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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s436856298
|
p03943
|
u533679935
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 106
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a =[int(i)for i in input().split()]
a.sort()
if a[0] == a[1]+a[2]:
print("Yes")
else:
print("No")
|
s493721869
|
Accepted
| 18
| 2,940
| 115
|
a,b,c = map(int,input().split())
if a + b == c or a + c == b or b + c == a:
print("Yes")
else:
print("No")
|
s122601619
|
p03644
|
u663438907
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 97
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
N = int(input())
l = [64, 32, 16, 8, 4, 2]
for i in range(len(l)):
if l[i] <= N:
print(N)
|
s114161806
|
Accepted
| 17
| 2,940
| 115
|
N = int(input())
l = [64, 32, 16, 8, 4, 2, 1]
for i in range(len(l)):
if l[i] <= N:
print(l[i])
break
|
s409436950
|
p04043
|
u460615319
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,136
| 73
|
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.
|
a, b, c = map(int, input().split())
print('Yes' if a+b+c == 17 else 'No')
|
s501648660
|
Accepted
| 28
| 9,092
| 73
|
a, b, c = map(int, input().split())
print('YES' if a+b+c == 17 else 'NO')
|
s283631722
|
p03494
|
u374765578
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 199
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N=int(input())
L=list(map(int,input().split()))
counter=0
t=0
for i in range(N):
if L[i] % 2 == 0:
t += 1
if t==N:
for i in range(N):
L[i]=int(L[i]/2)
counter+=1
print(counter)
|
s696105364
|
Accepted
| 20
| 3,060
| 225
|
N=int(input())
L=list(map(int,input().split()))
counter=0
t=0
for p in range (30):
t=0
for i in range(N):
if L[i]%2==0:
t+=1
if t==N:
for i in range(N):
L[i]=int(L[i]/2)
counter+=1
print(counter)
|
s461100731
|
p03163
|
u997113115
| 2,000
| 1,048,576
|
Wrong Answer
| 203
| 14,580
| 216
|
There are N items, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), Item i has a weight of w_i and a value of v_i. Taro has decided to choose some of the N items and carry them home in a knapsack. The capacity of the knapsack is W, which means that the sum of the weights of items taken must be at most W. Find the maximum possible sum of the values of items that Taro takes home.
|
import numpy as np
N, W = map(int, input().split())
dp = np.zeros(W+1, dtype=int)
for i in range(N):
w, v = map(int, input().split())
np.maximum(dp[:W-w+1]+v, dp[w:], out=dp[w:])
print(dp)
print(dp[-1])
|
s257549150
|
Accepted
| 172
| 14,600
| 202
|
import numpy as np
N, W = map(int, input().split())
dp = np.zeros(W+1, dtype=int)
for i in range(N):
w, v = map(int, input().split())
np.maximum(dp[:W-w+1]+v, dp[w:], out=dp[w:])
print(dp[-1])
|
s100442285
|
p03377
|
u152671129
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 81
|
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())
print('No' if x < a or a + b < x else 'Yes')
|
s298842069
|
Accepted
| 18
| 2,940
| 81
|
a, b, x = map(int, input().split())
print('NO' if x < a or a + b < x else 'YES')
|
s000795328
|
p03738
|
u210827208
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 110
|
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('GRETER')
elif a<b:
print('LESS')
else:
print('EQUAL')
|
s647999681
|
Accepted
| 17
| 2,940
| 111
|
a=int(input())
b=int(input())
if a>b:
print('GREATER')
elif a<b:
print('LESS')
else:
print('EQUAL')
|
s297525989
|
p02612
|
u195272001
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,160
| 86
|
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())
n = int(N // 1000)
if (n==0):
print(n)
else:
print(n+1)
|
s893706710
|
Accepted
| 27
| 9,172
| 108
|
N = int(input())
n = int(N // 1000)
m = int(N % 1000)
if (m==0):
print(0)
else:
print(1000*(n+1)-N)
|
s412948101
|
p03377
|
u239342230
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 67
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
a,b,c=map(int,input().split())
print(['No','Yes'][(a+b>=c)*(a<=c)])
|
s304570843
|
Accepted
| 17
| 2,940
| 67
|
a,b,c=map(int,input().split())
print(['NO','YES'][(a+b>=c)*(a<=c)])
|
s244173081
|
p03494
|
u054935796
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 234
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
A = list(map(int, input().split()))
t = 0
def exist_odd(A):
for i in A:
if i % 2 == 1:
break
return 1
else:
return 0
while exist_odd(A) ==1:
A = list(map(lambda x: x / 2, A))
t = t+1
print(t)
|
s046423295
|
Accepted
| 18
| 2,940
| 214
|
N = int(input())
A = list(map(int, input().split()))
t = 0
def exist_odd(A):
B = list(map(lambda x : x%2, A))
return any(B)
while exist_odd(A) != True:
A = list(map(lambda x: x // 2, A))
t = t+1
print(t)
|
s651644858
|
p02612
|
u048844044
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,152
| 32
|
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)
|
s369173936
|
Accepted
| 30
| 9,164
| 74
|
n = int(input())
x = 0
while x <= n:
x += 1000
print((x - n) % 1000)
|
s241004932
|
p02420
|
u131984977
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,572
| 169
|
Your task is to shuffle a deck of n cards, each of which is marked by a alphabetical letter. A single shuffle action takes out h cards from the bottom of the deck and moves them to the top of the deck. The deck of cards is represented by a string as follows. abcdeefab The first character and the last character correspond to the card located at the bottom of the deck and the card on the top of the deck respectively. For example, a shuffle with h = 4 to the above deck, moves the first 4 characters "abcd" to the end of the remaining characters "eefab", and generates the following deck: eefababcd You can repeat such shuffle operations. Write a program which reads a deck (a string) and a sequence of h, and prints the final state (a string).
|
while True:
s = input()
if s == '-':
break
m = int(input())
for mi in range(m):
h = int(input())
s = s[:h] + s[h:]
print(s)
|
s188561509
|
Accepted
| 20
| 7,648
| 169
|
while True:
s = input()
if s == '-':
break
m = int(input())
for mi in range(m):
h = int(input())
s = s[h:] + s[:h]
print(s)
|
s146360684
|
p03478
|
u186893542
| 2,000
| 262,144
|
Wrong Answer
| 34
| 2,940
| 146
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n, a, b = map(int, input().split())
res = 0
for i in range(a, n+1):
s = str(i)
sum = 0
for j in s:
sum += int(s)
res += sum
print(res)
|
s625671960
|
Accepted
| 32
| 3,060
| 166
|
n, a, b = map(int, input().split())
res = 0
for i in range(a, n+1):
s = str(i)
sum = 0
for j in s:
sum += int(j)
if a <= sum <= b:
res += i
print(res)
|
s235055941
|
p02646
|
u376812964
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,120
| 282
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
import sys
orga = list(map(float, input().split()))
child = list(map(float, input().split()))
time = float(input())
if orga[1] <= child[1]:
print("No")
sys.exit()
dis = abs(orga[0] - child[0])
if dis / (orga[1] - child[1]) <= time:
print("Yes")
else:
print("No")
|
s715166837
|
Accepted
| 24
| 9,208
| 280
|
import sys
orga = list(map(int, input().split()))
child = list(map(int, input().split()))
time = int(input())
speed = orga[1] - child[1]
dis = abs(orga[0] - child[0])
if speed <= 0:
print("NO")
sys.exit()
if dis / speed <= time:
print("YES")
else:
print("NO")
|
s404946648
|
p02255
|
u901205536
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 200
|
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 = int(input())
A = list(map(int, input().split(" ")))
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(A)
|
s819021100
|
Accepted
| 30
| 5,984
| 333
|
N = int(input())
A = list(map(int, input().split(" ")))
for k in range(len(A)-1):
print(A[k], end= " ")
print(A[-1])
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
for k in range(len(A)-1):
print(A[k], end= " ")
print(A[-1])
|
s650802193
|
p03369
|
u150641538
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 87
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
s = input()
cnt = 0
for t in s:
if(s=="o"):
cnt += 1
print(700 + cnt * 100)
|
s597382166
|
Accepted
| 17
| 2,940
| 87
|
s = input()
cnt = 0
for t in s:
if(t=="o"):
cnt += 1
print(700 + cnt * 100)
|
s492642511
|
p03565
|
u593567568
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 512
|
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
S = input()
T = input()
len_s = len(S)
len_t = len(T)
match = -1
for i in range((len_s - len_t), 0, -1):
print("i", i)
for j in range(len_t):
if T[j] != S[i+j] and S[i+j] != "?":
break
if j == len_t-1:
match = i
if match != -1:
break
if match == -1:
print("UNRESTORABLE")
exit()
list_s = list(S)
for i in range(match, match+len_t):
list_s[i] = T[i - match]
list_s = [x if x != "?" else "a" for x in list_s]
print("".join(list_s))
|
s030373063
|
Accepted
| 17
| 3,064
| 495
|
S = input()
T = input()
len_s = len(S)
len_t = len(T)
match = -1
for i in range((len_s - len_t), -1, -1):
for j in range(len_t):
if T[j] != S[i+j] and S[i+j] != "?":
break
if j == len_t-1:
match = i
if match != -1:
break
if match == -1:
print("UNRESTORABLE")
exit()
list_s = list(S)
for i in range(match, match+len_t):
list_s[i] = T[i - match]
list_s = [x if x != "?" else "a" for x in list_s]
print("".join(list_s))
|
s611080924
|
p02806
|
u051684204
| 2,525
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 238
|
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())
ls1=[]
for _ in range(N):
title,time=input().split()
ls1.append([title,float(time)])
X=input()
i=0
a=0
flag=False
while flag==False:
if ls1[i][0]==X:
i+=1
break
i+=1
while i<N:
a+=ls1[i][1]
i+=1
print(a)
|
s058532949
|
Accepted
| 17
| 3,060
| 236
|
N=int(input())
ls1=[]
for _ in range(N):
title,time=input().split()
ls1.append([title,int(time)])
X=input()
i=0
a=0
flag=False
while flag==False:
if ls1[i][0]==X:
i+=1
break
i+=1
while i<N:
a+=ls1[i][1]
i+=1
print(a)
|
s896055398
|
p03476
|
u074220993
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,206
| 31,592
| 455
|
We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime. You are given Q queries. In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.
|
import numpy as np
def isP(x):
for i in range(2,int(np.sqrt(x))+1):
if x % i == 0:
return False
return True
MAX = int(10e5+1)
P = [isP(x) for x in range(MAX)]
Ans = [0] * MAX
cnt = 0
for i in range(1,MAX):
if P[i] and P[(i+1)//2]:
cnt += 1
Ans[i] = cnt
q = int(input())
for i in range(q):
l, r = map(int, input().split())
print(Ans[r]-Ans[l-1])
|
s688024401
|
Accepted
| 346
| 47,588
| 763
|
import numpy as np
from math import sqrt
def main():
with open(0) as f:
Q = int(f.readline())
Query = [tuple(map(int, line.split())) for line in f.readlines()]
p_table = makePtable(10**5)
p_set = set(p_table)
Like2017 = [p for p in p_table if (p+1)//2 in p_set]
database = np.array([0] * (10**5+1))
for i in Like2017:database[i] = 1
database = database.cumsum(axis=0)
for l,r in Query:
print(database[r+1] - database[l-1])
def makePtable(N):
table = list(range(2,N+1))
result = []
p = 2
while p < sqrt(N):
p = table[0]
result.append(p)
table = [x for x in table if x % p != 0]
return result + table
main()
|
s067817182
|
p03377
|
u865383026
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
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())
print('Yes' if (A + B) > X and X >= A else 'No')
|
s639786786
|
Accepted
| 17
| 2,940
| 84
|
A, B, X = map(int, input().split())
print('YES' if (A + B) > X and X >= A else 'NO')
|
s733636022
|
p03644
|
u853900545
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 51
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n=int(input())
i=1
while i**2<=n:
i+=1
print(i)
|
s234160441
|
Accepted
| 20
| 2,940
| 63
|
n=int(input())
i=0
while 2**i<=n:
s=2**i
i+=1
print(s)
|
s156061647
|
p03645
|
u778700306
| 2,000
| 262,144
|
Wrong Answer
| 646
| 6,132
| 370
|
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
n, m = map(int, input().split())
ok1 = [False] * n
ok2 = [False] * n
for _ in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
if a > b:
a, b = b, a
if a == 0:
ok1[b] = True
if b == n - 1:
ok2[b] = True
for m in range(n):
if ok1[m] and ok2[m]:
print("POSSIBLE")
exit(0)
print("IMPOSSIBLE")
|
s801899122
|
Accepted
| 657
| 6,132
| 336
|
n, m = map(int, input().split())
ok1 = [False] * n
ok2 = [False] * n
for _ in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
if a == 0:
ok1[b] = True
if b == n - 1:
ok2[a] = True
for m in range(n):
if ok1[m] and ok2[m]:
print("POSSIBLE")
exit(0)
print("IMPOSSIBLE")
|
s653043127
|
p02613
|
u478222049
| 2,000
| 1,048,576
|
Wrong Answer
| 194
| 16,324
| 602
|
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.
|
result = ["AC","WA","TLE","RE"]
result_num = [0,0,0,0]
case = input("")
case = int(case)
list = []
A = 0
while A < case:
inp = input("")
list.append(inp)
A = A + 1
A = 0
while A < case:
if list[A] == result[0]:
result_num[0] = result_num[0] + 1
elif list[A] == result[1]:
result_num[1] = result_num[1] + 1
elif list[A] == result[2]:
result_num[2] = result_num[2] + 1
elif list[A] == result[2]:
result_num[3] = result_num[3] + 1
A =A + 1
print("AC x " , result_num[0])
print("WA x " , result_num[1])
print("TLE x " , result_num[2])
print("RE x " , result_num[3])
|
s982409631
|
Accepted
| 228
| 16,328
| 684
|
result = ["AC","WA","TLE","RE"]
result_num = [0,0,0,0]
case = input("")
case = int(case)
list = []
A = 0
while A < case:
inp = input("")
if inp == result[0] or inp == result[1] or inp == result[2] or inp == result[3]:
list.append(inp)
A = A + 1
A = 0
while A < case:
if list[A] == result[0]:
result_num[0] = result_num[0] + 1
elif list[A] == result[1]:
result_num[1] = result_num[1] + 1
elif list[A] == result[2]:
result_num[2] = result_num[2] + 1
elif list[A] == result[3]:
result_num[3] = result_num[3] + 1
A =A + 1
print("AC x" , result_num[0])
print("WA x" , result_num[1])
print("TLE x" , result_num[2])
print("RE x" , result_num[3])
|
s794429734
|
p03997
|
u615323709
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 71
|
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)
|
s435751419
|
Accepted
| 17
| 2,940
| 80
|
a = int(input())
b = int(input())
h = int(input())
s = (a + b) * h // 2
print(s)
|
s523464637
|
p03679
|
u093033848
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 143
|
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
x, a, b = map(int, input().split())
if b <= a:
print("delicious")
if b >= a and a+x <= b :
print("safe")
if a+b < b :
print("dangerous")
|
s590528702
|
Accepted
| 17
| 2,940
| 147
|
x, a, b = map(int, input().split())
if b <= a:
print("delicious")
elif b > a and a+x >= b :
print("safe")
elif a+x < b :
print("dangerous")
|
s589704344
|
p03150
|
u623687794
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 176
|
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()
l=len(s)
flag=0
for i in range(7):
if s[:i]+s[-(7-i):]=="keyence":
flag=1
break
if s[:7]=="keyence":
flag=1
if flag==1:
print("Yes")
else:
print("No")
|
s890668384
|
Accepted
| 17
| 2,940
| 168
|
s=input()
flag=0
for i in range(7):
if s[:i]+s[-(7-i):]=="keyence":
flag=1
break
if s[:7]=="keyence":
flag=1
if flag==1:
print("YES")
else:
print("NO")
|
s211855529
|
p03457
|
u260469505
| 2,000
| 262,144
|
Wrong Answer
| 522
| 36,244
| 417
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
n = int(input())
lists = [[0,0,0]]
for i in range(n):
a = list(map(int,input().split()))
lists.append(a)
diffs = []
for i in range(1,n+1,1):
diff = [ x - y for (x,y) in zip(lists[i], lists[i-1])]
diffs.append(diff)
for diff in diffs:
if diff[0] < abs(diff[1]) + abs(diff[2]):
print('No')
exit()
elif diff[0] % 2 != (abs(diff[1]) + abs(diff[2])) % 2:
print('No')
exit()
print('yes')
|
s733346582
|
Accepted
| 515
| 36,244
| 418
|
n = int(input())
lists = [[0,0,0]]
for i in range(n):
a = list(map(int,input().split()))
lists.append(a)
diffs = []
for i in range(1,n+1,1):
diff = [ x - y for (x,y) in zip(lists[i], lists[i-1])]
diffs.append(diff)
for diff in diffs:
if diff[0] < abs(diff[1]) + abs(diff[2]):
print('No')
exit()
elif diff[0] % 2 != (abs(diff[1]) + abs(diff[2])) % 2:
print('No')
exit()
print('Yes')
|
s202282955
|
p03861
|
u813174766
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a,b,c=map(int,input().split())
print((a-1)//c+b//c)
|
s120162406
|
Accepted
| 17
| 2,940
| 51
|
a,b,c=map(int,input().split())
print(b//c-(a-1)//c)
|
s340212569
|
p04029
|
u112523623
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 42
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n = int(input())
print(((1 + n) * n) / 2)
|
s580657642
|
Accepted
| 17
| 2,940
| 46
|
n = int(input())
print(int(((1 + n) * n) / 2))
|
s670624644
|
p03759
|
u994521204
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 97
|
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
A=list(map(int, input().split()))
A.sort()
if A[0]+A[2]==A[1]:
print('YES')
else:
print('NO')
|
s724939615
|
Accepted
| 17
| 2,940
| 77
|
a, b, c = map(int, input().split())
print("YES" if b - a == c - b else "NO")
|
s429846690
|
p02612
|
u930177016
| 2,000
| 1,048,576
|
Wrong Answer
| 34
| 9,092
| 24
|
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)
|
s068803781
|
Accepted
| 28
| 9,108
| 36
|
print((1000-int(input())%1000)%1000)
|
s226518430
|
p02396
|
u825994660
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,632
| 203
|
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
a = list(map(int,input().split()))
if a[-1] != 0:
for i in range(len(a)):
print("Case {}: {}".format(i, a[i]))
else:
for i in range(len(a)-1):
print("Case {}: {}".format(i, a[i]))
|
s491734594
|
Accepted
| 60
| 7,388
| 150
|
import sys
cnt = 1
while True:
x = sys.stdin.readline().strip()
if x == "0":
break
print("Case {}: {}".format(cnt,x))
cnt += 1
|
s356123604
|
p02646
|
u749742659
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 9,188
| 260
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a,v = map(int, input().split())
b,w = map(int, input().split())
t = int(input())
if(b > a):
if(b + w*t < a + v*t):
print('Yes')
else:
print('No')
if(b < a):
if(b - w*t > a - v*t):
print('Yes')
else:
print('No')
|
s665203118
|
Accepted
| 22
| 9,212
| 262
|
a,v = map(int, input().split())
b,w = map(int, input().split())
t = int(input())
if(b > a):
if(b + w*t <= a + v*t):
print('YES')
else:
print('NO')
if(b < a):
if(b - w*t >= a - v*t):
print('YES')
else:
print('NO')
|
s481927931
|
p03457
|
u476418095
| 2,000
| 262,144
|
Wrong Answer
| 334
| 3,060
| 179
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
n = int(input())
t = 0
for i in range(n):
a, b, c = map(int, input().split())
if abs(a) + abs(b) > c:
t += 1
if t != 0:
print('No')
else:
print('Yes')
|
s060598786
|
Accepted
| 325
| 3,060
| 210
|
n = int(input())
f = True
for i in range(n):
t, x, y = map(int, input().split())
if t < x + y or t % 2 != (x+y) % 2:
f = False
break
if f == True:
print('Yes')
else:
print('No')
|
s779890253
|
p03474
|
u102242691
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 318
|
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
a,b = map(int,input().split())
s = list(input())
num = [0,1,2,3,4,5,6,7,8,9]
status = False
if s[a] != "-":
print("No")
exit()
else:
for i in range(len(s)):
if i == a:
pass
else:
if s[i] not in num:
print("No")
exit()
print("Yes")
|
s804919781
|
Accepted
| 17
| 3,064
| 188
|
a,b = map(int,input().split())
s = list(input())
ans = 0
if s[a] == "-":
s.pop(a)
if s.count("-") == 0:
print("Yes")
else:
print("No")
else:
print("No")
|
s000972466
|
p03997
|
u309141201
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 75
|
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)
|
s378523816
|
Accepted
| 16
| 2,940
| 80
|
a = int(input())
b = int(input())
h = int(input())
print(int(((a + b) * h) / 2))
|
s783408114
|
p03695
|
u085510145
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 268
|
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
n = int(input())
scorelist = list(map(int, input().split(" ")))
color1 = len(list([int(i/400) for i in scorelist if i < 3200]))
color2 = len([i for i in scorelist if i >= 3200])
ans_min = max(1, color1)
ans_max = color1 + color2
print("{} {}".format(ans_min, ans_max))
|
s832743112
|
Accepted
| 17
| 3,060
| 267
|
n = int(input())
scorelist = list(map(int, input().split(" ")))
color1 = len(set([int(i/400) for i in scorelist if i < 3200]))
color2 = len([i for i in scorelist if i >= 3200])
ans_min = max(1, color1)
ans_max = color1 + color2
print("{} {}".format(ans_min, ans_max))
|
s945824020
|
p00424
|
u546285759
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,620
| 338
|
与えられた変換表にもとづき，データを変換するプログラムを作成しなさい． データに使われている文字は英字か数字で，英字は大文字と小文字を区別する．変換表に現れる文字の順序に規則性はない． 変換表は空白をはさんで前と後ろの 2 つの文字がある(文字列ではない)．変換方法は，変換表のある行の前の文字がデータに現れたら，そのたびにその文字を後ろの文字に変換し出力する．変換は 1 度だけで，変換した文字がまた変換対象の文字になっても変換しない．変換表に現れない文字は変換せず，そのまま出力する． 入力ファイルには，変換表（最初の n + 1 行）に続き変換するデータ（n + 2 行目以降）が書いてある． 1 行目に変換表の行数 n，続く n 行の各行は，空白をはさんで 2 つの文字，さらに続けて， n + 2 行目に変換するデータの行数 m，続く m 行の各行は 1 文字である． m ≤ 105 とする．出力は，出力例のように途中に空白や改行は入れず 1 行とせよ． 入力例 --- 3 A a 0 5 5 4 10 A B C 0 1 4 5 a b A 出力例 aBC5144aba
|
x = int(input())
i = 0
dcl = dict()
while True:
inp = list(map(str, input().split()))
dcl[inp[0]] = inp[1]
i += 1
if i == x:
break
x = int(input())
i = 0
ans = ""
while True:
y = str(input())
if y in dcl:
ans += dcl[y]
else:
ans += y
i += 1
if i == x:
break
print(ans)
|
s227093344
|
Accepted
| 270
| 6,572
| 364
|
while True:
n = int(input())
if n == 0:
break
converter = {}
for _ in range(n):
k, v = input().split()
converter[k] = v
m = int(input())
ans = []
for _ in range(m):
v = input().strip()
try:
ans.append(converter[v])
except:
ans.append(v)
print("".join(ans))
|
s304244854
|
p03599
|
u743229067
| 3,000
| 262,144
|
Time Limit Exceeded
| 3,156
| 3,064
| 766
|
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())
maxDensity = 0
ans_water = 0
ans_sugar = 0
for i in range(31):
for j in range(31):
for k in range(101):
for l in range(101):
water = (i * A + j * B) * 100
sugar = k * C + l * D
if water + sugar > F:
continue
if sugar * 100 > water * E:
continue
if sugar + water == 0:
density = 0
else:
density = sugar / (sugar + water)
if maxDensity < density:
ans_water = water
ans_sugar = sugar
maxDensity = density
print(ans_water + ans_sugar, ans_sugar)
|
s153211098
|
Accepted
| 1,338
| 3,064
| 884
|
A, B, C, D, E, F = map(int, input().split())
maxDensity = 0
ans_water = A * 100
ans_sugar = 0
for i in range(F // (100 * A) + 1):
for j in range(F // (100 * B) + 1):
water = (i * A + j * B) * 100
rest = F - water
if water > F:
continue
for k in range(rest // C + 1):
for l in range(rest // D + 1):
sugar = k * C + l * D
if water + sugar > F:
continue
if sugar * 100 > water * E:
continue
if sugar + water == 0:
density = 0
else:
density = sugar / (sugar + water)
if maxDensity < density:
ans_water = water
ans_sugar = sugar
maxDensity = density
print(ans_water + ans_sugar, ans_sugar)
|
s547141248
|
p02742
|
u171821586
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 132
|
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
import math
def bishop(h,w):
if h == 1:
ans = 1
else:
ans = int(w/2)*h + (w%2)*math.ceil(h/2)
return ans
|
s652421549
|
Accepted
| 17
| 3,060
| 183
|
from sys import stdin
import math
h,w = [int(x) for x in stdin.readline().rstrip().split()]
if (h==1)|(w==1):
ans = 1
else:
ans = int(w/2)*h + (w%2)*math.ceil(h/2)
print(ans)
|
s246098330
|
p03455
|
u763280125
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,008
| 96
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = input().split()
if (int(a) * int(b)) // 2 == 0:
print('Even')
else:
print('Odd')
|
s787846333
|
Accepted
| 32
| 9,152
| 95
|
a, b = input().split()
if (int(a) * int(b)) % 2 == 0:
print('Even')
else:
print('Odd')
|
s696490347
|
p03565
|
u930705402
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 367
|
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
S=input()
T=input()
li=[]
for i in reversed(range(len(S)-len(T)+1)):
st=S[i:i+len(T)]
t=True
for j in range(len(T)):
if(st[j]!='?' and st[j]!=T[j]):
t=False
break
if(t):
tmp=S.replace('?','a')
li.append(tmp[:i]+T+tmp[i+len(T):])
if(li):
print(sorted(li)[0])
else:
print("UNRESTORABLE")
print(li)
|
s232476379
|
Accepted
| 18
| 3,064
| 357
|
S=input()
T=input()
li=[]
for i in reversed(range(len(S)-len(T)+1)):
st=S[i:i+len(T)]
t=True
for j in range(len(T)):
if(st[j]!='?' and st[j]!=T[j]):
t=False
break
if(t):
tmp=S.replace('?','a')
li.append(tmp[:i]+T+tmp[i+len(T):])
if(li):
print(sorted(li)[0])
else:
print("UNRESTORABLE")
|
s858553334
|
p03433
|
u535102705
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 92
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
n = int(input())
a = int(input())
if (n % 500) < a:
print('YES')
else:
print('NO')
|
s549565239
|
Accepted
| 17
| 2,940
| 121
|
n = int(input())
a = int(input())
if (n % 500) <= a:
print('Yes')
else:
print('No')
|
s066201309
|
p03944
|
u496744988
| 2,000
| 262,144
|
Wrong Answer
| 595
| 6,540
| 801
|
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
#import pprint
w,h,n=map(int,input().split())
x=[]
y=[]
a=[]
for i in range(n):
xi,yi,ai=map(int,input().split())
x.append(xi)
y.append(yi)
a.append(ai)
dot = [[0] * w for i in range(h)]
#pprint.pprint(dot,width=60)
for i in range(n):
if a[i] == 1:
for j in range(x[i]):
for k in range(h):
print(k,j)
dot[k][j]=1
if a[i] == 2:
for j in range(x[i],w):
for k in range(h):
dot[k][j]=1
if a[i] == 3:
for j in range(y[i]):
for k in range(w):
dot[j][k]=1
if a[i] == 4:
for j in range(y[i],h):
for k in range(w):
print(j,k)
dot[j][k]=1
#pprint.pprint(dot,width=60)
print(sum(dot,[]).count(0))
|
s505482549
|
Accepted
| 73
| 3,408
| 747
|
#import pprint
w,h,n=map(int,input().split())
x=[]
y=[]
a=[]
for i in range(n):
xi,yi,ai=map(int,input().split())
x.append(xi)
y.append(yi)
a.append(ai)
dot = [[0] * w for i in range(h)]
#pprint.pprint(dot,width=60)
for i in range(n):
if a[i] == 1:
for j in range(x[i]):
for k in range(h):
dot[k][j]=1
if a[i] == 2:
for j in range(x[i],w):
for k in range(h):
dot[k][j]=1
if a[i] == 3:
for j in range(y[i]):
for k in range(w):
dot[j][k]=1
if a[i] == 4:
for j in range(y[i],h):
for k in range(w):
dot[j][k]=1
#pprint.pprint(dot,width=60)
print(sum(dot,[]).count(0))
|
s931931985
|
p00227
|
u314832372
| 1,000
| 131,072
|
Wrong Answer
| 140
| 5,720
| 561
|
悪天候が続き野菜の価格が高騰する中、セブンマートではお客様に野菜のまとめ買いセールを実施しています。 日ごろなかなか店頭に並ばない野菜もお手頃価格で手に入るとあって、 店内はとても賑わっています。 ある日、松長団地に住む仲良し 3 人組がセブンマートの広告を手に話に花を咲かせていました。今回のセールは「お客様大感謝祭」と銘打っただけに、袋詰めした野菜の中で最も安いものが無料になるのが目玉となっています。広告を読んでみると、どうやら以下のようなセールのようです。 * 1 つの袋には m 個まで野菜を詰められる。 * 野菜が m 個詰めてある袋については、その中で最も安い野菜が無料となる。 * 野菜の個数が m 個に達しない袋は割引の対象外。 3人は早速セブンマートへ買い物に行きました。 買い物が終わり、 お店の外で待ち合わせた 3 人は安くてたくさん購入できたことに満足した様子で話をしていると、どうやら 3 人とも同じ野菜を購入していたことが分かりました。ある一人が、「本当に安いわよねぇ。これでXXX円だもの!」と言うと、もう一人は、「え?私はそれより**円高かったわ!どうして?」と驚き、また、残りの一人はレシートを見て自分が一番安く購入したことに気付きました。 さて、どのように袋詰めすれば購入価格を一番安くできるでしょうか。 購入する野菜の個数、袋に入る野菜の個数、各野菜の値段を入力とし、最低購入価格を出力するプログラムを作成してください。
|
for i in range(1, 101):
count = 0
SetCount = 0
try:
str1 = input()
list1 = str1.split(" ")
str2 = input()
list2 = str2.split(" ")
list2.sort(reverse=True)
for m in range(0, int(list1[0])):
if (m+1) % int(list1[1]) == 0:
count = count + int(list2[m])
SetCount = SetCount + 1
elif (m+1) % int(list1[1]) != 0 | SetCount == int(list1[0])//int(list1[1]):
count = count + int(list2[m])
print(count)
except:
break
|
s932521935
|
Accepted
| 90
| 5,716
| 604
|
for i in range(1, 101):
count = 0
temp = 0
SetCount = 0
try:
str1 = input()
list1 = str1.split(" ")
str2 = input()
list2 = str2.split(" ")
list2.sort(key=int, reverse=True)
for m in range(0, int(list1[0])):
if (m+1) % int(list1[1]) == 0:
SetCount = SetCount + 1
#elif (m+1) % int(list1[1]) != 0 | SetCount == int(list1[0])//int(list1[1]):
# count = count + int(list2[m])
else:
count = count + int(list2[m])
print(count)
except:
break
|
s853305661
|
p03455
|
u953241727
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,008
| 93
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b = map(int, input().split())
if (a * b %2 == 0) :
print('even')
else:
print('odd')
|
s023196003
|
Accepted
| 24
| 9,144
| 144
|
a,b = map(int, input().split())
if a % 2 != 0:
if b % 2 != 0:
print("Odd")
else:
print("Even")
else:
print("Even")
|
s679026927
|
p02613
|
u869937227
| 2,000
| 1,048,576
|
Wrong Answer
| 367
| 9,208
| 332
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(n):
s = input()
print(s)
if s == 'AC':
ac += 1
elif s == 'WA':
wa += 1
elif s == 'TLE':
tle += 1
else:
re += 1
print('AC x ' + str(ac))
print('WA x ' + str(wa))
print('TLE x ' + str(tle))
print('RE x ' + str(re))
|
s737608878
|
Accepted
| 143
| 9,204
| 319
|
n = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(n):
s = input()
if s == 'AC':
ac += 1
elif s == 'WA':
wa += 1
elif s == 'TLE':
tle += 1
else:
re += 1
print('AC x ' + str(ac))
print('WA x ' + str(wa))
print('TLE x ' + str(tle))
print('RE x ' + str(re))
|
s983848455
|
p03469
|
u450147945
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 45
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
S = list(map(str, input()))
S[3]='8'
print(S)
|
s014901955
|
Accepted
| 17
| 2,940
| 54
|
S = list(map(str, input()))
S[3]='8'
print(''.join(S))
|
s345415884
|
p03610
|
u732870425
| 2,000
| 262,144
|
Wrong Answer
| 44
| 3,188
| 101
|
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s = input()
ans = ""
for i in range(len(s)):
if i%2 != 0:
ans = "".join(s[i])
print(ans)
|
s189335549
|
Accepted
| 44
| 3,188
| 97
|
s = input()
ans = ""
for i in range(len(s)):
if (i+1)%2 != 0:
ans += s[i]
print(ans)
|
s568660069
|
p03997
|
u870518235
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 69
|
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)
|
s356745231
|
Accepted
| 17
| 2,940
| 76
|
a = int(input())
b = int(input())
h = int(input())
S = ((a+b)*h)//2
print(S)
|
s954074822
|
p03455
|
u779599374
| 2,000
| 262,144
|
Wrong Answer
| 153
| 12,500
| 162
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
ab = input().replace(" ", "")
ab = int(ab)
import numpy as np
s = np.array(range(1,101))
s = s**2
s = list(s)
if ab in s:
print("Yes")
else:
print("No")
|
s842013076
|
Accepted
| 17
| 2,940
| 88
|
a, b = map(int, input().split())
if a*b%2 == 0:
print("Even")
else:
print("Odd")
|
s023614447
|
p03479
|
u699699071
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 100
|
As a token of his gratitude, Takahashi has decided to give his mother an integer sequence. The sequence A needs to satisfy the conditions below: * A consists of integers between X and Y (inclusive). * For each 1\leq i \leq |A|-1, A_{i+1} is a multiple of A_i and strictly greater than A_i. Find the maximum possible length of the sequence.
|
X,Y=map(int,input().split())
result=0
while X<=Y :
print(X)
X*=2
result+=1
print(result)
|
s707211047
|
Accepted
| 17
| 2,940
| 102
|
X,Y=map(int,input().split())
result=0
while X<=Y :
# print(X)
X*=2
result+=1
print(result)
|
s489137539
|
p03696
|
u017050982
| 2,000
| 262,144
|
Wrong Answer
| 32
| 9,248
| 443
|
You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.
|
import copy
N = int(input())
S = list(input())
S_ans = copy.deepcopy(S)
app = 0
q = [0]
save = 0
for i in range(N):
print(S)
if S[i] == '(':
q.append(q[-1] + 1)
if q[-1] == 1:
save = i
else:
q.append(q[-1] - 1)
if q[-1] == -1:
S_ans.insert(app + save,"(")
app += 1
q[-1] = 0
print(q)
for i in range(q[-1]):
S_ans.append(")")
print("".join(S_ans))
|
s117532988
|
Accepted
| 29
| 9,352
| 386
|
import copy
N = int(input())
S = list(input())
S_ans = copy.deepcopy(S)
app = 0
q = [0]
save = 0
for i in range(N):
if S[i] == '(':
q.append(q[-1] + 1)
else:
q.append(q[-1] - 1)
if q[-1] == -1:
S_ans.insert(app + save,"(")
app += 1
q[-1] = 0
for i in range(q[-1]):
S_ans.append(")")
print("".join(S_ans))
|
s405655590
|
p04043
|
u943657163
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 95
|
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.
|
t = [int(_) for _ in input().split()]
print('YES' if t[0] == t[2] == 5 and t[1] == 7 else 'NO')
|
s106276028
|
Accepted
| 17
| 2,940
| 86
|
t = sorted([int(_) for _ in input().split()])
print('YES' if t == [5, 5, 7] else 'NO')
|
s707636114
|
p03711
|
u957872856
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 141
|
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
x, y = map(int,input().split())
a = {'1','3','5','7','8','10','12'}
b = {'4','6','9','11'}
print("Yes" if {x,y} <= a or {x,y} <= b else "No")
|
s188841702
|
Accepted
| 17
| 2,940
| 132
|
x, y = input().split()
a = {'1','3','5','7','8','10','12'}
b = {'4','6','9','11'}
print("Yes" if {x,y} <= a or {x,y} <= b else "No")
|
s069894098
|
p03548
|
u313498252
| 2,000
| 262,144
|
Wrong Answer
| 37
| 3,060
| 215
|
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
x, y, z = input().split()
x = int(x)
y = int(y)
z = int(z)
cnt = 0
rest = x
while(True):
if(rest - (y + z) > z):
x -= y + z
cnt += 1
rest -= (y + z)
else:
break
print(cnt)
|
s443243010
|
Accepted
| 36
| 3,060
| 216
|
x, y, z = input().split()
x = int(x)
y = int(y)
z = int(z)
cnt = 0
rest = x
while(True):
if(rest - (y + z) >= z):
x -= y + z
cnt += 1
rest -= (y + z)
else:
break
print(cnt)
|
s950561171
|
p03448
|
u322187839
| 2,000
| 262,144
|
Wrong Answer
| 48
| 3,060
| 251
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
A=int(input())
B=int(input())
C=int(input())
X=int(input())
count=0
for i in range(A):
for j in range(B):
for k in range(C):
if 500*A+100*B+50*C==X:
count+=1
else:
pass
print(count)
|
s846210145
|
Accepted
| 54
| 3,060
| 276
|
A=int(input())
B=int(input())
C=int(input())
X=int(input())
count=0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
if 500*i+100*j+50*k==X:
count+=1
else:
pass
print(count)
|
s178037405
|
p02390
|
u781194524
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,572
| 59
|
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
S = int(input())
print(S//3600,(S%3600)//60,S//60,sep=':')
|
s651746037
|
Accepted
| 20
| 5,580
| 58
|
S = int(input())
print(S//3600,(S%3600)//60,S%60,sep=":")
|
s197755386
|
p02262
|
u508054630
| 6,000
| 131,072
|
Wrong Answer
| 20
| 5,604
| 826
|
Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+g] = A[j] 7 j = j - g 8 cnt++ 9 A[j+g] = v 10 11 shellSort(A, n) 12 cnt = 0 13 m = ? 14 G[] = {?, ?,..., ?} 15 for i = 0 to m-1 16 insertionSort(A, n, G[i]) A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$. Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements: * $1 \leq m \leq 100$ * $0 \leq G_i \leq n$ * cnt does not exceed $\lceil n^{1.5}\rceil$
|
#Shell sort
def insertionSort(A, n, g, count):
Acopy = A.copy()
for i in range(g, n):
v = Acopy[i]
j = i - g
while (j >= 0) and (Acopy[j] > v):
Acopy[j + g] = Acopy[j]
j = j - g
count += 1
Acopy[j + g] = v
return Acopy, count
def shellSort(A, n):
G = []
h = 1
while h < n:
G.append(h)
h = 3 * h + 1
Gr = G.copy()
Gr.reverse()
count = 0
for g in Gr:
A, count = insertionSort(A, n, g, count)
return A, G, count
n = int(input())
q = 0
A = []
while q < n:
A.append(int(input()))
q += 1
A, G, count = shellSort(A, n)
ans = str(len(G)) + '\n' + ' '.join(map(str, G)) + '\n' + str(count) + '\n'
ans += '\n'.join(map(str, A))
print(ans)
|
s865593588
|
Accepted
| 18,120
| 132,220
| 816
|
def insertionSort(A, n, g, count):
Acopy = A.copy()
for i in range(g, n):
v = Acopy[i]
j = i - g
while (j >= 0) and (Acopy[j] > v):
Acopy[j + g] = Acopy[j]
j = j - g
count += 1
Acopy[j + g] = v
return Acopy, count
def shellSort(A, n):
G = [1]
h = 4
while h < n:
G.append(h)
h = 3 * h + 1
Gr = G.copy()
Gr.reverse()
count = 0
for g in Gr:
A, count = insertionSort(A, n, g, count)
return A, Gr, count
n = int(input())
q = 0
A = []
while q < n:
A.append(int(input()))
q += 1
A, G, count = shellSort(A, n)
ans = str(len(G)) + '\n' + ' '.join(map(str, G)) + '\n' + str(count) + '\n'
ans += '\n'.join(map(str, A))
print(ans)
|
s371257097
|
p03080
|
u919734978
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 92
|
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.
|
n = int(input())
a = str(input())
if a.count("r") > n /2:
print("Yes")
else:
print("No")
|
s706572149
|
Accepted
| 18
| 2,940
| 93
|
n = int(input())
a = str(input())
if a.count("R") > n /2:
print("Yes")
else:
print("No")
|
s827675412
|
p03659
|
u934868410
| 2,000
| 262,144
|
Wrong Answer
| 156
| 24,808
| 175
|
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
n = int(input())
a = list(map(int,input().split()))
s = sum(a)
snuke = a[0]
ans = abs(s - snuke * 2)
for x in a[1:n-1]:
snuke += x
ans = min(ans, s - snuke * 2)
print(ans)
|
s792061393
|
Accepted
| 176
| 24,812
| 180
|
n = int(input())
a = list(map(int,input().split()))
s = sum(a)
snuke = a[0]
ans = abs(s - snuke * 2)
for x in a[1:n-1]:
snuke += x
ans = min(ans, abs(s - snuke * 2))
print(ans)
|
s941709588
|
p02271
|
u728137020
| 5,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 426
|
Write a program which reads a sequence _A_ of _n_ elements and an integer _M_ , and outputs "yes" if you can make _M_ by adding elements in _A_ , otherwise "no". You can use an element only once. You are given the sequence _A_ and _q_ questions where each question contains _M i_.
|
n=int(input())
A=list(map(int, input().split()))
m=int(input())
B=list(map(int, input().split()))
A.sort()
for i in range(m):
cnt=0
for j in range(n):
if cnt==0:
for k in range(n-j-1):
m=A[j]+A[n-k-1]
if m==B[i]:
print("yes")
cnt=1
elif m<B[i]:
break
if cnt==0:
print("no")
|
s860815281
|
Accepted
| 3,850
| 6,956
| 485
|
n=int(input())
A=[int(i) for i in input().split()]
m=int(input())
B=[int(i) for i in input().split()]
for i in range(len(B)):
C=[[0 for j in range(B[i]+1)]for k in range(len(A)+1)]
for j in range(1,len(A)+1):
for k in range(1,B[i]+1):
if A[j-1]<=k:
C[j][k]=max(C[j-1][k-A[j-1]]+A[j-1],C[j-1][k])
else:
C[j][k]=C[j-1][k]
if C[len(A)][B[i]]==B[i]:
print("yes")
else:
print("no")
|
s038227423
|
p03737
|
u655975843
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 83
|
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
|
a,b,c = map(str, input().split())
print(a[0].lower() + b[0].lower() + c[0].lower())
|
s204739816
|
Accepted
| 17
| 2,940
| 83
|
a,b,c = map(str, input().split())
print(a[0].upper() + b[0].upper() + c[0].upper())
|
s660262895
|
p03693
|
u265118937
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,156
| 93
|
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?
|
r, g, b = map(int, input().split())
ans = r*100 + g*10 + b
print("Yes" if ans%4==0 else "No")
|
s481502982
|
Accepted
| 30
| 9,152
| 93
|
r, g, b = map(int, input().split())
ans = r*100 + g*10 + b
print("YES" if ans%4==0 else "NO")
|
s168985018
|
p03679
|
u960513073
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 135
|
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
x,a,b = list(map(int, input().split()))
if b <= a:
print("delicious")
elif a<b and b <= x:
print("safe")
else:
print("dangerous")
|
s316597695
|
Accepted
| 20
| 3,060
| 145
|
x,a,b = list(map(int, input().split()))
if b <= a:
print("delicious")
elif a<b and b <= a+x:
print("safe")
elif a+x < b:
print("dangerous")
|
s990268054
|
p03377
|
u159994501
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 93
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A, B, X =map(int,input().split())
if A <= X <= A + B:
print("Yes")
else:
print("No")
|
s158463155
|
Accepted
| 17
| 2,940
| 95
|
A, B, X = map(int, input().split())
if A <= X <= A + B:
print("YES")
else:
print("NO")
|
s524469926
|
p03598
|
u131264627
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 148
|
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
n = int(input())
k = int(input())
xxx = list(map(int, input().split()))
ans = 0
for x in xxx:
ans += 2 * (x if k < 2 * x else k - x)
print(ans)
|
s156214674
|
Accepted
| 18
| 2,940
| 148
|
n = int(input())
k = int(input())
xxx = list(map(int, input().split()))
ans = 0
for x in xxx:
ans += 2 * (x if k > 2 * x else k - x)
print(ans)
|
s782607846
|
p03494
|
u086051538
| 2,000
| 262,144
|
Wrong Answer
| 160
| 12,468
| 252
|
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 as np
n=int(input())
alist=np.array(list(map(int,input().split())))
print(alist)
counter=0
flag=0
while flag==0:
for a in alist:
if a%2==1:
flag=1
break
if flag==0:
alist=alist/2
counter=counter+1
print(counter)
|
s485254820
|
Accepted
| 158
| 12,488
| 239
|
import numpy as np
n=int(input())
alist=np.array(list(map(int,input().split())))
counter=0
flag=0
while flag==0:
for a in alist:
if a%2==1:
flag=1
break
if flag==0:
alist=alist/2
counter=counter+1
print(counter)
|
s378516569
|
p03854
|
u690781906
| 2,000
| 262,144
|
Wrong Answer
| 41
| 3,188
| 217
|
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()
t = ''
add_list = ['dream', 'dreamer', 'erase', 'eraser']
cur = len(s)
for i in range(len(s)):
if s[-1 - i:cur] in add_list:
cur = i
if cur == len(s) - 1:
print('YES')
else:
print('No')
|
s162770300
|
Accepted
| 198
| 3,188
| 219
|
s = input()
t = ''
add_list = ['dream', 'dreamer', 'erase', 'eraser']
cur = len(s)
for i in range(len(s)):
if s[-1 - i:cur] in add_list:
cur = -1 - i
if -cur == len(s):
print('YES')
else:
print('NO')
|
s012525564
|
p02612
|
u486467381
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,144
| 87
|
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.
|
str = input()
# print(str)
a = int(str)
b = a /1000
c = a % 1000
# print(b)
print(c)
|
s089094263
|
Accepted
| 28
| 9,160
| 79
|
str = input()
a = int(str)
c = a % 1000
if c == 0:
c = 1000
print(1000 - c)
|
s167441878
|
p03351
|
u980492406
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 173
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,d = map(int,input().split())
AB = abs(a-b)
AC = abs(a-c)
BC = abs(b-c)
if AC <= d :
print('YES')
elif AB <= d and BC <= d :
print('YES')
else :
print('NO')
|
s830944227
|
Accepted
| 17
| 3,060
| 173
|
a,b,c,d = map(int,input().split())
AB = abs(a-b)
AC = abs(a-c)
BC = abs(b-c)
if AC <= d :
print('Yes')
elif AB <= d and BC <= d :
print('Yes')
else :
print('No')
|
s552673347
|
p02690
|
u091307273
| 2,000
| 1,048,576
|
Wrong Answer
| 42
| 9,280
| 266
|
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
|
def main():
x = int(input())
a, b = 1, 0
while True:
for b in reversed(range(-a + 1, a)):
print(a, b)
q = a**5 - b**5
if q == x:
print(f'{a} {b}')
return
a += 1
main()
|
s394016133
|
Accepted
| 35
| 9,096
| 267
|
def main():
x = int(input())
a, b = 1, 0
while True:
for b in reversed(range(-a + 1, a)):
# print(a, b)
q = a**5 - b**5
if q == x:
print(f'{a} {b}')
return
a += 1
main()
|
s242993259
|
p02277
|
u195186080
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,708
| 808
|
Let's arrange a deck of cards. Your task is to sort totally n cards. A card consists of a part of a suit (S, H, C or D) and an number. Write a program which sorts such cards based on the following pseudocode: Partition(A, p, r) 1 x = A[r] 2 i = p-1 3 for j = p to r-1 4 do if A[j] <= x 5 then i = i+1 6 exchange A[i] and A[j] 7 exchange A[i+1] and A[r] 8 return i+1 Quicksort(A, p, r) 1 if p < r 2 then q = Partition(A, p, r) 3 run Quicksort(A, p, q-1) 4 run Quicksort(A, q+1, r) Here, A is an array which represents a deck of cards and comparison operations are performed based on the numbers. Your program should also report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
s = True
def checher(A,start,end):
global s
for i in range(start+1,end+1):
if A[start][1]==A[i][1]:
if A[start][0]!=A[i][0]:
s = False
return
return
def partition(A,p,r):
x = A[r][1]
i=p-1
for j in range(p,r):
if A[j][1]<=x:
i=i+1
if s:
checher(A,i,j)
(A[i],A[j])=(A[j],A[i])
(A[i+1],A[r])=(A[r],A[i+1])
return i+1
def quickSort(A, p, r):
if p<r:
q = partition(A,p,r)
quickSort(A,p,q-1)
quickSort(A,q+1,r)
n = int(input())
l = []
for i in range(n):
card = input().split()
l.append((card[0],int(card[1])))
quickSort(l,0,n-1)
if s:
print("Stable")
else:
print("Not Stable")
for s,n in l:
print("{} {}".format(s,n))
|
s296888945
|
Accepted
| 960
| 22,152
| 695
|
def partition(A,p,r):
x = A[r][1]
i=p-1
for j in range(p,r):
if A[j][1]<=x:
i=i+1
(A[i],A[j])=(A[j],A[i])
(A[i+1],A[r])=(A[r],A[i+1])
return i+1
def quickSort(A, p, r):
if p<r:
q = partition(A,p,r)
quickSort(A,p,q-1)
quickSort(A,q+1,r)
def checker(A,start,end):
for i in range(start,end-1):
if A[i][1]==A[i+1][1]:
if A[i][2]>A[i+1][2]:
return "Not stable"
return "Stable"
n = int(input())
l = []
for i in range(n):
card = input().split()
l.append((card[0],int(card[1]),i))
quickSort(l,0,n-1)
print(checker(l,0,n))
for s,n,c in l:
print("{} {}".format(s,n))
|
s322700505
|
p03361
|
u062459048
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 316
|
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.
|
H, W = map(int, input().split())
S = [input() for i in range(H)]
for i in range(H):
for j in range(W):
if S[i][j] == "#":
bk = sum([t[j].count("#") for t in S[max(0,i-1):min(H,i+1)]])
bk += S[i][max(0,j-1):min(W,j+1)].count("#")-1
if bk == 1:
print("No")
exit()
print("Yes")
|
s656138627
|
Accepted
| 23
| 3,064
| 317
|
H, W = map(int, input().split())
S = [input() for i in range(H)]
for i in range(H):
for j in range(W):
if S[i][j] == "#":
bk = sum([t[j].count("#") for t in S[max(0,i-1):min(H,i+2)]])
bk += S[i][max(0,j-1):min(W,j+2)].count("#")-1
if bk == 1:
print("No")
exit()
print("Yes")
|
s402202023
|
p03377
|
u160244242
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,100
| 85
|
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())
print('Yes') if a <= x and x <= a+b else print('No')
|
s119594915
|
Accepted
| 22
| 9,052
| 85
|
a,b,x = map(int,input().split())
print('YES') if a <= x and x <= a+b else print('NO')
|
s947016015
|
p02744
|
u861141787
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 73
|
In this problem, we only consider strings consisting of lowercase English letters. Strings s and t are said to be **isomorphic** when the following conditions are satisfied: * |s| = |t| holds. * For every pair i, j, one of the following holds: * s_i = s_j and t_i = t_j. * s_i \neq s_j and t_i \neq t_j. For example, `abcac` and `zyxzx` are isomorphic, while `abcac` and `ppppp` are not. A string s is said to be in **normal form** when the following condition is satisfied: * For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison. For example, `abcac` is in normal form, but `zyxzx` is not since it is isomorphic to `abcac`, which is lexicographically smaller than `zyxzx`. You are given an integer N. Print all strings of length N that are in normal form, in lexicographically ascending order.
|
n = int(input())
if n == 3:
print("aaa", "aab", "aba", "abb", "abc")
|
s025326382
|
Accepted
| 145
| 4,340
| 305
|
n = int(input())
def dfs(s, mx):
if len(s) == n:
print(s)
else:
m = ord(mx) - ord("a")
for i in range(m+1):
c = chr(i + ord("a"))
if c == mx:
dfs(s + c, chr(ord(mx)+1))
else:
dfs(s + c, mx)
dfs("", "a")
|
s727896550
|
p03400
|
u800058906
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,048
| 135
|
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
n=int(input())
d,x=map(int,input().split())
a=[int(input()) for i in range(n)]
c=0
for i in a:
p=(d-1)//i+1
print(c+x)
|
s897952320
|
Accepted
| 27
| 9,052
| 146
|
import sys
n=int(input())
d,x=map(int,input().split())
a=[int(input()) for i in range(n)]
c=0
for i in a:
p=(d-1)//i+1
c+=p
print(c+x)
|
s199410190
|
p03434
|
u207799478
| 2,000
| 262,144
|
Wrong Answer
| 26
| 3,832
| 736
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
import math
import string
def readints():
return list(map(int, input().split()))
def nCr(n, r):
return math.factorial(n)//(math.factorial(n-r)*math.factorial(r))
def has_duplicates2(seq):
seen = []
for item in seq:
if not(item in seen):
seen.append(item)
return len(seq) != len(seen)
def divisor(n):
divisor = []
for i in range(1, n+1):
if n % i == 0:
divisor.append(i)
return divisor
# coordinates
dx = [-1, -1, -1, 0, 0, 1, 1, 1]
dy = [-1, 0, 1, -1, 1, -1, 0, 1]
N = int(input())
a = readints()
# print(a)
aa = sorted(a, reverse=True)
print(aa)
x = 0
for i in range(len(a)):
if i % 2 == 0:
x += aa[i]
else:
x -= aa[i]
print(x)
|
s755987386
|
Accepted
| 24
| 3,832
| 738
|
import math
import string
def readints():
return list(map(int, input().split()))
def nCr(n, r):
return math.factorial(n)//(math.factorial(n-r)*math.factorial(r))
def has_duplicates2(seq):
seen = []
for item in seq:
if not(item in seen):
seen.append(item)
return len(seq) != len(seen)
def divisor(n):
divisor = []
for i in range(1, n+1):
if n % i == 0:
divisor.append(i)
return divisor
# coordinates
dx = [-1, -1, -1, 0, 0, 1, 1, 1]
dy = [-1, 0, 1, -1, 1, -1, 0, 1]
N = int(input())
a = readints()
# print(a)
aa = sorted(a, reverse=True)
# print(aa)
x = 0
for i in range(len(a)):
if i % 2 == 0:
x += aa[i]
else:
x -= aa[i]
print(x)
|
s310678825
|
p03733
|
u458617779
| 2,000
| 262,144
|
Wrong Answer
| 153
| 25,200
| 268
|
In a public bath, there is a shower which emits water for T seconds when the switch is pushed. If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds. N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it. How long will the shower emit water in total?
|
while True:
try:
ins = input().split(" ")
N = int(ins[0])
T = int(ins[1])
a = []
inx = input().split(" ")
for i in range(0, N):
a.append(int(inx[i]))
ans = T
for x in range(1, N):
if a[x]+T > ans:
ans = a[x] + T
print(ans)
except:
break
|
s469173066
|
Accepted
| 187
| 25,840
| 396
|
while True:
try:
ins = input().split(" ")
N = int(ins[0])
T = int(ins[1])
a = []
inx = input().split(" ")
for i in range(0, N):
a.append(int(inx[i]))
ans = 0
end = 0
for x in range(0, N):
if a[x] < end:
ans = ans + a[x] + T - end
end = a[x] + T
elif a[x] > end:
ans += T
end = a[x] + T
else:
ans += T
end += T
print(ans)
except:
break
|
s681676161
|
p03044
|
u346812984
| 2,000
| 1,048,576
|
Wrong Answer
| 555
| 45,904
| 1,032
|
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
import heapq
import sys
sys.setrecursionlimit(10 ** 6)
INF = float("inf")
MOD = 10 ** 9 + 7
def input():
return sys.stdin.readline().strip()
def dijkstra(edges, n_nodes, start=0):
dist = [float("inf")] * n_nodes
dist[start] = 0
q = [(dist[start], start)]
heapq.heapify(q)
while q:
d, i = heapq.heappop(q)
if dist[i] < d:
continue
for weight, j in edges[i]:
tmp = dist[i] + weight
if dist[j] > tmp:
dist[j] = tmp
heapq.heappush(q, (dist[j], j))
return dist
def main():
N = int(input())
edges = [[] for _ in range(N)]
for _ in range(N - 1):
u, v, w = map(int, input().split())
u -= 1
v -= 1
edges[u].append((w, v))
edges[v].append((w, u))
dist = dijkstra(edges, N)
print(dist)
for d in dist:
if d % 2 == 0:
print(0)
else:
print(1)
if __name__ == "__main__":
main()
|
s701784813
|
Accepted
| 557
| 42,956
| 1,016
|
import heapq
import sys
sys.setrecursionlimit(10 ** 6)
INF = float("inf")
MOD = 10 ** 9 + 7
def input():
return sys.stdin.readline().strip()
def dijkstra(edges, n_nodes, start=0):
dist = [float("inf")] * n_nodes
dist[start] = 0
q = [(dist[start], start)]
heapq.heapify(q)
while q:
d, i = heapq.heappop(q)
if dist[i] < d:
continue
for weight, j in edges[i]:
tmp = dist[i] + weight
if dist[j] > tmp:
dist[j] = tmp
heapq.heappush(q, (dist[j], j))
return dist
def main():
N = int(input())
edges = [[] for _ in range(N)]
for _ in range(N - 1):
u, v, w = map(int, input().split())
u -= 1
v -= 1
edges[u].append((w, v))
edges[v].append((w, u))
dist = dijkstra(edges, N)
for d in dist:
if d % 2 == 0:
print(0)
else:
print(1)
if __name__ == "__main__":
main()
|
s027862598
|
p03441
|
u532966492
| 2,000
| 262,144
|
Wrong Answer
| 757
| 56,588
| 906
|
We have a tree with N vertices. The vertices are numbered 0 through N - 1, and the i-th edge (0 ≤ i < N - 1) comnnects Vertex a_i and b_i. For each pair of vertices u and v (0 ≤ u, v < N), we define the distance d(u, v) as the number of edges in the path u-v. It is expected that one of the vertices will be invaded by aliens from outer space. Snuke wants to immediately identify that vertex when the invasion happens. To do so, he has decided to install an antenna on some vertices. First, he decides the number of antennas, K (1 ≤ K ≤ N). Then, he chooses K different vertices, x_0, x_1, ..., x_{K - 1}, on which he installs Antenna 0, 1, ..., K - 1, respectively. If Vertex v is invaded by aliens, Antenna k (0 ≤ k < K) will output the distance d(x_k, v). Based on these K outputs, Snuke will identify the vertex that is invaded. Thus, in order to identify the invaded vertex no matter which one is invaded, the following condition must hold: * For each vertex u (0 ≤ u < N), consider the vector (d(x_0, u), ..., d(x_{K - 1}, u)). These N vectors are distinct. Find the minumum value of K, the number of antennas, when the condition is satisfied.
|
def main():
n = int(input())
ab = [list(map(int, input().split())) for _ in [0]*(n-1)]
g = [[] for _ in [0]*n]
[g[a-1].append(b-1) for a, b in ab]
[g[b-1].append(a-1) for a, b in ab]
root = 0
d = [-1]*n からの距離
d[root] = 0
q = [root]
cnt = 0
while q: # BFS
cnt += 1
qq = []
while q:
i = q.pop()
for j in g[i]:
if d[j] == -1:
d[j] = cnt
qq.append(j)
q = qq
d2 = sorted([(j, i) for i, j in enumerate(d)])[::-1]
stock = [0]*n
ans = 1
for _, i in d2:
dist = d[i]
s = 0
cnt = 0
for j in g[i]:
if dist < d[j]:
s += stock[j]
cnt += 1
ans += max(cnt-s-1, 0)
if s > 0 or cnt > 1:
stock[i] = 1
print(ans)
main()
|
s664920310
|
Accepted
| 704
| 50,004
| 1,037
|
def main():
n = int(input())
ab = [list(map(int, input().split())) for _ in [0]*(n-1)]
g = [[] for _ in [0]*n]
[g[a].append(b) for a, b in ab]
[g[b].append(a) for a, b in ab]
for i in range(n):
if len(g[i]) > 2:
root = i
break
else:
print(1)
return
d = [-1]*n
d[root] = 0
q = [root]
cnt = 0
while q: # BFS
cnt += 1
qq = []
while q:
i = q.pop()
for j in g[i]:
if d[j] == -1:
d[j] = cnt
qq.append(j)
q = qq
d2 = sorted([(j, i) for i, j in enumerate(d)])[::-1]
stock = [0]*n
ans = 0
for _, i in d2:
dist = d[i]
s = 0
cnt = 0
for j in g[i]:
if dist < d[j]:
s += stock[j]
cnt += 1
ans += max(cnt-s-1, 0)
s += max(cnt-s-1, 0)
if s > 0 or cnt > 1:
stock[i] = 1
print(ans)
main()
|
s010154178
|
p03456
|
u411544692
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 129
|
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
heihousuu = [i*i for i in range(int(100100**0.5))]
a,b= input().split()
ab = int(a+b)
print('YES' if ab in heihousuu else 'NO')
|
s838008434
|
Accepted
| 17
| 2,940
| 129
|
heihousuu = [i*i for i in range(int(100100**0.5))]
a,b= input().split()
ab = int(a+b)
print('Yes' if ab in heihousuu else 'No')
|
s765819899
|
p03729
|
u432805419
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 96
|
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 = input().split()
if a[-1] == b[-1] and b[-1] == c[-1]:
print("YES")
else:
print("NO")
|
s718929088
|
Accepted
| 17
| 2,940
| 94
|
a,b,c = input().split()
if a[-1] == b[0] and b[-1] == c[0]:
print("YES")
else:
print("NO")
|
s407919457
|
p03371
|
u652656291
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 135
|
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
a,b,c,x,y = map(int,input().split())
A_B = a*x+b*y
A_AB = a*x+c*y
AB_B = c*x+b*y
AB = c*max(x,y)
ans = min(A_B,A_AB,AB_B,AB)
print(ans)
|
s172401593
|
Accepted
| 142
| 3,060
| 198
|
A,B,C,X,Y = map(int,input().split())
answer = 10**18
for c in range(10**5+1):
a = X-c
b = Y-c
cost = C*c*2
cost += max(a,0)*A
cost += max(b,0)*B
answer = min(answer,cost)
print(answer)
|
s149231867
|
p03359
|
u760961723
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 71
|
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a, b = map(int, input().split())
if a<b:
print(a)
else:
print(a-1)
|
s764720720
|
Accepted
| 17
| 2,940
| 73
|
a, b = map(int, input().split())
if a<=b:
print(a)
else:
print(a-1)
|
s469072084
|
p03861
|
u422104747
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,064
| 81
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
s=input().split()
a=int(s[0])
b=int(s[1])
x=int(s[2])
print(-int(a//x)+int(b//x))
|
s757960051
|
Accepted
| 23
| 3,064
| 102
|
s=input().split()
a=int(s[0])
b=int(s[1])
x=int(s[2])
c=-int(a//x)+int(b//x)
if a%x==0:
c+=1
print(c)
|
s543501364
|
p02261
|
u929141425
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,332
| 464
|
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
import copy
N = int(input())
A = input().split()
B = copy.copy(A)
boo = 1
while boo:
boo = 0
for i in range(N-1):
if A[i][1] > A[i+1][1]:
A[i], A[i+1] = A[i+1], A[i]
boo = 1
print(*A)
print("Stable")
for i in range(N-1):
mi = i
for j in range(i,N):
if B[i][1] > B[j][1]:
mi = j
B[i], B[mi] = B[mi], B[i]
if A==B:
print(*B)
print("Stable")
else:
print(*B)
print("Not stable")
|
s388549970
|
Accepted
| 30
| 6,336
| 463
|
import copy
N = int(input())
A = input().split()
B = copy.copy(A)
boo = 1
while boo:
boo = 0
for i in range(N-1):
if A[i][1] > A[i+1][1]:
A[i], A[i+1] = A[i+1], A[i]
boo = 1
print(*A)
print("Stable")
for i in range(N):
mi = i
for j in range(i,N):
if B[mi][1] > B[j][1]:
mi = j
B[i], B[mi] = B[mi], B[i]
if A==B:
print(*B)
print("Stable")
else:
print(*B)
print("Not stable")
|
s088893912
|
p04045
|
u906481659
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 374
|
Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K. She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change). However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money. Find the amount of money that she will hand to the cashier.
|
N, L = map(str, input().split())
P = list(map(str, input().split()))
U = sorted(list(set([str(i) for i in range(10)])-set(P)))
#print(U)
for i in range(len(N)):
#print(N[i])
if N[i] in P:
print(N[i])
for j in U:
print(j)
if j >=N[i]:
N=N.replace(N[i],j)
#print(N)
break
print(N)
|
s969631449
|
Accepted
| 21
| 3,064
| 530
|
import itertools
N, L = map(str, input().split())
P = list(map(str, input().split()))
U = itertools.product(sorted(list(set([str(i) for i in range(10)])-set(P))),repeat=len(N))
F = itertools.product(sorted(list(set([str(i) for i in range(10)])-set(P))),repeat=len(N)+1)
K = str()
cnt=0
for i in U:
#print(''.join(i))
K = ''.join(i)
if int(K)>=int(N):
cnt = 1
print(K)
break
#print(cnt)
if cnt == 0:
for j in F:
K=''.join(j)
if int(K)>=int(N):
print(K)
break
|
s656635255
|
p01094
|
u284260266
| 8,000
| 262,144
|
Wrong Answer
| 480
| 5,640
| 952
|
The citizens of TKB City are famous for their deep love in elections and vote counting. Today they hold an election for the next chairperson of the electoral commission. Now the voting has just been closed and the counting is going to start. The TKB citizens have strong desire to know the winner as early as possible during vote counting. The election candidate receiving the most votes shall be the next chairperson. Suppose for instance that we have three candidates _A_ , _B_ , and _C_ and ten votes. Suppose also that we have already counted six of the ten votes and the vote counts of _A_ , _B_ , and _C_ are four, one, and one, respectively. At this moment, every candidate has a chance to receive four more votes and so everyone can still be the winner. However, if the next vote counted is cast for _A_ , _A_ is ensured to be the winner since _A_ already has five votes and _B_ or _C_ can have at most four votes at the end. In this example, therefore, the TKB citizens can know the winner just when the seventh vote is counted. Your mission is to write a program that receives every vote counted, one by one, identifies the winner, and determines when the winner gets ensured.
|
def secMax(a):
mx = 0
smx = 0
for i in range(0,len(a)):
if a[i] > mx:
smx = mx
mx = a[i]
elif a[i] > smx:
smx = a[i]
else:
return smx
while True:
n = int(input())
if n == 0:
break
c = list(map(str,input().split()))
alph = ["A","B","C","D","E","F","G","H",
"I","J","K","L","M","N","O","P","Q",
"R","S","T","U","V","W","X","Y","Z"]
abcCount = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
for i in range(0, n):
for j in range(0, 26):
if c[i] == alph[j]:
abcCount[j] = abcCount[j] + 1
break
print(abcCount)
if max(abcCount) > n-1-i+secMax(abcCount):
print(alph[abcCount.index(max(abcCount))], sum(abcCount))
break
else:
print("TIE")
|
s891027633
|
Accepted
| 340
| 5,628
| 953
|
def secMax(a):
mx = 0
smx = 0
for i in range(0,len(a)):
if a[i] > mx:
smx = mx
mx = a[i]
elif a[i] > smx:
smx = a[i]
else:
return smx
while True:
n = int(input())
if n == 0:
break
c = list(map(str,input().split()))
alph = ["A","B","C","D","E","F","G","H",
"I","J","K","L","M","N","O","P","Q",
"R","S","T","U","V","W","X","Y","Z"]
abcCount = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
for i in range(0, n):
for j in range(0, 26):
if c[i] == alph[j]:
abcCount[j] = abcCount[j] + 1
break
#print(abcCount)
if max(abcCount) > n-1-i+secMax(abcCount):
print(alph[abcCount.index(max(abcCount))], sum(abcCount))
break
else:
print("TIE")
|
s821266691
|
p03377
|
u620084012
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,316
| 71
|
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())
print("Yes" if A<=X<=A+B else "No")
|
s013465873
|
Accepted
| 21
| 3,316
| 71
|
A, B, X = map(int, input().split())
print("YES" if A<=X<=A+B else "NO")
|
s005250487
|
p00057
|
u350804311
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,604
| 117
|
無限に広い平面の上に、無限に長い直線を数本引くと、この平面はいくつかの領域に分割されます。たとえば、直線を１本引くと、平面は２つの領域に分割されます。同じ数の直線を引いても、引き方によって得られる領域の数は異なります。たとえば、2 本の直線を平行に引けば得られる領域は 3 つになり、互いに垂直に引けば得られる領域は 4 つになります。 n 本の直線を引くことで得られる最大の領域の数を出力するプログラムを作成してください。
|
while True:
try:
a = int(input())
print(((a * a) + a + 2) / 2)
except EOFError:
break
|
s796289679
|
Accepted
| 30
| 7,556
| 127
|
while True:
try:
a = int(input())
print(str(int(((a * a) + a + 2) / 2)))
except EOFError:
break
|
s609635118
|
p03861
|
u991604406
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,160
| 123
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a,b,x = map(int,input().split())
a2 = a
if a % x != 0:
a2 = (x - (a % x)) + a
b2 = b - (b % x)
print((b2 - a2) / x + 1)
|
s192797548
|
Accepted
| 28
| 9,008
| 141
|
a,b,x = map(int,input().split())
def f(n,x):
if n == -1:
return 0
else:
return n // x + 1
print(int(f(b,x)-f(a-1,x)))
|
s490156571
|
p03673
|
u602740328
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 25,156
| 143
|
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
n = int(input())
a = list(map(int, input().split()))
b = []
for i in range(n):
if i%2: b.insert(0, a[i])
else: b.append(a[i])
print(b)
|
s623714111
|
Accepted
| 74
| 27,716
| 193
|
n=int(input())
a=input().split()
b1=[a[2*i] for i in range(0,n//2)]
b2=[a[2*i+1] for i in range(0,n//2)]
if n%2: b=[a[-1]]+list(reversed(b1))+b2
else: b=list(reversed(b2))+b1
print(" ".join(b))
|
s118955828
|
p02255
|
u222257547
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,600
| 287
|
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
return A
N = int(input())
A = [int(x) for x in input().split(' ')]
A = insertionSort(A, N)
print(A)
|
s916434702
|
Accepted
| 20
| 5,980
| 397
|
def showList(A, N):
for i in range(N-1):
print(A[i],end=' ')
print(A[N-1])
def insertionSort(A, N):
showList(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
showList(A,N)
N = int(input())
A = [int(x) for x in input().split(' ')]
A = insertionSort(A, N)
|
s270993649
|
p03644
|
u312666261
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 208
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n = int(input())
ans = 0
anss = 0
for i in range(1,n+1):
t = 0
while i%2 == 0:
t +=1
i = i/2
if t > ans:
ans = t
anss = i*(2**t)
else:
pass
print(anss)
|
s982944288
|
Accepted
| 17
| 3,060
| 237
|
n = int(input())
ans = 0
anss = 0
for i in range(1,n+1):
t = 0
while i%2 == 0:
t +=1
i = i/2
if t > ans:
ans = t
anss = i*(2**t)
else:
pass
if n == 1:
anss = 1
print(int(anss))
|
s717565476
|
p03050
|
u667024514
| 2,000
| 1,048,576
|
Wrong Answer
| 144
| 2,940
| 134
|
Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied: * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds. Find all favorite numbers and print the sum of those.
|
import math
n = int(input())
ans = 0
for i in range(1,math.floor(math.sqrt(n))):
if n % i == 0:
ans += (n//i)-1
print(ans)
|
s769891244
|
Accepted
| 143
| 3,060
| 218
|
import math
n = int(input())
ans = 0
for i in range(1,math.floor(math.sqrt(n))+1):
if n % i == 0:
num = (n//i)-1
if num > 0:
if n // num == n % num:
ans += num
print(ans)
|
s281959449
|
p00004
|
u462831976
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,496
| 231
|
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.
|
# -*- coding: utf-8 -*-
import sys
import os
for s in sys.stdin:
a, b, c, d, e, f = list(map(int, s.split()))
delta = a * e - b * d
x = 1 / delta * (e * c - b * f)
y = 1 / delta * (-d * c + a * f)
print(x, y)
|
s617084464
|
Accepted
| 20
| 7,668
| 319
|
# -*- coding: utf-8 -*-
import sys
import os
for s in sys.stdin:
a, b, c, d, e, f = list(map(int, s.split()))
delta = a * e - b * d
if delta == 0:
break
else:
x = 1 / delta * (e * c - b * f)
y = 1 / delta * (-d * c + a * f)
print('{:.3f} {:.3f}'.format(x + 0, y + 0))
|
s314922025
|
p03379
|
u136553578
| 2,000
| 262,144
|
Wrong Answer
| 278
| 26,016
| 238
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
def main():
n = int(input())
a = sorted(map(int, input().split()))
c = n // 2
for i in range(n):
if i < c:
print(a[c])
else:
print(a[c - 1])
if __name__ == '__main__':
main()
|
s579627618
|
Accepted
| 276
| 25,220
| 228
|
def main():
c = int(input()) // 2
elem = list(map(int, input().split()))
left, right = sorted(elem)[c - 1:c + 1]
for val in elem:
print((left, right)[val <= left])
if __name__ == '__main__':
main()
|
s501805123
|
p03698
|
u681110193
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 82
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s=list(input())
if len(set(s))==len(list(s)):
print('Yes')
else:print('No')
|
s519472214
|
Accepted
| 17
| 2,940
| 82
|
s=list(input())
if len(set(s))==len(list(s)):
print('yes')
else:print('no')
|
s956102023
|
p02646
|
u062754605
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 9,168
| 155
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
if abs(A-B) >= abs((V - W) * T):
print("YES")
else:
print("NO")
|
s051391435
|
Accepted
| 21
| 9,116
| 167
|
A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
if V > W and abs(A - B) <= abs((W - V) * T):
print("YES")
else:
print("NO")
|
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