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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s483711067
|
p03006
|
u619458041
| 2,000 | 1,048,576 |
Wrong Answer
| 157 | 4,084 | 716 |
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i). We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation: * Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1. Find the minimum total cost required to collect all the balls when we optimally choose p and q.
|
import sys
from itertools import combinations
def main():
input = sys.stdin.readline
N = int(input())
cord = []
for _ in range(N):
x, y = map(int, input().split())
cord.append((x, y))
ans = 50
for a in combinations(cord, 2):
x1, y1 = a[0]
x2, y2 = a[1]
if x1 == x2:
slope = 0
else:
slope = (y2 - y1) / (x2 - x1)
intercepts = set()
for x, y in cord:
intercept = y - slope * x
intercept = round(intercept, 5)
intercepts.add(intercept)
print(intercepts)
ans = min(ans, len(intercepts))
return ans
if __name__ == '__main__':
print(main())
|
s047874862
|
Accepted
| 56 | 3,064 | 706 |
import sys
from itertools import combinations
def main():
input = sys.stdin.readline
N = int(input())
cord = set()
for _ in range(N):
x, y = map(int, input().split())
cord.add((x, y))
if N <= 2:
return 1
ans = 50
for a in combinations(cord, 2):
x1, y1 = a[0]
x2, y2 = a[1]
p = x2 - x1
q = y2 - y1
cnt1 = 0
cnt2 = 0
if p == q == 0:
continue
A = set()
B = set()
for x, y in cord:
A.add((x-p, y-q))
B.add((x+p, y+q))
ans = min(ans, len(A - cord), len(B - cord))
return ans
if __name__ == '__main__':
print(main())
|
s083290075
|
p04029
|
u023077142
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 731 |
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?
|
from itertools import combinations, product
def gcd(a, b):
if a < b:
a, b = b, a
while b != 0:
a, b = b, a % b
return a
def lcm(a, b):
return a * b // gcd(a, b)
def get(fmt):
ps = input().split()
r = []
for p, t in zip(ps, fmt):
if t == "i":
r.append(int(p))
if t == "f":
r.append(float(p))
if t == "s":
r.append(p)
if len(r) == 1:
r = r[0]
return r
def put(*args, **kwargs):
print(*args, **kwargs)
def rep(n, f, *args, **kwargs):
return [f(*args, **kwargs) for _ in range(n)]
def rep_im(n, v):
return rep(n, lambda: v)
YES_NO = ["NO", "YES"]
N = get("i")
put(N * (N - 1) // 2)
|
s198664630
|
Accepted
| 18 | 3,064 | 731 |
from itertools import combinations, product
def gcd(a, b):
if a < b:
a, b = b, a
while b != 0:
a, b = b, a % b
return a
def lcm(a, b):
return a * b // gcd(a, b)
def get(fmt):
ps = input().split()
r = []
for p, t in zip(ps, fmt):
if t == "i":
r.append(int(p))
if t == "f":
r.append(float(p))
if t == "s":
r.append(p)
if len(r) == 1:
r = r[0]
return r
def put(*args, **kwargs):
print(*args, **kwargs)
def rep(n, f, *args, **kwargs):
return [f(*args, **kwargs) for _ in range(n)]
def rep_im(n, v):
return rep(n, lambda: v)
YES_NO = ["NO", "YES"]
N = get("i")
put(N * (N + 1) // 2)
|
s649676757
|
p03623
|
u894694822
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 87 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x, a, b=map(int, input().split())
if abs(x-a)>abs(x-b):
print("A")
else:
print("B")
|
s392923502
|
Accepted
| 17 | 2,940 | 87 |
x, a, b=map(int, input().split())
if abs(x-a)<abs(x-b):
print("A")
else:
print("B")
|
s030212791
|
p03434
|
u824237520
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 140 |
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
n = int(input())
a = list(map(int, input().split()))
a = sorted(a)
ans = 0
flag = 1
for x in a:
ans += x * flag
flag *= -1
print(ans)
|
s840082611
|
Accepted
| 17 | 2,940 | 157 |
n = int(input())
a = list(map(int, input().split()))
a = list(reversed(sorted(a)))
ans = 0
flag = 1
for x in a:
ans += x * flag
flag *= -1
print(ans)
|
s413625983
|
p02600
|
u882564128
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,216 | 305 |
M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * From 1000 through 1199: 5-kyu * From 1200 through 1399: 4-kyu * From 1400 through 1599: 3-kyu * From 1600 through 1799: 2-kyu * From 1800 through 1999: 1-kyu What kyu does M-kun have?
|
x = int(input())
if x >= 400 and x <= 599:
s = 8
elif x >= 600 and x <= 799:
s = 7
elif x >= 800 and x <= 999:
s = 6
elif x >= 1000 and x <= 1199:
s = 5
elif x >= 1200 and x <= 1399:
s = 4
elif x >= 1400 and x <= 1599:
s = 3
elif x >= 1600 and x <= 1799:
s = 2
else:
s = 1
|
s266490319
|
Accepted
| 28 | 9,176 | 315 |
x = int(input())
if x >= 400 and x <= 599:
s = 8
elif x >= 600 and x <= 799:
s = 7
elif x >= 800 and x <= 999:
s = 6
elif x >= 1000 and x <= 1199:
s = 5
elif x >= 1200 and x <= 1399:
s = 4
elif x >= 1400 and x <= 1599:
s = 3
elif x >= 1600 and x <= 1799:
s = 2
else:
s = 1
print(s)
|
s925843199
|
p03151
|
u349091349
| 2,000 | 1,048,576 |
Wrong Answer
| 280 | 25,956 | 526 |
A university student, Takahashi, has to take N examinations and pass all of them. Currently, his _readiness_ for the i-th examination is A_{i}, and according to his investigation, it is known that he needs readiness of at least B_{i} in order to pass the i-th examination. Takahashi thinks that he may not be able to pass all the examinations, and he has decided to ask a magician, Aoki, to change the readiness for as few examinations as possible so that he can pass all of them, while not changing the total readiness. For Takahashi, find the minimum possible number of indices i such that A_i and C_i are different, for a sequence C_1, C_2, ..., C_{N} that satisfies the following conditions: * The sum of the sequence A_1, A_2, ..., A_{N} and the sum of the sequence C_1, C_2, ..., C_{N} are equal. * For every i, B_i \leq C_i holds. If such a sequence C_1, C_2, ..., C_{N} cannot be constructed, print -1.
|
import numpy as np
from collections import deque
import sys
N=int(input())
A=np.array([int(i) for i in input().split()])
B=np.array([int(i) for i in input().split()])
C = A-B
minus_idx=np.where(C<0)
plus_idx=np.where(C>0)
minus = np.sum(C[minus_idx])
plus = deque(np.sort(C[plus_idx]))
len_plus = len(plus)
if len_plus == 0:
print(-1)
sys.exit()
count = 0
while minus<0:
minus += plus.pop()
count += 1
len_plus -= 1
if len_plus == 0:
print(-1)
sys.exit()
print(len(minus_idx[0])+count)
|
s804659463
|
Accepted
| 286 | 23,976 | 594 |
import numpy as np
from collections import deque
import sys
N=int(input())
A=np.array([int(i) for i in input().split()])
B=np.array([int(i) for i in input().split()])
C = A-B
minus_idx=np.where(C<0)
plus_idx=np.where(C>=0)
minus = np.sum(C[minus_idx])
plus = deque(np.sort(C[plus_idx]))
len_plus = len(plus)
if len(minus_idx[0]) == 0:
print(0)
sys.exit()
if len_plus == 0:
print(-1)
sys.exit()
count = 0
while minus<0:
try:
minus += plus.pop()
count += 1
len_plus -= 1
except:
print(-1)
sys.exit()
print(len(minus_idx[0])+count)
|
s018237166
|
p03738
|
u035445296
| 2,000 | 262,144 |
Wrong Answer
| 26 | 8,908 | 104 |
You are given two positive integers A and B. Compare the magnitudes of these numbers.
|
a = int(input())
b = int(input())
ans = 'GREATER'
if a == b:
ans = 'EQUAL'
elif a < b:
ans = 'LESS'
|
s843697358
|
Accepted
| 30 | 9,060 | 114 |
a = int(input())
b = int(input())
ans = 'GREATER'
if a == b:
ans = 'EQUAL'
elif a < b:
ans = 'LESS'
print(ans)
|
s670756918
|
p03478
|
u531599639
| 2,000 | 262,144 |
Wrong Answer
| 57 | 3,060 | 210 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
N, A, B = map(int, input().split())
sums=0
for i in range(1, N+1):
m=0
sum=0
for j in range(len(str(i))-1, -1, -1):
m=i//(10**j)
i-=10**j*m
sum+=m
if A<=sum<=B:
sums+=sum
print(sums)
|
s116657684
|
Accepted
| 31 | 2,940 | 124 |
N, A, B = map(int, input().split())
sums=0
for i in range(1, N+1):
if A<=sum(map(int, str(i)))<=B:
sums+=i
print(sums)
|
s592843349
|
p00728
|
u661290476
| 1,000 | 131,072 |
Wrong Answer
| 30 | 5,604 | 350 |
The International Clown and Pierrot Competition (ICPC), is one of the most distinguished and also the most popular events on earth in the show business. One of the unique features of this contest is the great number of judges that sometimes counts up to one hundred. The number of judges may differ from one contestant to another, because judges with any relationship whatsoever with a specific contestant are temporarily excluded for scoring his/her performance. Basically, scores given to a contestant's performance by the judges are averaged to decide his/her score. To avoid letting judges with eccentric viewpoints too much influence the score, the highest and the lowest scores are set aside in this calculation. If the same highest score is marked by two or more judges, only one of them is ignored. The same is with the lowest score. The average, which may contain fractions, are truncated down to obtain final score as an integer. You are asked to write a program that computes the scores of performances, given the scores of all the judges, to speed up the event to be suited for a TV program.
|
while True:
try:
n = int(input())
MIN = 1000
MAX = 0
points = 0
for i in range(n):
p = int(input())
if MIN > p:
MIN = p
if MAX < p:
MAX = p
points += p
print((points - MIN - MAX) // (n - 2))
except:
break
|
s695727577
|
Accepted
| 30 | 5,592 | 296 |
while True:
n = int(input())
if n == 0:
break
MIN = 1000
MAX = 0
points = 0
for i in range(n):
p = int(input())
if MIN > p:
MIN = p
if MAX < p:
MAX = p
points += p
print((points - MIN - MAX) // (n - 2))
|
s214641800
|
p03353
|
u077337864
| 2,000 | 1,048,576 |
Wrong Answer
| 2,159 | 928,152 | 421 |
You are given a string s. Among the **different** substrings of s, print the K-th lexicographically smallest one. A substring of s is a string obtained by taking out a non-empty contiguous part in s. For example, if s = `ababc`, `a`, `bab` and `ababc` are substrings of s, while `ac`, `z` and an empty string are not. Also, we say that substrings are different when they are different as strings. Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
|
s = input().strip()
k = int(input())
subs = set()
ds = {}
for i, c in enumerate(s):
if c in ds:
ds[c].append(i)
else:
ds[c] = [i]
for c, il in sorted(ds.items(), key=lambda x: x[0]):
print(c, il)
subs |= set([c])
i = 2
while il[0]+i <= len(s):
for j in il:
if j + i > len(s):
break
subs.add(s[j:j+i])
i += 1
if len(subs) >= k:
break
print(list(sorted(subs))[k-1])
|
s518277484
|
Accepted
| 41 | 4,588 | 163 |
s = input().strip()
k = int(input())
subs = set()
for i in range(1, 6):
for j in range(0, len(s)-i+1):
subs |= set([s[j:j+i]])
print(list(sorted(subs))[k-1])
|
s867917065
|
p03433
|
u775652851
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,024 | 76 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
a=int(input())
b=int(input())
if a%500 <b:
print("No")
else:
print("No")
|
s293866996
|
Accepted
| 25 | 9,016 | 80 |
a=int(input())
b=int(input())
if a%500 <= b:
print("Yes")
else:
print("No")
|
s940056871
|
p03435
|
u401487574
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 289 |
We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j. Determine if he is correct.
|
nums = [list(map(int,input().split())) for i in range(3)]
a1 = 0
b1,b2,b3 = map(int,nums[0])
a2 = nums[1][0] -b2
a3 = nums[2][0] -b3
print(a2,a3)
print(nums)
if nums[1][1] == a2+b2 and nums[1][2] == a2+b3 and nums[2][1] == a3+b2 and nums[2][2] == a3+b3:
print("Yes")
else:print("No")
|
s781224742
|
Accepted
| 17 | 3,060 | 264 |
nums = [list(map(int,input().split())) for i in range(3)]
a1 = 0
b1,b2,b3 = map(int,nums[0])
a2 = nums[1][0] -b1
a3 = nums[2][0] -b1
if nums[1][1] == a2+b2 and nums[1][2] == a2+b3 and nums[2][1] == a3+b2 and nums[2][2] == a3+b3:
print("Yes")
else:print("No")
|
s620749289
|
p04043
|
u627691992
| 2,000 | 262,144 |
Wrong Answer
| 27 | 8,988 | 80 |
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.
|
S = input()
print('Yes' if(S.count('5') == 2 and S.count('7') == 1) else 'No')
|
s508637007
|
Accepted
| 24 | 8,872 | 80 |
S = input()
print('YES' if(S.count('5') == 2 and S.count('7') == 1) else 'NO')
|
s140242376
|
p03971
|
u638795007
| 2,000 | 262,144 |
Wrong Answer
| 112 | 4,796 | 1,288 |
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.
|
def examA():
S = SI(); N = len(S)
T = "CODEFESTIVAL2016"
ans = 0
for i in range(N):
if S[i]!=T[i]:
ans +=1
print(ans)
return
def examB():
N, A, B = LI()
S = SI()
d = defaultdict(int)
for s in S:
if s=="a":
if d["a"]+d["b"]<A+B:
print("YES")
d[s] += 1
else:
print("NO")
elif s=="b":
if d["a"]+d["b"]<A+B and d["b"]<B:
print("YES")
d[s] += 1
else:
print("NO")
else:
print("NO")
return
def examC():
return
import sys,copy,bisect,itertools,heapq,math
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,ALPHABET, _ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 1e-9
alphabet = [chr(ord('a') + i) for i in range(26)]
ALPHABET = [chr(ord('A') + i) for i in range(26)]
if __name__ == '__main__':
examB()
|
s460524377
|
Accepted
| 114 | 4,564 | 1,289 |
def examA():
S = SI(); N = len(S)
T = "CODEFESTIVAL2016"
ans = 0
for i in range(N):
if S[i]!=T[i]:
ans +=1
print(ans)
return
def examB():
N, A, B = LI()
S = SI()
d = defaultdict(int)
for s in S:
if s=="a":
if d["a"]+d["b"]<A+B:
print("Yes")
d[s] += 1
else:
print("No")
elif s=="b":
if d["a"]+d["b"]<A+B and d["b"]<B:
print("Yes")
d[s] += 1
else:
print("No")
else:
print("No")
return
def examC():
return
import sys,copy,bisect,itertools,heapq,math
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,ALPHABET, _ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 1e-9
alphabet = [chr(ord('a') + i) for i in range(26)]
ALPHABET = [chr(ord('A') + i) for i in range(26)]
if __name__ == '__main__':
examB()
|
s726320556
|
p03643
|
u642015143
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 21 |
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
print("abc"+input())
|
s285951372
|
Accepted
| 17 | 2,940 | 21 |
print("ABC"+input())
|
s057923321
|
p03485
|
u847165882
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 50 |
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())
print(-(-((a+b)/2)))
|
s826024179
|
Accepted
| 17 | 2,940 | 49 |
a,b=map(int,input().split())
print(-(-(a+b)//2))
|
s990267422
|
p04030
|
u411858517
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 114 |
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
s = input()
res = ''
for i in range(len(s)):
res += s[i]
if s[i] == 'B':
res = res[:i-2]
print(res)
|
s852477224
|
Accepted
| 18 | 2,940 | 258 |
s = input()
res = ''
for i in range(len(s)):
if res == '':
if s[i] == 'B':
pass
else:
res += s[i]
else:
if s[i] == 'B':
res = res[:-1]
else:
res += s[i]
print(res)
|
s758913098
|
p02614
|
u127643144
| 1,000 | 1,048,576 |
Wrong Answer
| 29 | 9,176 | 298 |
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
nCr = {}
def cmb(n, r):
if r == 0 or r == n: return 1
if r == 1: return n
if (n,r) in nCr: return nCr[(n,r)]
nCr[(n,r)] = cmb(n-1,r) + cmb(n-1,r-1)
return nCr[(n,r)]
h, w, k = map(int, input().split())
c = [list(input()) for _ in range(h)]
ans = 0
a = cmb(h*w,k)
print(a)
|
s729160091
|
Accepted
| 41 | 9,360 | 602 |
import copy
from itertools import combinations
h, w, k = map(int, input().split())
c =[input() for i in range(h)]
ans = 0
y = []
for i in range(h):
y.append(i)
x = []
for i in range(w):
x.append(i)
for i in range(h+1):
ty = list(combinations(y, i))
for j in range(w+1):
tx = list(combinations(x, j))
for l in ty:
for m in tx:
cnt = 0
for p in l:
for q in m:
if c[p][q] == '#':
cnt += 1
if cnt == k:
ans += 1
print(ans)
|
s254921087
|
p02694
|
u261082314
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,048 | 115 |
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?
|
import math
k=int(input())
#r = math.floor(t/100)
t = 100
n = 0
while t<= k:
t+=math.floor(t/100)
n+=1
print(n)
|
s484247724
|
Accepted
| 23 | 9,156 | 114 |
import math
k=int(input())
#r = math.floor(t/100)
t = 100
n = 0
while t< k:
t+=math.floor(t/100)
n+=1
print(n)
|
s601880118
|
p03160
|
u479353489
| 2,000 | 1,048,576 |
Wrong Answer
| 125 | 14,716 | 374 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
a = [int(x) for x in input().split()]
def dp(N,a):
F = [0]*(len(a)+1)
F[0]=0
F[1]=a[0]
F[2]=abs(a[1]-a[0])
if N == 1: return F[1]
elif N == 2: return F[2]
#F[3] = min(F[1]+abs(a[2]))
for n in range(3,N+1):
F[n] = min(F[n-1] + abs(a[n-1]-a[n-2]),F[n-2]+abs(a[n-1]-a[n-3]))
print(F)
return F[-1]
print(dp(N,a))
|
s791336311
|
Accepted
| 114 | 13,980 | 307 |
N = int(input())
a = [int(x) for x in input().split()]
def dp(N,a):
F = [0]*(len(a))
F[0]=0
F[1]=abs(a[1]-a[0])
if N == 1: return F[0]
elif N == 2: return F[1]
for n in range(2,N):
F[n] = min(F[n-1] + abs(a[n]-a[n-1]),F[n-2]+abs(a[n]-a[n-2]))
return F[-1]
print(dp(N,a))
|
s300608770
|
p03024
|
u219503971
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 103 |
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
s=input()
ct=0
for i in s:
if i=='o':
ct+=1
if 15-len(s)+ct>8:
print('YES')
else:
print('NO')
|
s361765179
|
Accepted
| 17 | 2,940 | 105 |
s=input()
ct=0
for i in s:
if i=='o':
ct+=1
if 15-len(s)+ct>=8:
print('YES')
else:
print('NO')
|
s907527760
|
p03457
|
u329006798
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 108,284 | 608 |
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.
|
def test():
n = int(input())
t = [0 for _ in range(n+1)]
x = [0 for _ in range(n+1)]
y = [0 for _ in range(n+1)]
for i in range(n):
t[i+1], x[i+1], y[i+1] = [int(z) for z in input().split()]
print(t,x,y)
dt = [t[i+1] - t[i] for i in range(n)]
dist = [abs(x[i+1] - x[i]) + abs(y[i+1] - y[i]) for i in range(n)]
print(dt, dist)
result = "Yes"
for i in range(len(t)):
if not t[i] % 2 == (abs(x[i])+abs(y[i])) % 2:
result = "No"
for i in range(len(dt)):
if dt < dist:
result = "No"
print(result)
test()
|
s635151909
|
Accepted
| 471 | 11,824 | 350 |
n = int(input())
t = [0 for _ in range(n+1)]
x = [0 for _ in range(n+1)]
y = [0 for _ in range(n+1)]
for i in range(n):
t[i+1], x[i+1], y[i+1] = [int(z) for z in input().split()]
if t[i+1] - t[i] < abs(x[i + 1] - x[i]) + abs(y[i + 1] - y[i]) or (t[i+1] % 2 != (abs(x[i+1]) + abs(y[i+1])) % 2):
print('No')
exit()
print('Yes')
|
s902256785
|
p03311
|
u089230684
| 2,000 | 1,048,576 |
Wrong Answer
| 306 | 39,024 | 393 |
Snuke has an integer sequence A of length N. He will freely choose an integer b. Here, he will get sad if A_i and b+i are far from each other. More specifically, the _sadness_ of Snuke is calculated as follows: * abs(A_1 - (b+1)) + abs(A_2 - (b+2)) + ... + abs(A_N - (b+N)) Here, abs(x) is a function that returns the absolute value of x. Find the minimum possible sadness of Snuke.
|
n = input()
arr = [int(i) for i in input().split(" ")]
sum = {}
for i, v in enumerate(arr):
if sum.get(abs(v - (i + 1)), False):
sum[abs(v - (i + 1))] += 1
else:
sum[abs(v - (i + 1))] = 1
max_b = 0
max_times = 0
for key, value in sum.items():
if value > max_times:
value = max_times
max_b = key
sum = 0
for i, v in enumerate(arr):
sum += abs(v - (-max_b + (i + 1)))
print(sum)
|
s989043736
|
Accepted
| 226 | 26,128 | 259 |
while True:
try:
n = int(input())
num = list(map(int, input().split()))
for ind in range(len(num)):
num[ind] -= ind+1
num.sort()
print(sum(abs(i-num[len(num)//2])for i in num))
except:
break
|
s617303393
|
p03485
|
u863442865
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,316 | 85 |
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:
print((a+b+1)/2)
else:
print((a+b)/2)
|
s423350721
|
Accepted
| 17 | 2,940 | 96 |
a,b = map(int, input().split())
if (a+b)%2:
print(int((a+b+1)/2))
else:
print(int((a+b)/2))
|
s658275275
|
p04029
|
u677991935
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 31 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N=int(input())
print(N*(N+1)/2)
|
s257047999
|
Accepted
| 17 | 2,940 | 36 |
N=int(input())
print(int(N*(N+1)/2))
|
s956326822
|
p03067
|
u148551245
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 93 |
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 = map(int, input().split())
if b < a or b > c:
print('No')
else:
print('Yes')
|
s272836223
|
Accepted
| 17 | 2,940 | 101 |
a, b, c = map(int, input().split())
if a < c < b or b < c < a:
print('Yes')
else:
print('No')
|
s574414848
|
p02865
|
u958693198
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 43 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
int = int(input())
print(int*(int-1)/2+int)
|
s161295900
|
Accepted
| 17 | 2,940 | 32 |
n = int(input())
print((n-1)//2)
|
s227083976
|
p00003
|
u386731818
| 1,000 | 131,072 |
Wrong Answer
| 40 | 8,760 | 502 |
Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so.
|
import string
import sys
import math
#??????????????\????????????ip?????\??????
ip = sys.stdin.readlines()
ip_list = {}
for i in range(len(ip)):
ip_list[i] = ip[i].strip("\n").split()
for i in range(len(ip)):
if len(ip_list[i]) == 1:
print("YES")
else:
if int(ip_list[i][0]) != int(ip_list[i][1]):
print("NO")
elif int(ip_list[i][0]) != int(ip_list[i][2]):
print("NO")
else:
print("YES")
|
s212667647
|
Accepted
| 100 | 8,692 | 732 |
import string
import sys
import math
#??????????????\????????????ip?????\??????
ip = sys.stdin.readlines()
ip_list = {}
for i in range(len(ip)):
ip_list[i] = ip[i].strip("\n").split()
for i in range(1,len(ip)):
for j in range(3):
#?????????????????????
for t in range(3):
for k in range(3):
if int(ip_list[i][t]) > int(ip_list[i][k]):
tmp = ip_list[i][t]
ip_list[i][t] = ip_list[i][k]
ip_list[i][k] = tmp
for i in range(1,len(ip)):
if int(ip_list[i][0])**2 == int(ip_list[i][1])**2 + int(ip_list[i][2])**2:
print("YES")
else:
print("NO")
|
s417868271
|
p00024
|
u299798926
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 105 |
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 sys
for s in sys.stdin:
n=float(s[:-1])
t=n/9.8
y=4.9*t**2
N=(y+5)//5
print(N)
|
s197170170
|
Accepted
| 20 | 5,640 | 171 |
import math
while(1):
try:
n=float(input())
t=n/9.8
y=4.9*t**2
N=(y+5)/5
print(math.ceil(N))
except EOFError:
break
|
s830927808
|
p02400
|
u023471147
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,632 | 88 |
Write a program which calculates the area and circumference of a circle for given radius r.
|
from math import pi
r = float(input())
print("{0:.10} {1:.10}".format(pi*r**2, 2*pi*r))
|
s270581277
|
Accepted
| 20 | 5,632 | 88 |
from math import pi
r = float(input())
print("{0:.6f} {1:.6f}".format(pi*r**2, 2*pi*r))
|
s014575304
|
p03162
|
u163501259
| 2,000 | 1,048,576 |
Wrong Answer
| 603 | 31,348 | 492 |
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
import sys
input = sys.stdin.readline
def main():
N = int(input())
HAPPY = [list(map(int, input().split())) for i in range(N)]
dp = [[0]*3]*(N+1)
for i in range(N):
for j in range(3):
for k in range(3):
if j == k:
continue
dp[i+1][k] = max(dp[i+1][k], HAPPY[i][k] + dp[i][j])
print(max(max(dp[N][0], dp[N][1]), dp[N][2]))
if __name__ == '__main__':
main()
|
s041548198
|
Accepted
| 673 | 47,332 | 466 |
import sys
input = sys.stdin.readline
def main():
N = int(input())
HAPPY = [list(map(int, input().split())) for i in range(N)]
dp = [[0]*3 for i in range(N+1)]
for i in range(N):
for j in range(3):
for k in range(3):
if j == k:
continue
dp[i+1][k] = max(dp[i+1][k], HAPPY[i][k] + dp[i][j])
print(max(max(dp[N][0], dp[N][1]), dp[N][2]))
if __name__ == '__main__':
main()
|
s198835065
|
p03657
|
u781025961
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 126 |
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
A,B = map(int,input().split())
if A % 3 ==0 or B % 3 ==0 or (A+B) % 3 ==0:
print("Possibe")
else:
print("Impossible")
|
s161697041
|
Accepted
| 21 | 3,316 | 127 |
A,B = map(int,input().split())
if A % 3 ==0 or B % 3 ==0 or (A+B) % 3 ==0:
print("Possible")
else:
print("Impossible")
|
s661509607
|
p03251
|
u884323674
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 182 |
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
N, M, X, Y = map(int, input().split())
X = [int(i) for i in input().split()]
Y = [int(i) for i in input().split()]
if min(Y) - max(X) > 1:
print("No War")
else:
print("War")
|
s415629719
|
Accepted
| 17 | 3,060 | 300 |
N, M, X, Y = map(int, input().split())
X_list = [int(i) for i in input().split()]
Y_list = [int(i) for i in input().split()]
found = False
for Z in range(X+1, Y+1):
if max(X_list) < Z and min(Y_list) >= Z:
found = True
print("No War")
break
if not found:
print("War")
|
s102493863
|
p02384
|
u244742296
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,764 | 1,282 |
Construct a dice from a given sequence of integers in the same way as
|
class Dice(object):
def __init__(self, s1, s2, s3, s4, s5, s6):
self.s1 = s1
self.s2 = s2
self.s3 = s3
self.s4 = s4
self.s5 = s5
self.s6 = s6
def east(self):
prev_s6 = self.s6
self.s6 = self.s3
self.s3 = self.s1
self.s1 = self.s4
self.s4 = prev_s6
def west(self):
prev_s6 = self.s6
self.s6 = self.s4
self.s4 = self.s1
self.s1 = self.s3
self.s3 = prev_s6
def north(self):
prev_s6 = self.s6
self.s6 = self.s5
self.s5 = self.s1
self.s1 = self.s2
self.s2 = prev_s6
def south(self):
prev_s6 = self.s6
self.s6 = self.s2
self.s2 = self.s1
self.s1 = self.s5
self.s5 = prev_s6
def top(self):
return self.s1
def front(self):
return self.s2
def right(self):
return self.s3
s1, s2, s3, s4, s5, s6 = map(int, input().split())
dice = Dice(s1, s2, s3, s4, s5, s6)
q = int(input())
for i in range(q):
t, f = map(int, input().split())
for j in range(4):
for k in range(4):
if dice.top() == t and dice.front() == f:
break
dice.east()
dice.north()
print(dice.right())
|
s453750045
|
Accepted
| 40 | 6,788 | 1,581 |
class Dice(object):
def __init__(self, s1, s2, s3, s4, s5, s6):
self.s1 = s1
self.s2 = s2
self.s3 = s3
self.s4 = s4
self.s5 = s5
self.s6 = s6
def east(self):
prev_s6 = self.s6
self.s6 = self.s3
self.s3 = self.s1
self.s1 = self.s4
self.s4 = prev_s6
def west(self):
prev_s6 = self.s6
self.s6 = self.s4
self.s4 = self.s1
self.s1 = self.s3
self.s3 = prev_s6
def north(self):
prev_s6 = self.s6
self.s6 = self.s5
self.s5 = self.s1
self.s1 = self.s2
self.s2 = prev_s6
def south(self):
prev_s6 = self.s6
self.s6 = self.s2
self.s2 = self.s1
self.s1 = self.s5
self.s5 = prev_s6
def rotate(self):
prev_s2 = self.s2
self.s2 = self.s4
self.s4 = self.s5
self.s5 = self.s3
self.s3 = prev_s2
def top(self):
return self.s1
def front(self):
return self.s2
def right(self):
return self.s3
s1, s2, s3, s4, s5, s6 = map(int, input().split())
dice = Dice(s1, s2, s3, s4, s5, s6)
q = int(input())
for i in range(q):
t, f = map(int, input().split())
flag = False
for j in range(6):
if j % 2 == 0:
dice.north()
else:
dice.west()
for k in range(4):
dice.rotate()
if dice.top() == t and dice.front() == f:
flag = True
break
if flag:
break
print(dice.right())
|
s062270219
|
p02257
|
u792145349
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,764 | 257 |
A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list.
|
N = int(input())
numbers = [int(input()) for i in range(N)]
ans = []
def check_prime(n):
for i in range(2, N):
if n % i == 0:
return 0
else:
return 1
for n in numbers:
ans.append(check_prime(n))
print(sum(ans))
|
s738164016
|
Accepted
| 570 | 8,472 | 269 |
N = int(input())
numbers = [int(input()) for i in range(N)]
ans = []
def check_prime(n):
for i in range(2, int(n**0.5)+1):
if n % i == 0:
return 0
else:
return 1
for n in numbers:
ans.append(check_prime(n))
print(sum(ans))
|
s441046867
|
p03814
|
u031146664
| 2,000 | 262,144 |
Wrong Answer
| 24 | 4,840 | 108 |
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
s = list(input())
fastA = s.index("A")
s.reverse()
lastZ = len(s) - s.index("Z")+1
print(lastZ - fastA + 1)
|
s530520743
|
Accepted
| 24 | 4,840 | 138 |
s = list(input())
len = len(s)
fastA = s.index("A")+1
s.reverse()
Ztsal = s.index("Z")+1
lastZ = len - Ztsal + 1
print(lastZ - fastA + 1)
|
s242001254
|
p03719
|
u473023730
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 84 |
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a,b,c=map(int,input().split())
if a<=b and b<=c:
print("YES")
else:
print("NO")
|
s852854980
|
Accepted
| 17 | 2,940 | 84 |
a,b,c=map(int,input().split())
if a<=c and c<=b:
print("Yes")
else:
print("No")
|
s081753769
|
p02645
|
u105146481
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,044 | 34 |
When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him.
|
s = input()
print(s[0],s[1],s[2])
|
s730407727
|
Accepted
| 21 | 8,888 | 34 |
s = input()
print(s[0]+s[1]+s[2])
|
s446991372
|
p03760
|
u702786238
| 2,000 | 262,144 |
Wrong Answer
| 350 | 21,264 | 67 |
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
import numpy as np
print(np.array([list(input()), list(input())]))
|
s933153472
|
Accepted
| 17 | 3,060 | 262 |
S1 = list(input())
S2 = list(input())
ans = []
if len(S1) != len(S2):
for s1, s2 in zip(S1[:-1], S2):
ans.append(s1)
ans.append(s2)
ans.append(S1[-1])
else:
for s1, s2 in zip(S1, S2):
ans.append(s1)
ans.append(s2)
print("".join(ans))
|
s537617404
|
p03695
|
u244836567
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,300 | 562 |
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
a=int(input())
b=list(map(int,input().split()))
c=0
d=0
e=0
f=0
g=0
h=0
i=0
j=0
k=0
x=0
for z in range(a):
if b[z]<=399:
c=c+1
if 400<=b[z]<799:
d=d+1
if 800<=b[z]<=1199:
e=e+1
if 1200<=b[z]<=1599:
f=f+1
if 1600<=b[z]<=1999:
g=g+1
if 2000<=b[z]<=2399:
h=h+1
if 2400<=b[z]<=2799:
i=i+1
if 2800<=b[z]<=3199:
j=j+1
if b[z]>=3200:
k=k+1
if c!=0:
x=x+1
if d!=0:
x=x+1
if e!=0:
x=x+1
if f!=0:
x=x+1
if g!=0:
x=x+1
if h!=0:
x=x+1
if i!=0:
x=x+1
if j!=0:
x=x+1
if k+x>a:
print(a)
else:
print(k+x)
|
s769491786
|
Accepted
| 31 | 9,304 | 615 |
a=int(input())
b=list(map(int,input().split()))
c=0
d=0
e=0
f=0
g=0
h=0
i=0
j=0
k=0
x=0
for z in range(a):
if b[z]<=399:
c=c+1
if 400<=b[z]<799:
d=d+1
if 800<=b[z]<=1199:
e=e+1
if 1200<=b[z]<=1599:
f=f+1
if 1600<=b[z]<=1999:
g=g+1
if 2000<=b[z]<=2399:
h=h+1
if 2400<=b[z]<=2799:
i=i+1
if 2800<=b[z]<=3199:
j=j+1
if b[z]>=3200:
k=k+1
if c!=0:
x=x+1
if d!=0:
x=x+1
if e!=0:
x=x+1
if f!=0:
x=x+1
if g!=0:
x=x+1
if h!=0:
x=x+1
if i!=0:
x=x+1
if j!=0:
x=x+1
if k+x>a:
print(x,a)
elif (c+d+e+f+g+h+i+j)==0 and k!=0:
print(1,k)
else:
print(x,k+x)
|
s195813857
|
p02690
|
u112629957
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,120 | 197 |
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.
|
X=int(input())
for a in range(X):
if (a+1)**5-a**5 <= X <= (a+1)**+a**5:
A=a
break
for b in range((a+1)**5-a**5,(a+1)**+a**5+1):
if (a+1)**5-b**5 == X:
B=b
break
print(a+1, b)
|
s996507176
|
Accepted
| 208 | 9,176 | 142 |
X=int(input())
I=J=0
for i in range(-300,300):
for j in range(-300,300):
if i**5-j**5==X:
I=i
J=j
break
print(I,J)
|
s369579265
|
p03047
|
u729173935
| 2,000 | 1,048,576 |
Wrong Answer
| 65 | 3,064 | 8 |
Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi. How many ways are there to choose K consecutive integers?
|
print(1)
|
s152760473
|
Accepted
| 17 | 2,940 | 48 |
x,y = map(int,input().split())
print( x - y + 1)
|
s033650084
|
p03698
|
u229518917
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 59 |
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
S=input()
L=set(S)
print('Yes' if len(S)==len(L) else "No")
|
s899468962
|
Accepted
| 17 | 2,940 | 59 |
S=input()
L=set(S)
print('yes' if len(S)==len(L) else "no")
|
s172460626
|
p04029
|
u399721252
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 34 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n = int(input())
print(n*(n-1)//2)
|
s516924948
|
Accepted
| 17 | 2,940 | 36 |
n = int(input())
print((n*(n+1))//2)
|
s016940377
|
p02694
|
u944325914
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,160 | 116 |
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?
|
import math
n=int(input())
c=100
for i in range(10**6):
c=math.floor(c*1.01)
if c>n:
print(i+1)
break
|
s576792057
|
Accepted
| 28 | 9,180 | 107 |
import math
n=int(input())
c=100
for i in range(4000):
c=(c*101)//100
if c>=n:
print(i+1)
break
|
s747411737
|
p02612
|
u234454594
| 2,000 | 1,048,576 |
Wrong Answer
| 32 | 9,136 | 32 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n=int(input())
x=n%1000
print(x)
|
s731451886
|
Accepted
| 28 | 9,156 | 65 |
n=int(input())
x=n%1000
if x!=0:
print(1000-x)
else:
print(0)
|
s429182681
|
p03227
|
u678980622
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 2,940 | 87 |
You are given a string S of length 2 or 3 consisting of lowercase English letters. If the length of the string is 2, print it as is; if the length is 3, print the string after reversing it.
|
import sys
s = sys.stdin.read().lstrip()
if len(s) % 2 == 1:
s = s[::-1]
print(s)
|
s928657018
|
Accepted
| 17 | 2,940 | 87 |
import sys
s = sys.stdin.read().rstrip()
if len(s) % 2 == 1:
s = s[::-1]
print(s)
|
s275691337
|
p03437
|
u000349418
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 27,252 | 232 |
You are given positive integers X and Y. If there exists a positive integer not greater than 10^{18} that is a multiple of X but not a multiple of Y, choose one such integer and print it. If it does not exist, print -1.
|
n=int(1.0e+18)
x,y=map(int,input().split(' '))
X = n//x
if x%y==0:
print('-1')
else:
i=2
while i < X:
if x*i % y != 0:
print(str(x*i))
else:
i+=1
if i >= X:
print('-1')
|
s018024450
|
Accepted
| 17 | 3,060 | 230 |
n=int(1.0e+18)
x,y=map(int,input().split(' '))
X = n//x
if x%y==0:
print('-1')
else:
i=2
while i < X:
if (x*i)%y!=0:
print(str(x*i))
break
i+=1
if i >= X:
print('-1')
|
s602175908
|
p03455
|
u399721252
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b = map(int, input().split(" "))
if (a * b) % 2 == 0:
print("even")
else:
print("odd")
|
s238598378
|
Accepted
| 18 | 2,940 | 90 |
a,b = map(int, input().split(" "))
if (a * b) % 2 == 0:
print("Even")
else:
print("Odd")
|
s882571410
|
p03139
|
u373295322
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 52 |
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(min(a, b))
|
s732182312
|
Accepted
| 24 | 2,940 | 87 |
n, a, b = map(int, input().split())
print("{} {}".format(min(a, b), max(a + b - n, 0)))
|
s787457055
|
p03637
|
u588341295
| 2,000 | 262,144 |
Wrong Answer
| 76 | 15,020 | 562 |
We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective.
|
# -*- coding: utf-8 -*-
N = int(input())
aN = list(map(int, input().split()))
ok = 0
ng = 0
excp = 0
for i in range(N):
if aN[i] % 4 == 0:
ok += 1
elif (aN[i] + 2) % 4 == 0:
excp += 1
else:
ng += 1
ng += excp // 2
ng += excp % 2
if ok + 1 >= ng:
print("yes")
else:
print("No")
|
s325866449
|
Accepted
| 76 | 15,020 | 577 |
# -*- coding: utf-8 -*-
N = int(input())
aN = list(map(int, input().split()))
ok = 0
ng = 0
excp = 0
for i in range(N):
if aN[i] % 4 == 0:
ok += 1
elif (aN[i] + 2) % 4 == 0:
excp += 1
else:
ng += 1
if excp > 0:
ng += 1
if ok + 1 >= ng:
print("Yes")
else:
print("No")
|
s907880369
|
p03609
|
u437638594
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand. How many grams of sand will the upper bulb contains after t seconds?
|
X, t = map(int, input().split())
ans = min(0, X-t)
print(ans)
|
s184801442
|
Accepted
| 17 | 2,940 | 61 |
X, t = map(int, input().split())
ans = max(0, X-t)
print(ans)
|
s290933405
|
p02678
|
u573272932
| 2,000 | 1,048,576 |
Wrong Answer
| 652 | 34,076 | 379 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
from collections import deque
N, M = map(int, input().split())
net = [[] for _ in range(N+1)]
vect = [0]*(N+1)
for _ in range(M):
A, B = map(int, input().split())
net[A].append(B)
net[B].append(A)
q = deque()
q.append(1)
while len(q)>0:
t = q.popleft()
for i in net[t]:
if t != i and vect[i] == 0:
vect[i] = t
q.append(i)
for i in range(2, N+1):
print(vect[i])
|
s797381168
|
Accepted
| 633 | 33,940 | 393 |
from collections import deque
print("Yes")
N, M = map(int, input().split())
net = [[] for _ in range(N+1)]
vect = [0]*(N+1)
for _ in range(M):
A, B = map(int, input().split())
net[A].append(B)
net[B].append(A)
q = deque()
q.append(1)
while len(q)>0:
t = q.popleft()
for i in net[t]:
if t != i and vect[i] == 0:
vect[i] = t
q.append(i)
for i in range(2, N+1):
print(vect[i])
|
s866510199
|
p02408
|
u286589639
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,604 | 569 |
Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.
|
n = int(input())
S = [1]*13
H = [1]*13
C = [1]*13
D = [1]*13
for i in range(n):
mark, number = map(str, input().split())
num = int(number)
if mark == "S":
S[num-1] = 0
if mark == "H":
H[num-1] = 0
if mark == "C":
C[num-1] = 0
if mark == "H":
H[num-1] = 0
for i in range(13):
if S[i]:
print("S " + str(i+1))
for i in range(13):
if H[i]:
print("H " + str(i+1))
for i in range(13):
if C[i]:
print("C " + str(i+1))
for i in range(13):
if D[i]:
print("D " + str(i+1))
|
s399991249
|
Accepted
| 30 | 7,648 | 632 |
n = int(input())
S = [1]*13
H = [1]*13
C = [1]*13
D = [1]*13
mark = [""]*n
num = [""]*n
for i in range(n):
mark[i], num[i] = input().split()
for i in range(n):
if mark[i] == "S":
S[int(num[i])-1] = 0
if mark[i] == "H":
H[int(num[i])-1] = 0
if mark[i] == "C":
C[int(num[i])-1] = 0
if mark[i] == "D":
D[int(num[i])-1] = 0
for i in range(13):
if S[i]:
print("S " + str(i+1))
for i in range(13):
if H[i]:
print("H " + str(i+1))
for i in range(13):
if C[i]:
print("C " + str(i+1))
for i in range(13):
if D[i]:
print("D " + str(i+1))
|
s262663472
|
p00071
|
u811733736
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,788 | 2,980 |
縦 8、横 8 のマスからなる図1 のような平面があります。その平面上に、いくつかの爆弾が置かれています。図2 にその例を示します（● = 爆弾）。 | □| □| □| □| □| □| □| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ | □| □| □| ●| □| □| ●| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| ●| □| □ ●| □| □| □| ●| □| □| ● □| □| ●| □| □| □| ●| □ □| ●| □| □| □| □| □| □ □| □| □| □| ●| □| □| □ ●| □| ●| □| □| □| ●| □ □| ●| □| ●| □| □| ●| □ 図1| 図2 爆弾が爆発すると、その爆弾の上下左右 3 マスに爆風の影響が及び、それらのマスに置かれている爆弾も連鎖的に爆発します。たとえば、図 3 に示す爆弾が爆発すると図 4 の■のマスに爆風の影響が及びます。 | □| □| □| □| □| □| □| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| ●| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ | □| □| □| □| □| □| □| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| □| □| □ □| □| □| ■| □| □| □| □ □| □| □| ■| □| □| □| □ □| □| □| ■| □| □| □| □ ■| ■| ■| ●| ■| ■| ■| □ □| □| □| ■| □| □| □| □ □| □| □| ■| □| □| □| □ 図3| 図4 爆弾が置かれている状態と最初に爆発する爆弾の位置を読み込んで、最終的な平面の状態を出力するプログラムを作成してください。
|
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0071
"""
import sys
def check_bombs(map, pos, dir, range=3):
if dir == 'up':
x, y = pos
while range > 0 and y > 0:
if map[y-1][x] == '1':
return (x, y-1)
else:
y -= 1
range -= 1
elif dir == 'down':
x, y = pos
while range > 0 and y < len(map)-1:
if map[y+1][x] == '1':
return (x, y+1)
else:
y += 1
range -= 1
elif dir == 'left':
x, y = pos
while range > 0 and x > 0:
if map[y][x-1] == '1':
return (x-1, y)
else:
x -= 1
range -= 1
elif dir == 'right':
x, y = pos
while range > 0 and x < len(map[0])-1:
if map[y][x+1] == '1':
return (x+1, y)
else:
x += 1
range -= 1
else:
return None
def chain_bombs(map, init_pos):
lmap = map[:]
detonated = [init_pos]
while detonated:
x, y = detonated.pop()
lmap[y][x] = '0'
res = check_bombs(lmap, (x, y), 'up')
if res:
detonated.append(res)
res = check_bombs(lmap, (x, y), 'down')
if res:
detonated.append(res)
res = check_bombs(lmap, (x, y), 'left')
if res:
detonated.append(res)
res = check_bombs(lmap, (x, y), 'right')
if res:
detonated.append(res)
return lmap
def main(args):
data_set = int(input())
maps = []
init_pos = []
for i in range(data_set):
_ = input()
maps.append([list(input().strip()) for _ in range(8)])
init_pos.append([int(input())-1, int(input())-1])
count = 1
for map, pos in zip(maps, init_pos):
result = chain_bombs(map, pos)
print('Data {}:'.format(count))
for row in result:
print(''.join(row))
if __name__ == '__main__':
main(sys.argv[1:])
|
s696452791
|
Accepted
| 20 | 7,904 | 2,999 |
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0071
"""
import sys
def check_bombs(map, pos, dir, range=3):
if dir == 'up':
x, y = pos
while range > 0 and y > 0:
if map[y-1][x] == '1':
return (x, y-1)
else:
y -= 1
range -= 1
elif dir == 'down':
x, y = pos
while range > 0 and y < len(map)-1:
if map[y+1][x] == '1':
return (x, y+1)
else:
y += 1
range -= 1
elif dir == 'left':
x, y = pos
while range > 0 and x > 0:
if map[y][x-1] == '1':
return (x-1, y)
else:
x -= 1
range -= 1
elif dir == 'right':
x, y = pos
while range > 0 and x < len(map[0])-1:
if map[y][x+1] == '1':
return (x+1, y)
else:
x += 1
range -= 1
else:
return None
def chain_bombs(map, init_pos):
lmap = map[:]
detonated = [init_pos]
while detonated:
x, y = detonated.pop()
lmap[y][x] = '0'
res = check_bombs(lmap, (x, y), 'up')
if res:
detonated.append(res)
res = check_bombs(lmap, (x, y), 'down')
if res:
detonated.append(res)
res = check_bombs(lmap, (x, y), 'left')
if res:
detonated.append(res)
res = check_bombs(lmap, (x, y), 'right')
if res:
detonated.append(res)
return lmap
def main(args):
data_set = int(input())
maps = []
init_pos = []
for i in range(data_set):
_ = input()
maps.append([list(input().strip()) for _ in range(8)])
init_pos.append([int(input())-1, int(input())-1])
count = 1
for map, pos in zip(maps, init_pos):
result = chain_bombs(map, pos)
print('Data {}:'.format(count))
for row in result:
print(''.join(row))
count += 1
if __name__ == '__main__':
main(sys.argv[1:])
|
s743725122
|
p03712
|
u393512980
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 107 |
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=['#']*(h+2)+['#'+input()+'#' for _ in range(h)]+['#']
for x in a:
print(x)
|
s577073739
|
Accepted
| 17 | 3,060 | 114 |
h,w=map(int,input().split())
a=['#'*(w+2)]+['#'+input()+'#' for _ in range(h)]+['#'*(w+2)]
for x in a:
print(x)
|
s551391631
|
p02843
|
u706330549
| 2,000 | 1,048,576 |
Time Limit Exceeded
| 2,104 | 3,060 | 446 |
AtCoder Mart sells 1000000 of each of the six items below: * Riceballs, priced at 100 yen (the currency of Japan) each * Sandwiches, priced at 101 yen each * Cookies, priced at 102 yen each * Cakes, priced at 103 yen each * Candies, priced at 104 yen each * Computers, priced at 105 yen each Takahashi wants to buy some of them that cost exactly X yen in total. Determine whether this is possible. (Ignore consumption tax.)
|
import sys
x = int(input())
total = 0
for a in range(0,1000):
for b in range(0,1000):
for c in range(0,1000):
for d in range(0,1000):
for e in range(0,1000):
for f in range(0,1000):
total = 100*a + 101*b + 102*c + 103*d + 104*e +105*f
if x == total:
print(1)
sys.exit()
print(0)
|
s895212834
|
Accepted
| 17 | 2,940 | 330 |
x = int(input())
# i=1 100-105
# i=2 200-210
# i=3 300-315
# i=4 400-420
# i=5 500-525
# i=6 600-630
# i=7 700-735
# i=8 800-840
# i=9 900-945
# ...
# i=19 1900-1995
i = x // 100
if 100*i <= x <= 105*i:
print(1)
else:
print(0)
|
s576401716
|
p03160
|
u975389469
| 2,000 | 1,048,576 |
Wrong Answer
| 178 | 43,132 | 652 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
arr = [int(i) for i in input().split(" ")]
dp = []
for i in range(N-1):
diff1 = abs(arr[i+1]-arr[i])
if i == N-2: diff2 = diff1
else: diff2 = abs(arr[i+2]-arr[i])
dp.append({
1: diff1,
2: diff2
})
def getMinSteps(i):
total_cost = 0
while i < len(dp):
if dp[i][1] > dp[i][2]:
total_cost += dp[i][2]
i += 2
else:
total_cost += dp[i][1]
i += 1
return total_cost
print(min(dp[0][1] + getMinSteps(1), dp[0][2] + getMinSteps(2)))
|
s619919588
|
Accepted
| 525 | 21,076 | 666 |
from copy import deepcopy
N = int(input())
arr = [int(i) for i in input().split(" ")]
def getMinSteps():
dp_2 = [0, ]
dp_1 = [abs(arr[1] - arr[0]), ]
total_cost = 0
i = 2
while i < N:
diff1 = abs(arr[i] - arr[i-1])
diff2 = abs(arr[i] - arr[i-2])
temp = deepcopy(dp_2)
dp_2 = deepcopy(dp_1)
for j in range(len(dp_1)):
dp_1[j] += diff1
for j in range(len(temp)):
dp_1.append(temp[j] + diff2)
dp_1 = sorted(dp_1)[0:2]
i += 1
return min(dp_1)
print(getMinSteps())
|
s259687983
|
p03478
|
u265118937
| 2,000 | 262,144 |
Wrong Answer
| 47 | 9,040 | 330 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n, a, b = map(int, input().split())
ans = 0
def digitSum(n):
s = str(n)
array = list(map(int, s))
return sum(array)
for i in range(1, n+1):
if a <= digitSum(i) <= b:
ans += digitSum(i)
print(ans)
|
s261206257
|
Accepted
| 36 | 8,908 | 320 |
n, a, b = map(int, input().split())
ans = 0
def digitSum(n):
s = str(n)
array = list(map(int, s))
return sum(array)
for i in range(1, n+1):
if a <= digitSum(i) <= b:
ans += i
print(ans)
|
s603048327
|
p03545
|
u814986259
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 400 |
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
abcd=list(input())
abcd=list(map(int,abcd))
for i in range(2**3):
ret=abcd[0]
for j in range(3):
if (i>>j)%2==1:
ret-=abcd[j+1]
else:
ret+=abcd[j+1]
if ret==7:
ans=[str(abcd[0])]
for j in range(3):
if (i>>j)%2==1:
ans.append("-")
else:
ans.append(str(abcd[j+1]))
ans.append("=")
ans.append("7")
print("".join(ans))
exit(0)
|
s826760666
|
Accepted
| 20 | 3,064 | 427 |
ABCD = list(map(int, list(input())))
ans = ['']*9
for i in range(4):
ans[i*2] = str(ABCD[i])
for i in range(2**3):
ret = ABCD[0]
for j in range(3):
if (i >> j) % 2 == 1:
ret += ABCD[j+1]
ans[j*2 + 1] = '+'
else:
ret -= ABCD[j+1]
ans[j*2 + 1] = '-'
if ret == 7:
ans[7] = '='
ans[8] = '7'
print(''.join(ans))
exit(0)
|
s711691739
|
p03472
|
u022215787
| 2,000 | 262,144 |
Wrong Answer
| 354 | 11,684 | 417 |
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?
|
N, H = map(int, input().split())
a_list = []
b_list = []
for _ in range(N):
a,b = map(int, input().split())
a_list.append(a)
b_list.append(b)
max_a = max(a_list)
b_list = [ i for i in b_list if i > max_a ]
b_list.sort(reverse = True)
total = 0
count = 0
for b in b_list:
total += b
count += 1
if total > H:
break
if total < H:
count += int(-(-(H - total)) // max_a)
print(count)
|
s770579441
|
Accepted
| 354 | 11,792 | 449 |
import math
N, H = map(int, input().split())
a_list = []
b_list = []
for _ in range(N):
a,b = map(int, input().split())
a_list.append(a)
b_list.append(b)
max_a = max(a_list)
b_list = [ i for i in b_list if i > max_a ]
b_list.sort(reverse = True)
total = 0
count = 0
for d in b_list:
total += d
count += 1
if total >= H:
break
if total < H:
x = math.ceil((H - total) / float(max_a))
count += x
print(count)
|
s437354427
|
p02612
|
u362855700
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,128 | 90 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
price = int(input())
a = price % 1000
result = a if a < 0 else 1000 - price
print(result)
|
s297323671
|
Accepted
| 29 | 9,144 | 124 |
price = int(input())
a = price % 1000
if a == 0:
print(0)
elif a < 0:
print(1000 - price)
else:
print(1000 - a)
|
s442711898
|
p02389
|
u948918088
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,584 | 116 |
Write a program which calculates the area and perimeter of a given rectangle.
|
arg = input().split()
a = arg[0]
b = arg[1]
o = str(int(a)*int(b))
o2 = str(2*(int(a)*int(b)))
print(o + " " + o2)
|
s285532269
|
Accepted
| 20 | 5,588 | 117 |
arg = input().split()
a = arg[0]
b = arg[1]
o = str(int(a)*int(b))
o2 = str(2*(int(a)+int(b)))
print(o + " " + o2)
|
s578669008
|
p03069
|
u781535828
| 2,000 | 1,048,576 |
Wrong Answer
| 177 | 9,636 | 379 |
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
N = int(input())
S = list(input())
zeros = ["." for i in range(N)]
ones = ["
lastone = ["." for i in range(N - 1)]
lastone.append("#")
z = 0
o = 0
l = 0
for i in range(N):
if S[i] != zeros[i]:
z += 1
for i in range(N):
if S[i] != ones[i]:
o += 1
for i in range(N):
if S[i] != lastone[i]:
l += 1
print(min([z, o, l]))
|
s224666508
|
Accepted
| 119 | 12,836 | 298 |
N = int(input())
S = list(input())
bcount = 0
wcount = 0
mins = []
for i in range(N):
if S[i] == ".":
wcount += 1
mins.append(wcount)
for i in range(N):
if S[i] == '#':
bcount += 1
else:
wcount -= 1
mins.append(bcount + wcount)
print(min(mins))
|
s218018555
|
p04029
|
u526818929
| 2,000 | 262,144 |
Wrong Answer
| 29 | 9,036 | 41 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N = int(input())
print((N ** 2 + N) / 2)
|
s377998756
|
Accepted
| 25 | 9,152 | 42 |
N = int(input())
print((N ** 2 + N) // 2)
|
s245754686
|
p03623
|
u846150137
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,316 | 93 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b=[int(i) for i in input().split()]
if abs(x-a)>abs(x-b):
print("A")
else:
print("B")
|
s151256589
|
Accepted
| 17 | 2,940 | 93 |
n,a,b=[int(i) for i in input().split()]
if abs(n-a)<abs(n-b):
print("A")
else:
print("B")
|
s177920777
|
p03658
|
u187205913
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 130 |
Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.
|
n,k = map(int,input().split())
l = list(map(int,input().split()))
l.sort()
ans = 0
for i in range(k):
ans += l[-i]
print(ans)
|
s803682484
|
Accepted
| 18 | 2,940 | 131 |
n,k = map(int,input().split())
l = list(map(int,input().split()))
l.sort()
ans = 0
for i in range(k):
ans += l[-i-1]
print(ans)
|
s843182342
|
p02646
|
u966542724
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,192 | 244 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a = input().split()
b = input().split()
t = input()
d = abs(int(a[0]) - int(b[0]))
if (int(a[1]) <= int(b[1])):
print('No')
else:
sek = int(a[1]) - int(b[1])
if d <= int(t)*sek:
print('Yes')
else :
print('No')
|
s589316635
|
Accepted
| 23 | 9,196 | 244 |
a = input().split()
b = input().split()
t = input()
d = abs(int(a[0]) - int(b[0]))
if (int(a[1]) <= int(b[1])):
print('NO')
else:
sek = int(a[1]) - int(b[1])
if d <= int(t)*sek:
print('YES')
else :
print('NO')
|
s832465742
|
p03251
|
u875769753
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,116 | 183 |
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
N,M,X,Y = map(int,input().split())
lsx = list(map(int,input().split()))+[X]
lsy = list(map(int,input().split()))+[Y]
if max(lsx) < min(lsy):
print('No war')
else:
print('War')
|
s350456568
|
Accepted
| 33 | 9,116 | 183 |
N,M,X,Y = map(int,input().split())
lsx = list(map(int,input().split()))+[X]
lsy = list(map(int,input().split()))+[Y]
if max(lsx) < min(lsy):
print('No War')
else:
print('War')
|
s464778326
|
p03408
|
u844697453
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 312 |
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
ans={}
n=int(input())
for i in range(n):
a = input()
if a in ans:
ans[a]+=1
else:
ans[a] = 1
m=int(input())
for i in range(m):
a = input()
if a in ans:
ans[a]-=1
else:
ans[a] = -1
ans = sorted(ans.items(), key=lambda x: x[1], reverse=True)
print(ans[0][0])
|
s744203374
|
Accepted
| 17 | 3,064 | 353 |
ans={}
n=int(input())
for i in range(n):
a = input()
if a in ans:
ans[a]+=1
else:
ans[a] = 1
m=int(input())
for i in range(m):
a = input()
if a in ans:
ans[a]-=1
else:
ans[a] = -1
ans = sorted(ans.items(), key=lambda x: x[1], reverse=True)
if ans[0][1] < 0:
print(0)
else:
print(ans[0][1])
|
s793340762
|
p03623
|
u045408189
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 60 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b=map(int,input().split())
print(min(abs(a-x),abs(b-x)))
|
s601055129
|
Accepted
| 17 | 2,940 | 87 |
x,a,b=map(int,input().split())
print('A' if min(abs(a-x),abs(b-x))==abs(a-x) else 'B')
|
s236457008
|
p03739
|
u958506960
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 14,332 | 296 |
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()))
cnt = 0
s = a[0]
for i in range(1, n):
if a[0] > 0:
while s + a[i] >= 0:
a[i] -= 1
cnt += 1
if a[0] < 0:
while s + a[i] <= 0:
a[i] += 1
cnt += 1
s += a[i]
print(cnt)
|
s318325228
|
Accepted
| 68 | 14,332 | 434 |
n = int(input())
a = list(map(int, input().split()))
def f(isPosi):
total = 0
cnt = 0
for i in a:
total += i
if isPosi:
if total >= 0:
cnt += 1 + total
total = -1
isPosi = False
else:
if total <= 0:
cnt += 1 - total
total = 1
isPosi = True
return cnt
print(min(f(True), f(False)))
|
s013042661
|
p02795
|
u007808656
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 68 |
We have a grid with H rows and W columns, where all the squares are initially white. You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions: * Choose one row, then paint all the squares in that row black. * Choose one column, then paint all the squares in that column black. At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations.
|
h=int(input())
w=int(input())
n=int(input())
print(min(h//n,w//n)+1)
|
s375046320
|
Accepted
| 17 | 2,940 | 69 |
h=int(input())
w=int(input())
n=int(input())
print(-max(-n//h,-n//w))
|
s132004484
|
p02413
|
u343748576
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,636 | 396 |
Your task is to perform a simple table calculation. Write a program which reads the number of rows r, columns c and a table of r × c elements, and prints a new table, which includes the total sum for each row and column.
|
r, c = [int(x) for x in input().split()]
matrix = []
for i in range(r):
matrix.append([int(x) for x in input().split()])
for i in range(r):
for j in range(c):
print(matrix[i][j], '', end='')
print(sum(matrix[i]))
for j in range(c):
csum = 0
for i in range(r):
csum += matrix[i][j]
print(csum, '', end='')
crsum = 0
crsum += csum
print(crsum)
|
s832547245
|
Accepted
| 40 | 8,788 | 392 |
r, c = [int(x) for x in input().split()]
matrix = []
for i in range(r):
matrix.append([int(x) for x in input().split()])
for i in range(r):
for j in range(c):
print(matrix[i][j], '', end='')
print(sum(matrix[i]))
crsum = 0
for j in range(c):
csum = 0
for i in range(r):
csum += matrix[i][j]
print(csum, '', end='')
crsum += csum
print(crsum)
|
s840818903
|
p03371
|
u578953945
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 413 |
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
A,B,AB,X,Y=map(int,input().split())
ANS=0
if A + B >= AB * 2:
if min(X,Y) == X:
# X < Y
ANS = AB * X * 2
R = Y - X if Y - X > 0 else 0
if B <= AB * 2:
ANS += B * R
else:
ANS += AB * 2 * R
else:
# X > Y
ANS = AB * Y * 2 if Y - X > 0 else 0
R = X - Y
if A <= AB * 2:
ANS += A * R
else:
ANS += AB * 2 * R
else:
ANS = A * X + B * Y
print(ANS)
|
s571143874
|
Accepted
| 19 | 3,064 | 413 |
A,B,AB,X,Y=map(int,input().split())
ANS=0
if A + B >= AB * 2:
if min(X,Y) == X:
# X < Y
ANS = AB * X * 2
R = Y - X if Y - X > 0 else 0
if B <= AB * 2:
ANS += B * R
else:
ANS += AB * 2 * R
else:
# X > Y
ANS = AB * Y * 2
R = X - Y if X - Y > 0 else 0
if A <= AB * 2:
ANS += A * R
else:
ANS += AB * 2 * R
else:
ANS = A * X + B * Y
print(ANS)
|
s150201349
|
p03556
|
u896741788
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,220 | 69 |
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
n=int(input())
for i in range(1,10**3+2):
if i**2>n:print((i-1)**2)
|
s943855604
|
Accepted
| 33 | 9,084 | 77 |
n=int(input())
for i in range(1,10**5+2):
if i**2>n:print((i-1)**2);exit()
|
s043692447
|
p02602
|
u233032582
| 2,000 | 1,048,576 |
Wrong Answer
| 188 | 24,792 | 288 |
M-kun is a student in Aoki High School, where a year is divided into N terms. There is an exam at the end of each term. According to the scores in those exams, a student is given a grade for each term, as follows: * For the first through (K-1)-th terms: not given. * For each of the K-th through N-th terms: the multiplication of the scores in the last K exams, including the exam in the graded term. M-kun scored A_i in the exam at the end of the i-th term. For each i such that K+1 \leq i \leq N, determine whether his grade for the i-th term is **strictly** greater than the grade for the (i-1)-th term.
|
terms, graded_terms = input().split(" ")
grade_list = [i for i in input().split(" ")]
x1 = 0
x2 = int(graded_terms)
for i in range(int(terms)-int(graded_terms)):
if grade_list[int(x1)] < grade_list[int(x2)]:
print("yes2")
else:
print("no")
x1 += 1
x2 += 1
|
s687390362
|
Accepted
| 203 | 24,912 | 287 |
terms, graded_terms = input().split(" ")
grade_list = [i for i in input().split(" ")]
x1 = 0
x2 = int(graded_terms)
for i in range(int(terms)-int(graded_terms)):
if int(grade_list[x2]) > int(grade_list[x1]):
print("Yes")
else:
print("No")
x1 += 1
x2 += 1
|
s973906547
|
p02401
|
u202430481
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 303 |
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_list = [str(x) for x in input().split()]
a = int(a_list[0])
b = int(a_list[2])
op = a_list[1]
if op == '+':
ab = a + b
print(ab)
elif op == '-':
ab = a - b
print(ab)
elif op == '*':
ab = a * b
print(ab)
elif op == '/':
ab = a // b
print(ab)
elif op == '-':
exit()
|
s558579852
|
Accepted
| 20 | 5,600 | 543 |
i = 100
for i in range (i):
a_list = [str(x) for x in input().split()]
a = int(a_list[0])
b = int(a_list[2])
op = a_list[1]
if op == '+':
ab = a + b
print(ab)
i = i + 1
continue
elif op == '-':
ab = a - b
print(ab)
i = i + 1
continue
elif op == '*':
ab = a * b
print(ab)
i = i + 1
continue
elif op == '/':
ab = a // b
print(ab)
i = i + 1
continue
elif op == '?':
exit()
|
s350033346
|
p03854
|
u240096083
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 142 |
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 = str(input())
l = ['dreamer', 'dream', 'eraser', 'erase']
for i in l:
s = s.replace(i,'')
if s == "":
print('OK')
else:
print('NG')
|
s126091767
|
Accepted
| 69 | 3,188 | 421 |
s = str(input())
s = s[::-1]
l = ['resare', 'esare', 'remaerd', 'maerd']
f = 0
if len(s) == 0:
f = 1
while len(s) != 0:
if s[:7] == 'remaerd':
s = s[7::]
elif s[:6] == 'resare':
s = s[6::]
elif s[:5] == 'esare':
s = s[5:]
elif s[:5] == 'maerd':
s = s[5:]
else:
f = 1
break
if f == 1:
print("NO")
else:
print("YES")
|
s828440080
|
p03455
|
u403355272
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 99 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b = map(int,input().split())
if not (a * b // 2 == 0):
print('Odd')
else:
print('Even')
|
s527783884
|
Accepted
| 20 | 3,060 | 73 |
a,b = map(int,input().split())
print("Even" if a * b % 2 == 0 else "Odd")
|
s957803814
|
p03447
|
u617225232
| 2,000 | 262,144 |
Wrong Answer
| 31 | 9,160 | 66 |
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?
|
z = int(input())
a = int(input())
b = int(input())
print((z-a)//b)
|
s152541734
|
Accepted
| 27 | 9,112 | 66 |
z = int(input())
a = int(input())
b = int(input())
print((z-a)%b)
|
s832521060
|
p03023
|
u995102075
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 39 |
Given an integer N not less than 3, find the sum of the interior angles of a regular polygon with N sides. Print the answer in degrees, but do not print units.
|
N = int(input())
print(180 * (N - 3))
|
s053434965
|
Accepted
| 17 | 2,940 | 39 |
N = int(input())
print(180 * (N - 2))
|
s710348706
|
p03380
|
u638456847
| 2,000 | 262,144 |
Wrong Answer
| 79 | 14,052 | 364 |
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
import bisect
def main():
N = int(input())
a = [int(i) for i in input().split()]
a.sort()
n = a[-1]
m = n // 2
right = bisect.bisect_right(a, m)
if a[right-1] == m:
r = m
elif m - a[right-1] <= a[right] - m:
r = a[right-1]
else:
r = a[right]
print(n, r)
if __name__ == "__main__":
main()
|
s241340041
|
Accepted
| 101 | 14,052 | 284 |
def main2():
N = int(input())
a = [int(i) for i in input().split()]
a.sort()
n = a[-1]
m = n / 2
tmp = 10**10
for i in a:
if abs(m - i) < tmp:
tmp = abs(m - i)
r = i
print(n, r)
if __name__ == "__main__":
main2()
|
s756292982
|
p03149
|
u892796322
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 113 |
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
n = input()
print(n)
if '1' in n and '7' in n and '9' in n and '4' in n:
print("YES")
else:
print("NO")
1
|
s452501847
|
Accepted
| 18 | 2,940 | 102 |
n = input()
if '1' in n and '7' in n and '9' in n and '4' in n:
print("YES")
else:
print("NO")
|
s916616952
|
p03643
|
u114954806
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 47 |
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
print("ABC" if 1<=int(input())<=999 else "ABD")
|
s489851919
|
Accepted
| 17 | 2,940 | 20 |
print("ABC"+input())
|
s704067761
|
p03469
|
u839270538
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 31 |
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.
|
a = input()
print("2017"+a[4:])
|
s242256712
|
Accepted
| 17 | 2,940 | 32 |
a = input()
print("2018"+a[4:])
|
s817937548
|
p03798
|
u996252264
| 2,000 | 262,144 |
Wrong Answer
| 133 | 4,904 | 787 |
Snuke, who loves animals, built a zoo. There are N animals in this zoo. They are conveniently numbered 1 through N, and arranged in a circle. The animal numbered i (2≤i≤N-1) is adjacent to the animals numbered i-1 and i+1. Also, the animal numbered 1 is adjacent to the animals numbered 2 and N, and the animal numbered N is adjacent to the animals numbered N-1 and 1. There are two kinds of animals in this zoo: honest sheep that only speak the truth, and lying wolves that only tell lies. Snuke cannot tell the difference between these two species, and asked each animal the following question: "Are your neighbors of the same species?" The animal numbered i answered s_i. Here, if s_i is `o`, the animal said that the two neighboring animals are of the same species, and if s_i is `x`, the animal said that the two neighboring animals are of different species. More formally, a sheep answered `o` if the two neighboring animals are both sheep or both wolves, and answered `x` otherwise. Similarly, a wolf answered `x` if the two neighboring animals are both sheep or both wolves, and answered `o` otherwise. Snuke is wondering whether there is a valid assignment of species to the animals that is consistent with these responses. If there is such an assignment, show one such assignment. Otherwise, print `-1`.
|
N = int(input())
t = list(input())
Ans = ['n' for _ in t]
def judge(prevSW, SW, ox):
if SW == 'S':
if ox == 'o':
return prevSW
else:
if prevSW == 'S':
return 'W'
else: return 'S'
elif SW == 'W':
if ox == 'x':
return prevSW
else:
if prevSW == 'S':
return 'W'
else: return 'S'
Ans[0] = 'S'
for i in range(1,N):
if i == 1:
Ans[i] = 'S'
else:
Ans[i] = judge(Ans[i-2], Ans[i-1], t[i-1])
if judge(Ans[-1], Ans[0], t[-1]) is 'S':
print("".join(Ans))
else:
for i in range(1,N):
if i == 1:
Ans[i] = 'W'
else:
Ans[i] = judge(Ans[i-2], Ans[i-1], t[i-1])
print("".join(Ans))
|
s793420136
|
Accepted
| 174 | 4,900 | 1,095 |
N = int(input())
t = list(input())
Ans = ['n' for _ in t]
def judge(prevSW, SW, ox):
if SW == 'S':
if ox == 'o':
return prevSW
else:
if prevSW == 'S':
return 'W'
else: return 'S'
elif SW == 'W':
if ox == 'x':
return prevSW
else:
if prevSW == 'S':
return 'W'
else: return 'S'
def decideorder(startSW):
Ans[1] = startSW
for i in range(2,N):
Ans[i] = judge(Ans[i-2], Ans[i-1], t[i-1])
def check():
if judge(Ans[-2], Ans[-1], t[-1]) is Ans[0] and judge(Ans[-1], Ans[0], t[0]) is Ans[1]:
return True
else:
return False
Ans[0] = 'S'
decideorder('S')
if check():
print("".join(Ans))
else:
decideorder('W')
if check():
print("".join(Ans))
else:
Ans[0] = 'W'
decideorder('S')
if check():
print("".join(Ans))
else:
decideorder('W')
if check():
print("".join(Ans))
else:
print(-1)
|
s590460904
|
p03557
|
u969190727
| 2,000 | 262,144 |
Wrong Answer
| 2,105 | 23,092 | 286 |
The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower. He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i. To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar. How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.
|
import bisect
n=int(input())
A=sorted(list(map(int,input().split())))
B=sorted(list(map(int,input().split())))
C=sorted(list(map(int,input().split())))
ans=0
for a in A:
b=bisect.bisect_left(B,a+1)
print(b)
for i in range(n-b):
ans+=n-bisect.bisect_left(C,B[b+i]+1)
print(ans)
|
s127837205
|
Accepted
| 331 | 23,200 | 260 |
n=int(input())
A=[int(i) for i in input().split()]
B=[int(i) for i in input().split()]
C=[int(i) for i in input().split()]
import bisect
A.sort()
B.sort()
C.sort()
ans=0
for b in B:
ans+=bisect.bisect_left(A,b)*(n-bisect.bisect_right(C,b))
print(ans)
|
s305158260
|
p03730
|
u152614052
| 2,000 | 262,144 |
Wrong Answer
| 30 | 8,948 | 131 |
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())
ans = "No"
for i in range(b):
if a*i % b == c:
ans = "Yes"
break
print(ans)
|
s241962298
|
Accepted
| 29 | 8,980 | 132 |
a, b, c = map(int, input().split())
ans = "NO"
for i in range(b):
if a*i % b == c:
ans = "YES"
break
print(ans)
|
s798553112
|
p02390
|
u971748390
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,796 | 106 |
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
import math
S=int(input()) ;
h = S // 3600 ;
m= (S % 3600)/60 ;
s= m % 60 ;
print ("%d:%d:%d"% (h,m,s))
|
s517374667
|
Accepted
| 30 | 7,672 | 74 |
t=int(input())
h=t//3600
m=(t%3600)//60
s=(t%3600)%60
print(h,m,s,sep=":")
|
s217708219
|
p03719
|
u744898490
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 123 |
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
abc = input().split(' ')
if int(abc[2]) > int(abc[1]) and int(abc[2]) < int(abc[1]):
print('YES')
else:
print('NO')
|
s054192493
|
Accepted
| 17 | 2,940 | 127 |
abc = input().split(' ')
if int(abc[2]) <= int(abc[1]) and int(abc[2]) >= int(abc[0]):
print('Yes')
else:
print('No')
|
s480776300
|
p03352
|
u859210968
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 152 |
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
x = int(input())
a = 1
for i in range(1, 10) :
for b in range(1, 32) :
if( b**i<=x and a<b**i ) :
a = b**i
print("{}".format(a))
|
s877970291
|
Accepted
| 17 | 2,940 | 152 |
x = int(input())
a = 1
for i in range(2, 10) :
for b in range(1, 32) :
if( b**i<=x and a<b**i ) :
a = b**i
print("{}".format(a))
|
s486710474
|
p03377
|
u826929627
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 113 |
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A,B,X = map(int, input().split())
_ = X-A
if (_ >= 0) and (_ <= B):
ans = 'Yes'
else:
ans = 'No'
print(ans)
|
s341015674
|
Accepted
| 17 | 2,940 | 113 |
A,B,X = map(int, input().split())
_ = X-A
if (_ >= 0) and (_ <= B):
ans = 'YES'
else:
ans = 'NO'
print(ans)
|
s982728779
|
p03671
|
u397563544
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 117 |
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.
|
s = input()
n = len(s)//2
for i in range(1,n):
if s[:n-i] == s[n-i:2*(n-i)]:
print(2*(n-i))
break
|
s077848002
|
Accepted
| 18 | 2,940 | 60 |
a = list(map(int,input().split()))
a.sort()
print(a[0]+a[1])
|
s598578999
|
p02240
|
u726330006
| 1,000 | 131,072 |
Time Limit Exceeded
| 9,990 | 6,000 | 971 |
Write a program which reads relations in a SNS (Social Network Service), and judges that given pairs of users are reachable each other through the network.
|
from collections import deque
user_num,rer_num=map(int,input().split())
graph=[[] for loop in range(user_num)]
for loop in range(rer_num):
user1,user2=map(int,input().split())
graph[user1].append(user2)
graph[user2].append(user1)
group=[-1]*user_num
que=deque([])
que.append(0)
tmp_group=0
for user in range(user_num):
if(group[user]==-1):
que.append(user)
group[user]=tmp_group
while(1):
user1=que.popleft()
for user2 in graph[user1]:
if(group[user2]!=-1):
que.append(user2)
group[user2]=tmp_group
if(len(que)==0):
tmp_group+=1
break
for loop in range(int(input())):
user1,user2=map(int(input()))
if(group[user1]==group[user2]):
print("yes")
else:
print("no")
|
s637993814
|
Accepted
| 610 | 23,556 | 977 |
from collections import deque
user_num,rer_num=map(int,input().split())
graph=[[] for loop in range(user_num)]
for loop in range(rer_num):
user1,user2=map(int,input().split())
graph[user1].append(user2)
graph[user2].append(user1)
group=[-1]*user_num
que=deque([])
que.append(0)
tmp_group=0
for user in range(user_num):
if(group[user]==-1):
que.append(user)
group[user]=tmp_group
while(1):
user1=que.popleft()
for user2 in graph[user1]:
if(group[user2]==-1):
que.append(user2)
group[user2]=tmp_group
if(len(que)==0):
tmp_group+=1
break
for loop in range(int(input())):
user1,user2=map(int,input().split())
if(group[user1]==group[user2]):
print("yes")
else:
print("no")
|
s828345930
|
p03228
|
u067983636
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 3,060 | 224 |
In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.
|
A, B, K = map(int, input().split())
for k in range(K):
if k % 2 == 0:
if A % 2 == 1:
A -= 1
A //= 2
B += A // 2
else:
if B % 2 == 1:
B -= 1
B //= 2
A += B // 2
print(A, B)
|
s861370076
|
Accepted
| 17 | 2,940 | 237 |
A, B, K = map(int, input().split())
for k in range(K):
if k % 2 == 0:
if A % 2 == 1:
A -= 1
B += A // 2
A -= A // 2
else:
if B % 2 == 1:
B -= 1
A += B // 2
B -= B // 2
print(A, B)
|
s780058558
|
p03659
|
u411923565
| 2,000 | 262,144 |
Wrong Answer
| 228 | 32,660 | 259 |
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
N = int(input())
A = list(map(int,input().split()))
s = [0]*(N+1)
for i in range(N-1):
s[i+1] = s[i] + A[i]
print(s)
ans = 10**10
for i in range(1,N):
diff = abs((s[N-1] + A[-1] - s[i]) - s[i])
ans = min(ans,diff)
print(ans)
|
s362966414
|
Accepted
| 215 | 30,536 | 250 |
N = int(input())
A = list(map(int,input().split()))
s = [0]*(N+1)
for i in range(N-1):
s[i+1] = s[i] + A[i]
ans = 10**10
for i in range(1,N):
diff = abs((s[N-1] + A[-1] - s[i]) - s[i])
ans = min(ans,diff)
print(ans)
|
s408107526
|
p02608
|
u380815663
| 2,000 | 1,048,576 |
Wrong Answer
| 823 | 9,320 | 295 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
N = int(input())
ans = list()
for i in range(N):
ans.append(0)
for x in range(1,100):
for y in range(1,100):
for z in range(1,100):
tmp = x**2 + y**2 + z**2 + x*y + y*z + z*x
if tmp < N:
ans[tmp] += 1
for i in range(N):
print(ans[i])
|
s447543286
|
Accepted
| 935 | 9,348 | 306 |
N = int(input())
ans = [0]*10050
for i in range(N+1):
ans.append(0)
for x in range(1,105):
for y in range(1,105):
for z in range(1,105):
tmp = x**2 + y**2 + z**2 + x*y + y*z + z*x
if tmp < 10050:
ans[tmp] += 1
for i in range(N):
print(ans[i+1])
|
s049300233
|
p03339
|
u853952087
| 2,000 | 1,048,576 |
Wrong Answer
| 314 | 15,392 | 386 |
There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions.
|
n=int(input())
s=input()
r=len([i for i in range(1,n) if s[i]=='E'])
l=0
print(l,r)
i=1
x=n
while i<=n-1:
if s[i]=='E' and s[i-1]=='E':
r=r-1
l=l
elif s[i]=='W' and s[i-1]=='W':
r=r
l=l+1
elif s[i]=='W' and s[i-1]=='E':
r=r
l=l
elif s[i]=='E' and s[i-1]=='W':
r=r-1
l=l+1
x=min(l+r,x)
i+=1
print(x)
|
s609422811
|
Accepted
| 317 | 15,520 | 377 |
n=int(input())
s=input()
r=len([i for i in range(1,n) if s[i]=='E'])
l=0
i=1
x=r+l
while i<=n-1:
if s[i]=='E' and s[i-1]=='E':
r=r-1
l=l
elif s[i]=='W' and s[i-1]=='W':
r=r
l=l+1
elif s[i]=='W' and s[i-1]=='E':
r=r
l=l
elif s[i]=='E' and s[i-1]=='W':
r=r-1
l=l+1
x=min(l+r,x)
i+=1
print(x)
|
s565429687
|
p03486
|
u277802731
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 172 |
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
#82b
s=sorted(list(input()))
t=sorted(list(input()))
ss=''.join(s)
tt=''.join(t)
ans=sorted([tt,ss])
if ss==tt:
print('NO')
else:
print(['No','Yes'][ans[0]==ss])
|
s361302696
|
Accepted
| 17 | 3,060 | 182 |
#82b
s=sorted(list(input()))
t=sorted(list(input()))
t=t[::-1]
ss=''.join(s)
tt=''.join(t)
ans=sorted([tt,ss])
if ss==tt:
print('No')
else:
print(['No','Yes'][ans[0]==ss])
|
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