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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s866345341
|
p03672
|
u814986259
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 167 |
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
S=input()
for i in reversed(range(len(S))):
if i % 2 == 1:
moji1 = S[0:i//2 + 1]
moji2 = S[i//2 + 1: i+ 1]
if moji1==moji2:
print(i+1)
break
|
s341367537
|
Accepted
| 17 | 2,940 | 107 |
S = input()
for i in range(len(S)-2, -1, -2):
if S[:i//2] == S[i//2:i]:
print(i)
break
|
s118668180
|
p03729
|
u231875465
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 164 |
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a, b, c = input().split()
def check(a, b, c):
if a[-1] == b[0] and b[-1] == c[0]:
return True
return False
print('Yes' if check(a, b, c) else 'No')
|
s508958089
|
Accepted
| 17 | 3,064 | 164 |
a, b, c = input().split()
def check(a, b, c):
if a[-1] == b[0] and b[-1] == c[0]:
return True
return False
print('YES' if check(a, b, c) else 'NO')
|
s432689004
|
p03568
|
u291988695
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 370 |
We will say that two integer sequences of length N, x_1, x_2, ..., x_N and y_1, y_2, ..., y_N, are _similar_ when |x_i - y_i| \leq 1 holds for all i (1 \leq i \leq N). In particular, any integer sequence is similar to itself. You are given an integer N and an integer sequence of length N, A_1, A_2, ..., A_N. How many integer sequences b_1, b_2, ..., b_N are there such that b_1, b_2, ..., b_N is similar to A and the product of all elements, b_1 b_2 ... b_N, is even?
|
i=int(input())
s=input().split()
j=[]
k=[]
a=0
for h in range(i):
j.append(int(s[h])%2)
for g in range(i):
plus=1
if j[g]==0:
if len(k)>0:
for h in k:
plus=plus*h
plus=plus*(3^(9-g))
a+=plus
k.append(2)
else:
if len(k)>0:
for h in k:
plus=plus*h
plus=2*plus*(3^(9-g))
a+=plus
k.append(1)
print(str(a))
|
s059393469
|
Accepted
| 28 | 9,080 | 188 |
n=int(input())
ls=list(map(int,input().split()))
prev=1
ans=0
for i in range(n):
k=ls[i]
if k%2==1:
ans+=prev*2*(3**(n-i-1))
else:
ans+=prev*3**(n-i-1)
prev*=2
print(ans)
|
s447826183
|
p03140
|
u945460548
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,064 | 572 |
You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective?
|
N = int(input())
A = input()
B = input()
C = input()
cnt = 0
for idx in range(N):
if A[idx] == B[idx] and A[idx] == C[idx] and B[idx] == C[idx]:
print(0)
continue
elif A[idx] == B[idx]:
C = C[:idx] + A[idx] + C[idx+1:]
cnt += 1
elif A[idx] == C[idx]:
B = B[:idx] + A[idx] + B[idx+1:]
cnt += 1
elif B[idx] == C[idx]:
A = A[:idx] + B[idx] + A[idx+1:]
cnt += 1
else:
B = B[:idx] + A[idx] + B[idx+1:]
C = C[:idx] + A[idx] + C[idx+1:]
cnt += 2
#print(A,B,C)
print(cnt)
|
s478579730
|
Accepted
| 18 | 3,064 | 555 |
N = int(input())
A = input()
B = input()
C = input()
cnt = 0
for idx in range(N):
if A[idx] == B[idx] and A[idx] == C[idx] and B[idx] == C[idx]:
continue
elif A[idx] == B[idx]:
C = C[:idx] + A[idx] + C[idx+1:]
cnt += 1
elif A[idx] == C[idx]:
B = B[:idx] + A[idx] + B[idx+1:]
cnt += 1
elif B[idx] == C[idx]:
A = A[:idx] + B[idx] + A[idx+1:]
cnt += 1
else:
B = B[:idx] + A[idx] + B[idx+1:]
C = C[:idx] + A[idx] + C[idx+1:]
cnt += 2
#print(A,B,C)
print(cnt)
|
s896805942
|
p02255
|
u554198876
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,684 | 275 |
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 = [int(i) for i in input().split()]
def insertionSort(A, N):
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(A)
insertionSort(A, N)
|
s927969645
|
Accepted
| 40 | 7,720 | 303 |
N = int(input())
A = [int(i) for i in input().split()]
def insertionSort(A, N):
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([str(i) for i in A]))
insertionSort(A, N)
|
s949886147
|
p03416
|
u157085392
| 2,000 | 262,144 |
Wrong Answer
| 89 | 2,940 | 172 |
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
a,b = [int(x) for x in input().split()]
def ip(n):
s = list(str(n))
r = s
r.reverse()
return r == s
ans = 0
for i in range(a,b+1):
if ip(i): ans += 1
print(ans)
|
s239373470
|
Accepted
| 114 | 3,060 | 175 |
a,b = [int(x) for x in input().split()]
def ip(n):
r = list(str(n))
r.reverse()
return r == list(str(n))
ans = 0
for i in range(a,b+1):
if ip(i): ans += 1
print(ans)
|
s221209637
|
p03477
|
u507116804
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 123 |
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())
m=a+b
n=c+d
if m>n:
print("left")
elif m==n:
print("balanced")
else:
print("left")
|
s501259967
|
Accepted
| 17 | 3,060 | 124 |
a,b,c,d=map(int,input().split())
m=a+b
n=c+d
if m>n:
print("Left")
elif m==n:
print("Balanced")
else:
print("Right")
|
s424440521
|
p03556
|
u146714526
| 2,000 | 262,144 |
Wrong Answer
| 35 | 3,060 | 440 |
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
def main():
N = int(input())
num = 1
for i in range(1,6):
if N >= (10**i)**2 and N < (10**(i+1))**2:
for j in range(10**i, 10**(i+1)):
if j**2 > 10**9 or j**2 > N:
print (num)
return
else:
num = j**2
print (num)
if __name__ == "__main__":
# global stime
# stime = time.clock()
main()
|
s690276980
|
Accepted
| 35 | 2,940 | 439 |
def main():
N = int(input())
num = 1
for i in range(6):
if N >= (10**i)**2 and N < (10**(i+1))**2:
for j in range(10**i, 10**(i+1)):
if j**2 > 10**9 or j**2 > N:
print (num)
return
else:
num = j**2
print (num)
if __name__ == "__main__":
# global stime
# stime = time.clock()
main()
|
s515190176
|
p03598
|
u252828980
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 146 |
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
n = int(input())
k = int(input())
li = list(map(int,input().split()))
num = 0
for i in range(n):
num += max(abs(li[i]),abs(k-li[i]))
print(num)
|
s586457402
|
Accepted
| 17 | 2,940 | 147 |
n = int(input())
k = int(input())
li = list(map(int,input().split()))
num = 0
for i in range(n):
num += min(abs(li[i]),abs(k-li[i]))
print(num*2)
|
s849141159
|
p02613
|
u392441504
| 2,000 | 1,048,576 |
Wrong Answer
| 149 | 16,228 | 269 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
S = []
for i in range(N):
S.append((input()))
ac = str(S.count('AC'))
wa = str(S.count('WA'))
tle = str(S.count('TLE'))
re = str(S.count('TLE'))
ans = 'AC x'+' '+ ac + "\n"+'WA x'+' '+ wa + "\n"+'TLE x' +' '+ tle +"\n" + 'RE x'+' '+re
print(ans)
|
s452357511
|
Accepted
| 148 | 16,108 | 269 |
N = int(input())
S = []
for i in range(N):
S.append((input()))
ac = str(S.count('AC'))
wa = str(S.count('WA'))
tle = str(S.count('TLE'))
re = str(S.count('RE'))
ans = 'AC x'+' '+ ac + "\n"+'WA x'+' '+ wa + "\n"+'TLE x' +' '+ tle +"\n" + 'RE x'+' '+re
print(ans)
|
s925558995
|
p02806
|
u477977638
| 2,525 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 408 |
Niwango created a playlist of N songs. The title and the duration of the i-th song are s_i and t_i seconds, respectively. It is guaranteed that s_1,\ldots,s_N are all distinct. Niwango was doing some work while playing this playlist. (That is, all the songs were played once, in the order they appear in the playlist, without any pause in between.) However, he fell asleep during his work, and he woke up after all the songs were played. According to his record, it turned out that he fell asleep at the very end of the song titled X. Find the duration of time when some song was played while Niwango was asleep.
|
import sys
input = sys.stdin.readline
# mod=10**9+7
# rstrip().decode('utf-8')
# map(int,input().split())
# import numpy as np
def main():
n=int(input())
s=[]
t=[]
for i in range(n):
a,b=input().split()
s.append(a)
t.append(int(b))
x=input()
for i in range(n):
if x==s[i]:
print(sum(t[i+1:]))
exit(0)
if __name__ == "__main__":
main()
|
s076773368
|
Accepted
| 18 | 3,060 | 358 |
import sys
# mod=10**9+7
# rstrip().decode('utf-8')
# map(int,input().split())
# import numpy as np
def main():
n=int(input())
s=[]
t=[]
for i in range(n):
a,b=input().split()
s.append(a)
t.append(int(b))
x=input()
for i in range(n):
if x==s[i]:
print(sum(t[i+1:]))
if __name__ == "__main__":
main()
|
s763965514
|
p03435
|
u881116515
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 208 |
We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j. Determine if he is correct.
|
a = [list(map(int,input().split())) for i in range(3)]
if a[0][0]-a[1][0] == a[0][1]-a[1][1] == a[0][2]-a[1][2] and a[1][0]-a[2][0] == a[1][1]-a[2][1] == a[1][2] == a[2][2]:
print("Yes")
else:
print("No")
|
s181624842
|
Accepted
| 17 | 3,060 | 206 |
a = [list(map(int,input().split())) for i in range(3)]
if a[0][0]-a[1][0] == a[0][1]-a[1][1] == a[0][2]-a[1][2] and a[1][0]-a[2][0] == a[1][1]-a[2][1] == a[1][2]-a[2][2]:
print("Yes")
else:
print("No")
|
s400763706
|
p03386
|
u226912938
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 186 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a, b, k = map(int, input().split())
X = []
for i in range(a, min(a+k, b)):
X.append(i)
for j in range(max(b-k+1, a+1), b+1):
X.append(j)
X_s = set(X)
for k in X_s:
print(k)
|
s354002883
|
Accepted
| 17 | 3,064 | 220 |
a, b, k = map(int, input().split())
X = []
for i in range(a, min(a+k, b+1)):
X.append(i)
for j in range(max(b-k+1, a), b+1):
X.append(j)
X = sorted(X)
X_s = sorted(set(X), key=X.index)
for k in X_s:
print(k)
|
s242487742
|
p03409
|
u444856278
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,064 | 619 |
On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i). A red point and a blue point can form a _friendly pair_ when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point. At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.
|
from bisect import bisect
N = int(input())
red, blue = [], []
for _ in range(N):
red.append(tuple(int(i) for i in input().split()))
for _ in range(N):
blue.append(tuple(int(i) for i in input().split()))
pair = []
red.sort(key=lambda x: x[0])
blue.sort(reverse=True,key=lambda x: x[0])
r = [po[0] for po in red]
for point in blue:
mxred = bisect(r,point[0])
re = red[:mxred]
re.sort(reverse=True, key=lambda x: x[1])
for i in range(len(re)):
if re[i][1] < point[1] and re[i] not in pair:
pair.append(re[i])
print(re[i],point)
break
print(len(pair))
|
s070249329
|
Accepted
| 19 | 3,064 | 423 |
N = int(input())
red, blue = [], []
for _ in range(N):
red.append(tuple(int(i) for i in input().split()))
for _ in range(N):
blue.append(tuple(int(i) for i in input().split()))
ans = 0
red.sort(reverse=True,key=lambda x: x[0])
blue.sort(key=lambda x: x[1])
for bl in blue:
for re in red:
if re[0] < bl[0] and re[1] < bl[1]:
ans += 1
red.remove(re)
break
print(ans)
|
s723518359
|
p02261
|
u613534067
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,608 | 838 |
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
def bubble_sort(c, n):
x = c[:]
for i in range(n):
for k in range(n-1, i, -1):
if x[k][1] < x[k-1][1]:
x[k], x[k-1] = x[k-1], x[k]
return x
def selection_sort(c, n):
x = c[:]
for i in range(n):
mink = i
for k in range(i, n):
if x[k][1] < x[mink][1]:
mink = k
x[i], x[mink] = x[mink], x[i]
return x
n = int(input())
c = []
for i in list(input().split()):
c.append((i[0], int(i[1])))
b = bubble_sort(c, n)
print(" ".join(map(str, [i[0]+str(i[1]) for i in b])))
stable = True
print("Stable")
s = selection_sort(c, n)
print(" ".join(map(str, [i[0]+str(i[1]) for i in s])))
if b != s:
stable = False
if stable:
print("Stable")
else:
print("Not Stable")
|
s584045275
|
Accepted
| 30 | 5,616 | 838 |
def bubble_sort(c, n):
x = c[:]
for i in range(n):
for k in range(n-1, i, -1):
if x[k][1] < x[k-1][1]:
x[k], x[k-1] = x[k-1], x[k]
return x
def selection_sort(c, n):
x = c[:]
for i in range(n):
mink = i
for k in range(i, n):
if x[k][1] < x[mink][1]:
mink = k
x[i], x[mink] = x[mink], x[i]
return x
n = int(input())
c = []
for i in list(input().split()):
c.append((i[0], int(i[1])))
b = bubble_sort(c, n)
print(" ".join(map(str, [i[0]+str(i[1]) for i in b])))
stable = True
print("Stable")
s = selection_sort(c, n)
print(" ".join(map(str, [i[0]+str(i[1]) for i in s])))
if b != s:
stable = False
if stable:
print("Stable")
else:
print("Not stable")
|
s802692344
|
p02600
|
u499061721
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 8,992 | 194 |
M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * From 1000 through 1199: 5-kyu * From 1200 through 1399: 4-kyu * From 1400 through 1599: 3-kyu * From 1600 through 1799: 2-kyu * From 1800 through 1999: 1-kyu What kyu does M-kun have?
|
X = int(input())
up = 599
down = 400
i=8
while(X <= 400 and X >= 1999):
if(X <= up and X >= down):
print(i)
break
else:
up += 200
down += 200
i-=1
|
s228003999
|
Accepted
| 31 | 9,124 | 178 |
X = int(input())
up = 599
down = 400
i=8
while(i>=1):
if(X <= up and X >= down):
print(i)
break
else:
up += 200
down += 200
i -= 1
|
s975288222
|
p03760
|
u598229387
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 66 |
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
o = input()
e = input()
for i,j in zip(o,e):
print(i+j,end='')
|
s475650832
|
Accepted
| 17 | 3,064 | 292 |
o=list(input())
e=list(input())
ans=[]
if len(o)==len(e):
for i in range(len(o)):
ans.append(o[i])
ans.append(e[i])
print(''.join(ans))
else:
for i in range(len(e)):
ans.append(o[i])
ans.append(e[i])
ans.append(o[-1])
print(''.join(ans))
|
s659451240
|
p03141
|
u102960641
| 2,000 | 1,048,576 |
Wrong Answer
| 639 | 33,616 | 245 |
There are N dishes of cuisine placed in front of Takahashi and Aoki. For convenience, we call these dishes Dish 1, Dish 2, ..., Dish N. When Takahashi eats Dish i, he earns A_i points of _happiness_ ; when Aoki eats Dish i, she earns B_i points of happiness. Starting from Takahashi, they alternately choose one dish and eat it, until there is no more dish to eat. Here, both of them choose dishes so that the following value is maximized: "the sum of the happiness he/she will earn in the end" minus "the sum of the happiness the other person will earn in the end". Find the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".
|
n = int(input())
a = []
for i in range(n):
a1,a2 = map(int, input().split())
a.append([a1+a2,a1,a2])
a.sort()
a.reverse()
ans = 0
print(a)
for i in range(n):
if i % 2 == 0:
ans += a[i//2][1]
else:
ans -= a[i//2 + 1][2]
print(ans)
|
s526694071
|
Accepted
| 574 | 23,260 | 226 |
n = int(input())
a = []
for i in range(n):
a1,a2 = map(int, input().split())
a.append([a1+a2,a1,a2])
a.sort()
a.reverse()
ans = 0
for i in range(n):
if i % 2 == 0:
ans += a[i][1]
else:
ans -= a[i][2]
print(ans)
|
s559295110
|
p02612
|
u981884699
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,136 | 41 |
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())
v = (N % 1000)
print(v)
|
s287186826
|
Accepted
| 29 | 9,148 | 71 |
N = int(input())
v = (N % 1000)
if v != 0:
v = 1000 - v
print(v)
|
s116135496
|
p03635
|
u272377260
| 2,000 | 262,144 |
Wrong Answer
| 20 | 2,940 | 45 |
In _K-city_ , there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city?
|
n, m = map(int, input().split())
print(n * m)
|
s422734990
|
Accepted
| 17 | 2,940 | 54 |
n, m = map(int, input().split())
print((n-1) * (m -1))
|
s131822703
|
p03352
|
u222634810
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 2,940 | 89 |
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
X = int(input())
for i in range(32):
if X < i * i:
print((i-1) * (i-1))
|
s729256123
|
Accepted
| 17 | 2,940 | 210 |
import sys
X = int(input())
array = []
for i in range(2, 11):
for j in range(1, 33):
if pow(j, i) > X:
break
else:
array.append(pow(j, i))
print(max(array))
|
s680888870
|
p02615
|
u525589885
| 2,000 | 1,048,576 |
Wrong Answer
| 112 | 31,608 | 183 |
Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?
|
n = int(input())
c = list(map(int, input().split()))
c.sort(reverse = True)
a = 0
for i in range(int(n//2+1)):
a += c[i]
if n%2==0:
print(a*2-c[0])
else:
print(a*2-c[0]-c[n//2])
|
s250872841
|
Accepted
| 123 | 31,584 | 185 |
n = int(input())
c = list(map(int, input().split()))
c.sort(reverse = True)
a = 0
for i in range(int((n+1)//2)):
a += c[i]
if n%2==0:
print(a*2-c[0])
else:
print(a*2-c[0]-c[n//2])
|
s666061030
|
p02854
|
u985408358
| 2,000 | 1,048,576 |
Wrong Answer
| 208 | 26,220 | 319 |
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters. Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut: * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan). * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen. Find the minimum amount of money needed before cutting the bar into two parts with the same length.
|
n = int(input())
a= [int(x) for x in input().split(" ",n-1)]
print(a)
s = 0
for x in range(n):
s += a[x-1]
r = 0
l = 0
x = 0
while r < s/2:
r += a[x]
x += 1
y = 1
while l < s/2:
l += a[n-y]
y += 1
if r == s//2:
print(0)
else:
k = min(r,l)
v = k - s/2
m = v // 0.5
print(int(m))
|
s026476837
|
Accepted
| 185 | 26,220 | 310 |
n = int(input())
a= [int(x) for x in input().split(" ",n-1)]
s = 0
for x in range(n):
s += a[x-1]
r = 0
l = 0
x = 0
while r < s/2:
r += a[x]
x += 1
y = 1
while l < s/2:
l += a[n-y]
y += 1
if r == s//2:
print(0)
else:
k = min(r,l)
v = k - s/2
m = v // 0.5
print(int(m))
|
s229818205
|
p03447
|
u474925961
| 2,000 | 262,144 |
Wrong Answer
| 22 | 3,316 | 134 |
You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?
|
import sys
if sys.platform =='ios':
sys.stdin=open('input_file.txt')
x,a,b=[int(input()) for i in range(3)]
print((x-a)//b)
|
s561546314
|
Accepted
| 17 | 2,940 | 144 |
import sys
if sys.platform =='ios':
sys.stdin=open('input_file.txt')
x,a,b=[int(input()) for i in range(3)]
print((x-a)-((x-a)//b)*b)
|
s358914980
|
p02388
|
u433250944
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,576 | 70 |
Write a program which calculates the cube of a given integer x.
|
x = input()
n = (int(x)**3)
print("n =" ,n)
|
s936890257
|
Accepted
| 20 | 5,576 | 63 |
x = input()
n = (int(x)**3)
print(n)
|
s629537484
|
p02613
|
u630237503
| 2,000 | 1,048,576 |
Wrong Answer
| 142 | 9,208 | 450 |
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.
|
def main():
N = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(0, N):
res = input()
if (res == 'AC'):
ac += 1
if (res == 'WA'):
wa += 1
if (res == 'TLE'):
tle += 1
if (res == 'RE'):
re += 1
print(f"AC x {ac}")
print(f"WA x {wa}")
print(f"TEL x {tle}")
print(f"RE x {re}")
if __name__ == '__main__':
main()
|
s997105153
|
Accepted
| 146 | 9,212 | 450 |
def main():
N = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(0, N):
res = input()
if (res == 'AC'):
ac += 1
if (res == 'WA'):
wa += 1
if (res == 'TLE'):
tle += 1
if (res == 'RE'):
re += 1
print(f"AC x {ac}")
print(f"WA x {wa}")
print(f"TLE x {tle}")
print(f"RE x {re}")
if __name__ == '__main__':
main()
|
s916230863
|
p03130
|
u658915215
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 8,944 | 177 |
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.
|
l = []
for i in range(3):
l += list(map(int, input().split()))
print(l)
for i in range(4):
if l.count(i) == 3:
print('NO')
break
else:
print('YES')
|
s917825209
|
Accepted
| 28 | 9,164 | 172 |
l = []
for _ in range(3):
l += list(map(int, input().split()))
for i in range(4):
if l.count(i + 1) == 3:
print('NO')
break
else:
print('YES')
|
s956346016
|
p03637
|
u466105944
| 2,000 | 262,144 |
Wrong Answer
| 74 | 14,252 | 348 |
We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective.
|
n = int(input())
a_list = list(map(int,input().split()))
zero = 0
one = 0
two = 0
for a in a_list:
remain = a%2
if remain == 0:
if a/2 >= 2:
two += 1
elif a/2 == 1:
one += 1
else:
zero += 1
if one:
zero += 1
if zero <= two+1 or zero <= two:
print('yes')
else:
print('no')
|
s502222172
|
Accepted
| 53 | 15,020 | 217 |
n = int(input())
a_list = list(map(int,input().split()))
divs = [a%4 for a in a_list]
mp_of_four = divs.count(0)
mp_of_two = divs.count(2)
if mp_of_four+mp_of_two//2 >= n//2:
print('Yes')
else:
print('No')
|
s684686837
|
p04030
|
u064408584
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 109 |
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
s=input()
a=''
for i in s:
if i=='B':
if len(a)!=0:a=a[:-1]
else:a+=i
print(a,i)
print(a)
|
s085496371
|
Accepted
| 17 | 2,940 | 90 |
a=''
for i in input():
if i=='B':
if len(a)!=0:a=a[:-1]
else:a+=i
print(a)
|
s027460958
|
p03142
|
u368796742
| 2,000 | 1,048,576 |
Wrong Answer
| 359 | 28,352 | 510 |
There is a rooted tree (see Notes) with N vertices numbered 1 to N. Each of the vertices, except the root, has a directed edge coming from its parent. Note that the root may not be Vertex 1. Takahashi has added M new directed edges to this graph. Each of these M edges, u \rightarrow v, extends from some vertex u to its descendant v. You are given the directed graph with N vertices and N-1+M edges after Takahashi added edges. More specifically, you are given N-1+M pairs of integers, (A_1, B_1), ..., (A_{N-1+M}, B_{N-1+M}), which represent that the i-th edge extends from Vertex A_i to Vertex B_i. Restore the original rooted tree.
|
from collections import deque
n,m = map(int,input().split())
e = [[] for i in range(n)]
d = [1]*n
s = [0]*n
for i in range(n+m-1):
a,b = map(int,input().split())
e[a-1].append(b-1)
s[b-1] += 1
if 0 in s:
ans = [-1]*n
p = s.index(0)
ans[p] = 0
q = deque([])
q.append(p)
while q:
now = q.popleft()
for nex in e[now]:
if ans[nex] == -1:
ans[nex] = now+1
q.append(nex)
for i in ans:
print(i)
exit()
|
s401514495
|
Accepted
| 487 | 46,940 | 701 |
from collections import deque
n,m = map(int,input().split())
e = [[] for i in range(n)]
d = [1]*n
s = [0]*n
t = [[] for i in range(n)]
for i in range(n+m-1):
a,b = map(int,input().split())
a -= 1
b -= 1
e[a].append(b)
t[b].append(a)
s[b] += 1
ans = [-1]*n
p = s.index(0)
ans[p] = 0
q = deque([])
q.append(p)
s = set()
while q:
now = q.popleft()
s.add(now)
for nex in e[now]:
if ans[nex] == -1:
check = True
for j in t[nex]:
if j not in s:
check = False
break
if check:
ans[nex] = now+1
q.append(nex)
for i in ans:
print(i)
|
s924510497
|
p02388
|
u140168530
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,392 | 26 |
Write a program which calculates the cube of a given integer x.
|
def f(x):
return x**3
|
s544098268
|
Accepted
| 40 | 7,688 | 132 |
def cube(x):
return x**3
def main():
x = int(input())
ans = cube(x)
print(ans)
if __name__=='__main__':
main()
|
s231430661
|
p00016
|
u436634575
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,848 | 218 |
When a boy was cleaning up after his grand father passing, he found an old paper: In addition, other side of the paper says that "go ahead a number of steps equivalent to the first integer, and turn clockwise by degrees equivalent to the second integer". His grand mother says that Sanbonmatsu was standing at the center of town. However, now buildings are crammed side by side and people can not walk along exactly what the paper says in. Your task is to write a program which hunts for the treature on the paper. For simplicity, 1 step is equivalent to 1 meter. Input consists of several pairs of two integers d (the first integer) and t (the second integer) separated by a comma. Input ends with "0, 0". Your program should print the coordinate (x, y) of the end point. There is the treature where x meters to the east and y meters to the north from the center of town. You can assume that d ≤ 100 and -180 ≤ t ≤ 180\.
|
from math import radians
from cmath import rect
z = 0
deg = 90
while True:
r, d = map(int, input().split(','))
if r == d == 0: break
z += rect(r, radians(deg))
deg += d
print(-int(z.real), int(z.imag))
|
s725657823
|
Accepted
| 30 | 6,864 | 225 |
from math import radians
from cmath import rect
z = 0
deg = 90
while True:
r, d = map(float, input().split(','))
if r == d == 0: break
z += rect(r, radians(deg))
deg -= d
print(int(z.real))
print(int(z.imag))
|
s777794310
|
p03910
|
u153094838
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,572 | 335 |
The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems. The score allocated to the i-th (1≦i≦N) problem is i points. Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve. As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him. Determine the set of problems that should be solved.
|
# -*- coding: utf-8 -*-
"""
Created on Fri Jun 12 20:20:11 2020
@author: NEC-PCuser
"""
N = int(input())
i = 1
li = []
while (N >= (i * (i + 1)) // 2):
li.append(i)
i += 1
value = (i - 1) * i // 2
print(value)
index = len(li) - 1
for i in range(value, N):
li[index] += 1
index -= 1
for i in li:
print(i)
|
s773721143
|
Accepted
| 22 | 3,572 | 322 |
# -*- coding: utf-8 -*-
"""
Created on Fri Jun 12 20:20:11 2020
@author: NEC-PCuser
"""
N = int(input())
i = 1
li = []
while (N >= (i * (i + 1)) // 2):
li.append(i)
i += 1
value = (i - 1) * i // 2
index = len(li) - 1
for i in range(value, N):
li[index] += 1
index -= 1
for i in li:
print(i)
|
s104696765
|
p03729
|
u181709987
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 137 |
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
A,B,C = input().split()
X = False
if A[-1]==B[0]:
if B[-1] == C[0]:
X = True
if X:
print('Yes')
else:
print('No')
|
s134995447
|
Accepted
| 17 | 2,940 | 137 |
A,B,C = input().split()
X = False
if A[-1]==B[0]:
if B[-1] == C[0]:
X = True
if X:
print('YES')
else:
print('NO')
|
s877209381
|
p03659
|
u163320134
| 2,000 | 262,144 |
Wrong Answer
| 255 | 24,824 | 203 |
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
n=int(input())
arr=list(map(int,input().split()))
for i in range(n-1):
arr[i+1]+=arr[i]
print(arr)
ans=10**18
for i in range(n-1):
x=arr[i]
y=arr[n-1]-x
tmp=abs(x-y)
ans=min(ans,tmp)
print(ans)
|
s108587995
|
Accepted
| 236 | 24,824 | 193 |
n=int(input())
arr=list(map(int,input().split()))
for i in range(n-1):
arr[i+1]+=arr[i]
ans=10**18
for i in range(n-1):
x=arr[i]
y=arr[n-1]-x
tmp=abs(x-y)
ans=min(ans,tmp)
print(ans)
|
s741854044
|
p04029
|
u050034744
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 68 |
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())
sum=0
for i in range(n):
sum+=(n+1)**2
print(sum)
|
s311645905
|
Accepted
| 17 | 2,940 | 65 |
n=int(input())
sum=0
for i in range(n):
sum+=(i+1)
print(sum)
|
s248028126
|
p02843
|
u088553842
| 2,000 | 1,048,576 |
Wrong Answer
| 38 | 10,252 | 242 |
AtCoder Mart sells 1000000 of each of the six items below: * Riceballs, priced at 100 yen (the currency of Japan) each * Sandwiches, priced at 101 yen each * Cookies, priced at 102 yen each * Cakes, priced at 103 yen each * Candies, priced at 104 yen each * Computers, priced at 105 yen each Takahashi wants to buy some of them that cost exactly X yen in total. Determine whether this is possible. (Ignore consumption tax.)
|
n = int(input())
if n <= 99:
print('No')
exit()
dp = [True] + [False] * n
for i in range(100, n + 1):
if dp[i - 100] or dp[i - 101] or dp[i - 102] or dp[i - 103] or dp[i - 104] or dp[i - 105]:
dp[i] = True
print(['No','Yes'][dp[n]])
|
s848906009
|
Accepted
| 44 | 10,112 | 273 |
n = int(input())
if n <= 99:
print(0)
exit()
if 100 <= n <= 105:
print(1)
exit()
dp = [True] + [False] * n
for i in range(100, n + 1):
if dp[i - 100] or dp[i - 101] or dp[i - 102] or dp[i - 103] or dp[i - 104] or dp[i - 105]:
dp[i] = True
print([0, 1][dp[n]])
|
s558453565
|
p02401
|
u276050131
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,600 | 273 |
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while True:
a,op,b = input().split()
a = int(a)
b = int(b)
if (op == "+"):
print(a + b)
elif(op == "-"):
print(a + b)
elif(op == "/"):
print(a // b)
elif(op == "*"):
print(a + b)
elif(op == "?"):
break
|
s295592649
|
Accepted
| 20 | 7,696 | 273 |
while True:
a,op,b = input().split()
a = int(a)
b = int(b)
if (op == "+"):
print(a + b)
elif(op == "-"):
print(a - b)
elif(op == "/"):
print(a // b)
elif(op == "*"):
print(a * b)
elif(op == "?"):
break
|
s308488582
|
p03478
|
u104442591
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 259 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n, a, b = map(int, input().split())
def divide(num):
rest = num
sum = 0
while rest > 0:
num /= 10
rest %= 10
sum += rest
return sum
for i in range(n):
temp = divide(i+1)
ans = 0
if temp>= a and temp <= b:
ans += temp
print(ans)
|
s802264458
|
Accepted
| 26 | 3,060 | 234 |
n, a, b = map(int, input().split())
def divide(num):
sum = 0
while num > 0:
sum += num % 10
num //= 10
return sum
ans = 0
for i in range(n):
temp = divide(i+1)
if temp>= a and temp <= b:
ans += (i+1)
print(ans)
|
s484706927
|
p03611
|
u410717334
| 2,000 | 262,144 |
Wrong Answer
| 164 | 24,176 | 232 |
You are given an integer sequence of length N, a_1,a_2,...,a_N. For each 1≤i≤N, you have three choices: add 1 to a_i, subtract 1 from a_i or do nothing. After these operations, you select an integer X and count the number of i such that a_i=X. Maximize this count by making optimal choices.
|
from collections import defaultdict
N=int(input())
a=list(map(int,input().split()))
d=defaultdict(int)
for item in a:
d[item-1]+=1
d[item]+=1
d[item+1]+=1
b=sorted(d.items(),key=lambda x:x[1],reverse=True)
print(b[0][0])
|
s185168476
|
Accepted
| 163 | 24,176 | 232 |
from collections import defaultdict
N=int(input())
a=list(map(int,input().split()))
d=defaultdict(int)
for item in a:
d[item-1]+=1
d[item]+=1
d[item+1]+=1
b=sorted(d.items(),key=lambda x:x[1],reverse=True)
print(b[0][1])
|
s141382910
|
p02678
|
u111473084
| 2,000 | 1,048,576 |
Wrong Answer
| 1,288 | 33,744 | 412 |
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.
|
n, m = map(int, input().split())
graph = [[] for _ in range(n+1)]
for i in range(m):
a, b = map(int, input().split())
graph[a].append(b)
graph[b].append(a)
# bfs
milestone = [0] * (n+1)
q = []
q.append(1)
while len(q) != 0:
for i in graph[q[0]]:
if milestone[i] == 0:
milestone[i] = q[0]
q.append(i)
q.pop(0)
for i in range(2, n+1):
print(milestone[i])
|
s623127703
|
Accepted
| 1,303 | 33,808 | 425 |
n, m = map(int, input().split())
graph = [[] for _ in range(n+1)]
for i in range(m):
a, b = map(int, input().split())
graph[a].append(b)
graph[b].append(a)
# bfs
milestone = [0] * (n+1)
q = []
q.append(1)
while len(q) != 0:
for i in graph[q[0]]:
if milestone[i] == 0:
milestone[i] = q[0]
q.append(i)
q.pop(0)
print("Yes")
for i in range(2, n+1):
print(milestone[i])
|
s339363453
|
p03407
|
u977642052
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 202 |
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
def main(A: int, B: int, C: int):
if A + B <= C:
return 'Yes'
else:
return 'No'
if __name__ == "__main__":
A, B, C = map(int, input().split(' '))
print(main(A, B, C))
|
s610916886
|
Accepted
| 17 | 2,940 | 204 |
def main(A: int, B: int, C: int):
if (A + B) >= C:
return 'Yes'
else:
return 'No'
if __name__ == "__main__":
A, B, C = map(int, input().split(' '))
print(main(A, B, C))
|
s552933510
|
p03796
|
u823458368
| 2,000 | 262,144 |
Wrong Answer
| 230 | 3,980 | 58 |
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.
|
import math
n = int(input())
math.factorial(n)%(10**9 + 7)
|
s777959591
|
Accepted
| 42 | 2,940 | 87 |
ans=1
n=int(input())
for i in range(1,n+1):
ans*=i
ans=ans%(10**9+7)
print(ans)
|
s412407708
|
p02692
|
u367548378
| 2,000 | 1,048,576 |
Wrong Answer
| 287 | 104,476 | 796 |
There is a game that involves three variables, denoted A, B, and C. As the game progresses, there will be N events where you are asked to make a choice. Each of these choices is represented by a string s_i. If s_i is `AB`, you must add 1 to A or B then subtract 1 from the other; if s_i is `AC`, you must add 1 to A or C then subtract 1 from the other; if s_i is `BC`, you must add 1 to B or C then subtract 1 from the other. After each choice, none of A, B, and C should be negative. Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.
|
import sys
sys.setrecursionlimit(1000000)
def F(S,curr,D,res):
if len(S)==curr:
print("Yes")
for c in res:
print(c)
#print("".join(res))
return True
s=S[curr]
a=False
b=False
if D[s[0]]>=1:
D[s[0]]-=1
D[s[1]]+=1
res.append(s[0])
a=F(S,curr+1,D,res)
res.pop(-1)
D[s[0]]+=1
D[s[1]]-=1
else:
if D[s[1]]>=1:
D[s[1]]-=1
D[s[0]]+=1
res.append(s[1])
b=F(S,curr+1,D,res)
res.pop()
D[s[1]]+=1
D[s[0]]-=1
return a or b
N,A,B,C=map(int,input().strip().split())
D={"A":A,"B":B,"C":C}
S=[]
res=[]
for i in range(N):
S.append(input())
if not (F(S,0,D,res)):
print("No")
|
s327443991
|
Accepted
| 304 | 104,388 | 867 |
import sys
sys.setrecursionlimit(1000000)
def F(S,curr,D,res):
if D["A"]<0 or D["B"]<0 or D["C"]<0:
return False
if len(S)==curr:
print("Yes")
for c in res:
print(c)
#print("".join(res))
return True
s=S[curr]
a=False
b=False
if D[s[0]]>=1:
D[s[0]]-=1
D[s[1]]+=1
res.append(s[1])
a=F(S,curr+1,D,res)
res.pop(-1)
D[s[0]]+=1
D[s[1]]-=1
if a:
return a
if D[s[1]]>=1:
D[s[1]]-=1
D[s[0]]+=1
res.append(s[0])
b=F(S,curr+1,D,res)
res.pop()
D[s[1]]+=1
D[s[0]]-=1
return a or b
N,A,B,C=map(int,input().strip().split())
D={"A":A,"B":B,"C":C}
S=[]
res=[]
for i in range(N):
S.append(input())
if not (F(S,0,D,res)):
print("No")
|
s270763842
|
p03547
|
u073139376
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`. When X and Y are seen as hexadecimal numbers, which is larger?
|
X, Y = input().split()
if X > Y:
print('<')
elif X == Y:
print('=')
else:
print('>')
|
s531224275
|
Accepted
| 17 | 3,064 | 91 |
X, Y = input().split()
if X > Y:
print('>')
elif X == Y:
print('=')
else:
print('<')
|
s154533348
|
p02390
|
u193025715
| 1,000 | 131,072 |
Wrong Answer
| 40 | 6,724 | 97 |
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
n = int(input())
h = n % 3600
n -= h * 3600
m = n % 60
n -= m
print(str(h)+':'+str(m)+':'+str(n))
|
s229799635
|
Accepted
| 30 | 6,732 | 122 |
n = int(input())
h = int(n / 3600)
m = int((n-h*3600) / 60)
s = n - h * 3600 - m * 60
print(':'.join(map(str, [h,m,s])))
|
s646139288
|
p03457
|
u955907183
| 2,000 | 262,144 |
Wrong Answer
| 444 | 27,324 | 497 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
c = input()
data = []
for s in range(0,int(c)):
data.append(list(map(int, input().split())) )
nowtime = 0
nowx = 0
nowy = 0
nexttime = 0
nextx = 0
nexty = 0
costtime = 0
diffx = 0
diffy = 0
for s in range(0, len(data)):
nexttime = data[s][0]
nextx = data[s][1]
nexty = data[s][2]
costtime = nexttime - nowtime
diffx = nextx - nowx
diffy = nexty - nowy
if ( (costtime < diffx + diffy) or (((costtime - (diffx + diffy)) % 2) == 1 ) ):
print("NO")
exit()
print("YES")
|
s914548940
|
Accepted
| 417 | 27,400 | 354 |
c = input()
data = []
for s in range(0,int(c)):
data.append(list(map(int, input().split())) )
nexttime = 0
nextx = 0
nexty = 0
for s in range(0, len(data)):
nexttime = data[s][0]
nextx = data[s][1]
nexty = data[s][2]
if ( (nexttime < nextx+nexty) or ((((nextx + nexty) ) % 2) != (nexttime % 2) ) ):
print("No")
exit()
print("Yes")
|
s799794172
|
p03636
|
u736479342
| 2,000 | 262,144 |
Wrong Answer
| 25 | 9,028 | 69 |
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
s = list(input())
print(s)
l = len(s)-2
print(s[0] + str(l) + s[-1])
|
s146857943
|
Accepted
| 25 | 9,032 | 60 |
s = list(input())
l = len(s)-2
print(s[0] + str(l) + s[-1])
|
s368545813
|
p03400
|
u474270503
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 189 |
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
N=int(input())
D, X=map(int,input().split())
A=[]
for i in range(N):
A.append(int(input()))
ans=0
for i in range(N):
if A[i]<=D:
ans+=D//A[i]+1
print(ans)
print(ans+X)
|
s787152906
|
Accepted
| 18 | 3,060 | 203 |
N=int(input())
D, X=map(int,input().split())
A=[]
for i in range(N):
A.append(int(input()))
ans=0
for i in range(N):
if A[i]<=D:
ans+=(D-1)//A[i]+1
else:
ans+=1
print(ans+X)
|
s833023235
|
p03943
|
u048956777
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 91 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b ,c = map(int, input().split())
if(a+b+c % 3 == 0):
print('Yes')
else:
print('No')
|
s480647777
|
Accepted
| 17 | 2,940 | 108 |
a, b ,c = map(int, input().split())
if (a+b==c) or (b+c==a) or (a+c==b):
print('Yes')
else:
print('No')
|
s328866750
|
p03971
|
u077337864
| 2,000 | 262,144 |
Wrong Answer
| 2,102 | 4,528 | 354 |
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
N, A, B = map(int, input().split())
S = input()
passed = []
for i in range(N):
if S[i] == 'c':
print('No')
elif S[i] == 'a':
if len(passed) <= A+B:
print('Yes')
passed.append('a')
else: print('No')
else:
if len(passed) <= A+B and passed.count('b') <= B:
print('Yes')
passed.append('b')
else: print('No')
|
s656707362
|
Accepted
| 107 | 4,016 | 343 |
n, a, b = map(int, input().split())
s = input().strip()
sumy = 0
sumf = 0
for _s in s:
if _s == 'c':
print('No')
elif _s == 'a':
if sumy < a + b:
print('Yes')
sumy += 1
else:
print('No')
else:
if sumy < a + b and sumf < b:
print('Yes')
sumy += 1
sumf += 1
else:
print('No')
|
s215015116
|
p03814
|
u047197186
| 2,000 | 262,144 |
Wrong Answer
| 38 | 9,360 | 131 |
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
s = input()
end = 0
for i in range(len(s) - 1, -1, -1):
if s[i] == 'Z':
end = i + 1
break
print(s[s.index('A'):end])
|
s829874876
|
Accepted
| 36 | 9,256 | 136 |
s = input()
end = 0
for i in range(len(s) - 1, -1, -1):
if s[i] == 'Z':
end = i + 1
break
print(len(s[s.index('A'):end]))
|
s824004989
|
p02742
|
u068944955
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 173 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
H, W = map(int, input().split())
if H == 1 or W == 1:
print(1)
exit()
if H % 2 == 0 or W % 2 == 0:
print(H * W / 2)
else:
print(H * (W // 2) + H // 2 + 1)
|
s090191878
|
Accepted
| 21 | 3,064 | 174 |
H, W = map(int, input().split())
if H == 1 or W == 1:
print(1)
exit()
if H % 2 == 0 or W % 2 == 0:
print(H * W // 2)
else:
print(H * (W // 2) + H // 2 + 1)
|
s051563454
|
p03963
|
u576434377
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 61 |
There are N balls placed in a row. AtCoDeer the deer is painting each of these in one of the K colors of his paint cans. For aesthetic reasons, any two adjacent balls must be painted in different colors. Find the number of the possible ways to paint the balls.
|
N,K = map(int,input().split())
print(K * ((K - 1) ** N - 1))
|
s540665628
|
Accepted
| 17 | 2,940 | 63 |
N,K = map(int,input().split())
print(K * ((K - 1) ** (N - 1)))
|
s660051126
|
p03495
|
u813450984
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 26,148 | 390 |
Takahashi has N balls. Initially, an integer A_i is written on the i-th ball. He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls. Find the minimum number of balls that Takahashi needs to rewrite the integers on them.
|
n, k = map(int, input().split())
nums = list(map(int, input().split()))
def judge(n, k, nums):
original = nums
only = set(nums)
diff = len(only) - k
count = []
ans = 0
if len(only) <= k:
return 0
for i in only:
count.append([original.count(i), i])
count.sort()
print(count)
for i in range(diff):
ans += count[i][0]
return ans
print(judge(n, k, nums))
|
s938724240
|
Accepted
| 226 | 25,644 | 360 |
n, k = map(int, input().split())
l = sorted(list(map(int, input().split())))
counts = []
now = l[0]
count = 1
for i in range(1, n):
if l[i] == now:
count += 1
else:
counts.append(count)
now = l[i]
count = 1
if i == n-1:
counts.append(count)
if len(counts) <= k:
print(0)
else:
counts = sorted(counts)
print(sum(counts[:-k]))
|
s571254286
|
p03644
|
u994988729
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 155 |
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
def two(n):
count=0
while n%2==0:
count+=1
n//=2
return count
n=int(input())
ans=0
for i in range(1, n+1):
ans=max(ans, two(i))
print(ans)
|
s699099929
|
Accepted
| 28 | 9,152 | 88 |
N = int(input())
x = 1
while True:
if x * 2 > N:
break
x *= 2
print(x)
|
s916065193
|
p03861
|
u884601206
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 88 |
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a,b,x=map(int,input().split())
if a!=0:
print(b%x - (a-1)%x)
else:
print(b%x+1)
|
s803422912
|
Accepted
| 17 | 2,940 | 95 |
a,b,x=map(int,input().split())
if a!=0:
print((b//x) - (a-1)//x)
else:
print((b//x)+1)
|
s600839913
|
p03998
|
u770077083
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 268 |
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
cards = [list(input().replace("a", "0").replace("b","1").replace("c","2")) for _ in range(3)]
next = 0
while len(cards[0]) * len(cards[1]) * len(cards[2]) != 0:
next = int(cards[next].pop(0))
print("A" if len(cards[0]) == 0 else ("B" if len(cards[1]) == 0 else "C"))
|
s299614856
|
Accepted
| 17 | 3,060 | 247 |
cards = [list(input().replace("a", "0").replace("b","1").replace("c","2")) for _ in range(3)]
next = 0
while True:
if len(cards[next]) == 0:
print("A" if next == 0 else ("B" if next == 1 else "C"))
exit()
next = int(cards[next].pop(0))
|
s203433256
|
p03455
|
u262481526
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 86 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
if a * b % 2 == 0:
print('Yes')
else:
print('No')
|
s586640354
|
Accepted
| 17 | 2,940 | 88 |
a, b = map(int, input().split())
if a * b % 2 == 0:
print('Even')
else:
print('Odd')
|
s287924777
|
p02865
|
u586577600
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 28 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n = int(input())
print(n//2)
|
s113196881
|
Accepted
| 17 | 2,940 | 67 |
n = int(input())
if n%2==0:
print(n//2-1)
else:
print(n//2)
|
s668123064
|
p02612
|
u697658632
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 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.
|
print((10000 - int(input())) // 1000)
|
s001135616
|
Accepted
| 27 | 9,140 | 36 |
print((10000 - int(input())) % 1000)
|
s627301498
|
p03408
|
u644907318
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 251 |
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
N = int(input())
S = [input().strip() for _ in range(N)]
M = int(input())
T = [input().strip() for _ in range(M)]
cntmax = 0
for s in S:
cnt = 0
for t in T:
if t==s:
cnt += 1
if cnt==0:
cntmax += 1
print(cntmax)
|
s326072864
|
Accepted
| 17 | 3,064 | 254 |
N = int(input())
S = [input().strip() for _ in range(N)]
M = int(input())
T = [input().strip() for _ in range(M)]
cntmax = 0
for s in S:
cnt = S.count(s)
for t in T:
if t==s:
cnt -= 1
cntmax = max(cntmax,cnt)
print(cntmax)
|
s565453527
|
p03574
|
u466105944
| 2,000 | 262,144 |
Wrong Answer
| 26 | 3,188 | 684 |
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
h,w = map(int,input().split())
board = [input() for _ in range(h)]
dx = [-1,-1,-1, 0,0, 1,1,1]
dy = [-1, 0, 1,-1,1,-1,0,1]
ans_board = [[0 for _ in range(w)] for _ in range(h)]
for x in range(h):
cnt = 0
for y in range(w):
if board[x][y] == '#':
ans_board[x][y] = '#'
else:
for d in range(len(dx)):
n_x = x+dx[d]
n_y = y+dy[d]
if 0 < n_x <= h or 0 < n_y <= w:
continue
elif board[n_x][n_y] == '#':
cnt += 1
ans_board[x][y] = str(cnt)
for ans in ans_board:
print('' .join(ans))
|
s369783371
|
Accepted
| 31 | 3,188 | 630 |
h,w = map(int,input().split())
board = [input() for _ in range(h)]
dx = [-1,-1,-1, 0,0, 1,1,1]
dy = [-1, 0, 1,-1,1,-1,0,1]
ans_board = [[0 for _ in range(w)] for _ in range(h)]
for x in range(h):
for y in range(w):
cnt = 0
if board[x][y] == '#':
ans_board[x][y] = '#'
else:
for d in range(len(dx)):
n_x = x+dx[d]
n_y = y+dy[d]
if 0 <= n_x < h and 0 <= n_y < w:
if board[n_x][n_y] == '#':
cnt += 1
ans_board[x][y] = str(cnt)
for ans in ans_board:
print('' .join(ans))
|
s857210231
|
p03860
|
u241190800
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 72 |
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.
|
data = [i for i in input().split(" ")]
print('A#{}C'.format(data[1][0]))
|
s121842541
|
Accepted
| 18 | 2,940 | 68 |
data = [i for i in input().split()]
print('A{}C'.format(data[1][0]))
|
s761618603
|
p03448
|
u500207661
| 2,000 | 262,144 |
Wrong Answer
| 47 | 3,060 | 333 |
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().strip())
b = int(input().strip())
c = int(input().strip())
x = int(input().strip())
count = 0
for a_n in range(a):
for b_n in range(b):
for c_n in range(c):
if x - (a_n * 500) - (b_n * 100) - (c_n * 50) is 0:
count += 1
else:
continue
print(count)
|
s394615788
|
Accepted
| 51 | 2,940 | 424 |
a = int(input().strip())
b = int(input().strip())
c = int(input().strip())
x = int(input().strip())
if a * 500 + b *100 + c * 50 < x:
print(0)
else:
count = 0
for a_n in range(a+1):
for b_n in range(b+1):
for c_n in range(c+1):
if (a_n * 500) + (b_n * 100) + (c_n * 50) == x:
count += 1
else:
continue
print(count)
|
s918611376
|
p03720
|
u013408661
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 201 |
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
n,m=map(int,input().split())
city=[[] for i in range(n+1)]
for i in range(m):
a,b=map(int,input().split())
city[a].append(b)
city[b].append(a)
for i in range(1,n+1):
print(len(set(city[i]))-1)
|
s686245637
|
Accepted
| 17 | 3,060 | 193 |
n,m=map(int,input().split())
city=[[] for i in range(n+1)]
for i in range(m):
a,b=map(int,input().split())
city[a].append(b)
city[b].append(a)
for i in range(1,n+1):
print(len(city[i]))
|
s328048387
|
p03796
|
u296150111
| 2,000 | 262,144 |
Wrong Answer
| 42 | 3,064 | 75 |
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())
ans=1
for i in range(n):
ans*=i
ans=ans%10**9+7
print(ans)
|
s437640870
|
Accepted
| 41 | 2,940 | 81 |
n=int(input())
ans=1
for i in range(1,n+1):
ans*=i
ans=ans%(10**9+7)
print(ans)
|
s863069925
|
p03720
|
u827202523
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 234 |
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
n,m = list(map(int, input().split()))
cities = [0 for i in range(n)]
for i in range(m):
bridges = list(map(int, input().split()))
cities = [i+1 if i+1 in bridges else i for i in cities]
print("\n".join(list(map(str, cities))))
|
s206919010
|
Accepted
| 17 | 3,060 | 247 |
n,m = list(map(int, input().split()))
cities = [0 for i in range(n)]
for i in range(m):
bridges = list(map(int, input().split()))
cities = [x+1 if i+1 in bridges else x for i,x in enumerate(cities)]
print("\n".join(list(map(str, cities))))
|
s438589915
|
p03359
|
u287880059
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 55 |
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a,b = map(int,input().split())
if a>b:
a -1
else:
a
|
s650427997
|
Accepted
| 17 | 2,940 | 56 |
a, b = map(int,input().split())
print(a-1 if a>b else a)
|
s583064697
|
p03433
|
u728307690
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,124 | 95 |
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())
r = n % 500
if a >= r:
print("YES")
else:
print("NO")
|
s110316273
|
Accepted
| 30 | 9,168 | 94 |
n = int(input())
a = int(input())
r = n % 500
if a < r:
print('No')
else:
print('Yes')
|
s914474042
|
p02659
|
u921352252
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,024 | 50 |
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
x,y=map(float,input().split())
print(round(x*y))
|
s172118393
|
Accepted
| 22 | 9,032 | 80 |
x,y=map(float,input().split())
x=int(x)
y1=int(round(y*100))
print((x*y1//100))
|
s437256967
|
p03555
|
u502731482
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 69 |
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.
|
s = input()
t = input()
s = s[::-1]
print("Yes" if s == t else "No")
|
s632067506
|
Accepted
| 17 | 2,940 | 69 |
s = input()
t = input()
s = s[::-1]
print("YES" if s == t else "NO")
|
s965561174
|
p02394
|
u791170614
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,556 | 237 |
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
W, H, x, y, r = map(int, input().split())
def YesOrNo(W, H , x, y, r):
Xr = x + r
Xl = x - r
Yu = y + r
Yd = y - r
if (W >= Xr and H >= Yu and 0 >= Xl and 0 >= Yd):
return "Yes"
else:
return "No"
print(YesOrNo(W, H, x, y, r))
|
s081168635
|
Accepted
| 20 | 7,676 | 237 |
W, H, x, y, r = map(int, input().split())
def YesOrNo(W, H , x, y, r):
Xr = x + r
Xl = x - r
Yu = y + r
Yd = y - r
if (W >= Xr and H >= Yu and 0 <= Xl and 0 <= Yd):
return "Yes"
else:
return "No"
print(YesOrNo(W, H, x, y, r))
|
s882041417
|
p03545
|
u689322583
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,316 | 806 |
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
#coding: utf-8
S = str(input())
A = int(S[0])
B = int(S[1])
C = int(S[2])
D = int(S[3])
shiki = ''
for i in range(2):
for j in range(2):
for k in range(2):
B = B * (-1) ** i
C = C * (-1) ** j
D = D * (-1) ** k
if(A + B + C + D == 7):
shiki += str(A)
if(B<0):
shiki += '-' + str(B)
else:
shiki += '+' + str(B)
if(C<0):
shiki += '-' + str(C)
else:
shiki += '+' + str(C)
if(D<0):
shiki += '-' + str(D)
else:
shiki += '+' + str(D)
print(shiki,end='')
break
break
break
|
s095828117
|
Accepted
| 17 | 3,064 | 898 |
#coding: utf-8
S = str(input())
A = int(S[0])
B = int(S[1])
C = int(S[2])
D = int(S[3])
for i in range(2):
for j in range(2):
for k in range(2):
A = int(S[0])
B = int(S[1])
C = int(S[2])
D = int(S[3])
B *= (-1) ** i
C *= (-1) ** j
D *= (-1) ** k
if(A + B + C + D == 7):
shiki = ''
shiki += str(A)
if(B<0):
shiki += str(B)
else:
shiki += '+' + str(B)
if(C<0):
shiki += str(C)
else:
shiki += '+' + str(C)
if(D<0):
shiki += str(D)
else:
shiki += '+' + str(D)
shiki += '=7'
print(shiki)
exit()
|
s755150058
|
p02260
|
u370086573
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,656 | 354 |
Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠ mini.
|
def selectionSort(n, A):
cnt =0
for i in range(n):
minj = i
for j in range(i,n):
if A[minj] > A[j]:
minj = j
A[i],A[minj] = A[minj],A[i]
cnt += 1
print(*A)
print(cnt)
if __name__ == '__main__':
n = int(input())
A = list(map(int, input().split()))
selectionSort(n,A)
|
s626922865
|
Accepted
| 20 | 7,736 | 384 |
def selectionSort(n, A):
cnt =0
for i in range(n):
minj = i
for j in range(i,n):
if A[minj] > A[j]:
minj = j
if i != minj:
A[i],A[minj] = A[minj],A[i]
cnt += 1
print(*A)
print(cnt)
if __name__ == '__main__':
n = int(input())
A = list(map(int, input().split()))
selectionSort(n,A)
|
s562072067
|
p03067
|
u453526259
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 3,060 | 82 |
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
a,b,c = map(int,input().split())
if a >= c >= b:
print("Yes")
else:
print("No")
|
s600462655
|
Accepted
| 17 | 2,940 | 117 |
a,b,c = map(int,input().split())
if a >= c >= b:
print("Yes")
elif a <= c <= b:
print("Yes")
else:
print("No")
|
s865378247
|
p03024
|
u209619667
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 130 |
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.
|
A = input()
count = 0
for i in A:
if i == 'x':
count += 1
b = len(A) - count
if 15-b > 7:
print('Yes')
else:
print('No')
|
s030956808
|
Accepted
| 17 | 2,940 | 132 |
A = input()
count = 0
for i in A:
if i == 'o':
count += 1
b = len(A) - count
if 15 - b > 7:
print('YES')
else:
print('NO')
|
s322469638
|
p03501
|
u672316981
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,056 | 51 |
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
n, a, b = map(int, input().split())
print(n * a, b)
|
s218421547
|
Accepted
| 26 | 9,160 | 56 |
n, a, b = map(int, input().split())
print(min(n * a, b))
|
s969109098
|
p02646
|
u833492079
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,188 | 194 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a,v = map(int, input().split())
b,w = map(int, input().split())
t = int(input())
ans = "No"
if v <= w:
print(ans)
exit()
x = abs(v-w)
y = abs(a-b)
if y <= x*t:
ans = "Yes"
print(ans)
|
s034855645
|
Accepted
| 23 | 9,128 | 401 |
#-------------------------------------------------------------------
import sys
def p(*_a):
#return
_s=" ".join(map(str,_a))
#print(_s)
sys.stderr.write(_s+"\n")
#-------------------------------------------------------------------
a,v = map(int, input().split())
b,w = map(int, input().split())
t = int(input())
ans = "NO"
x = v-w
y = abs(a-b)
p(x,y)
if y <= x*t:
ans = "YES"
print(ans)
|
s989048110
|
p03090
|
u352623442
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 3,740 | 374 |
You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem.
|
n = int(input())
if n%2 == 0:
su = n+1
print((n*(n-1)//2)-n//2)
for i in range(1,n+1):
for k in range(1,n+1):
if i+k != su and i < k:
print(i,k)
if n%2 == 1:
su = n+1
print((n*(n-1)//2)-(n-1)//2)
for i in range(1,n+1):
for k in range(1,n+1):
if i+k != su and i < k:
print(i,k)
|
s521307604
|
Accepted
| 24 | 3,612 | 372 |
n = int(input())
if n%2 == 0:
su = n+1
print((n*(n-1)//2)-n//2)
for i in range(1,n+1):
for k in range(1,n+1):
if i+k != su and i < k:
print(i,k)
if n%2 == 1:
su = n
print((n*(n-1)//2)-(n-1)//2)
for i in range(1,n+1):
for k in range(1,n+1):
if i+k != su and i < k:
print(i,k)
|
s488946517
|
p02690
|
u290886932
| 2,000 | 1,048,576 |
Time Limit Exceeded
| 2,205 | 9,004 | 367 |
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
|
X = int(input())
def modpow(a, n, mod):
ret = 1
while n > 0:
if n % 2 == 1:
ret = ret * a % mod
a *= a
a = a % mod
return ret
i = 0
L = list()
while True:
num = modpow(i,5,X)
if num in L:
j = L.index(num)
ret = '{} {}'.format(i,j)
print(ret)
break
else:
L.append(num)
|
s992276574
|
Accepted
| 24 | 9,208 | 453 |
X = int(input())
L = list()
i = 0
while True:
if i ** 5 > 10 ** 11:
break
L.append(i ** 5)
i += 1
flag = False
ret = ''
for i in range(len(L)):
if flag:
break
for j in range(i+1, len(L)):
if L[j] - L[i] == X :
ret ='{} {}'.format(j,i)
flag = True
break
if L[j] + L[i] == X :
ret ='{} {}'.format(j,-i)
flag = True
break
print(ret)
|
s324853508
|
p02601
|
u781543416
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,120 | 441 |
M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.
|
abc=list(map(int, input().split()))
k=int(input())
a=abc[0]
b=abc[1]
c=abc[2]
num=0
if(a < b):
if(b < c):
print("Yes")
else:
while num<k:
c=2*c
num=num+1
if(b<c):
print("Yes")
break
print("No")
else:
while num<k:
b=2*b
num=num+1
if(a<b):
while num<k:
c=2*c
num=num+1
if(b<c):
print("Yes")
break
print("No")
print("No")
|
s492835711
|
Accepted
| 30 | 9,060 | 492 |
abc=list(map(int, input().split()))
k=int(input())
a=abc[0]
b=abc[1]
c=abc[2]
num=0
if(a < b):
if(b < c):
print("Yes")
else:
while num<k:
c=2*c
num=num+1
if(b<c):
print("Yes")
break
if(b>=c):
print("No")
else:
while num<k:
b=2*b
num=num+1
if(a<b):
break
if(a<b):
while num<k:
c=2*c
num=num+1
if(b<c):
print("Yes")
break
if(b>=c):
print("No")
else:
print("No")
|
s028956156
|
p03679
|
u368796742
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 126 |
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
x,a,b = map(int,input().split())
if x+a >= b:
print("delicious")
elif x+a+1 == b:
print("safe")
else:
print("dangerous")
|
s710038901
|
Accepted
| 19 | 2,940 | 123 |
x,a,b = map(int,input().split())
if a >= b:
print("delicious")
elif x+a >= b:
print("safe")
else:
print("dangerous")
|
s529712959
|
p03386
|
u614314290
| 2,000 | 262,144 |
Wrong Answer
| 2,238 | 4,084 | 99 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
A, B, K = map(int, input().split())
L = [x for x in range(A, B + 1)]
print(*set((L[:K] + L[-K:])))
|
s757865420
|
Accepted
| 17 | 3,060 | 107 |
A, B, K = map(int, input().split())
L = range(A, B + 1)
print(*sorted(set(L[:K]) | set(L[-K:])), sep="\n")
|
s864090133
|
p03380
|
u672220554
| 2,000 | 262,144 |
Wrong Answer
| 144 | 15,712 | 320 |
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
n =int(input())
a = list(map(int,input().split()))
seta = sorted(list(set(a)))
resn = seta[-1]
resr = seta[0]
for sa in seta:
sabun = abs(resn / 2 - sa)
if sabun == 0 or sabun == 0.5:
resr == sa
break
elif sabun < abs(resn / 2 - resr):
resr = sa
print(str(resn) + " " + str(resr))
|
s131109641
|
Accepted
| 146 | 15,712 | 319 |
n =int(input())
a = list(map(int,input().split()))
seta = sorted(list(set(a)))
resn = seta[-1]
resr = seta[0]
for sa in seta:
sabun = abs(resn / 2 - sa)
if sabun == 0 or sabun == 0.5:
resr = sa
break
elif sabun < abs(resn / 2 - resr):
resr = sa
print(str(resn) + " " + str(resr))
|
s073022871
|
p02389
|
u498041957
| 1,000 | 131,072 |
Wrong Answer
| 40 | 7,416 | 30 |
Write a program which calculates the area and perimeter of a given rectangle.
|
ab = input().split()
print(ab)
|
s222769708
|
Accepted
| 20 | 7,672 | 66 |
a, b = [int(i) for i in input().split()]
print(a * b, (a + b) * 2)
|
s280821018
|
p03827
|
u762420987
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 121 |
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
N = int(input())
S = input()
num = 0
for s in S:
if s == "I":
num += 1
else:
num -= 1
print(num)
|
s355641221
|
Accepted
| 17 | 2,940 | 169 |
N = int(input())
S = input()
max_num = 0
num = 0
for s in S:
if s == "I":
num += 1
else:
num -= 1
max_num = max(max_num, num)
print(max_num)
|
s513530873
|
p03131
|
u221345507
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,064 | 324 |
Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the maximum possible number of biscuits in Snuke's pocket after K operations.
|
K, A, B =map(int,input().split())
import sys
should_actAB = (B-A)
if should_actAB<=2:
print(K+1)
sys.exit()
else:
ans_cookie=1
K_=K-(A-1)
print(K_)
ans_cookie+=(A-1)
print(ans_cookie)
if K_%2==0:
ans_cookie+=should_actAB*(K_)//2
else:
ans_cookie+=should_actAB*(K_-1)//2+1
|
s231928131
|
Accepted
| 18 | 3,060 | 307 |
K, A, B =map(int,input().split())
import sys
should_actAB = (B-A)
if should_actAB<=2:
print(K+1)
sys.exit()
else:
ans_cookie=1
K_=K-(A-1)
ans_cookie+=(A-1)
if K_%2==0:
ans_cookie+=should_actAB*(K_)//2
else:
ans_cookie+=should_actAB*(K_-1)//2+1
print(ans_cookie)
|
s791965939
|
p03408
|
u926678805
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 259 |
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
dic={}
for i in range(int(input())):
s=input()
if s in dic:
dic[s]+=1
else:
dic[s]=1
for i in range(int(input())):
s=input()
if s in dic:
dic[s]-=1
else:
dic[s]=-1
print(min(sorted(dic.values())[-1],0))
|
s118767799
|
Accepted
| 18 | 3,064 | 284 |
import sys
sys.setrecursionlimit(100000)
dic={}
for i in range(int(input())):
s=input()
try:
dic[s]+=1
except:
dic[s]=1
for i in range(int(input())):
s=input()
try:
dic[s]-=1
except:
dic[s]=-1
print(max(list(dic.values())+[0]))
|
s449048933
|
p02315
|
u887884261
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,616 | 647 |
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi. You want to find a subset of items to put such that: * The total value of the items is as large as possible. * The items have combined weight at most W, that is capacity of the knapsack. Find the maximum total value of items in the knapsack.
|
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
n, max_w = [int(i) for i in input().split()]#print(n, max_w)
w = []
v = []
for j in range(n):
weight,value=[int(i) for i in input().split()]
w.append(weight)
v.append(value)
dp = [[0 for i in range(max_w+1)] for j in range(n+1)]
for i in range(n):
for cw in range(max_w+1):
if cw >= w[i]:
if cw == 8:
print(i+1,cw,cw-w[i],i)
print(dp[i+1][cw])
print(dp[i][cw-w[i]])
print(dp[i][cw])
dp[i+1][cw] = max(dp[i][cw-w[i]]+v[i], dp[i][cw])
else:
dp[i+1][cw] = dp[i][cw]
|
s430034555
|
Accepted
| 940 | 42,268 | 496 |
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
n, max_w = [int(i) for i in input().split()]#print(n, max_w)
v = []
w = []
for j in range(n):
value,weight=[int(i) for i in input().split()]
v.append(value)
w.append(weight)
dp = [[0 for i in range(max_w+1)] for j in range(n+1)]
for i in range(n):
for cw in range(max_w+1):
if cw >= w[i]:
dp[i+1][cw] = max(dp[i][cw-w[i]]+v[i], dp[i][cw])
else:
dp[i+1][cw] = dp[i][cw]
print(dp[n][max_w])
|
s413808469
|
p02612
|
u450147945
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,016 | 36 |
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 = input()
N = int(N)
print(N%1000)
|
s234962382
|
Accepted
| 30 | 8,856 | 37 |
N = input()
N = int(N)
print(-N%1000)
|
s201304121
|
p02613
|
u024343432
| 2,000 | 1,048,576 |
Wrong Answer
| 149 | 9,296 | 499 |
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.
|
from sys import stdin,stdout
import math
from collections import Counter,defaultdict
LI=lambda:list(map(int,input().split()))
MAP=lambda:map(int,input().split())
IN=lambda:int(input())
S=lambda:input()
def case():
a=b=c=d=0
for i in range(IN()):
s=S()
if s=="AC":a+=1
elif s=="TLE":b+=1
elif s=="WA":c+=1
else:d+=1
print("AC","x",a)
print("AC","x",c)
print("AC","x",b)
print("AC","x",d)
for _ in range(1):
case()
|
s747387035
|
Accepted
| 144 | 9,436 | 500 |
from sys import stdin,stdout
import math
from collections import Counter,defaultdict
LI=lambda:list(map(int,input().split()))
MAP=lambda:map(int,input().split())
IN=lambda:int(input())
S=lambda:input()
def case():
a=b=c=d=0
for i in range(IN()):
s=S()
if s=="AC":a+=1
elif s=="TLE":b+=1
elif s=="WA":c+=1
else:d+=1
print("AC","x",a)
print("WA","x",c)
print("TLE","x",b)
print("RE","x",d)
for _ in range(1):
case()
|
s780943101
|
p03846
|
u598699652
| 2,000 | 262,144 |
Wrong Answer
| 69 | 13,880 | 408 |
There are N people, conveniently numbered 1 through N. They were standing in a row yesterday, but now they are unsure of the order in which they were standing. However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person. According to their reports, the difference above for person i is A_i. Based on these reports, find the number of the possible orders in which they were standing. Since it can be extremely large, print the answer modulo 10^9+7. Note that the reports may be incorrect and thus there may be no consistent order. In such a case, print 0.
|
N = int(input())
A = list(map(int, input().split()))
ans = pow(2, N // 2)
# 0*1 2*2 4*2 2^N/2*2
check = [0 for p in range(0, N // 2 + N % 2)]
for o in A:
check[o // 2] += 1
for q in range(0, N // 2 + N % 2):
if q == 0 & N % 2 == 1:
if check[q] != 1:
ans = 0
break
else:
if check[q] != 2:
ans = 0
break
print(ans % (pow(10, 9) + 7))
|
s609392518
|
Accepted
| 65 | 14,008 | 410 |
N = int(input())
A = list(map(int, input().split()))
ans = pow(2, N // 2)
# 0*1 2*2 4*2 2^N/2*2
check = [0 for p in range(0, N // 2 + N % 2)]
for o in A:
check[o // 2] += 1
for q in range(0, N // 2 + N % 2):
if q == 0 and N % 2 == 1:
if check[q] != 1:
ans = 0
break
else:
if check[q] != 2:
ans = 0
break
print(ans % (pow(10, 9) + 7))
|
s326459897
|
p03130
|
u811000506
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 161 |
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.
|
li = []
for i in range(3):
a,b = map(int,input().split())
li.append(a)
li.append(b)
li.sort()
if li==[1, 2, 2, 2, 3, 4]:
print("YES")
else:
print("NO")
|
s106486243
|
Accepted
| 17 | 3,060 | 162 |
li = []
for i in range(3):
a,b = map(int,input().split())
li.append(a)
li.append(b)
li.sort()
if li==[1, 2, 2, 3, 3, 4]:
print("YES")
else:
print("NO")
|
s181049289
|
p02413
|
u853619096
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,588 | 166 |
Your task is to perform a simple table calculation. Write a program which reads the number of rows r, columns c and a table of r × c elements, and prints a new table, which includes the total sum for each row and column.
|
r, c = map(int, input().split())
for j in range(r):
a = [int(i) for i in input().split()]
for k in a:
print('{} '.format(k),end='')
print(sum(a))
|
s774522051
|
Accepted
| 30 | 7,736 | 338 |
r,c=map(int,input().split())
a=[]
for i in range(r):
a+=[list(map(int,input().split()))]
a[i].append(sum(a[i]))
for i in range(len(a)):
print(" ".join(list(map(str,a[i]))))
b=[]
t=[]
d=0
for i in range(c+1):
for j in range(r):
b+=[a[j][i]]
t+=[sum(b)]
b=[]
print(" ".join(list(map(str,t))))
|
s168920784
|
p03999
|
u633548583
| 2,000 | 262,144 |
Wrong Answer
| 30 | 3,060 | 217 |
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion. All strings that can be obtained in this way can be evaluated as formulas. Evaluate all possible formulas, and print the sum of the results.
|
s=input()
cnt=len(s)-1
for i in range(2**cnt):
sum=s[0]
for j in range(cnt):
if ((i>>j)&1):
sum+="+"
else:
sum+="-"
sum+=s[j+1]
print(eval(sum))
|
s050463837
|
Accepted
| 27 | 2,940 | 229 |
s=input()
cnt=len(s)-1
ans=0
for i in range(2**cnt):
sum=s[0]
for j in range(cnt):
if ((i>>j)&1):
sum+="+"
else:
pass
sum+=s[j+1]
ans+=eval(sum)
print(ans)
|
s730136885
|
p02613
|
u140084432
| 2,000 | 1,048,576 |
Wrong Answer
| 154 | 17,464 | 407 |
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.
|
def main():
N = int(input())
judges = [input() for _ in range(N)]
print(judges)
res_dic = {"AC": 0, "WA": 0, "TLE": 0, "RE": 0}
for judge in judges:
if judge in res_dic:
res_dic[judge] += 1
else:
res_dic[judge] = 0
for k, v in res_dic.items():
print("{} x {}".format(k, v))
if __name__ == '__main__':
main()
|
s627228796
|
Accepted
| 141 | 16,244 | 384 |
def main():
N = int(input())
judges = [input() for _ in range(N)]
res_dic = {"AC": 0, "WA": 0, "TLE": 0, "RE": 0}
for judge in judges:
if judge in res_dic:
res_dic[judge] += 1
else:
res_dic[judge] = 0
for k, v in res_dic.items():
print("{} x {}".format(k, v))
if __name__ == '__main__':
main()
|
s001859164
|
p03493
|
u894348341
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 25 |
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
print(input().count('a'))
|
s478304496
|
Accepted
| 17 | 2,940 | 25 |
print(input().count('1'))
|
s985046159
|
p03853
|
u325282913
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 95 |
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())
for _ in range(H-1):
s = input()
print(s)
print(s)
|
s299416874
|
Accepted
| 18 | 3,060 | 157 |
H, W = map(int, input().split())
array = []
for _ in range(H):
s = input()
array.append(s)
for i in range(H):
print(array[i])
print(array[i])
|
s700611300
|
p03139
|
u379692329
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 99 |
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
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N, A, B = map(int, input().split())
ResMax = max(A, B)
ResMin = max(0, A+B-N)
print(ResMax, ResMin)
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s397107553
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Accepted
| 17 | 2,940 | 99 |
N, A, B = map(int, input().split())
ResMax = min(A, B)
ResMin = max(0, A+B-N)
print(ResMax, ResMin)
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