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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s373614454
|
p03844
|
u119714109
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,064 | 304 |
Joisino wants to evaluate the formula "A op B". Here, A and B are integers, and the binary operator op is either `+` or `-`. Your task is to evaluate the formula instead of her.
|
# coding=utf-8
import sys
stdin = sys.stdin
def na(): return map(int, stdin.readline().split())
def ns(): return stdin.readline().strip()
def main():
a,op,b = stdin.readline().split()
if op == '+':
print(a+b)
else:
print(a-b)
pass
if __name__ == '__main__':
main()
|
s169466330
|
Accepted
| 24 | 3,188 | 332 |
# coding=utf-8
import sys
stdin = sys.stdin
def na(): return map(int, stdin.readline().split())
def ns(): return stdin.readline().strip()
def main():
a,op,b = stdin.readline().strip().split()
if op == '+':
print(int(a)+int(b))
else:
print(int(a)-int(b))
pass
if __name__ == '__main__':
main()
|
s234039130
|
p03658
|
u757274384
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 147 |
Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.
|
n,k = map(int, input().split())
L = list(map(int, input().split()))
L.sort()
L.reverse()
ans = 0
for i in range(0,n):
ans += L[i]
print(ans)
|
s464934926
|
Accepted
| 17 | 2,940 | 148 |
n,k = map(int, input().split())
L = list(map(int, input().split()))
L.sort()
L.reverse()
ans = 0
for i in range(0,k):
ans += L[i]
print(ans)
|
s928255316
|
p03477
|
u717993780
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 120 |
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 = map(int,input().split())
if a+b<c+d:
print("Right")
elif a+b>c+d:
print("Light")
else:
print("Balanced")
|
s403548589
|
Accepted
| 17 | 2,940 | 119 |
a,b,c,d = map(int,input().split())
if a+b<c+d:
print("Right")
elif a+b>c+d:
print("Left")
else:
print("Balanced")
|
s481299934
|
p00006
|
u747479790
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,464 | 36 |
Write a program which reverses a given string str.
|
a=list(input())
a.reverse()
print(a)
|
s126139349
|
Accepted
| 30 | 7,336 | 20 |
print(input()[::-1])
|
s544397731
|
p00010
|
u572790226
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,956 | 1,210 |
Write a program which prints the central coordinate $(p_x, p_y)$ and the radius $r$ of a circumscribed circle of a triangle which is constructed by three points $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ on the plane surface.
|
from math import sqrt
def circle(x1, y1, x2, y2, x3, y3):
if x1 == 0:
dx = 1
x1 = x1 + dx
x2 = x2 + dx
x3 = x3 + dx
else:
dx = 0
if y2 == 0:
dy = 1
y1 = y1 + dy
y2 = y2 + dy
y3 = y3 + dy
else:
dy = 0
A = [[x1, y1, 1, 1, 0, 0],[x2, y2, 1, 0, 1, 0],[x3, y3, 1, 0, 0, 1]]
# print(A)
for i in range(3):
A[0] = [x/A[0][0] for x in A[0]]
A[1] = [A[1][j] - A[1][0] * A[0][j] for j in range(6)]
A[2] = [A[2][j] - A[2][0] * A[0][j] for j in range(6)]
# print(A)
for j in range(3):
A[j] = A[j][1:] + A[j][:1]
A = A[1:] + A[:1]
# print(A)
for i in range(3):
A[i] = A[i][:3]
# print(A)
V = [-x1**2-y1**2, -x2**2-y2**2, -x3**2-y3**2]
M = [(A[i][0] * V[0] + A[i][1] * V[1] + A[i][2] * V[2]) for i in range(3)]
xcenter = -0.5 * M[0] - dx
ycenter = -0.5 * M[1] - dy
radius = sqrt((M[0]**2) /4 + (M[1]**2) /4 - M[2])
return xcenter, ycenter, radius
n = int(input())
for line in range(n):
x1, y1, x2, y2, x3, y3 = map(float, input().split())
xc, yc, ra = circle(x1, y1, x2, y2, x3, y3)
print(xc, yc,ra)
|
s595740441
|
Accepted
| 30 | 7,820 | 1,232 |
from math import sqrt
def circle(x1, y1, x2, y2, x3, y3):
if x1 == 0:
dx = 1
x1 = x1 + dx
x2 = x2 + dx
x3 = x3 + dx
else:
dx = 0
if y2 == 0:
dy = 1
y1 = y1 + dy
y2 = y2 + dy
y3 = y3 + dy
else:
dy = 0
A = [[x1, y1, 1, 1, 0, 0],[x2, y2, 1, 0, 1, 0],[x3, y3, 1, 0, 0, 1]]
# print(A)
for i in range(3):
A[0] = [x/A[0][0] for x in A[0]]
A[1] = [A[1][j] - A[1][0] * A[0][j] for j in range(6)]
A[2] = [A[2][j] - A[2][0] * A[0][j] for j in range(6)]
# print(A)
for j in range(3):
A[j] = A[j][1:] + A[j][:1]
A = A[1:] + A[:1]
# print(A)
for i in range(3):
A[i] = A[i][:3]
# print(A)
V = [-x1**2-y1**2, -x2**2-y2**2, -x3**2-y3**2]
M = [(A[i][0] * V[0] + A[i][1] * V[1] + A[i][2] * V[2]) for i in range(3)]
xcenter = -0.5 * M[0] - dx
ycenter = -0.5 * M[1] - dy
radius = sqrt((M[0]**2) /4 + (M[1]**2) /4 - M[2])
return xcenter, ycenter, radius
n = int(input())
for line in range(n):
x1, y1, x2, y2, x3, y3 = map(float, input().split())
xc, yc, ra = circle(x1, y1, x2, y2, x3, y3)
print('%.3f %.3f %.3f' % (xc, yc, ra))
|
s866997502
|
p03796
|
u319612498
| 2,000 | 262,144 |
Wrong Answer
| 33 | 2,940 | 106 |
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
n=int(input())
sum=0
for i in range(n):
sum*=(i+1)
if sum>10**9+7:
sum%=10**9+7
print(sum)
|
s387824020
|
Accepted
| 48 | 2,940 | 107 |
n=int(input())
sum=1
for i in range(n):
sum*=(i+1)
if sum>10**9+7:
sum%=10**9+7
print(sum)
|
s276565426
|
p03853
|
u672898046
| 2,000 | 262,144 |
Wrong Answer
| 24 | 3,956 | 247 |
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).
|
h, w = map(int, input().split())
r = [list(input()) for i in range(h)]
new_h = [0] * w
new_r = [new_h for i in range(2*h)]
for i in range(0, 2*h, 2):
new_r[i] = r[(i+1)//2][:]
new_r[i+1] = r[(i+1)//2][:]
for i in new_r:
print(*i, end="\n")
|
s660781155
|
Accepted
| 26 | 4,596 | 246 |
h, w = map(int, input().split())
r = [list(input()) for i in range(h)]
new_h = [0] * w
new_r = [new_h for i in range(2*h)]
for i in range(0, 2*h, 2):
new_r[i] = r[(i+1)//2]
new_r[i+1] = r[(i+1)//2]
for i in new_r:
print(*i,sep="", end="\n")
|
s534831280
|
p03417
|
u882209234
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 139 |
There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region. The front and back sides of these cards can be distinguished, and initially every card faces up. We will perform the following operation once for each square contains a card: * For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square. It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed. Find the number of cards that face down after all the operations.
|
N,M = sorted(map(int,input().split()))
print(N,M)
if N == 1:
if M == 1: print(1)
else: print(max(0,M-2))
else: print(N*M-2*N-2*M+4)
|
s517975333
|
Accepted
| 17 | 2,940 | 128 |
N,M = sorted(map(int,input().split()))
if N == 1:
if M == 1: print(1)
else: print(max(0,M-2))
else: print(N*M-2*N-2*M+4)
|
s671947533
|
p02743
|
u022658079
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 82 |
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
a,b,c=map(int, input().split())
if a^2+b^2<c^2:
print("Yes")
else:
print("No")
|
s824584641
|
Accepted
| 17 | 2,940 | 99 |
a,b,c=map(int, input().split())
r=c-a-b
if r > 0 and r**2>4*a*b:
print("Yes")
else:
print("No")
|
s317793048
|
p03494
|
u909716307
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,112 | 183 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n=int(input())
l=list(map(int,input().split()))
s=float('inf')
for a in l:
c=0
while 1:
if a%2!=0:
break
a/=2
c+=1
s=min(s,c)
print(c)
|
s092029167
|
Accepted
| 29 | 9,052 | 151 |
n=int(input())
l=list(map(int,input().split()))
s=float('inf')
for a in l:
c=0
while a%2==0:
c+=1
a/=2
s=min(s,c)
print(s)
|
s582896660
|
p02386
|
u440180827
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,836 | 1,229 |
Write a program which reads $n$ dices constructed in the same way as [Dice I](description.jsp?id=ITP1_11_A), and determines whether they are all different. For the determination, use the same way as [Dice III](description.jsp?id=ITP1_11_C).
|
n = int(input())
x = [[] for i in range(n)]
for i in range(n):
x[i] = list(map(int, input().split()))
z = [[0, 1, 2, 3], [0, 2, 1, 1], [1, 0, 2, 1], [2, 0, 1, 0], [1, 2, 0, 0], [2, 1, 0, 1]]
def compare(x, y):
succeed = False
for i in range(6):
match = True
order = [0, 0, 0]
for j in range(3):
if {x[j], x[-j-1]} != {y[z[i][j]], y[-z[i][j]-1]}:
match = False
break
if x[j] == y[z[i][j]]:
order[j] = 1
if x[j] == x[-j-1]:
order[j] = 2
if match:
if 2 in order:
succeed = True
break
if (z[i][3] == 0 or z[i][3] == 3) and sum(order) % 2:
succeed = True
break
if z[i][3] == 1 and not sum(order) % 2:
succeed = True
break
if succeed:
return True
else:
return False
different = True
for i in range(n):
for j in range(i+1, n, 1):
if compare(x[i], x[j]):
print(i, j)
different = False
break
if not different:
break
if different:
print('Yes')
else:
print('No')
|
s386587662
|
Accepted
| 60 | 7,800 | 768 |
import sys
def check(x, y):
for i in range(6):
order = [0, 0, 0]
for j in range(3):
k = a[i][j]
if {x[j], x[-j-1]} != {y[k], y[-k-1]}:
break
if x[j] == y[k]:
order[j] = 1
if x[j] == x[-j-1]:
order[j] = 2
else:
if 2 in order or (i < 3) == sum(order) % 2:
return True
return False
a = [[0, 1, 2], [2, 0, 1], [1, 2, 0], [0, 2, 1], [2, 1, 0], [1, 0, 2]]
n = int(sys.stdin.readline())
x = [sys.stdin.readline().split() for i in range(n)]
for i in range(n):
for j in range(i+1, n):
if check(x[i], x[j]):
break
else:
continue
print('No')
break
else:
print('Yes')
|
s808268256
|
p03719
|
u119982147
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 96 |
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a, b, c = map(int, input().split())
if a <= c and b >= c:
print("YES")
else:
print("NO")
|
s527279729
|
Accepted
| 18 | 2,940 | 96 |
a, b, c = map(int, input().split())
if a <= c and b >= c:
print("Yes")
else:
print("No")
|
s280293101
|
p03957
|
u782654209
| 1,000 | 262,144 |
Wrong Answer
| 21 | 3,188 | 91 |
This contest is `CODEFESTIVAL`, which can be shortened to the string `CF` by deleting some characters. Mr. Takahashi, full of curiosity, wondered if he could obtain `CF` from other strings in the same way. You are given a string s consisting of uppercase English letters. Determine whether the string `CF` can be obtained from the string s by deleting some characters.
|
import re
print('Yes') if len(re.compile(r'.+C.+F.+').findall(input()))>=1 else print('No')
|
s313117144
|
Accepted
| 20 | 3,188 | 91 |
import re
print('Yes') if len(re.compile(r'.*C.*F.*').findall(input()))>=1 else print('No')
|
s935361205
|
p03636
|
u227254381
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
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.
|
sent = input()
print(sent[1] + str(len(sent) - 2) + sent[-1])
|
s756367532
|
Accepted
| 17 | 2,940 | 61 |
sent = input()
print(sent[0] + str(len(sent) - 2) + sent[-1])
|
s234766055
|
p03024
|
u178192749
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 59 |
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
s = input()
n = s.count('o')
print('YES' if n>=8 else 'NO')
|
s476937049
|
Accepted
| 17 | 2,940 | 72 |
s = input()
d = len(s)
n = s.count('o')
print('YES' if d-n<=7 else 'NO')
|
s295257335
|
p02393
|
u186524656
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,724 | 281 |
Write a program which reads three integers, and prints them in ascending order.
|
a, b, c = map(int, input().split())
if a < b and b < c and a < c:
print(a, b, c)
if a > b and a > c and b > c:
print(c, b, a)
if a > b and b < c and a < c:
print(b, a, c)
if a > b and a > c and b < c:
print(b, c, a)
if a < b and c < b and b > a:
print(a, c, b)
|
s445495848
|
Accepted
| 30 | 6,724 | 60 |
print(' '.join(map(str, sorted(map(int, input().split())))))
|
s807362754
|
p03448
|
u252964975
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 218 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
A=int(input())
B=int(input())
C=int(input())
X=int(input())
count = 0
for a in range(min([A,X//500])):
for b in range(min([B,(X-500*A)//100])):
if (X-500*A-100*B) % 50 == 0:
count = count + 1
print(count)
|
s932314264
|
Accepted
| 18 | 3,064 | 427 |
A=int(input())
B=int(input())
C=int(input())
X=int(input())
count = 0
for a in range(min([A,X//500])+1):
for b in range(min([B,(X-500*a)//100])+1):
# print(a*500,b*100,X-500*a-100*b, (X-500*a-100*b)%50==0,(X-500*a-100*b)/50<=C)
if (X-500*a-100*b) % 50 == 0 and (X-500*a-100*b) / 50 <= C:
# print(a*500,b*100,X-500*a-100*b, (X-500*a-100*b)%50==0,(X-500*a-100*b)/50<=C)
count = count + 1
print(count)
|
s036242058
|
p02396
|
u405758695
| 1,000 | 131,072 |
Wrong Answer
| 130 | 5,572 | 117 |
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.
|
cnt=0
while(True):
x=input()
cnt+=1
if cnt>=10000 or x==0: break
print('Case {}: {}'.format(cnt,x))
|
s920253458
|
Accepted
| 150 | 5,596 | 107 |
cnt=0
while True:
x=int(input())
cnt+=1
if x==0: break
print('Case {}: {}'.format(cnt,x))
|
s643417197
|
p03089
|
u893828545
| 2,000 | 1,048,576 |
Wrong Answer
| 150 | 12,432 | 384 |
Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
|
import numpy as np
import math
def ordering(N,b):
a=[]
comb=math.factorial(N)
x=[[]]*comb
y=[[]]*comb
for i in range(1,N):
a=a+[i]
for n,m in enumerate(a):
x[n].insert(m-1,m)
y[n].append(m)
resnum=[x.index(i) for i in x if i==b]
try:
for i in y[resnum[0]]:
print(i)
except:
return(-1)
|
s985361550
|
Accepted
| 18 | 3,060 | 237 |
N=int(input())
b=[int(i) for i in input().split()]
a=[]
try:
while len(b)>0:
a2=max([j for i,j in enumerate(b) if b[i]==i+1])
a.insert(0,a2)
x=b.pop(a2-1)
for n in a:
print(n)
except:
print(-1)
|
s512534504
|
p04043
|
u293825440
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 434 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
A,B,C = map(int,input().split())
lis=[A,B,C]
def hantei(list):
if int(list[0]) == 7:
if int(list[1]) == 5 and int(list[2]) == 5:
return True
if int(list[1]) == 7:
if int(list[0]) == 5 and int(list[2]) == 5:
return True
if int(list[2]) == 7:
if int(list[0]) == 5 and int(list[1]) == 5:
return True
if hantei(lis) == True:
print("Yes")
else:
print("No")
|
s337927539
|
Accepted
| 17 | 3,064 | 434 |
A,B,C = map(int,input().split())
lis=[A,B,C]
def hantei(list):
if int(list[0]) == 7:
if int(list[1]) == 5 and int(list[2]) == 5:
return True
if int(list[1]) == 7:
if int(list[0]) == 5 and int(list[2]) == 5:
return True
if int(list[2]) == 7:
if int(list[0]) == 5 and int(list[1]) == 5:
return True
if hantei(lis) == True:
print("YES")
else:
print("NO")
|
s163735502
|
p03612
|
u142415823
| 2,000 | 262,144 |
Wrong Answer
| 42 | 14,004 | 293 |
You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this.
|
import sys
def main():
N = int(sys.stdin.readline())
p = list(map(int, input().split()))
count = 0
for i in p:
if i == p:
count += 1
if count % 2 == 0:
print(count/2)
else:
print((count+1)/2)
if __name__ == '__main__':
main()
|
s338900471
|
Accepted
| 60 | 14,008 | 319 |
import sys
def main():
N = int(sys.stdin.readline())
p = list(map(int, input().split()))
count = 0
for i, n in enumerate(p, 1):
if i == n:
count += 1
if len(p) != i:
p[i - 1], p[i] = p[i], p[i - 1]
print(count)
if __name__ == '__main__':
main()
|
s466317900
|
p02612
|
u047816928
| 2,000 | 1,048,576 |
Wrong Answer
| 33 | 9,144 | 37 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print((N+999)//1000)
|
s963576284
|
Accepted
| 29 | 9,152 | 50 |
N = int(input())
M = (N+999)//1000
print(M*1000-N)
|
s280536764
|
p02613
|
u619819312
| 2,000 | 1,048,576 |
Wrong Answer
| 141 | 9,116 | 121 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
c={"AC":0,"WA":0,"TLE":0,"RE":0}
for i in range(int(input())):
c[input()]+=1
for i,j in c.items():
print(i,"X",j)
|
s399889979
|
Accepted
| 139 | 9,184 | 121 |
c={"AC":0,"WA":0,"TLE":0,"RE":0}
for i in range(int(input())):
c[input()]+=1
for i,j in c.items():
print(i,"x",j)
|
s990703004
|
p02865
|
u350997995
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 34 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
N = int(input())-1
print(N//2-N%2)
|
s133392632
|
Accepted
| 17 | 2,940 | 30 |
N = int(input())-1
print(N//2)
|
s625362494
|
p02928
|
u455533363
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 3,188 | 507 |
We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.
|
n,k=map(int,input().split())
List=list(map(int,input().split()))
li = []
for i in range(n):
count=0
for j in range(n):
if List[i]>List[j]:
count += 1
li.append(count)
print(str(sum(li))+"li")
count2=0
for i in range(1,n):
for j in range(i+1,n+1):
#print(i,j)
if int(List[i-1]) > int(List[j-1]):
count2 += 1
#print(str(count2)+"count2")
for m in range(k,0,-1):
counted=count2+count2*(m)
ans = counted % 1000000007
print(ans)
|
s968483968
|
Accepted
| 1,662 | 3,188 | 416 |
n,k=map(int,input().split())
List=list(map(int,input().split()))
li = []
for i in range(n):
count=0
for j in range(n):
if List[i]>List[j]:
count += 1
li.append(count)
count2=0
for i in range(1,n):
for j in range(i+1,n+1):
if int(List[i-1]) > int(List[j-1]):
count2 += 1
counted = k*count2 + (sum(li))*k*(k-1)//2
ans = counted % (10**9+7)
print(int(ans))
|
s191216422
|
p03044
|
u451017206
| 2,000 | 1,048,576 |
Wrong Answer
| 423 | 4,976 | 753 |
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.
|
from collections import defaultdict
N = int(input())
g = defaultdict(list)
ans = [None] * N
for i in range(N-1):
u, v, w = map(int, input().split())
if w % 2 == 0:
if ans[u-1] is None and ans[v-1] is None:
ans[u-1] = ans[v-1] = 1
elif ans[u-1] is not None:
ans[v-1] = ans[u-1]
elif ans[v-1] is not None:
ans[u-1] = ans[v-1]
else:
pass
else:
if ans[u-1] is None and ans[v-1] is None:
ans[u-1] = 1
ans[v-1] = 0
elif ans[u-1] is not None:
ans[v-1] = (ans[u-1] + 1) % 2
elif ans[v-1] is not None:
ans[u-1] = (ans[v-1] + 1) % 2
else:
pass
for a in ans:
print(a)
|
s641028944
|
Accepted
| 849 | 102,552 | 508 |
from sys import setrecursionlimit, getrecursionlimit
setrecursionlimit(getrecursionlimit()*10000)
from collections import defaultdict
g = defaultdict(list)
visited = defaultdict(bool)
N = int(input())
ans = [0] * N
def dfs(v, d=0):
if visited[v]:
return
visited[v] = True
if d % 2:
ans[v-1] = 1
for c,w in g[v]:
dfs(c, d+w)
for j in range(N-1):
u,v,w = map(int, input().split())
g[u].append((v, w))
g[v].append((u, w))
dfs(1)
for a in ans:
print(a)
|
s920429779
|
p03555
|
u951672936
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 121 |
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
s1 = input()
s2 = input()
ok = True
for i in range(3):
if s1[i] != s2[2-i]:
ok = False
print("Yes" if ok else "No")
|
s817045786
|
Accepted
| 17 | 2,940 | 122 |
s1 = input()
s2 = input()
ok = True
for i in range(3):
if s1[i] != s2[2-i]:
ok = False
print("YES" if ok else "NO")
|
s139913032
|
p03700
|
u785578220
| 2,000 | 262,144 |
Wrong Answer
| 2,053 | 7,072 | 346 |
You are going out for a walk, when you suddenly encounter N monsters. Each monster has a parameter called _health_ , and the health of the i-th monster is h_i at the moment of encounter. A monster will vanish immediately when its health drops to 0 or below. Fortunately, you are a skilled magician, capable of causing explosions that damage monsters. In one explosion, you can damage monsters as follows: * Select an alive monster, and cause an explosion centered at that monster. The health of the monster at the center of the explosion will decrease by A, and the health of each of the other monsters will decrease by B. Here, A and B are predetermined parameters, and A > B holds. At least how many explosions do you need to cause in order to vanish all the monsters?
|
N,A,B = map(int,input().split())
H = [int(input()) for _ in range(N)]
from math import ceil
ng,ok = 0,10**9+1
while abs(ok-ng)>1:
mid = (ng+ok)//2
cnt = 0
for i in range(N):
cnt += max(ceil((H[i] - B*mid)/(A-B)) ,0)
if cnt == mid:
break
if cnt <= mid:
ok = mid
else:
ng = mid
print(ok)
|
s474505278
|
Accepted
| 2,000 | 7,072 | 350 |
#####################################
N,A,B = map(int,input().split())
H = [int(input()) for _ in range(N)]
from math import ceil
ng,ok = 0,10**9+1
while abs(ok-ng)>1:
mid = (ng+ok)//2
cnt = 0
for i in range(N):
cnt += max(ceil((H[i] - B*mid)/(A-B)) ,0)
if cnt <= mid:
ok = mid
else:
ng = mid
print(ok)
|
s062163304
|
p00007
|
u696166817
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,756 | 551 |
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.
|
if __name__ == "__main__":
n = int(input())
# print(n)
y = 100000
y2 = y / 1000
for i in range(0, n):
y4 = y2 * 1.05
# y3 = round(y2 * 1.05 -0.5, 0)
y3 = float(int(y2 * 1.05))
print(" {} {}".format(y4, y3))
if y4 - y3 > 0:
y4 += 1
y2 = int(y4)
print("{} {}".format(i, y2 * 1000))
print(y2 * 1000)
# 0 105,000
# 1 110,250 -> 111,000
# 2 116,550 -> 117,000
# 3 122,850 -> 123,000
#4 129,150 -> 130,000
|
s727758267
|
Accepted
| 30 | 6,752 | 553 |
if __name__ == "__main__":
n = int(input())
# print(n)
y = 100000
y2 = y / 1000
for i in range(0, n):
y4 = y2 * 1.05
# y3 = round(y2 * 1.05 -0.5, 0)
y3 = float(int(y2 * 1.05))
#print(" {} {}".format(y4, y3))
if y4 - y3 > 0:
y4 += 1
y2 = int(y4)
#print("{} {}".format(i, y2 * 1000))
print(y2 * 1000)
# 0 105,000
# 1 110,250 -> 111,000
# 2 116,550 -> 117,000
# 3 122,850 -> 123,000
#4 129,150 -> 130,000
|
s828996466
|
p03160
|
u690781906
| 2,000 | 1,048,576 |
Wrong Answer
| 133 | 13,928 | 201 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
n = int(input())
h = list(map(int, input().split()))
dp = [0] * n
dp[1] = abs(h[1]-h[0])
for i in range(2, n):
dp[i] = min(dp[i-1] + abs(h[i] - h[i-2]), dp[i-1] + abs(h[i] - h[i-1]))
print(dp[-1])
|
s612360716
|
Accepted
| 130 | 13,928 | 201 |
n = int(input())
h = list(map(int, input().split()))
dp = [0] * n
dp[1] = abs(h[1]-h[0])
for i in range(2, n):
dp[i] = min(dp[i-2] + abs(h[i] - h[i-2]), dp[i-1] + abs(h[i] - h[i-1]))
print(dp[-1])
|
s789905945
|
p03563
|
u562120243
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 181 |
Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it.
|
ans = 1
n = int(input())
k = int(input())
cnt = 0
while ans * 2 < ans + k:
if cnt > n:
break
ans *= 2
cnt +=1
for i in range(cnt,n):
ans += k
print(ans)
|
s081303566
|
Accepted
| 18 | 2,940 | 46 |
r = int(input())
g = int(input())
print(2*g-r)
|
s480565815
|
p02678
|
u425068548
| 2,000 | 1,048,576 |
Wrong Answer
| 621 | 39,936 | 919 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
import collections
def get_input():
return list(map(lambda x:int(x), input().split()))
def main():
inp = get_input()
n, m = inp[0], inp[1]
adj_list = [[] for i in range(n)]
for i in range(m):
inpu = get_input()
a, b = inpu[0], inpu[1]
adj_list[a - 1].append(b - 1)
adj_list[b - 1].append(a - 1)
q = collections.deque()
visited = set()
q.append(0)
visited.add(0)
p = [-1] * n
#print(adj_list)
while q:
u = q.popleft()
for i in adj_list[u]:
#print("checking", u, i)
if i not in visited:
q.append(i)
visited.add(i)
p[i] = u
#print(p)
for i in range(1, len(p)):
if p[i] == -1:
print("NO")
return
print("YES")
for i in range(1, n):
print(p[i] + 1)
if __name__ == '__main__':
main()
|
s347689034
|
Accepted
| 740 | 40,252 | 919 |
import collections
def get_input():
return list(map(lambda x:int(x), input().split()))
def main():
inp = get_input()
n, m = inp[0], inp[1]
adj_list = [[] for i in range(n)]
for i in range(m):
inpu = get_input()
a, b = inpu[0], inpu[1]
adj_list[a - 1].append(b - 1)
adj_list[b - 1].append(a - 1)
q = collections.deque()
visited = set()
q.append(0)
visited.add(0)
p = [-1] * n
#print(adj_list)
while q:
u = q.popleft()
for i in adj_list[u]:
#print("checking", u, i)
if i not in visited:
q.append(i)
visited.add(i)
p[i] = u
#print(p)
for i in range(1, len(p)):
if p[i] == -1:
print("No")
return
print("Yes")
for i in range(1, n):
print(p[i] + 1)
if __name__ == '__main__':
main()
|
s943926615
|
p03548
|
u304318875
| 2,000 | 262,144 |
Wrong Answer
| 156 | 13,568 | 1,550 |
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
import numpy as np
import math
import fractions
class KyoPro:
AtCoder_Mod = 1000 * 1000 * 1000 + 7
def ReadListOfNumbers():
return list(map(int, input().split()))
@staticmethod
def MakeStringFromNumbers(a):
if len(a) == 0:
return
s = str(a[0])
for i in range(1, len(a)):
s += ' ' + str(a[i])
return s
@staticmethod
def BinarySearch(ls, target, func):
if(func is None):
def func(a, b): return a < b
def eq(a, b): return (not func(a, b)) and (not func(b, a))
left = 0
right = len(ls) - 1
while(True):
if(right - left < 10):
for i in range(left, right + 1):
if(eq(ls[i], target)):
return True
return False
center = (left + right) // 2
if(eq(ls[center], target)):
return True
if(func(ls[center], target)):
left = center
if(func(target, ls[center])):
right = center
class cin:
InputList = []
def __init__(self):
pass
def Read(self):
cin.InputList += input().split()
def ReadInt(self):
if(len(cin.InputList) == 0):
self.Read
def main():
temp = KyoPro.ReadListOfNumbers()
x = temp[0]
y = temp[1]
z = temp[2]
y += z
x += z
print(x / y)
main()
|
s826396501
|
Accepted
| 158 | 13,568 | 1,551 |
import numpy as np
import math
import fractions
class KyoPro:
AtCoder_Mod = 1000 * 1000 * 1000 + 7
def ReadListOfNumbers():
return list(map(int, input().split()))
@staticmethod
def MakeStringFromNumbers(a):
if len(a) == 0:
return
s = str(a[0])
for i in range(1, len(a)):
s += ' ' + str(a[i])
return s
@staticmethod
def BinarySearch(ls, target, func):
if(func is None):
def func(a, b): return a < b
def eq(a, b): return (not func(a, b)) and (not func(b, a))
left = 0
right = len(ls) - 1
while(True):
if(right - left < 10):
for i in range(left, right + 1):
if(eq(ls[i], target)):
return True
return False
center = (left + right) // 2
if(eq(ls[center], target)):
return True
if(func(ls[center], target)):
left = center
if(func(target, ls[center])):
right = center
class cin:
InputList = []
def __init__(self):
pass
def Read(self):
cin.InputList += input().split()
def ReadInt(self):
if(len(cin.InputList) == 0):
self.Read
def main():
temp = KyoPro.ReadListOfNumbers()
x = temp[0]
y = temp[1]
z = temp[2]
y += z
x -= z
print(x // y)
main()
|
s055948431
|
p03377
|
u597047658
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 88 |
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A, B, X = map(int, input().split())
if A+B > X:
print('NO')
else:
print("YES")
|
s419457991
|
Accepted
| 19 | 2,940 | 103 |
A, B, X = map(int, input().split())
if (A+B > X) and (A <= X):
print('YES')
else:
print('NO')
|
s003682170
|
p02742
|
u679089074
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,076 | 97 |
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:
|
#6
H,W = map(int,input().split())
k = H*W
if k%2 == 0:
print(k/2)
else:
print(int(k/2)+1)
|
s565427377
|
Accepted
| 25 | 9,132 | 163 |
#6
import sys
H,W = map(int,input().split())
k = H*W
if H ==1 or W == 1:
print(1)
sys.exit()
if k%2 == 0:
print(int(k/2))
else:
print(int(k/2)+1)
|
s556579580
|
p03645
|
u368796742
| 2,000 | 262,144 |
Wrong Answer
| 690 | 61,108 | 257 |
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
n,m = map(int,input().split())
l = [list(map(int,input().split())) for i in range(m)]
a = []
b = []
for i in l:
if i[0] == 1:
a.append(i[1])
if i[1] == n:
b.append(i[0])
print("Possible" if len(set(a)&set(b)) > 0 else "Impossible")
|
s856604117
|
Accepted
| 595 | 18,892 | 217 |
n,m = map(int,input().split())
a = set()
b = set()
for i in range(m):
a_,b_ = map(int,input().split())
if a_ == 1:
a.add(b_)
if b_ == n:
b.add(a_)
print("POSSIBLE" if a&b else "IMPOSSIBLE")
|
s291371859
|
p02612
|
u472039178
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,028 | 59 |
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())
n = (N-(N%1000))/1000
print(1000*(n+1)-N)
|
s354904808
|
Accepted
| 28 | 9,040 | 98 |
N=int(input())
if N%1000==0:
print("0")
else:
n=((N-(N%1000))/1000)+1
print(int((1000*n)-N))
|
s343367117
|
p04030
|
u227082700
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 145 |
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
def b(x):
if x=='':return''
else:return x[:-1]
s,ans=input(),''
for i in range(len(s)):
if s[i]!='B':ans=b(ans)
else:ans+=s[i]
print(ans)
|
s422844695
|
Accepted
| 17 | 2,940 | 106 |
ans=[]
for i in input():
if i=="B":
if len(ans):del ans[-1]
else:ans.append(i)
print("".join(ans))
|
s207033710
|
p02613
|
u554237650
| 2,000 | 1,048,576 |
Wrong Answer
| 153 | 15,988 | 339 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
import os
N = int(input())
S = [0] * N
for i in range(0, N):
S[i] = input()
AC = 0
WA = 0
TLE = 0
RE = 0
for i in S:
if i == "AC":
AC+=1
elif i == "WA":
WA+=1
elif i == "TLE":
TLE+=1
elif i == "RE":
RE+=1
print("AC × ",AC)
print("WA × ",WA)
print("TLE × ",TLE)
print("RE × ",RE)
|
s740173758
|
Accepted
| 161 | 15,992 | 341 |
import os
N = int(input())
S = [0] * N
for i in range(0, N):
S[i] = input()
AC = 0
WA = 0
TLE = 0
RE = 0
for i in S:
if i == "AC":
AC+=1
elif i == "WA":
WA+=1
elif i == "TLE":
TLE+=1
elif i == "RE":
RE+=1
print("AC x",AC)
print("WA x",WA)
print("TLE x",TLE)
print("RE x",RE)
|
s028094627
|
p03456
|
u320567105
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,188 | 240 |
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.
|
ri = lambda: int(input())
rl = lambda: list(map(int,input().split()))
rr = lambda N: [ri() for _ in range(N)]
YN = lambda b: print('YES') if b else print('NO')
INF = 10**18
a,b = rl()
c = int(str(a)+str(b))
c_ = int(c**0.5)
YN(c==c_*c_)
|
s904570111
|
Accepted
| 17 | 3,060 | 279 |
ri = lambda: int(input())
rl = lambda: list(map(int,input().split()))
rr = lambda N: [ri() for _ in range(N)]
YN = lambda b: print('YES') if b else print('NO')
INF = 10**18
a,b = rl()
c = int(str(a)+str(b))
c_ = int(c**0.5)
if c==c_*c_:
print('Yes')
else:
print('No')
|
s029276194
|
p02659
|
u131638468
| 2,000 | 1,048,576 |
Wrong Answer
| 22 | 9,092 | 44 |
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
a,b=map(float,input().split())
print(a*b//1)
|
s750764198
|
Accepted
| 23 | 9,180 | 127 |
a,b=map(str,input().split())
a=int(a)
c=''
for i in b:
if i=='.':
continue
else:
c+=i
c=int(c)
print(int(a*c//100))
|
s067467699
|
p02842
|
u585963419
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 105 |
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
n=int(input())
step0=(n/108*100)
step1=(step0/100*108)
if step1!=n:
print(':(')
else:
print(step0)
|
s025085632
|
Accepted
| 17 | 3,064 | 466 |
n=int(input())
step00=int((n/108*100))
step01=int((n/108*100))-1
step02=int((n/108*100))+1
step10=int((step00/100*108))
step11=int((step01/100*108))
step12=int((step02/100*108))
nocnt=0
ok=0
if step10!=n:
nocnt=nocnt+1
else:
print(step00)
ok=1
if step11!=n:
nocnt=nocnt+1
else:
if ok==0:
print(step10)
ok=1
if step12!=n:
nocnt=nocnt+1
else:
if ok==0:
print(step02)
ok=1
if nocnt>=3:
print(':(')
|
s364401219
|
p04029
|
u637824361
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 38 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N = int(input())
print(N * (N+1) / 2)
|
s265353743
|
Accepted
| 17 | 2,940 | 43 |
N = int(input())
print(int(N * (N+1) / 2))
|
s856175361
|
p02616
|
u218623249
| 2,000 | 1,048,576 |
Wrong Answer
| 147 | 31,744 | 173 |
Given are N integers A_1,\ldots,A_N. We will choose exactly K of these elements. Find the maximum possible product of the chosen elements. Then, print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).
|
suus,cleck = map(int,input().split()); cuc = 1
fef = sorted(map(int,input().split()), reverse = True)
for joj in range(cleck):
cuc = (cuc * fef[joj])%1000000007
print(cuc)
|
s323372718
|
Accepted
| 271 | 31,868 | 1,176 |
N,K=map(int, input().split())
A=list(map(int, input().split()))
D,E=[],[]
zcnt,scnt,fcnt=0,0,0
for i in A:
if i==0:
zcnt+=1
D.append(0)
elif i>0:
D.append(i)
scnt+=1
else:
E.append(i)
fcnt+=1
mod=10**9+7
ans=1
if K==N:
for i in A:
ans*=i
ans%=mod
print(ans)
exit()
if K%2==1 and max(A)<0:
A=sorted(A)[::-1]
for i in range(K):
ans*=A[i]
ans%=mod
print(ans)
exit()
if K>scnt+fcnt:
print(0)
exit()
D,E=sorted(D)[::-1],sorted(E)
#print(D,E)
ans=1
cnt=0
a,b=0,0
while K-cnt>1:
if a+1<=len(D)-1 and b+1<=len(E)-1:
d,e=D[a]*D[a+1],E[b]*E[b+1]
if d>e:
ans*=D[a]
a+=1
cnt+=1
ans%=mod
else:
ans*=e
b+=2
ans%=mod
cnt+=2
elif a+1<=len(D)-1:
d=D[a]*D[a+1]
ans*=D[a]
a+=1
cnt+=1
ans%=mod
elif b+1<=len(E)-1:
e=E[b]*E[b+1]
ans*=e
b+=2
cnt+=2
ans%=mod
if K-cnt==1:
Z=[]
if a!=scnt:
Z.append(D[a])
if b!=fcnt:
Z.append(E[-1])
if 0 in A:
Z.append(0)
ans*=max(Z)
ans%=mod
print(ans)
|
s801985030
|
p03372
|
u562935282
| 2,000 | 262,144 |
Wrong Answer
| 704 | 48,692 | 1,021 |
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter. Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories. Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter. Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
|
def solve(xv_sorted):
x = [False for _ in range(n)]
y = [False for _ in range(n)]
s = 0
buf = (- 1) * float('inf')
for i in range(n):
s += xv_sorted[i][1]##cal
x[i] = max(buf, s - 2 * xv_sorted[i][0])##pos
buf = x[i]
print(x)
s = 0
buf = (- 1) * float('inf')
for i in sorted(range(n), reverse=True):
s += xv_sorted[i][1]##cal
y[i] = max(buf, s - (c - xv_sorted[i][0]))##pos
buf = y[i]
print(y)
ans = max(x[n - 1], y[0])
for i in range(n - 1):
ans = max(ans, x[i] + y[i + 1])
return ans
if __name__ == '__main__':
n, c = map(int, input().split())
xv = []
for i in range(n):
xv.append(list(map(int, input().split())))
xv_sorted = sorted(xv, key=lambda x:x[0])
ans = 0
ans = max(ans, solve(xv_sorted))
for i in range(n):
xv_sorted[i][0] = abs(c - xv_sorted[i][0])
xv_sorted = sorted(xv, key=lambda x:x[0])
ans = max(ans, solve(xv_sorted))
print(ans)
|
s980518099
|
Accepted
| 470 | 30,224 | 1,800 |
def main():
from collections import namedtuple
import sys
input = sys.stdin.readline
Sushi = namedtuple('Sushi', 'x cal')
n, c = map(int, input().split())
a = []
for _ in range(n):
x, v = map(int, input().split())
a.append(Sushi(x=x, cal=v))
clock = [0] * (n + 1)
clock_to_0 = [0] * (n + 1)
ma = 0
ma0 = 0
curr = 0
for i, s in enumerate(a, start=1):
curr += s.cal
ma = max(ma, curr - s.x)
ma0 = max(ma0, curr - s.x * 2)
clock[i] = ma
clock_to_0[i] = ma0
anti = [0] * (n + 1)
anti_to_0 = [0] * (n + 1)
ma = 0
ma0 = 0
curr = 0
for i, s in zip(range(n, -1, -1), reversed(a)):
curr += s.cal
ma = max(ma, curr - (c - s.x))
ma0 = max(ma0, curr - (c - s.x) * 2)
anti[i] = ma
anti_to_0[i] = ma0
ans = 0
for exit_pos in range(1, n + 1):
ans = max(
ans,
clock_to_0[exit_pos - 1] + anti[exit_pos],
anti_to_0[(exit_pos + 1) % (n + 1)] + clock[exit_pos]
)
print(ans)
if __name__ == '__main__':
main()
|
s518554713
|
p02842
|
u860338101
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 119 |
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
import math
N = int(input())
M = N / 1.08
if math.floor(M) * 1.08 == N:
print(math.floor(M*1.08))
else:
print(':(')
|
s937044670
|
Accepted
| 17 | 3,060 | 240 |
import math
N = int(input())
flag = False
Min = math.floor(N / 1.08)
Max = math.floor( ( N + 1 ) / 1.08 + 1)
for i in range(Min, Max + 1):
if math.floor( i * 1.08 ) == N:
print( i )
flag = True
break
if not flag:
print(':(')
|
s361253693
|
p02619
|
u370562440
| 2,000 | 1,048,576 |
Wrong Answer
| 130 | 27,036 | 613 |
Let's first write a program to calculate the score from a pair of input and output. You can know the total score by submitting your solution, or an official program to calculate a score is often provided for local evaluation as in this contest. Nevertheless, writing a score calculator by yourself is still useful to check your understanding of the problem specification. Moreover, the source code of the score calculator can often be reused for solving the problem or debugging your solution. So it is worthwhile to write a score calculator unless it is very complicated.
|
#B
import numpy as np
D = int(input())
C = list(map(int,input().split()))
S = np.empty((D,26),dtype=np.int8)
for i in range(D):
l = list(map(int,input().split()))
for j in range(26):
S[i][j] = l[j]
R = np.zeros(26)
sc = 0
for i in range(1,D+1):
con = int(input()) - 1
sc += S[i-1][con]
R[con] = i
for j in range(26):
sc -= C[j]*(i-R[j])
print(sc)
|
s552282601
|
Accepted
| 136 | 26,992 | 644 |
#B
import numpy as np
D = int(input())
C = list(map(int,input().split()))
S = np.empty((D,26))
for i in range(D):
l = list(map(int,input().split()))
for j in range(26):
S[i][j] = l[j]
R = np.zeros(26)
sc = 0
for i in range(1,D+1):
con = int(input()) - 1
sc += S[i-1][con]
R[con] = i
for j in range(26):
sc -= C[j]*(i-R[j])
print(int(sc))
|
s901905643
|
p03162
|
u006883624
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 421 |
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
def main():
n = int(input())
table = []
for _ in range(n):
table.append([int(v) for v in input().split()])
dp = [0] * 3
for i in range(n):
dp_n = [0] * 3
a, b, c = table[i]
a0 = dp[0]
b0 = dp[1]
c0 = dp[2]
dp_n[0] = max(b0, c0) + a
dp_n[1] = max(c0, a0) + b
dp_n[2] = max(a0, b0) + c
dp = dp_n
print(max(dp))
|
s780357990
|
Accepted
| 396 | 20,536 | 400 |
def main():
N = int(input())
table = []
for _ in range(N):
a, b, c = map(int, input().split())
table.append((a, b, c))
dp = [0] * 3
for i in range(N):
a, b, c = table[i]
a0 = dp[0]
b0 = dp[1]
c0 = dp[2]
dp[0] = max(b0, c0) + a
dp[1] = max(c0, a0) + b
dp[2] = max(a0, b0) + c
print(max(dp))
main()
|
s015599991
|
p03470
|
u165318982
| 2,000 | 262,144 |
Wrong Answer
| 25 | 9,156 | 82 |
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
N = int(input())
A = list(map(int, input().split()))
A = set(A)
print(str(len(A)))
|
s763453177
|
Accepted
| 27 | 9,100 | 83 |
N = int(input())
A = [int(input()) for _ in range(N)]
A = set(A)
print(str(len(A)))
|
s285498522
|
p02865
|
u190580703
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 29 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
a=int(input())
print((a-1)/2)
|
s362592869
|
Accepted
| 17 | 2,940 | 30 |
a=int(input())
print((a-1)//2)
|
s110568846
|
p03853
|
u870841038
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 117 |
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).
|
h, w = map(int, input().split())
marks = list(input() for i in range(h))
for i in range(h):
print(marks[i] * 2)
|
s904906287
|
Accepted
| 17 | 3,060 | 140 |
h, w = map(int, input().split())
marks = list(input() for i in range(h))
for i in range(h):
for j in range(2):
print(marks[i])
|
s222529809
|
p02288
|
u912143677
| 2,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 441 |
A binary heap which satisfies max-heap property is called max-heap. In a max- heap, for every node $i$ other than the root, $A Write a program which reads an array and constructs a max-heap from the array based on the following pseudo code. $maxHeapify(A, i)$ move the value of $A[i]$ down to leaves to make a sub-tree of node $i$ a max-heap. Here, $H$ is the size of the heap. 1 maxHeapify(A, i) 2 l = left(i) 3 r = right(i) 4 // select the node which has the maximum value 5 if l ≤ H and A[l] > A[i] 6 largest = l 7 else 8 largest = i 9 if r ≤ H and A[r] > A[largest] 10 largest = r 11 12 if largest ≠ i // value of children is larger than that of i 13 swap A[i] and A[largest] 14 maxHeapify(A, largest) // call recursively The following procedure buildMaxHeap(A) makes $A$ a max-heap by performing maxHeapify in a bottom-up manner. 1 buildMaxHeap(A) 2 for i = H/2 downto 1 3 maxHeapify(A, i)
|
n = int(input())
a = list(map(int, input().split()))
def maxheapify(a, i):
l = i*2 + 1
r = i*2 + 2
if l <= n and a[l] > a[i]:
largest = l
else:
largest = i
if r <= n and a[r] > a[largest]:
largest = r
if largest != i:
memo = a[i]
a[i] = a[largest]
a[largest] = memo
maxheapify(a, largest)
h = (n - 1) // 2
for i in reversed(range(h)):
maxheapify(a, i)
|
s885587886
|
Accepted
| 880 | 63,696 | 721 |
n = int(input())
a = list(map(int, input().split()))
def maxheapify(a, i):
l = i*2 + 1
r = i*2 + 2
if r < n:
if a[l] > a[i]:
largest = l
else:
largest = i
if a[r] > a[largest]:
largest = r
if largest != i:
memo = a[i]
a[i] = a[largest]
a[largest] = memo
maxheapify(a, largest)
elif l < n:
if a[l] > a[i]:
largest = l
else:
largest = i
if largest != i:
memo = a[i]
a[i] = a[largest]
a[largest] = memo
maxheapify(a, largest)
for i in reversed(range(n)):
maxheapify(a, i)
print("", *a)
|
s067095304
|
p03610
|
u021548497
| 2,000 | 262,144 |
Wrong Answer
| 29 | 3,572 | 69 |
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s = input()
print("".join([str(s[2*i+1]) for i in range(len(s)//2)]))
|
s140574560
|
Accepted
| 39 | 3,188 | 91 |
s = input()
ans = ""
for i in range(len(s)):
if i%2 == 0:
ans = ans + s[i]
print(ans)
|
s545952328
|
p03493
|
u160659351
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 57 |
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
data = [x for x in input().rstrip()]
print(data.count(1))
|
s416545159
|
Accepted
| 19 | 3,060 | 59 |
data = [x for x in input().rstrip()]
print(data.count("1"))
|
s602942238
|
p02612
|
u325660636
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,088 | 48 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
otsuri = N % 1000
print(otsuri)
|
s458585999
|
Accepted
| 29 | 9,164 | 149 |
N = int(input())
a = N // 1000
b = N % 1000
if b ==0:
print(0)
else:
if N<=1000:
print(1000-b)
else:
print(1000*(a+1)-N)
|
s282457628
|
p02865
|
u459774442
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 79 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
N = int(input())
ans=0
if N%2==0:
ans=N/2
else:
ans=int(N/2)
print(ans)
|
s616999285
|
Accepted
| 17 | 2,940 | 86 |
N = int(input())
ans=0
if N%2==0:
ans=N/2-1
elif N!=1:
ans=N/2
print(int(ans))
|
s031140633
|
p03455
|
u247465867
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 586 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
# -*- codinf: utf-8 -*-
numList = [int(num) for num in input().split()]
productNumList = numList[0] * numList[1]
if ((productNumList % 2) == 0 ):
print("even")
else:
print("odd")
|
s068512090
|
Accepted
| 17 | 2,940 | 90 |
#2019/10/10
a, b = map(int, open(0).read().split())
print("Odd" if (a*b)%2!=0 else "Even")
|
s791283241
|
p02255
|
u716198574
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,516 | 529 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
N = int(input())
A = list(map(int, input().split()))
for i in range(1, N):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j + 1] = A[j]
j -= 1
A[j + 1] = v
print(' '.join(map(str, A)))
|
s823383675
|
Accepted
| 30 | 7,608 | 526 |
N = int(input())
A = list(map(int, input().split()))
for i in range(N):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j + 1] = A[j]
j -= 1
A[j + 1] = v
print(' '.join(map(str, A)))
|
s482205755
|
p03386
|
u015187377
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,316 | 224 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
import collections
a,b,k=map(int,input().split())
l1=list(range(a,a+k))
l1=[str(i) for i in l1]
l2=list(range(b-k+1,b+1))
l2=[str(i) for i in l2]
ans=l1+l2
ans=collections.Counter(ans)
ans=sorted(ans)
print(" ".join(ans))
|
s030908283
|
Accepted
| 17 | 3,060 | 165 |
a,b,k=map(int,input().split())
ans=[]
for i in range(k):
if a+i<=b:
ans.append(a+i)
if b-i>=a:
ans.append(b-i)
ans=sorted(set(ans))
print(*ans, sep='\n')
|
s497086785
|
p02406
|
u706217959
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,588 | 164 |
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; }
|
def main():
n = int(input())
for i in range(n+1):
if i % 3 == 0 or i % 3 == 10:
print(i,end=" ")
if __name__ == "__main__":
main()
|
s821012088
|
Accepted
| 30 | 5,880 | 337 |
def call(a):
if a % 3 == 0:
return True
while a != 0:
if a % 10 == 3:
return True
a //= 10
return False
def main():
n = int(input())
for i in range(1, n + 1):
if call(i):
print(" {0}".format(i), end="")
print("")
if __name__ == "__main__":
main()
|
s859887047
|
p03149
|
u721316601
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 346 |
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".
|
s = 'keyence'
S = input()
ans = 'NO'
for i in range(len(S)):
if S[i] == 'k':
for l in range(1, len(s)):
if S[i+l] != s[l]:
break
for n in range(len(S)-l):
if S[i:l] + S[l+n:l+n+7-3] == s:
ans = 'YES'
break
if ans == 'YES':
break
print(ans)
|
s684718397
|
Accepted
| 17 | 2,940 | 103 |
N = input()
if '1' in N and '9' in N and '7' in N and '4' in N:
print('YES')
else:
print('NO')
|
s110705689
|
p02608
|
u255943004
| 2,000 | 1,048,576 |
Wrong Answer
| 950 | 11,812 | 236 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
N = int(input())
from collections import defaultdict
ans = defaultdict(int)
for a in range(101):
for b in range(101):
for c in range(101):
ans[a**2 + b**2 + c**2 + a*b + a*c + b*c] += 1
for n in range(1,N+1):
print(ans[n])
|
s596397944
|
Accepted
| 949 | 11,956 | 243 |
N = int(input())
from collections import defaultdict
ans = defaultdict(int)
for a in range(1,101):
for b in range(1,101):
for c in range(1,101):
ans[a**2 + b**2 + c**2 + a*b + a*c + b*c] += 1
for n in range(1,N+1):
print(ans[n])
|
s447328485
|
p03485
|
u766407523
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,316 | 117 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
ab = input().split()
a = int(ab[0])
b = int(ab[1])
if (a+b)%2 == 0:
print((a+b)/2)
else:
print((a+b)//2 + 1)
|
s357470931
|
Accepted
| 17 | 2,940 | 117 |
ab = input().split()
a, b = int(ab[0]), int(ab[1])
if (a+b)%2 == 0:
print((a+b)//2)
else:
print((a+b)//2 + 1)
|
s629997350
|
p03762
|
u614875193
| 2,000 | 262,144 |
Wrong Answer
| 68 | 25,476 | 529 |
On a two-dimensional plane, there are m lines drawn parallel to the x axis, and n lines drawn parallel to the y axis. Among the lines parallel to the x axis, the i-th from the bottom is represented by y = y_i. Similarly, among the lines parallel to the y axis, the i-th from the left is represented by x = x_i. For every rectangle that is formed by these lines, find its area, and print the total area modulo 10^9+7. That is, for every quadruple (i,j,k,l) satisfying 1\leq i < j\leq n and 1\leq k < l\leq m, find the area of the rectangle formed by the lines x=x_i, x=x_j, y=y_k and y=y_l, and print the sum of these areas modulo 10^9+7.
|
m,n=map(int,input().split())
X=list(map(int,input().split()))
Y=list(map(int,input().split()))
mod=10**9+7
if m>2 and n>2:
X2=[(X[1]-X[0])%mod,(X[-2]-X[1])%mod,(X[-1]-X[-2])%mod]
Y2=[(Y[1]-Y[0])%mod,(Y[-2]-Y[1])%mod,(Y[-1]-Y[-2])%mod]
ans=0
a,b1,b2,c=m*n,(2*m-2)*n,(2*n-2)*m,(2*m-2)*(2*n-2)
ans+=(X2[0]*Y2[0] + X2[2]*Y2[0] + X2[0]*Y2[2] + X2[2]*Y2[2])*a%mod
ans+=(X2[1]*Y2[0]+X2[1]*Y2[2])*b1%mod
ans+=(X2[0]*Y2[1]+X2[2]*Y2[1])*b2%mod
ans+=X2[1]*Y2[1]*c%mod
print(ans%mod)
else:
print(mod)
|
s479956090
|
Accepted
| 136 | 24,356 | 270 |
n,m=map(int,input().split())
X=list(map(int,input().split()))
Y=list(map(int,input().split()))
mod=10**9+7
ans1=0
for k in range(1,n+1):
ans1+=(2*k-1-n)*X[k-1]
ans1%=mod
ans2=0
for k in range(1,m+1):
ans2+=(2*k-1-m)*Y[k-1]
ans2%=mod
print(ans1*ans2%mod)
|
s411830680
|
p02975
|
u627234757
| 2,000 | 1,048,576 |
Wrong Answer
| 51 | 14,704 | 825 |
Snuke has N hats. The i-th hat has an integer a_i written on it. There are N camels standing in a circle. Snuke will put one of his hats on each of these camels. If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print `Yes`; otherwise, print `No`. * The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself. What is XOR? The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non- negative integers x_1, x_2, \ldots, x_n is defined as follows: - When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even. For example, 3 \oplus 5 = 6.
|
n = int(input())
nums = list(map(int, input().split()))
unq = list(set(nums))
if len(unq) > 3:
print("No")
if len(unq) == 3:
a,b,c = unq
if a ^ b ^ c != 0:
if nums.count(a) == nums.count(b) == nums.count(c):
print("Yes")
else:
print("No")
else:
print("No")
if len(unq) == 2:
a,b = sorted(unq)
if a == 0:
if nums.count(a) * 2 == nums.count(b):
print("Yes")
else:
print("No")
else:
print("No")
if len(unq) == 1:
a = unq[0]
if a != 0:
print("No")
else:
print("Yes")
|
s020483479
|
Accepted
| 52 | 14,704 | 826 |
n = int(input())
nums = list(map(int, input().split()))
unq = list(set(nums))
if len(unq) > 3:
print("No")
if len(unq) == 3:
a,b,c = unq
if a ^ b ^ c == 0:
if nums.count(a) == nums.count(b) == nums.count(c):
print("Yes")
else:
print("No")
else:
print("No")
if len(unq) == 2:
a,b = sorted(unq)
if a == 0:
if nums.count(a) * 2 == nums.count(b):
print("Yes")
else:
print("No")
else:
print("No")
if len(unq) == 1:
a = unq[0]
if a != 0:
print("No")
else:
print("Yes")
|
s740716548
|
p03494
|
u932868243
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 96 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n=int(input())
a=list(map(int,input().split()))
n=0
if all(aa%2==0 for aa in a):
n+=1
print(n)
|
s053903365
|
Accepted
| 18 | 3,060 | 129 |
n=int(input())
a=list(map(int,input().split()))
ans=0
while all(aa%2==0 for aa in a):
ans+=1
a=[aa//2 for aa in a]
print(ans)
|
s054796309
|
p04043
|
u823885866
| 2,000 | 262,144 |
Wrong Answer
| 25 | 8,808 | 52 |
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.
|
print('YNEOS'[eval(input().replace(*' *'))!=245::2])
|
s746443854
|
Accepted
| 27 | 8,904 | 52 |
print('YNEOS'[eval(input().replace(*' *'))!=175::2])
|
s344992929
|
p02388
|
u976860528
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,352 | 14 |
Write a program which calculates the cube of a given integer x.
|
print(input())
|
s658510354
|
Accepted
| 20 | 7,556 | 29 |
x = int(input())
print(x*x*x)
|
s333038762
|
p02744
|
u537905693
| 2,000 | 1,048,576 |
Wrong Answer
| 726 | 19,104 | 665 |
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.
|
#!/usr/bin/env python
# coding: utf-8
def ri():
return int(input())
def rl():
return list(input().split())
def rli():
return list(map(int, input().split()))
words = set()
alp = list("abcdefghij")
def dfs(n, s):
if len(s) == n:
return
for i in range(0, len(s)):
for c in alp:
if ord(c)>ord(s[i])+1:
break
ns = s[:i+1]+c+s[i+1:]
if ns in words:
continue
words.add(ns)
dfs(n, ns)
def main():
n = ri()
dfs(n, 'a')
for w in sorted(words):
if len(w) == n:
print(w)
if __name__ == '__main__':
main()
|
s946236745
|
Accepted
| 768 | 19,104 | 714 |
#!/usr/bin/env python
# coding: utf-8
def ri():
return int(input())
def rl():
return list(input().split())
def rli():
return list(map(int, input().split()))
words = set()
alp = list("abcdefghij")
def dfs(n, s):
if len(s) == n:
return
for i in range(0, len(s)):
for c in alp:
if ord(c)>ord(s[i])+1:
break
ns = s[:i+1]+c+s[i+1:]
if ns in words:
continue
words.add(ns)
dfs(n, ns)
def main():
n = ri()
if n == 1:
print("a")
return
dfs(n, 'a')
for w in sorted(words):
if len(w) == n:
print(w)
if __name__ == '__main__':
main()
|
s701838070
|
p03860
|
u500279510
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 47 |
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()
x = s[0]
ans = 'A'+x+'C'
print(ans)
|
s443694314
|
Accepted
| 17 | 2,940 | 47 |
s = input()
x = s[8]
ans = 'A'+x+'C'
print(ans)
|
s927158869
|
p03433
|
u068538925
| 2,000 | 262,144 |
Wrong Answer
| 30 | 8,972 | 94 |
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())
temp = n%500
if temp<=a:
print("YES")
else:
print("NO")
|
s900060069
|
Accepted
| 28 | 9,100 | 94 |
n = int(input())
a = int(input())
temp = n%500
if temp<=a:
print("Yes")
else:
print("No")
|
s311949263
|
p03721
|
u543954314
| 2,000 | 262,144 |
Wrong Answer
| 480 | 21,524 | 234 |
There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3.
|
n, k = map(int, input().split())
l = [tuple(map(int, input().split())) for i in range(n)]
l.sort()
l.append((0, k))
d = [l[0][1]]
for i in range(1,n+1):
d.append(d[i-1] + l[i][1])
t = 0
while d[t] <= k:
t += 1
print(l[t-1][0])
|
s301243528
|
Accepted
| 443 | 21,524 | 231 |
n, k = map(int, input().split())
l = [tuple(map(int, input().split())) for i in range(n)]
l.sort()
l.append((0, k))
d = [l[0][1]]
for i in range(1,n+1):
d.append(d[i-1] + l[i][1])
t = 0
while d[t] < k:
t += 1
print(l[t][0])
|
s303877168
|
p04043
|
u653005308
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 122 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a,b,c=map(int,input().split())
if (a,b,c).count("5")==2 and (a,b,c).count("7")==1:
print('YES')
else:
print('NO')
|
s601302471
|
Accepted
| 17 | 2,940 | 117 |
a,b,c=map(int,input().split())
if (a,b,c).count(5)==2 and (a,b,c).count(7)==1:
print('YES')
else:
print('NO')
|
s004890530
|
p03110
|
u118147328
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 2,940 | 208 |
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
N = int(input())
X = [input().split() for _ in range(N)]
Y = 0
for i in X:
print(i[1])
if i[1] == 'JPY':
Y += int(i[0])
elif i[1] == 'BTC':
Y += (380000.0 * float(i[0]))
print(Y)
|
s027271218
|
Accepted
| 17 | 2,940 | 192 |
N = int(input())
X = [input().split() for _ in range(N)]
Y = 0
for i in X:
if i[1] == 'JPY':
Y += int(i[0])
elif i[1] == 'BTC':
Y += (380000.0 * float(i[0]))
print(Y)
|
s528125916
|
p02842
|
u769551032
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 97 |
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
N = int(input())
untax_N = round(N/1.08)
if round(untax_N*1.08) != N:
print(":(")
else:
pass
|
s933944094
|
Accepted
| 33 | 2,940 | 148 |
N = int(input())
count = 0
for i in range(N+1):
if N <= int(i*1.08) < N+1:
print(i)
count += 1
break
if count == 0:
print(":(")
|
s876034870
|
p03434
|
u676707608
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 92 |
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
n=int(input())
a=list(map(int,input().split()))
a.sort()
print((sum(a[::2])-(sum(a[1::2]))))
|
s918654234
|
Accepted
| 17 | 2,940 | 104 |
n=int(input())
a=list(map(int,input().split()))
a.sort(reverse=True)
print((sum(a[::2])-(sum(a[1::2]))))
|
s121452557
|
p03477
|
u168416324
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,148 | 125 |
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=map(int,input().split())
x=a+b
y=c+d
if x>y:
print("Left")
if x>y:
print("Right")
if x==y:
print("Balanced")
|
s862743402
|
Accepted
| 29 | 9,072 | 126 |
a,b,c,d=map(int,input().split())
x=a+b
y=c+d
if x>y:
print("Left")
if x<y:
print("Right")
if x==y:
print("Balanced")
|
s542615192
|
p03575
|
u499259667
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,572 | 777 |
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges.
|
N , M = map(int,input().split())
dotlist = [[] for i in range(N)]
sidlist = []
count = 0
for m in range(M):
a,b = map(int,input().split())
dotlist[a -1].append(b)
dotlist[b -1].append(a)
sidlist.append([a,b])
if N -1 == M:
print(0)
exit()
def check(anum,bnum,cnum):
for i in dotlist[cnum - 1]:
if (anum == cnum and i == bnum) or (bnum == cnum and i == anum):
pass
elif donedot[i - 1] == False:
donedot[i - 1] = True
print(cnum,i)
check(anum,bnum,i)
donedot = None
for m in range(M):
donedot = [False for i in range(M)]
a,b = sidlist[m]
check(a,b,1)
TorF = True
for i in donedot:
TorF = TorF and i
if not TorF:
count += 1
print(count)
|
s522218832
|
Accepted
| 22 | 3,316 | 705 |
N , M = map(int,input().split())
dotlist = [[] for i in range(N)]
sidlist = []
count = 0
for m in range(M):
a,b = map(int,input().split())
dotlist[a -1].append(b)
dotlist[b -1].append(a)
sidlist.append([a,b])
def check(anum,bnum,cnum):
donedot[cnum-1] = True
for i in dotlist[cnum - 1]:
if (anum == cnum and i == bnum) or (bnum == cnum and i == anum):
pass
elif donedot[i - 1] == False:
check(anum,bnum,i)
donedot = None
for m in range(M):
donedot = [False for i in range(N)]
a,b = sidlist[m]
check(a,b,1)
TorF = True
for i in donedot:
TorF = TorF and i
if not TorF:
count += 1
print(count)
|
s641817464
|
p03911
|
u600402037
| 2,000 | 262,144 |
Wrong Answer
| 2,107 | 73,484 | 893 |
On a planet far, far away, M languages are spoken. They are conveniently numbered 1 through M. For _CODE FESTIVAL 20XX_ held on this planet, N participants gathered from all over the planet. The i-th (1≦i≦N) participant can speak K_i languages numbered L_{i,1}, L_{i,2}, ..., L_{i,{}K_i}. Two participants A and B can _communicate_ with each other if and only if one of the following conditions is satisfied: * There exists a language that both A and B can speak. * There exists a participant X that both A and B can communicate with. Determine whether all N participants can communicate with all other participants.
|
import sys
import itertools
stdin = sys.stdin
ri = lambda: int(rs())
rl = lambda: list(map(int, stdin.readline().split())) # applies to numbers only
rs = lambda: stdin.readline().rstrip()
N, M = rl()
L = [rl() for _ in range(N)]
translate = [set() for _ in range(M+1)]
for i, x in enumerate(L):
k, *language = x
for l in language:
translate[l].add(i)
for x, y in itertools.combinations(range(N), 2):
x_use_language = set(L[x][1:])
y_use_language = set(L[y][1:])
x_can_speak = set([x])
y_can_speak = set([y])
for xuse in x_use_language:
x_can_speak = x_can_speak.union(translate[xuse])
for yuse in y_use_language:
y_can_speak = y_can_speak.union(translate[yuse])
if len(x_can_speak & y_can_speak) == 0:
print('NO')
exit()
print('YES')
# 26
|
s005674584
|
Accepted
| 309 | 41,544 | 895 |
import sys
sys.setrecursionlimit(10 ** 7)
stdin = sys.stdin
ri = lambda: int(rs())
rl = lambda: list(map(int, stdin.readline().split())) # applies to numbers only
rs = lambda: stdin.readline().rstrip()
N, M = rl()
KL = [rl() for _ in range(N)]
speak_lan = [[] for _ in range(M+1)]
for i, x in enumerate(KL):
k, *lan = x
for l in lan:
speak_lan[l].append(i)
communicate = set()
used_language = set()
def dfs(person):
global communicate
global used_language
lan = KL[person][1:]
for l in lan:
if l in used_language:
continue
used_language.add(l)
for p in speak_lan[l]:
if p not in communicate:
communicate.add(p)
dfs(p)
dfs(0)
if len(communicate) == N:
print('YES')
else:
print('NO')
|
s543979071
|
p03860
|
u761989513
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 62 |
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.
|
a, x, c = input().split()
print("{} {} {}".format(a, x[0], c))
|
s256225416
|
Accepted
| 17 | 2,940 | 67 |
a, x, c = input().split()
print("{}{}{}".format(a[0], x[0], c[0]))
|
s043743585
|
p03485
|
u803647747
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 101 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
input_list = list(map(int,input().split()))
import math
math.ceil((input_list[0] + input_list[1])/2)
|
s080079830
|
Accepted
| 17 | 2,940 | 108 |
input_list = list(map(int,input().split()))
import math
print(math.ceil((input_list[0] + input_list[1])/2))
|
s668764348
|
p02865
|
u498620941
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 98 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
N = int(input())
if N // 2 == 0 :
print("{}".format(N//2-1))
else:
print("{}".format(N//2))
|
s071384511
|
Accepted
| 18 | 2,940 | 94 |
N = int(input())
if N % 2 == 0 :
print("{}".format(N//2-1))
else:
print("{}".format(N//2))
|
s541616385
|
p02972
|
u322229918
| 2,000 | 1,048,576 |
Wrong Answer
| 2,109 | 51,848 | 298 |
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.
|
import numpy as np
N = int(input())
As = np.array([0] + list(map(int, input().split())))
res = np.zeros_like(As)
for idx, ai in enumerate(As[::-1]):
idx = N - idx
if idx == 0:
break
oe = res[::idx].sum() % 2
res[idx] = ai ^ oe
print(N)
print(" ".join(res[1:].astype(str)))
|
s632697974
|
Accepted
| 516 | 14,120 | 305 |
N = int(input())
As = list(map(int, input().split()))
res = [0] * N
for idx, ai in zip(range(N-1, -1, -1), As[::-1]):
oe = sum([res[i] for i in range(idx, N, idx+1)]) % 2#
res[idx] = ai ^ oe
print(sum(res))
idxs = []
for i, val in enumerate(res):
if val:
idxs += [i + 1]
print(*idxs)
|
s252594209
|
p03612
|
u797016134
| 2,000 | 262,144 |
Wrong Answer
| 80 | 14,008 | 161 |
You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this.
|
n = int(input())
p = list(map(int, input().split()))
ans = 0
for i in range(n-1):
if i+1 == p[i]:
p[i],p[i+1] = p[i+1],p[i]
ans += 1
print(ans)
print(p)
|
s541460221
|
Accepted
| 71 | 14,008 | 183 |
n = int(input())
p = list(map(int, input().split()))
ans = 0
for i in range(n-1):
if i+1 == p[i]:
p[i],p[i+1] = p[i+1],p[i]
ans += 1
if p[-1] == len(p):
ans += 1
print(ans)
|
s685388820
|
p02262
|
u508732591
| 6,000 | 131,072 |
Wrong Answer
| 30 | 8,096 | 615 |
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 $\lceil n^{1.5}\rceil$
|
import sys
import math
from collections import deque
def insertion_sort(a, n, g):
ct = 0
for i in range(g,n):
v = a[i]
j = i-g
while j >= 0 and a[j] > v:
a[j+g] = a[j]
j = j-g
ct += 1
a[j+g] = v
return ct
n = int(input())
a = list(map(int, sys.stdin.readlines()))
b = 701
ct= 0
g = deque([x for x in [701,301,132,57,23,10,1] if x < n])
while True:
b = math.floor(2.25*b)
if b > n:
break
g.appendleft(b)
for i in g:
ct += insertion_sort(a, n, i)
print(len(g))
print(*g, sep=" ")
print(ct)
print(*a, sep="\n")
|
s467267252
|
Accepted
| 10,980 | 118,484 | 551 |
import sys
import math
ct= 0
def insertion_sort(a, n, g):
global ct
for j in range(0,n-g):
v = a[j+g]
while j >= 0 and a[j] > v:
a[j+g] = a[j]
j = j-g
ct += 1
a[j+g] = v
n = int(input())
a = list(map(int, sys.stdin.readlines()))
b = math.floor(2.25*701)
g = [x for x in [1,4,10,23,57,132,301,701] if x <= n]
while b<=n:
g.append(b)
b = math.floor(2.25*b)
g = g[::-1]
for i in g:
insertion_sort(a, n, i)
print(len(g))
print(*g, sep=" ")
print(ct)
print(*a, sep="\n")
|
s227435789
|
p03623
|
u011872685
| 2,000 | 262,144 |
Wrong Answer
| 29 | 8,944 | 244 |
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|.
|
data=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t',
'u','v','w','x','y','z',]
s=list(input())
t=sorted(s)
i=0
while (t[i]==data[i] and i<25):
i=i+1
if i<25:
print(data[i])
else:
print('None')
|
s060409598
|
Accepted
| 29 | 9,024 | 88 |
x,a,b=map(int,input().split())
if abs(a-x)<abs(b-x):
print('A')
else:
print('B')
|
s926531852
|
p03369
|
u328755070
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 44 |
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 = list(input())
print(700 + S.count("o"))
|
s960748935
|
Accepted
| 18 | 3,064 | 52 |
S = list(input())
print(700 + S.count("o") * 100)
|
s023701495
|
p02678
|
u700526568
| 2,000 | 1,048,576 |
Wrong Answer
| 2,206 | 33,448 | 607 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
import time
start = time.time()
# ----------------------------------------
n, m = map(int, input().split())
e = [[] for _ in range(n+1)]
for i in range(0, m):
a, b = map(int, input().split())
e[a].append(b)
e[b].append(a)
depth = {1:0}
while len(depth) < n:
for i in range(2, n+1):
if i in depth: continue
for v in e[i]:
if v in list(depth.keys()) and i not in depth:
depth[i] = depth[v] + 1
if time.time() - start > 1.8: break
if len(depth) == n:
print("yes")
sorted_depth = sorted(depth.items(), key=lambda x: x[0])
for i in range(1, n):
print(sorted_depth[i][1])
else:
print("no")
|
s613106524
|
Accepted
| 647 | 34,204 | 498 |
from collections import deque
n, m = map(int, input().split())
e = [[] for _ in range(n+1)]
lndmrk = [-1] * (n + 1)
for i in range(m):
a, b = map(int, input().split())
e[a].append(b)
e[b].append(a)
q = deque()
lndmrk[1] = 0
q.append(1)
while(len(q) != 0):
v = q.popleft()
for i in e[v]:
if lndmrk[i] != -1:
continue
lndmrk[i] = v
q.append(i)
f = 1
for i in range(2, n+1):
if lndmrk[i] == -1:
f = 0
if f:
print("Yes")
for i in range(2, n+1):
print(lndmrk[i])
else:
print("No")
|
s085803805
|
p03730
|
u040298438
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,084 | 145 |
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())
for i in range(a, a * b + 1, a):
if i % b == c:
print("Yes")
break
else:
print("No")
|
s409029696
|
Accepted
| 27 | 9,072 | 145 |
a, b, c = map(int, input().split())
for i in range(a, a * b + 1, a):
if i % b == c:
print("YES")
break
else:
print("NO")
|
s142163200
|
p03609
|
u519939795
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 62 |
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()
for i in range(0,len(s),2):
print(s[i],end="")
|
s791667421
|
Accepted
| 17 | 2,940 | 73 |
x,t=map(int,input().split())
if x-t>=0:
print(x-t)
else:
print(0)
|
s894833476
|
p03251
|
u211160392
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 200 |
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
N,M,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
if max(x) < min(y) and (max(x) < Y and min(y) > X):
print("War")
else:
print("No War")
|
s868718845
|
Accepted
| 18 | 3,060 | 200 |
N,M,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
if max(x) < min(y) and (max(x) < Y and min(y) > X):
print("No War")
else:
print("War")
|
s933775413
|
p03485
|
u792856505
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 85 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
import math
a,b = [int(i) for i in input().split()]
ans = a + b
print(math.ceil(ans))
|
s748567894
|
Accepted
| 19 | 3,188 | 89 |
import math
a,b = [int(i) for i in input().split()]
ans = a + b
print(math.ceil(ans / 2))
|
s266623344
|
p03693
|
u422272120
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 77 |
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 = input().split()
print ("Yes") if int(r+g+b)%4 == 0 else print ("No")
|
s628731041
|
Accepted
| 17 | 2,940 | 77 |
r,g,b = input().split()
print ("YES") if int(r+g+b)%4 == 0 else print ("NO")
|
s238433692
|
p02263
|
u947762778
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,348 | 328 |
An expression is given in a line. Two consequtive symbols (operand or operator) are separated by a space character. You can assume that +, - and * are given as the operator and an operand is a positive integer less than 106
|
stack = input().split()
for i in stack:
if i in ['+', '-', '*']:
b, a = stack.pop(), stack.pop()
if i == '+':
stack.append(b + a)
if i == '-':
stack.append(b - a)
if i == '*':
stack.append(b * a)
else:
stack.append(i)
print(stack.pop())
|
s501210874
|
Accepted
| 20 | 7,720 | 338 |
inList = input().split()
stack = []
for i in inList:
if i in ['+', '-', '*']:
b, a = stack.pop(), stack.pop()
if i == '+':
stack.append(b + a)
if i == '-':
stack.append(a - b)
if i == '*':
stack.append(b * a)
else:
stack.append(int(i))
print(stack.pop())
|
s965153225
|
p04029
|
u434630332
| 2,000 | 262,144 |
Wrong Answer
| 32 | 9,036 | 57 |
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())
answer = (n + 1) * n / 2
print(answer)
|
s150665520
|
Accepted
| 22 | 9,100 | 44 |
n = int(input())
print(int((n + 1) * n / 2))
|
s522226828
|
p03659
|
u260764548
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 24,832 | 401 |
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
# -*- coding:utf-8 -*-
N = int(input())
a = list(map(int,input().split()))
SUM_CARD = sum(a)
print(SUM_CARD)
i=1
diff_sum = 0
sunuke = sum(a[0:i])
diff_sum = abs(sunuke*2-SUM_CARD)
i+=1
while(i<N):
sunuke = sum(a[0:i])
print(sunuke)
diff_sum_new = abs(sunuke*2-SUM_CARD)
print(diff_sum_new)
if diff_sum_new<diff_sum:
diff_sum = diff_sum_new
i += 1
print(diff_sum)
|
s040038992
|
Accepted
| 187 | 24,832 | 347 |
# -*- coding:utf-8 -*-
N = int(input())
a = list(map(int,input().split()))
SUM_CARD = sum(a)
i=0
sum_before = a[i]
diff_sum = abs(sum_before*2-SUM_CARD)
i+=1
while(i<N-1):
sum_before = sum_before + a[i]
diff_sum_new = abs(sum_before*2-SUM_CARD)
if diff_sum_new<diff_sum:
diff_sum = diff_sum_new
i += 1
print(diff_sum)
|
s758415443
|
p03455
|
u741261352
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 83 |
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())
r = 'Odd' if (a * b % 2) == 0 else 'Even'
print(r)
|
s901027654
|
Accepted
| 17 | 2,940 | 83 |
a, b = map(int, input().split())
r = 'Even' if (a * b % 2) == 0 else 'Odd'
print(r)
|
s381305146
|
p03024
|
u136090046
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 55 |
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
s = input()
print("YES" if s.count("o") >= 8 else "NO")
|
s494290582
|
Accepted
| 17 | 2,940 | 68 |
s = input()
print("YES" if s.count("o") + 15-len(s) >= 8 else "NO")
|
s255747997
|
p03475
|
u561870477
| 3,000 | 262,144 |
Wrong Answer
| 72 | 3,060 | 286 |
A railroad running from west to east in Atcoder Kingdom is now complete. There are N stations on the railroad, numbered 1 through N from west to east. Tomorrow, the opening ceremony of the railroad will take place. On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i to Station i+1 in C_i seconds. No other trains will be operated. The first train from Station i to Station i+1 will depart Station i S_i seconds after the ceremony begins. Thereafter, there will be a train that departs Station i every F_i seconds. Here, it is guaranteed that F_i divides S_i. That is, for each Time t satisfying S_i≤t and t％F_i=0, there will be a train that departs Station i t seconds after the ceremony begins and arrives at Station i+1 t+C_i seconds after the ceremony begins, where A％B denotes A modulo B, and there will be no other trains. For each i, find the earliest possible time we can reach Station N if we are at Station i when the ceremony begins, ignoring the time needed to change trains.
|
n = int(input())
l = [list(map(int, input().split())) for _ in range(n-1)]
ans = [0]*n
for i in range(n):
time = 0
for j in range(i, n-1):
if l[j][1] - time > 0:
time += l[j][1] - time
time += time % l[j][2] + l[j][0]
ans[i] += time
print(ans)
|
s570021820
|
Accepted
| 113 | 3,188 | 362 |
n = int(input())
l = [list(map(int, input().split())) for _ in range(n-1)]
ans = [0]*n
for i in range(n):
time = 0
for j in range(i, n-1):
if l[j][1] - time > 0:
time += l[j][1] - time
if time % l[j][2] != 0:
time += l[j][2] - time % l[j][2]
time += l[j][0]
ans[i] += time
for i in ans:
print(i)
|
s425238380
|
p03129
|
u589881693
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 86 |
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
a,b = map(int, input().split(" "))
if (a-1)//b >=2:
print("YES")
else:
print("NO")
|
s324314634
|
Accepted
| 18 | 2,940 | 122 |
a,b = map(int, input().split(" "))
b = b -1
if b==0:
print("YES")
elif (a-1)//b >=2:
print("YES")
else:
print("NO")
|
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