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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s046421464
|
p02866
|
u738898077
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 83,092
| 536
|
Given is an integer sequence D_1,...,D_N of N elements. Find the number, modulo 998244353, of trees with N vertices numbered 1 to N that satisfy the following condition: * For every integer i from 1 to N, the distance between Vertex 1 and Vertex i is D_i.
|
import sys
pre_strk = 1
now_strk = 1
ans = 1
n = int(input())
d_list = list(map(int,input().split()))
d_list.sort()
print(d_list)
if d_list[0] != 0:
print(0)
sys.exit()
for i in range(1,n):
if d_list[i] == d_list[i-1]:
now_strk += 1
elif d_list[i] != d_list[i-1] and (d_list[i-1] != 1 or d_list[i-1] != 0):
ans *= pre_strk ** now_strk
pre_strk = now_strk
now_strk = 1
print(ans)
else:
pre_strk = now_strk
now_strk = 1
ans *= pre_strk ** now_strk
print(ans)
|
s329286414
|
Accepted
| 871
| 14,396
| 727
|
import sys
pre_strk = 1
now_strk = 1
ans = 1
n = int(input())
d_list = list(map(int,input().split()))
# print(d_list)
if d_list[0] != 0:
print(0)
sys.exit()
d_list.sort()
if d_list[1] == 0:
print(0)
sys.exit()
for i in range(1,n):
if d_list[i] == d_list[i-1]:
now_strk += 1
elif d_list[i] > d_list[i-1] + 1:
print(0)
sys.exit()
elif d_list[i] != d_list[i-1] and d_list[i-1] != 1:
for j in range(now_strk):
ans *= pre_strk
pre_strk = now_strk
now_strk = 1
# print(ans)
ans = ans % 998244353
else:
pre_strk = now_strk
now_strk = 1
for i in range(now_strk):
ans *= pre_strk
print(ans%998244353)
|
s001094691
|
p02416
|
u536089081
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 138
|
Write a program which reads an integer and prints sum of its digits.
|
from sys import stdin
for line in stdin:
if line == '0':
break
num = list(map(int, line.strip('\n')))
print(sum(num))
|
s693093604
|
Accepted
| 20
| 5,608
| 140
|
from sys import stdin
for line in stdin:
if line == '0\n':
break
num = list(map(int, line.strip('\n')))
print(sum(num))
|
s696365738
|
p03971
|
u883459277
| 2,000
| 262,144
|
Wrong Answer
| 73
| 9,892
| 282
|
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
count1 = 0
count2 = 0
N, A, B = map(int, input().split())
S = list(input())
for i in S:
if i == "a" and A + B >= count1:
print("Yes")
count1 += 1
elif i == "b" and A + B >= count1 and B >= count2:
print("Yes")
count1 += 1
count2 += 1
else :
print("No")
|
s442156611
|
Accepted
| 70
| 9,988
| 279
|
count1 = 0
count2 = 0
N, A, B = map(int, input().split())
S = list(input())
for i in S:
if i == "a" and A + B > count1:
print("Yes")
count1 += 1
elif i == "b" and A + B > count1 and B > count2:
print("Yes")
count1 += 1
count2 += 1
else :
print("No")
|
s592961592
|
p03672
|
u327248573
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 152
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
S = input()
for i in range(len(S) - 1, 0, -1):
S = S[:i]
print(S)
if i % 2 == 0 and S[:(i//2)] == S[(i//2):]:
print(i)
break
|
s910042408
|
Accepted
| 17
| 2,940
| 139
|
S = input()
for i in range(len(S) - 1, 0, -1):
S = S[:i]
if i % 2 == 0 and S[:(i//2)] == S[(i//2):]:
print(i)
break
|
s725040291
|
p03862
|
u136395536
| 2,000
| 262,144
|
Wrong Answer
| 115
| 14,548
| 343
|
There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies. Snuke can perform the following operation any number of times: * Choose a box containing at least one candy, and eat one of the candies in the chosen box. His objective is as follows: * Any two neighboring boxes contain at most x candies in total. Find the minimum number of operations required to achieve the objective.
|
N,x = (int(i) for i in input().split())
a = [int(i) for i in input().split()]
count = 0
if a[1] > x:
difference = x-a[1]
count += difference
a[1] -= difference
for i in range(N-1):
difference = x - (a[i] + a[i+1])
if difference < 0:
count += abs(difference)
a[i+1] += difference
print(a)
print(count)
|
s473542986
|
Accepted
| 115
| 14,068
| 407
|
N,x = (int(i) for i in input().split())
a = [int(i) for i in input().split()]
count = 0
if a[1] > x:
difference = x-a[1]
count += difference
a[1] -= difference
for i in range(N-1):
difference = x - (a[i] + a[i+1])
if difference < 0:
count += abs(difference)
a[i+1] += difference
if a[i+1] < 0:
a[i] += a[i]
a[i+1] = 0
print(count)
|
s253130147
|
p03739
|
u820351940
| 2,000
| 262,144
|
Wrong Answer
| 99
| 14,212
| 455
|
You are given an integer sequence of length N. The i-th term in the sequence is a_i. In one operation, you can select a term and either increment or decrement it by one. At least how many operations are necessary to satisfy the following conditions? * For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero. * For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.
|
n = int(input())
a = list(map(int, input().split()))
now = a[0]
m = now < 0 and -1 or 1
result = 0
for i, v in enumerate(a[1:]):
tmp = now * -1
if m < 0:
tmp += 1
if tmp > v:
now += v
else:
result += abs(tmp - v)
now = -1
else:
tmp -= 1
if tmp < v:
now += v
else:
result += abs(tmp - v)
now = 1
m *= -1
print(result)
|
s874047129
|
Accepted
| 233
| 23,244
| 752
|
import numpy as np
input()
a = list(map(int, input().split()))
def calc(a):
value = a[0]
sign = value >= 0
result = 0
for i in a[1:]:
value += i
diff = 0
if sign and value >= 0:
diff = value + 1
elif not sign and value < 0:
diff = value - 1
result += abs(diff)
sign = not sign
value -= diff
if value == 0:
diff += sign
value += diff
result += diff
return result
if a[0] == 0:
a[0] = 1
r1 = calc(a[:])
a[0] = -1
r2 = calc(a[:])
print(min(r1, r2) + 1)
else:
r1 = calc(a[:])
diff = -a[0] - a[0] // abs(a[0])
a[0] += diff
r2 = calc(a[:]) + abs(diff)
print(min(r1, r2))
|
s592665092
|
p03679
|
u333190709
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 99
|
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 x >= (b-a):
print('delicious')
else:
print('dangerous')
|
s687311088
|
Accepted
| 17
| 2,940
| 131
|
x, a, b = map(int, input().split())
if a >= b:
print('delicious')
elif x >= (b - a):
print('safe')
else:
print('dangerous')
|
s504541125
|
p04011
|
u611090896
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 64
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
n,k,x,y = [int(input()) for i in range(4)]
print(n*x+abs(n-k)*y)
|
s981012860
|
Accepted
| 17
| 2,940
| 70
|
n,k,x,y = [int(input()) for i in range(4)]
print(n*x-(x-y)*max(n-k,0))
|
s891888680
|
p03543
|
u002459665
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 69
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
n = input()
if n == n[0] * 4:
print("Yes")
else:
print("No")
|
s612962343
|
Accepted
| 20
| 3,060
| 91
|
s = input()
if (s[0] * 3 in s) or (s[3] * 3 in s):
print("Yes")
else:
print("No")
|
s412896962
|
p03478
|
u089230684
| 2,000
| 262,144
|
Wrong Answer
| 41
| 2,940
| 256
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
def digit_sum(x):
string_x = str(x)
total = 0
for char in string_x:
total += int(char)
return total
a, b, c = map(int, input().split())
sum=0
for i in range(0,a):
if(digit_sum(i)>=b and digit_sum(i)<=c):
sum=sum+i
print(sum)
|
s658216829
|
Accepted
| 25
| 2,940
| 211
|
n, A, B = map(int, input().split())
def ok(a):
s = 0
while a > 0:
s += a % 10
a //= 10
return (A <= s and s <= B)
s = 0
for i in range(1, n+1):
if ok(i):
s += i
print(s)
|
s694633257
|
p02419
|
u971748390
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,328
| 87
|
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
|
sentence=input()
snt=sentence.lower()
word=input()
cnt=sentence.count(word)
print(cnt)
|
s749988048
|
Accepted
| 20
| 7,432
| 198
|
w= input().casefold()
cnt = 0
while(1):
snt = input()
if snt == "END_OF_TEXT":
break
sntt = snt.casefold()
words = sntt.split()
cnt += words.count(w)
print(cnt)
|
s332239462
|
p03696
|
u563824531
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 408
|
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.
|
n = input()
s = input()
count = 0
countl = 0
for c in s:
if c=='(':
count += 1
else:
count -= 1
if count < 0:
countl += 1
count = 0
count = 0
countr = 0
for c in s[::-1]:
if c=='(':
count -= 1
else:
count += 1
if count < 0:
countr += 1
count = 0
print(countl)
print(countr)
print("("*countl + s + ")"*countr)
|
s963161849
|
Accepted
| 18
| 3,064
| 372
|
n = input()
s = input()
count = 0
countl = 0
for c in s:
if c=='(':
count += 1
else:
count -= 1
if count < 0:
countl += 1
count = 0
count = 0
countr = 0
for c in s[::-1]:
if c=='(':
count -= 1
else:
count += 1
if count < 0:
countr += 1
count = 0
print('('*countl + s + ')'*countr)
|
s470214554
|
p02613
|
u427231601
| 2,000
| 1,048,576
|
Wrong Answer
| 148
| 16,196
| 319
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n = int(input())
s = [input() for _ in range(n)]
a = 0
w = 0
t = 0
r = 0
for i in range(n):
if s[i] == "AC":
a += 1
elif s[i] == "WA":
w += 1
elif s[i] == "TLE":
t += 1
else:
r += 1
print("AC x" +str(a))
print("WA x" +str(w))
print("TLE x" +str(t))
print("RE x" +str(r))
|
s772107748
|
Accepted
| 157
| 16,332
| 315
|
n = int(input())
s = [input() for _ in range(n)]
a = 0
w = 0
t = 0
r = 0
for i in range(n):
if s[i] == "AC":
a += 1
elif s[i] == "WA":
w += 1
elif s[i] == "TLE":
t += 1
else:
r += 1
print("AC x",str(a))
print("WA x",str(w))
print("TLE x",str(t))
print("RE x",str(r))
|
s076058714
|
p04012
|
u551109821
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 168
|
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
w = list(input())
W = list(set(w))
cnt = [0]*len(W)
for i in w:
cnt[W.index(i)] += 1
for i in cnt:
if i%2!=0:
print('NO')
exit()
print('YES')
|
s620568424
|
Accepted
| 18
| 3,064
| 167
|
w = list(input())
W = list(set(w))
cnt = [0]*len(W)
for i in w:
cnt[W.index(i)] += 1
for i in cnt:
if i%2!=0:
print('No')
exit()
print('Yes')
|
s506438740
|
p02357
|
u073562076
| 2,000
| 262,144
|
Wrong Answer
| 40
| 8,308
| 1,819
|
For a given array $a_1, a_2, a_3, ... , a_N$ of $N$ elements and an integer $L$, find the minimum of each possible sub-arrays with size $L$ and print them from the beginning. For example, for an array $\\{1, 7, 7, 4, 8, 1, 6\\}$ and $L = 3$, the possible sub-arrays with size $L = 3$ includes $\\{1, 7, 7\\}$, $\\{7, 7, 4\\}$, $\\{7, 4, 8\\}$, $\\{4, 8, 1\\}$, $\\{8, 1, 6\\}$ and the minimum of each sub-array is 1, 4, 4, 1, 1 respectively.
|
import collections, numbers
import random
def compute(array, window, maximize):
if not isinstance(window, numbers.Integral):
raise TypeError()
if not isinstance(maximize, bool):
raise TypeError()
if window <= 0:
raise ValueError("Window size must be positive")
result = []
deque = collections.deque()
for i, val in enumerate(array):
val = array[i]
while len(deque) > 0 and ((not maximize and val < deque[-1]) or (maximize and val > deque[-1])):
deque.pop()
deque.append(val)
j = i + 1 - window
if j >= 0:
result.append(deque[0])
if array[j] == deque[0]:
deque.popleft()
return result
class SlidingWindowMinMax(object):
def __init__(self):
self.mindeque = collections.deque()
self.maxdeque = collections.deque()
def get_minimum(self):
return self.mindeque[0]
def get_maximum(self):
return self.maxdeque[0]
def add_tail(self, val):
while len(self.mindeque) > 0 and val < self.mindeque[-1]:
self.mindeque.pop()
self.mindeque.append(val)
while len(self.maxdeque) > 0 and val > self.maxdeque[-1]:
self.maxdeque.pop()
self.maxdeque.append(val)
def remove_head(self, val):
if val < self.mindeque[0]:
raise ValueError("Wrong value")
elif val == self.mindeque[0]:
self.mindeque.popleft()
if val > self.maxdeque[0]:
raise ValueError("Wrong value")
elif val == self.maxdeque[0]:
self.maxdeque.popleft()
if __name__ == "__main__":
n, l = (int(x) for x in input().split())
array = list((int(x) for x in input().split()))
answer = compute(array, l, False)
print(answer)
|
s040689837
|
Accepted
| 1,790
| 122,036
| 916
|
import collections, numbers
import random
def compute(array, window, maximize):
if not isinstance(window, numbers.Integral):
raise TypeError()
if not isinstance(maximize, bool):
raise TypeError()
if window <= 0:
raise ValueError("Window size must be positive")
result = []
deque = collections.deque()
for i, val in enumerate(array):
val = array[i]
while len(deque) > 0 and ((not maximize and val < deque[-1]) or (maximize and val > deque[-1])):
deque.pop()
deque.append(val)
j = i + 1 - window
if j >= 0:
result.append(deque[0])
if array[j] == deque[0]:
deque.popleft()
return result
if __name__ == "__main__":
n, l = (int(x) for x in input().split())
array = list((int(x) for x in input().split()))
answer = compute(array, l, False)
print(*answer)
|
s800341068
|
p03795
|
u328755070
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 52
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
N = int(input())
ans = 800 * N - N // 15
print(ans)
|
s504391514
|
Accepted
| 17
| 2,940
| 60
|
N = int(input())
ans = 800 * N - (N // 15) * 200
print(ans)
|
s002929744
|
p02747
|
u316175331
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 3,188
| 99
|
A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.
|
import re
pattern = re.compile("(hi)+")
s = input()
print("YES" if pattern.fullmatch(s) else "NO")
|
s084066119
|
Accepted
| 22
| 3,188
| 100
|
import re
pattern = re.compile("(hi)+")
s = input()
print("Yes" if pattern.fullmatch(s) else "No")
|
s820358459
|
p02936
|
u244416763
| 2,000
| 1,048,576
|
Wrong Answer
| 2,107
| 57,224
| 655
|
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
N,Q = map(int,input().split())
a = [0 for _ in range(N-1)]
b = [0 for _ in range(N-1)]
for i in range(N-1):
a[i],b[i] = map(int,input().split())
p = [0 for _ in range(Q)]
x = [0 for _ in range(Q)]
for i in range(Q):
p[i],x[i] = map(int,input().split())
parent = [0 for _ in range(N)]
counter = [0 for _ in range(N)]
for i in range(Q):
counter[p[i]-1] += x[i]
for i in range(N-1):
parent[b[i]-1] = a[i]-1
point = [counter[i] for i in range(N)]
for i in range(1,N):
now = i
while(1):
point[i] += counter[parent[now]]
if parent[now] == 0:
break
else:
now = parent[now]
print(point)
|
s765857564
|
Accepted
| 1,553
| 56,088
| 631
|
from collections import deque
def main():
n,q = map(int,input().split())
node = [[]for _ in range(n)]
looked = [1]*n
for i in range(n-1):
a,b = map(int,input().split())
node[b-1].append(a-1)
node[a-1].append(b-1)
counter = [0]*n
for i in range(q):
p,x = map(int,input().split())
counter[p-1] += x
queue = deque([0])
looked[0] = 0
while(queue):
now = queue.pop()
for i in node[now]:
if looked[i]:
queue.append(i)
counter[i] += counter[now]
looked[i] = 0
print(*counter)
main()
|
s759272386
|
p03854
|
u079625320
| 2,000
| 262,144
|
Wrong Answer
| 25
| 4,976
| 143
|
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`.
|
import re
S = input()
t = re.findall('dream|dremer|eraser|erase', S)
print(t)
T = ''.join(t)
if S == T:
print('YES')
else:
print('NO')
|
s351229801
|
Accepted
| 25
| 4,512
| 159
|
import re
S = ''.join(list(reversed(input())))
t = re.findall("maerd|remaerd|resare|esare",S)
T = ''.join(t)
if S == T:
print("YES")
else:
print("NO")
|
s484356724
|
p03447
|
u875924067
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 86
|
You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?
|
x = int(input(""))
a = int(input(""))
b = int(input(""))
ans = x - a - b
print(ans)
|
s166236048
|
Accepted
| 17
| 2,940
| 132
|
x = int(input(""))
a = int(input(""))
b = int(input(""))
x = x - a
i = 0
while(x - b * i >= 0):
i+=1
i-=1
print(x - b * i)
|
s368605043
|
p03605
|
u916743460
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 86
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
n = int(input())
if n / 10 == 9 or n % 10 == 9:
print("yes")
else :
print("no")
|
s110845106
|
Accepted
| 17
| 2,940
| 97
|
n = int(input())
if n / 10 == 9 or n % 10 == 9 or n >= 90:
print("Yes")
else :
print("No")
|
s515103869
|
p03699
|
u268318377
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 317
|
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
import sys
read = sys.stdin.buffer.read
N = int(input())
S = list(map(int, read().rstrip().split()))
sumS = sum(S)
if all(s % 10 == 0 for s in S):
print(0)
elif sumS % 10 != 0:
print(sumS)
else:
S = sorted(S)
for s in S:
if s % 10 != 0:
t = s
break
print(sumS - t)
|
s648823757
|
Accepted
| 17
| 3,064
| 269
|
N = int(input())
S = [int(input()) for _ in range(N)]
sumS = sum(S)
if all(s % 10 == 0 for s in S):
print(0)
elif sumS % 10 != 0:
print(sumS)
else:
S = sorted(S)
for s in S:
if s % 10 != 0:
t = s
break
print(sumS - t)
|
s016496616
|
p02690
|
u864069774
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,352
| 426
|
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.
|
import itertools
z = []
for i in range(100):
a =i**5
# print(a)
z.append(a)
if len(str(a)) > 9:
break
X = int(input())
z_p = list(itertools.permutations(z, 2))
for i in range(len(z_p)):
if z_p[i][0] - z_p[i][1] == X:
print(z.index(z_p[i][0]),z.index(z_p[i][1]))
exit()
elif z_p[i][0] + z_p[i][1] == X:
print(z.index(z_p[i][0]),-(z.index(z_p[i][1])))
exit()
|
s832870360
|
Accepted
| 173
| 80,344
| 389
|
import itertools
z = []
for i in range(1000):
a =i**5
# print(a)
z.append(a)
X = int(input())
z_p = list(itertools.permutations(z, 2))
for i in range(len(z_p)):
if z_p[i][0] - z_p[i][1] == X:
print(z.index(z_p[i][0]),z.index(z_p[i][1]))
exit()
elif z_p[i][0] + z_p[i][1] == X:
print(z.index(z_p[i][1]),-(z.index(z_p[i][0])))
exit()
|
s676595063
|
p00295
|
u797673668
| 1,000
| 131,072
|
Wrong Answer
| 80
| 7,860
| 1,269
|
フロッピーキューブをプログラミングで解いてみましょう。フロッピーキューブは図のように表面に色のついた９個の立方体から構成されている立体パズルで、キューブの列を回転させることによって、６つの各面の色をそろえます。 フロッピーキューブに対しては下図のような４種類の操作を行うことができ、一回の操作で、端にある３つの隣接したキューブを１８０度回転することができます。わかりやすいように、図では、上面に＋(赤色)、下面に＊(緑色)、右前面に□(黄色)、左前面に●(青色)、右奥面に○(水色)、左奥面に■紫色) の記号が付いている状態を初期状態としています。 フロッピーキューブの初期状態が与えられるので、パズルを解くために必要な最小の操作回数を求めるプログラムを作成してください。
|
import heapq
swap_list = [
{(6, 21), (7, 22), (8, 23), (9, 11), (12, 17)},
{(0, 27), (1, 28), (2, 29), (14, 15), (18, 20)},
{(0, 23), (3, 26), (6, 29), (15, 17), (9, 20)},
{(2, 21), (5, 24), (8, 27), (12, 14), (11, 18)}
]
def check_complete(cube):
for surface in (cube[:9], cube[9:12], cube[12:15], cube[15:18], cube[18:21]):
val = surface[0]
if any(block - val for block in surface):
return False
return True
def next_cubes(cube, prev):
nc = [None] * 4
for i in range(4):
if i != prev:
nci = cube[:]
for a, b in swap_list[i]:
nci[a], nci[b] = cube[b], cube[a]
nc[i] = nci
return nc
n = int(input())
while n:
cube0 = list(map(int, input().split()))
checked = []
queue = [(0, cube0, -1)]
while queue:
cnt, cube, prev = heapq.heappop(queue)
print(cnt, cube, prev)
if cube in checked:
continue
if check_complete(cube):
print(cnt)
break
checked.append(cube)
for i, next_cube in enumerate(next_cubes(cube, prev)):
if next_cube and not next_cube in checked:
heapq.heappush(queue, (cnt + 1, next_cube, i))
n -= 1
|
s914747134
|
Accepted
| 240
| 7,884
| 1,238
|
import heapq
swap_list = [
{(6, 21), (7, 22), (8, 23), (9, 11), (12, 17)},
{(0, 27), (1, 28), (2, 29), (14, 15), (18, 20)},
{(0, 23), (3, 26), (6, 29), (15, 17), (9, 20)},
{(2, 21), (5, 24), (8, 27), (12, 14), (11, 18)}
]
def check_complete(cube):
for surface in (cube[:9], cube[9:12], cube[12:15], cube[15:18], cube[18:21]):
val = surface[0]
if any(block - val for block in surface):
return False
return True
def next_cubes(cube, prev):
nc = [None] * 4
for i in range(4):
if i != prev:
nci = cube[:]
for a, b in swap_list[i]:
nci[a], nci[b] = cube[b], cube[a]
nc[i] = nci
return nc
n = int(input())
while n:
cube0 = list(map(int, input().split()))
checked = []
queue = [(0, cube0, -1)]
while queue:
cnt, cube, prev = heapq.heappop(queue)
if cube in checked:
continue
if check_complete(cube):
print(cnt)
break
checked.append(cube)
for i, next_cube in enumerate(next_cubes(cube, prev)):
if next_cube and not next_cube in checked:
heapq.heappush(queue, (cnt + 1, next_cube, i))
n -= 1
|
s439656645
|
p03999
|
u378691508
| 2,000
| 262,144
|
Wrong Answer
| 31
| 9,168
| 194
|
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion. All strings that can be obtained in this way can be evaluated as formulas. Evaluate all possible formulas, and print the sum of the results.
|
S=input()
n=len(S)
ans=0
for paint in range((n-1)**2):
temp_S=S[0]
for i in range(n-1):
if (paint>>i)&1==1:
temp_S+="+"
temp_S+=S[i]
ans+=sum(map(int, temp_S.split("+")))
print(ans)
|
s329036050
|
Accepted
| 33
| 9,192
| 199
|
S=input()
n=len(S)
ans=0
for paint in range(2**(n-1)):
temp_S=S[0]
for i in range(n-1):
if (paint>>i)&1==1:
temp_S+="+"
temp_S+=S[i+1]
ans+=sum(map(int, temp_S.split("+")))
print(ans)
|
s782956395
|
p03854
|
u928784113
| 2,000
| 262,144
|
Wrong Answer
| 89
| 3,188
| 200
|
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()
i = 0
while S:
S = S.replace("eraser","").replace("erase","").replace("dreamer","").replace("dream","")
i += 1
if i == 10**5+1:
print("NO")
exit()
print("Yes")
|
s895050287
|
Accepted
| 91
| 3,188
| 200
|
S = input()
i = 0
while S:
S = S.replace("eraser","").replace("erase","").replace("dreamer","").replace("dream","")
i += 1
if i == 10**5+1:
print("NO")
exit()
print("YES")
|
s684287963
|
p03471
|
u557523358
| 2,000
| 262,144
|
Wrong Answer
| 2,103
| 3,060
| 248
|
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
n, y = map(int, input().split())
x, y, z = -1, -1, -1
for i in range(n + 1):
for j in range(n + 1 - i):
for k in range(n + 1 - i - j):
if 10000 * i + 5000 * j + 1000 * k == y:
x, y, z = i, j, k
print(x, y, z)
|
s876616588
|
Accepted
| 769
| 3,060
| 231
|
n, total = map(int, input().split())
x, y, z = -1, -1, -1
for i in range(n + 1):
for j in range(n + 1 - i):
k = n - i - j
if 10000 * i + 5000 * j + 1000 * k == total:
x, y, z = i, j, k
print(x, y, z)
|
s735913477
|
p04043
|
u788023488
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 140
|
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.
|
List = [int(i) for i in input().split()]
List.sort()
ans='No'
if List[0]==5:
if List[1]==5:
if List[2]==7:
ans='Yes'
print(ans)
|
s296724307
|
Accepted
| 17
| 2,940
| 140
|
List = [int(i) for i in input().split()]
List.sort()
ans='NO'
if List[0]==5:
if List[1]==5:
if List[2]==7:
ans='YES'
print(ans)
|
s848182053
|
p00024
|
u777299405
| 1,000
| 131,072
|
Wrong Answer
| 50
| 7,380
| 98
|
Ignoring the air resistance, velocity of a freely falling object $v$ after $t$ seconds and its drop $y$ in $t$ seconds are represented by the following formulas: $ v = 9.8 t $ $ y = 4.9 t^2 $ A person is trying to drop down a glass ball and check whether it will crack. Your task is to write a program to help this experiment. You are given the minimum velocity to crack the ball. Your program should print the lowest possible floor of a building to crack the ball. The height of the $N$ floor of the building is defined by $5 \times N - 5$.
|
import math
m = float(input())
t = m / 9.8
y = 4.9 * t * t
n = math.ceil(y / 5)
print(int(n + 1))
|
s581577826
|
Accepted
| 40
| 7,504
| 129
|
import math
import sys
for s in sys.stdin:
d = float(s)
y = 4.9 * (d / 9.8)**2
n = math.ceil(y / 5) + 1
print(n)
|
s594160481
|
p03657
|
u772588522
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 96
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a, b = map(int, input().split())
if a+b % 3:
print('Impossible')
else:
print('Possible')
|
s198291471
|
Accepted
| 17
| 2,940
| 274
|
a, b = map(int, input().split())
if not (a+b) % 3 or not a % 3 or not b % 3:
print('Possible')
else:
print('Impossible')
#if not a % 3:
# print('Impossible')
#elif not b % 3:
# print('Impossible')
#elif not a+b % 3:
# print('Impossible')
#else:
# print('Possible')
|
s425121062
|
p03385
|
u734805215
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 108
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
s=input()
flag = s.count('a')*s.count('b')*s.count('c')
if flag >=1:
print('yes')
else:
print('no')
|
s413100324
|
Accepted
| 17
| 2,940
| 108
|
s=input()
flag = s.count('a')*s.count('b')*s.count('c')
if flag >=1:
print('Yes')
else:
print('No')
|
s746635511
|
p03543
|
u921773161
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 80
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
N = input()
if N[0] == N[1] == N[2] == N[3] :
print('Yes')
else:
print('No')
|
s078655434
|
Accepted
| 17
| 2,940
| 117
|
N = input()
if N[0] == N[1] == N[2] :
print('Yes')
elif N[1] == N[2] == N[3] :
print('Yes')
else:
print('No')
|
s711256214
|
p03853
|
u732844340
| 2,000
| 262,144
|
Wrong Answer
| 44
| 4,472
| 225
|
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
hw_ = list(map(int, input().split()))
h = hw_[0]
w = hw_[1]
c = []
for i in range(0, h):
c.append(input())
print(c)
for i in range(0, h * 2):
for j in range(0, w):
print(c[int(i / 2)][j],end='')
print()
|
s377613779
|
Accepted
| 45
| 4,472
| 216
|
hw_ = list(map(int, input().split()))
h = hw_[0]
w = hw_[1]
c = []
for i in range(0, h):
c.append(input())
for i in range(0, h * 2):
for j in range(0, w):
print(c[int(i / 2)][j],end='')
print()
|
s068618851
|
p03854
|
u283929013
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 405
|
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()
marker = 0
flag = True
while flag and (marker < len(S)):
if S[marker:marker+5] == "dream":
marker += 5
if S[marker+5:marker+7] == "er":
marker += 2
elif S[marker:marker+5] == "erase":
marker += 5
if S[marker+5:marker+6] == "r":
marker += 1
else:
flag = False
if flag:
print("Yes")
else:
print("No")
|
s743354726
|
Accepted
| 28
| 3,188
| 391
|
S = input()
marker = len(S)
flag = True
while flag and (marker > 0):
if S[marker-7:marker] == "dreamer":
marker -= 7
elif S[marker-6:marker] == "eraser":
marker -= 6
elif S[marker-5:marker] == "dream":
marker -= 5
elif S[marker-5:marker] == "erase":
marker -= 5
else:
flag = False
if flag:
print("YES")
else:
print("NO")
|
s751713817
|
p03555
|
u354773735
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 309
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
# coding: utf-8
#a = int(input())
#b = int(input())
#a, b, c = map(int, input().split())
s1 = input()
s2 = input()
if s1[0] == s2[2] and s1[1] == s2[1] and s1[2] == s2[0]:
ans = "Yes"
else :
ans = "No"
print(ans)
|
s104696445
|
Accepted
| 17
| 3,060
| 309
|
# coding: utf-8
#a = int(input())
#b = int(input())
#a, b, c = map(int, input().split())
s1 = input()
s2 = input()
if s1[0] == s2[2] and s1[1] == s2[1] and s1[2] == s2[0]:
ans = "YES"
else :
ans = "NO"
print(ans)
|
s612745183
|
p04012
|
u640603056
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 462
|
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
import sys
def ILI(): return list(map(int, sys.stdin.readline().rstrip().split()))
def ISI(): return map(int, sys.stdin.readline().rstrip().split())
def II(): return int(sys.stdin.readline().rstrip())
def ISS(): return sys.stdin.readline().rstrip().split()
def IS(): return sys.stdin.readline().rstrip()
w = list(IS())
set_w = set(w)
for s in set_w:
if w.count(s)%2 != 0:
print("NO")
break
else:
print("YES")
|
s860904319
|
Accepted
| 17
| 3,064
| 462
|
import sys
def ILI(): return list(map(int, sys.stdin.readline().rstrip().split()))
def ISI(): return map(int, sys.stdin.readline().rstrip().split())
def II(): return int(sys.stdin.readline().rstrip())
def ISS(): return sys.stdin.readline().rstrip().split()
def IS(): return sys.stdin.readline().rstrip()
w = list(IS())
set_w = set(w)
for s in set_w:
if w.count(s)%2 != 0:
print("No")
break
else:
print("Yes")
|
s704647567
|
p03069
|
u626468554
| 2,000
| 1,048,576
|
Wrong Answer
| 84
| 8,036
| 268
|
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
#!/usr/bin/env python3
n = int(input())
s = input()
S =[]
for i in s:
S.append(i)
w = 0
b = 0
an = n
for i in range(n-1,-1,-1):
if i == "#":
b += 1
elif i == ".":
w += 1
if w > b:
an = n-i
print(S)
print(i)
|
s093266170
|
Accepted
| 157
| 12,936
| 378
|
#!/usr/bin/env python3
n = int(input())
s = input()
S =[]
for i in s:
S.append(i)
w = 0
b = 0
for i in range(n):
if S[i] == "#":
b += 1
elif S[i] == ".":
w += 1
ans = [0 for i in range(n+1)]
ans[0] = w
for i in range(n):
if S[i] == "#":
ans[i+1] = ans[i]+1
elif S[i] == ".":
ans[i+1] = ans[i]-1
print(min(ans))
|
s900974826
|
p03386
|
u163874353
| 2,000
| 262,144
|
Wrong Answer
| 2,153
| 798,488
| 158
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a, b, k = map(int, input().split())
l =[]
for i in range(a, b):
l.append(i)
#print(l)
f = l[:k]
b = l[len(l) - k:]
n = f + b
for j in l:
print(j)
|
s411082380
|
Accepted
| 17
| 3,060
| 190
|
a, b, k = map(int, input().split())
l = []
for i in range(a, min(a + k, b)):
l.append(i)
for i in range(max(a, b - k + 1), b + 1):
l.append(i)
for i in sorted(set(l)):
print(i)
|
s585371828
|
p02692
|
u159723084
| 2,000
| 1,048,576
|
Wrong Answer
| 232
| 16,844
| 1,016
|
There is a game that involves three variables, denoted A, B, and C. As the game progresses, there will be N events where you are asked to make a choice. Each of these choices is represented by a string s_i. If s_i is `AB`, you must add 1 to A or B then subtract 1 from the other; if s_i is `AC`, you must add 1 to A or C then subtract 1 from the other; if s_i is `BC`, you must add 1 to B or C then subtract 1 from the other. After each choice, none of A, B, and C should be negative. Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.
|
# -*- coding: utf-8 -*-
N,A,B,C=input().split()
D=dict()
D["A"]=int(A)
D["B"]=int(B)
D["C"]=int(C)
S=[]
for i in range(int(N)):
S.append(input())
ans="Yes"
F=[]
for i,s in enumerate(S):
if D[s[0]]==0 and D[s[1]]==0:
ans="No"
break
if D[s[0]]==0 and D[s[1]]==1:
D[s[0]]=1
D[s[1]]=0
F.append(s[0])
if D[s[0]]==1 and D[s[1]]==0:
D[s[0]]=0
D[s[1]]=1
F.append(s[1])
if D[s[0]]==1 and D[s[1]]==1:
if D["A"]+D["B"]+D["C"]==2:
if i!=int(N)-1:
s0=S[i+1][0]
s1=S[i+1][1]
if s[0]==s0 or s[0]==s1:
D[s[0]]=2
D[s[1]]=0
F.append(s[0])
if s[1]==s0 or s[1]==s1:
D[s[0]]=0
D[s[1]]=2
F.append(s[1])
if len(F)!=i+1:
D[s[0]]+=1
D[s[1]]-=1
F.append(s[0])
print(ans)
if ans=="Yes":
for f in F:
print(f)
|
s535585994
|
Accepted
| 286
| 16,860
| 976
|
# -*- coding: utf-8 -*-
N,A,B,C=input().split()
D=dict()
D["A"]=int(A)
D["B"]=int(B)
D["C"]=int(C)
S=[]
for i in range(int(N)):
S.append(input())
ans="Yes"
F=[]
for i,s in enumerate(S):
if D[s[0]]==0 and D[s[1]]==0:
ans="No"
break
if D[s[0]]==1 and D[s[1]]==1:
if D["A"]+D["B"]+D["C"]==2:
if i!=int(N)-1:
if S[i+1] != s:
s0=S[i+1][0]
s1=S[i+1][1]
if s[0]==s0 or s[0]==s1:
F.append(s[0])
if s[1]==s0 or s[1]==s1:
F.append(s[1])
if len(F)!=i+1:
if D[s[0]]<D[s[1]]:
F.append(s[0])
if D[s[0]]>D[s[1]]:
F.append(s[1])
if D[s[0]]==D[s[1]]:
F.append(s[0])
D[F[-1]]+=1
if s[0]==F[-1]: D[s[1]]-=1
if s[1]==F[-1]: D[s[0]]-=1
#print(D)
#print(F)
print(ans)
if ans=="Yes":
for f in F:
print(f)
|
s679966304
|
p03361
|
u062147869
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 422
|
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 = [list(input()) for i in range(H)]
T =[['.']*(W+2)]
for y in range (H):
T.append(['.']+S[y]+['.'])
T.append(['.']*(W+2))
ans = 'Yes'
for y in range(1,H+1):
for x in range(1,W+1):
if ((T[y][x]=='#') and (T[y-1][x-1]=='.')
and (T[y+1][x-1]=='.') and (T[y-1][x+1]=='.') and (T[y+1][x+1]=='.') ):
ans = 'No'
print(ans)
|
s480581036
|
Accepted
| 22
| 3,064
| 414
|
H,W = map(int, input().split())
S = [list(input()) for i in range(H)]
T =[['.']*(W+2)]
for y in range (H):
T.append(['.']+S[y]+['.'])
T.append(['.']*(W+2))
ans = 'Yes'
for y in range(1,H+1):
for x in range(1,W+1):
if ((T[y][x]=='#') and (T[y-1][x]=='.')
and (T[y+1][x]=='.') and (T[y][x+1]=='.') and (T[y][x-1]=='.') ):
ans = 'No'
print(ans)
|
s926070035
|
p03485
|
u204800924
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 99
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a,b = map(int, input().split())
if (a+b)%2 == 0:
print((a+b)/2)
else :
print((a+b)/2 + 1)
|
s405191080
|
Accepted
| 18
| 2,940
| 109
|
a,b = map(int, input().split())
if (a+b)%2 == 0:
print(int((a+b)/2))
else :
print(int((a+b)/2) + 1)
|
s916494362
|
p03671
|
u516579758
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 62
|
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.
|
l=list(map(int,input().split()))
d=min(l)
e=min(l)
print(d+e)
|
s801698408
|
Accepted
| 17
| 2,940
| 73
|
l=list(map(int,input().split()))
d=min(l)
l.remove(d)
e=min(l)
print(d+e)
|
s133904456
|
p02663
|
u308914480
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,100
| 71
|
In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?
|
H1,H2,M1,M2,K=map(int, input().split())
print(((60*(H2-H1))+(M2-M1)-K))
|
s053715761
|
Accepted
| 25
| 9,156
| 73
|
a,b,c,d,e=map(int,input().split())
l=(60*c+d)-(60*a+b)
ans=l-e
print(ans)
|
s096731474
|
p04031
|
u751047721
| 2,000
| 262,144
|
Wrong Answer
| 53
| 3,064
| 277
|
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
n = int(input())
list_n = [int(n) for n in input().split(' ')]
print(list_n)
A = 0
for n in list_n:
A += (100-n)**2
for n in range(-100,100):
B = 0
for num in list_n:
B += (n-num)**2
if A > B:
A = B
print(A)
|
s789503793
|
Accepted
| 51
| 3,064
| 263
|
n = int(input())
list_n = [int(n) for n in input().split(' ')]
A = 0
for n in list_n:
A += (100-n)**2
for n in range(-100,100):
B = 0
for num in list_n:
B += (n-num)**2
if A > B:
A = B
print(A)
|
s036577390
|
p02612
|
u857657465
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,140
| 31
|
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)
|
s217649811
|
Accepted
| 29
| 9,144
| 76
|
N = int(input())
if(N%1000 == 0):
print(0)
else:
print(1000-N%1000)
|
s812291106
|
p02613
|
u867320886
| 2,000
| 1,048,576
|
Wrong Answer
| 147
| 9,236
| 326
|
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 _ in range(n):
s = input()
if s=="AC":
ac += 1
elif s=="WA":
wa += 1
elif tle=="TLE":
tle += 1
else:
re += 1
print("AC x {}".format(ac))
print("WA x {}".format(wa))
print("TLE x {}".format(tle))
print("RE x {}".format(re))
|
s335687784
|
Accepted
| 150
| 9,204
| 324
|
n = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for _ 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 {}".format(ac))
print("WA x {}".format(wa))
print("TLE x {}".format(tle))
print("RE x {}".format(re))
|
s189591877
|
p02865
|
u278183305
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 22
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
print(int(input())//2)
|
s526733206
|
Accepted
| 17
| 2,940
| 26
|
print((int(input())-1)//2)
|
s423761940
|
p03378
|
u007550226
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 169
|
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
_,M,X = list(map(int,input().split()))
a = list(map(int,input().split()))
for i,c in enumerate(a):
print(i,c,X)
if X < c:
print(min(i,M-i))
break
|
s515452331
|
Accepted
| 18
| 3,060
| 221
|
_,M,X = list(map(int,input().split()))
a = list(map(int,input().split()))
'''
for i,c in enumerate(a):
if X < c:
print(min(i,M-i))
break
'''
import bisect
i = bisect.bisect_left(a,X)
print(min(i,M-i))
|
s161101837
|
p03679
|
u766566560
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 146
|
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 X - A >= 0:
print('delicious')
elif A - X < 0 and B - A <= X:
print('safe')
else:
print('dangerous')
|
s115877747
|
Accepted
| 18
| 2,940
| 136
|
X, A, B = map(int, input().split())
if A - B >= 0:
print('delicious')
elif A < B <= A + X:
print('safe')
else:
print('dangerous')
|
s888342828
|
p03196
|
u319245933
| 2,000
| 1,048,576
|
Wrong Answer
| 56
| 9,208
| 712
|
There are N integers a_1, a_2, ..., a_N not less than 1. The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P. Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.
|
def prime_factorize(n):
a = []
f = 2
while f * f <= n:
count = 0
while n % f == 0:
count += 1
n //= f
if count:
a.append([f, count])
f += 2 if f != 2 else 1
if n != 1:
a.append([n, 1])
return a
N, P = map(int, input().split())
res = 1
for p in prime_factorize(P):
for _ in range(p[1] // 4):
res *= p[0]
print(res)
|
s943288250
|
Accepted
| 64
| 9,192
| 712
|
def prime_factorize(n):
a = []
f = 2
while f * f <= n:
count = 0
while n % f == 0:
count += 1
n //= f
if count:
a.append([f, count])
f += 2 if f != 2 else 1
if n != 1:
a.append([n, 1])
return a
N, P = map(int, input().split())
res = 1
for p in prime_factorize(P):
for _ in range(p[1] // N):
res *= p[0]
print(res)
|
s336599764
|
p03469
|
u028973125
| 2,000
| 262,144
|
Wrong Answer
| 27
| 8,892
| 66
|
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.
|
import sys
S = sys.stdin.readline().strip()
print("2018" + S[5:])
|
s878176647
|
Accepted
| 26
| 9,016
| 67
|
import sys
S = sys.stdin.readline().strip()
print("2018" + S[4:])
|
s853078557
|
p02618
|
u940533000
| 2,000
| 1,048,576
|
Wrong Answer
| 180
| 27,300
| 620
|
AtCoder currently hosts three types of contests: ABC, ARC, and AGC. As the number of users has grown, in order to meet the needs of more users, AtCoder has decided to increase the number of contests to 26 types, from AAC to AZC. For convenience, we number these 26 types as type 1 through type 26. AtCoder wants to schedule contests for D days so that user satisfaction is as high as possible. For every day, AtCoder will hold exactly one contest, and each contest will end on that day. The satisfaction is calculated as follows. * The satisfaction at the beginning of day 1 is 0. Satisfaction can be negative. * Holding contests increases satisfaction. The amount of increase will vary depending on a variety of factors. Specifically, we know in advance that holding a contest of type i on day d will increase the satisfaction by s_{d,i}. * If a particular type of contest is not held for a while, the satisfaction decreases. Each contest type i has an integer c_i, and at the end of each day d=1,2,...,D, the satisfaction decreases as follows. Let \mathrm{last}(d,i) be the last day before day d (including d) on which a contest of type i was held. If contests of type i have never been held yet, we define \mathrm{last}(d,i)=0. At the end of day d, the satisfaction decreases by \sum _{i=1}^{26}c_i \times (d-\mathrm{last}(d,i)). Please schedule contests on behalf of AtCoder. If the satisfaction at the end of day D is S, you will get a score of \max(10^6 + S, 0). There are 50 test cases, and the score of a submission is the total scores for each test case. You can make submissions multiple times, and the highest score among your submissions will be your score.
|
import numpy as np
D = int(input())
c = list(map(int, input().split()))
s = []
for i in range(D):
tmp = list(map(int, input().split()))
s.append(tmp)
penalty = np.zeros(26)
date = np.zeros(26)
for i in range(D):
for j in range(26):
penalty[j] = c[j]*((i+1)-date[j])
max_performance = 0
max_idx = 0
for j in range(26):
tmp_performance = s[i][j] - sum(penalty) + c[j]*((i+1)-date[j])
if max_performance < tmp_performance:
max_performance = tmp_performance
max_idx = j+1
date[max_idx-1] = i+1
print(max_idx)
|
s600668131
|
Accepted
| 121
| 27,380
| 310
|
import numpy as np
D = int(input())
c = list(map(int, input().split()))
s = []
for i in range(D):
tmp = list(map(int, input().split()))
s.append(tmp)
for i in range(D):
val_max = 0
idx_max = 0
for j in range(26):
if val_max < s[i][j]:
val_max = s[i][j]
idx_max = j+1
print(idx_max)
|
s817030497
|
p03712
|
u065099501
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,092
| 147
|
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
h,w=map(int,input().split())
a=[input() for _ in range(h)]
print('*'*len(max(a))+'**')
for s in a:
print('*'+s+'*')
print('*'*len(max(a))+'**')
|
s073445215
|
Accepted
| 25
| 9,104
| 147
|
h,w=map(int,input().split())
a=[input() for _ in range(h)]
print('#'*len(max(a))+'##')
for s in a:
print('#'+s+'#')
print('#'*len(max(a))+'##')
|
s697074049
|
p03433
|
u830162518
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 131
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N=int(input())
A=int(input())
if N<=A:
print('Yes')
if N>A:
if (N-A)%500==0:
print('Yes')
else:
print('No')
|
s836629504
|
Accepted
| 17
| 2,940
| 92
|
N=int(input())
A=int(input())
b=N//500
c=N-500*b
if c<=A:
print('Yes')
else:
print('No')
|
s669836232
|
p03378
|
u729300713
| 2,000
| 262,144
|
Wrong Answer
| 20
| 2,940
| 106
|
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
n, m, x = map(int, input().split())
a = list(map(int, input().split()))
print(max(sum(a[:m]), sum(a[m:])))
|
s573904001
|
Accepted
| 17
| 2,940
| 154
|
n, m, x = map(int, input().split())
a = list(map(int, input().split()))
for i,j in enumerate(a):
if j > x:
cnt = i
break
print(min(cnt, m-cnt))
|
s826389171
|
p02936
|
u368796742
| 2,000
| 1,048,576
|
Wrong Answer
| 1,557
| 52,404
| 534
|
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
from collections import deque
n,q = map(int,input().split())
E = [[] for i in range(n)]
for i in range(n-1):
a,b = map(int,input().split())
E[a-1].append(b-1)
score = [0]*n
for i in range(q):
a,b = map(int,input().split())
score[a-1] += b
def bfs(i):
s = score[i-1]
q = deque([])
q.append((i-1,s))
while q:
num,sco = q.popleft()
for j in E[num]:
score[j] += sco
sco = score[j]
q.append((j,sco))
return score
print(*bfs(1))
|
s196632704
|
Accepted
| 1,767
| 62,236
| 614
|
from collections import deque
n,q = map(int,input().split())
E = [[] for i in range(n)]
for i in range(n-1):
a,b = map(int,input().split())
E[a-1].append(b-1)
E[b-1].append(a-1)
score = [0]*n
for i in range(q):
a,b = map(int,input().split())
score[a-1] += b
def bfs(i):
s = -1
q = deque([])
q.append((i-1,s))
while q:
num,bef = q.pop()
for j in E[num]:
if j == bef:
continue
else:
score[j] += score[num]
q.append((j,num))
return score
print(*bfs(1))
|
s079330913
|
p03503
|
u020373088
| 2,000
| 262,144
|
Wrong Answer
| 306
| 3,064
| 367
|
Joisino is planning to open a shop in a shopping street. Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods. There are already N stores in the street, numbered 1 through N. You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2. Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}. Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
|
import itertools
n = int(input())
f = [list(map(int, input().split())) for i in range(n)]
p = [list(map(int, input().split())) for i in range(n)]
ans = -float("inf")
for i in itertools.product((0,1), repeat=10):
a = 0
for j in range(n):
c = 0
for k in range(10):
if i[k] == f[j][k]:
c += 1
a += p[j][c]
ans = max(ans, a)
print(ans)
|
s592692630
|
Accepted
| 317
| 3,064
| 411
|
import itertools
n = int(input())
f = [list(map(int, input().split())) for i in range(n)]
p = [list(map(int, input().split())) for i in range(n)]
ans = -float("inf")
for i in itertools.product((0,1), repeat=10):
if list(i) == [0] * 10:
continue
a = 0
for j in range(n):
c = 0
for k in range(10):
if i[k] == f[j][k] == 1:
c += 1
a += p[j][c]
ans = max(ans, a)
print(ans)
|
s939042856
|
p02263
|
u728700495
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,788
| 388
|
An expression is given in a line. Two consequtive symbols (operand or operator) are separated by a space character. You can assume that +, - and * are given as the operator and an operand is a positive integer less than 106
|
import re
data = input().split()
stack = []
for elem in data:
if re.match('\d+', elem):
stack.append(int(elem))
else:
a = stack.pop()
b = stack.pop()
if elem == "+":
stack.append(b+a)
elif elem == "-":
stack.append(b-a)
elif elem == "*":
stack.append(b*a)
print(elem, stack)
print(stack[0])
|
s211044022
|
Accepted
| 30
| 7,808
| 366
|
import re
data = input().split()
stack = []
for elem in data:
if re.match('\d+', elem):
stack.append(int(elem))
else:
a = stack.pop()
b = stack.pop()
if elem == "+":
stack.append(b+a)
elif elem == "-":
stack.append(b-a)
elif elem == "*":
stack.append(b*a)
print(stack[0])
|
s794105353
|
p03448
|
u766566560
| 2,000
| 262,144
|
Wrong Answer
| 53
| 3,060
| 218
|
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())
ans = 0
for i in range(A):
for j in range(B):
for k in range(C):
if X == 500 * i + 100 * j + 50 * k:
ans += 1
print(ans)
|
s056852614
|
Accepted
| 50
| 3,060
| 216
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
ans = 0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
if X == 500 * i + 100 * j + 50 * k:
ans += 1
print(ans)
|
s646623784
|
p03761
|
u667084803
| 2,000
| 262,144
|
Wrong Answer
| 32
| 3,956
| 255
|
Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string. He will receive a headline which contains one of the strings S_1,...,S_n tomorrow. He is excited and already thinking of what string he will create. Since he does not know the string on the headline yet, he is interested in strings that can be created regardless of which string the headline contains. Find the longest string that can be created regardless of which string among S_1,...,S_n the headline contains. If there are multiple such strings, find the lexicographically smallest one among them.
|
import string
# -*- coding: utf-8 -*-
n=int(input())
S=[]
for i in range (0,n):
S.append(input())
count=0
for c in range(97,97+26):
a=S[0].count(chr(c))
for i in range(0,n):
a=min(a,S[i].count(chr(c)))
count+=a
print(count)
|
s932068510
|
Accepted
| 25
| 3,768
| 289
|
import string
# -*- coding: utf-8 -*-
n=int(input())
S=[]
for i in range (0,n):
S.append(input())
count=0
ans=""
for c in range(97,97+26):
a=S[0].count(chr(c))
for i in range(0,n):
a=min(a,S[i].count(chr(c)))
for i in range (0,a):
ans+=chr(c)
print(ans)
|
s251026553
|
p02612
|
u604398799
| 2,000
| 1,048,576
|
Wrong Answer
| 138
| 31,152
| 479
|
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.
|
# -*- coding: utf-8 -*-
import bisect
import copy
import collections
import heapq
import itertools
import math
import numpy as np
from operator import itemgetter
import scipy.misc
import sys
N=int(input())
def main():
print(N%1000)
return
if __name__ == '__main__':
main()
|
s756666823
|
Accepted
| 131
| 31,052
| 490
|
# -*- coding: utf-8 -*-
import bisect
import copy
import collections
import heapq
import itertools
import math
import numpy as np
from operator import itemgetter
import scipy.misc
import sys
N=int(input())
def main():
print((1000-N%1000)%1000)
return
if __name__ == '__main__':
main()
|
s651370384
|
p03129
|
u966000628
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 86
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
n,m = list(map(int,input().split()))
if m*2+1 <= n:
print("YES")
else:
print("NO")
|
s862833690
|
Accepted
| 18
| 2,940
| 87
|
n,m = list(map(int,input().split()))
if m*2-1 <= n:
print("YES")
else:
print("NO")
|
s028465916
|
p03067
|
u065025264
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 124
|
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
[a, b, c] = list(map(int, input().split(' ')))
print(a, b, c)
if a < b and c < b:
print("Yes")
else:
print("No")
|
s361308644
|
Accepted
| 17
| 2,940
| 147
|
[a, b, c] = list(map(int, input().split(' ')))
if a < c and c < b:
print("Yes")
elif b < c and c < a:
print("Yes")
else:
print("No")
|
s316163562
|
p02853
|
u776864893
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 364
|
We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned.
|
code,implement = input().split(" ")
def prize_dict(rank):
if rank == "1":
prize = 300000
elif rank == "2":
prize = 200000
elif rank == "3":
prize = 100000
else:
prize = 0
return prize
if code == "1" and implement == "1":
prize = 400000
else:
prize = prize_dict(code)+prize_dict(implement)
print(prize)
|
s665969685
|
Accepted
| 17
| 3,064
| 365
|
code,implement = input().split(" ")
def prize_dict(rank):
if rank == "1":
prize = 300000
elif rank == "2":
prize = 200000
elif rank == "3":
prize = 100000
else:
prize = 0
return prize
if code == "1" and implement == "1":
prize = 1000000
else:
prize = prize_dict(code)+prize_dict(implement)
print(prize)
|
s569124294
|
p02396
|
u987701388
| 1,000
| 131,072
|
Wrong Answer
| 80
| 5,900
| 145
|
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= []
while True:
x=int(input())
if x==0: break
a.append(x)
j=0
for i in a:
j+=1
print('case {0}: {1}'.format(j,i))
|
s424863373
|
Accepted
| 140
| 5,600
| 103
|
i=1
while True:
x=int(input())
if x==0: break
print("Case {0}: {1}".format(i,x))
i+=1
|
s174557616
|
p02612
|
u366996583
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,136
| 48
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
X=int(input())
while X>=1000:
X-=1000
print(X)
|
s867336554
|
Accepted
| 32
| 9,144
| 52
|
X=int(input())
while X>1000:
X-=1000
print(1000-X)
|
s140833364
|
p03139
|
u782654209
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 69
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
N,A,B=map(int,input().split(' '))
print(str(A+B-N)+' '+str(min(A,B)))
|
s950912693
|
Accepted
| 17
| 2,940
| 78
|
N,A,B=map(int,input().split(' '))
print(str(min(A,B))+' '+str(max(0,(A+B-N))))
|
s756628501
|
p02421
|
u986478725
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 496
|
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each. Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
|
# AOJ ITP1_9_C
def numinput():
a = input().split()
for i in range(len(a)):
a[i] = int(a[i])
return a
def main():
SCORE_taro = 0
SCORE_hanako = 0
n = int(input())
for i in range(n):
words = input()
if words[0] > words[1]: SCORE_taro += 3
elif words[0] < words[1]: SCORE_hanako += 3
else:
SCORE_taro += 1; SCORE_hanako += 1
print(str(SCORE_taro) + " " + str(SCORE_hanako))
if __name__ == "__main__":
main()
|
s422672348
|
Accepted
| 20
| 5,600
| 504
|
# AOJ ITP1_9_C
def numinput():
a = input().split()
for i in range(len(a)):
a[i] = int(a[i])
return a
def main():
SCORE_taro = 0
SCORE_hanako = 0
n = int(input())
for i in range(n):
words = input().split()
if words[0] > words[1]: SCORE_taro += 3
elif words[0] < words[1]: SCORE_hanako += 3
else:
SCORE_taro += 1; SCORE_hanako += 1
print(str(SCORE_taro) + " " + str(SCORE_hanako))
if __name__ == "__main__":
main()
|
s568013159
|
p03605
|
u723721005
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 129
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
a=int(input())
b=[19.29,39,49,59,69,79,89,90,91,92,93,94,95,96,97,98,99]
if b.count(a)!=0:
print("Yes")
else:
print("No")
|
s614794141
|
Accepted
| 17
| 2,940
| 58
|
if "9" in input():
print("Yes")
else:
print("No")
|
s162763429
|
p00033
|
u661290476
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,612
| 297
|
図のように二股に分かれている容器があります。1 から 10 までの番号が付けられた10 個の玉を容器の開口部 A から落とし、左の筒 B か右の筒 C に玉を入れます。板 D は支点 E を中心に左右に回転できるので、板 D を動かすことで筒 B と筒 C のどちらに入れるか決めることができます。 開口部 A から落とす玉の並びを与えます。それらを順番に筒 B 又は筒 Cに入れていきます。このとき、筒 B と筒 C のおのおのが両方とも番号の小さい玉の上に大きい玉を並べられる場合は YES、並べられない場合は NO と出力するプログラムを作成してください。ただし、容器の中で玉の順序を入れ替えることはできないものとします。また、続けて同じ筒に入れることができるものとし、筒 B, C ともに 10 個の玉がすべて入るだけの余裕があるものとします。
|
def solve(n,left,right):
if n==10:
return 1
if a[n]>left:
return solve(n+1,a[n],right)
if a[n]>right:
return solve(n+1,left,a[n])
return 0
n=int(input())
for _ in range(n):
a=[int(i) for i in input().split()]
print("YES" if solve(10,0,0) else "NO")
|
s088723729
|
Accepted
| 20
| 7,620
| 346
|
ans=0
def solve(n,left,right):
global ans
if n==10:
ans=1
else:
if a[n]>left:
ans=solve(n+1,a[n],right)
if a[n]>right:
ans=solve(n+1,left,a[n])
return ans
n=int(input())
for _ in range(n):
a=[int(i) for i in input().split()]
print("YES" if solve(0,0,0) else "NO")
ans=0
|
s195953904
|
p02402
|
u518939641
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,544
| 92
|
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
n=int(input())
l=list(map(int,input().split()))
print('%d %d %d'%(min(l), max(l), sum(l)/n))
|
s541521124
|
Accepted
| 60
| 8,580
| 90
|
n=int(input())
l=list(map(int,input().split()))
print('%d %d %d'%(min(l), max(l), sum(l)))
|
s631050520
|
p03129
|
u401093859
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 91
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
N,K = map(int,input().split())
if N <= 1+(2*(K-1)):
print("No")
else:
print("Yes")
|
s313808513
|
Accepted
| 17
| 2,940
| 91
|
N,K = map(int,input().split())
if N >= 1+(2*(K-1)):
print("YES")
else:
print("NO")
|
s944567362
|
p00050
|
u150984829
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,532
| 33
|
福島県は果物の産地としても有名で、その中でも特に桃とりんごは全国でも指折りの生産量を誇っています。ところで、ある販売用の英文パンフレットの印刷原稿を作ったところ、手違いでりんごに関する記述と桃に関する記述を逆に書いてしまいました。 あなたは、apple と peach を修正する仕事を任されましたが、なにぶん面倒です。１行の英文を入力して、そのなかの apple という文字列を全て peach に、peach という文字列を全てapple に交換した英文を出力するプログラムを作成してください。
|
input().replace('apple','peach')
|
s732179054
|
Accepted
| 20
| 5,544
| 80
|
a,p,t='apple','peach','_'
print(input().replace(a,t).replace(p,a).replace(t,p))
|
s293496170
|
p03434
|
u769411997
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 196
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
n = int(input())
a = list(map(int, input().split()))
ans = 0
a.sort(reverse = True)
print(a)
for i in range(len(a)):
if i%2 == 0:
ans += a[i]
else:
ans -= a[i]
print(ans)
|
s990848115
|
Accepted
| 17
| 3,060
| 187
|
n = int(input())
a = list(map(int, input().split()))
ans = 0
a.sort(reverse = True)
for i in range(len(a)):
if i%2 == 0:
ans += a[i]
else:
ans -= a[i]
print(ans)
|
s171483828
|
p03472
|
u499259667
| 2,000
| 262,144
|
Wrong Answer
| 354
| 11,316
| 375
|
You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?
|
import math
N,H = map(int,input().split())
alis=[];blis=[]
for c in range(N):
a,b = map(int,input().split())
alis.append(a);blis.append(b)
a = max(alis)
blis.sort();blis.reverse()
count = 0
num = 0
for i in blis:
if num >= H:
print(count)
exit()
if i < a:
break
num += i
count += 1
count += math.floor((H-num)/a)
print(count)
|
s263092313
|
Accepted
| 349
| 11,332
| 374
|
import math
N,H = map(int,input().split())
alis=[];blis=[]
for c in range(N):
a,b = map(int,input().split())
alis.append(a);blis.append(b)
a = max(alis)
blis.sort();blis.reverse()
count = 0
num = 0
for i in blis:
if num >= H:
print(count)
exit()
if i < a:
break
num += i
count += 1
count += math.ceil((H-num)/a)
print(count)
|
s972066294
|
p02928
|
u922952729
| 2,000
| 1,048,576
|
Wrong Answer
| 1,129
| 3,188
| 355
|
We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.
|
N,K=[int(i) for i in input().split(" ")]
A=[int(i) for i in input().split(" ")]
inside=0
for i in range(N):
for j in range(i,N):
if A[i]>A[j]:
inside+=1
outside=0
for i in range(N):
for j in range(N):
if A[i]>A[j]:
outside+=1
print(inside,outside)
s=inside*K+outside*((K*(K-1))//2)
print(s%(10**9+7))
|
s137060028
|
Accepted
| 1,039
| 3,188
| 332
|
N,K=[int(i) for i in input().split(" ")]
A=[int(i) for i in input().split(" ")]
inside=0
for i in range(N):
for j in range(i,N):
if A[i]>A[j]:
inside+=1
outside=0
for i in range(N):
for j in range(N):
if A[i]>A[j]:
outside+=1
s=inside*K+outside*((K*(K-1))//2)
print(s%(10**9+7))
|
s498726153
|
p03457
|
u282412844
| 2,000
| 262,144
|
Wrong Answer
| 346
| 17,324
| 594
|
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())
travels = []
for i in range(N):
t, x, y = map(int, input().split())
travels.append((t, x, y))
def is_possible(travels):
current = (0, 0, 0)
for t, x, y in travels:
prev_t, prev_x, prev_y = current
dt = t - prev_t
dx = x - prev_x
dy = y - prev_y
dxy = dx + dy
if dt < dxy:
return False
if dt == dxy:
continue
if (dt - dxy) % 2 != 0:
return False
current = (t, x, y)
return True
if is_possible(travels):
print('YES')
else:
print('NO')
|
s105396487
|
Accepted
| 350
| 17,320
| 594
|
N = int(input())
travels = []
for i in range(N):
t, x, y = map(int, input().split())
travels.append((t, x, y))
def is_possible(travels):
current = (0, 0, 0)
for t, x, y in travels:
prev_t, prev_x, prev_y = current
dt = t - prev_t
dx = x - prev_x
dy = y - prev_y
dxy = dx + dy
if dt < dxy:
return False
if dt == dxy:
continue
if (dt - dxy) % 2 != 0:
return False
current = (t, x, y)
return True
if is_possible(travels):
print('Yes')
else:
print('No')
|
s899051792
|
p03997
|
u823885866
| 2,000
| 262,144
|
Wrong Answer
| 114
| 27,192
| 468
|
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.
|
import sys
import math
import itertools
import collections
import heapq
import re
import numpy as np
from functools import reduce
rr = lambda: sys.stdin.readline().rstrip()
rs = lambda: sys.stdin.readline().split()
ri = lambda: int(sys.stdin.readline())
rm = lambda: map(int, sys.stdin.readline().split())
rf = lambda: map(float, sys.stdin.readline().split())
rl = lambda: list(map(int, sys.stdin.readline().split()))
inf = float('inf')
mod = 10**9 + 7
print('test')
|
s974563660
|
Accepted
| 28
| 8,956
| 50
|
print((int(input())+int(input()))*int(input())//2)
|
s704068804
|
p02273
|
u631142478
| 2,000
| 131,072
|
Wrong Answer
| 30
| 6,156
| 717
|
Write a program which reads an integer _n_ and draws a Koch curve based on recursive calles of depth _n_. The Koch curve is well known as a kind of You should start (0, 0), (100, 0) as the first segment.
|
import math
import collections
N = int(input())
Point = collections.namedtuple('Point', ['x', 'y'])
def koch(d, p1, p2):
if d == 0:
return
th = math.pi * 60 / 180
s = Point(x=(2 * p1.x + p2.x) / 3, y=(2 * p1.y + p2.y) / 3)
t = Point(x=(p1.x + 2 * p2.x) / 3, y=(p1.y + 2 * p2.y) / 3)
u = Point(x=(t.x - s.x) * math.cos(th) - (t.y - s.y) * math.sin(th) + s.x,
y=(t.x - s.x) * math.sin(th) + (t.y - s.y) * math.cos(th) - s.y)
koch(d - 1, p1, s)
print(s.x, s.y)
koch(d - 1, s, u)
print(u.x, u.y)
koch(d - 1, u, t)
print(t.x, t.y)
koch(d - 1, t, p2)
p1 = Point(x=0, y=0)
p2 = Point(x=100, y=0)
print(p1.x, p1.y)
koch(N, p1, p2)
print(p2.x, p2.y)
|
s295074722
|
Accepted
| 40
| 6,204
| 837
|
import math
import collections
N = int(input())
Point = collections.namedtuple('Point', ['x', 'y'])
def koch(d, p1, p2):
if d == 0:
return
th = math.pi * 60 / 180
s = Point(x=(2 * p1.x + p2.x) / 3, y=(2 * p1.y + p2.y) / 3)
t = Point(x=(p1.x + 2 * p2.x) / 3, y=(p1.y + 2 * p2.y) / 3)
u = Point(x=(t.x - s.x) * math.cos(th) - (t.y - s.y) * math.sin(th) + s.x,
y=(t.x - s.x) * math.sin(th) + (t.y - s.y) * math.cos(th) + s.y)
koch(d - 1, p1, s)
print('{:.8f} {:.8f}'.format(s.x, s.y))
koch(d - 1, s, u)
print('{:.8f} {:.8f}'.format(u.x, u.y))
koch(d - 1, u, t)
print('{:.8f} {:.8f}'.format(t.x, t.y))
koch(d - 1, t, p2)
p1 = Point(x=0, y=0)
p2 = Point(x=100, y=0)
print('{:.8f} {:.8f}'.format(p1.x, p1.y))
koch(N, p1, p2)
print('{:.8f} {:.8f}'.format(p2.x, p2.y))
|
s622882531
|
p03401
|
u713368238
| 2,000
| 262,144
|
Wrong Answer
| 240
| 14,048
| 290
|
There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.
|
n = int(input())
a = list((map(int, input().split())))
a = [0] + a + [0]
print(a)
totalcost = 0
for i in range(len(a) - 1):
totalcost += abs(a[i] - a[i+1])
for i in range(1, n+1):
icost = totalcost + abs(a[i-1] - a[i+1]) - abs(a[i-1] - a[i]) - abs(a[i] - a[i+1])
print(icost)
|
s448819583
|
Accepted
| 233
| 13,920
| 281
|
n = int(input())
a = list((map(int, input().split())))
a = [0] + a + [0]
totalcost = 0
for i in range(len(a) - 1):
totalcost += abs(a[i] - a[i+1])
for i in range(1, n+1):
icost = totalcost + abs(a[i-1] - a[i+1]) - abs(a[i-1] - a[i]) - abs(a[i] - a[i+1])
print(icost)
|
s667771853
|
p03548
|
u667949809
| 2,000
| 262,144
|
Wrong Answer
| 29
| 2,940
| 100
|
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 = map(int,input().split())
X=x-z
num = 0
while X>(y+z):
X=X-(y+z)
num=num+1
print(num)
|
s019827716
|
Accepted
| 30
| 2,940
| 101
|
x,y,z = map(int,input().split())
X=x-z
num = 0
while X>=(y+z):
X=X-(y+z)
num=num+1
print(num)
|
s085869300
|
p03698
|
u411237324
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 85
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s = list(input())
a = len(s)
b = len(set(s))
print('Yes') if a == b else print('No')
|
s868917256
|
Accepted
| 18
| 2,940
| 85
|
s = list(input())
a = len(s)
b = len(set(s))
print('yes') if a == b else print('no')
|
s442610788
|
p04029
|
u942028688
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 55
|
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())
result = (N + 1) * N / 2
print(result)
|
s872528970
|
Accepted
| 17
| 2,940
| 60
|
N = int(input())
result = int((N + 1) * N / 2)
print(result)
|
s666252154
|
p03997
|
u030726788
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 61
|
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)
|
s031886955
|
Accepted
| 18
| 2,940
| 62
|
a=int(input())
b=int(input())
h=int(input())
print((a+b)*h//2)
|
s972540108
|
p03574
|
u897328029
| 2,000
| 262,144
|
Wrong Answer
| 49
| 3,604
| 1,024
|
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
hw_list = list(map(int, input().split()))
h_max = hw_list[0]
w_max = hw_list[1]
s_list = []
for i in range(h_max):
work_list = input()
s_list.append(work_list)
answer_list = []
for h in range(h_max):
work_list = s_list[h]
work_answer = []
for w, target in enumerate(work_list):
if target == '.':
tonari_list = [(w-1, h-1), (w, h-1), (w+1, h-1),
(w-1, h), (w+1, h),
(w-1, h+1), (w, h+1), (w+1, h+1)]
count = 0
for x, y in tonari_list:
if 0 <= x < w_max and 0 <= y < h_max:
print('{}, {}'.format(x, y))
if s_list[y][x] == '#':
count += 1
else:
continue
work_answer.append(str(count))
else:
work_answer.append('#')
answer_list.append(work_answer)
for a_list in answer_list:
print(''.join(a_list))
|
s308052444
|
Accepted
| 27
| 3,188
| 975
|
hw_list = list(map(int, input().split()))
h_max = hw_list[0]
w_max = hw_list[1]
s_list = []
for i in range(h_max):
work_list = input()
s_list.append(work_list)
answer_list = []
for h in range(h_max):
work_list = s_list[h]
work_answer = []
for w, target in enumerate(work_list):
if target == '.':
tonari_list = [(w-1, h-1), (w, h-1), (w+1, h-1),
(w-1, h), (w+1, h),
(w-1, h+1), (w, h+1), (w+1, h+1)]
count = 0
for x, y in tonari_list:
if 0 <= x < w_max and 0 <= y < h_max:
if s_list[y][x] == '#':
count += 1
else:
continue
work_answer.append(str(count))
else:
work_answer.append('#')
answer_list.append(work_answer)
for a_list in answer_list:
print(''.join(a_list))
|
s364632212
|
p03448
|
u991087410
| 2,000
| 262,144
|
Wrong Answer
| 51
| 3,060
| 254
|
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.
|
C500 = int(input())
C100 = int(input())
C50 = int(input())
X = int(input())
count = 0
for i in range(C500):
for j in range(C100):
for k in range(C50):
TOTAL = 500*i+100*j+50*k
if TOTAL == X:
count += 1
break
print(count)
|
s084994473
|
Accepted
| 51
| 3,060
| 260
|
C500 = int(input())
C100 = int(input())
C50 = int(input())
X = int(input())
count = 0
for i in range(C500+1):
for j in range(C100+1):
for k in range(C50+1):
TOTAL = 500*i+100*j+50*k
if TOTAL == X:
count += 1
break
print(count)
|
s073955893
|
p03657
|
u505420467
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 112
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a,b=map(int,input().split())
if a%3==0 and b%3==0 or (a+b)%3==0:
print('possible')
else:
print('impossible')
|
s155740234
|
Accepted
| 18
| 2,940
| 94
|
a,b=map(int,input().split());print(["Impossible","Possible"][a%3==0 or b%3==0 or (a+b)%3==0])
|
s522574154
|
p03543
|
u765815947
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 85
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
n = input()
for i in range(10):
if 'iii' in n:
print('Yes')
else:
print('No')
|
s373227284
|
Accepted
| 18
| 2,940
| 117
|
n = input()
if (n[0] == n[1] and n[1] == n[2]) or (n[1] == n[2] and n[2] == n[3]):
print('Yes')
else:
print('No')
|
s608210711
|
p03524
|
u103902792
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,188
| 146
|
Snuke has a string S consisting of three kinds of letters: `a`, `b` and `c`. He has a phobia for palindromes, and wants to permute the characters in S so that S will not contain a palindrome of length 2 or more as a substring. Determine whether this is possible.
|
s = input()
a = s.count('a')
b = s.count('b')
c = s.count('c')
if abs(a-b)<=1 and abs(b-c)<=1 and abs(c-a)<=1:
print('Yes')
else:
print('No')
|
s449284869
|
Accepted
| 18
| 3,188
| 148
|
s = input()
a = s.count('a')
b = s.count('b')
c = s.count('c')
if abs(a-b)<=1 and abs(b-c)<=1 and abs(c-a)<=1:
print('YES')
else:
print('NO')
|
s581043946
|
p04043
|
u144029820
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 147
|
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=list(map(int,input().split()))
#a=[map(int,input().split())]
print(a)
if a.count(5)==2 and a.count(7)==1:
print('YES')
else:
print('NO')
|
s923068170
|
Accepted
| 17
| 2,940
| 148
|
a=list(map(int,input().split()))
#a=[map(int,input().split())]
#print(a)
if a.count(5)==2 and a.count(7)==1:
print('YES')
else:
print('NO')
|
s056261806
|
p03730
|
u740767776
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 162
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
a,b,c=map(int,input().split())
f = False
for i in range(1,b+1):
if a * i % b == c:
f = True
break
if f:
print("Yes")
else:
print("No")
|
s219845736
|
Accepted
| 17
| 2,940
| 162
|
a,b,c=map(int,input().split())
f = False
for i in range(1,b+1):
if a * i % b == c:
f = True
break
if f:
print("YES")
else:
print("NO")
|
s220954682
|
p03555
|
u357050210
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
s = input()
t = input()
if t[::-1] == s:
print("Yes")
else:
print("No")
|
s837348202
|
Accepted
| 17
| 2,940
| 80
|
s = input()
t = input()
if t[::-1] == s:
print("YES")
else:
print("NO")
|
s052182849
|
p02401
|
u298999032
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,660
| 211
|
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
a,op,b=input().split()
a,b=map(int,(a,b))
if op=='?':
None
elif op=='+':
print(str(int(a+b)))
elif op=='-':
print(str(int(a-b)))
elif op=='*':
print(str(int(a*b)))
else:
print(str(int(a//b)))
|
s819455379
|
Accepted
| 30
| 7,724
| 272
|
while True:
a,op,b=input().split()
a,b=map(int,(a,b))
if op=='?':
break
elif op=='+':
print(str(int(a+b)))
elif op=='-':
print(str(int(a-b)))
elif op=='*':
print(str(int(a*b)))
else:
print(str(int(a//b)))
|
s422542042
|
p03433
|
u729217226
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 93
|
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')
|
s017908192
|
Accepted
| 17
| 2,940
| 92
|
N = int(input())
A = int(input())
if (N % 500) <= A:
print('Yes')
else:
print('No')
|
s978841033
|
p02612
|
u018846452
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,152
| 72
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
if N % 1000 == 0:
print(0)
else:
print(N//1000 +1 )
|
s203722255
|
Accepted
| 28
| 9,140
| 48
|
N = int(input())
print((N+1000-1)//1000* 1000-N)
|
s566310844
|
p03693
|
u910756197
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
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?
|
n = int("".join(input().split()))
if n % 4 == 0:
print("Yes")
else:
print("No")
|
s243479393
|
Accepted
| 17
| 2,940
| 88
|
n = int("".join(input().split()))
if n % 4 == 0:
print("YES")
else:
print("NO")
|
s065955930
|
p03565
|
u413659369
| 2,000
| 262,144
|
Wrong Answer
| 36
| 9,232
| 466
|
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
from copy import deepcopy
S = list(input())
T = list(input())
s_num = len(S)
t_num = len(T)
S_copy = deepcopy(S)
for i in range(s_num-t_num, -1, -1):
s_key = S[i:i+t_num]
judge = 0
for s, t in zip(s_key, T):
if s=='?':
judge += 1
elif s == t:
judge += 1
if judge == t_num:
S_copy[i:i+t_num] = T
if S == S_copy:
print('UNRESTORABLE')
else:
for i in S_copy:
if i == '?':
print('a', end='')
else:
print(i, end='')
|
s285376688
|
Accepted
| 31
| 9,272
| 498
|
from copy import deepcopy
S = input()
T = input()
s_num = len(S)
t_num = len(T)
ans = []
if s_num < t_num:
print('UNRESTORABLE')
else:
for i in range(s_num-t_num+1):
s_key = S[i:i+t_num]
judge = 0
for s, t in zip(s_key, T):
if s=='?':
judge += 1
elif s == t:
judge += 1
if judge == t_num:
tmp = S[:i] + T + S[i+t_num:]
tmp = tmp.replace('?', 'a')
ans.append(tmp)
if ans:
print(sorted(ans)[0])
else:
print('UNRESTORABLE')
|
s254234366
|
p03378
|
u840592162
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 161
|
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
n,m,x = map(int,input().split())
A = list(map(int,input().split()))
ans1 = len([a for a in A if a <x])
ans2 = len([a for a in A if a < x])
print(min(ans1,ans2))
|
s611009656
|
Accepted
| 19
| 3,060
| 162
|
n,m,x = map(int,input().split())
A = list(map(int,input().split()))
ans1 = len([a for a in A if a < x])
ans2 = len([a for a in A if a > x])
print(min(ans1,ans2))
|
s357895647
|
p02694
|
u375006224
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 9,156
| 63
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
a=int(input())
b=100
c=0
while b>=a:
c+=1
b=b*1.01
print(c)
|
s881410816
|
Accepted
| 25
| 9,160
| 67
|
a=int(input())
b=100
c=0
while b<a:
c+=1
b=int(b*1.01)
print(c)
|
s893784344
|
p03970
|
u960080897
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,092
| 117
|
CODE FESTIVAL 2016 is going to be held. For the occasion, Mr. Takahashi decided to make a signboard. He intended to write `CODEFESTIVAL2016` on it, but he mistakenly wrote a different string S. Fortunately, the string he wrote was the correct length. So Mr. Takahashi decided to perform an operation that replaces a certain character with another in the minimum number of iterations, changing the string to `CODEFESTIVAL2016`. Find the minimum number of iterations for the rewrite operation.
|
s = input()
t = "CODEFESTIVAL2016"
cnt = 0
for i, z in zip(s, t):
if s != z:
cnt += 1
print(cnt)
|
s357186244
|
Accepted
| 26
| 9,092
| 117
|
s = input()
t = "CODEFESTIVAL2016"
cnt = 0
for i, z in zip(s, t):
if i != z:
cnt += 1
print(cnt)
|
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