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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s267910497
|
p02390
|
u058933891
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,540 | 104 |
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 - s % 3600
m = s% 3600 - (s % 3600) % 60
s = (s % 3600) % 60
print ( "h:m:s")
|
s904451058
|
Accepted
| 30 | 7,664 | 166 |
s = int(input())
h = str(int((s - s % 3600 ) / 3600 ))
m = str(int((s% 3600 - (s % 3600) % 60) /60 ))
s = str(int((s % 3600) % 60 ))
print ( h + ":" + m + ":" + s)
|
s275942900
|
p03480
|
u762557532
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 6,264 | 543 |
You are given a string S consisting of `0` and `1`. Find the maximum integer K not greater than |S| such that we can turn all the characters of S into `0` by repeating the following operation some number of times. * Choose a contiguous segment [l,r] in S whose length is at least K (that is, r-l+1\geq K must be satisfied). For each integer i such that l\leq i\leq r, do the following: if S_i is `0`, replace it with `1`; if S_i is `1`, replace it with `0`.
|
from collections import Counter
S = input()
boundary = []
pre_state = '0'
for i in range(len(S)):
if S[i] != pre_state:
boundary.append(i)
pre_state = S[i]
else:
if S[-1] == '1':
boundary.append(len(S))
k = 2
while True:
lst_mod = list(map(lambda x: x % k, boundary))
count_mod = list(Counter(lst_mod).values())
if not any(list(map(lambda x: x % 2, count_mod))):
k += 1
else:
print(k-1)
exit()
|
s850012745
|
Accepted
| 63 | 6,388 | 458 |
S = input()
boundary = []
pre_state = '0'
for i in range(len(S)):
if S[i] != pre_state:
boundary.append(i)
pre_state = S[i]
else:
if S[-1] == '1':
boundary.append(len(S))
if len(boundary) == 0:
print(len(S))
exit()
lst = []
for i in range(len(boundary)):
lst.append(max(boundary[i], len(S)-boundary[i]))
ans = min(lst)
print(ans)
|
s013557974
|
p03623
|
u500297289
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 91 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x, a, b = map(int, input().split())
if abs(x - a) < abs(x - b):
print(a)
else:
print(b)
|
s756575343
|
Accepted
| 18 | 2,940 | 95 |
x, a, b = map(int, input().split())
if abs(x - a) < abs(x - b):
print('A')
else:
print('B')
|
s241978660
|
p02600
|
u347600233
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,152 | 47 |
M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * From 1000 through 1199: 5-kyu * From 1200 through 1399: 4-kyu * From 1400 through 1599: 3-kyu * From 1600 through 1799: 2-kyu * From 1800 through 1999: 1-kyu What kyu does M-kun have?
|
x = int(input())
print(8 - (1800 - 400) // 200)
|
s752289484
|
Accepted
| 26 | 9,152 | 44 |
x = int(input())
print(8 - (x - 400) // 200)
|
s036946679
|
p02865
|
u343850880
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 70 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n = int(input())
if n%2==0:
print(n/2-1)
else:
print((n-1)/2)
|
s255798094
|
Accepted
| 17 | 2,940 | 80 |
n = int(input())
if n%2==0:
print(int(n/2-1))
else:
print(int((n-1)/2))
|
s220983464
|
p02690
|
u042113240
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 9,188 | 154 |
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
|
X = int(input())
x = 1
while ((x-1)**4)*5 > X:
x += 1
for i in range(x):
for j in range(-x, i):
if i**5-j**5 == X:
a = [i,j]
print(*a)
|
s887421935
|
Accepted
| 33 | 9,116 | 182 |
X = int(input())
x = 1
while ((x-1)**4)*5 < X:
x += 1
for i in range(x):
for j in range(-x, i):
if i**5-j**5 == X:
a = [i,j]
break
print(*a)
|
s622091212
|
p03943
|
u894694822
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,060 | 105 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b, c =map(int,input().split())
if a+b == c or a+c == b or b+c == a:
print("YES")
else:
print("NO")
|
s106399646
|
Accepted
| 20 | 2,940 | 105 |
a, b, c =map(int,input().split())
if a+b == c or a+c == b or b+c == a:
print("Yes")
else:
print("No")
|
s521404114
|
p02743
|
u692687119
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 144 |
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
a, b, c = map(int, input().split())
A = a * a
B = b * b
C = c * c
X = ((C - A - B))-(4 * a * b)
if X > 0:
print('Yes')
else:
print('No')
|
s183578998
|
Accepted
| 32 | 2,940 | 164 |
a, b, c = map(int, input().split())
m = c - a - b
if m <= 0:
print('No')
else:
X = (m * m) - (4 * a * b)
if X > 0:
print('Yes')
else:
print('No')
|
s911398510
|
p03944
|
u143278390
| 2,000 | 262,144 |
Wrong Answer
| 342 | 12,540 | 816 |
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
import numpy as np
W,H,N=[int(i) for i in input().split()]
lst=[]
for j in range(N):
lst.append([int(i) for i in input().split()])
print(lst)
square=np.zeros(H*W,dtype=np.int64).reshape(H,W)
print(square)
for i in lst:
x=i[0]
y=i[1]
a=i[2]
if(a==1):
for j in range(x):
for k in range(H):
square[k][j]=1
#print(square)
elif(a==2):
for j in range(x,W):
for k in range(H):
square[k][j]=1
#print(square)
elif(a==3):
for j in range(H-y,H):
for k in range(W):
square[j][k]=1
#print(square)
else:
for j in range(H-y-1,-1,-1):
for k in range(W):
square[j][k]=1
#print(square)
print(np.count_nonzero(square ==0))
|
s349237183
|
Accepted
| 341 | 12,512 | 818 |
import numpy as np
W,H,N=[int(i) for i in input().split()]
lst=[]
for j in range(N):
lst.append([int(i) for i in input().split()])
#print(lst)
square=np.zeros(H*W,dtype=np.int64).reshape(H,W)
#print(square)
for i in lst:
x=i[0]
y=i[1]
a=i[2]
if(a==1):
for j in range(x):
for k in range(H):
square[k][j]=1
#print(square)
elif(a==2):
for j in range(x,W):
for k in range(H):
square[k][j]=1
#print(square)
elif(a==3):
for j in range(H-y,H):
for k in range(W):
square[j][k]=1
#print(square)
else:
for j in range(H-y-1,-1,-1):
for k in range(W):
square[j][k]=1
#print(square)
print(np.count_nonzero(square ==0))
|
s334394245
|
p02396
|
u566311709
| 1,000 | 131,072 |
Wrong Answer
| 80 | 5,904 | 153 |
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
a = []
while True:
n = int(input())
if n == 0:
break
a.append(n)
for i in range(len(a)):
print("Case {0}: {1}".format(i, a[i]))
|
s149474044
|
Accepted
| 80 | 5,908 | 144 |
a = []
while True:
n = int(input())
if n == 0:
break
a.append(n)
for i in range(len(a)):
print("Case {0}: {1}".format(i + 1, a[i]))
|
s689085092
|
p03605
|
u629540524
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 63 |
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
n = input()
if n in '9':
print('Yes')
else:
print('No')
|
s671245202
|
Accepted
| 17 | 2,940 | 63 |
n = input()
if '9' in n:
print('Yes')
else:
print('No')
|
s843771964
|
p00007
|
u350804311
| 1,000 | 131,072 |
Wrong Answer
| 40 | 7,620 | 126 |
Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week. Write a program which computes the amount of the debt in n weeks.
|
import math
n = int(input())
debt = 100
for i in range(n):
debt = math.ceil(debt * 1.05)
print(str(debt * 1000))
|
s995242092
|
Accepted
| 40 | 7,648 | 122 |
import math
n = int(input())
debt = 100
for i in range(n):
debt = math.ceil(debt * 1.05)
print(str(debt * 1000))
|
s336378846
|
p03574
|
u240793404
| 2,000 | 262,144 |
Wrong Answer
| 27 | 3,188 | 543 |
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())
mine = []
for i in range(h):
mine.append([])
inp = input()
for j in range(w):
mine[i].append(inp[j])
print(mine)
for i in range(h):
for j in range(w):
if mine[i][j] == '.':
cnt = 0
for k in range(3):
for l in range(3):
try:
if mine[i-1+k][j+l-1] == '#':
cnt+= 1
except:
continue
mine[i][j] = str(cnt)
print(mine)
|
s412259539
|
Accepted
| 29 | 3,188 | 467 |
import itertools
h,w=map(int,input().split())
s=[list(input()) for i in range(h)]
l=list(itertools.product([-1,0,1],repeat=2))
for i in range(h):
for j in range(w):
if s[i][j]=='.':
cnt = 0
for k in l:
x = i+k[0]
y = j+k[1]
if 0<=x<=h-1 and 0<=y<=w-1:
if s[x][y]=='#':
cnt+=1
s[i][j]=str(cnt)
for i in s:
print(''.join(i))
|
s830292021
|
p03478
|
u687044304
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 309 |
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).
|
# -*- coding:utf-8 -*-
def solve():
N, A, B = list(map(int, input().split()))
ans = 0
for i in range(N):
bit0 = i%10
bit1 = i//10
total = (bit0 + bit1)
if total >= A and total <= B:
ans += 1
print(ans)
if __name__ == "__main__":
solve()
|
s673789735
|
Accepted
| 23 | 3,060 | 411 |
# -*- coding:utf-8 -*-
def solve():
N, A, B = list(map(int, input().split()))
ans = 0
for i in range(1, N+1):
keta_sum = 0
num = i
while True:
if num == 0:
break
keta_sum += num%10
num = num//10
if keta_sum >= A and keta_sum <= B:
ans += i
print(ans)
if __name__ == "__main__":
solve()
|
s374628866
|
p02749
|
u023229441
| 2,000 | 1,048,576 |
Wrong Answer
| 1,046 | 167,732 | 1,422 |
We have a tree with N vertices. The vertices are numbered 1 to N, and the i-th edge connects Vertex a_i and Vertex b_i. Takahashi loves the number 3. He is seeking a permutation p_1, p_2, \ldots , p_N of integers from 1 to N satisfying the following condition: * For every pair of vertices (i, j), if the distance between Vertex i and Vertex j is 3, the sum or product of p_i and p_j is a multiple of 3. Here the distance between Vertex i and Vertex j is the number of edges contained in the shortest path from Vertex i to Vertex j. Help Takahashi by finding a permutation that satisfies the condition.
|
import sys
import math
mod=10**9+7
inf=float("inf")
from math import sqrt
from collections import deque
from collections import Counter
input=lambda: sys.stdin.readline().strip()
sys.setrecursionlimit(11451419)
from functools import lru_cache
n=int(input())
G=[[] for i in range(n)]
for i in range(n-1):
a,b=map(int,input().split())
G[a-1].append(b-1)
G[b-1].append(a-1)
color=[-1 for i in range(n)]
color[0]=0
def coloring(x):
for i in G[x]:
if color[i] != -1:continue
color[i]=1-color[x]
coloring(i)
coloring(0)
A=[];B=[]
ans=[0 for i in range(n)]
for i in range(n):
if color[i]==0:A.append(i)
else:B.append(i)
if len(A)<=n//3 or len(B)<=n//3:
if len(A)>len(B):
A,B=B,A
qq=3
for i in A:
ans[i]=qq
qq+=3
qq=deque([3*i+1 for i in range(int(n/3+0.9999))]+[3*i+2 for i in range(int(n/3+0.9999))]+[3*i for i in range(int(n//3),-1,-1)])
for t in range(n):
if ans[t]!=0:
ans[t]=qq.popleft()
print(*ans)
exit()
qq=deque([3*i+1 for i in range(int(n/3+0.9999))]+[3*i for i in range(int(n//3),0,-1)])
for i in A:
ans[i]=qq.popleft()
qq.extend([3*i+2 for i in range(int(n/3+0.9999))])
for j in B:
ans[j]=qq.pop()
print(*ans)
|
s912147142
|
Accepted
| 940 | 62,152 | 1,447 |
n=int(input())
dist=[-1]*n
G=[[] for i in range(n)]
for i in range(n-1):
a,b=map(int,input().split())
a-=1;b-=1
G[a].append(b)
G[b].append(a)
from collections import deque
queue=deque([[0,i] for i in G[0]])
dist[0]=0
while queue:
now,to=queue.pop()
# print(now,to)
if dist[to]!=-1:continue
dist[to]=1-dist[now]
for i in G[to]:
queue.appendleft([to,i])
# print(dist)
q=sum(dist)
LAST=[0]*n
if n/3 <= q and n/3 <= n-q:
A=[i for i in range(1,n+1,3)] + [0]*n
B=[i for i in range(2,n+1,3)] + [0]*n
now1=0
now2=0
for i in range(n):
if dist[i]:
LAST[i]=B[now2]
now2+=1
else:
LAST[i]=A[now1]
now1+=1
w=3
for i in range(n):
if LAST[i]==0:
LAST[i]=w
w+=3
print(*LAST)
elif n/3 >q:
w=3
for i in range(n):
if dist[i]:
LAST[i]=w
w+=3
A=[i for i in range(w,n+1,3)]+[i for i in range(1,n+1) if i%3!=0]
now=0
for i in range(n):
if LAST[i]==0:
LAST[i]=A[now]
now+=1
print(*LAST)
else:
w=3
for i in range(n):
if dist[i]==0:
LAST[i]=w
w+=3
A=[i for i in range(w,n+1,3)]+[i for i in range(1,n+1) if i%3!=0]
now=0
for i in range(n):
if LAST[i]==0:
LAST[i]=A[now]
now+=1
print(*LAST)
|
s733883849
|
p03351
|
u636162168
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 156 |
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,d=map(int,input().split())
if abs(a-c)<d:
print("Yes")
else:
if abs(a-b)<d and abs(b-c)<d:
print("Yes")
else: print("No")
|
s522320648
|
Accepted
| 17 | 2,940 | 159 |
a,b,c,d=map(int,input().split())
if abs(a-c)<=d:
print("Yes")
else:
if abs(a-b)<=d and abs(b-c)<=d:
print("Yes")
else: print("No")
|
s915962113
|
p03719
|
u346308892
| 2,000 | 262,144 |
Wrong Answer
| 26 | 2,940 | 102 |
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a,b,c=list(map(int,input().split(" ")))
if c>=a and c<=b:
print("yes")
else:
print("no")
|
s526063043
|
Accepted
| 18 | 3,064 | 98 |
a,b,c=list(map(int,input().split(" ")))
if c>=a and c<=b:
print("Yes")
else:
print("No")
|
s939502995
|
p03853
|
u781288689
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,064 | 127 |
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
m,n=input().split()
a=[]
for i in range(int(m)) :
a.append(input())
for i in a :
print(i)
for i in a :
print(i)
|
s583297587
|
Accepted
| 23 | 3,192 | 111 |
m,n=input().split()
a=[]
for i in range(int(m)) :
a.append(input())
for i in a :
print(i)
print(i)
|
s213047052
|
p03487
|
u729133443
| 2,000 | 262,144 |
Wrong Answer
| 42 | 14,692 | 47 |
You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N). Your objective is to remove some of the elements in a so that a will be a **good sequence**. Here, an sequence b is a **good sequence** when the following condition holds true: * For each element x in b, the value x occurs exactly x times in b. For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not. Find the minimum number of elements that needs to be removed so that a will be a good sequence.
|
n=int(input())
a=list(map(int,input().split()))
|
s658212626
|
Accepted
| 94 | 17,704 | 109 |
n,*a=map(int,open(0).read().split())
d={}
for i in a:d[i]=d.get(i,0)+1
print(sum(d[i]-i*(d[i]>=i)for i in d))
|
s686283952
|
p03447
|
u117193815
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 71 |
You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?
|
a = int(input())
if a%2==0:
print(int(a/2))
else:
print(a//2+1)
|
s129180405
|
Accepted
| 17 | 2,940 | 77 |
x = int(input())
a = int(input())
b = int(input())
print((x-a)%b)
|
s841873637
|
p03698
|
u856232850
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 89 |
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
a = input()
ans = 'no'
for i in a:
if a.count(i) >= 2:
ans = 'yes'
print(ans)
|
s476396770
|
Accepted
| 17 | 2,940 | 89 |
a = input()
ans = 'yes'
for i in a:
if a.count(i) >= 2:
ans = 'no'
print(ans)
|
s363062533
|
p03862
|
u923659712
| 2,000 | 262,144 |
Wrong Answer
| 92 | 14,252 | 146 |
There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies. Snuke can perform the following operation any number of times: * Choose a box containing at least one candy, and eat one of the candies in the chosen box. His objective is as follows: * Any two neighboring boxes contain at most x candies in total. Find the minimum number of operations required to achieve the objective.
|
n,x=map(int,input().split())
a=list(map(int,input().split()))
c=0
for i in range(n-1):
if a[i]+a[i+1]>x:
c+=(a[i]+a[i+1]-x)
print(c)
|
s816180957
|
Accepted
| 124 | 14,060 | 212 |
n,x=map(int,input().split())
a=list(map(int,input().split()))
c=0
if a[0] > x:
c= a[0] - x
a[0] = x
for i in range(0,n-1):
if a[i]+a[i+1]>x:
c+=(a[i]+a[i+1]-x)
a[i+1]-=(a[i]+a[i+1]-x)
print(c)
|
s973841587
|
p03644
|
u706414019
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,028 | 105 |
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())
a = 0
for i in range(N):
a = max(a,len(bin(64))-(bin(64)).rfind('1')-1)
print(2**a)
|
s818039581
|
Accepted
| 29 | 9,052 | 108 |
N = int(input())
counter = 0
for i in range(8):
if N // 2**i ==0:
print(2**(i-1))
break
|
s248132735
|
p03007
|
u201234972
| 2,000 | 1,048,576 |
Wrong Answer
| 247 | 14,092 | 1,126 |
There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.
|
N = int( input())
A = list( map( int, input().split()))
A.sort(reverse=True)
if N%2 == 0:
print( sum(A[:N//2]) - sum(A[N//2:]))
now = A[0]
print(now, A[N-1])
now -= A[N-1]
for i in range(2,N):
if i%2 == 0:
print(A[N-1-i//2], now)
now = A[N-1-i//2] - now
else:
print(A[i//2], now)
now = A[i//2] - now
else:
k = N//2
t = 1
if A[k] < 0:
t = 0
print( sum(A[:k+t]) - sum(A[k+t:]))
if t == 1:
now = A[N-1]
print(now, A[0])
now = now - A[0]
for i in range(2, N):
if i%2 == 0:
print(A[i//2], now)
now = A[i//2] - now
else:
print(A[N-1-i//2], now)
now = A[N-1-i//2] - now
else:
now = A[0]
print(now, A[N-1])
now -= A[N-1]
for i in range(2, N-1):
if i%2 == 0:
print(A[N-1-i//2], now)
now = A[N-1-i//2] - now
else:
print(A[i//2], now)
now = A[i//2] - now
print(now, A[k])
|
s391954554
|
Accepted
| 262 | 14,260 | 1,001 |
N = int( input())
A = list( map( int, input().split()))
A.sort(reverse = True)
plus = 0
zero = 0
for i in range(N):
if A[i] > 0:
plus += 1
elif A[i] == 0:
zero += 1
minus = N - plus - zero
if plus + zero== N:
print(sum(A)-A[-1]*2)
now = A[-1]
for i in range(N-2):
print(now, A[i])
now -= A[i]
print(A[N-2], now)
elif minus + zero== N:
print(A[0]*2 - sum(A))
now = A[0]
for i in range(1,N):
print(now, A[i])
now -= A[i]
elif plus+zero == 1:
print( sum( list( map( abs, A))))
now = A[0]
for i in range(1,N):
print(now, A[i])
now -= A[i]
else:
print( sum( list( map( abs, A))))
now = A[N-1]
k = N-1
for i in range(1,N-1):
if A[i] <= 0:
k = i
break
print(now, A[i])
now -= A[i]
print(A[0], now)
now = A[0] - now
for i in range(k,N-1):
if A[i] > 0:
pass
print(now, A[i])
now -= A[i]
|
s587759720
|
p03599
|
u382303205
| 3,000 | 262,144 |
Wrong Answer
| 3,156 | 3,444 | 1,144 |
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
a,b,c,d,e,f=map(int,input().split())
ans1 = 100
ans2 = 0
if a>b:
a, b = b, a
for _a in range(f//(a*100) + 1):
for _b in range((f - _a*100)//(b*100) + 1):
for _c in range((f - _a*100 - _b*100)//c + 1):
for _d in range((f - _a*100 - _b*100 - _c * c)//d + 1):
if a*100*_a+b*100*_b+c*_c+d*_d == 0:
continue
if a*100*_a+b*100*_b+c*_c+d*_d > f:
continue
if (c*_c+d*_d)/(a*100*_a+b*100*_b+c*_c+d*_d) == e/(100+e):
print(a*100*_a+b*100*_b+c*_c+d*_d, c*_c+d*_d)
break
elif (c*_c+d*_d)/(a*100*_a+b*100*_b+c*_c+d*_d) > e/(100+e):
continue
else:
if ans2/ans1 < (c*_c+d*_d)/(a*100*_a+b*100*_b+c*_c+d*_d):
ans1 = a*100*_a+b*100*_b+c*_c+d*_d
ans2 = c*_c+d*_d
print(_a)
else:
print(ans1, ans2)
|
s288696508
|
Accepted
| 468 | 3,064 | 1,062 |
import sys
a,b,c,d,e,f=map(int,input().split())
if a>b:
a, b = b, a
ans1 = a*100
ans2 = 0
for _a in range(f//(a*100) + 1):
for _b in range((f - _a*100)//(b*100) + 1):
max_sugar = min(f-(_a+_b)*100, (f-(_a+_b)*100)//100*e)
for _c in range(max_sugar//c + 1):
for _d in range((max_sugar-c*_c)//d + 1):
if a*100*_a+b*100*_b+c*_c+d*_d == 0:
continue
if a*100*_a+b*100*_b+c*_c+d*_d > f:
break
if (c*_c+d*_d)/(a*100*_a+b*100*_b+c*_c+d*_d) == e/(100+e):
print(a*100*_a+b*100*_b+c*_c+d*_d, c*_c+d*_d)
sys.exit(0)
elif (c*_c+d*_d)/(a*100*_a+b*100*_b+c*_c+d*_d) > e/(100+e):
continue
else:
if ans2/ans1 < (c*_c+d*_d)/(a*100*_a+b*100*_b+c*_c+d*_d):
ans1 = a*100*_a+b*100*_b+c*_c+d*_d
ans2 = c*_c+d*_d
else:
print(ans1, ans2)
|
s283435884
|
p03672
|
u223904637
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 305 |
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
s=list(input())
def ans(s):
n=len(s)//2
f=0
for i in range(n):
if s[i]!=s[i+n]:
f+=1
break
if f==0:
return True
else:
return False
s.pop(-1)
s.pop(-1)
kai=2
while not ans(s):
s.pop(-1)
s.pop(-1)
kai+=2
print(kai)
|
s944123057
|
Accepted
| 17 | 3,064 | 310 |
s=list(input())
h=len(s)
def ans(s):
n=len(s)//2
f=0
for i in range(n):
if s[i]!=s[i+n]:
f+=1
break
if f==0:
return True
else:
return False
s.pop(-1)
s.pop(-1)
kai=2
while not ans(s):
s.pop(-1)
s.pop(-1)
kai+=2
print(h-kai)
|
s994600985
|
p03149
|
u505547600
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 120 |
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
list = input().split() #
if 1 in list and 9 in list and 7 in list and 4 in list:
print("YES")
else:
print("NO")
|
s605732601
|
Accepted
| 17 | 2,940 | 240 |
list =input().split() #
var=[]
for a in list:
a = int(a)
var.append(a)
flg = False
if 1 in var:
if 9 in var:
if 7 in var:
if 4 in var:
flg = True
if flg:
print("YES")
else:
print("NO")
|
s369845362
|
p03860
|
u401183062
| 2,000 | 262,144 |
Wrong Answer
| 27 | 8,972 | 130 |
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.
|
def iroha():
string = input()
Capital = string[0]
print("A" + Capital + "C")
if __name__ == "__main__":
iroha()
|
s558269628
|
Accepted
| 25 | 9,072 | 204 |
def iroha():
head, string, heel = map(str, input().split())
headC = head[0]
Capital = string[0]
heelC = heel[0]
print(headC + Capital + heelC)
if __name__ == "__main__":
iroha()
|
s108551719
|
p03730
|
u924182136
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 166 |
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())
flag = False
for i in range(B):
tmp = i*A%B
if tmp == C:
flag = True
if flag:
print("Yes")
else:
print("No")
|
s299967916
|
Accepted
| 17 | 2,940 | 168 |
A,B,C = map(int,input().split())
flag = False
for i in range(B):
tmp = i*A%B
if tmp == C:
flag = True
if flag:
print("YES")
else:
print("NO")
|
s373176128
|
p03760
|
u310381103
| 2,000 | 262,144 |
Wrong Answer
| 24 | 9,128 | 135 |
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
o=input()
e=input()
print("".join([o[i]+e[i] for i in range(len(o)-1)]),end="")
print(o[-1],end="")
if len(o) != len(e):
print(e[-1])
|
s399684453
|
Accepted
| 29 | 9,128 | 135 |
o=input()
e=input()
print("".join([o[i]+e[i] for i in range(len(o)-1)]),end="")
print(o[-1],end="")
if len(o) == len(e):
print(e[-1])
|
s976515966
|
p03860
|
u023077142
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 38 |
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()
print("A{}C".format(S[0]))
|
s591963558
|
Accepted
| 17 | 2,940 | 52 |
_, S, _ = input().split()
print("A{}C".format(S[0]))
|
s656509168
|
p02744
|
u007627455
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 335 |
In this problem, we only consider strings consisting of lowercase English letters. Strings s and t are said to be **isomorphic** when the following conditions are satisfied: * |s| = |t| holds. * For every pair i, j, one of the following holds: * s_i = s_j and t_i = t_j. * s_i \neq s_j and t_i \neq t_j. For example, `abcac` and `zyxzx` are isomorphic, while `abcac` and `ppppp` are not. A string s is said to be in **normal form** when the following condition is satisfied: * For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison. For example, `abcac` is in normal form, but `zyxzx` is not since it is isomorphic to `abcac`, which is lexicographically smaller than `zyxzx`. You are given an integer N. Print all strings of length N that are in normal form, in lexicographically ascending order.
|
# coding: utf-8
n = int(input())
alphas = "abcdefghij"
init_str = ""
def standard(moji, x):
if len(moji) == n:
print(moji)
else:
for j in range(x):
if j == x:
standard(moji + alphas[j], i + 1)
else:
standard(moji + alphas[j], i)
standard(init_str, 0)
|
s031186915
|
Accepted
| 121 | 4,340 | 451 |
# coding: utf-8
n = int(input())
alphas = "abcdefghij"
init_str = ""
def standard(moji, x):
if len(moji) == n:
print(moji)
else:
for j in range(x+1):
if j == x:
# print("j == x")
standard(moji + alphas[j], x + 1)
else:
# print("j != x")
standard(moji + alphas[j], x)
standard(init_str, 0)
|
s326081303
|
p00012
|
u957840591
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,784 | 2,477 |
There is a triangle formed by three points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ on a plain. Write a program which prints "YES" if a point $P$ $(x_p, y_p)$ is in the triangle and "NO" if not.
|
class vector(object):
def __init__(self,a,b):
self.x=b.x-a.x
self.y=b.y-a.y
@staticmethod
def cross_product(a,b):
return a.x*b.y-a.y*b.x
class vertex(object):
def __init__(self,a):
self.x=a[0]
self.y=a[1]
class circle(object):
def __init__(self,p,r):
self.px=p.x
self.py=p.y
self.r=r
class triangle(object):
def __init__(self,a,b,c):
self.a=a
self.b=b
self.c=c
import math
self.ab=math.sqrt((self.a.x-self.b.x)**2+(self.a.y-self.b.y)**2)
self.bc=math.sqrt((self.b.x-self.c.x)**2+(self.b.y-self.c.y)**2)
self.ca=math.sqrt((self.c.x-self.a.x)**2+(self.c.y-self.a.y)**2)
c=self.ab
a=self.bc
b=self.ca
self.cosA=(b**2+c**2-a**2)/(2*b*c)
self.cosB=(a**2+c**2-b**2)/(2*a*c)
self.cosC=(b**2+a**2-c**2)/(2*b*a)
self.sinA=math.sqrt(1-self.cosA**2)
self.sinB=math.sqrt(1-self.cosB**2)
self.sinC=math.sqrt(1-self.cosC**2)
self.sin2A=2*self.sinA*self.cosA
self.sin2B=2*self.sinB*self.cosB
self.sin2C=2*self.sinC*self.cosC
def area(self):
import math
s=(self.ab+self.bc+self.ca)/2
S=math.sqrt(s*(s-self.ab)*(s-self.bc)*(s-self.ca))
return S
def circumscribed(self):
R=self.ab/(2*self.sinC)
px=(self.sin2A*self.a.x+self.sin2B*self.b.x+self.sin2C*self.c.x)/(self.sin2A+self.sin2B+self.sin2C)
py=(self.sin2A*self.a.y+self.sin2B*self.b.y+self.sin2C*self.c.y)/(self.sin2A+self.sin2B+self.sin2C)
px=round(px,3)
py=round(py,3)
R=round(R,3)
p=vertex((px,py))
return circle(p,R)
def isin(self,p):
AB=vector(self.a,self.b)
BC=vector(self.b,self.c)
CA=vector(self.c,self.a)
AP=vector(self.a,p)
BP=vector(self.b,p)
CP=vector(self.c,p)
if (vector.cross_product(AB,AP)>0 and vector.cross_product(BC,BP)>0 and vector.cross_product(CA,CP)>0)or(vector.cross_product(AB,AP)<0 and vector.cross_product(BC,BP)<0 and vector.cross_product(CA,CP)<0):
return 'Yes'
else:return 'No'
A=[]
B=[]
C=[]
p=[]
import sys
for line in sys.stdin:
a,b,c,d,e,f,g,h=list(map(float,line.split()))
A.append(vertex((a,b)))
B.append(vertex((c,d)))
C.append(vertex((e,f)))
p.append(vertex((g,h)))
for i in range(len(A)):
Triangle=triangle(A[i],B[i],C[i])
print(Triangle.isin(p[i]))
|
s732955672
|
Accepted
| 30 | 7,744 | 2,477 |
class vector(object):
def __init__(self,a,b):
self.x=b.x-a.x
self.y=b.y-a.y
@staticmethod
def cross_product(a,b):
return a.x*b.y-a.y*b.x
class vertex(object):
def __init__(self,a):
self.x=a[0]
self.y=a[1]
class circle(object):
def __init__(self,p,r):
self.px=p.x
self.py=p.y
self.r=r
class triangle(object):
def __init__(self,a,b,c):
self.a=a
self.b=b
self.c=c
import math
self.ab=math.sqrt((self.a.x-self.b.x)**2+(self.a.y-self.b.y)**2)
self.bc=math.sqrt((self.b.x-self.c.x)**2+(self.b.y-self.c.y)**2)
self.ca=math.sqrt((self.c.x-self.a.x)**2+(self.c.y-self.a.y)**2)
c=self.ab
a=self.bc
b=self.ca
self.cosA=(b**2+c**2-a**2)/(2*b*c)
self.cosB=(a**2+c**2-b**2)/(2*a*c)
self.cosC=(b**2+a**2-c**2)/(2*b*a)
self.sinA=math.sqrt(1-self.cosA**2)
self.sinB=math.sqrt(1-self.cosB**2)
self.sinC=math.sqrt(1-self.cosC**2)
self.sin2A=2*self.sinA*self.cosA
self.sin2B=2*self.sinB*self.cosB
self.sin2C=2*self.sinC*self.cosC
def area(self):
import math
s=(self.ab+self.bc+self.ca)/2
S=math.sqrt(s*(s-self.ab)*(s-self.bc)*(s-self.ca))
return S
def circumscribed(self):
R=self.ab/(2*self.sinC)
px=(self.sin2A*self.a.x+self.sin2B*self.b.x+self.sin2C*self.c.x)/(self.sin2A+self.sin2B+self.sin2C)
py=(self.sin2A*self.a.y+self.sin2B*self.b.y+self.sin2C*self.c.y)/(self.sin2A+self.sin2B+self.sin2C)
px=round(px,3)
py=round(py,3)
R=round(R,3)
p=vertex((px,py))
return circle(p,R)
def isin(self,p):
AB=vector(self.a,self.b)
BC=vector(self.b,self.c)
CA=vector(self.c,self.a)
AP=vector(self.a,p)
BP=vector(self.b,p)
CP=vector(self.c,p)
if (vector.cross_product(AB,AP)>0 and vector.cross_product(BC,BP)>0 and vector.cross_product(CA,CP)>0)or(vector.cross_product(AB,AP)<0 and vector.cross_product(BC,BP)<0 and vector.cross_product(CA,CP)<0):
return 'YES'
else:return 'NO'
A=[]
B=[]
C=[]
p=[]
import sys
for line in sys.stdin:
a,b,c,d,e,f,g,h=list(map(float,line.split()))
A.append(vertex((a,b)))
B.append(vertex((c,d)))
C.append(vertex((e,f)))
p.append(vertex((g,h)))
for i in range(len(A)):
Triangle=triangle(A[i],B[i],C[i])
print(Triangle.isin(p[i]))
|
s277513023
|
p02936
|
u591503175
| 2,000 | 1,048,576 |
Wrong Answer
| 2,115 | 190,244 | 1,030 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
N, Q = [int(item) for item in input().split()]
tree_list = [input() for _ in range(1,N)]
tree_list_int = [list(map(int,i.split())) for i in tree_list]
action_list = [input() for _ in range(Q)]
action_list_int = [list(map(int, i.split())) for i in action_list]
print('tree', tree_list_int)
print('act', action_list_int)
class Node:
def __init__(self, data):
self.data = data
self.cnt = 0
self.child = []
class tree:
def __init__(self, N, tree_list_int):
self.node_list = [Node(i) for i in range(1,N+1)]
for i, j in tree_list_int:
self.node_list[i-1].child.append(self.node_list[j-1])
def adder(self, node, val):
node.cnt += val
if not node.child:
return node.cnt
for next_child in node.child:
self.adder(next_child, val)
ins=tree(N, tree_list_int)
for pos, val in action_list_int:
ins.adder(ins.node_list[pos-1], val)
for i in range(N):
print(i+1, ins.node_list[i].cnt)
|
s473717662
|
Accepted
| 1,919 | 208,132 | 762 |
N, Q = [int(item) for item in input().split()]
tree_list = [input().split() for j in range(1, N)]
query_list = [input().split() for k in range(Q)]
query_list_int = [[int(k) for k in i] for i in query_list]
val_list = [0 for _ in range(N)]
linked_node_list = [[] for _ in range(N)]
for a, b in tree_list:
a, b = int(a)-1, int(b) -1
linked_node_list[a].append(b)
linked_node_list[b].append(a)
for index, val in query_list_int:
val_list[index-1] += val
stack = [0]
parent = [0] * (N+1)
while True:
v=stack.pop()
for child in linked_node_list[v]:
if child != parent[v]:
parent[child] = v
stack.append(child)
val_list[child] += val_list[v]
if not stack:
break
print(*val_list)
|
s822364099
|
p00050
|
u868716420
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,716 | 210 |
福島県は果物の産地としても有名で、その中でも特に桃とりんごは全国でも指折りの生産量を誇っています。ところで、ある販売用の英文パンフレットの印刷原稿を作ったところ、手違いでりんごに関する記述と桃に関する記述を逆に書いてしまいました。 あなたは、apple と peach を修正する仕事を任されましたが、なにぶん面倒です。１行の英文を入力して、そのなかの apple という文字列を全て peach に、peach という文字列を全てapple に交換した英文を出力するプログラムを作成してください。
|
from collections import deque
result = deque()
for _ in input().split():
if 'apple' in _ : result.append('peach')
elif 'peach' in _ : result.append('apple')
else : result.append(_)
print(*result)
|
s851499367
|
Accepted
| 30 | 7,804 | 379 |
from collections import deque
result = deque()
for _ in input().split():
if 'apple' in _ :
if _ == 'apple' : result.append('peach')
else : result.append(_.replace('apple', 'peach'))
elif 'peach' in _ :
if _ == 'peach' : result.append('apple')
else : result.append(_.replace('peach', 'apple'))
else : result.append(_)
print(*result)
|
s794765007
|
p02646
|
u180704972
| 2,000 | 1,048,576 |
Wrong Answer
| 2,206 | 9,144 | 295 |
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.
|
AV = list(map(int,input().split()))
BW = list(map(int,input().split()))
T = int(input())
A = AV[0]
V = AV[1]
B = BW[0]
W = BW[1]
for i in range(T):
if A <= B:
B += W
A += V
else:
B -= W
A -= V
if abs(A) >= abs(B):
print('Yes')
else:
print('No')
|
s224622249
|
Accepted
| 23 | 9,216 | 396 |
AV = list(map(int,input().split()))
BW = list(map(int,input().split()))
T = int(input())
A = AV[0]
V = AV[1]
B = BW[0]
W = BW[1]
pos_A = A
pos_B = B
if A <= B:
pos_B = B + T*W
pos_A = A + T*V
if pos_A >= pos_B:
print('YES')
else:
print('NO')
else:
pos_B = B - T*W
pos_A = A - T*V
if pos_A <= pos_B:
print('YES')
else:
print('NO')
|
s033333439
|
p03694
|
u690536347
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 52 |
It is only six months until Christmas, and AtCoDeer the reindeer is now planning his travel to deliver gifts. There are N houses along _TopCoDeer street_. The i-th house is located at coordinate a_i. He has decided to deliver gifts to all these houses. Find the minimum distance to be traveled when AtCoDeer can start and end his travel at any positions.
|
l=sorted(map(int,input().split()))
print(l[-1]-l[0])
|
s507346682
|
Accepted
| 17 | 2,940 | 68 |
N=int(input())
l=sorted(map(int,input().split()))
print(l[-1]-l[0])
|
s688451157
|
p02614
|
u562662744
| 1,000 | 1,048,576 |
Wrong Answer
| 238 | 27,244 | 710 |
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.
|
import numpy as np
import itertools
h,w,k=map(int,input().split())
g=np.ones((h,w))
l=['' for _ in range(h)]
for i in range(h):
l[i]=str(input())
for i in range(h):
for j in range(w):
if l[i][j]=='.': g[i][j]=0
s=np.sum(g)
ans=0
for i in range(h):
for j in range(w):
u=list(itertools.combinations(range(h),i))
v=list(itertools.combinations(range(w),j))
for x in u:
for y in v:
a=0
b=0
c=0
for p in x:
a+=np.sum(g[p,:])
print(a)
for q in y:
b+=np.sum(g[:,q])
print(b)
for p in x:
for q in y:
c+=g[p][q]
print(s-a-b+c)
if s-a-b+c==k:
ans+=1
print(ans)
|
s656623737
|
Accepted
| 225 | 26,984 | 653 |
import numpy as np
import itertools
h,w,k=map(int,input().split())
g=np.ones((h,w))
l=['' for _ in range(h)]
for i in range(h):
l[i]=str(input())
for i in range(h):
for j in range(w):
if l[i][j]=='.': g[i][j]=0
s=np.sum(g)
ans=0
for i in range(h):
for j in range(w):
u=list(itertools.combinations(range(h),i))
v=list(itertools.combinations(range(w),j))
for x in u:
for y in v:
a=0
b=0
c=0
for p in x:
a+=np.sum(g[p,:])
for q in y:
b+=np.sum(g[:,q])
for p in x:
for q in y:
c+=g[p][q]
if s-a-b+c==k:
ans+=1
print(ans)
|
s277982275
|
p03478
|
u881116515
| 2,000 | 262,144 |
Wrong Answer
| 28 | 3,060 | 262 |
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(2):
for j in range(10):
for k in range(10):
for l in range(10):
for m in range(10):
if i*10000+j*1000+k*100+l*10+m <= n and a <= i+j+k+l+m <= b:
ans += 1
print(ans)
|
s071081707
|
Accepted
| 30 | 3,060 | 288 |
n,a,b = map(int,input().split())
ans = 0
for i in range(2):
for j in range(10):
for k in range(10):
for l in range(10):
for m in range(10):
if i*10000+j*1000+k*100+l*10+m <= n and a <= i+j+k+l+m <= b:
ans += i*10000+j*1000+k*100+l*10+m
print(ans)
|
s865062059
|
p02578
|
u304593245
| 2,000 | 1,048,576 |
Wrong Answer
| 174 | 32,188 | 190 |
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
N = int(input())
A = list(map(int, input().split()))
ans = 0
for i in range(1,N):
tmp = A[i-1] - A[i]
if A[i-1] > A[i]:
A[i] += tmp
ans += tmp
print(A)
print(ans)
|
s942886207
|
Accepted
| 159 | 32,372 | 191 |
N = int(input())
A = list(map(int, input().split()))
ans = 0
for i in range(1,N):
tmp = A[i-1] - A[i]
if A[i-1] > A[i]:
A[i] += tmp
ans += tmp
#print(A)
print(ans)
|
s748307212
|
p03455
|
u834157170
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 87 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
if (a*b)//2 == 0:
print('Even')
else:
print('Odd')
|
s711797023
|
Accepted
| 17 | 2,940 | 88 |
a, b = map(int, input().split())
if (a*b) % 2 == 0:
print('Even')
else:
print('Odd')
|
s602427267
|
p03698
|
u684305751
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 55 |
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s=input()
print("Yes" if len(s)==len(set(s)) else "No")
|
s888973108
|
Accepted
| 17 | 2,940 | 55 |
s=input()
print("yes" if len(s)==len(set(s)) else "no")
|
s082893182
|
p03998
|
u278356323
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 522 |
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.
|
#ABC045a
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**6)
a = input()[:-1]
b = input()[:-1]
c = input()[:-1]
nextPerson = "a"
while (True):
if (nextPerson == "a"):
if (len(a) == 1):
break
nextPerson = a[0]
a = a[1:]
elif (nextPerson == "b"):
if (len(b) == 1):
break
nextPerson = b[0]
b = b[1:]
else:
if (len(c) == 1):
break
nextPerson = "c"
c = c[1:]
print(nextPerson.capitalize())
|
s521763637
|
Accepted
| 17 | 3,064 | 580 |
#ABC045a
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**6)
a = input()[:-1]
b = input()[:-1]
c = input()[:-1]
aa = 0
bb = 0
cc = 0
nextPerson = "a"
while (True):
#print(nextPerson)
if (nextPerson == "a"):
if (len(a) - aa == 0):
break
nextPerson = a[aa]
aa += 1
elif (nextPerson == "b"):
if (len(b) - bb == 0):
break
nextPerson = b[bb]
bb += 1
else:
if (len(c) - cc == 0):
break
nextPerson = c[cc]
cc += 1
print(nextPerson.capitalize())
|
s022447860
|
p03369
|
u226779434
| 2,000 | 262,144 |
Wrong Answer
| 24 | 9,024 | 43 |
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
s = input()
print(700 + s.count("○")*100)
|
s267103814
|
Accepted
| 27 | 8,984 | 41 |
s = input()
print(700 + s.count("o")*100)
|
s761461202
|
p03486
|
u595375942
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 145 |
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s = input()
t = input()
ss = ''.join(sorted(s))
tt = ''.join(sorted(t, reverse=True))
if ss > tt:
ans = 'Yes'
else:
ans = 'No'
print(ans)
|
s264594899
|
Accepted
| 18 | 2,940 | 129 |
S = ''.join(sorted(list(input())))
T = ''.join(reversed(sorted(list(input()))))
if S < T:
print('Yes')
else:
print('No')
|
s573553173
|
p03477
|
u494748969
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 150 |
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
a, b, c, d = list(map(int, input().split()))
if a + b > c + d:
print("left")
elif a + b == c + d:
print("Balanced")
else:
print("Right")
|
s093269424
|
Accepted
| 17 | 2,940 | 150 |
a, b, c, d = list(map(int, input().split()))
if a + b > c + d:
print("Left")
elif a + b == c + d:
print("Balanced")
else:
print("Right")
|
s653002147
|
p03854
|
u588526762
| 2,000 | 262,144 |
Wrong Answer
| 72 | 3,188 | 226 |
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()
while 1:
for x in ('dream', 'dreamer', 'erase', 'eraser'):
if s.endswith(x):
s = s[:-len(x)]
break
else:
print('NO')
break
if not s:
print('YES')
|
s271078865
|
Accepted
| 71 | 3,316 | 240 |
s = input()
while 1:
for x in ('dream', 'dreamer', 'erase', 'eraser'):
if s.endswith(x):
s = s[:-len(x)]
break
else:
print('NO')
break
if not s:
print('YES')
break
|
s356393071
|
p03997
|
u432805419
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 62 |
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()) for i in range(3)]
print((a[0]+a[1])*a[2]/2)
|
s964037557
|
Accepted
| 18 | 2,940 | 67 |
a = [int(input()) for _ in range(3)]
print(int((a[0]+a[1])*a[2]/2))
|
s909905497
|
p04043
|
u457901067
| 2,000 | 262,144 |
Wrong Answer
| 16 | 2,940 | 84 |
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.
|
X = list(map(int, input().split())).sort()
print("YES" if X == [5, 5, 7] else "NO")
|
s800731055
|
Accepted
| 17 | 2,940 | 85 |
X = sorted(list(map(int, input().split())))
print("YES" if X == [5, 5, 7] else "NO")
|
s746391181
|
p02411
|
u777299405
| 1,000 | 131,072 |
Wrong Answer
| 40 | 6,724 | 435 |
Write a program which reads a list of student test scores and evaluates the performance for each student. The test scores for a student include scores of the midterm examination m (out of 50), the final examination f (out of 50) and the makeup examination r (out of 100). If the student does not take the examination, the score is indicated by -1. The final performance of a student is evaluated by the following procedure: * If the student does not take the midterm or final examination, the student's grade shall be F. * If the total score of the midterm and final examination is greater than or equal to 80, the student's grade shall be A. * If the total score of the midterm and final examination is greater than or equal to 65 and less than 80, the student's grade shall be B. * If the total score of the midterm and final examination is greater than or equal to 50 and less than 65, the student's grade shall be C. * If the total score of the midterm and final examination is greater than or equal to 30 and less than 50, the student's grade shall be D. However, if the score of the makeup examination is greater than or equal to 50, the grade shall be C. * If the total score of the midterm and final examination is less than 30, the student's grade shall be F.
|
while True:
m, f, r = map(int, (input().split()))
score = m + f
if m == f == r == -1:
break
else:
if m == -1 or f == -1:
print("F")
elif score >= 80:
print("A")
elif 80 > score >= 65:
print("B")
elif 65 > score >= 50:
print("C")
elif 50 > score >= 35 and r >= 50:
print("C")
else:
print("D")
|
s090028462
|
Accepted
| 30 | 6,724 | 500 |
while True:
m, f, r = map(int, (input().split()))
score = m + f
if m == f == r == -1:
break
else:
if m == -1 or f == -1:
print("F")
elif score >= 80:
print("A")
elif 80 > score >= 65:
print("B")
elif 65 > score >= 50:
print("C")
elif 50 > score >= 30 and r >= 50:
print("C")
elif 50 > score >= 30 and r < 50:
print("D")
else:
print("F")
|
s352624787
|
p02669
|
u189479417
| 2,000 | 1,048,576 |
Wrong Answer
| 441 | 14,016 | 1,201 |
You start with the number 0 and you want to reach the number N. You can change the number, paying a certain amount of coins, with the following operations: * Multiply the number by 2, paying A coins. * Multiply the number by 3, paying B coins. * Multiply the number by 5, paying C coins. * Increase or decrease the number by 1, paying D coins. You can perform these operations in arbitrary order and an arbitrary number of times. What is the minimum number of coins you need to reach N? **You have to solve T testcases.**
|
def solve(N, A, B, C, D):
d = {0:0, 1:D}
s = set()
return dfs(N, A, B, C, D, d, s)
def dfs(n, A, B, C, D, d, s):
if n in d:
return d[n]
if n in s:
return float('inf')
s.add(n)
ans = float('inf')
p = n % 2
if p == 0:
ans = min(ans, dfs(n // 2, A, B, C, D, d, s) + A)
else:
ans = min(ans, dfs((n - p) // 2, A, B, C, D, d, s) + A + D * p)
ans = min(ans, dfs((n + (2 - p)) // 2, A, B, C, D, d, s) + A + D * (2 - p))
q = n % 3
if q == 0:
ans = min(ans, dfs(n // 3, A, B, C, D, d, s) + B)
else:
ans = min(ans, dfs((n - q) // 3, A, B, C, D, d, s) + B + D * q)
ans = min(ans, dfs(n + (3 - q), A, B, C, D, d, s) + B + D * (3 - q))
r = n % 5
if r == 0:
ans = min(ans, dfs(n // 5, A, B, C, D, d, s) + C)
else:
ans = min(ans, dfs((n - r) // 5, A, B, C, D, d, s) + C + D * r)
ans = min(ans, dfs((n + (5 - r)) // 5, A, B, C, D, d, s) + C + D * (5 - r))
ans = min(ans, D * n)
d[n] = ans
return ans
T = int(input())
for _ in range(T):
N, A, B, C, D = map(int,input().split())
ans = solve(N, A, B, C, D)
print(ans)
|
s783722762
|
Accepted
| 315 | 11,712 | 1,208 |
def solve(N, A, B, C, D):
d = {0:0, 1:D}
s = set()
return dfs(N, A, B, C, D, d, s)
def dfs(n, A, B, C, D, d, s):
if n in d:
return d[n]
if n in s:
return float('inf')
s.add(n)
ans = float('inf')
p = n % 2
if p == 0:
ans = min(ans, dfs(n // 2, A, B, C, D, d, s) + A)
else:
ans = min(ans, dfs((n - p) // 2, A, B, C, D, d, s) + A + D * p)
ans = min(ans, dfs((n + (2 - p)) // 2, A, B, C, D, d, s) + A + D * (2 - p))
q = n % 3
if q == 0:
ans = min(ans, dfs(n // 3, A, B, C, D, d, s) + B)
else:
ans = min(ans, dfs((n - q) // 3, A, B, C, D, d, s) + B + D * q)
ans = min(ans, dfs((n + (3 - q)) // 3, A, B, C, D, d, s) + B + D * (3 - q))
r = n % 5
if r == 0:
ans = min(ans, dfs(n // 5, A, B, C, D, d, s) + C)
else:
ans = min(ans, dfs((n - r) // 5, A, B, C, D, d, s) + C + D * r)
ans = min(ans, dfs((n + (5 - r)) // 5, A, B, C, D, d, s) + C + D * (5 - r))
ans = min(ans, D * n)
d[n] = ans
return ans
T = int(input())
for _ in range(T):
N, A, B, C, D = map(int,input().split())
ans = solve(N, A, B, C, D)
print(ans)
|
s484070762
|
p02694
|
u373703188
| 2,000 | 1,048,576 |
Wrong Answer
| 22 | 9,012 | 590 |
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
# import math
x = int(input())
cnt = 0
answer = 100
# fv = 100 (1 + 1)
# x = x / 100
# print(math.log(2, x))
# # ans += 100 * 0.01
# if x <= ans:
# break
# print(answer)
while answer <= x:
cnt += 1
answer += answer // 100
print(cnt)
|
s980863784
|
Accepted
| 20 | 9,168 | 679 |
# import math
x = int(input())
cnt = 0
answer = 100
# fv = 100 (1 + 1)
# x = x / 100
# print(math.log(2, x))
# # ans += 100 * 0.01
# if x <= ans:
# break
# print(answer)
while answer < x:
cnt += 1
answer += answer // 100
# print(answer)
# print(x)
print(cnt)
|
s448280371
|
p03609
|
u617010143
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 134 |
We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand. How many grams of sand will the upper bulb contains after t seconds?
|
s = input()
hoge =list(s)
moji = ''
for num in range(len(s)):
if num%2 == 0:
moji += hoge[num]
continue
print(moji)
|
s999635759
|
Accepted
| 17 | 2,940 | 74 |
X,t =map(int,input().split())
if X >= t:
print(X-t)
else:
print(0)
|
s798556006
|
p03067
|
u864650257
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 119 |
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
A,B,C = map (int,input().split())
if A < C < B:
print("YES")
elif A > C > B:
print("YES")
else:
print("NO")
|
s906704671
|
Accepted
| 17 | 2,940 | 119 |
A,B,C = map (int,input().split())
if A < C < B:
print("Yes")
elif A > C > B:
print("Yes")
else:
print("No")
|
s501030841
|
p03636
|
u750651325
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 46 |
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
S = input()
l = len(S)-2
print(S[0]+"l"+S[-1])
|
s595418048
|
Accepted
| 17 | 2,940 | 49 |
S = input()
l = str(len(S)-2)
print(S[0]+l+S[-1])
|
s897538989
|
p03068
|
u254745082
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 224 |
You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.
|
n = int(input())
s = input()
k = int(input())
str_list = list(s)
x = str_list[k-1]
print(n,s,k,x)
for i in range(len(s)):
if str_list[i] != x:
str_list[i] = "*"
str_changed = "".join(str_list)
print(str_changed)
|
s221606455
|
Accepted
| 18 | 3,060 | 209 |
n = int(input())
s = input()
k = int(input())
str_list = list(s)
x = str_list[k-1]
for i in range(len(s)):
if str_list[i] != x:
str_list[i] = "*"
str_changed = "".join(str_list)
print(str_changed)
|
s867253601
|
p02239
|
u255317651
| 1,000 | 131,072 |
Wrong Answer
| 30 | 5,624 | 752 |
Write a program which reads an directed graph $G = (V, E)$, and finds the shortest distance from vertex $1$ to each vertex (the number of edges in the shortest path). Vertices are identified by IDs $1, 2, ... n$.
|
# -*- coding: utf-8 -*-
"""
Created on Sun Jul 8 10:56:33 2018
@author: maezawa
"""
n = int(input())
adj = [[] for _ in range(n)]
d = [-1 for _ in range(n)]
d[0] = 0
for j in range(n):
ain = list(map(int, input().split()))
u = ain[0]
k = ain[1]
for i in range(k):
adj[u-1].append(ain[i+2]-1)
adj[u-1].sort()
#print(*adj[u-1])
stack = [0]
while stack:
print(stack)
current = stack[-1]
flag = 0
for k in adj[current]:
if d[k] <= 0:
d[k] = d[current]+1
stack.append(k)
flag = 1
if flag == 0:
stack.pop()
#print('current = {}, t = {}'.format(current, t))
#print('stack', *stack)
for i in range(n):
print('{} {}'.format(i+1, d[i]))
|
s603417613
|
Accepted
| 20 | 5,632 | 867 |
# -*- coding: utf-8 -*-
"""
Created on Sun Jul 8 10:56:33 2018
@author: maezawa
"""
n = int(input())
adj = [[] for _ in range(n)]
d = [-1 for _ in range(n)]
d[0] = 0
for j in range(n):
ain = list(map(int, input().split()))
u = ain[0]
k = ain[1]
for i in range(k):
adj[u-1].append(ain[i+2]-1)
adj[u-1].sort()
#print(*adj[u-1])
stack = [0]
while stack:
#print(stack)
current = stack[-1]
flag = 0
for k in adj[current]:
if d[k] < 0:
d[k] = d[current]+1
stack.append(k)
flag = 1
elif d[k] > d[current]+1:
d[k] = d[current]+1
stack.append(k)
flag = 1
if flag == 0:
stack.pop()
#print('current = {}, t = {}'.format(current, t))
#print('stack', *stack)
for i in range(n):
print('{} {}'.format(i+1, d[i]))
|
s737619085
|
p02748
|
u690184681
| 2,000 | 1,048,576 |
Wrong Answer
| 506 | 39,340 | 335 |
You are visiting a large electronics store to buy a refrigerator and a microwave. The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen. You have M discount tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time. You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required.
|
A,B,M = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
discount = []
for i in [0]*M:
discount.append(list(map(int,input().split())))
min_comb = min(a)+min(b)
for i in range(M):
min_comb = min(min_comb,a[discount[i][0]-1]+b[discount[i][1]-1]-discount[i][2])
ans = 0
print(ans)
|
s203860012
|
Accepted
| 521 | 39,268 | 342 |
A,B,M = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
discount = []
for i in [0]*M:
discount.append(list(map(int,input().split())))
min_comb = min(a)+min(b)
for i in range(M):
min_comb = min(min_comb,a[discount[i][0]-1]+b[discount[i][1]-1]-discount[i][2])
ans = min_comb
print(ans)
|
s210762201
|
p03369
|
u247211039
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 41 |
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
S = input()
print(700 + S.count("o")*300)
|
s640928612
|
Accepted
| 17 | 2,940 | 41 |
S = input()
print(700 + S.count("o")*100)
|
s611510042
|
p03456
|
u320098990
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,184 | 189 |
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
import math
input_list = input().split(' ')
a = input_list[0]
b = input_list[1]
N = math.sqrt(int(a+b))
print(a+b)
print(N)
if isinstance(N, float):
print('Yes')
else:
print('No')
|
s683397888
|
Accepted
| 25 | 9,092 | 165 |
import math
input_list = input().split(' ')
a = input_list[0]
b = input_list[1]
N = math.sqrt(int(a+b))
if N.is_integer():
print('Yes')
else:
print('No')
|
s213681292
|
p03433
|
u280552586
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 91 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
n = int(input())
a = int(input())
if n % 500 <= a:
print('YES')
else:
print('NO')
|
s561362923
|
Accepted
| 17 | 3,064 | 79 |
n, a = [int(input()) for _ in range(2)]
print('Yes' if n % 500 <= a else 'No')
|
s617429148
|
p03657
|
u841153348
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 300 |
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
def yagi(cookie_1,cookie_2):
if cookie_1 % 3 == 0:
return "possible"
elif cookie_2 % 3 == 0:
return "possible"
elif cookie_1 + cookie_2 % 3 == 0:
return "possible"
else:
return "impossible"
cookie = (input())
unti = cookie.split()
print(yagi(int(unti[0]),int(unti[1])))
|
s332081942
|
Accepted
| 18 | 2,940 | 272 |
cookie = (input())
unti = cookie.split()
cookie_1 = int(unti[0])
cookie_2 = int(unti[1])
if cookie_1 % 3 == 0:
print( "Possible")
elif cookie_2 % 3 == 0:
print( "Possible")
elif ((cookie_1 + cookie_2) % 3 == 0):
print( "Possible")
else :
print( "Impossible")
|
s009154663
|
p03997
|
u387456967
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 67 |
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,b,h = [int(input()) for i in range(3)]
ans = (a+b)*h/2
print(ans)
|
s230689261
|
Accepted
| 17 | 2,940 | 72 |
a,b,h = [int(input()) for i in range(3)]
ans = int((a+b)*h/2)
print(ans)
|
s875333981
|
p00135
|
u032662562
| 1,000 | 131,072 |
Wrong Answer
| 40 | 7,644 | 386 |
原始スローライフ主義組織「アカルイダ」から、いたずらの予告状が届きました。アカルイダといえば、要人の顔面にパイを投げつけたりするいたずらで有名ですが、最近では火薬を用いてレセプション会場にネズミ花火をまき散らすなど、より過激化してきました。予告状は次の文面です。 ---パソコン ヒトの時間を奪う。良くない。 時計の短い針と長い針 出会うころ、アカルイダ 正義行う。 スローライフ 偉大なり。 たどたどしくてよく解らないのですが、時計の短針と長針とが重なったころにいたずらを決行するという意味のようです。 このいたずらを警戒するため、時刻を入力として、短針と長針が近い場合は "alert"、遠い場合は "safe"、それ以外の場合は "warning" と出力するプログラムを作成してください。ただし、「近い」とは短針と長針の角度が 0° 以上 30° 未満の場合をいい、「遠い」とは 90° 以上 180° 以下の場合をいいます。なお、時刻は 00:00 以上 11:59 以下とします。
|
def solve(h,m):
dm = m / 60.0 * 360.0
dh = h / 12.0 * 360.0 + 360.0/ 12.0 * m / 60.0
print(dh, dm)
if 0 <= abs(dh-dm) < 30.0:
return("alert")
elif 90.0 <= abs(dh-dm) < 180.0:
return("safe")
else:
return("warning")
n = int(input().strip())
for i in range(n):
v = list(map(int, input().strip().split(':')))
print(solve(v[0],v[1]))
|
s729476031
|
Accepted
| 40 | 7,628 | 406 |
def solve(h,m):
dh = h / 12.0 * 360.0 + 360.0/ 12.0 * m / 60.0
dm = m / 60.0 * 360.0
x = abs(dh-dm)
if x > 180:
x = 360 - x
if 0 <= x < 30.0:
return("alert")
elif 90.0 <= x <= 180.0:
return("safe")
else:
return("warning")
n = int(input().strip())
for i in range(n):
v = list(map(int, input().strip().split(':')))
print(solve(v[0],v[1]))
|
s166179612
|
p02748
|
u371409687
| 2,000 | 1,048,576 |
Wrong Answer
| 419 | 18,608 | 224 |
You are visiting a large electronics store to buy a refrigerator and a microwave. The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen. You have M discount tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time. You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required.
|
a,b,m=map(int,input().split())
A=list(map(int,input().split()))
B=list(map(int,input().split()))
ans=min(A)+min(B)
for i in range(m):
x,y,c=map(int,input().split())
tmp=A[x-1]+B[y-1]+c
ans=min(ans,tmp)
print(ans)
|
s780526058
|
Accepted
| 417 | 18,608 | 224 |
a,b,m=map(int,input().split())
A=list(map(int,input().split()))
B=list(map(int,input().split()))
ans=min(A)+min(B)
for i in range(m):
x,y,c=map(int,input().split())
tmp=A[x-1]+B[y-1]-c
ans=min(ans,tmp)
print(ans)
|
s505936604
|
p03779
|
u145600939
| 2,000 | 262,144 |
Wrong Answer
| 30 | 2,940 | 64 |
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
x = int(input())
i = 0
while i*(i-1)//2 < x:
i += 1
print(i)
|
s713496410
|
Accepted
| 31 | 2,940 | 64 |
x = int(input())
i = 0
while i*(i+1)//2 < x:
i += 1
print(i)
|
s616666751
|
p02846
|
u918714262
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 404 |
Takahashi and Aoki are training for long-distance races in an infinitely long straight course running from west to east. They start simultaneously at the same point and moves as follows **towards the east** : * Takahashi runs A_1 meters per minute for the first T_1 minutes, then runs at A_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever. * Aoki runs B_1 meters per minute for the first T_1 minutes, then runs at B_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever. How many times will Takahashi and Aoki meet each other, that is, come to the same point? We do not count the start of the run. If they meet infinitely many times, report that fact.
|
t1, t2 = map(int, input().split())
a1, a2 = map(int, input().split())
b1, b2 = map(int, input().split())
a1 -= b1
a2 -= b2
if a1*t1+a2*t2 == 0:
print("infinity")
quit()
if a1 < 0:
a1 = -a1
a2 = -a2
if a2 > 0:
print(0)
quit()
d = a1*t1+a2*t2
if d > 0:
print(0)
quit()
print(a1*t1, -d)
ans = ((a1*t1-1)//(-d)+1)*2-1
if a1*t1%(-d) == 0:
ans += 1
print(ans)
|
s167453708
|
Accepted
| 17 | 3,064 | 386 |
t1, t2 = map(int, input().split())
a1, a2 = map(int, input().split())
b1, b2 = map(int, input().split())
a1 -= b1
a2 -= b2
if a1*t1+a2*t2 == 0:
print("infinity")
quit()
if a1 < 0:
a1 = -a1
a2 = -a2
if a2 > 0:
print(0)
quit()
d = a1*t1+a2*t2
if d > 0:
print(0)
quit()
ans = ((a1*t1-1)//(-d)+1)*2-1
if a1*t1%(-d) == 0:
ans += 1
print(ans)
|
s273088685
|
p03024
|
u826557401
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 78 |
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
s = str(input())
if s.count('x') >= 8:
print("No")
else:
print("Yes")
|
s551667460
|
Accepted
| 17 | 2,940 | 78 |
s = str(input())
if s.count('x') >= 8:
print("NO")
else:
print("YES")
|
s989496552
|
p02841
|
u539805724
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 220 |
In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month.
|
def main():
M = []
D = []
for i in range(2):
m, d = [int(i) for i in input().split()]
M.append(m)
D.append(d)
if (M[0] != M[1]):
print(0)
else:
print(1)
if __name__ == "__main__":
main()
|
s667705009
|
Accepted
| 17 | 3,060 | 191 |
def main():
M = []
D = []
for i in range(2):
m, d = [int(i) for i in input().split()]
M.append(m)
D.append(d)
if (M[0] != M[1]):
print(1)
else:
print(0)
main()
|
s948579143
|
p03150
|
u854685063
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,016 | 688 |
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
import sys
import math
import itertools as it
def I():return int(sys.stdin.readline().replace("\n",""))
def I2():return map(int,sys.stdin.readline().replace("\n","").split())
def S():return str(sys.stdin.readline().replace("\n",""))
def L():return list(sys.stdin.readline().replace("\n",""))
def Intl():return [int(k) for k in sys.stdin.readline().replace("\n","").split()]
def Lx(k):return list(map(lambda x:int(x)*-k,sys.stdin.readline().replace("\n","").split()))
if __name__ == "__main__":
s = S()
cnt = 0
tgt = "keyence"
for i in range(len(tgt)):
a,b = tgt[:i],tgt[i:]
if a in s and b in s:
print("Yes")
exit()
print("No")
|
s712379482
|
Accepted
| 26 | 9,068 | 663 |
import sys
import math
import itertools as it
def I():return int(sys.stdin.readline().replace("\n",""))
def I2():return map(int,sys.stdin.readline().replace("\n","").split())
def S():return str(sys.stdin.readline().replace("\n",""))
def L():return list(sys.stdin.readline().replace("\n",""))
def Intl():return [int(k) for k in sys.stdin.readline().replace("\n","").split()]
def Lx(k):return list(map(lambda x:int(x)*-k,sys.stdin.readline().replace("\n","").split()))
if __name__ == "__main__":
s = S()
cnt = 0
tgt = "keyence"
for i in range(8):
if s[:i] + s[len(s)-7+i:] == tgt:
print("YES")
exit()
print("NO")
|
s900811798
|
p03472
|
u454760747
| 2,000 | 262,144 |
Wrong Answer
| 1,202 | 23,216 | 884 |
You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?
|
N, H = map(int, input().split())
alist = []
blist = []
for i in range(N):
A, B = map(int, input().split())
alist.append(A)
blist.append(B)
maxa = max(alist)
blist = list(filter(lambda el: el > maxa, blist))
def mergesort(l):
def merge(left, right):
result = []
while left and right:
if left[0] <= right[0]:
result.append(left.pop(0))
else:
result.append(right.pop(0))
return result + left + right
length = len(l)
if length <= 1:
return l
middle = length // 2
return merge(mergesort(l[:middle]), mergesort(l[middle:]))
blist = mergesort(blist)
blist.reverse()
sum_for_h = 0
ans = 0
for i in blist:
sum_for_h += i
ans += 1
if sum_for_h >= H:
print(ans , flush=True)
exit()
anum = (H-sum_for_h) // maxa
print(ans + anum, flush=True)
|
s407873069
|
Accepted
| 1,179 | 22,244 | 919 |
N, H = map(int, input().split())
alist = []
blist = []
for i in range(N):
A, B = map(int, input().split())
alist.append(A)
blist.append(B)
maxa = max(alist)
blist = list(filter(lambda el: el > maxa, blist))
def mergesort(l):
def merge(left, right):
result = []
while left and right:
if left[0] <= right[0]:
result.append(left.pop(0))
else:
result.append(right.pop(0))
return result + left + right
length = len(l)
if length <= 1:
return l
middle = length // 2
return merge(mergesort(l[:middle]), mergesort(l[middle:]))
blist = mergesort(blist)
blist.reverse()
sum_for_h = 0
ans = 0
for i in blist:
sum_for_h += i
ans += 1
if sum_for_h >= H:
print(ans , flush=True)
exit()
anum = (H-sum_for_h) / maxa
import math
anum = math.ceil(anum)
print(ans + anum, flush=True)
|
s845440753
|
p02388
|
u040556632
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,320 | 14 |
Write a program which calculates the cube of a given integer x.
|
print(input())
|
s125914556
|
Accepted
| 30 | 7,580 | 34 |
x = int(input())
x = x**3
print(x)
|
s823521628
|
p02390
|
u527389300
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,608 | 137 |
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())
temp = s % 60
m = (s - temp) / 60
temp = m % 60
h = (m - temp) / 60
print(str(h) + ':' + str(m) + ':' + str(s))
|
s981398719
|
Accepted
| 20 | 5,588 | 163 |
time = int(input())
h = time - (time % 3600)
time = time - h
m = time - (time % 60)
s = time - m
print(str(int(h / 3600))+':'+str(int(m / 60))+':'+str(s))
|
s236186552
|
p02972
|
u278864208
| 2,000 | 1,048,576 |
Wrong Answer
| 478 | 18,852 | 242 |
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
#D
N = int(input())
box = [0] * N
for i in range(N-1, 0, -1):
aisum = 0
for j in range(i, N, i):
aisum += box[j]
if aisum%2 == 0:
box[i] = 1
else:
box[i] = 0
print(' '.join(map(str, box[::-1])))
|
s701910568
|
Accepted
| 611 | 19,856 | 366 |
#D
N = int(input())
a = list(map(int,input().split()))
box = [0] * N
result = []
for i in range(N-1, -1, -1):
aisum = 0
for j in range(i, N, i+1):
aisum += box[j]
if aisum%2 != a[i]%2:
box[i] = 1
result.append(i+1)
else:
box[i] = 0
print(len(result))
if len(result) != 0:
print(' '.join(map(str, result)))
|
s940301763
|
p03623
|
u752552310
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 91 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x, a, b = map(int, input().split())
if abs(x-a) > abs(x-b):
print('A')
else:
print('B')
|
s581326484
|
Accepted
| 17 | 2,940 | 91 |
x, a, b = map(int, input().split())
if abs(x-a) > abs(x-b):
print('B')
else:
print('A')
|
s609845812
|
p03693
|
u982594421
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 89 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r, g, b = map(int, input().split())
if g * b % 4 == 0:
print('YES')
else:
print('NO')
|
s104405679
|
Accepted
| 19 | 2,940 | 96 |
r, g, b = map(int, input().split())
if (g * 10 + b) % 4 == 0:
print('YES')
else:
print('NO')
|
s829765549
|
p02697
|
u945228737
| 2,000 | 1,048,576 |
Wrong Answer
| 72 | 9,276 | 145 |
You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given.
|
def solve():
N, M = map(int, input().split())
for i in range(M):
print(i + 1, N - i)
if __name__ == '__main__':
solve()
|
s113569289
|
Accepted
| 75 | 9,232 | 449 |
def solve():
N, M = map(int, input().split())
m = 0
oddN = M
oddN += (oddN + 1) % 2
for i in range(oddN // 2):
print(i + 1, oddN - i)
m += 1
if m == M:
return
for i in range(M):
print(oddN + i + 1, M * 2 + 1 - i)
m += 1
if m == M:
return
if __name__ == '__main__':
solve()
|
s140659710
|
p03047
|
u801512570
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,064 | 67 |
Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi. How many ways are there to choose K consecutive integers?
|
N,K=map(int, input().split())
print(sum([i for i in range(1,N+1)]))
|
s583063684
|
Accepted
| 17 | 2,940 | 44 |
N,K=map(int,input().split())
print(N-(K-1))
|
s893473076
|
p03469
|
u347600233
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 33 |
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('2017' + s[4:])
|
s442861466
|
Accepted
| 17 | 2,940 | 33 |
s = input()
print('2018' + s[4:])
|
s547493816
|
p03713
|
u557494880
| 2,000 | 262,144 |
Wrong Answer
| 328 | 3,064 | 349 |
There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle. Snuke is trying to divide the bar as evenly as possible. More specifically, he is trying to minimize S_{max} \- S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece. Find the minimum possible value of S_{max} - S_{min}.
|
W,H = map(int,input().split())
ans = 10**100
for i in range(1,W):
a = H*i
w = W - i
b = min((w//2)*H,w*(H//2))
c = w*H - b
x = max(a,b,c) - min(a,b,c)
ans = min(ans,x)
for i in range(1,H):
a = W*i
h = H - i
b = min((W//2)*h,W*(h//2))
c = h*W - b
x = max(a,b,c) - min(a,b,c)
ans = min(ans,x)
print(ans)
|
s738827403
|
Accepted
| 324 | 3,064 | 349 |
W,H = map(int,input().split())
ans = 10**30
for i in range(1,W):
a = H*i
w = W - i
b = max((w//2)*H,w*(H//2))
c = w*H - b
x = max(a,b,c) - min(a,b,c)
ans = min(ans,x)
for i in range(1,H):
a = W*i
h = H - i
b = max((W//2)*h,W*(h//2))
c = h*W - b
x = max(a,b,c) - min(a,b,c)
ans = min(ans,x)
print(ans)
|
s215446545
|
p03992
|
u374802266
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 113 |
This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s.
|
a=input()
for i in range(4):
print(a[i],end='')
print(' ',end='')
for i in range(8):
print(a[i+4],end='')
|
s290482999
|
Accepted
| 31 | 8,940 | 28 |
s=input()
print(s[:4],s[4:])
|
s943499674
|
p04029
|
u989326345
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 42 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N=int(input())
ans=(N*(N-1))//2
print(ans)
|
s450609312
|
Accepted
| 17 | 2,940 | 43 |
N=int(input())
ans=(N*(N+1))//2
print(ans)
|
s072656772
|
p02262
|
u630566146
| 6,000 | 131,072 |
Wrong Answer
| 20 | 5,616 | 882 |
Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+g] = A[j] 7 j = j - g 8 cnt++ 9 A[j+g] = v 10 11 shellSort(A, n) 12 cnt = 0 13 m = ? 14 G[] = {?, ?,..., ?} 15 for i = 0 to m-1 16 insertionSort(A, n, G[i]) A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$. Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements: * $1 \leq m \leq 100$ * $0 \leq G_i \leq n$ * cnt does not exceed $\lceil n^{1.5}\rceil$
|
def insertion_sort(l, g):
cnt = 0
for i in range(g, len(l)):
tmp = l[i]
j = i - g
while j >= 0 and l[j] > tmp:
l[j+g] = l[j]
cnt += 1
j -= g
l[j+g] = tmp
return l, cnt
def shell_sort(l, lg):
tot = 0
for g in lg:
sl, cnt = insertion_sort(l, g)
tot += cnt
return sl, tot
if __name__ == '__main__':
N = int(input())
l = []
for _ in range(N):
l.append(int(input()))
#l = list(map(int, input().split()))
#sl = insertion_sort(l, 1)
#lg = [4, 1]
lg = [int(1/2*(3**(i+1)-1)) for i in range((N//3 + 1), -1, -1)]
print(len(lg))
print(' '.join(map(str, lg)))
sl, tot = shell_sort(l, lg)
print(tot)
#print(' '.join(map(str, sl)))
for e in sl:
print(e)
|
s408084529
|
Accepted
| 19,330 | 45,504 | 767 |
def insertion_sort(l, g):
cnt = 0
for i in range(g, len(l)):
tmp = l[i]
j = i - g
while j >= 0 and l[j] > tmp:
l[j+g] = l[j]
cnt += 1
j -= g
l[j+g] = tmp
return l, cnt
def shell_sort(l, lg):
tot = 0
for g in lg:
sl, cnt = insertion_sort(l, g)
tot += cnt
return sl, tot
if __name__ == '__main__':
N = int(input())
l = []
for _ in range(N):
l.append(int(input()))
lg = [1]
while True:
gc = 3 * lg[-1] + 1
if gc > N:
break
lg.append(gc)
lg = lg[::-1]
print(len(lg))
print(' '.join(map(str, lg)))
sl, tot = shell_sort(l, lg)
print(tot)
for e in sl:
print(e)
|
s146228229
|
p02406
|
u261533743
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,720 | 121 |
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
N = int(input())
result = ' '.join([str(c) for c in range(1, N + 1)
if c == 3 or '3' in str(c)])
print(' ' + result)
|
s896282108
|
Accepted
| 20 | 7,772 | 116 |
N = int(input())
print(' {}'.format(' '.join([str(c) for c in range(1, N + 1)
if c % 3 == 0 or '3' in str(c)])))
|
s192838173
|
p04043
|
u039002256
| 2,000 | 262,144 |
Wrong Answer
| 22 | 8,968 | 212 |
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.
|
l = list(map(int, input().split()))
flag_5 = flag_7 = 0
for i in l:
if i == 5:
flag_5 += 1
if i == 7:
flag_7 += 1
if flag_5 == 2 and flag_7 == 1:
print("Yes")
else:
print("No")
|
s763732082
|
Accepted
| 28 | 8,964 | 212 |
l = list(map(int, input().split()))
flag_5 = flag_7 = 0
for i in l:
if i == 5:
flag_5 += 1
if i == 7:
flag_7 += 1
if flag_5 == 2 and flag_7 == 1:
print("YES")
else:
print("NO")
|
s456224943
|
p04029
|
u263824932
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 303 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
S=input()
words=[a for a in S]
answers=[]
for n in words:
if n=='1':
answers.append('1')
elif n=='0':
answers.append('0')
else:
if not answers:
continue
else:
answers.pop()
mojiretu=''
for x in answers:
mojiretu+=x
print(mojiretu)
|
s037828491
|
Accepted
| 17 | 2,940 | 70 |
N=int(input())
C=[]
for n in range(N+1):
C.append(n)
print(sum(C))
|
s041749251
|
p03944
|
u530786533
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 299 |
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
w, h, n = map(int, input().split())
p, q, r, s = 0, 0, w, h
for i in range(n):
x, y, a = map(int, input().split())
if (a == 1):
p = x
elif (a == 2):
r = x
elif (a == 3):
q = y
else:
s = y
if (r - p < 0):
r = p
if (s - q < 0):
s = q
print(r - p, s - q)
|
s504490158
|
Accepted
| 18 | 3,064 | 389 |
w, h, n = map(int, input().split())
p, q, r, s = 0, 0, w, h
for i in range(n):
x, y, a = map(int, input().split())
if (a == 1):
if (x > p):
p = x
elif (a == 2):
if (x < r):
r = x
elif (a == 3):
if (y > q):
q = y
else:
if (y < s):
s = y
if (r < p or s < q):
print(0)
else:
print((r - p)* (s - q))
|
s555736482
|
p04029
|
u461954362
| 2,000 | 262,144 |
Wrong Answer
| 40 | 3,064 | 63 |
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?
|
#!/usr/bin/env python
N = int(input())
print(N * (N + 1) / 2)
|
s479802725
|
Accepted
| 38 | 3,064 | 64 |
#!/usr/bin/env python
N = int(input())
print(N * (N + 1) // 2)
|
s800605725
|
p02612
|
u731362892
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,148 | 50 |
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())
while n<=0:
n-=1000
print(abs(n))
|
s347276810
|
Accepted
| 28 | 9,152 | 49 |
n=int(input())
while n>0:
n-=1000
print(abs(n))
|
s431086340
|
p03549
|
u333945892
| 2,000 | 262,144 |
Wrong Answer
| 29 | 4,212 | 332 |
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
|
from collections import defaultdict
import sys,heapq,bisect,math,itertools,string,queue,datetime
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inpl(): return list(map(int, input().split()))
def inpl_s(): return list(input().split())
N,M = inpl()
T = (N-M)*100 + 1900*M
print(T)
k = 2**M
print(k*T)
|
s831234825
|
Accepted
| 29 | 4,204 | 323 |
from collections import defaultdict
import sys,heapq,bisect,math,itertools,string,queue,datetime
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inpl(): return list(map(int, input().split()))
def inpl_s(): return list(input().split())
N,M = inpl()
T = (N-M)*100 + 1900*M
k = 2**M
print(k*T)
|
s718850549
|
p02694
|
u434609232
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,164 | 107 |
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())
money = 100
year = 0
while money <= X:
money += money // 100
year += 1
print(year)
|
s012041762
|
Accepted
| 27 | 9,100 | 106 |
X = int(input())
money = 100
year = 0
while money < X:
money += money // 100
year += 1
print(year)
|
s880205653
|
p02422
|
u957840591
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,680 | 615 |
Write a program which performs a sequence of commands to a given string $str$. The command is one of: * print a b: print from the a-th character to the b-th character of $str$ * reverse a b: reverse from the a-th character to the b-th character of $str$ * replace a b p: replace from the a-th character to the b-th character of $str$ with p Note that the indices of $str$ start with 0.
|
str=input()
q=eval(input())
command=[]
replace=[]
numdata=[]
for i in range(q):
x=input()
if 'replace' in x:
a,b,c,d=x.split()
replace.append(d)
numdata.append((int(b),int(c)))
else:
a,b,c=x.split()
numdata.append((int(b),int(c)))
command.append(a)
Re=iter(replace)
for i in range(q):
if command[i]=='print':
print(str[numdata[i][0]:numdata[i][1]+1])
elif command[i]=='reverse':
str=str[:numdata[i][0]]+str[numdata[i][1]:numdata[i][0]-1:-1]+str[numdata[i][1]+1:]
else:
str=str[:numdata[i][0]]+next(Re)+str[numdata[i][1]+1:]
|
s958571872
|
Accepted
| 30 | 7,708 | 636 |
str=input()
q=eval(input())
command=[]
replace=[]
numdata=[]
for i in range(q):
x=input()
if 'replace' in x:
a,b,c,d=x.split()
replace.append(d)
numdata.append((int(b),int(c)))
else:
a,b,c=x.split()
numdata.append((int(b),int(c)))
command.append(a)
Re=iter(replace)
for i in range(q):
if command[i]=='print':
print(str[numdata[i][0]:numdata[i][1]+1])
elif command[i]=='reverse':
temp=str[numdata[i][0]:numdata[i][1]+1]
str=str[:numdata[i][0]]+temp[::-1]+str[numdata[i][1]+1:]
else:
str=str[:numdata[i][0]]+next(Re)+str[numdata[i][1]+1:]
|
s796173869
|
p03401
|
u806855121
| 2,000 | 262,144 |
Wrong Answer
| 194 | 14,048 | 420 |
There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.
|
N = int(input())
A = list(map(int, input().split()))
s = 0
p = 0
for i in range(N):
s += abs(p - A[i])
p = A[i]
if p != 0:
s += abs(p)
A.append(0)
A.insert(0, 0)
for i in range(1, N+1):
if A[i] < 0:
if A[i] < A[i+1]:
print(s - abs(A[i-1] - A[i])*2)
continue
else:
if A[i] > A[i+1]:
print(s - abs(A[i-1]-A[i])*2)
continue
print(s)
|
s155100734
|
Accepted
| 1,346 | 13,536 | 262 |
from collections import deque
N = int(input())
A = deque(map(int, input().split()))
A.appendleft(0)
A.append(0)
S = 0
for i in range(1, N+2):
S += abs(A[i-1] - A[i])
for i in range(1, N+1):
print(S+abs(A[i-1]-A[i+1])-(abs(A[i-1]-A[i])+abs(A[i]-A[i+1])))
|
s125906715
|
p03827
|
u324549724
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 134 |
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
a = input()
maxi = 0
now = 0
for c in a:
if c == 'I':
now += 1
else:
now -= 1
if now > maxi:
maxi = now
print(maxi)
|
s801676084
|
Accepted
| 17 | 2,940 | 134 |
input()
a = input()
maxi = 0
now = 0
for c in a:
if c == 'I':
now += 1
else:
now -= 1
maxi = max(now, maxi)
print(maxi)
|
s377827443
|
p02612
|
u220560107
| 2,000 | 1,048,576 |
Wrong Answer
| 32 | 9,148 | 28 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N=int(input())
print(1000-N)
|
s366845115
|
Accepted
| 28 | 9,120 | 40 |
N=int(input())
print((1000-N%1000)%1000)
|
s009034441
|
p03369
|
u737508101
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 38 |
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
s = input()
print(s.count("o") * 100)
|
s159366523
|
Accepted
| 17 | 2,940 | 43 |
s = input()
print(s.count("o") * 100 + 700)
|
s001705377
|
p04012
|
u594956556
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 172 |
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()
cdict = {}
for c in w:
if c in cdict:
cdict[c] += 1
else:
cdict[c] = 0
if all(cdict[c]%2==0 for c in cdict):
print('Yes')
else:
print('No')
|
s979712926
|
Accepted
| 17 | 2,940 | 173 |
w = input()
cdict = {}
for c in w:
if c in cdict:
cdict[c] += 1
else:
cdict[c] = 1
if all(cdict[c]%2==0 for c in cdict):
print('Yes')
else:
print('No')
|
s531864276
|
p03044
|
u652057333
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 16,432 | 959 |
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.
|
n = int(input())
u = [0] * (n-1)
v = [0] * (n-1)
w = [0] * (n-1)
node = [i for i in range(n)]
group1 = []
group2 = []
for i in range(n-1):
u[i], v[i], w[i] = map(int, input().split())
u[i] -= 1
v[i] -= 1
w[i] = w[i] % 2
if w[i] == 0:
if node[u[i]] == u[i]:
node[u[i]] = node[v[i]]
elif node[u[i]] != u[i]:
node[v[i]] = node[u[i]]
for i in range(n-1):
if w[i] == 1:
if u[i] in group1:
group2.append(node[v[i]])
elif u[i] in group2:
group1.append(node[v[i]])
elif v[i] in group1:
group2.append(node[u[i]])
elif v[i] in group2:
group1.append(node[u[i]])
else:
group1.append(node[v[i]])
group2.append(node[u[i]])
for i in range(n):
if node[i] in group1:
node[i] = 0
else:
node[i] = 1
for color in node:
print(color)
|
s488447670
|
Accepted
| 1,304 | 51,348 | 480 |
import queue
n = int(input())
links = [set() for _ in [0] * n]
for i in range(n-1):
u, v, w = map(int, input().split())
u -= 1
v -= 1
links[u].add((v, w))
links[v].add((u, w))
ans = [-1] * n
q = queue.Queue()
q.put((0, 0, -1))
while not q.empty():
v, d, p = q.get()
if d % 2 == 0:
ans[v] = 0
else:
ans[v] = 1
for u, w in links[v]:
if u == p:
continue
q.put((u, d + w, v))
for i in ans:
print(i)
|
s640649260
|
p03145
|
u675918663
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 2,940 | 97 |
There is a right triangle ABC with ∠ABC=90°. Given the lengths of the three sides, |AB|,|BC| and |CA|, find the area of the right triangle ABC. It is guaranteed that the area of the triangle ABC is an integer.
|
import sys
for line in sys.stdin:
nbs = line.split(' ')
print(int(nbs[0]) * int(nbs[1]) / 2)
|
s337716392
|
Accepted
| 18 | 2,940 | 99 |
import sys
for line in sys.stdin:
nbs = line.split(' ')
print(int(nbs[0]) * int(nbs[1]) // 2)
|
s477727503
|
p02742
|
u767995501
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 56 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
H, W = map(int, input().split())
print((H / 2)*(W / 2))
|
s647403902
|
Accepted
| 17 | 2,940 | 122 |
H,W = map(int, input().split())
if H > W:
H,W = W,H
if H == 1:
print(1)
else:
N = H * W
print((N+1)//2)
|
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