if moda == 0:
if modb == 0:
print("Even")
else:
print("Odd")
else:
print("Odd")
|
s492344496
|
Accepted
| 17
| 2,940
| 198
|
a, b = map(int, input().split())
moda = a % 2
modb = b % 2
if moda == 0:
print("Even")
else:
if modb == 0:
print("Even")
else:
print("Odd")
|
s979068577
|
p03150
|
u212328220
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,068
| 240
|
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
s = input()
if s == 'keyence':
print('YES')
exit()
for i in range(0,len(s)-1):
for j in range(i+1,len(s)):
print(s[:i]+s[j:])
if s[:i]+s[j:] == 'keyence':
print('YES')
exit()
print('NO')
|
s328290680
|
Accepted
| 31
| 9,060
| 213
|
s = input()
if s == 'keyence':
print('YES')
exit()
for i in range(0,len(s)-1):
for j in range(i+1,len(s)):
if s[:i]+s[j:] == 'keyence':
print('YES')
exit()
print('NO')
|
s126348649
|
p03386
|
u492030100
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 262
|
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())
ans_array = []
for i in range(A, A + K if B > (A + K) else B):
ans_array.append(i)
for i in range(B, B - K if A < (B - K) else A, -1):
ans_array.append(i)
ans_array = sorted(list(set(ans_array)))
print(ans_array)
|
s215446830
|
Accepted
| 17
| 3,060
| 285
|
A, B, K = map(int, input().split())
ans_array = []
for i in range(A, A + K if B > (A + K) else (B+1)):
ans_array.append(i)
for i in range(B, B - K if A < (B - K) else (A-1), -1):
ans_array.append(i)
ans_array = sorted(list(set(ans_array)))
for i in ans_array:
print(i)
|
s374414946
|
p03502
|
u969211566
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 172
|
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())
total = 0
flag = 1
nn = n
while flag:
total += nn % 10
nn = nn // 10
if nn <= 0:
flag = 0
if total % n == 0:
print("Yes")
else:
print("No")
|
s721538129
|
Accepted
| 18
| 2,940
| 172
|
n = int(input())
total = 0
flag = 1
nn = n
while flag:
total += nn % 10
nn = nn // 10
if nn <= 0:
flag = 0
if n % total == 0:
print("Yes")
else:
print("No")
|
s778024877
|
p03828
|
u934868410
| 2,000
| 262,144
|
Wrong Answer
| 76
| 3,436
| 239
|
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.
|
from collections import defaultdict
n = int(input())
f = defaultdict(int)
for i in range(2, n+1):
for j in range(2, i+1):
if i % j == 0:
f[j] += 1
ans = 1
mod = 10**9 + 7
for k,v in f.items():
ans = (ans * v) % mod
print(ans)
|
s554964191
|
Accepted
| 22
| 3,316
| 403
|
from collections import defaultdict
n = int(input())
f = defaultdict(int)
def fact(n, d):
div = 2
while n > 1 and div * div <= n:
if n % div == 0:
n //= div
d[div] += 1
else:
div += 1
if n > 1:
d[n] += 1
for i in range(2, n+1):
fact(i, f)
ans = 1
mod = 10**9 + 7
for k,v in f.items():
ans = (ans * (v+1)) % mod
print(ans)
|
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