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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s029447191
|
p03385
|
u279493135
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 69 |
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
S = input()
if sorted(S) == "abc":
print("Yes")
else:
print("No")
|
s628080388
|
Accepted
| 17 | 2,940 | 78 |
S = input()
if "".join(sorted(S)) == "abc":
print("Yes")
else:
print("No")
|
s898357660
|
p02392
|
u248424983
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,512 | 82 |
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
a,b,c = input().split(' ')
a=int(a)
b=int(b)
c=int(c)
ret="Yes" if a<b<c else "No"
|
s258868827
|
Accepted
| 30 | 7,660 | 93 |
a,b,c = input().split(' ')
a=int(a)
b=int(b)
c=int(c)
ret="Yes" if a<b<c else "No"
print(ret)
|
s955210711
|
p04012
|
u958506960
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 75 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
w = input()
c = w.count('w')
print('Yes' if c % 2 == 0 and c > 0 else 'No')
|
s508129480
|
Accepted
| 21 | 3,316 | 163 |
from collections import Counter
w = input()
c = Counter(w)
cnt = 0
for i in c.values():
if i % 2 == 0:
cnt += 1
print('Yes' if cnt == len(c) else 'No')
|
s515781677
|
p02256
|
u853619096
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,640 | 124 |
Write a program which finds the greatest common divisor of two natural numbers _a_ and _b_
|
a,b=map(int,input().split())
z=[]
for i in range(min(a,b)):
if a%(i+1)==0 and b%(i+1)==0:
z+=[i+1]
print(min(z))
|
s915406775
|
Accepted
| 20 | 5,600 | 132 |
def gcd(a,b):
if b==0:
return a
else:
return gcd(b,a%b)
a,b=list(map(int,input().split()))
print(gcd(a,b))
|
s571629600
|
p04043
|
u265118937
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 132 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a,b,c=map(int,input().split())
List=[a,b,c]
if List.count("5") == 2 and List.count("7") ==1:
print("YES")
else:
print("NO")
|
s776438103
|
Accepted
| 17 | 3,060 | 326 |
a, b, c = map(int, input().split())
if b == 7:
if a == 5 and c == 5:
print("YES")
else:
print("NO")
elif c == 7:
if a == 5 and b == 5:
print("YES")
else:
print("NO")
elif a == 7 :
if b == 5 and c == 5:
print("YES")
else:
print("NO")
else:
print("NO")
|
s369418241
|
p03719
|
u239342230
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 52 |
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
print(['No','Yes'][eval(input().replace(' ','<='))])
|
s017013050
|
Accepted
| 18 | 2,940 | 59 |
a,b,c=map(int,input().split());print(['No','Yes'][a<=c<=b])
|
s922608133
|
p03826
|
u679817762
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,144 | 251 |
There are two rectangles. The lengths of the vertical sides of the first rectangle are A, and the lengths of the horizontal sides of the first rectangle are B. The lengths of the vertical sides of the second rectangle are C, and the lengths of the horizontal sides of the second rectangle are D. Print the area of the rectangle with the larger area. If the two rectangles have equal areas, print that area.
|
lst = input().split()
for i in range(len(lst)):
lst[i] = int(lst[i])
area1 = lst[0] * lst[1]
area2 = lst[2] * lst[3]
areas = [area1, area2]
areas.sort()
print(areas[0])
|
s113190914
|
Accepted
| 29 | 9,152 | 251 |
lst = input().split()
for i in range(len(lst)):
lst[i] = int(lst[i])
area1 = lst[0] * lst[1]
area2 = lst[2] * lst[3]
areas = [area1, area2]
areas.sort()
print(areas[1])
|
s985390984
|
p02600
|
u107267797
| 2,000 | 1,048,576 |
Wrong Answer
| 33 | 9,184 | 346 |
M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * From 1000 through 1199: 5-kyu * From 1200 through 1399: 4-kyu * From 1400 through 1599: 3-kyu * From 1600 through 1799: 2-kyu * From 1800 through 1999: 1-kyu What kyu does M-kun have?
|
N = int(input())
if N >=400 or N <= 599:
print(8)
elif N >=600 or N <= 799:
print(7)
elif N >=800 or N <= 999:
print(6)
elif N >=1000 or N <= 1199:
print(5)
elif N >=1200 or N <= 1399:
print(4)
elif N >=1400 or N <= 1599:
print(3)
elif N >=1600 or N <= 1799:
print(2)
elif N >=1800 or N <= 1999:
print(1)
|
s947984565
|
Accepted
| 33 | 9,180 | 354 |
N = int(input())
if N >= 400 and N <= 599:
print(8)
elif N >= 600 and N <= 799:
print(7)
elif N >= 800 and N <= 999:
print(6)
elif N >= 1000 and N <= 1199:
print(5)
elif N >= 1200 and N <= 1399:
print(4)
elif N >= 1400 and N <= 1599:
print(3)
elif N >= 1600 and N <= 1799:
print(2)
elif N >= 1800 and N <= 1999:
print(1)
|
s847706567
|
p02390
|
u798565376
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,572 | 41 |
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
s = int(input())
h = (s % 3600)
print(h)
|
s926199650
|
Accepted
| 20 | 5,584 | 135 |
input_s = int(input())
h = input_s // 3600
rest_s = input_s % 3600
m = rest_s // 60
s = rest_s % 60
print('{}:{}:{}'.format(h, m, s))
|
s208200797
|
p03563
|
u207707177
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 701 |
Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it.
|
S2 = list(input())
T = list(input())
S = []
Temp = 0
for _ in range(len(S2)-len(T)+1):
for j in range(len(T)):
if T[j] == S2[Temp + j] or S2[Temp + j] == "?":
if j == len(T) - 1:
S.append([S2[x] for x in range(Temp)] + [T[x] for x in range(len(T))] + [S2[x] for x in range(Temp + len(T), len(S2))])
Temp += 1
break
else:
continue
else:
Temp += 1
break
#print(S2)
#print(S)
for i in range(len(S)):
for j in range(len(S2)):
if S[i][j] == '?':
S[i][j] = 'a'
S[i] = ''.join(S[i])
if S != []:
print(sorted(S)[0])
else: print('UNRESTORABLE')
|
s727187568
|
Accepted
| 19 | 2,940 | 54 |
R = int(input())
G = int(input())
x = 2*G - R
print(x)
|
s852725243
|
p03436
|
u088552457
| 2,000 | 262,144 |
Wrong Answer
| 38 | 4,072 | 684 |
We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j). Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player repeatedly moves Kenus up, down, left or right by one square. The game is completed when Kenus reaches square (H, W) passing only white squares. Before Snuke starts the game, he can change the color of some of the white squares to black. However, he cannot change the color of square (1, 1) and (H, W). Also, changes of color must all be carried out before the beginning of the game. When the game is completed, Snuke's score will be the number of times he changed the color of a square before the beginning of the game. Find the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed, that is, Kenus can never reach square (H, W) regardless of how Snuke changes the color of the squares. The color of the squares are given to you as characters s_{i, j}. If square (i, j) is initially painted by white, s_{i, j} is `.`; if square (i, j) is initially painted by black, s_{i, j} is `#`.
|
from collections import deque
H, W = map(int, input().split())
f_inf = float('inf')
C = [input() for _ in range(H)]
def bfs():
dist = [[f_inf] * W for _ in range(H)]
D = ((1,0), (0,1), (-1,0), (0,-1))
que = deque([])
que.append((0,0))
dist[0][0] = 0
while que:
p = que.popleft()
for d in D:
nx = p[0] + d[0]
ny = p[1] + d[1]
print(nx, ny)
if H > nx >= 0 and W > ny >= 0 and C[nx][ny] != '#' and dist[nx][ny] == f_inf:
que.append((nx,ny))
dist[nx][ny] = dist[p[0]][p[1]]+1
return dist[H-1][W-1]
cnt = 0
for c in C:
cnt += c.count('.')
ans = bfs()
if ans == f_inf:
print(-1)
else:
print(cnt - 1 - ans)
|
s951229585
|
Accepted
| 26 | 3,316 | 660 |
from collections import deque
H, W = map(int, input().split())
f_inf = float('inf')
C = [input() for _ in range(H)]
def bfs():
dist = [[False]*W for _ in range(H)]
D = ((1,0), (0,1), (-1,0), (0,-1))
que = deque([])
que.append((0,0))
dist[0][0] = 0
while que:
p = que.popleft()
for d in D:
nx = p[0] + d[0]
ny = p[1] + d[1]
if H > nx >= 0 and W > ny >= 0 and C[nx][ny] != '#' and dist[nx][ny] is False:
que.append((nx,ny))
dist[nx][ny] = dist[p[0]][p[1]]+1
return dist[H-1][W-1]
cnt = 0
for c in C:
cnt += c.count('.')
ans = bfs()
if ans is False:
print(-1)
else:
print(cnt - 1 - ans)
|
s142182394
|
p03999
|
u809819902
| 2,000 | 262,144 |
Wrong Answer
| 29 | 9,156 | 191 |
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion. All strings that can be obtained in this way can be evaluated as formulas. Evaluate all possible formulas, and print the sum of the results.
|
s=input()
n=len(s)-1
ans=0
for i in range(2**n):
k=0
for j in range(n):
if i>>j&1:
ans+=int(s[k:j+1])
k=j+1
ans+=int(s[k:])
print(ans)
|
s236093843
|
Accepted
| 31 | 9,080 | 189 |
s=input()
n=len(s)-1
ans=0
for i in range(2**n):
k=0
for j in range(n):
if i>>j & 1:
ans+=int(s[k:j+1])
k=j+1
ans+=int(s[k:])
print(ans)
|
s166529169
|
p03369
|
u580236524
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 59 |
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
t = input()
ans=700
for x in t:
if x=='o':
ans+=100
|
s856951882
|
Accepted
| 17 | 2,940 | 69 |
t = input()
ans=700
for x in t:
if x=='o':
ans+=100
print(ans)
|
s997443996
|
p02613
|
u697386253
| 2,000 | 1,048,576 |
Wrong Answer
| 144 | 16,276 | 249 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n = int(input())
s = []
for i in range(n):
s.append(input())
ac = s.count('AC')
wa = s.count('WA')
tle = s.count('TLE')
re = s.count('RE')
print('AC x ' + str(ac))
print('WA × ' + str(wa))
print('TLE × ' + str(tle))
print('RE × ' + str(re))
|
s638875750
|
Accepted
| 147 | 16,180 | 247 |
n = int(input())
s = []
for i in range(n):
s.append(input())
ac = s.count('AC')
wa = s.count('WA')
tle = s.count('TLE')
re = s.count('RE')
print('AC x ' + str(ac))
print('WA x ' + str(wa))
print('TLE x ' + str(tle))
print('RE x ' + str(re))
|
s961329745
|
p04029
|
u169221932
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 55 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
num = int(input())
sum = (num + 1) * num / 2
print(sum)
|
s660908102
|
Accepted
| 17 | 2,940 | 60 |
num = int(input())
sum = (num + 1) * num / 2
print(int(sum))
|
s187861231
|
p03836
|
u707808519
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,316 | 465 |
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 = tx-sx, ty-sy
ans = []
for _ in range(dy):
ans.append('U')
for _ in range(dx):
ans.append('R')
for _ in range(dy):
ans.append('D')
for _ in range(dx):
ans.append('L')
ans.append('L')
for _ in range(dy+1):
ans.append('U')
for _ in range(dx+1):
ans.append('R')
ans.append('D')
ans.append('R')
for _ in range(dy+1):
ans.append('D')
for _ in range(dx+1):
ans.append('L')
ans.append('U')
|
s185380303
|
Accepted
| 17 | 3,060 | 187 |
sx, sy, tx, ty = map(int, input().split())
dx, dy = tx-sx, ty-sy
S = 'U'*dy + 'R'*dx + 'D'*dy + 'L'*dx + 'L' + 'U'*(dy+1) + 'R'*(dx+1) + 'D' + 'R' + 'D'*(dy+1) + 'L'*(dx+1) + 'U'
print(S)
|
s810383354
|
p03712
|
u302292660
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 227 |
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
h,w = map(int,input().split())
L = []
L.append(["#"]*(w+2))
for i in range(h):
l=["#"]
l.append(input())
l.append("#")
print(l)
L.append(l)
L.append(["#"]*(w+2))
for i in range(h+2):
print("".join(L[i]))
|
s373219494
|
Accepted
| 17 | 3,060 | 214 |
h,w = map(int,input().split())
L = []
L.append(["#"]*(w+2))
for i in range(h):
l=["#"]
l.append(input())
l.append("#")
L.append(l)
L.append(["#"]*(w+2))
for i in range(h+2):
print("".join(L[i]))
|
s652817159
|
p03400
|
u903596281
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 108 |
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
N=int(input())
D,X=map(int,input().split())
ans=X
for i in range(N):
ans+=1+(N-1)//int(input())
print(ans)
|
s401134921
|
Accepted
| 17 | 2,940 | 108 |
N=int(input())
D,X=map(int,input().split())
ans=X
for i in range(N):
ans+=1+(D-1)//int(input())
print(ans)
|
s830818201
|
p02417
|
u100813820
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,348 | 10 |
Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.
|
import sys
|
s386947446
|
Accepted
| 40 | 7,392 | 1,060 |
# 17-Character-Counting_Characters.py
# ?????????????????????
# Input
# Output
# a : a????????°
# .
# .
# Constraints
# Sample Input
# This is a pen.
# Sample Output
# a : 1
# b : 0
# c : 0
# d : 0
# f : 0
# g : 0
# h : 1
# j : 0
# l : 0
# m : 0
# n : 1
# p : 1
# r : 0
# t : 1
# w : 0
# x : 0
import sys
s = str()
alphabet = "abcdefghijklmnopqrstuvwxyz"
for line in sys.stdin:
s += line.lower()
# print(list(s))
for c in alphabet:
print("{} : {}".format(c, s.count(c)) )
|
s361774488
|
p03957
|
u181431922
| 1,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 113 |
This contest is `CODEFESTIVAL`, which can be shortened to the string `CF` by deleting some characters. Mr. Takahashi, full of curiosity, wondered if he could obtain `CF` from other strings in the same way. You are given a string s consisting of uppercase English letters. Determine whether the string `CF` can be obtained from the string s by deleting some characters.
|
s = input()
a = s.find('C')
if a == -1:
print("No")
else:
if s[a:].find('F') == -1:
print("Yes")
|
s825248107
|
Accepted
| 18 | 2,940 | 135 |
s = input()
a = s.find('C')
if a == -1:
print("No")
else:
if s[a:].find('F') == -1:
print("No")
else: print("Yes")
|
s687847524
|
p02678
|
u252828980
| 2,000 | 1,048,576 |
Wrong Answer
| 2,268 | 2,213,304 | 648 |
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 scipy.sparse.csgraph import shortest_path, floyd_warshall, dijkstra
n,m = map(int,input().split())
d = [[0]*n for i in range(n)]
for i in range(m):
a,b = map(int,input().split())
a,b = a-1,b-1
d[a][b] = 1
d[b][a] = 1
#print(d)
d = dijkstra(d)
#print(d)
li = []
for i in range(1,n):
for j in range(1,n):
for k in range(j+1,n):
if d[i][j] > d[i][k]:
if d[i][k] + d[0][k] == d[i][0] and k-1!=i:
li.append(k)
break
break
# print(i,j,k,li,d[i][k],d[0][k],d[i][j])
if len(li) == n-1:
for i in range(len(li)):
print(li[i])
|
s048021853
|
Accepted
| 646 | 35,160 | 572 |
n,m = map(int,input().split())
d = [[] for _ in range(n+1)]
for i in range(m):
a,b = map(int,input().split())
#a,b = a-1,b-1
d[a].append(b)
d[b].append(a)
#print(d)
from collections import deque
q = deque([1])
visited = [False]*(n+1)
visited[1] = True
signpost = [False]*(n+1)
while q:
k = q.popleft()
for j in d[k]:
if visited[j] != False:
continue
visited[j] = True
signpost[j] = k
q.append(j)
#print(visited,signpost)
if signpost[2:]:
print("Yes")
for i in range(2,n+1):
print(signpost[i])
|
s525758277
|
p03612
|
u903005414
| 2,000 | 262,144 |
Wrong Answer
| 302 | 23,416 | 293 |
You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this.
|
import numpy as np
N = int(input())
p = np.array(list(map(int, input().split())))
v = p == np.arange(1, N + 1)
print('v', v)
ans = 0
for i in range(N):
if i == N - 1 and v[i]:
ans += 1
continue
if v[i]:
v[i], v[i + 1] = False, False
ans += 1
print(ans)
|
s817069851
|
Accepted
| 227 | 23,416 | 295 |
import numpy as np
N = int(input())
p = np.array(list(map(int, input().split())))
v = p == np.arange(1, N + 1)
# print('v', v)
ans = 0
for i in range(N):
if i == N - 1 and v[i]:
ans += 1
continue
if v[i]:
v[i], v[i + 1] = False, False
ans += 1
print(ans)
|
s608198282
|
p03535
|
u785989355
| 2,000 | 262,144 |
Wrong Answer
| 25 | 3,536 | 1,221 |
In CODE FESTIVAL XXXX, there are N+1 participants from all over the world, including Takahashi. Takahashi checked and found that the _time gap_ (defined below) between the local times in his city and the i-th person's city was D_i hours. The time gap between two cities is defined as follows. For two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is defined to be min(d,24-d) hours. Here, we are using 24-hour notation. That is, the local time in the i-th person's city is either d o'clock or 24-d o'clock at the moment when the local time in Takahashi's city is 0 o'clock, for example. Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours. Find the maximum possible value of s.
|
N=int(input())
D=list(map(int,input().split()))
flg =False
A = [0 for i in range(13)]
A[0]=1
for i in range(N):
j = min(i,24-i)
if j==0:
flg=True
break
elif j==12:
if A[j]==1:
flg=True
break
else:
A[j]+=1
else:
if A[j]==2:
flg=True
break
else:
A[j]+=1
import copy
if flg:
print(0)
else:
count=0
one_times=[]
times=[0]
for i in range(1,12):
if A[i]==1:
count+=1
one_times.append(i)
if A[i]==2:
times.append(i)
times.append(24-i)
if A[12]==1:
times.append(12)
score_max=0
for i in range(2**count):
t=copy.copy(times)
for j in range(count):
if (i//(2**j))%2==0:
t.append(one_times[j])
else:
t.append(24-one_times[j])
t.sort()
score_min=24
for j in range(len(t)):
if j==len(t)-1:
score_min=min(t[0]+24-t[len(t)-1],score_min)
else:
score_min=min(t[j+1]-t[j],score_min)
score_max=(score_min,score_max)
print(score_max)
|
s868602829
|
Accepted
| 28 | 3,536 | 1,238 |
N=int(input())
D=list(map(int,input().split()))
flg =False
A = [0 for i in range(13)]
A[0]=1
for i in range(N):
j = min(D[i],24-D[i])
if j==0:
flg=True
break
elif j==12:
if A[j]==1:
flg=True
break
else:
A[j]+=1
else:
if A[j]==2:
flg=True
break
else:
A[j]+=1
import copy
if flg:
print(0)
else:
count=0
one_times=[]
times=[0]
for i in range(1,12):
if A[i]==1:
count+=1
one_times.append(i)
if A[i]==2:
times.append(i)
times.append(24-i)
if A[12]==1:
times.append(12)
score_max=0
for i in range(2**count):
t=copy.copy(times)
for j in range(count):
if (i//(2**j))%2==0:
t.append(one_times[j])
else:
t.append(24-one_times[j])
t.sort()
score_min=24
for j in range(len(t)):
if j==len(t)-1:
score_min=min(t[0]+24-t[len(t)-1],score_min)
else:
score_min=min(t[j+1]-t[j],score_min)
score_max=max(score_min,score_max)
print(score_max)
|
s723663163
|
p02357
|
u686180487
| 2,000 | 262,144 |
Wrong Answer
| 20 | 5,600 | 322 |
For a given array $a_1, a_2, a_3, ... , a_N$ of $N$ elements and an integer $L$, find the minimum of each possible sub-arrays with size $L$ and print them from the beginning. For example, for an array $\\{1, 7, 7, 4, 8, 1, 6\\}$ and $L = 3$, the possible sub-arrays with size $L = 3$ includes $\\{1, 7, 7\\}$, $\\{7, 7, 4\\}$, $\\{7, 4, 8\\}$, $\\{4, 8, 1\\}$, $\\{8, 1, 6\\}$ and the minimum of each sub-array is 1, 4, 4, 1, 1 respectively.
|
# -*- coding: utf-8 -*-
N,L = list(map(int, input().split()))
alist = list(map(int, input().split()))
def printMax(arr, n, k):
max = 0
for i in range(n - k + 1):
max = arr[i]
for j in range(1, k):
if arr[i + j] > max:
max = arr[i + j]
print(str(max) + " ", end = "")
printMax(alist, N, L)
|
s612640323
|
Accepted
| 2,890 | 120,916 | 401 |
# -*- coding: utf-8 -*-
N,L = list(map(int, input().split()))
alist = list(map(int, input().split()))
arr = []
for i in range(L):
while arr and alist[i] <= alist[arr[-1]]:
arr.pop()
arr.append(i)
for i in range(L, N):
print(alist[arr[0]], end=' ')
while arr and arr[0] <= i-L:
arr.pop(0)
while arr and alist[i] <= alist[arr[-1]]:
arr.pop()
arr.append(i)
print(alist[arr[0]])
|
s781232537
|
p03485
|
u426964396
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 54 |
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=input().split()
a=int(a)
b=int(b)
print((a+b+1)/2)
|
s879898027
|
Accepted
| 17 | 2,940 | 75 |
import math
print(int(math.ceil(sum([int(x) for x in input().split()])/2)))
|
s561850179
|
p04043
|
u883048396
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 83 |
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.
|
if input().rstrip in ("5 5 7","5 7 5","7 5 5"):
print("YES")
else:
print("NO")
|
s978501465
|
Accepted
| 17 | 2,940 | 90 |
if input().rstrip() in ("5 5 7","5 7 5","7 5 5"):
print("YES")
else:
print("NO")
|
s023324910
|
p03471
|
u225642513
| 2,000 | 262,144 |
Wrong Answer
| 977 | 3,060 | 286 |
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
N,Y = map(int, input().split())
a = Y//10000
b = (Y%10000)//5000
c = (Y%5000)//1000
for i in range(a+1):
x = a-i
y = b + i * 2
for i in range(y+1):
z = c + i * 5
if x+y+z == N:
print("{} {} {}".format(x,y,z))
exit()
print("-1 -1 -1")
|
s746719603
|
Accepted
| 37 | 3,060 | 332 |
N,Y = map(int, input().split())
a = Y//10000
b = (Y%10000)//5000
c = (Y%5000)//1000
for i in range(a+1):
x = a-i
y = b + i * 2
for j in range(y+1):
z = c + j * 5
if x+y-j+z == N:
print("{} {} {}".format(x,y-j,z))
exit()
if x+y-j+z > N:
break
print("-1 -1 -1")
|
s358570155
|
p02608
|
u646130340
| 2,000 | 1,048,576 |
Wrong Answer
| 2,206 | 19,056 | 418 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
N = int(input())
def perfect_number(x, y, z):
n = x**2 + y**2 + z**2 + x*y + y*z + z*x
return n
l = []
for x in range(1, N):
for y in range(1, N):
for z in range(1, N):
n = perfect_number(x,y,z)
if n > N:
break
else:
l.append(n)
counts = [l.count(x) for x in sorted(set(l))]
d = dict(zip(l, counts))
for i in range(1, N+1):
if i in d:
print(d[i])
else:
print(0)
|
s482646214
|
Accepted
| 300 | 19,856 | 479 |
N = int(input())
def calculate_n(x, y, z):
n = x**2 + y**2 + z**2 + x*y + y*z + z*x
return n
l = []
for x in range(1, N):
for y in range(1, N):
for z in range(1, N):
n = calculate_n(x,y,z)
if n > N:
break
else:
l.append(n)
if calculate_n(x, y+1, 1) > N:
break
if calculate_n(x+1, 1, 1) > N:
break
import collections
c = collections.Counter(l)
for i in range(1, N+1):
if i in c:
print(c[i])
else:
print(0)
|
s869351663
|
p03433
|
u717993780
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 113 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
n = int(input())
li = sorted(map(int,input().split()),reverse=True)
ans = sum(li[::2]) - sum(li[1::2])
print(ans)
|
s097889092
|
Accepted
| 17 | 2,940 | 86 |
n = int(input())
a = int(input())
if n % 500 <= a:
print("Yes")
else:
print("No")
|
s537040418
|
p03471
|
u096983897
| 2,000 | 262,144 |
Wrong Answer
| 779 | 3,064 | 385 |
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():
num, money = map(int, input().split())
array = [-1,-1,-1]
for i in range(num+1):
for j in range(num+1):
if i + j > num:
continue
sum = i*10000 + j*5000 + (num-i-j)*1000
if sum == money:
array[0]=i
array[1]=j
array[2]=num-i-j
break
print(array)
print("{0[0]} {0[1]} {0[2]}".format(array))
main()
|
s068247001
|
Accepted
| 772 | 3,060 | 370 |
def main():
num, money = map(int, input().split())
array = [-1,-1,-1]
for i in range(num+1):
for j in range(num+1):
if i + j > num:
continue
sum = i*10000 + j*5000 + (num-i-j)*1000
if sum == money:
array[0]=i
array[1]=j
array[2]=num-i-j
break
print("{0[0]} {0[1]} {0[2]}".format(array))
main()
|
s643067437
|
p03574
|
u027622859
| 2,000 | 262,144 |
Wrong Answer
| 28 | 3,064 | 536 |
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
h, w = map(int, input().split())
s = [input() for i in range(h)]
print(s)
way = ((1,1),(1,0),(1,-1),(0,1),(0,-1),(-1,0),(-1,1),(-1,-1))
ans = [[0 for j in range(w)] for i in range(h)]
for i in range(h):
for j in range(w):
if s[i][j] == "#":
ans[i][j] = "#"
continue
for k in way:
if not (0 <= i+k[0] < h and 0 <= j+k[1] < w):
continue
if s[i+k[0]][j+k[1]] == "#":
ans[i][j] += 1
for i in range(h):
print(''.join(map(str, ans[i])))
|
s894339016
|
Accepted
| 32 | 3,444 | 688 |
H, W = map(int, input().split())
s = [list(input())for _ in range(H)]
grid = [(-1,-1),(-1,1),(1,-1),(1,1),(1,0),(0,1),(-1,0),(0,-1)]
for y in range(H):
for x in range(W):
count = 0
if s[y][x] == '#':
print('#', end='')
else:
for g in grid:
try:
sy = y+g[0]
sx = x+g[1]
except IndexError:
continue
if sy < 0 or sy >= H:
continue
elif sx < 0 or sx >= W:
continue
if s[sy][sx] == '#':
count += 1
print(count,end='')
print()
|
s242460944
|
p03880
|
u729133443
| 2,000 | 262,144 |
Wrong Answer
| 111 | 7,068 | 220 |
A cheetah and a cheater are going to play the game of Nim. In this game they use N piles of stones. Initially the i-th pile contains a_i stones. The players take turns alternately, and the cheetah plays first. In each turn, the player chooses one of the piles, and takes one or more stones from the pile. The player who can't make a move loses. However, before the game starts, the cheater wants to cheat a bit to make sure that he can win regardless of the moves by the cheetah. From each pile, the cheater takes zero or one stone and eats it before the game. In case there are multiple ways to guarantee his winning, he wants to minimize the number of stones he eats. Compute the number of stones the cheater will eat. In case there is no way for the cheater to win the game even with the cheating, print `-1` instead.
|
n,*a=map(int,open(0))
x=0
c=[0]*30
for i in a:
x^=i
c[bin(i^i-1)[::-1].rfind('1')]=1
a=0
for i in range(29,-1,-1):
if x&2**i:
if c[i]:
a+=1
x^=2**i-1
else:
print(-1)
break
else:
print(a)
|
s004252308
|
Accepted
| 110 | 7,068 | 226 |
n,*a=map(int,open(0))
x=0
c=[0]*30
for i in a:
x^=i
c[bin(i^i-1)[::-1].rfind('1')]=1
a=0
for i in range(29,-1,-1):
if x&2**i:
if c[i]:
a+=1
x^=2**i-1
else:
print(-1)
break
else:
print(a)
|
s940693591
|
p03163
|
u864013199
| 2,000 | 1,048,576 |
Wrong Answer
| 2,120 | 169,868 | 351 |
There are N items, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), Item i has a weight of w_i and a value of v_i. Taro has decided to choose some of the N items and carry them home in a knapsack. The capacity of the knapsack is W, which means that the sum of the weights of items taken must be at most W. Find the maximum possible sum of the values of items that Taro takes home.
|
import sys
input = sys.stdin.readline
N,W = map(int,input().split())
dp = [[0]*(W+1) for _ in range(N+1)]
for i in range(N):
w,v = map(int,input().split())
for j in range(W+1):
if j+w <= W:
dp[i+1][j+w] = dp[i][j]+v
dp[i+1][j] = max(dp[i+1][j],dp[i][j])
print(dp)
print(dp[N][W])
|
s897498032
|
Accepted
| 172 | 14,684 | 275 |
import numpy as np
import sys
input = sys.stdin.readline
N,W = map(int,input().split())
dp = np.zeros(W+1, dtype=np.int64)
for i in range(N):
w,v = map(int,input().split())
np.maximum(dp[:-w]+v, dp[w:], out = dp[w:])
print(dp[-1])
|
s534282640
|
p03739
|
u268516119
| 2,000 | 262,144 |
Wrong Answer
| 110 | 14,332 | 648 |
You are given an integer sequence of length N. The i-th term in the sequence is a_i. In one operation, you can select a term and either increment or decrement it by one. At least how many operations are necessary to satisfy the following conditions? * For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero. * For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.
|
import sys
fastinput=sys.stdin.readline
n=int(fastinput())
ai=[int(i) for i in fastinput().split()]
goukei=0
sousa=0
for a in ai:
goukei+=a
if a%2:
if goukei<=0:
sousa+=1-goukei
goukei=1
else:#odd:minus
if goukei>=0:
sousa+=goukei+1
goukei=-1
ans1=sousa
goukei=0
sousa=0
for a in ai:
goukei+=a
if not a%2:
if goukei<=0:
sousa+=1-goukei
goukei=1
else:#even:minus
if goukei>=0:
sousa+=goukei+1
goukei=-1
ans2=sousa
print(min(ans1,ans2))
|
s274727772
|
Accepted
| 125 | 15,100 | 674 |
import sys
fastinput=sys.stdin.readline
n=int(fastinput())
ai=[int(i) for i in fastinput().split()]
goukei=0
sousa=0
for k,a in enumerate(ai):
goukei+=a
if not k%2:
if goukei<=0:
sousa+=1-goukei
goukei=1
else:#odd:minus
if goukei>=0:
sousa+=goukei+1
goukei=-1
ans1=sousa
goukei=0
sousa=0
for k,a in enumerate(ai):
goukei+=a
if k%2:
if goukei<=0:
sousa+=1-goukei
goukei=1
else:#even:minus
if goukei>=0:
sousa+=goukei+1
goukei=-1
ans2=sousa
print(min(ans1,ans2))
|
s187030040
|
p02608
|
u991619971
| 2,000 | 1,048,576 |
Wrong Answer
| 1,034 | 27,068 | 653 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
import numpy as np
#import functools
#from itertools import combinations as comb
#from itertools import combinations_with_replacement as comb_with
#from itertools import permutations as perm
#import collections as C #most_common
#import math
#import sympy
N = int(input())
#N,K,d= map(int,input().split())
#A = list(map(int,input().split()))
#S = str(input())
#T = str(input())
num=np.zeros(10**4)
for x in range(1,100):
for y in range(1,100):
for z in range(1,100):
ans = x**2 + y**2 + z**2 + x*y + y*z + z*x
if ans < 10**4:
num[ans]+=1
for i in range(N):
print(int(num[i]))
|
s096664546
|
Accepted
| 974 | 26,544 | 405 |
import numpy as np
N = int(input())
#N,K,d= map(int,input().split())
#A = list(map(int,input().split()))
#S = str(input())
#T = str(input())
num=np.zeros(10**4+1)
for x in range(1,100):
for y in range(1,100):
for z in range(1,100):
ans = x**2 + y**2 + z**2 + x*y + y*z + z*x
if ans < 10**4+1:
num[ans]+=1
for i in range(N):
print(int(num[i+1]))
|
s626199453
|
p03338
|
u113255362
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,100 | 241 |
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
N=int(input())
S=input()
s=""
m=""
ss=[]
mm=[]
s1=set()
m1=set()
res = 0
for i in range(N):
s=S[0:i]
m=S[i+1:N]
ss=list(s)
mm=list(m)
s1=set(ss)
m1=set(mm)
s_intersection = s1 & m1
res= max(res,len(s_intersection))
print(res)
|
s802285797
|
Accepted
| 30 | 9,192 | 243 |
N=int(input())
S=input()
s=""
m=""
ss=[]
mm=[]
s1=set()
m1=set()
res = 0
for i in range(N):
s=S[0:i+1]
m=S[i+1:N]
ss=list(s)
mm=list(m)
s1=set(ss)
m1=set(mm)
s_intersection = s1 & m1
res= max(res,len(s_intersection))
print(res)
|
s355985456
|
p03854
|
u744034042
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,188 | 204 |
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()
u = ['dreamer', 'eraser', 'dream', 'erase']
while len(s) > 0:
for i in u:
if s[0:len(i)] == i:
s = s.lstrip(i)
else:
print("NO")
exit()
print("YES")
|
s081403770
|
Accepted
| 18 | 3,188 | 139 |
s = input().replace('eraser','').replace('erase','').replace('dreamer','').replace('dream','')
if s:
print('NO')
else:
print('YES')
|
s372240669
|
p03359
|
u229518917
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 71 |
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a,b= map(int,input().split())
if b>a:
print(a)
else:
print(a-1)
|
s087600591
|
Accepted
| 17 | 2,940 | 72 |
a,b= map(int,input().split())
if b>=a:
print(a)
else:
print(a-1)
|
s798447091
|
p03998
|
u729133443
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,188 | 419 |
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
sa=list(input())
sb=list(input())
sc=list(input())
ia=0
ib=0
ic=0
a=len(sa)
b=len(sb)
c=len(sc)
t=0
while True:
if t==0:
if ia == a:
print('a')
break
t=abs(int(ord('a')-ord(sa[ia])))
ia+=1
elif t==1:
if ib == b:
print('b')
break
t=abs(int(ord('a')-ord(sb[ib])))
ib+=1
else:
if ic == c:
print('c')
break
t=abs(int(ord('a')-ord(sc[ic])))
ic+=1
|
s068264626
|
Accepted
| 18 | 2,940 | 80 |
d={i:list(input())for i in'abc'};i='a'
while d[i]:i=d[i].pop(0)
print(i.upper())
|
s772229565
|
p03813
|
u367130284
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 98 |
Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise.
|
i=int(input())
ans=i//11*2
if 0<i%11<=6:
ans+=1
elif i%11==0:
ans+=0
else:
ans+=2
print(ans)
|
s740985746
|
Accepted
| 17 | 2,940 | 35 |
print("A"+"RB"[input()<"1200"]+"C")
|
s343293735
|
p02614
|
u679390859
| 1,000 | 1,048,576 |
Wrong Answer
| 80 | 9,352 | 642 |
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
H,W,K = map(int,input().split())
L = []
for _ in range(H):
l = list(input())
L.append(l)
import copy
ans =0
for x in range(2 ** H):
for y in range(2 ** W):
c = copy.copy(L)
for i in range(H):
if x & (1 << i):
for yoko in range(W):
c[i][yoko] = '.'
for j in range(W):
if y & (1 << j):
for tate in range(H):
c[tate][j] = '.'
#count
cnt = 0
for i in range(H):
for j in range(W):
if c[i][j] == '#': cnt += 1
if cnt == K:
ans += 1
print(ans)
|
s717559985
|
Accepted
| 163 | 9,264 | 646 |
H,W,K = map(int,input().split())
L = []
for _ in range(H):
l = list(input())
L.append(l)
import copy
ans =0
for x in range(2 ** H):
for y in range(2 ** W):
c = copy.deepcopy(L)
for i in range(H):
if x & (1 << i):
for yoko in range(W):
c[i][yoko] = '.'
for j in range(W):
if y & (1 << j):
for tate in range(H):
c[tate][j] = '.'
#count
cnt = 0
for i in range(H):
for j in range(W):
if c[i][j] == '#': cnt += 1
if cnt == K:
ans += 1
print(ans)
|
s591421073
|
p03385
|
u269969976
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 107 |
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
# -*- coding: utf-8 -*-
print("yes" if "".join(sorted([i for i in input().rstrip()])) == "abc" else "no")
|
s871992197
|
Accepted
| 17 | 2,940 | 107 |
# -*- coding: utf-8 -*-
print("Yes" if "".join(sorted([i for i in input().rstrip()])) == "abc" else "No")
|
s159726984
|
p03379
|
u816587940
| 2,000 | 262,144 |
Wrong Answer
| 897 | 60,448 | 368 |
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
import numpy as np
n = int(input())
a = list(map(int, input().split()))
b = []
for i in range(n):
b.append([a[i], i])
b = sorted(b, key=lambda x: x[0], reverse=False)
for i in range(n):
if i<n//2:
b[i].append(b[n//2-1][0])
else:
b[i].append(b[n//2][0])
b = sorted(b, key=lambda x: x[1], reverse=False)
for i in range(n):
print(b[i][2])
|
s432447003
|
Accepted
| 716 | 51,392 | 349 |
n = int(input())
a = list(map(int, input().split()))
b = []
for i in range(n):
b.append([a[i], i])
b = sorted(b, key=lambda x: x[0], reverse=False)
for i in range(n):
if i<n//2:
b[i].append(b[n//2][0])
else:
b[i].append(b[n//2-1][0])
b = sorted(b, key=lambda x: x[1], reverse=False)
for i in range(n):
print(b[i][2])
|
s466448083
|
p03644
|
u555356625
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 78 |
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())
cnt = 0
while n%2 == 0:
n = n / 2
cnt += 1
print(cnt)
|
s350564873
|
Accepted
| 17 | 2,940 | 78 |
n = int(input())
for i in range(7):
if n >= 2**i:
ans = 2**i
print(ans)
|
s616905914
|
p02612
|
u925478395
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,144 | 42 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
a = int(input())
ans = a % 1000
print(ans)
|
s217521392
|
Accepted
| 33 | 9,164 | 83 |
a = int(input())
b = a % 1000
if b != 0:
ans = 1000 - b
else:
ans = b
print(ans)
|
s366968843
|
p03730
|
u776311944
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,164 | 167 |
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
A, B, C = map(int, input().split())
calc = 0
for i in range(1000):
calc += A
rem = calc % B
if rem == C:
print('Yes')
exit()
print('No')
|
s770204073
|
Accepted
| 28 | 9,160 | 167 |
A, B, C = map(int, input().split())
calc = 0
for i in range(1000):
calc += A
rem = calc % B
if rem == C:
print('YES')
exit()
print('NO')
|
s038260698
|
p02264
|
u604774382
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,716 | 333 |
_n_ _q_ _name 1 time1_ _name 2 time2_ ... _name n timen_ In the first line the number of processes _n_ and the quantum _q_ are given separated by a single space. In the following _n_ lines, names and times for the _n_ processes are given. _name i_ and _time i_ are separated by a single space.
|
n, q = [ int( val ) for val in input( ).split( " " ) ]
ps = [0]*n
t = [0]*n
for i in range( n ):
ps[i], t[i] = input( ).split( " " )
output = []
qsum = 0
while t:
psi = ps.pop( 0 )
ti = int( t.pop( 0 ) )
if ti <= q:
qsum += ti
output.append( psi+" "+str( qsum ) )
else:
t.append( ti - q )
ps.append( psi )
qsum += q
|
s332497154
|
Accepted
| 440 | 21,232 | 527 |
from collections import deque
n, q = [ int( val ) for val in input( ).split( " " ) ]
processes = deque( )
for i in range( n ):
name, time = input( ).split( " " )
processes.append( ( name, int( time ) ) )
qsum = 0
output = []
while len( processes ):
process = processes.popleft( )
if process[1] <= q:
qsum += process[1]
output.append( "{:s} {:d}".format( process[0], qsum ) )
else:
processes.append( ( process[0], process[1]- q ) )
qsum += q
print( "\n".join( output ) )
|
s288724810
|
p03795
|
u739480378
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
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())
power=math.factorial(N)
v=10**9+7
answer=power%v
print(answer)
|
s228872409
|
Accepted
| 17 | 2,940 | 50 |
N=int(input())
x=800*N
v=N//15
y=200*v
print(x-y)
|
s742020473
|
p04011
|
u987164499
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 253 |
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
from sys import stdin
from itertools import combinations
n = int(stdin.readline().rstrip())
k = int(stdin.readline().rstrip())
x = int(stdin.readline().rstrip())
y = int(stdin.readline().rstrip())
if n <= k:
print(x*n)
else:
print(x*n+y*(n-k))
|
s697240129
|
Accepted
| 17 | 2,940 | 219 |
from sys import stdin
n = int(stdin.readline().rstrip())
k = int(stdin.readline().rstrip())
x = int(stdin.readline().rstrip())
y = int(stdin.readline().rstrip())
if n <= k:
print(x*n)
else:
print(x*k+y*(n-k))
|
s521057598
|
p03455
|
u844196583
| 2,000 | 262,144 |
Wrong Answer
| 24 | 9,012 | 100 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
if a*b-int(a*b/2)*2 == 0:
print('Odd')
else:
print('Even')
|
s830047842
|
Accepted
| 27 | 9,004 | 100 |
a, b = map(int, input().split())
if a*b-int(a*b/2)*2 == 0:
print('Even')
else:
print('Odd')
|
s200430976
|
p03860
|
u223646582
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 44 |
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
s=input().split()[1]
print('A{}C'.format(s))
|
s181735173
|
Accepted
| 17 | 2,940 | 47 |
s=input().split()[1][0]
print('A{}C'.format(s))
|
s668282650
|
p03448
|
u630666565
| 2,000 | 262,144 |
Wrong Answer
| 314 | 3,440 | 646 |
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.
|
#-*- coding: utf-8 -*-
a = int(input())
b = int(input())
c = int(input())
x = int(input())
def combi(x, a, b):
ans = 0
b_num = []
if a == 0 and b != 0:
if x/b == 0:
ans += 1
b_num.append(x)
elif a != 0 and b == 0:
if x/a == 0:
ans += 1
elif a == 0 and b == 0:
ans = 0
else:
for i in range(int(x/a)+1):
if (x - i*a)%b == 0:
ans += 1
b_num.append(x - i*a)
return ans, b_num
ans1, test = combi(x, a, b)
sum1 = 0
for i in range(len(test)):
ans, val = combi(test[i], b, c)
sum1 += ans
print(sum1)
|
s475174459
|
Accepted
| 49 | 3,060 | 258 |
# -*- coding: utf-8 -*-
A = int(input())
B = int(input())
C = int(input())
X = int(input())
ans = 0
for i in range(A + 1):
for j in range(B + 1):
for k in range(C + 1):
if 500*i + 100*j + 50*k == X:
ans += 1
print(ans)
|
s235664382
|
p03370
|
u189487046
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 99 |
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
N,X = map(int, input().split())
li = [int(input()) for i in range(N)]
print(N+(sum(li)//min(li)))
|
s129987963
|
Accepted
| 17 | 2,940 | 101 |
N,X = map(int, input().split())
li = [int(input()) for i in range(N)]
print(N+(X-sum(li))//min(li))
|
s019940477
|
p02608
|
u229518917
| 2,000 | 1,048,576 |
Wrong Answer
| 41 | 9,752 | 479 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
N=int(input())
ans=0
mul=1
while ans<=N:
ans=6*(mul**2)
mul+=1
ANS={}
for i in range(1,mul+1):
for j in range(i,mul+1):
for k in range(j,mul+1):
if i!=j and j!=k:
ANS[i*i+j*j+k*k+i*j+j*k+i*k]=6
elif i==j and j==k:
ANS[i*i+j*j+k*k+i*j+j*k+i*k]=1
else:
ANS[i*i+j*j+k*k+i*j+j*k+i*k]=3
for i in range(N+1):
if i in ANS.keys():
print(ANS[i])
else:
print(0)
|
s281439257
|
Accepted
| 222 | 11,336 | 861 |
N=int(input())
ans=0
mul=1
while ans<=N:
ans=6*(mul**2)
mul+=1
ANS={}
for i in range(1,101):
for j in range(i,101):
for k in range(j,101):
if (i*i+j*j+k*k+i*j+j*k+i*k) not in ANS.keys():
if i!=j and j!=k and k!=i:
ANS[i*i+j*j+k*k+i*j+j*k+i*k]=6
elif i==j and j==k and k==i:
ANS[i*i+j*j+k*k+i*j+j*k+i*k]=1
else:
ANS[i*i+j*j+k*k+i*j+j*k+i*k]=3
else:
if i!=j and j!=k and k!=i:
ANS[i*i+j*j+k*k+i*j+j*k+i*k]+=6
elif i==j and j==k and k==i:
ANS[i*i+j*j+k*k+i*j+j*k+i*k]+=1
else:
ANS[i*i+j*j+k*k+i*j+j*k+i*k]+=3
for Z in range(1,N+1):
if Z in ANS.keys():
print(ANS[Z])
else:
print(0)
|
s932375497
|
p03455
|
u648011094
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 128 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
# -*- coding: utf-8 -*-
a, b = map(int, input().split())
if a * b % 2 == 0:
print("even")
else:
print("odd")
|
s647272450
|
Accepted
| 17 | 2,940 | 128 |
# -*- coding: utf-8 -*-
a, b = map(int, input().split())
if a * b % 2 == 0:
print("Even")
else:
print("Odd")
|
s098487359
|
p03079
|
u231905444
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 104 |
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
arr=list(map(int,input().split()))
if(arr.count(arr[0])==3):
print('yes')
else:
print('no')
|
s091627244
|
Accepted
| 17 | 2,940 | 100 |
arr=list(map(int,input().split()))
if(arr.count(arr[0])==3):
print('Yes')
else:
print('No')
|
s402147339
|
p03836
|
u405256066
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 325 |
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.
|
from sys import stdin
sx,sy,tx,ty = [int(x) for x in stdin.readline().rstrip().split()]
ans = ""
ans += (ty-sy)*"U"
ans += (tx-sx)*"R"
ans2 = ans.replace("R","tmp").replace("L","R").replace("tmp","L")
ans2 = ans2.replace("U","tmp").replace("D","U").replace("tmp","D")
ans = ans + "LU"+ans+"RD" + "RD" + ans2 + "LU"
print(ans)
|
s850590239
|
Accepted
| 17 | 3,064 | 331 |
from sys import stdin
sx,sy,tx,ty = [int(x) for x in stdin.readline().rstrip().split()]
ans = ""
ans += (ty-sy)*"U"
ans += (tx-sx)*"R"
ans2 = ans.replace("R","tmp").replace("L","R").replace("tmp","L")
ans2 = ans2.replace("U","tmp").replace("D","U").replace("tmp","D")
ans = ans + ans2 +"LU"+ans+"RD" + "RD" + ans2 + "LU"
print(ans)
|
s838388455
|
p02646
|
u746154235
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,208 | 208 |
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 A == B:
print('YES')
exit(0)
print(T*V+A)
print(T*W+B)
if ((T*V+A) - (T*W+B) >=0):
print("YES")
else:
print("NO")
|
s782138942
|
Accepted
| 22 | 9,188 | 231 |
A,V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
if A < B:
if (T*V+A) >= (T*W+B):
print("YES")
else:
print("NO")
else:
if (A-T*V) <= (B-T*W):
print("YES")
else:
print("NO")
|
s989963732
|
p03359
|
u644778646
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 81 |
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a,b = map(int,input().split())
if a >= b:
print(a-1)
else:
print(a)
|
s587811959
|
Accepted
| 17 | 2,940 | 80 |
a,b = map(int,input().split())
if a > b:
print(a-1)
else:
print(a)
|
s468720697
|
p02831
|
u185424824
| 2,000 | 1,048,576 |
Wrong Answer
| 34 | 2,940 | 108 |
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
A,B = map(int,input().split())
C = 0
while C == 0:
if A % B == 0:
C = 1
else:
A += 1
print(A)
|
s428649614
|
Accepted
| 35 | 2,940 | 111 |
A,B = map(int,input().split())
C = 0
for i in range(B):
if A * (i+1) % B == 0:
print(A*(i+1))
break
|
s187757503
|
p03730
|
u046592970
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 141 |
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
import sys
a,b,c = map(int,input().split())
for i in range(1,b+1):
if a*i % b == c:
print("Yes")
sys.exit()
print("No")
|
s858034123
|
Accepted
| 17 | 2,940 | 141 |
import sys
a,b,c = map(int,input().split())
for i in range(1,b+1):
if a*i % b == c:
print("YES")
sys.exit()
print("NO")
|
s699232812
|
p02261
|
u362520072
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 885 |
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).
|
def bubble_sort(c, n):
for i in range(0,n):
for j in range(n-1,i-1,-1):
if c[j][1] < c[j-1][1]:
w = c[j]
c[j] = c[j-1]
c[j-1] = w
def selection_sort(c, n):
for i in range(0,n):
minj = i
for j in range(i, n):
if c[j][1] < c[minj][1]:
minj = j
w = c[i]
c[i] = c[minj]
c[minj] = w
def isStable(inn, out, n):
for i in range(0,n):
for j in range(i+1, n):
for a in range(0, n):
for b in range(a+1, n):
if inn[i] == inn[j] and inn[i] == out[b] and inn[j] == out[a]:
return print("Not stable")
return print("Stable")
n = int(input())
c = list(map(str, input().split()))
before_c = list(c)
bubble_sort(c,n)
print(" ".join(map(str, c)))
isStable(before_c, bubble_sort(c,n), n)
selection_sort(c,n)
print(" ".join(map(str, c)))
isStable(before_c, selection_sort(c,n), n)
|
s533617556
|
Accepted
| 20 | 5,608 | 652 |
def bubble_sort(a, n):
for i in range(0,n):
for j in range(n-1,i,-1):
if a[j][1] < a[j-1][1]:
a[j], a[j-1] = a[j-1], a[j]
return a
def selection_sort(b, n):
for i in range(0,n):
minj = i
for j in range(i, n):
if b[j][1] < b[minj][1]:
minj = j
b[i], b[minj] = b[minj], b[i]
return b
if __name__ == '__main__':
n = int(input())
a = list(map(str, input().split()))
b = a[:]
buble = bubble_sort(a,n)
select = selection_sort(b,n)
print(" ".join(map(str, buble)))
print("Stable")
print(" ".join(map(str, select)))
if select == buble:
print("Stable")
else:
print("Not stable")
|
s854110901
|
p03738
|
u426572476
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
You are given two positive integers A and B. Compare the magnitudes of these numbers.
|
123456789012345678901234567890
234567890123456789012345678901
|
s462822958
|
Accepted
| 17 | 2,940 | 112 |
import math
a, b = [int(input()) for i in range(2)]
print("GREATER" if a > b else "LESS" if a < b else "EQUAL")
|
s604307284
|
p03737
|
u177398299
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 47 |
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.
|
print(*(s[0].upper() for s in input().split()))
|
s013008118
|
Accepted
| 17 | 2,940 | 55 |
print(*(s[0].upper() for s in input().split()), sep='')
|
s086164534
|
p03971
|
u820357030
| 2,000 | 262,144 |
Wrong Answer
| 132 | 4,204 | 446 |
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.
|
nums = list(map(int, input().split()))
s=input()
print(nums[0], nums[1], nums[2])
print(s)
N=nums[0]
A=nums[1]
B=nums[2]
cA=0
cB=0
for i in range(N):
if(s[i:i+1] == "a"):
# print(s[i:i+1])
if((cA+cB)<(A+B)):
print("Yes")
else:
print("No")
cA+=1
elif(s[i:i+1]=="b"):
# print(s[i:i+1])
if((cA+cB)<(A+B)):
if(cB<B):
print("Yes")
else:
print("No")
else:
print("No")
cB += 1
elif(s[i:i+1]=="c"):
print("No")
|
s961522554
|
Accepted
| 137 | 4,040 | 452 |
nums = list(map(int, input().split()))
s=input()
#print(nums[0], nums[1], nums[2])
#print(s)
N=nums[0]
A=nums[1]
B=nums[2]
cA=0
cB=0
for i in range(N):
if(s[i:i+1] == "a"):
# print(s[i:i+1])
if((cA+cB)<(A+B)):
print("Yes")
cA+=1
else:
print("No")
elif(s[i:i+1]=="b"):
# print(s[i:i+1])
if((cA+cB)<(A+B)):
if(cB<B):
print("Yes")
cB+=1
else:
print("No")
else:
print("No")
elif(s[i:i+1]=="c"):
print("No")
|
s657518325
|
p02833
|
u816631826
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 137 |
For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
|
p=int(input())
if p%2 == 0:
num=10
c=0;
while num <= p:
c += (p/num)
num *= 5
print(c)
else:
print(0)
|
s546537590
|
Accepted
| 17 | 3,060 | 285 |
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
l= int(readline())
if l% 2 == 1:
print(0)
else:
l //= 2
ans = 0
while l:
l //= 5
ans += l
print(ans)
|
s135147664
|
p03971
|
u767664985
| 2,000 | 262,144 |
Wrong Answer
| 112 | 4,016 | 412 |
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()
num = 0
num_b = 0
for i in range(N):
if S[i] == "a":
if num < A + B:
print("Yes")
num += 1
else:
print("No")
elif S[i] == "b":
if num < A + B and num_b < B:
print("Yes")
num += 1
num_b += 1
else:
print("Np")
else:
print("No")
|
s521539218
|
Accepted
| 117 | 4,016 | 412 |
N, A, B = map(int, input().split())
S = input()
num = 0
num_b = 0
for i in range(N):
if S[i] == "a":
if num < A + B:
print("Yes")
num += 1
else:
print("No")
elif S[i] == "b":
if num < A + B and num_b < B:
print("Yes")
num += 1
num_b += 1
else:
print("No")
else:
print("No")
|
s768815112
|
p03303
|
u366886346
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 88 |
You are given a string S consisting of lowercase English letters. We will write down this string, starting a new line after every w letters. Print the string obtained by concatenating the letters at the beginnings of these lines from top to bottom.
|
s=input()
w=int(input())
ans=""
for i in range(-len(s)//-w):
ans+=s[i*w]
print(ans)
|
s714606452
|
Accepted
| 17 | 2,940 | 92 |
s=input()
w=int(input())
ans=""
for i in range(-1*(len(s)//-w)):
ans+=s[i*w]
print(ans)
|
s270615333
|
p02743
|
u892308039
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 182 |
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
import math
i = list(map(int, input().split()))
a = i[0]
b = i[1]
c = i[2]
a = math.sqrt(a)
b = math.sqrt(b)
c = math.sqrt(c)
if(a + b) > c:
print('Yes')
else:
print('No')
|
s994373256
|
Accepted
| 35 | 5,076 | 200 |
import math
from decimal import *
i = list(map(int, input().split()))
a = Decimal(i[0])
b = Decimal(i[1])
c = Decimal(i[2])
if(a.sqrt() + b.sqrt()) < c.sqrt():
print('Yes')
else:
print('No')
|
s972596377
|
p03163
|
u167681750
| 2,000 | 1,048,576 |
Wrong Answer
| 171 | 14,592 | 278 |
There are N items, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), Item i has a weight of w_i and a value of v_i. Taro has decided to choose some of the N items and carry them home in a knapsack. The capacity of the knapsack is W, which means that the sum of the weights of items taken must be at most W. Find the maximum possible sum of the values of items that Taro takes home.
|
import numpy as np
n, w = map(int, input().split())
wv = [list(map(int, input().split())) for i in range(n)]
table = np.zeros(w+1, dtype=np.int64)
for i in range(n):
l = table[:-wv[i][0]]
r = table[wv[i][0]:]
np.maximum(l + wv[i][1], r, out = r)
print(table[w-1])
|
s308321741
|
Accepted
| 174 | 14,604 | 276 |
import numpy as np
n, w = map(int, input().split())
wv = [list(map(int, input().split())) for i in range(n)]
table = np.zeros(w+1, dtype=np.int64)
for i in range(n):
l = table[:-wv[i][0]]
r = table[wv[i][0]:]
np.maximum(l + wv[i][1], r, out = r)
print(table[w])
|
s238503992
|
p03730
|
u932719058
| 2,000 | 262,144 |
Wrong Answer
| 34 | 2,940 | 159 |
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
a, b, c = map(int, input().split())
if a % b == 0 :
print('No')
exit()
for i in range(1, 10**5) :
if (a * i) % b == c :
print('Yes')
exit()
print('No')
|
s714061961
|
Accepted
| 34 | 2,940 | 159 |
a, b, c = map(int, input().split())
if a % b == 0 :
print('NO')
exit()
for i in range(1, 10**5) :
if (a * i) % b == c :
print('YES')
exit()
print('NO')
|
s586430113
|
p02646
|
u160224209
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 9,024 | 163 |
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())
oni = a + v*t
nige = b + w*t
if oni >= nige:
print("Yes")
else:
print("No")
|
s199835285
|
Accepted
| 23 | 9,116 | 326 |
a,v = map(int,input().split())
b,w = map(int,input().split())
t = int(input())
if a > b:
oni = a + (-1*v) * t
nige = b + (-1*w) * t
if oni <= nige:
print("YES")
else:
print("NO")
else:
oni = a + v*t
nige = b + w * t
if oni >= nige:
print("YES")
else:
print("NO")
|
s631883740
|
p03449
|
u678009529
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 368 |
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel?
|
n = int(input())
a = []
ans = 0
is_move = True
for i in range(2):
a.append(list(map(int, input().split())))
ans += a[0][0]
for i in range(n):
if i+1 == n:
ans += a[1][i]
else:
if sum(a[0][i:]) >= sum(a[1][i:]) and is_move == True:
ans += a[0][i+1]
else:
ans += a[1][i]
is_move = False
print(ans)
|
s178501836
|
Accepted
| 18 | 3,060 | 176 |
n = int(input())
a = []
ans = 0
for i in range(2):
a.append(list(map(int, input().split())))
for i in range(n):
ans = max(ans, sum(a[0][:(i+1)] + a[1][i:]))
print(ans)
|
s107500739
|
p02678
|
u764399371
| 2,000 | 1,048,576 |
Wrong Answer
| 665 | 34,764 | 594 |
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())
graph = [[] for _ in range(n)]
for i in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
graph[a].append(b)
graph[b].append(a)
que = deque()
que.append(0)
visited = [0 for _ in range(n)]
visited[0] = 1
ans = [None for _ in range(n)]
while len(que) > 0:
c = que.popleft()
for ne in graph[c]:
if visited[ne] == 1:
continue
que.append(ne)
ans[ne] = c
# print(f'ans[{ne}] = {c}')
visited[ne] = 1
for i in range(1, n):
print(ans[i] + 1)
|
s661138400
|
Accepted
| 656 | 34,976 | 607 |
from collections import deque
n, m = map(int, input().split())
graph = [[] for _ in range(n)]
for i in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
graph[a].append(b)
graph[b].append(a)
que = deque()
que.append(0)
visited = [0 for _ in range(n)]
visited[0] = 1
ans = [None for _ in range(n)]
print('Yes')
while len(que) > 0:
c = que.popleft()
for ne in graph[c]:
if visited[ne] == 1:
continue
que.append(ne)
ans[ne] = c
# print(f'ans[{ne}] = {c}')
visited[ne] = 1
for i in range(1, n):
print(ans[i] + 1)
|
s497966459
|
p04030
|
u764401543
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 265 |
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
s = list(input())
s.reverse()
print(''.join(s))
b = 0
ans = ''
for c in s:
if c == 'B':
b += 1
else:
if b > 0:
b -= 1
continue
else:
ans += c
ans = list(ans)
ans.reverse()
print(''.join(ans))
|
s255117149
|
Accepted
| 17 | 2,940 | 224 |
s = list(input())
s.reverse()
b = 0
ans = ''
for c in s:
if c == 'B':
b += 1
else:
if b > 0:
b -= 1
else:
ans += c
ans = list(ans)
ans.reverse()
print(''.join(ans))
|
s059476824
|
p03997
|
u760767494
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 73 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a + b) * h / 2)
|
s108663915
|
Accepted
| 17 | 2,940 | 79 |
a = int(input())
b = int(input())
h = int(input())
print(int((a + b) * h / 2))
|
s427361939
|
p03478
|
u750651325
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,136 | 147 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
N, A, B = map(int, input().split())
ans = 0
for i in range(1,N+1):
if B >= i >= A:
ans += i
elif i < B:
break
print(ans)
|
s350445307
|
Accepted
| 35 | 9,220 | 655 |
N, A, B = map(int, input().split())
ans = 0
for i in range(1,N+1):
if i < 10:
if A<=i<=B:
ans += i
elif i < 100:
a = i // 10
b = i - a*10
s = a+b
if A<=s<=B:
ans+=i
elif i < 1000:
a = i // 100
b = (i-a*100)//10
c = i-a*100-b*10
s = a+b+c
if A<=s<=B:
ans+=i
elif i < 10000:
a = i // 1000
b = (i-a*1000)//100
c = (i-a*1000-b*100)//10
d = i-a*1000-b*100-c*10
s = a+b+c+d
if A<=s<=B:
ans+=i
else:
s = 1
if A<=s<=B:
ans+=i
print(ans)
|
s656452073
|
p00037
|
u024715419
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,576 | 1,260 |
上から見ると図 1 のような形の格子状の広場があります。この格子の各辺に「壁」があるかないかを 0 と 1 の並びで表します。点 A に立って壁に右手をつき、壁に右手をついたまま、矢印の方向に歩き続けて再び点 A に戻ってくるまでの経路を出力するプログラムを作成してください。 --- 図1 ---
|
def move(position):
x = position[0]
y = position[1]
d = position[2]
if d == "L":
p = "DLUR"
elif d == "R":
p = "URDL"
elif d == "U":
p = "LURD"
else:
p = "RDLU"
for i in range(4):
if p[i] in grid[y][x]:
d = p[i]
if d == "L":
print("L",end="")
x -= 1
break
elif d == "R":
print("R",end="")
x += 1
break
elif d == "U":
print("U",end="")
y -= 1
break
else:
print("D",end="")
y += 1
break
return [x, y, d]
pos = [1,1,"R"]
grid = [["" for i in range(6)] for j in range(6)]
for i in range(9):
inp = input()
if i%2 == 0:
for j in range(4):
if inp[j] == "1":
grid[i//2 + 1][j + 1] += "R"
grid[i//2 + 1][j + 2] += "L"
else:
for j in range(5):
if inp[j] == "1":
grid[i//2 + 1][j + 1] += "D"
grid[i//2 + 2][j + 1] += "U"
print(*grid,sep="\n")
while True:
pos = move(pos)
if pos[0] == 1 and pos[1] == 1:
break
|
s781289891
|
Accepted
| 20 | 5,572 | 1,247 |
def move(position):
x = position[0]
y = position[1]
d = position[2]
if d == "L":
p = "DLUR"
elif d == "R":
p = "URDL"
elif d == "U":
p = "LURD"
else:
p = "RDLU"
for i in range(4):
if p[i] in grid[y][x]:
d = p[i]
if d == "L":
print("L",end="")
x -= 1
break
elif d == "R":
print("R",end="")
x += 1
break
elif d == "U":
print("U",end="")
y -= 1
break
else:
print("D",end="")
y += 1
break
return [x, y, d]
pos = [1,1,"R"]
grid = [["" for i in range(6)] for j in range(6)]
for i in range(9):
inp = input()
if i%2 == 0:
for j in range(4):
if inp[j] == "1":
grid[i//2 + 1][j + 1] += "R"
grid[i//2 + 1][j + 2] += "L"
else:
for j in range(5):
if inp[j] == "1":
grid[i//2 + 1][j + 1] += "D"
grid[i//2 + 2][j + 1] += "U"
while True:
pos = move(pos)
if pos[0] == 1 and pos[1] == 1:
break
print("")
|
s091949065
|
p03469
|
u779728630
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 34 |
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
s = input()
print("2018" + s[:4])
|
s466074955
|
Accepted
| 17 | 2,940 | 35 |
s = input()
print("2018" + s[4:])
|
s960679147
|
p02410
|
u474232743
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,684 | 203 |
Write a program which reads a $ n \times m$ matrix $A$ and a $m \times 1$ vector $b$, and prints their product $Ab$. A column vector with m elements is represented by the following equation. \\[ b = \left( \begin{array}{c} b_1 \\\ b_2 \\\ : \\\ b_m \\\ \end{array} \right) \\] A $n \times m$ matrix with $m$ column vectors, each of which consists of $n$ elements, is represented by the following equation. \\[ A = \left( \begin{array}{cccc} a_{11} & a_{12} & ... & a_{1m} \\\ a_{21} & a_{22} & ... & a_{2m} \\\ : & : & : & : \\\ a_{n1} & a_{n2} & ... & a_{nm} \\\ \end{array} \right) \\] $i$-th element of a $m \times 1$ column vector $b$ is represented by $b_i$ ($i = 1, 2, ..., m$), and the element in $i$-th row and $j$-th column of a matrix $A$ is represented by $a_{ij}$ ($i = 1, 2, ..., n,$ $j = 1, 2, ..., m$). The product of a $n \times m$ matrix $A$ and a $m \times 1$ column vector $b$ is a $n \times 1$ column vector $c$, and $c_i$ is obtained by the following formula: \\[ c_i = \sum_{j=1}^m a_{ij}b_j = a_{i1}b_1 + a_{i2}b_2 + ... + a_{im}b_m \\]
|
n, m = map(int, input().split())
a = [[v for v in list(map(int, input().split()))] for _ in range(n)]
b = [int(input()) for _ in range(m)]
print(sum([a[i][j] * b[j] for j in range(m)]) for i in range(n))
|
s361437672
|
Accepted
| 50 | 8,060 | 208 |
n, m = map(int, input().split())
a = [[v for v in list(map(int, input().split()))] for i in range(n)]
b = [int(input()) for j in range(m)]
for i in range(n):
print(sum([a[i][j] * b[j] for j in range(m)]))
|
s529831016
|
p03000
|
u094932051
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 241 |
A ball will bounce along a number line, making N + 1 bounces. It will make the first bounce at coordinate D_1 = 0, and the i-th bounce (2 \leq i \leq N+1) at coordinate D_i = D_{i-1} + L_{i-1}. How many times will the ball make a bounce where the coordinate is at most X?
|
while True:
try:
N, X = map(int, input().split())
L = list(map(int, input().split()))
D = 0
i = 0
while (D <= X):
D += L[i]
i += 1
print(i)
except:
break
|
s485842518
|
Accepted
| 17 | 3,060 | 347 |
while True:
try:
N, X = map(int, input().split())
L = list(map(int, input().split()))
D = [0]
for i in range(1, N+1):
tmp = D[i-1] + L[i-1]
D.append(tmp)
if tmp > X:
print(i)
break
else:
print(N+1)
except:
break
|
s152209372
|
p04011
|
u785578220
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 136 |
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
a = int(input())
b = int(input())
c = int(input())
d = int(input())
res = 0
if a<b:
res += c*a + d*(b-a)
else:res+=c*a
print(res)
|
s034238691
|
Accepted
| 18 | 2,940 | 136 |
a = int(input())
b = int(input())
c = int(input())
d = int(input())
res = 0
if a>b:
res += c*b + d*(a-b)
else:res+=c*a
print(res)
|
s928555935
|
p02392
|
u126478680
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,588 | 126 |
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
#! python3
# range.py
a, b, c = [int(x) for x in input().split()]
if a < b and b < c:
print('YES')
else:
print('NO')
|
s769098455
|
Accepted
| 20 | 5,592 | 127 |
#! python3
# range.py
a, b, c = [int(x) for x in input().split()]
if a < b and b < c:
print('Yes')
else:
print('No')
|
s490393480
|
p02257
|
u918276501
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,708 | 210 |
A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list.
|
n = int(input())
for i in range(n):
p = int(input())
if not p%2:
n -= 1
continue
for j in range(3,int(p**0.5),2):
if not p%j:
n -= 1
continue
print(n)
|
s813015608
|
Accepted
| 410 | 7,720 | 232 |
n = int(input())
for i in range(n):
p = int(input())
if not p%2:
if p != 2:
n -= 1
continue
for j in range(3,int(p**0.5)+1,2):
if not p%j:
n -= 1
break
print(n)
|
s526436315
|
p03433
|
u548545174
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 96 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N = int(input())
A = int(input())
if (N - A) % 500 == 0:
print('Yes')
else:
print('No')
|
s965766195
|
Accepted
| 18 | 2,940 | 91 |
N = int(input())
A = int(input())
if N % 500 <= A:
print('Yes')
else:
print('No')
|
s600709558
|
p03449
|
u594956556
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 247 |
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel?
|
N = int(input())
A1 = list(map(int, input().split()))
A2 = list(map(int, input().split()))
A2.reverse()
for i in range(N-1):
A1[i+1] += A1[i]
A2[i+1] += A2[i]
ans = 100000
for i in range(N):
ans = max(ans, A1[i]+A2[N-1-i])
print(ans)
|
s555126247
|
Accepted
| 17 | 3,060 | 242 |
N = int(input())
A1 = list(map(int, input().split()))
A2 = list(map(int, input().split()))
A2.reverse()
for i in range(N-1):
A1[i+1] += A1[i]
A2[i+1] += A2[i]
ans = 0
for i in range(N):
ans = max(ans, A1[i]+A2[N-1-i])
print(ans)
|
s674834768
|
p02646
|
u194472175
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,172 | 217 |
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())
dif = abs(B-A)
speed = V-W
if speed <= 0:
print('No')
else:
if speed*T >= dif:
print('Yes')
else:
print('No')
|
s101599980
|
Accepted
| 26 | 9,172 | 217 |
A,V = map(int,input().split())
B,W = map(int,input().split())
T = int(input())
dif = abs(B-A)
speed = V-W
if speed <= 0:
print('NO')
else:
if speed*T >= dif:
print('YES')
else:
print('NO')
|
s755240231
|
p03814
|
u224224351
| 2,000 | 262,144 |
Wrong Answer
| 41 | 6,180 | 208 |
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
s = list(input())
r = list(reversed(s))
for i in range(len(s)):
if s[i] == "A":
a = i +1
break
for i in range(len(r)):
if s[i] == "Z":
z = i +1
break
print(z-a+2)
|
s665364652
|
Accepted
| 41 | 6,180 | 209 |
s = list(input())
r = list(reversed(s))
for i in range(len(s)):
if s[i] == "A":
a = i +1
break
for j in range(len(r)):
if r[j] == "Z":
z = len(r) - j
break
print(z-a+1)
|
s642974324
|
p02694
|
u408791346
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,256 | 106 |
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())
m = 100
ans = 0
while m < x:
m = int(m*1.01)
print(m)
ans += 1
print(ans)
|
s348905034
|
Accepted
| 24 | 9,160 | 92 |
x = int(input())
m = 100
ans = 0
while m < x:
m = int(m*1.01)
ans += 1
print(ans)
|
s943681797
|
p03044
|
u387774811
| 2,000 | 1,048,576 |
Wrong Answer
| 910 | 43,228 | 503 |
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
import sys
input = sys.stdin.readline
N=int(input())
ki=[[] for f in range(N)]
for i in range(N-1):
u,v,w = map(int,input().split())
ki[u-1].append([v-1,w%2])
ki[v-1].append([u-1,w%2])
stack = [[0,0]]
check = [0]*N
ans=[2]*N
ans[0]=0
while stack != [] :
a=stack.pop()
if check[a[0]]==0:
check[a[0]]=1
for i in range(len(ki[a[0]])):
stack.append([ki[a[0]][i][0],(ki[a[0]][i][1]+ans[i])%2])
ans[ki[a[0]][i][0]]=(ki[a[0]][i][1]+ans[i])%2
for i in ans:
print(i)
|
s166399738
|
Accepted
| 846 | 43,008 | 584 |
from collections import deque
import sys
input = sys.stdin.readline
N=int(input())
ki=[[] for f in range(N)]
for i in range(N-1):
u,v,w = map(int,input().split())
ki[u-1].append([v-1,w%2])
ki[v-1].append([u-1,w%2])
stack = deque([[0,0]])
check = [0]*N
ans=[2]*N
ans[0]=0
while stack != deque([]) :
a=stack.pop()
ne=a[0]
if check[ne]==0:
check[ne]=1
for i in range(len(ki[ne])):
eda=ki[ne][i][0]
edanum=ki[ne][i][1]
edaans=(edanum+ans[ne])%2
if check[eda]==0:
stack.append([eda,edaans])
ans[eda]=edaans
for i in ans:
print(i)
|
s587006246
|
p03080
|
u758973277
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 84 |
There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat.
|
N = int(input())
s = input()
if s.count('R')>N/2:
print('YES')
else:
print('NO')
|
s305400587
|
Accepted
| 17 | 2,940 | 84 |
N = int(input())
s = input()
if s.count('R')>N/2:
print('Yes')
else:
print('No')
|
s148535372
|
p04029
|
u674190122
| 2,000 | 262,144 |
Wrong Answer
| 16 | 2,940 | 172 |
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?
|
given = [x for x in input()]
rets = ''
for i, v in enumerate(given):
if v == "B" and len(rets) > 0:
rets = rets[:-1]
else:
rets += v
print(rets)
|
s571740502
|
Accepted
| 19 | 3,060 | 48 |
N = int(input())
a = (N * (N + 1)) // 2
print(a)
|
s113633965
|
p02392
|
u507758132
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,588 | 108 |
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
a,b,c = map(int,input().split())
if a> b:
if b>c:
print("Yes")
else:
print("No")
else:
print("No")
|
s503038846
|
Accepted
| 20 | 5,588 | 90 |
a,b,c = map(int,input().split())
if a < b and b < c:
print("Yes")
else:
print("No")
|
s893763596
|
p02418
|
u692415695
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,432 | 107 |
Write a program which finds a pattern $p$ in a ring shaped text $s$.
|
# Belongs to : midandfeed aka asilentvoice
s = str(input())*2
q = str(input())
print(["NO", "YES"][q in s])
|
s742855997
|
Accepted
| 60 | 7,512 | 107 |
# Belongs to : midandfeed aka asilentvoice
s = str(input())*2
q = str(input())
print(["No", "Yes"][q in s])
|
s259334115
|
p02601
|
u353797797
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,228 | 564 |
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.
|
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
a,b,c=MI()
k=II()
while b<=a:
b*=2
k-=1
if k<0:
print("NO")
exit()
while c<=b:
c*=2
k-=1
if k<0:
print("NO")
exit()
print("YES")
|
s876558481
|
Accepted
| 31 | 9,208 | 564 |
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
a,b,c=MI()
k=II()
while b<=a:
b*=2
k-=1
if k<0:
print("No")
exit()
while c<=b:
c*=2
k-=1
if k<0:
print("No")
exit()
print("Yes")
|
s364225087
|
p03679
|
u202570162
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 130 |
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
x,a,b = map(int,input().split())
if a-b <= 0:
print('delicious')
elif a-b <= x:
print('safe')
else:
print('dangerous')
|
s325809793
|
Accepted
| 17 | 2,940 | 132 |
x,a,b = map(int,input().split())
if -a+b <= 0:
print('delicious')
elif -a+b <= x:
print('safe')
else:
print('dangerous')
|
s564058377
|
p02277
|
u918276501
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,776 | 929 |
Let's arrange a deck of cards. Your task is to sort totally n cards. A card consists of a part of a suit (S, H, C or D) and an number. Write a program which sorts such cards based on the following pseudocode: Partition(A, p, r) 1 x = A[r] 2 i = p-1 3 for j = p to r-1 4 do if A[j] <= x 5 then i = i+1 6 exchange A[i] and A[j] 7 exchange A[i+1] and A[r] 8 return i+1 Quicksort(A, p, r) 1 if p < r 2 then q = Partition(A, p, r) 3 run Quicksort(A, p, q-1) 4 run Quicksort(A, q+1, r) Here, A is an array which represents a deck of cards and comparison operations are performed based on the numbers. Your program should also report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
def swap(A,i,j):
A[i],A[j] = A[j],A[i]
return A
def isStable(A):
for i in range(1, len(A)):
if A[i][1] == A[i-1][1]:
if A[i][2] < A[i-1][2]:
return False
return True
def partition(A,p=0, r=None):
if r is None:
r = len(A)-1
x = A[r]
i = p-1
for j in range(p,r):
if A[j] <= x:
i += 1
swap(A,i,j)
swap(A,i+1,r)
return i+1
def quick(A,p=0,r=None):
"16c"
if r is None:
r = len(A)-1
if p < r:
q = partition(A,p,r)
quick(A,p,q-1)
quick(A,q+1,r)
import sys
if __name__ == "__main__":
n = int(input())
A = []
for i in range(n):
card = sys.stdin.readline().split()
A.append([card[0], int(card[1]), i])
quick(A)
if isStable(A):
print("Stable")
else:
print("Not Stable")
for card in A:
print(card[0], card[1])
|
s420761190
|
Accepted
| 1,120 | 24,720 | 944 |
def swap(A,i,j):
A[i],A[j] = A[j],A[i]
return A
def isStable(A):
for i in range(1, len(A)):
if A[i][1] == A[i-1][1]:
if A[i][2] < A[i-1][2]:
return False
return True
def partition(A,p=0, r=None):
if r is None:
r = len(A)-1
x = A[r]
i = p-1
for j in range(p,r):
if A[j][1] <= x[1]:
i += 1
swap(A,i,j)
swap(A,i+1,r)
return i+1
def quick(A,p=0,r=None):
if r is None:
r = len(A)-1
if p < r:
q = partition(A,p,r)
quick(A,p,q-1)
quick(A,q+1,r)
import sys
if __name__ == "__main__":
n = int(sys.stdin.readline())
A = []
for i in range(n):
card = sys.stdin.readline().split()
A.append([card[0], int(card[1]), i])
quick(A,0,n-1)
if isStable(A):
print("Stable")
else:
print("Not stable")
for card in A:
print(card[0], card[1])
|
s481843078
|
p03543
|
u168416324
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,028 | 102 |
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
x=list(input())
for i in range(10):
if x.count(i)>=3:
print("Yes")
break
else:
print("No")
|
s033353776
|
Accepted
| 25 | 9,000 | 124 |
x=list(input())
for i in range(10):
if x.count(str(i))>=3 and x[1]==x[2]:
print("Yes")
break
else:
print("No")
|
s551278881
|
p02612
|
u369796672
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,076 | 48 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
a = N//1000
print(N - 1000*a)
|
s264114697
|
Accepted
| 29 | 9,160 | 86 |
N = int(input())
a = N//1000
if 1000*a != N:
print(1000*(a+1)-N)
else:
print(0)
|
s302204855
|
p03909
|
u623349537
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 277 |
There is a grid with H rows and W columns. The square at the i-th row and j-th column contains a string S_{i,j} of length 5. The rows are labeled with the numbers from 1 through H, and the columns are labeled with the uppercase English letters from `A` through the W-th letter of the alphabet. Exactly one of the squares in the grid contains the string `snuke`. Find this square and report its location. For example, the square at the 6-th row and 8-th column should be reported as `H6`.
|
abc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
H, W = map(int, input().split())
S = [[] for i in range(H)]
for i in range(H):
S[i] = list(input().split())
for i in range(H):
for j in range(W):
if S[i][j] == "Snuke":
print(abc[j] + str(i))
break
|
s833651274
|
Accepted
| 18 | 3,060 | 281 |
abc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
H, W = map(int, input().split())
S = [[] for i in range(H)]
for i in range(H):
S[i] = list(input().split())
for i in range(H):
for j in range(W):
if S[i][j] == "snuke":
print(abc[j] + str(i + 1))
break
|
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