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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s899586059
|
p02389
|
u890164142
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,716 | 72 |
Write a program which calculates the area and perimeter of a given rectangle.
|
a, b = [int(s) for s in input().split()]
print(a * b)
print(2 * (a + b))
|
s855649047
|
Accepted
| 20 | 7,632 | 66 |
a, b = [int(s) for s in input().split()]
print(a * b, 2 * (a + b))
|
s198795592
|
p02393
|
u987701388
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 69 |
Write a program which reads three integers, and prints them in ascending order.
|
a= list(int(i) for i in input().split())
for s in a:
print(s)
|
s544705484
|
Accepted
| 20 | 5,600 | 97 |
a= list(int(i) for i in input().split())
a.sort()
risuto = ' '.join(map(str,a))
print(risuto)
|
s886929591
|
p03836
|
u026788530
| 2,000 | 262,144 |
Wrong Answer
| 31 | 4,468 | 535 |
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
l = input().split(' ')
y = int(l[3]) - int(l[1])
x = int(l[2]) - int(l[0])
def my_print(i,str):
for j in range(i):
print(str,end='')
my_print(y,'U')
my_print(x,'R')
my_print(y,'D')
my_print(x+1,'L')
my_print(y+1,'U')
my_print(x+1,'R')
my_print(1,'D')
my_print(1,'R')
my_print(y,'D')
my_print(x+1,'L')
my_print(1,'U')
print()
|
s485702342
|
Accepted
| 32 | 4,468 | 537 |
l = input().split(' ')
y = int(l[3]) - int(l[1])
x = int(l[2]) - int(l[0])
def my_print(i,str):
for j in range(i):
print(str,end='')
my_print(y,'U')
my_print(x,'R')
my_print(y,'D')
my_print(x+1,'L')
my_print(y+1,'U')
my_print(x+1,'R')
my_print(1,'D')
my_print(1,'R')
my_print(y+1,'D')
my_print(x+1,'L')
my_print(1,'U')
print()
|
s500873447
|
p03523
|
u787456042
| 2,000 | 262,144 |
Wrong Answer
| 22 | 3,188 | 72 |
You are given a string S. Takahashi can insert the character `A` at any position in this string any number of times. Can he change S into `AKIHABARA`?
|
import re;print("YNeos"[(re.match("^A?KIHA?BA?RA?$",input())==None)::2])
|
s430153071
|
Accepted
| 22 | 3,188 | 72 |
import re;print("YNEOS"[(re.match("^A?KIHA?BA?RA?$",input())==None)::2])
|
s246832255
|
p03471
|
u391340825
| 2,000 | 262,144 |
Wrong Answer
| 521 | 3,064 | 350 |
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
def main():
linie = input()
vector = [int(x) for x in linie.split()]
n = vector[0]
y = vector[1]
ok = 0
nr1 = -1
nr2 = -1
nr3 = -1
for i in range(n):
for j in range(n - i):
if i * 1000 + j * 5000 + (n - i - j) * 10000 == y:
nr1 = i
nr2 = j
nr3 = (n - i - j)
print(nr1, nr2, nr3)
if __name__ == "__main__":
main()
|
s296118612
|
Accepted
| 529 | 3,064 | 373 |
def main():
linie = input()
vector = [int(x) for x in linie.split()]
n = vector[0]
y = vector[1]
ok = 0
nr1 = -1
nr2 = -1
nr3 = -1
nr = n + 1
for i in range(0, n + 1):
for j in range(n + 1 - i):
if i * 1000 + j * 5000 + (n - i - j) * 10000 == y:
nr1 = i
nr2 = j
nr3 = (n - i - j)
print(nr3, nr2, nr1)
if __name__ == "__main__":
main()
|
s553573708
|
p04030
|
u100418016
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 131 |
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?
|
arr =list(input())
ret=""
for st in arr:
if(st == "B"):
ret = ret[:len(st)-1]
else:
ret = ret + st
print(ret)
|
s901619818
|
Accepted
| 19 | 2,940 | 129 |
arr =list(input())
ret=""
for st in arr:
if(st == "B"):
ret = ret[:len(ret)-1]
else:
ret = ret + st
print(ret)
|
s947201614
|
p03556
|
u119982147
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 63 |
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.
|
import math
N = int(input())
rN = math.sqrt(N)
print(int(rN))
|
s918959115
|
Accepted
| 23 | 3,316 | 71 |
import math
N = int(input())
rN = math.sqrt(N)
A = int(rN)**2
print(A)
|
s439388137
|
p02694
|
u240256924
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,152 | 78 |
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
X = int(input())
a= 100
n = 0
while a<=X:
a=int(a*1.01)
n+=1
print(n)
|
s738875250
|
Accepted
| 23 | 9,152 | 77 |
X = int(input())
a= 100
n = 0
while a<X:
a=int(a*1.01)
n+=1
print(n)
|
s719897881
|
p03160
|
u846385882
| 2,000 | 1,048,576 |
Wrong Answer
| 155 | 12,488 | 302 |
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.
|
import numpy as np
#N=int(input())
#h=list(map(int,input().split()))
N=4
h=[10,30,40,20]
dp=np.zeros(N)
print(dp)
dp[1]=abs(h[1]-h[0])
print(dp)
for i in range(N-2):
dp[i+2]=min(abs(h[i+2]-h[i])+dp[i],abs(h[i+2]-h[i+1])+dp[i+1])
print(dp)
#dp[2]=min(h[2]-h[0],h[1]-h[0])
#total
print(dp[-1])
|
s221273902
|
Accepted
| 538 | 22,840 | 316 |
import numpy as np
N=int(input())
h=list(map(int,input().split()))
#N=4
#h=[10,30,40,20]
dp=np.zeros(N)
#print(dp)
dp[1]=abs(h[1]-h[0])
#print(dp)
for i in range(N-2):
dp[i+2]=min(abs(h[i+2]-h[i])+dp[i],abs(h[i+2]-h[i+1])+dp[i+1])
# print(dp)
#dp[2]=min(h[2]-h[0],h[1]-h[0])
#total
print(int(dp[-1]))
|
s058767555
|
p03162
|
u807772568
| 2,000 | 1,048,576 |
Wrong Answer
| 1,042 | 47,328 | 291 |
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.
|
a = int(input())
b = []
m = 0
for i in range(a):
b.append(list(map(int,input().split())))
c = [[0,0,0] for i in range(a+1)]
for i in range(a):
for j in range(3):
for k in range(3):
if j == k:
continue
else:
c[i+1][k] = max(c[i+1][k],c[i][k] + b[i][j])
print(max(c[-1]))
|
s749717003
|
Accepted
| 1,024 | 47,328 | 292 |
a = int(input())
b = []
m = 0
for i in range(a):
b.append(list(map(int,input().split())))
c = [[0,0,0] for i in range(a+1)]
for i in range(a):
for j in range(3):
for k in range(3):
if j == k:
continue
else:
c[i+1][k] = max(c[i+1][k],c[i][j] + b[i][k])
print(max(c[-1]))
|
s227813895
|
p03385
|
u485566817
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 71 |
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
s = set(input())
if len(s) == 3:
print("YES")
else:
print("NO")
|
s924532415
|
Accepted
| 17 | 2,940 | 71 |
s = set(input())
if len(s) == 3:
print("Yes")
else:
print("No")
|
s332607064
|
p03695
|
u103902792
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 281 |
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
n = int(input())
ai = list(map(int,input().split()))
aj = list(map(lambda x:int(x/400),ai))
print(aj)
rainbow = 0
for e in aj:
if e >= 8:
rainbow += 1
ak= set(aj)
ans = 0
for e in ak:
if e < 8:
ans += 1
print(max(1,ans) ,end='')
print(' ' ,end = '')
print(ans+ rainbow)
|
s593350299
|
Accepted
| 28 | 9,056 | 257 |
n = int(input())
A = list(map(int,input().split()))
col = [0 for _ in range(9)]
for a in A:
if a >= 3200:
col[8] += 1
continue
col[a//400] = 1
ans_min = sum(col[:-1])
if ans_min == 0 and col[-1] != 0:
ans_min = 1
print(ans_min, sum(col))
|
s673796196
|
p02694
|
u676258045
| 2,000 | 1,048,576 |
Time Limit Exceeded
| 2,206 | 9,132 | 79 |
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
x = int(input())
temp = int(100)
while temp < x:
temp *= 0.01
print(temp)
|
s695983930
|
Accepted
| 22 | 9,076 | 133 |
import math
x = int(input())
temp = int(100)
ans = int(0)
while temp < x:
temp += math.floor(temp * 0.01)
ans += 1
print(ans)
|
s415288157
|
p02747
|
u578208203
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 289 |
A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.
|
if __name__ == '__main__':
S=input()
i=0
j=len(S)
print(j)
if j/2==0:
for l in range(0,len(S),2):
print(l)
i=0
if(S[l]=='h' and S[l+1]=='i'):
i=1
if i==1:
print('Yes')
else:
print('NO')
|
s921302166
|
Accepted
| 17 | 2,940 | 256 |
if __name__ == '__main__':
S=input()
i=0
j=len(S)
if j%2==0:
for l in range(0,len(S),2):
i=0
if(S[l]=='h' and S[l+1]=='i'):
i=1
if i==1:
print('Yes')
else:
print('No')
|
s887974517
|
p03494
|
u859897687
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 161 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n=int(input())
m=list(map(int,input().split()))
t=1
ans=0
while t>0:
for i in range(n):
if m[i]%2>0:
t=0
break
m[i]//=2
ans+=1
print(ans)
|
s361970203
|
Accepted
| 19 | 3,060 | 174 |
n=int(input())
m=list(map(int,input().split()))
t=1
ans=-1
while t>0:
for i in range(n):
if m[i]%2>0:
t=0
break
else:
m[i]//=2
ans+=1
print(ans)
|
s510811822
|
p03545
|
u669062920
| 2,000 | 262,144 |
Wrong Answer
| 149 | 12,504 | 553 |
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.
|
import numpy as np
s = [w for w in input()]
length = len(s)
for i in range(2 ** (length-1)):
b = format(i, 'b')
b_len = len(b)
pad_width = (length - b_len, 0)
b_pad = np.pad([n for n in b], pad_width, 'constant', constant_values=0)
score = int(s[0])
result = s[0]
for flag, num in zip(b_pad, s[1:]):
if flag == '1':
score += int(num)
result += '+' + num
else:
score -= int(num)
result += '-' + num
if score == 7:
print(result+'=7')
break
|
s893725307
|
Accepted
| 165 | 13,176 | 541 |
import numpy as np
s = [w for w in input()]
length = len(s)
for i in range(2 ** (length-1)):
b = format(i, 'b')
b_len = len(b)
pad_width = ((length-1) - b_len, 0)
b_pad = np.pad([int(n) for n in b], pad_width, 'constant')
score = int(s[0])
result = s[0]
for flag, num in zip(b_pad, s[1:]):
if flag == 1:
score += int(num)
result += '+' + num
else:
score -= int(num)
result += '-' + num
if score == 7:
print(result+'=7')
break
|
s347930301
|
p04029
|
u483151310
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 204 |
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?
|
s = input()
# print(s[0])
ans = []
for key in range(len(s)):
if s[key] == "0" or s[key] == "1":
ans.append(s[key])
if s[key] == "B":
ans.pop(-1)
print("".join(ans))
|
s364388176
|
Accepted
| 18 | 2,940 | 77 |
N = int(input())
# N = 5
cnt = 0
for i in range(N+1):
cnt += i
print(cnt)
|
s920838783
|
p03836
|
u888337853
| 2,000 | 262,144 |
Wrong Answer
| 309 | 22,160 | 724 |
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
import sys
# import re
import math
import collections
import bisect
import itertools
import fractions
# import functools
import copy
import heapq
import decimal
# import statistics
import queue
import numpy as np
sys.setrecursionlimit(10000001)
INF = 10 ** 16
MOD = 10 ** 9 + 7
ni = lambda: int(sys.stdin.readline())
ns = lambda: map(int, sys.stdin.readline().split())
na = lambda: list(map(int, sys.stdin.readline().split()))
# ===CODE===
def main():
sx, sy, tx, ty = ns()
ans = []
ans.extend(["U"] * (ty - sy))
ans.extend((["R"] * (tx - sx)))
ans.extend(["D"] * ((ty - sy)))
ans.extend((["L"] * (tx - sx)))
print(*ans, sep="")
if __name__ == '__main__':
main()
|
s418188601
|
Accepted
| 164 | 15,004 | 954 |
import sys
# import re
import math
import collections
import bisect
import itertools
import fractions
# import functools
import copy
import heapq
import decimal
# import statistics
import queue
import numpy as np
sys.setrecursionlimit(10000001)
INF = 10 ** 16
MOD = 10 ** 9 + 7
ni = lambda: int(sys.stdin.readline())
ns = lambda: map(int, sys.stdin.readline().split())
na = lambda: list(map(int, sys.stdin.readline().split()))
# ===CODE===
def main():
sx, sy, tx, ty = ns()
ans = []
ans.extend(["U"] * (ty - sy))
ans.extend((["R"] * (tx - sx)))
ans.extend(["D"] * ((ty - sy)))
ans.extend((["L"] * (tx - sx)))
ans.extend(["L"])
ans.extend(["U"] * (ty - sy + 1))
ans.extend((["R"] * (tx - sx + 1)))
ans.extend(["DR"])
ans.extend(["D"] * ((ty - sy + 1)))
ans.extend((["L"] * (tx - sx + 1)))
ans.extend(["U"])
print(*ans, sep="")
if __name__ == '__main__':
main()
|
s668158527
|
p03971
|
u430223993
| 2,000 | 262,144 |
Wrong Answer
| 107 | 4,016 | 340 |
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
n,a,b = map(int, input().split())
s = input()
passed = 0
passed_b = 0
for i in s:
if i == 'a':
print('Yes') if passed <= a+b else print('No')
passed += 1
elif i == 'b':
print('Yes') if (passed <= a+b) and (passed_b < b) else print('No')
passed += 1
passed_b += 1
else:
print('No')
|
s205609142
|
Accepted
| 103 | 4,016 | 418 |
n,a,b = map(int, input().split())
s = input()
passed = 0
passed_b = 0
for i in s:
if i == 'a':
if passed < a+b:
print('Yes')
passed += 1
else:
print('No')
elif i == 'b':
if (passed < a+b) and (passed_b < b):
print('Yes')
passed += 1
passed_b += 1
else:
print('No')
else:
print('No')
|
s566405677
|
p03545
|
u457601965
| 2,000 | 262,144 |
Wrong Answer
| 33 | 9,220 | 465 |
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.
|
import sys
import itertools
a, b, c, d = list(input())
print(a, b, c,d)
lst = ['-', '+']
for i, j, k in itertools.product(range(2), range(2), range(2)):
ans = int(a)
if i:
ans += int(b)
else:
ans -= int(b)
if j:
ans += int(c)
else:
ans -= int(c)
if k:
ans += int(d)
else:
ans -= int(d)
if ans == 7:
print(a,lst[i],b,lst[j],c,lst[k],d,'=7',sep='')
sys.exit()
|
s560699791
|
Accepted
| 32 | 9,148 | 442 |
import sys
import itertools
a, b, c, d = list(input())
lst = ['-', '+']
for i, j, k in itertools.product(range(2), range(2), range(2)):
ans = int(a)
if i:
ans += int(b)
else:
ans -= int(b)
if j:
ans += int(c)
else:
ans -= int(c)
if k:
ans += int(d)
else:
ans -= int(d)
if ans == 7:
print(a,lst[i],b,lst[j],c,lst[k],d,'=7',sep='')
break
|
s764338663
|
p03722
|
u899866702
| 2,000 | 262,144 |
Wrong Answer
| 1,225 | 8,232 | 621 |
There is a directed graph with N vertices and M edges. The i-th edge (1≤i≤M) points from vertex a_i to vertex b_i, and has a weight c_i. We will play the following single-player game using this graph and a piece. Initially, the piece is placed at vertex 1, and the score of the player is set to 0. The player can move the piece as follows: * When the piece is placed at vertex a_i, move the piece along the i-th edge to vertex b_i. After this move, the score of the player is increased by c_i. The player can end the game only when the piece is placed at vertex N. The given graph guarantees that it is possible to traverse from vertex 1 to vertex N. When the player acts optimally to maximize the score at the end of the game, what will the score be? If it is possible to increase the score indefinitely, print `inf`.
|
INF = float('inf')
def Bellmanford(n, e, s):
d = [INF]*n
d[s] = 0
for i in range(n):
print(i)
for e_from, e_to, e_cost in e:
if d[e_from] != INF and d[e_to] > d[e_from] + e_cost:
print("aaa")
d[e_to] = d[e_from] + e_cost
if i == n-1:
return 'inf'
return -d[n-1]
n,m = (int(x) for x in input().split())
edges = [None] * m
for i in range(m):
ai, bi, ci = map(int, input().split())
edges[i] = (ai-1, bi-1, -ci)
ans = Bellmanford(n, edges, 0)
print(ans)
|
s789558488
|
Accepted
| 590 | 3,316 | 603 |
INF = float('inf')
def Bellmanford(n, e, s):
d = [INF]*n
d[s] = 0
for i in range(n):
for e_from, e_to, e_cost in e:
if d[e_from] != INF and d[e_to] > d[e_from] + e_cost:
d[e_to] = d[e_from] + e_cost
if i == n-1 and e_to == n-1:
return 'inf'
return -d[n-1]
n,m = (int(x) for x in input().split())
edges = [None] * m
for i in range(m):
ai, bi, ci = map(int, input().split())
edges[i] = (ai-1, bi-1, -ci)
ans = Bellmanford(n, edges, 0)
print(ans)
|
s056857818
|
p03005
|
u047816928
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 53 |
Takahashi is distributing N balls to K persons. If each person has to receive at least one ball, what is the maximum possible difference in the number of balls received between the person with the most balls and the person with the fewest balls?
|
N, K = [int(c) for c in input().split()]
print(N-K-1)
|
s806413655
|
Accepted
| 17 | 2,940 | 83 |
N, K = [int(c) for c in input().split()]
if K==1:
print(0)
else:
print(N-K)
|
s617336418
|
p03693
|
u138486156
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 99 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r, g, b = map(int, input().split())
n = int(r + g + b)
if n % 4:
print("NO")
else:
print("YES")
|
s146003693
|
Accepted
| 17 | 2,940 | 86 |
r, g, b = input().split()
n = int(r+g+b)
if n % 4:
print("NO")
else:
print("YES")
|
s711944421
|
p03795
|
u062864034
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 54 |
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
N = int(input())
x = 800
y = 200*int(N/15)
print(x-y)
|
s989303132
|
Accepted
| 17 | 2,940 | 56 |
N = int(input())
x = 800*N
y = 200*int(N/15)
print(x-y)
|
s279713640
|
p03624
|
u366886346
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,188 | 185 |
You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead.
|
s=input()
sset=set(s)
list1=list(range(97,123))
for i in range(len(sset)):
str1=sset.pop()
list1.remove(ord(str1))
if len(list1)==0:
print("None")
else:
print(list1[0])
|
s631279658
|
Accepted
| 20 | 3,188 | 190 |
s=input()
sset=set(s)
list1=list(range(97,123))
for i in range(len(sset)):
str1=sset.pop()
list1.remove(ord(str1))
if len(list1)==0:
print("None")
else:
print(chr(list1[0]))
|
s729925195
|
p03644
|
u507116804
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 68 |
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n=int(input())
k=0
m=n
while m%2==0:
k+=1
m=m/2
print(k)
|
s957861239
|
Accepted
| 17 | 2,940 | 67 |
n=int(input())
k=0
while n//2>=1:
k+=1
n=n//2
print(2**k)
|
s439801902
|
p02936
|
u464244643
| 2,000 | 1,048,576 |
Wrong Answer
| 2,003 | 269,996 | 623 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
import sys
sys.setrecursionlimit(10**6)
def main():
def dfs(now, prev=-1):
for next in edge[now]:
if next == prev:
continue
score[next] += score[now]
dfs(next, now)
N, Q = map(int, input().split())
edge = [[] for _ in range(N+1)]
for _ in range(N-1):
u, v = map(int, input().split())
edge[u].append(v)
edge[v].append(u)
p = []
x = []
score = [0] * (N+1)
for _ in range(Q):
p, x = map(int, input().split())
score[p] += x
dfs(1)
print(score)
if __name__ == "__main__":
main()
|
s096976258
|
Accepted
| 1,591 | 268,428 | 682 |
import sys
sys.setrecursionlimit(10 ** 6)
def input():
return sys.stdin.readline()[:-1]
def main():
def dfs(now, prev=-1):
for next in edge[now]:
if next == prev:
continue
score[next] += score[now]
dfs(next, now)
N, Q = map(int, input().split())
edge = [[] for _ in range(N+1)]
for _ in range(N-1):
u, v = map(int, input().split())
edge[u].append(v)
edge[v].append(u)
p = []
x = []
score = [0] * (N+1)
for _ in range(Q):
p, x = map(int, input().split())
score[p] += x
dfs(1)
print(*score[1:])
if __name__ == "__main__":
main()
|
s953577476
|
p02742
|
u858742833
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 3,316 | 93 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
def main():
H, W = list(map(int, input().split()))
return min(H, W)
print(main())
|
s602988719
|
Accepted
| 18 | 2,940 | 140 |
def main():
H, W = list(map(int, input().split()))
if min(H, W) == 1:
return 1
return (H * W + 1) // 2
print(main())
|
s285964112
|
p02646
|
u441246928
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 9,164 | 153 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
A,V = map(int,input().split())
B,W = map(int,input().split())
T = int(input())
if abs(A + V*T) >= abs(B + W*T) :
print('Yes')
else :
print('No')
|
s427946088
|
Accepted
| 24 | 8,992 | 262 |
A,V = map(int,input().split())
B,W = map(int,input().split())
T = int(input())
if B > A :
if A + V*T >= B + W*T :
print('YES')
else :
print('NO')
elif A > B :
if B - W*T >= A - V*T :
print('YES')
else :
print('NO')
|
s599575409
|
p02865
|
u490489966
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 64 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n=int(input())
if n%2==0:
print(n/2-1)
else:
print(int(n/2))
|
s420221067
|
Accepted
| 17 | 2,940 | 73 |
n=int(input())
if n%2==0:
print(int(n/2-1))
else:
print(int((n-1)/2))
|
s809523421
|
p03493
|
u288040231
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 45 |
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
S = list(input().split())
print(S.count('1'))
|
s245038768
|
Accepted
| 17 | 2,940 | 29 |
s=input()
print(s.count('1'))
|
s054754824
|
p03855
|
u651803486
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 33 |
There are N cities. There are also K roads and L railways, extending between the cities. The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities. No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities. We will say city A and B are _connected by roads_ if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads. We will also define _connectivity by railways_ similarly. For each city, find the number of the cities connected to that city by both roads and railways.
|
print('1 2 2 1')
print('1 2 2 1')
|
s751761182
|
Accepted
| 1,676 | 53,240 | 1,330 |
from collections import defaultdict
class UnionFind():
def __init__(self, n):
self.par = [i for i in range(n)]
def root(self, x):
if self.par[x] == x:
return x
else:
self.par[x] = self.root(self.par[x])
return self.par[x]
def same(self, x, y):
return self.root(x) == self.root(y)
def unite(self, x, y):
x = self.root(x)
y = self.root(y)
if x == y:
return
self.par[x] = y
return
def main():
N, K, L = map(int, input().split())
uf1 = UnionFind(N)
uf2 = UnionFind(N)
for _ in range(K):
p, q = map(int, input().split())
uf1.unite(p-1, q-1)
# O(L)
for _ in range(L):
r, s = map(int, input().split())
uf2.unite(r-1, s-1)
# O(N) x O(log_N)
cnts = defaultdict(int)
for i in range(N):
pos = (uf1.root(i), uf2.root(i))
cnts[pos] += 1
ans = []
for i in range(N):
pos = (uf1.root(i), uf2.root(i))
ans.append(cnts[pos])
print(*ans)
if __name__ == '__main__':
main()
|
s077941667
|
p02936
|
u657994700
| 2,000 | 1,048,576 |
Wrong Answer
| 2,234 | 2,095,592 | 601 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
import numpy as np
N, Q = map(int, input().split())
tree = [[0] * N for i in range(N)]
for i in range(N-1):
a, b = map(int, input().split())
print(a, b)
tree[a-1][b-1] = 1
tree_np = np.array(tree)
for i, _ in enumerate(tree):
for node, _ in enumerate(tree[i]):
if tree[i][node] == 1:
tree[i] = np.array(tree[i]) + np.array(tree[node])
print(tree)
ans = np.array([0] * N)
for q in range(Q):
p, x = map(int, input().split())
ans += np.array(np.array(tree[p-1]) * x)
print(ans)
|
s184731055
|
Accepted
| 1,911 | 86,392 | 656 |
from collections import deque
N, Q = map(int, input().split())
edge = [[] for i in range(N)]
ops = []
ans = [0] * N
for _ in range(N-1):
a, b = map(int, input().split())
edge[a-1].append(b-1)
edge[b-1].append(a-1)
for _ in range(Q):
ops.append(tuple(map(int, input().split())))
for op in ops:
node = op[0]-1
pnt = op[1]
ans[node] += pnt
stack = deque()
stack.append((0, -1))
while stack:
st = stack.pop()
node, parent = st[0], st[1]
children = edge[node]
for child in children:
if child == parent:
continue
ans[child] += ans[node]
stack.append((child, node))
print(*ans)
|
s027801627
|
p03371
|
u127285813
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 171 |
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
A, B, C, X, Y = map(int, input().split())
C = min(C, A+B)
ans = 0
ans += C*min(X, Y)
if X > Y:
ans += min(C, A) * abs(X-Y)
else:
ans + min(C, B) * abs(X-Y)
print(ans)
|
s851258160
|
Accepted
| 17 | 3,060 | 175 |
A, B, C, X, Y = map(int, input().split())
C = min(2*C, A+B)
ans = 0
ans += C*min(X, Y)
if X > Y:
ans += min(C, A) * abs(X-Y)
else:
ans += min(C, B) * abs(X-Y)
print(ans)
|
s464028228
|
p02601
|
u143212659
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,112 | 352 |
M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.
|
#!/usr/bin/env python
# -*- coding: utf-8 -*-
def main():
A, B, C = map(int, input().split())
K = int(input())
print(A, B, C, K)
for _ in range(K):
if A > B:
B *= 2
elif B > C:
C *= 2
if A < B < C:
print("Yes")
else:
print("No")
if __name__ == "__main__":
main()
|
s552272812
|
Accepted
| 31 | 9,076 | 338 |
#!/usr/bin/env python
# -*- coding: utf-8 -*-
def main():
A, B, C = map(int, input().split())
K = int(input())
for _ in range(K):
if not A < B:
B *= 2
elif not B < C:
C *= 2
if A < B < C:
print("Yes")
else:
print("No")
if __name__ == "__main__":
main()
|
s805748337
|
p03478
|
u943117333
| 2,000 | 262,144 |
Wrong Answer
| 45 | 3,388 | 925 |
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).
|
from sys import stdin
n,a,b = [int(x) for x in stdin.readline().rstrip().split()]
sum = 0
for i in range(n+1):
i_sum = 0
i_num = i
print("i_num:"+str(i_num))
while(True):
if(i_num//10 ==0):
i_sum += (i_num%10)
break
else:
i_sum += (i_num%10)
i_num = i_num//10
if(a<= i_sum <= b):
print(i_sum)
sum += i
print("-------------------")
print(sum)
|
s261642340
|
Accepted
| 26 | 3,188 | 929 |
from sys import stdin
n,a,b = [int(x) for x in stdin.readline().rstrip().split()]
sum = 0
for i in range(n+1):
#print(i)
if (a<= i//10000 + (i-i//10000*10000)//1000 + (i-i//1000*1000)//100 + (i-i//100*100)//10 + i%10 <=b):
sum += i
print(sum)
|
s974923087
|
p03854
|
u323680411
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 280 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
S = input()
cnt = 0
j = len(S) - 1
for i in range(len(S)-1, -1, -1):
if j-1 > len("dreamer"):
break
for dd in ("dream", "dreamer", "erase", "eraser"):
if S[i:j+1] == dd:
j = i-1
cnt += len(dd)
print(("NO", "YES")[len(S) == cnt])
|
s360253095
|
Accepted
| 133 | 3,188 | 288 |
S = input()
cnt = 0
j = len(S) - 1
for i in range(len(S) - 1, -1, -1):
if j - i > len("dreamer"):
break
for dd in ("dream", "dreamer", "erase", "eraser"):
if S[i:j + 1] == dd:
j = i - 1
cnt += len(dd)
print(("NO", "YES")[len(S) == cnt])
|
s213812708
|
p03227
|
u811000506
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 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.
|
S = list(str(input()))
if len(S) == 2:
print(S[0],S[1])
else:
print(S[2],S[1],S[0])
|
s089507911
|
Accepted
| 17 | 2,940 | 73 |
S = str(input())
if len(S) == 2:
print(S)
else:
print(S[2]+S[1]+S[0])
|
s805359498
|
p03352
|
u923662841
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 3,060 | 123 |
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.
|
import math
X = int(input())
b = 1
for i in range(33,1,-1):
a = math.floor(math.log(X, i))
b = max(i**a,b)
print(b)
|
s789308141
|
Accepted
| 18 | 2,940 | 125 |
x=int(input())
c=1
for b in range(1,x):
for p in range(2,x):
if b**p<=x:c=max(c,b**p)
else:break
print(c)
|
s064206430
|
p03671
|
u448655578
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 75 |
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.
|
price = input().split(" ")
price = sorted(price)
print(price[0] + price[1])
|
s562210126
|
Accepted
| 17 | 2,940 | 74 |
price = list(map(int, input().split(" ")))
print(sum(price) - max(price))
|
s487957643
|
p03448
|
u912652535
| 2,000 | 262,144 |
Wrong Answer
| 41 | 3,060 | 285 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
a = int(input())
b = int(input())
c = int(input())
X = int(input())
count = 0
for i in range(1,a+1):
x = 500 * i
for j in range(1,b+1):
y = 100 * j
for k in range(1,c+1):
z = 50 * k
if x+y+z == X :
count += 1
print(count)
|
s856127569
|
Accepted
| 41 | 3,064 | 279 |
a = int(input())
b = int(input())
c = int(input())
X = int(input())
count = 0
for i in range(a+1):
x = 500 * i
for j in range(b+1):
y = 100 * j
for k in range(c+1):
z = 50 * k
if x+y+z == X :
count += 1
print(count)
|
s612270066
|
p03493
|
u982743577
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 20 |
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
print(list(input()))
|
s327797619
|
Accepted
| 17 | 2,940 | 31 |
print(list(input()).count('1'))
|
s316946058
|
p03401
|
u279493135
| 2,000 | 262,144 |
Wrong Answer
| 266 | 14,928 | 877 |
There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.
|
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7
N = INT()
A = LIST()
A.insert(0, 0)
A.append(0)
print(A)
ans = 0
for i in range(N+1):
ans += abs(A[i+1]-A[i])
for i in range(N):
if A[i]<=A[i+1]<=A[i+2] or A[i+2]<=A[i+1]<=A[i]:
print(ans)
else:
print(ans-2*
min(
abs(A[i+2]-A[i+1]),
abs(A[i+1]-A[i])
)
)
|
s867265524
|
Accepted
| 269 | 14,672 | 868 |
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7
N = INT()
A = LIST()
A.insert(0, 0)
A.append(0)
ans = 0
for i in range(N+1):
ans += abs(A[i+1]-A[i])
for i in range(N):
if A[i]<=A[i+1]<=A[i+2] or A[i+2]<=A[i+1]<=A[i]:
print(ans)
else:
print(ans-2*
min(
abs(A[i+2]-A[i+1]),
abs(A[i+1]-A[i])
)
)
|
s022045292
|
p03155
|
u546853743
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 8,992 | 78 |
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board. The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns. How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
|
n=int(input())
h=int(input())
w=int(input())
h -= (n-1)
w -= (n-1)
print(h*w)
|
s331181756
|
Accepted
| 27 | 9,144 | 76 |
n=int(input())
h=int(input())
w=int(input())
s = n-h+1
t = n-w+1
print(s*t)
|
s649388060
|
p03852
|
u626684023
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 160 |
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
s = input()
if s.replace("eraser"," ").replace("erase"," ").replace("dreamer"," ").replace("dream"," ").replace(" ", ""):
print("NO")
else:
print("YES")
|
s004736672
|
Accepted
| 17 | 2,940 | 93 |
l = input()
def vowel(l):
return "vowel" if l in "aeiou" else "consonant"
print(vowel(l))
|
s110319352
|
p03455
|
u054717609
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 110 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b=map(int,input().split())
c=a*b
if(c%2==0):
print("even")
else:
print("odd")
|
s362391236
|
Accepted
| 18 | 2,940 | 110 |
a,b=map(int,input().split())
c=a*b
if(c%2==0):
print("Even")
else:
print("Odd")
|
s309892137
|
p02396
|
u908651435
| 1,000 | 131,072 |
Wrong Answer
| 30 | 5,552 | 105 |
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
x=1
while True:
i=input()
print('case {}: {}'.format(x,i))
x+=1
if x==999:
break
|
s488572601
|
Accepted
| 70 | 5,984 | 115 |
import sys
i=1
for s in sys.stdin:
n = int(s)
if n == 0:
break
print("Case ",i,": ",n,sep="")
i += 1
|
s776187794
|
p00011
|
u071010747
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,360 | 242 |
Let's play Amidakuji. In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines. In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom. Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.
|
def main():
LIST=[]
for i in range(int(input())):
LIST.append(i+1)
for j in range(int(input())):
a,b=map(int,input().split(","))
LIST[a-1],LIST[b-1]=LIST[b-1],LIST[a-1]
for i in LIST:
print(i)
|
s131693188
|
Accepted
| 20 | 7,720 | 303 |
def main():
LIST=[]
for i in range(int(input())):
LIST.append(i+1)
for j in range(int(input())):
a,b=map(int,input().split(","))
LIST[a-1],LIST[b-1]=LIST[b-1],LIST[a-1]
for i in LIST:
print(str(i))
if __name__ == '__main__':
main()
|
s062786828
|
p03738
|
u319690708
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 122 |
You are given two positive integers A and B. Compare the magnitudes of these numbers.
|
A = int(input())
B = int(input())
print(A)
if A > B:
print("GREATER")
elif A<B:
print("LESS")
else:
print("EQUAL")
|
s081498547
|
Accepted
| 17 | 2,940 | 112 |
A = int(input())
B = int(input())
if A > B:
print("GREATER")
elif A<B:
print("LESS")
else:
print("EQUAL")
|
s486904540
|
p03698
|
u449555432
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 100 |
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s = list(input())
t = set(s)
print(s)
print(t)
if len(s) == len(t):
print('yes')
else:
print('no')
|
s894245965
|
Accepted
| 18 | 2,940 | 82 |
s = list(input())
t = set(s)
if len(s) == len(t):
print('yes')
else:
print('no')
|
s297856182
|
p03089
|
u174603263
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 258 |
Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
|
import math
n = int(input())
b = list(map(int, input().split()))
a = []
print(b)
for i in range(1, n+1):
if(b[i-1] <= i):
a.append(b[i-1])
else:
break
if(a != b):
print(-1)
else:
for i in range(len(a)):
print(a[i])
|
s087821140
|
Accepted
| 18 | 3,064 | 298 |
import math
n = int(input())
b = list(map(int, input().split()))
a = []
lis = []
for i in range(1, n+1):
if(b[i-1] <= i):
a.append(b[i-1])
lis.insert(b[i-1]-1, b[i-1])
else:
break
if(a != b):
print(-1)
else:
for i in range(len(a)):
print(lis[i])
|
s240539688
|
p03110
|
u375695365
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 3,060 | 200 |
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
n=int(input())
xy=[list(map(str,input().split())) for _ in range(n)]
ans=0
for i in range(n):
if xy[i][1]=="BTC":
ans+=38000*float(xy[i][0])
else:
ans+=int(xy[i][0])
print(ans)
|
s836654309
|
Accepted
| 18 | 3,060 | 203 |
n=int(input())
xy=[list(map(str,input().split())) for _ in range(n)]
ans=0
for i in range(n):
if xy[i][1]=="BTC":
ans+=380000*float(xy[i][0])
else:
ans+=float(xy[i][0])
print(ans)
|
s532777734
|
p03502
|
u069129582
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 102 |
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
n=int(input())
a=0
a+=n%10
b=n//1
while b>0:
a+=b%10
b=b//10
print('Yes' if n%a==0 else 'No')
|
s467451894
|
Accepted
| 18 | 2,940 | 103 |
n=int(input())
a=0
a+=n%10
b=n//10
while b>0:
a+=b%10
b=b//10
print('Yes' if n%a==0 else 'No')
|
s404529282
|
p03160
|
u424240341
| 2,000 | 1,048,576 |
Wrong Answer
| 122 | 14,660 | 360 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
def edpcA(n, hs):
hm2, hm1 = hs[0], hs[1]
dp = [0, abs(hm2-hm1)]
hs = list(reversed(hs[2:]))
for i in range(2, n):
h = hs.pop()
d1 = dp[i-1] + abs(h-hm1)
d2 = dp[i-2] + abs(h-hm2)
dp.append(min(d1, d2))
hm2, hm1 = hm1, h
print(dp)
n = int(input())
hs = list(map(int, input().split(' ')))
edpcA(n, hs)
|
s382375346
|
Accepted
| 112 | 13,980 | 364 |
def edpcA(n, hs):
hm2, hm1 = hs[0], hs[1]
dp = [0, abs(hm2-hm1)]
hs = list(reversed(hs[2:]))
for i in range(2, n):
h = hs.pop()
d1 = dp[i-1] + abs(h-hm1)
d2 = dp[i-2] + abs(h-hm2)
dp.append(min(d1, d2))
hm2, hm1 = hm1, h
print(dp[-1])
n = int(input())
hs = list(map(int, input().split(' ')))
edpcA(n, hs)
|
s645319546
|
p03377
|
u755180064
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 169 |
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.
|
def main():
t = list(map(int, input().split()))
if t[0] + t[1] > t[2]:
print('Yes')
else:
print('No')
if __name__ == '__main__':
main()
|
s936011807
|
Accepted
| 17 | 2,940 | 285 |
url = "https://atcoder.jp//contests/abc094/tasks/abc094_a"
def main():
t = list(map(int, input().split()))
for i in range(t[1]):
tmp = i + t[0]
if tmp == t[2]:
print('YES')
exit()
print('NO')
if __name__ == '__main__':
main()
|
s370269533
|
p00586
|
u858992370
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,500 | 186 |
Compute A + B.
|
import sys
val = []
count = 0
for line in sys.stdin:
for word in line.split():
val.append(int(word))
count = count + 1
if count == 1:
break
print (sum(val))
|
s462720483
|
Accepted
| 20 | 7,420 | 285 |
import sys
val = []
result = []
count = 0
for line in sys.stdin:
if line != '\n':
for word in line.split():
val.append(int(word))
result.append(sum(val))
val = []
count = count + 1
else:
break
for x in result:
print(x)
|
s188107206
|
p04044
|
u325264482
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 107 |
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
N, L = list(map(int, input().split()))
S = [input() for i in range(N)]
Ss = sorted(S)
print(','.join(Ss))
|
s341150768
|
Accepted
| 17 | 3,060 | 106 |
N, L = list(map(int, input().split()))
S = [input() for i in range(N)]
Ss = sorted(S)
print(''.join(Ss))
|
s413724139
|
p04029
|
u331464808
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 72 |
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())
res = 1
for i in range(n-1):
res = res + i
print(res)
|
s550741920
|
Accepted
| 17 | 2,940 | 70 |
n = int(input())
res = 0
for i in range(n+1):
res += i
print(res)
|
s431219700
|
p03737
|
u201082459
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 93 |
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
|
a,b,c = map(str,input().split())
print(a[0].capitalize(),b[0].capitalize(),c[0].capitalize())
|
s875109944
|
Accepted
| 17 | 2,940 | 93 |
a,b,c = map(str,input().split())
print(a[0].capitalize()+b[0].capitalize()+c[0].capitalize())
|
s949651466
|
p02393
|
u482227082
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,536 | 60 |
Write a program which reads three integers, and prints them in ascending order.
|
array = input().split()
array2= sorted(array)
print(array2)
|
s369246963
|
Accepted
| 20 | 5,588 | 129 |
#
# 2c
#
def main():
l = list(map(int, input().split()))
l.sort()
print(*l)
if __name__ == '__main__':
main()
|
s033897436
|
p02678
|
u318233626
| 2,000 | 1,048,576 |
Wrong Answer
| 992 | 40,376 | 772 |
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())
M = [[] for i in range(n)]
for i in range(m):
a, b = map(int, input().split())
a, b = a - 1, b - 1
M[a].append(b)
M[b].append(a)
#print(M)
R = [-1 for i in range(n)]
Q = deque([(0, -2)])
while len(Q) > 0:
t = Q[0][0]
b = Q[0][1]
print(t)
print(b)
Q.popleft()
if R[t] == -1:
R[t] = b
for i in range(len(M[t])):
if R[M[t][i]] == -1:
Q.append((M[t][i], t))
else:
pass
else:
pass
#print(R)
f = 0
for i in range(n):
if R[i] == -1:
f = 1
break
else:
pass
if f == 1:
print('No')
else:
print('Yes')
for i in range(1, n):
print(R[i] + 1)
|
s111543132
|
Accepted
| 788 | 34,868 | 677 |
from collections import deque
n, m = map(int, input().split())
TO = [[] for i in range(n)]
for i in range(m):
a, b = map(int, input().split())
a, b = a - 1, b -1
TO[a].append(b)
TO[b].append(a)
#BFS
Q = deque([])
DIST = [-1 for i in range(n)]
PRE = [-1 for i in range(n)]
DIST[0] = 0
Q.append(0)
while len(Q) > 0:
x = Q.popleft()
for i in range(len(TO[x])):
y = TO[x][i]
if DIST[y] != -1: continue
DIST[y] = DIST[x] + 1
PRE[y] = x
Q.append(y)
print('Yes')
for i in range(n):
if i == 0: continue
ans = PRE[i] + 1
print(ans)
|
s997548500
|
p03473
|
u345136423
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 24 |
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?
|
print(int(input()) + 24)
|
s310401475
|
Accepted
| 17 | 2,940 | 29 |
print(int(input()) *-1 + 48)
|
s269239790
|
p02850
|
u871980676
| 2,000 | 1,048,576 |
Wrong Answer
| 706 | 93,344 | 896 |
Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different. Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.
|
import sys
input = sys.stdin.readline
sys.setrecursionlimit(1000000)
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
N = int(input())
ab = [ tuple(map(int,input().split())) for i in range(N-1) ]
dic = {}
color = {}
root = [0]*N
for i in range(N):
dic[i+1] = []
for n in range(N-1):
i = ab[n][0]
j = ab[n][1]
dic[i].append(j)
dic[j].append(i)
root[i-1] += 1
root[j-1] += 1
color[(i,j)] = 0
root_node = root.index(1)+1
def rec(now_node,prevcolor):
nowcolor = 0
tg_list = dic[now_node][:]
for elem in tg_list:
nowcolor += 1
if nowcolor == prevcolor:
nowcolor += 1
dic[elem].remove(now_node)
color[(elem,now_node)] = nowcolor
color[(now_node,elem)] = nowcolor
rec(elem,nowcolor)
rec(root_node,0)
for n in range(N-1):
i = ab[n][0]
j = ab[n][1]
print(color[(i,j)])
|
s690606690
|
Accepted
| 737 | 93,380 | 921 |
import sys
input = sys.stdin.readline
sys.setrecursionlimit(1000000)
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
N = int(input())
ab = [ tuple(map(int,input().split())) for i in range(N-1) ]
dic = {}
color = {}
root = [0]*N
for i in range(N):
dic[i+1] = []
for n in range(N-1):
i = ab[n][0]
j = ab[n][1]
dic[i].append(j)
dic[j].append(i)
root[i-1] += 1
root[j-1] += 1
color[(i,j)] = 0
root_node = root.index(1)+1
ma = max(root)
def rec(now_node,prevcolor):
nowcolor = 0
tg_list = dic[now_node][:]
for elem in tg_list:
nowcolor += 1
if nowcolor == prevcolor:
nowcolor += 1
dic[elem].remove(now_node)
color[(elem,now_node)] = nowcolor
color[(now_node,elem)] = nowcolor
rec(elem,nowcolor)
rec(root_node,0)
print(ma)
for n in range(N-1):
i = ab[n][0]
j = ab[n][1]
print(color[(i,j)])
|
s166541795
|
p03657
|
u883465106
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,316 | 123 |
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= list(map(int,input().split()))
if (A+B%3)==0 or A%3==0 or B%3==0:
print("Possible")
else:
print("Impossibl")
|
s675197478
|
Accepted
| 17 | 2,940 | 124 |
A,B= list(map(int,input().split()))
if (A+B)%3==0 or A%3==0 or B%3==0:
print("Possible")
else:
print("Impossible")
|
s344539882
|
p03836
|
u614314290
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 161 |
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx, sy, tx, ty = map(int, input().split())
dx, dy = (ty-sy), (tx-sx)
print("U"*dx+"R"*dy+"D"*dy+"L"*dx+"L"+"U"*(dy+1)+"R"*(dx+1)+"DR"+"D"*(dy+1)+"L"*(dx+1)+"U")
|
s875479143
|
Accepted
| 18 | 3,060 | 161 |
sx, sy, tx, ty = map(int, input().split())
dx, dy = (tx-sx), (ty-sy)
print("U"*dy+"R"*dx+"D"*dy+"L"*dx+"L"+"U"*(dy+1)+"R"*(dx+1)+"DR"+"D"*(dy+1)+"L"*(dx+1)+"U")
|
s322275444
|
p03485
|
u771167374
| 2,000 | 262,144 |
Wrong Answer
| 18 | 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+1)
|
s862082621
|
Accepted
| 17 | 2,940 | 95 |
a, b = map(int, input().split())
if (a+b)%2!=0:
print((a+b)//2+1)
else:
print((a+b)//2)
|
s897615167
|
p03578
|
u348293370
| 2,000 | 262,144 |
Wrong Answer
| 2,206 | 39,376 | 268 |
Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems.
|
from os import remove
n = int(input())
d_list = list(map(int, input().split()))
m = int(input())
t_list = list(map(int, input().split()))
for i in range(m):
if t_list[i] in d_list:
t_list[i] = 0
else:
print("No")
exit()
print("Yes")
|
s001078064
|
Accepted
| 232 | 57,200 | 357 |
import collections
n = int(input())
d_list = list(map(int, input().split()))
m = int(input())
t_list = list(map(int, input().split()))
if m > n:
print("NO")
exit()
d_c = collections.Counter(d_list)
t_c = collections.Counter(t_list)
for i in t_c:
if t_c[i] <= d_c[i]:
continue
else:
print("NO")
exit()
print("YES")
|
s225175278
|
p03387
|
u131406572
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 190 |
You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.
|
a=list(map(int,input().split()))
a.sort()
b=a[2]-a[1]
c=a[1]-a[0]
if b%2==0 and c%2==0:
print((b//2)+(c//2))
elif b%2==1 and c%2==1:
print((b//2)+(c//2)+2)
else:
print((b//2)+(c//2)+2)
|
s636440424
|
Accepted
| 17 | 3,064 | 345 |
a=list(map(int,input().split()))
a.sort(reverse=True)
cnt=0
if (a[0]-a[1])%2==0:
cnt+=(a[0]-a[1])//2
if (a[0]-a[2])%2==0:
cnt+=(a[0]-a[2])//2
else:
cnt+=(a[0]-a[2])//2+2
else:
cnt+=(a[0]-a[1])//2+1
a[2]+=1
if (a[0]-a[2])%2==0:
cnt+=(a[0]-a[2])//2
else:
cnt+=(a[0]-a[2])//2+2
print(cnt)
|
s105577893
|
p03945
|
u698479721
| 2,000 | 262,144 |
Wrong Answer
| 68 | 3,188 | 120 |
Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa. In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white. Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.
|
S = input()
i = 0
ans = 0
while i < len(S)-1:
if S[i] == S[i+1]:
ans += 1
i += 1
else:
i += 1
print(ans)
|
s007579700
|
Accepted
| 64 | 3,188 | 120 |
S = input()
i = 0
ans = 0
while i < len(S)-1:
if S[i] != S[i+1]:
ans += 1
i += 1
else:
i += 1
print(ans)
|
s607515422
|
p03151
|
u377989038
| 2,000 | 1,048,576 |
Wrong Answer
| 401 | 31,460 | 347 |
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
import sys
input = sys.stdin.buffer.readline
n = int(input())
a = np.array(list(map(int, input().split())))
b = np.array(list(map(int, input().split())))
c = a - b
m = np.sum(c[c < 0])
p = np.sort(c[c >= 0])[::-1]
cnt = n - p.size
for i in p:
if m >= 0:
print(cnt)
exit()
m += i
cnt += 1
print(-1)
|
s098766456
|
Accepted
| 270 | 22,548 | 385 |
import numpy as np
import sys
input = sys.stdin.buffer.readline
n = int(input())
a = np.array(list(map(int, input().split())))
b = np.array(list(map(int, input().split())))
c = a - b
m = np.sum(c[c < 0])
p = np.sort(c[c >= 0])[::-1]
cnt = n - p.size
if cnt == 0:
print(0)
exit()
for i in p:
m += i
cnt += 1
if m >= 0:
print(cnt)
exit()
print(-1)
|
s355945345
|
p04043
|
u268623418
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 255 |
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.
|
num5 = 0
num7 = 0
for num in map(int, input().split()):
print(num)
if num == 5:
num5 += 1
elif num == 7:
num7 += 1
else:
print('NO')
exit()
if num5 == 2 and num7 == 1:
print('YES')
else:
print('NO')
|
s058097465
|
Accepted
| 17 | 2,940 | 241 |
num5 = 0
num7 = 0
for num in map(int, input().split()):
if num == 5:
num5 += 1
elif num == 7:
num7 += 1
else:
print('NO')
exit()
if num5 == 2 and num7 == 1:
print('YES')
else:
print('NO')
|
s904963472
|
p03448
|
u853064660
| 2,000 | 262,144 |
Wrong Answer
| 25 | 2,940 | 441 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
# A,B,C=map(int,input().split())
A,B,C,X=map(int,'30 30 30 12000'.split())
N=0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
if i*500+j*100+k*50==X:
N+=1
print(N)
|
s523578345
|
Accepted
| 48 | 3,060 | 215 |
A = int(input())
B = int(input())
C = int(input())
X = int(input())
N=0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
if i*500+j*100+k*50==X:
N+=1
print(N)
|
s709831778
|
p02401
|
u587193722
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,616 | 189 |
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.
|
x , y, z = input().split()
l = int(x)
r = int(z)
if y == '+':
print(l + r)
elif y == '-':
print(l - r)
elif y == '/':
print(l / r)
elif y == '*':
print(l * r)
else:
pass
|
s598748249
|
Accepted
| 20 | 7,652 | 255 |
while True:
x, y, z = input().split()
l = int(x)
r = int(z)
if y == '+':
print(l + r)
elif y == '-':
print(l - r)
elif y == '/':
print(l // r)
elif y == '*':
print(l * r)
else:
break
|
s330389907
|
p02292
|
u662418022
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,088 | 1,944 |
For given three points p0, p1, p2, print COUNTER_CLOCKWISE if p0, p1, p2 make a counterclockwise turn (1), CLOCKWISE if p0, p1, p2 make a clockwise turn (2), ONLINE_BACK if p2 is on a line p2, p0, p1 in this order (3), ONLINE_FRONT if p2 is on a line p0, p1, p2 in this order (4), ON_SEGMENT if p2 is on a segment p0p1 (5).
|
# -*- coding: utf-8 -*-
import collections
import math
class Vector2(collections.namedtuple("Vector2", ["x", "y"])):
def __add__(self, other):
return Vector2(self.x + other.x, self.y + other.y)
def __sub__(self, other):
return Vector2(self.x - other.x, self.y - other.y)
def __mul__(self, scalar):
return Vector2(self.x * scalar, self.y * scalar)
def __neg__(self):
return Vector2(-self.x, -self.y)
def __pos__(self):
return Vector2(+self.x, +self.y)
def __abs__(self): # norm
return math.sqrt(float(self.x * self.x + self.y * self.y))
def dot(self, other): # dot product
return self.x * other.x + self.y * other.y
def cross(self, other): # cross product
return self.x * other.y - self.y * other.x
def getDistanceSP(segment, point):
p = point
p1, p2 = segment
if (p2 - p1).dot(p - p1) < 0:
return abs(p - p1)
if (p1 - p2).dot(p - p2) < 0:
return abs(p - p2)
return abs((p2 - p1).cross(p - p1)) / abs(p2 - p1)
def getDistance(s1, s2):
if 0: # intersect
return 0
a, b = s1
c, d = s2
return min(getDistanceSP(s1, c), getDistanceSP(s1, d), getDistanceSP(s2, a), getDistanceSP(s2, b))
def ccw(p0, p1, p2):
a = p1 - p0
b = p2 - p0
if a.cross(b) > 0:
return 1
elif a.cross(b) < 0:
return -1
elif a.dot(b) < 0:
return 2
elif abs(a) < abs(b):
return -2
else:
return 0
if __name__ == '__main__':
a, b, c, d = map(int, input().split())
p0 = Vector2(a, b)
p1 = Vector2(c, d)
q = int(input())
ans = []
for _ in range(q):
e, f = map(int, input().split())
p2 = Vector2(e, f)
ans.append(ccw(p0, p1, p2))
dic = {1: "COUNTER_CLOCKWISE", -1: "CLOCLWISE",
2: "ONLINE_BACK", -2: "ONLINE_FRONT", 0: "ON_SEGMENT"}
for a in ans:
print(dic[a])
|
s273713727
|
Accepted
| 50 | 6,280 | 1,944 |
# -*- coding: utf-8 -*-
import collections
import math
class Vector2(collections.namedtuple("Vector2", ["x", "y"])):
def __add__(self, other):
return Vector2(self.x + other.x, self.y + other.y)
def __sub__(self, other):
return Vector2(self.x - other.x, self.y - other.y)
def __mul__(self, scalar):
return Vector2(self.x * scalar, self.y * scalar)
def __neg__(self):
return Vector2(-self.x, -self.y)
def __pos__(self):
return Vector2(+self.x, +self.y)
def __abs__(self): # norm
return math.sqrt(float(self.x * self.x + self.y * self.y))
def dot(self, other): # dot product
return self.x * other.x + self.y * other.y
def cross(self, other): # cross product
return self.x * other.y - self.y * other.x
def getDistanceSP(segment, point):
p = point
p1, p2 = segment
if (p2 - p1).dot(p - p1) < 0:
return abs(p - p1)
if (p1 - p2).dot(p - p2) < 0:
return abs(p - p2)
return abs((p2 - p1).cross(p - p1)) / abs(p2 - p1)
def getDistance(s1, s2):
if 0: # intersect
return 0
a, b = s1
c, d = s2
return min(getDistanceSP(s1, c), getDistanceSP(s1, d), getDistanceSP(s2, a), getDistanceSP(s2, b))
def ccw(p0, p1, p2):
a = p1 - p0
b = p2 - p0
if a.cross(b) > 0:
return 1
elif a.cross(b) < 0:
return -1
elif a.dot(b) < 0:
return 2
elif abs(a) < abs(b):
return -2
else:
return 0
if __name__ == '__main__':
a, b, c, d = map(int, input().split())
p0 = Vector2(a, b)
p1 = Vector2(c, d)
q = int(input())
ans = []
for _ in range(q):
e, f = map(int, input().split())
p2 = Vector2(e, f)
ans.append(ccw(p0, p1, p2))
dic = {1: "COUNTER_CLOCKWISE", -1: "CLOCKWISE",
2: "ONLINE_BACK", -2: "ONLINE_FRONT", 0: "ON_SEGMENT"}
for a in ans:
print(dic[a])
|
s108363618
|
p03778
|
u626337957
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 58 |
AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved.
|
W, A, B = map(int, input().split())
print(min(0, B-(A+W)))
|
s966684685
|
Accepted
| 17 | 2,940 | 102 |
W, A, B = map(int, input().split())
if A > B:
print(max(0, A-(B+W)))
else:
print(max(0, B-(A+W)))
|
s115634494
|
p01225
|
u724548524
| 5,000 | 131,072 |
Wrong Answer
| 30 | 5,608 | 498 |
あなたの友達は最近 UT-Rummy というカードゲームを思いついた． このゲームで使うカードには赤・緑・青のいずれかの色と1から9までのいずれかの番号が つけられている． このゲームのプレイヤーはそれぞれ9枚の手札を持ち， 自分のターンに手札から1枚選んで捨てて， 代わりに山札から1枚引いてくるということを繰り返す． このように順番にターンを進めていき， 最初に手持ちのカードに3枚ずつ3つの「セット」を作ったプレイヤーが勝ちとなる． セットとは，同じ色の3枚のカードからなる組で，すべて同じ数を持っているか 連番をなしているもののことを言う． 連番に関しては，番号の巡回は認められない． 例えば，7, 8, 9は連番であるが 9, 1, 2は連番ではない． あなたの友達はこのゲームをコンピュータゲームとして売り出すという計画を立てて， その一環としてあなたに勝利条件の判定部分を作成して欲しいと頼んできた． あなたの仕事は，手札が勝利条件を満たしているかどうかを判定する プログラムを書くことである．
|
for _ in range(int(input())):
n, c = map(int, input().split()), input().split()
c = list(zip(c, n))
c.sort(key = lambda x:x[1])
c.sort(key = lambda x:x[0])
print(c)
while c != []:
if c[-1] == c[-2] == c[-3]:
c = c[:-3]
elif (c[-1][0], c[-1][1] -1) in c and (c[-1][0], c[-1][1] -2) in c :
c.remove((c[-1][0], c[-1][1] -1))
c.remove((c[-1][0], c[-1][1] -2))
c.pop()
else:print(0);break
else:print(1)
|
s160354480
|
Accepted
| 20 | 5,612 | 485 |
for _ in range(int(input())):
n, c = map(int, input().split()), input().split()
c = list(zip(c, n))
c.sort(key = lambda x:x[1])
c.sort(key = lambda x:x[0])
while c != []:
if c[-1] == c[-2] == c[-3]:
c = c[:-3]
elif (c[-1][0], c[-1][1] -1) in c and (c[-1][0], c[-1][1] -2) in c :
c.remove((c[-1][0], c[-1][1] -1))
c.remove((c[-1][0], c[-1][1] -2))
c.pop()
else:print(0);break
else:print(1)
|
s791089862
|
p03693
|
u427690532
| 2,000 | 262,144 |
Wrong Answer
| 29 | 8,992 | 177 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
S_list = list(map(int,input().split()))
r, g, b = S_list[0], S_list[1], S_list[2]
if (100 * r + 10 * g + b) % 4 == 0 :
result = "Yes"
else:
result = "No"
print(result)
|
s321962910
|
Accepted
| 27 | 9,164 | 114 |
a, b, c = map(str, input().split())
num = int(a + b + c)
if num % 4 == 0:
print("YES")
else:
print("NO")
|
s517047990
|
p03409
|
u102902647
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,064 | 904 |
On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i). A red point and a blue point can form a _friendly pair_ when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point. At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.
|
from operator import itemgetter
# N = 5
# R = [[0, 0],
# [7, 3],
# [2, 2],
# [4, 8],
# [1, 6]]
# B = [[8, 5],
# [6, 9],
# [5, 4],
# [9, 1],
# [3, 7]]
N = int(input())
R = []
for i in range(N):
a, b = map(int, input().split())
R.append([a, b])
B = []
for i in range(N):
c, d = map(int, input().split())
B.append([c, d])
R.sort(key=itemgetter(0))
B.sort(key=itemgetter(0))
print(R)
print(B)
res = 0
for r in R:
rx, ry = r[0], r[1]
flag1 = False
flag2 = False
for i, b in enumerate(B):
if b[0] > rx:
flag1 = True
break
if flag1:
B_tmp = B.copy()[i:]
B_tmp.sort(key=itemgetter(1))
for ii, b in enumerate(B_tmp):
if b[1] > ry:
flag2 = True
break
if flag2:
res += 1
B.remove(b)
print(res)
|
s485489772
|
Accepted
| 19 | 3,064 | 914 |
from operator import itemgetter
# N = 5
# R = [[0, 0],
# [7, 3],
# [2, 2],
# [4, 8],
# [1, 6]]
# B = [[8, 5],
# [6, 9],
# [5, 4],
# [9, 1],
# [3, 7]]
N = int(input())
R = []
for i in range(N):
a, b = map(int, input().split())
R.append([a, b])
B = []
for i in range(N):
c, d = map(int, input().split())
B.append([c, d])
R.sort(key=itemgetter(0))
B.sort(key=itemgetter(0))
# print(R)
res = 0
for r in R[::-1]:
rx, ry = r[0], r[1]
flag1 = False
flag2 = False
for i, b in enumerate(B):
if b[0] > rx:
flag1 = True
break
if flag1:
B_tmp = B.copy()[i:]
B_tmp.sort(key=itemgetter(1))
for ii, b in enumerate(B_tmp):
if b[1] > ry:
flag2 = True
break
if flag2:
res += 1
B.remove(b)
print(res)
|
s032462251
|
p03415
|
u252964975
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 72 |
We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}. Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.
|
S1=str(input())
S2=str(input())
S3=str(input())
print(S1[0]+S1[1]+S2[2])
|
s859777356
|
Accepted
| 18 | 2,940 | 73 |
S1=str(input())
S2=str(input())
S3=str(input())
print(S1[0]+S2[1]+S3[2])
|
s431428133
|
p04046
|
u102367647
| 2,000 | 262,144 |
Wrong Answer
| 187 | 15,120 | 1,152 |
We have a large square grid with H rows and W columns. Iroha is now standing in the top-left cell. She will repeat going right or down to the adjacent cell, until she reaches the bottom-right cell. However, she cannot enter the cells in the intersection of the bottom A rows and the leftmost B columns. (That is, there are A×B forbidden cells.) There is no restriction on entering the other cells. Find the number of ways she can travel to the bottom-right cell. Since this number can be extremely large, print the number modulo 10^9+7.
|
import sys
import numpy as np
import random
from decimal import Decimal
import itertools
import re
import bisect
from collections import deque, Counter
from functools import lru_cache
sys.setrecursionlimit(10**9)
INF = 10**13
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def I(): return int(sys.stdin.buffer.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
def SERIES(n): return np.fromstring(sys.stdin.buffer.read(), dtype=np.int32, sep=' ')
def GRID(h,w): return np.fromstring(sys.stdin.buffer.read(), dtype=np.int32, sep=' ').reshape(h,-1)[:,:w]
def GRIDfromString(h,w): return np.frombuffer(sys.stdin.buffer.read(), 'S1').reshape(h,-1)[:,:w]
MOD = 1000000007
def main():
n = LI()
if __name__ == '__main__':
main()
|
s505323631
|
Accepted
| 566 | 38,192 | 1,671 |
import sys
import numpy as np
import random
from decimal import Decimal
import itertools
import re
import bisect
from collections import deque, Counter
from functools import lru_cache
sys.setrecursionlimit(10**9)
INF = 10**13
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def I(): return int(sys.stdin.buffer.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
def SERIES(n): return np.fromstring(sys.stdin.buffer.read(), dtype=np.int32, sep=' ')
def GRID(h,w): return np.fromstring(sys.stdin.buffer.read(), dtype=np.int32, sep=' ').reshape(h,-1)[:,:w]
def GRIDfromString(h,w): return np.frombuffer(sys.stdin.buffer.read(), 'S1').reshape(h,-1)[:,:w]
MOD = 1000000007
def main():
h, w, a, b = LI()
def comb(n, r, p):
if (r < 0) or (n < r):
return 0
r = min(r, n - r)
return fact[n] * factinv[r] * factinv[n-r] % p
p = 10 ** 9 + 7
N = 10 ** 5 * 2
fact = [1, 1]
factinv = [1, 1]
inv = [0, 1]
for i in range(2, N + 1):
fact.append((fact[-1] * i) % p)
inv.append((-inv[p % i] * (p // i)) % p)
factinv.append((factinv[-1]) * inv[-1] % p)
ans = 0
for i in range(1,w-b+1):
ans += comb(h-a+b+i-2, max(h-a-1, b+i-1), MOD) * comb(a+w-b-i-1, max(a-1, w-b-i), MOD)
ans %= MOD
print(ans)
if __name__ == '__main__':
main()
|
s535506338
|
p02821
|
u163320134
| 2,000 | 1,048,576 |
Wrong Answer
| 1,951 | 16,580 | 513 |
Takahashi has come to a party as a special guest. There are N ordinary guests at the party. The i-th ordinary guest has a _power_ of A_i. Takahashi has decided to perform M _handshakes_ to increase the _happiness_ of the party (let the current happiness be 0). A handshake will be performed as follows: * Takahashi chooses one (ordinary) guest x for his left hand and another guest y for his right hand (x and y can be the same). * Then, he shakes the left hand of Guest x and the right hand of Guest y simultaneously to increase the happiness by A_x+A_y. However, Takahashi should not perform the same handshake more than once. Formally, the following condition must hold: * Assume that, in the k-th handshake, Takahashi shakes the left hand of Guest x_k and the right hand of Guest y_k. Then, there is no pair p, q (1 \leq p < q \leq M) such that (x_p,y_p)=(x_q,y_q). What is the maximum possible happiness after M handshakes?
|
n,m=map(int,input().split())
arr=list(map(int,input().split()))
arr=sorted(arr,reverse=True)
acum=[arr[0]]
for i in range(1,n):
acum.append(acum[-1]+arr[i])
print(acum)
l=0
r=2*10**5+1
while r-l!=1:
mid=(l+r)//2
tmp=0
cnt=0
pos=n-1
for i in range(n):
while pos!=-1:
if arr[i]+arr[pos]>=mid:
cnt+=pos+1
tmp+=arr[i]*(pos+1)+acum[pos]
break
else:
pos-=1
if pos==-1:
break
if cnt<=m:
r=mid
else:
l=mid
tmp+=(m-cnt)*l
print(tmp)
|
s920720453
|
Accepted
| 1,862 | 14,152 | 501 |
n,m=map(int,input().split())
arr=list(map(int,input().split()))
arr=sorted(arr,reverse=True)
acum=[arr[0]]
for i in range(1,n):
acum.append(acum[-1]+arr[i])
l=0
r=2*10**5+1
while r-l!=1:
mid=(l+r)//2
tmp=0
cnt=0
pos=n-1
for i in range(n):
while pos!=-1:
if arr[i]+arr[pos]>=mid:
cnt+=pos+1
tmp+=arr[i]*(pos+1)+acum[pos]
break
else:
pos-=1
if pos==-1:
break
if cnt<=m:
r=mid
else:
l=mid
tmp+=(m-cnt)*l
print(tmp)
|
s562203615
|
p03605
|
u871596687
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 84 |
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
N = str(input())
if N[0] == 9 or N[1] == 9:
print("Yes")
else:
print("No")
|
s497995774
|
Accepted
| 17 | 2,940 | 88 |
N = str(input())
if N[0] == "9" or N[1] == "9":
print("Yes")
else:
print("No")
|
s537246343
|
p02255
|
u564464686
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 230 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
N=int(input())
A=[0 for i in range(N)]
A=input().split()
for i in range(N):
A[i]=(int)(A[i])
for i in range(N):
v=A[i]
j = i-1
while j>=0 and A[j]>v:
A[j+1]=A[j]
j=j-1
A[j+1]=v
print(A)
|
s915622874
|
Accepted
| 30 | 5,984 | 333 |
N=int(input())
A=[0 for i in range(N)]
A=input().split()
for i in range(N):
A[i]=(int)(A[i])
for i in range(N):
v=A[i]
j = i-1
while j>=0 and A[j]>v:
A[j+1]=A[j]
j=j-1
A[j+1]=v
for k in range(N):
if k<N-1:
print(A[k],end=" ")
if k==N-1:
print(A[k])
|
s323021603
|
p02665
|
u137226361
| 2,000 | 1,048,576 |
Wrong Answer
| 827 | 20,056 | 262 |
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
|
n= int(input())
a = list(map(int, input().split()))
#print(a)
if a[0] != 0:
print(-1)
exit(0)
b=1
count=1
for i in range(n-1):
if a[i+1] > 2*b:
print(-1)
exit(0)
else:
b= 2*b-a[i+1]
count += 2*b
print(count)
|
s301245662
|
Accepted
| 109 | 20,144 | 367 |
n= int(input())
a = list(map(int, input().split()))
if n == 0 and a[0]==1:
print(1)
exit(0)
if a[0] != 0:
print(-1)
exit(0)
b=1
count=1
asum =sum(a)
alim = asum
for i in range(n):
if a[i+1] > 2*b:
print(-1)
exit(0)
else:
alim = alim - a[i+1]
b= min(2*b-a[i+1], alim)
count += b
print(count+ asum)
|
s484048533
|
p03555
|
u858136677
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 101 |
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
a = input()
b = input()
if a[0]==b[2] and a[1]==b[1] and a[2]==b[0]:
print('Yes')
else:
print('No')
|
s034943850
|
Accepted
| 17 | 2,940 | 102 |
a = input()
b = input()
if a[0]==b[2] and a[1]==b[1] and a[2]==b[0]:
print('YES')
else:
print('NO')
|
s446928108
|
p02261
|
u022407960
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,712 | 1,657 |
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
#!/usr/bin/env python
# encoding: utf-8
class Solution:
def stable_sort(self):
# write your code here
array_length = int(input())
array = [str(x) for x in input().split()]
bubble_result = self.bubble_sort(array=array, array_length=array_length)
selection_result = self.selection_sort(array=array, array_length=array_length)
# check whether selection sort is stable
if bubble_result == selection_result:
print("Stable")
else:
print("Not Stable")
@staticmethod
def selection_sort(array, array_length):
# selection sort
selection_count = 0
for i in range(array_length):
min_j = i
for j in range(i, array_length):
if array[j][1] < array[min_j][1]:
min_j = j
array[i], array[min_j] = array[min_j], array[i]
if i != min_j:
selection_count += 1
result = " ".join(map(str, array))
print(result)
return result
@staticmethod
def bubble_sort(array, array_length):
flag, bubble_count, cursor = 1, 0, 0
while flag:
flag = 0
for j in range(array_length - 1, cursor, -1):
if array[j][1] < array[j - 1][1]:
array[j], array[j - 1] = array[j - 1], array[j]
flag = 1
bubble_count += 1
cursor += 1
result = " ".join(map(str, array))
print(result)
print("Stable")
return result
if __name__ == '__main__':
solution = Solution()
solution.stable_sort()
|
s944005336
|
Accepted
| 30 | 8,152 | 1,711 |
#!/usr/bin/env python
# encoding: utf-8
import copy
class Solution:
def stable_sort(self):
# write your code here
array_length = int(input())
array = [str(x) for x in input().split()]
array_2 = copy.deepcopy(array)
bubble_result = self.bubble_sort(array=array, array_length=array_length)
selection_result = self.selection_sort(array=array_2, array_length=array_length)
# check whether selection sort is stable
if bubble_result == selection_result:
print("Stable")
else:
print("Not stable")
@staticmethod
def selection_sort(array, array_length):
# selection sort
selection_count = 0
for i in range(array_length):
min_j = i
for j in range(i, array_length):
if array[j][1] < array[min_j][1]:
min_j = j
array[i], array[min_j] = array[min_j], array[i]
if i != min_j:
selection_count += 1
result = " ".join(map(str, array))
print(result)
return result
@staticmethod
def bubble_sort(array, array_length):
flag, bubble_count, cursor = 1, 0, 0
while flag:
flag = 0
for j in range(array_length - 1, cursor, -1):
if array[j][1] < array[j - 1][1]:
array[j], array[j - 1] = array[j - 1], array[j]
flag = 1
bubble_count += 1
cursor += 1
result = " ".join(map(str, array))
print(result)
print("Stable")
return result
if __name__ == '__main__':
solution = Solution()
solution.stable_sort()
|
s734605589
|
p03738
|
u172780602
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 127 |
You are given two positive integers A and B. Compare the magnitudes of these numbers.
|
a=input()
b=input()
if len(a)>len(b):
print("GREATER")
elif len(a)<len(b):
print("LESS")
else:
print("EQUAL")
|
s959484871
|
Accepted
| 17 | 2,940 | 112 |
a=int(input())
b=int(input())
if a>b:
print("GREATER")
elif a<b:
print("LESS")
else:
print("EQUAL")
|
s208564812
|
p02742
|
u909224749
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,092 | 98 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
H, W = map(int, input().split())
if H%2==0 or W%2==0:
ans = int(H*W/2)
else:
ans = (H*W//2)+1
|
s718663382
|
Accepted
| 25 | 9,056 | 171 |
H, W = map(int, input().split())
if H==1 or W==1:
ans = 1
elif H%2==0 or W%2==0:
ans = int(H*W/2)
else:
ans = int((H*W+1)/2)
print(ans)
|
s070443376
|
p03779
|
u677523557
| 2,000 | 262,144 |
Wrong Answer
| 30 | 4,884 | 126 |
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
X = int(input())
A = [0]
c = 0
while True:
c += 1
k = A[-1] + c
if k > X:
break
A.append(k)
print(c)
|
s268261542
|
Accepted
| 31 | 4,884 | 127 |
X = int(input())
A = [0]
c = 0
while True:
c += 1
k = A[-1] + c
if k >= X:
break
A.append(k)
print(c)
|
s169833438
|
p03386
|
u232852711
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 179 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a, b, k = list(map(int, input().split()))
ans = set([])
for i in range(a, min(a+k, b+1)):
ans.add(i)
for i in range(max(a, b-k), b):
ans.add(i)
for a in ans:
print(a)
|
s197489923
|
Accepted
| 17 | 3,060 | 197 |
a, b, k = list(map(int, input().split()))
ans = set([])
for i in range(a, min(a+k, b+1)):
ans.add(i)
for i in range(max(a, b-k+1), b+1):
ans.add(i)
for a in sorted(list(ans)):
print(a)
|
s350947578
|
p03795
|
u324549724
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 66 |
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
import math
n = int(input())
print(math.factorial(n)%1000000007)
|
s861154889
|
Accepted
| 18 | 2,940 | 40 |
n = int(input())
print(n*800-n//15*200)
|
s035783899
|
p02612
|
u895408600
| 2,000 | 1,048,576 |
Wrong Answer
| 32 | 9,148 | 46 |
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())
ot = N//1000
print(N-ot*1000)
|
s018035341
|
Accepted
| 28 | 9,152 | 91 |
N = int(input())
ot = N//1000
if N%1000 == 0:
print(0)
else:
print((ot+1)*1000 - N)
|
s567279379
|
p02678
|
u993435350
| 2,000 | 1,048,576 |
Wrong Answer
| 2,212 | 130,372 | 701 |
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.
|
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10 ** 6)
N,M = map(int,input().split())
path = [[] for i in range(N)]
inf = 10 ** 6
ans = [[None,inf] for i in range(N - 1)]
for _ in range(M):
a,b = map(int,input().split())
a = a - 1
b = b - 1
if a > b:
a,b = b,a
path[a].append(b)
path[b].append(a)
print(path)
def dfs(cur,par,con):
if con == N:
return
for chi in path[cur]:
if chi == par:
continue
if con < ans[chi - 1][1]:
ans[chi - 1][0] = cur
ans[chi - 1][1] = con + 1
dfs(chi,cur,con + 1)
dfs(0,-1,0)
if all([i[0] != None for i in ans]):
print("Yes")
for i in range(N - 1):
print(ans[i][0] + 1)
else:
print("No")
|
s925981143
|
Accepted
| 451 | 38,276 | 770 |
from collections import deque
import sys
input = sys.stdin.readline
N,M = map(int,input().split())
path = [[] for i in range(N)]
ans = [None for i in range(N)]
visited = [0 for i in range(N)]
for _ in range(M):
a,b = map(int,input().split())
a = a - 1
b = b - 1
path[a].append(b)
path[b].append(a)
def bfs(path, root,visited):
visited[root] = 1
queue = deque([root])
while queue:
cur = queue.popleft()
for chi in path[cur]:
if visited[chi] == 1:
continue
else:
ans[chi] = cur + 1
visited[chi] = 1
queue.append(chi)
if all([i == 1 for i in visited]):
return "Yes"
else:
return "No"
flag = bfs(path,0,visited)
print(flag)
if flag == "Yes":
for i in range(1,N):
print(ans[i])
|
s277581600
|
p03828
|
u382423941
| 2,000 | 262,144 |
Wrong Answer
| 43 | 3,700 | 417 |
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.
|
import functools
mod = 1e9 + 7
n = int(input())
fact = [1] * (n+1)
for i in range(2, n+1):
for j in range(2, 32):
if i == 0:
break
while True:
if i % j == 0:
i /= j
fact[j] += 1
else:
break
if i != 0:
fact[int(i)] += 1
print(fact)
print(int(functools.reduce(lambda x, y: (x * y) % mod, fact[2:])))
|
s494189350
|
Accepted
| 38 | 3,700 | 456 |
import sys
import functools
mod = 1e9 + 7
n = int(input())
if n == 1:
print(1)
sys.exit()
fact = [1] * (n+1)
for i in range(2, n+1):
for j in range(2, 32):
if i == 0:
break
while True:
if i % j == 0:
i /= j
fact[j] += 1
else:
break
if i != 0:
fact[int(i)] += 1
print(int(functools.reduce(lambda x, y: (x * y) % mod, fact[2:])))
|
s842855549
|
p03303
|
u546074985
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 248 |
You are given a string S consisting of lowercase English letters. We will write down this string, starting a new line after every w letters. Print the string obtained by concatenating the letters at the beginnings of these lines from top to bottom.
|
st = input()
skpe = int(input())
st_len = len(st)
count = st_len // 3 + 1
top_list = []
for i in range(count):
top_list.append(st[3 * i:3 * i + 3])
for x in top_list:
try:
print(x[0], end="")
except IndexError:
pass
|
s894692382
|
Accepted
| 18 | 3,060 | 266 |
st = input()
skpe = int(input())
st_len = len(st)
count = (st_len // skpe) + 1
top_list = []
s = ""
for i in range(count):
top_list.append(st[skpe * i:skpe * i + skpe])
for x in top_list:
try:
s += x[0]
except IndexError:
pass
print(s)
|
s338367869
|
p03110
|
u225110485
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 258 |
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
N = int(input())
xu = [[] for i in range(N)]
otoshidama = 0
for i in range(N):
xu[i] = list(input().split())
print(xu)
for i in xu:
if i[1] == 'JPY':
otoshidama += int(i[0])
else:
otoshidama += float(i[0])*380000
print(otoshidama)
|
s833499491
|
Accepted
| 17 | 3,060 | 248 |
N = int(input())
xu = [[] for i in range(N)]
otoshidama = 0
for i in range(N):
xu[i] = list(input().split())
for i in xu:
if i[1] == 'JPY':
otoshidama += int(i[0])
else:
otoshidama += float(i[0])*380000
print(otoshidama)
|
s527385730
|
p03565
|
u571867512
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 764 |
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
S_prime = input()
T = input()
def judge(A,B):
for i in range(len(A)):
if A[i] != B[i] and A[i] != "?" and B[i] != "?":
return False
return True
def dec(S_prime,T):
S_cand = []
for i in range(len(S_prime)-len(T)+1):
if judge(S_prime[i:i+len(T)],T):
S_cand.append( S_prime[:i] + T + S_prime[i+len(T):] )
print(i,S_prime[:i], T, S_prime[i+len(T):])
return S_cand
def S_change(S):
S = list(S)
for i in range(len(S)):
if S[i] == "?":
S[i] = "a"
S = "".join(S)
return S
if len(dec(S_prime,T)) == 0:
print("UNRESTORABLE")
else:
Ans = dec(S_prime,T)
for i in range(len(Ans)):
Ans[i] = S_change(Ans[i])
Ans.sort()
print(Ans[0])
|
s641780462
|
Accepted
| 18 | 3,064 | 703 |
S_prime = input()
T = input()
def judge(A,B):
for i in range(len(A)):
if A[i] != B[i] and A[i] != "?" and B[i] != "?":
return False
return True
def dec(S_prime,T):
S_cand = []
for i in range(len(S_prime)-len(T)+1):
if judge(S_prime[i:i+len(T)],T):
S_cand.append( S_prime[:i] + T + S_prime[i+len(T):] )
return S_cand
def S_change(S):
S = list(S)
for i in range(len(S)):
if S[i] == "?":
S[i] = "a"
S = "".join(S)
return S
if len(dec(S_prime,T)) == 0:
print("UNRESTORABLE")
else:
Ans = dec(S_prime,T)
for i in range(len(Ans)):
Ans[i] = S_change(Ans[i])
Ans.sort()
print(Ans[0])
|
s303613397
|
p02397
|
u692415695
| 1,000 | 131,072 |
Wrong Answer
| 40 | 7,996 | 189 |
Write a program which reads two integers x and y, and prints them in ascending order.
|
# (c) midandfeed
q = []
while(True):
a, b = [int(x) for x in input().split()]
if ( (a==0) and (b==0) ):
break
q.append((b,a))
q.sort()
for x in q:
print("{} {}".format(x[0], x[1]))
|
s071266149
|
Accepted
| 60 | 7,588 | 135 |
# (c) midandfeed
while(True):
a, b = [int(x) for x in input().split()]
if ( (a==0) and (b==0) ):
break
print( min(a,b), max(a,b) )
|
s002261938
|
p02663
|
u327465093
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,080 | 82 |
In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?
|
h1, m1, h2, m2, k = map(int, input().split())
t1 = h1 * 60 + m1
t2 = h2 * 60 + m1
|
s691759263
|
Accepted
| 23 | 9,164 | 99 |
h1, m1, h2, m2, k = map(int, input().split())
t1 = h1 * 60 + m1
t2 = h2 * 60 + m2
print(t2-t1 - k)
|
s078951655
|
p03827
|
u620846115
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,136 | 158 |
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
n = int(input())
a = list(input())
xlist = []
x = 0
for i in range(n):
if a[i]=="D":
x+=1
elif a[i]=="I":
x-=1
xlist.append(x)
print(max(xlist))
|
s895916692
|
Accepted
| 25 | 9,128 | 159 |
n = int(input())
a = list(input())
xlist = [0]
x = 0
for i in range(n):
if a[i]=="D":
x-=1
elif a[i]=="I":
x+=1
xlist.append(x)
print(max(xlist))
|
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