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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s043564536 | p04030 | u842388336 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 191 | 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()
ans=[]
for c in s:
if s=="0":
ans.append(0)
if s=="i":
ans.append(1)
if s=="B":
if len(ans)==0:
pass
else:
ans.pop()
ans="".join(ans)
print(ans) | s010128987 | Accepted | 17 | 2,940 | 195 | s = input()
ans=[]
for c in s:
if c=="0":
ans.append("0")
if c=="1":
ans.append("1")
if c=="B":
if len(ans)==0:
pass
else:
ans.pop()
ans="".join(ans)
print(ans) |
s353279142 | p02414 | u671553883 | 1,000 | 131,072 | Wrong Answer | 30 | 7,624 | 202 | Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula: \\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\] where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively. | n, m, l = [int(i) for i in input().split()]
A = []
B = []
C = []
for ni in range(n):
A.append([int(i) for i in input().split()])
for mi in range(m):
B.append([int(i) for i in input().split()]) | s249507423 | Accepted | 520 | 8,988 | 422 | n, m, l = [int(i) for i in input().split()]
A = []
B = []
C = []
for ni in range(n):
A.append([int(i) for i in input().split()])
for mi in range(m):
B.append([int(i) for i in input().split()])
for i in range(n):
C.append([])
for j in range(l):
C[i].append(0)
for k in range(m):
C[i] [j] += A[i] [k] * B[k] [j]
for ni in range(n):
print(" ".join([str(s) for s in C[ni]])) |
s069768969 | p03251 | u477216059 | 2,000 | 1,048,576 | Wrong Answer | 19 | 3,060 | 307 | Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out. | if __name__ == '__main__':
n, m, X, Y = (int(i) for i in input().split())
x = [int(i) for i in input().split()]
y = [int(i) for i in input().split()]
result = "War"
for i in range(X + 1, Y + 1):
if max(x) < 16 and min(y) >= 16:
result = "No war"
print(result)
| s983895158 | Accepted | 18 | 3,060 | 323 | if __name__ == '__main__':
n, m, X, Y = (int(i) for i in input().split())
x = [int(i) for i in input().split()]
y = [int(i) for i in input().split()]
result = "War"
for i in range(-100, 101):
if X < i and Y >= i and max(x) < i and min(y) >= i:
result = "No War"
print(result)
|
s766191994 | p03680 | u777028980 | 2,000 | 262,144 | Wrong Answer | 2,104 | 7,084 | 164 | Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons. | n=int(input())
hoge=[]
for i in range(n):
hoge.append(int(input()))
ans=0
punch=1
while 1:
ans+=1
punch=hoge[punch-1]
if(punch==1):
print(ans)
break | s247859419 | Accepted | 212 | 7,084 | 211 | n=int(input())
hoge=[]
for i in range(n):
hoge.append(int(input()))
ans=0
punch=1
while 1:
ans+=1
punch=hoge[punch-1]
if(punch-1==1):
print(ans)
break
if(ans>=100001):
print(-1)
break
|
s601785171 | p00010 | u519227872 | 1,000 | 131,072 | Wrong Answer | 30 | 7,688 | 758 | Write a program which prints the central coordinate $(p_x, p_y)$ and the radius $r$ of a circumscribed circle of a triangle which is constructed by three points $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ on the plane surface. | from math import sqrt
n = int(input())
for i in range(n):
l = input()
x1,y1,x2,y2,x3,y3=map(float,l.split(' '))
a_2 = (x1 - x2)**2 + (y1 - y2)**2
b_2 = (x2 - x3)**2 + (y2 - y3)**2
c_2 = (x3 - x1)**2 + (y3 - y1)**2
cos_a_2 = (b_2 + c_2 - a_2)**2/(4* b_2 * c_2)
sin_a_2 = 1 - cos_a_2
r = round(sqrt(a_2/sin_a_2)/2,3)
a = (x1**2 - x3**2 + y1**2 - y3**2)*(x2 - x1)
b = (x1**2 - x2**2 + y1**2 - y2**2)*(x3 - x1)
c = (y2 - y1)*(x3 - x1)
d = (y3 - y1)*(x2 - x1)
py = round((a-b)/(c-d)/2,3)
a = (x1**2 - x3**2 + y1**2 - y3**2)*(y2 - y1)
b = (x1**2 - x2**2 + y1**2 - y2**2)*(y3 - y1)
c = (x2 - x1)*(y3 - y1)
d = (x3 - x1)*(y2 - y1)
px = round((a-b)/(c-d)/2,3)
print('%s %s %s' %(px, py, r)) | s903524426 | Accepted | 30 | 7,672 | 772 | from math import sqrt
n = int(input())
for i in range(n):
l = input()
x1,y1,x2,y2,x3,y3=map(float,l.split(' '))
a_2 = (x1 - x2)**2 + (y1 - y2)**2
b_2 = (x2 - x3)**2 + (y2 - y3)**2
c_2 = (x3 - x1)**2 + (y3 - y1)**2
cos_a_2 = (b_2 + c_2 - a_2)**2/(4* b_2 * c_2)
sin_a_2 = 1 - cos_a_2
r = round(sqrt(a_2/sin_a_2)/2,3)
a = (x1**2 - x3**2 + y1**2 - y3**2)*(x2 - x1)
b = (x1**2 - x2**2 + y1**2 - y2**2)*(x3 - x1)
c = (y2 - y1)*(x3 - x1)
d = (y3 - y1)*(x2 - x1)
py = round((a-b)/(c-d)/2,3) + 0
a = (x1**2 - x3**2 + y1**2 - y3**2)*(y2 - y1)
b = (x1**2 - x2**2 + y1**2 - y2**2)*(y3 - y1)
c = (x2 - x1)*(y3 - y1)
d = (x3 - x1)*(y2 - y1)
px = round((a-b)/(c-d)/2,3) + 0
print('%.3f %.3f %.3f' %(px, py, r)) |
s187255411 | p03450 | u765386817 | 2,000 | 262,144 | Wrong Answer | 2,110 | 151,448 | 748 | There are N people standing on the x-axis. Let the coordinate of Person i be x_i. For every i, x_i is an integer between 0 and 10^9 (inclusive). It is possible that more than one person is standing at the same coordinate. You will given M pieces of information regarding the positions of these people. The i-th piece of information has the form (L_i, R_i, D_i). This means that Person R_i is to the right of Person L_i by D_i units of distance, that is, x_{R_i} - x_{L_i} = D_i holds. It turns out that some of these M pieces of information may be incorrect. Determine if there exists a set of values (x_1, x_2, ..., x_N) that is consistent with the given pieces of information. | import queue
n,m = map(int,input().split())
dis = [[] for i in range(n)]
for i in range(m):
g1,g2,L = map(int,input().split())
dis[g1-1].append((g2-1,L))
dis[g2-1].append((g1-1,-L))
print(dis)
geo = dict()
geo[0] = 0
q = queue.Queue()
q.put(0)
#fig[0] = 1
while not q.empty():
g1 = q.get()
#if fig[g1] == 1:
for g2,L in dis[g1]:
print(geo)
if g2 not in geo:
geo[g2] = geo[g1] + L
#fig[g2] = 1
q.put(g2)
else:
if geo[g2] != geo[g1] + L:
print("no")
exit()
#fig[g1] = 2
print("yes")
| s915799712 | Accepted | 1,126 | 109,956 | 837 | import collections
from collections import deque
import sys
input = sys.stdin.readline
def bfs(s):
d = deque()
d.append(s)
dis[s]=0
flag=True
while len(d):
x = d.popleft()
for i,j in vec[x]:
if dis[i] == INF:
dis[i]=dis[x]+j
d.append(i)
else:
if dis[i]==dis[x]+j:
continue
else:
print('No')
exit()
return flag
n,m = map(int, input().split())
a = [list(map(int, input().split())) for i in range(m)]
vec= [[] for i in range(n)]
for i in range(m):
vec[a[i][0]-1].append((a[i][1]-1,a[i][2]))
vec[a[i][1]-1].append((a[i][0]-1,-a[i][2]))
INF=float('inf')
dis=[INF]*n
for i in range(n):
if dis[i]==INF:
bfs(i)
print('Yes') |
s963917640 | p03448 | u103341055 | 2,000 | 262,144 | Wrong Answer | 33 | 9,176 | 698 | You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different. | A = int(input())
B = int(input())
C = int(input())
X = int(input())
ans = 0
# for a in range(A+1):
# x = X - a*500
# for b in range(B+1):
# x -= b*100
# for c in range(C+1):
# x -= c*50
# print(a,b,c,x)
# if x == 0:
# ans += 1
# for a in range(A+1):
# for b in range(B+1):
# for c in range(C+1):
# if 500*a+100*b+50*c == X:
# ans += 1
for a in range(A+1):
for b in range(B+1):
x = X - 500*a -100*b
if x%50 == 0:
if x//50 <= C:
ans += 1
print(ans)
#
| s278320184 | Accepted | 27 | 9,096 | 783 | A = int(input())
B = int(input())
C = int(input())
X = int(input())
ans = 0
# for a in range(A+1):
# x = X - a*500
# for b in range(B+1):
# x -= b*100
# for c in range(C+1):
# x -= c*50
# print(a,b,c,x)
# if x == 0:
# ans += 1
# for a in range(A+1):
# for b in range(B+1):
# for c in range(C+1):
# if 500*a+100*b+50*c == X:
# ans += 1
for a in range(A+1):
if 500 * a > X:
break
for b in range(B+1):
if 500 * a + 100 * b > X:
break
x = X - 500*a -100*b
if x%50 == 0:
if x//50 <= C:
ans += 1
print(ans)
#
|
s317498514 | p03997 | u003505857 | 2,000 | 262,144 | Wrong Answer | 26 | 8,972 | 118 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. |
a = int(input())
b = int(input())
h = int(input())
answer = (a + b) / 2 * h
print(answer) | s048475518 | Accepted | 27 | 9,092 | 98 |
a = int(input())
b = int(input())
h = int(input())
print(int((a + b) / 2 * h)) |
s975282799 | p03359 | u846652026 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 56 | 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())
print(a if a<b else a-1) | s496131864 | Accepted | 17 | 2,940 | 57 | a,b = map(int, input().split())
print(a if a<=b else a-1) |
s657112821 | p03591 | u077685310 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 141 | Ringo is giving a present to Snuke. Ringo has found out that Snuke loves _yakiniku_ (a Japanese term meaning grilled meat. _yaki_ : grilled, _niku_ : meat). He supposes that Snuke likes grilled things starting with `YAKI` in Japanese, and does not like other things. You are given a string S representing the Japanese name of Ringo's present to Snuke. Determine whether S starts with `YAKI`. | a=list(input())
if len(a)<=3:
print("No")
else:
if a[0]=="T" and a[1]=="A" and a[2]=="K" and a[3]=="I":
print("Yes")
else:
print("No") | s087093878 | Accepted | 17 | 2,940 | 141 | a=list(input())
if len(a)<=3:
print("No")
else:
if a[0]=="Y" and a[1]=="A" and a[2]=="K" and a[3]=="I":
print("Yes")
else:
print("No") |
s530427192 | p02397 | u681232780 | 1,000 | 131,072 | Wrong Answer | 50 | 5,608 | 171 | Write a program which reads two integers x and y, and prints them in ascending order. | while True :
x,y = map(int,input().split())
if y < x :
print(y,x)
elif x < y :
print(x,y)
else :
print(x,y)
if x == 0 and y == 0 :
break
| s187257562 | Accepted | 50 | 5,608 | 143 | while True :
x,y = map(int,input().split())
if x == 0 and y == 0 :
break
elif y < x :
print(y,x)
else :
print(x,y)
|
s124424718 | p03853 | u264681142 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 139 | 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())
l = []
for i in range(h):
l.append(input().split())
for i in range(h):
print(l[i])
print(l[i]) | s642816613 | Accepted | 17 | 3,060 | 139 | h, w = map(int, input().split())
l = []
for i in range(h):
s = input()
l.append(s)
for i in range(h):
print(l[i])
print(l[i]) |
s152563248 | p02602 | u969081133 | 2,000 | 1,048,576 | Wrong Answer | 142 | 31,512 | 141 | M-kun is a student in Aoki High School, where a year is divided into N terms. There is an exam at the end of each term. According to the scores in those exams, a student is given a grade for each term, as follows: * For the first through (K-1)-th terms: not given. * For each of the K-th through N-th terms: the multiplication of the scores in the last K exams, including the exam in the graded term. M-kun scored A_i in the exam at the end of the i-th term. For each i such that K+1 \leq i \leq N, determine whether his grade for the i-th term is **strictly** greater than the grade for the (i-1)-th term. | n,k=map(int,input().split())
a=list(map(int,input().split()))
for i in range(k,n):
if a[k]>a[i-k]:
print('Yes')
else:
print('No') | s478849994 | Accepted | 141 | 31,444 | 141 | n,k=map(int,input().split())
a=list(map(int,input().split()))
for i in range(k,n):
if a[i]>a[i-k]:
print('Yes')
else:
print('No') |
s903422188 | p03693 | u037164984 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 140 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4? | # -*- coding: utf-8 -*-
r, g, b = map(int,input().split())
a = g+b
answer = a % 4
if answer == 0:
print("YES")
else:
print("NO") | s791338552 | Accepted | 17 | 2,940 | 145 | # -*- coding: utf-8 -*-
r, g, b = map(int,input().split())
a = 10*g + b
answer = a % 4
if answer == 0:
print("YES")
else:
print("NO") |
s310142145 | p03555 | u590048048 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 93 | You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise. | a = input()
b = input()
if "".join(reversed(a)) == b:
print("Yes")
else:
print("No")
| s080037389 | Accepted | 17 | 2,940 | 93 | a = input()
b = input()
if "".join(reversed(a)) == b:
print("YES")
else:
print("NO")
|
s676906496 | p04044 | u333917945 | 2,000 | 262,144 | Wrong Answer | 24 | 3,316 | 97 | Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m. | n, l = map(int, input().split())
s = [[i for i in input()] for j in range(n)]
s.sort()
print(s)
| s895480559 | Accepted | 24 | 3,064 | 178 | n, l = map(int, input().split())
s = [[i for i in input()] for j in range(n)]
#print(s)
s.sort()
#print(s)
for i in range(n):
s[i] = "".join(s[i])
s = "".join(s)
print(s)
|
s803187828 | p03861 | u325206354 | 2,000 | 262,144 | Wrong Answer | 2,104 | 3,068 | 118 | 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,c=list(map(int,input().split()))
count = 0
for i in range(a,b):
if i % c == 0:
count+=1
print(count)
| s515214975 | Accepted | 17 | 2,940 | 58 | a,b,c=list(map(int,input().split()))
print(b//c-(a-1)//c)
|
s071417942 | p03593 | u930705402 | 2,000 | 262,144 | Wrong Answer | 23 | 3,436 | 753 | We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter. Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition: * Every row and column in A' can be read as a palindrome. Determine whether he can create a matrix satisfying the condition. | from collections import Counter
H,W=map(int,input().split())
li=[]
for i in range(H):
a=list(input())
for j in range(W):
li.append(a[j])
c=Counter(li)
if(H%2==0 and W%2==0):
for x,y in c.items():
if(y%4!=0):
print('No')
exit()
print('Yes')
elif(H%2==1 and W%2==1):
f=(W//2)*(H//2)
t=(H//2)+(W//2)
o=1
fc,tc,oc=0,0,0
for x,y in c.items():
fc+=y//4
r=y%4
tc+=r//2
oc+=r%2
if(fc==f and tc==t and oc==o):
print('Yes')
else:
print('No')
else:
if(H%2==1 and W%2==0):
H,W=W,H
f=(W//2)*(H//2)
t=H//2
fc,tc,oc=0,0,0
for x,y in c.items():
fc+=y//4
r=y%4
tc+=r//2
oc+=r%2 | s463747340 | Accepted | 21 | 3,444 | 483 | from collections import Counter
H,W=map(int,input().split())
li=[]
for i in range(H):
a=list(input())
li+=a
c=Counter(li)
t,o=0,0
for x,y in c.items():
l=y%4
t+=l//2
o+=l%2
isok=True
if(H%2==0 and W%2==0):
if t!=0 or o!=0:
isok=False
elif(H%2==0 and W%2==1):
if(t>H//2 or o>0):
isok=False
elif(H%2==1 and W%2==0):
if(t>W//2 or o>0):
isok=False
else:
if(t>W//2+H//2 or o>1):
isok=False
print('Yes' if isok else 'No') |
s058713938 | p03623 | u440478998 | 2,000 | 262,144 | Wrong Answer | 26 | 9,028 | 83 | Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|. | x,a,b = map(int, input().split())
print("A") if abs(x-a) > abs(x-b) else print("B") | s103092475 | Accepted | 28 | 9,160 | 83 | x,a,b = map(int, input().split())
print("A") if abs(x-a) < abs(x-b) else print("B") |
s165341399 | p03387 | u798731634 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 697 | You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations. | n = list(map(int,input().split()))
n.sort(reverse=True)
a = n[0]-n[1]
b = n[0]-n[2]
if a == 0 and b == 0:
print(0)
if a != 0 and b == 0:
if a%2 == 0:
print(a/2)
elif a == 1:
print(2)
elif a%2 != 0:
print(1+(a-1)/2)
if a == 0 and b != 0:
if b%2 == 0:
print(b/2)
elif b == 1:
print(2)
elif b%2 != 0:
print(1+(b-1)/2)
if a != 0 and b != 0:
if a == 1 and b == 1:
print(1)
elif a%2 == 0 and b%2 == 0:
print(a/2+b/2)
elif a%2 != 0 and b%2 == 0:
print(2+(a-1)/2+b/2)
elif a%2 == 0 and b%2 != 0:
print(2+a/2+(b-1)/2)
elif a%2 != 0 and b%2 != 0:
print(1+(a-1)/2+(b-1)/2) | s793210239 | Accepted | 18 | 3,064 | 737 | n = list(map(int,input().split()))
n.sort(reverse=True)
a = n[0]-n[1]
b = n[0]-n[2]
if a == 0 and b == 0:
print(0)
if a != 0 and b == 0:
if a%2 == 0:
print(int(a/2))
elif a == 1:
print(2)
elif a%2 != 0:
print(int(2+(a-1)/2))
if a == 0 and b != 0:
if b%2 == 0:
print(int(b/2))
elif b == 1:
print(2)
elif b%2 != 0:
print(int(2+(b-1)/2))
if a != 0 and b != 0:
if a == 1 and b == 1:
print(1)
elif a%2 == 0 and b%2 == 0:
print(int(a/2+b/2))
elif a%2 != 0 and b%2 == 0:
print(int(2+(a-1)/2+b/2))
elif a%2 == 0 and b%2 != 0:
print(int(2+a/2+(b-1)/2))
elif a%2 != 0 and b%2 != 0:
print(int(1+(a-1)/2+(b-1)/2)) |
s232341985 | p03457 | u487392837 | 2,000 | 262,144 | Wrong Answer | 455 | 21,148 | 526 | 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. | nums = int(input())
tmp = list()
for num in range(nums):
tmp.append([int(_) for _ in str(input()).split(" ")])
for num, case in enumerate(tmp):
t, x, y = case
if t < abs(x) + abs(y):
print("No")
break
else:
if ((abs(x) + abs(y)) % 2) and (t % 2):
if num == len(tmp):
print("Yes")
elif ((abs(x) + abs(y)) % 2 == 0) and (t % 2 == 0):
if num == len(tmp):
print("Yes")
else:
print("No")
break
| s060131358 | Accepted | 450 | 21,108 | 532 | nums = int(input())
tmp = list()
for num in range(nums):
tmp.append([int(_) for _ in str(input()).split(" ")])
for num, case in enumerate(tmp):
t, x, y = case
if t < abs(x) + abs(y):
print("No")
break
else:
if ((abs(x) + abs(y)) % 2) and (t % 2):
if num +1 == len(tmp):
print("Yes")
elif ((abs(x) + abs(y)) % 2 == 0) and (t % 2 == 0):
if num +1 == len(tmp):
print("Yes")
else:
print("No")
break
|
s531626179 | p00516 | u352394527 | 8,000 | 131,072 | Wrong Answer | 20 | 5,600 | 328 | A global sports event will be held in Tokyo in 20XX. Programming contests are enjoyed all over the world as sports, and there is a possibility that they will be adopted as competitions. When I researched the judging committee that decides which competitions will be adopted, I found the following. For the judging panel, a list of his N possible competitions, ordered by funniest, was created. The i-th item from the top of the list contains the i-th interesting competition. Call it competition i. In addition, the cost Ai required to hold competition i is written. Also, the judging committee consists of her M members, from member 1 to member M. Member j has her own judging criteria Bj, and he votes one vote for the most interesting competition that costs less to run than her Bj. For any committee's judging criteria, he had at least one event that cost less than the judging criteria. Therefore, all members voted one vote. He had only one competition in which he received the most votes. Given a list of competitions and member information, write a program to find the number of the competition with the most votes. | def main():
n,m = map(int,input().split())
A = [int(input()) for _ in range(n)]
V = [0 for _ in range(n)]
m = 0
ans = 0
for i in range(m):
b = int(input())
for j in range(n):
if A[j] <= b:
V[j] += 1
if m < V[j]:
m += 1
ans = j
break
print(ans + 1)
main()
| s371700314 | Accepted | 60 | 5,620 | 328 | def main():
n,m = map(int,input().split())
A = [int(input()) for _ in range(n)]
V = [0 for _ in range(n)]
M = 0
ans = 0
for i in range(m):
b = int(input())
for j in range(n):
if A[j] <= b:
V[j] += 1
if M < V[j]:
M += 1
ans = j
break
print(ans + 1)
main()
|
s840251004 | p03379 | u780698286 | 2,000 | 262,144 | Wrong Answer | 191 | 30,656 | 143 | When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N. | n = int(input())
x = list(map(int, input().split()))
x.sort()
for i in range(n):
if i < n//2:
print(x[n//2])
else:
print(x[n//2-1]) | s864048256 | Accepted | 208 | 30,952 | 157 | n = int(input())
x = list(map(int, input().split()))
y = sorted(x)
for i in range(n):
if x[i] <= y[n//2-1]:
print(y[n//2])
else:
print(y[n//2-1]) |
s414045255 | p03067 | u729133443 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 62 | 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());print('YNeo s'[(c>a)^(c>b)::2]) | s134392576 | Accepted | 18 | 2,940 | 62 | a,b,c=map(int,input().split());print('NYoe s'[(c>a)^(c>b)::2]) |
s299943821 | p00637 | u546285759 | 1,000 | 131,072 | Wrong Answer | 30 | 7,536 | 458 | To write a research paper, you should definitely follow the structured format. This format, in many cases, is strictly defined, and students who try to write their papers have a hard time with it. One of such formats is related to citations. If you refer several pages of a material, you should enumerate their page numbers in ascending order. However, enumerating many page numbers waste space, so you should use the following abbreviated notation: When you refer all pages between page _a_ and page _b_ ( _a_ < _b_ ), you must use the notation " _a_ - _b_ ". For example, when you refer pages 1, 2, 3, 4, you must write "1-4" not "1 2 3 4". You must not write, for example, "1-2 3-4", "1-3 4", "1-3 2-4" and so on. When you refer one page and do not refer the previous and the next page of that page, you can write just the number of that page, but you must follow the notation when you refer successive pages (more than or equal to 2). Typically, commas are used to separate page numbers, in this problem we use space to separate the page numbers. You, a kind senior, decided to write a program which generates the abbreviated notation for your junior who struggle with the citation. | while True:
n = int(input())
if n == 0:
break
line = list(map(int, input().split()))
t = len(line)-1
ans = []
i = 0
while True:
ans.append(line[i])
while i < t:
if line[i]+1 == line[i+1]:
i += 1
else:
ans.append(line[i])
i += 1
break
if i == t:
ans.append(line[i])
break
print(*ans) | s702601982 | Accepted | 30 | 7,600 | 434 | while True:
n = int(input())
if n == 0:
break
line = list(map(int, input().split()))
i = 0
while i < n:
j = i
while j+1 < n and line[j+1]-line[j] == 1:
j += 1
s = "\n" if n-1 == j else " "
if i == j:
print("%d%s" % (line[i], s), end='')
i += 1
else:
print("%d-%d%s" % (line[i], line[j], s), end='')
i = j+1 |
s268522051 | p04043 | u329407311 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 177 | Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order. | a,b,c = map(int,input().split())
if a+b+c == 19:
if a == 7 or a == 5:
if b == 7 or b == 5:
if c == 7 or c == 5:
print("YES")
else:
print("NO") | s808064442 | Accepted | 17 | 2,940 | 201 | a,b,c = map(int,input().split())
d=0
if a+b+c == 17:
if a == 7 or a == 5:
if b == 7 or b == 5:
if c == 7 or c == 5:
print("YES")
d =1
if d == 0:print("NO")
|
s684154540 | p02238 | u094978706 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 557 | Depth-first search (DFS) follows the strategy to search ”deeper” in the graph whenever possible. In DFS, edges are recursively explored out of the most recently discovered vertex $v$ that still has unexplored edges leaving it. When all of $v$'s edges have been explored, the search ”backtracks” to explore edges leaving the vertex from which $v$ was discovered. This process continues until all the vertices that are reachable from the original source vertex have been discovered. If any undiscovered vertices remain, then one of them is selected as a new source and the search is repeated from that source. DFS timestamps each vertex as follows: * $d[v]$ records when $v$ is first discovered. * $f[v]$ records when the search finishes examining $v$’s adjacency list. Write a program which reads a directed graph $G = (V, E)$ and demonstrates DFS on the graph based on the following rules: * $G$ is given in an adjacency-list. Vertices are identified by IDs $1, 2,... n$ respectively. * IDs in the adjacency list are arranged in ascending order. * The program should report the discover time and the finish time for each vertex. * When there are several candidates to visit during DFS, the algorithm should select the vertex with the smallest ID. * The timestamp starts with 1. | n = int(input())
adj = [0]*n
for i in range(n):
tmp = list(map( int, input().split() ) )
adj[tmp[0]-1] = list(map(lambda x : x - 1, tmp[2:]))
c = ['w']*n # color of each vertex
d = [0]*n # discovery time for each vertex
f = [0]*n # finish time for each vertex
time = 0
def DFS_visit(u):
global time
time += 1
d[u] = time
c[u] = 'g'
for v in adj[u]:
if c[v] == 'w':
DFS_visit(v)
c[u] = 'b'
time += 1
f[u] = time
for u in range(n):
if c[u] == 'w':
DFS_visit(u)
| s717306292 | Accepted | 20 | 5,636 | 628 | n = int(input())
adj = [0]*n
for i in range(n):
tmp = list(map( int, input().split() ) )
adj[tmp[0]-1] = list(map(lambda x : x - 1, tmp[2:]))
c = ['w']*n # color of each vertex
d = [0]*n # discovery time for each vertex
f = [0]*n # finish time for each vertex
time = 0
def DFS_visit(u):
global time
time += 1
d[u] = time
c[u] = 'g'
for v in adj[u]:
if c[v] == 'w':
DFS_visit(v)
c[u] = 'b'
time += 1
f[u] = time
for u in range(n):
if c[u] == 'w':
DFS_visit(u)
for u in range(n):
print(f"{u+1} {d[u]} {f[u]}")
|
s947596126 | p02612 | u741579801 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,052 | 44 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | N = int(input())
# N = 1900
print(N % 1000) | s510127144 | Accepted | 29 | 9,004 | 53 | N = int(input())
# N = 1900
print((1000 - N) % 1000) |
s201901079 | p03370 | u757274384 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 125 | Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition. | n,x = map(int, input().split())
A = [int(input()) for i in range(n)]
A.sort()
A.reverse()
print(len(A) + (x - sum(A))//A[0]) | s098313956 | Accepted | 17 | 2,940 | 107 | n,x = map(int, input().split())
A = [int(input()) for i in range(n)]
print(len(A) + (x - sum(A))//min(A))
|
s949126172 | p03251 | u294385082 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,060 | 198 | Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out. | N,M,X,Y = map(int,input().split())
x = list(input().split())
y = list(input().split())
if X >= Y:
print("War")
exit()
else:
Z = Y
if int(max(x)) < Z and int(min(y)) >= Z:
print("No War") | s534358528 | Accepted | 18 | 3,060 | 177 | N,M,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
if max(max(x),X) < min(min(y),Y):
print("No War")
else:
print("War") |
s770895213 | p03997 | u620157187 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 68 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2) | s133520070 | Accepted | 17 | 2,940 | 73 | a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2)) |
s864252332 | p03997 | u628070051 | 2,000 | 262,144 | Wrong Answer | 22 | 8,856 | 103 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. |
a, b, h = [int(input())for i in range(3)]
S = (a + b) * h / 2
print(S) | s232791120 | Accepted | 27 | 9,156 | 104 |
a, b, h = [int(input())for i in range(3)]
S = (a + b) * h // 2
print(S) |
s426574804 | p03163 | u163907160 | 2,000 | 1,048,576 | Wrong Answer | 327 | 21,132 | 355 | There are N items, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), Item i has a weight of w_i and a value of v_i. Taro has decided to choose some of the N items and carry them home in a knapsack. The capacity of the knapsack is W, which means that the sum of the weights of items taken must be at most W. Find the maximum possible sum of the values of items that Taro takes home. | import numpy as np
N, W = map(int, input().split())
w, v = [], []
for i in range(N):
a, b = map(int, input().split())
w += [a]
v += [b]
DP = np.zeros(W + 1)
for i in range(N):
# print(DP[w[i]:])
# print(DP[:-w[i]])
# print(DP[:-w[i]] + v[i])
DP[w[i]:] = np.maximum(DP[:-w[i]] + v[i], DP[w[i]:])
# print(DP)
print(DP[-1]) | s034094403 | Accepted | 214 | 15,480 | 365 | import numpy as np
N, W = map(int, input().split())
w, v = [], []
for i in range(N):
a, b = map(int, input().split())
w += [a]
v += [b]
DP = np.zeros(W + 1,dtype=int)
for i in range(N):
# print(DP[w[i]:])
# print(DP[:-w[i]])
# print(DP[:-w[i]] + v[i])
DP[w[i]:] = np.maximum(DP[:-w[i]] + v[i], DP[w[i]:])
# print(DP)
print(DP[-1]) |
s842492564 | p03149 | u911206742 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 99 | You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974". | i = input()
set_1 = set(i)
if set_1 == {"1", "9", "7", "4"}:
print("YES")
else:
print("NO") | s487881888 | Accepted | 18 | 2,940 | 110 | i = input().split(" ")
set_1 = set(i)
if set_1 == {"1", "9", "7", "4"}:
print("YES")
else:
print("NO") |
s226718199 | p03852 | u951601135 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 133 | Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`. | s=input().replace('eraser','').replace('erase','').replace('dreamer','').replace('dream','')
if s:
print('NO')
else:
print('YES') | s199950913 | Accepted | 19 | 2,940 | 78 | if(input() in ['a','i','u','e','o']):
print('vowel')
else:print('consonant') |
s404054025 | p02562 | u401686269 | 5,000 | 1,048,576 | Wrong Answer | 537 | 54,800 | 761 | You are given a grid of N rows and M columns. The square at the i-th row and j-th column will be denoted as (i,j). A nonnegative integer A_{i,j} is written for each square (i,j). You choose some of the squares so that each row and column contains at most K chosen squares. Under this constraint, calculate the maximum value of the sum of the integers written on the chosen squares. Additionally, calculate a way to choose squares that acheives the maximum. | import networkx as nx
N,K=map(int,input().split())
A=[list(map(int,input().split())) for _ in range(N)]
maxA = max([max(A[i]) for i in range(N)])
G = nx.DiGraph()
G.add_node('source',demand=-K*N)
G.add_node('sink',demand=K*N)
for i in range(N):
for j in range(N):
G.add_edge(i,N+j,weight=maxA-A[i][j],capacity=1)
for i in range(N):
G.add_edge('source',i,weight=0,capacity=K)
G.add_edge(N+i,'sink',weight=0,capacity=K)
flowCost,flowDict = nx.network_simplex(G)
ans=[['.']*N for _ in range(N)]
for i in flowDict.keys():
if i in ('source','sink'):continue
for j in flowDict[i].keys():
if j in ('source','sink'):continue
if flowDict[i][j]:
ans[i][j-N] = 'x'
print(maxA*K*N-flowCost)
for i in range(N):
print(''.join(ans[i])) | s688537700 | Accepted | 465 | 54,964 | 815 | import networkx as nx
N,K=map(int,input().split())
A=[list(map(int,input().split())) for _ in range(N)]
maxA = max([max(A[i]) for i in range(N)])
G = nx.DiGraph()
G.add_node('source',demand=-K*N)
G.add_node('sink',demand=K*N)
G.add_edge('source','sink',weight=maxA, capacity=K*N)
for i in range(N):
for j in range(N):
G.add_edge(i,N+j,weight=maxA-A[i][j],capacity=1)
for i in range(N):
G.add_edge('source',i,weight=0,capacity=K)
G.add_edge(N+i,'sink',weight=0,capacity=K)
flowCost,flowDict = nx.network_simplex(G)
ans=[['.']*N for _ in range(N)]
for i in flowDict.keys():
if i in ('source','sink'):continue
for j in flowDict[i].keys():
if j in ('source','sink'):continue
if flowDict[i][j]:
ans[i][j-N] = 'X'
print(maxA*K*N-flowCost)
for i in range(N):
print(''.join(ans[i])) |
s621676340 | p02259 | u722558010 | 1,000 | 131,072 | Wrong Answer | 20 | 7,636 | 229 | Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode. | N=int(input())
A=list(map(int,input().split()))
count=0
i=N-1
while i>=0:
j=0
while j<=i:
if A[N-1-j]<A[N-2-j]:
(A[N-1-j],A[N-2-j])=(A[N-2-j],A[N-1-j])
count+=1
j+=1
i-=1
print(' '.join(map(str,A)))
print(count) | s184698479 | Accepted | 40 | 7,704 | 259 | N=int(input())
A=list(map(int,input().split()))
count=0
i=N-1
while i>0:
j=0
while j<i:
if A[N-1-j]<A[N-2-j]:
(A[N-1-j],A[N-2-j])=(A[N-2-j],A[N-1-j])
count+=1
#print(' '.join(map(str,A)))
j+=1
i-=1
print(' '.join(map(str,A)))
print(count) |
s716122991 | p03695 | u970809473 | 2,000 | 262,144 | Wrong Answer | 19 | 3,064 | 601 | In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users. | n = int(input())
a = list(map(int, input().split()))
color = [0] * 9
for i in range(n):
if a[i] < 400:
color[0] = 1
elif a[i] < 800:
color[1] = 1
elif a[i] < 1200:
color[2] = 1
elif a[i] < 1600:
color[3] = 1
elif a[i] < 2000:
color[4] = 1
elif a[i] < 2400:
color[5] = 1
elif a[i] < 2800:
color[6] = 1
elif a[i] < 3200:
color[7] = 1
else:
color[8] = color[8] + 1
if color[8] >= 1:
while int(0) in color and color[8] != 0:
color.remove(0)
color.insert(0,1)
color[8] = color[8] - 1
print(color.count(1))
else:
print(color.count(1)) | s046929043 | Accepted | 17 | 3,064 | 513 | n = int(input())
a = list(map(int, input().split()))
color = [0] * 8
tmp = 0
for i in range(n):
if a[i] < 400:
color[0] = 1
elif a[i] < 800:
color[1] = 1
elif a[i] < 1200:
color[2] = 1
elif a[i] < 1600:
color[3] = 1
elif a[i] < 2000:
color[4] = 1
elif a[i] < 2400:
color[5] = 1
elif a[i] < 2800:
color[6] = 1
elif a[i] < 3200:
color[7] = 1
else:
tmp = tmp + 1
l = color.count(1)
if l == 0:
print("1 {}".format(tmp))
else:
print("{} {}".format(l, l + tmp)) |
s305867019 | p03456 | u548545174 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 198 | AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number. | a, b = input().split()
target = a + b
heihousuu = [i ** 2 for i in range(1000) if i ** 2 <= 100100]
for h in heihousuu:
if target == h:
print('Yes')
break
else:
print('No') | s095349751 | Accepted | 17 | 3,060 | 109 | a, b = input().split()
ab = int(a + b)
if (ab ** 0.5).is_integer():
print("Yes")
else:
print("No") |
s176051638 | p02613 | u015019251 | 2,000 | 1,048,576 | Wrong Answer | 146 | 16,180 | 393 | 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. | # -*- coding: utf-8 -*-
"""
Created on Sat Aug 15 17:00:46 2020
@author: saito
"""
# %% import phase
# %% input phase
N = int(input())
S = [0]*N
for i in range(N):
S[i] = input()
# %% process phase
C = [0]*4
rslt = ['AC', 'WA', 'TLE', 'RE']
for i in range(4):
C[i] = S.count(rslt[i])
# %%output phase
for i in range(4):
print(rslt[i]+' '+'×'+' '+str(C[i])) | s725735092 | Accepted | 144 | 16,184 | 392 | # -*- coding: utf-8 -*-
"""
Created on Sat Aug 15 17:00:46 2020
@author: saito
"""
# %% import phase
# %% input phase
N = int(input())
S = [0]*N
for i in range(N):
S[i] = input()
# %% process phase
C = [0]*4
rslt = ['AC', 'WA', 'TLE', 'RE']
for i in range(4):
C[i] = S.count(rslt[i])
# %%output phase
for i in range(4):
print(rslt[i]+' '+'x'+' '+str(C[i])) |
s957752753 | p03338 | u102242691 | 2,000 | 1,048,576 | Wrong Answer | 22 | 3,064 | 280 | You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position. |
n = int(input())
s = list(input())
x = []
y = []
ans = []
for i in range(1,n):
x = s[0:i]
y = s[i:n]
print(x)
print(y)
for j in range(len(x)):
a = 0
if y.count(x[j]) >= 1:
a += 1
ans.append(a)
print(ans)
print(max(ans))
| s845017467 | Accepted | 18 | 3,060 | 253 | n = int(input())
s = input()
ans = 0
for i in range(1, n):
head = s[:i]
tail = s[i:]
cnt = 0
for i in range(26):
c = chr(ord('a') + i)
if c in head and c in tail:
cnt += 1
ans = max(ans, cnt)
print(ans) |
s104399106 | p02606 | u081075058 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,020 | 67 | How many multiples of d are there among the integers between L and R (inclusive)? | L,R,d = map(int,input().split())
r = R//4
l = (L-1)//4
print(r-l) | s706177420 | Accepted | 24 | 9,168 | 67 | L,R,d = map(int,input().split())
r = R//d
l = (L-1)//d
print(r-l) |
s320363711 | p02678 | u285257696 | 2,000 | 1,048,576 | Wrong Answer | 2,206 | 35,176 | 674 | There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists. | from collections import deque
N, M = map(int, input().rstrip().split())
G = [[] for i in range(N+1)]
for i in range(M):
A, B = map(int, input().rstrip().split())
G[A].append(B)
G[B].append(A)
sign = [-1]*(N+1)
sign[1] = 0
queue = deque([1])
while queue:
node = queue.popleft()
for next in G[node]:
if sign[next] > -1:
continue
sign[next] = node
queue.append(next)
ans = "Yes"
print(sign)
for i in range(2, N+1):
S = sign[i]
while S > 0:
S = sign[S]
if not S == 0:
ans = "No"
print(ans)
if ans == "Yes":
for i in range(2, N+1):
print(sign[i])
| s322541348 | Accepted | 622 | 34,168 | 572 | from collections import deque
N, M = map(int, input().rstrip().split())
G = [[] for i in range(N+1)]
for i in range(M):
A, B = map(int, input().rstrip().split())
G[A].append(B)
G[B].append(A)
sign = [-1]*(N+1)
sign[0] = None
sign[1] = 0
queue = deque([1])
while queue:
node = queue.popleft()
for next in G[node]:
if sign[next] > -1:
continue
sign[next] = node
queue.append(next)
if -1 in sign:
print("No")
else:
print("Yes")
for i in range(2, N+1):
print(sign[i])
|
s511064972 | p00015 | u957021485 | 1,000 | 131,072 | Wrong Answer | 30 | 7,692 | 881 | A country has a budget of more than 81 trillion yen. We want to process such data, but conventional integer type which uses signed 32 bit can represent up to 2,147,483,647. Your task is to write a program which reads two integers (more than or equal to zero), and prints a sum of these integers. If given integers or the sum have more than 80 digits, print "overflow". | import sys
import itertools
dataset_num = int(input())
class Overflow(Exception):
pass
for _ in range(dataset_num):
try:
k1 = input()
k2 = input()
k1_val = [0] * 80
k2_val = [0] * 80
if len(k1) >= 80 or len(k2) >= 80:
raise Overflow()
for i, x in enumerate(reversed(k1)):
k1_val[i] = int(x)
for i, x in enumerate(reversed(k2)):
k2_val[i] = int(x)
res = [0] * 80
for i in range(0, 80):
res[i] += k1_val[i] + k2_val[i]
if res[i] >= 10:
res[i+1] += 1
res[i] -= 10
sumstr = "".join([str(x) for x in itertools.dropwhile(lambda x: x == 0, reversed(res))])
if len(sumstr) >= 80:
raise Overflow()
print(sumstr)
except Overflow:
print("overflow") | s434659223 | Accepted | 30 | 7,656 | 937 | import sys
import itertools
dataset_num = int(input())
class Overflow(Exception):
pass
for _ in range(dataset_num):
try:
k1 = input()
k2 = input()
k1_val = [0] * 100
k2_val = [0] * 100
if len(k1) > 80 or len(k2) > 80:
raise Overflow()
for i, x in enumerate(reversed(k1)):
k1_val[i] = int(x)
for i, x in enumerate(reversed(k2)):
k2_val[i] = int(x)
res = [0] * 100
for i in range(0, 100):
res[i] += k1_val[i] + k2_val[i]
if res[i] >= 10:
res[i+1] += 1
res[i] -= 10
sumstr = "".join([str(x) for x in itertools.dropwhile(lambda x: x == 0, reversed(res))])
if len(sumstr) > 80:
raise Overflow()
if len(sumstr) == 0:
sumstr = "0"
print(sumstr)
except Overflow:
print("overflow") |
s414843703 | p02694 | u030856690 | 2,000 | 1,048,576 | Wrong Answer | 23 | 9,172 | 279 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time? | # -*- coding: utf-8 -*-
"""
Created on Sat May 2 19:45:18 2020
@author: harsh
"""
#%%
import math
x=int(input())
rate=100
i=1
while(True):
if(rate>=x):
print(i)
break
else:
rate=rate*1.01
i+=1
#print((math.log(x/100))/(math.log(1.01)))
| s392216002 | Accepted | 25 | 9,096 | 286 | # -*- coding: utf-8 -*-
"""
Created on Sat May 2 19:45:18 2020
@author: harsh
"""
#%%
import math
x=int(input())
rate=100
i=1
while(True):
if(rate>=x):
print(i-1)
break
else:
rate=int(rate*1.01)
i+=1
#print((math.log(x/100))/(math.log(1.01)))
|
s334475941 | p03698 | u277104886 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 93 | You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different. | s = input()
for i in s:
if s.count(i) != 1:
print('No')
exit
print('Yes') | s739672933 | Accepted | 18 | 2,940 | 95 | s = input()
for i in s:
if s.count(i) != 1:
print('no')
exit()
print('yes') |
s627637287 | p03719 | u550574002 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 59 | You are given three integers A, B and C. Determine whether C is not less than A and not greater than B. | a,b,c=map(int,input().split())
print(("NO","YES")[a<=c<=b]) | s217251817 | Accepted | 17 | 2,940 | 59 | a,b,c=map(int,input().split())
print(("No","Yes")[a<=c<=b]) |
s212968048 | p04043 | u464912173 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 132 | Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order. | a, b, c = map(int, input().split())
s = sorted([a, b, c])
print('Yes' if s[0] ==5 and s[1] == 5 and s[2]== 7 else 'No') | s043573551 | Accepted | 17 | 2,940 | 85 | a,b,c = sorted(map(int, input().split()))
print("YES" if a==b==5 and c ==7 else "NO") |
s233569252 | p03302 | u583276018 | 2,000 | 1,048,576 | Wrong Answer | 30 | 8,968 | 97 | You are given two integers a and b. Determine if a+b=15 or a\times b=15 or neither holds. Note that a+b=15 and a\times b=15 do not hold at the same time. | a, b = map(int, input().split())
if(a + b == 15 or a * b == 15):
print("*")
else:
print("x") | s589904855 | Accepted | 32 | 9,028 | 115 | a, b = map(int, input().split())
if(a + b == 15):
print("+")
elif(a * b == 15):
print("*")
else:
print("x")
|
s482414637 | p03720 | u652081898 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 214 | 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())
bridge = [list(map(int, input().split())) for _ in range(m)]
for i in range(n):
ans = 0
for x, y in bridge:
if x == i or y == i:
ans += 1
print(ans) | s337072818 | Accepted | 18 | 2,940 | 220 | n, m = map(int, input().split())
bridge = [list(map(int, input().split())) for _ in range(m)]
for i in range(1, n+1):
ans = 0
for x, y in bridge:
if x == i or y == i:
ans += 1
print(ans)
|
s146594177 | p03695 | u844789719 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 242 | In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users. | N = input()
a = [int(i) for i in input().split()]
scores = [0,0,0,0,0,0,0,0]
score8 = 0
for i in a:
checknum = int(i/400)
if checknum > 7:
score8 += 1
else:
scores[checknum] = 1
print(min([sum(scores) + score8,8])) | s809865057 | Accepted | 18 | 2,940 | 214 | N = int(input())
A = [int(_) for _ in input().split()]
colors = [0] * 9
for a in A:
c = a // 400
if c < 8:
colors[c] = 1
else:
colors[8] += 1
print(max(sum(colors[:8]), 1), sum(colors))
|
s740410065 | p03502 | u218843509 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 97 | An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number. | n = int(input())
t = sum(map(int, list(str(n))))
if n // t == 0:
print("Yes")
else:
print("No") | s107592426 | Accepted | 18 | 2,940 | 96 | n = int(input())
t = sum(map(int, list(str(n))))
if n % t == 0:
print("Yes")
else:
print("No") |
s104463653 | p03433 | u379424722 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 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())
M = N % 500
if M < A:
print("YES")
else:
print("NO") | s552009360 | Accepted | 17 | 2,940 | 96 | N = int(input())
A = int(input())
M = N % 500
if M <= A:
print("Yes")
else:
print("No") |
s533840023 | p02255 | u209940149 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 203 | Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step. | n = int(input())
A = list(map(int, input().split()))
for i in range(1, n):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j + 1] = A[j]
j -= 1
A[j + 1] = v
print(A)
| s196350000 | Accepted | 20 | 5,984 | 214 | n = int(input())
A = list(map(int, input().split()))
print(*A)
for i in range(1, n):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j + 1] = A[j]
j -= 1
A[j + 1] = v
print(*A)
|
s799429618 | p03486 | u870518235 | 2,000 | 262,144 | Wrong Answer | 26 | 9,072 | 701 | You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order. | # B - Two Anagrams
s = list(input())
t = list(input())
s = sorted(s)
s = "".join(s)
t = sorted(t, reverse=True)
t = "".join(t)
print(s,t)
ans = ""
N = min(len(s),len(t))
for i in range(N):
print(ord(s[i]), ord(t[i]))
if ord(s[i]) == ord(t[i]):
pass
elif ord(s[i]) < ord(t[i]):
ans = "Yes"
break
else:
ans = "No"
break
else:
if len(s) < len(t):
ans = "Yes"
else:
ans = "No"
print(ans)
| s924634682 | Accepted | 27 | 9,096 | 657 | # B - Two Anagrams
s = list(input())
t = list(input())
s = sorted(s)
s = "".join(s)
t = sorted(t, reverse=True)
t = "".join(t)
ans = ""
N = min(len(s),len(t))
for i in range(N):
if ord(s[i]) == ord(t[i]):
pass
elif ord(s[i]) < ord(t[i]):
ans = "Yes"
break
else:
ans = "No"
break
else:
if len(s) < len(t):
ans = "Yes"
else:
ans = "No"
print(ans)
|
s029844170 | p03720 | u117193815 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 212 | 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())
l=[list(map(int ,input().split())) for x in range(m)]
count=0
for i in range(1,m+1):
for j in range(m):
if i in l[j]:
count+=1
print(count)
count=0 | s403644698 | Accepted | 17 | 3,060 | 217 | n,m = map(int ,input().split())
l=[list(map(int ,input().split())) for x in range(m)]
count=0
for i in range(1,n+1):
for j in range(m):
if i in l[j]:
count+=1
print(count)
count=0 |
s298837505 | p00015 | u747479790 | 1,000 | 131,072 | Wrong Answer | 20 | 7,568 | 113 | A country has a budget of more than 81 trillion yen. We want to process such data, but conventional integer type which uses signed 32 bit can represent up to 2,147,483,647. Your task is to write a program which reads two integers (more than or equal to zero), and prints a sum of these integers. If given integers or the sum have more than 80 digits, print "overflow". | n=int(input())
for i in range(n):
a=input()
if a==1:
break
b=input()
print(str(a)+str(b)) | s882422220 | Accepted | 30 | 7,572 | 167 | n = int(input())
for i in range(n):
a = int(input())
b = int(input())
if (a + b) < (10 ** 80):
print(a + b)
else:
print("overflow") |
s040515409 | p03836 | u780962115 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 163 | Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him. | sx,sy,tx,ty=map(int,input().split())
string="U"*(ty-sy)+"R"*(tx-sx)+"L"*(tx-sx+1)+"U"*(ty-sy+1)+"R"*(tx-sx+1)+"D"+"R"+"D"*(ty-sy+1)+"L"*(tx-sx+1)+"U"
print(string) | s610125846 | Accepted | 17 | 3,060 | 175 | sx,sy,tx,ty=map(int,input().split())
string="U"*(ty-sy)+"R"*(tx-sx)+"D"*(ty-sy)+"L"*(tx-sx+1)+"U"*(ty-sy+1)+"R"*(tx-sx+1)+"D"+"R"+"D"*(ty-sy+1)+"L"*(tx-sx+1)+"U"
print(string) |
s546111829 | p03478 | u256801986 | 2,000 | 262,144 | Wrong Answer | 768 | 3,472 | 387 | 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). | def DigitsSUM(Num):
sumN = 0
while Num > 0:
sumN += Num % 10
Num/=10
return int(sumN)
N,A,B = input().split()
LowerLimit=int(A)
UpperLimit=int(B)
num = int(N)
sumary = []
counter = 0
for i in range(1,num+1):
sumN = DigitsSUM(i)
if LowerLimit <= sumN and sumN <= UpperLimit+1:
print(i, sumN)
counter+=int(i)
print(counter)
| s588062693 | Accepted | 887 | 3,064 | 371 | def DigitsSUM(Num):
sumN = 0
while Num > 0:
sumN += int(Num) % 10
Num /= 10
return int(sumN)
N,A,B = input().split()
LowerLimit=int(A)
UpperLimit=int(B)
num = int(N)
sumary = []
counter = 0
for i in range(1,num+1):
sumN = DigitsSUM(i)
if LowerLimit <= sumN and sumN <= UpperLimit:
counter+=int(i)
print(counter) |
s138067743 | p03813 | u614875193 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 110 | Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. | X=int(input())
ans=X//11
if X%11==0:
print(ans*2)
elif X%11<7:
print(ans*2+1)
else:
print(ans*2+2) | s268213519 | Accepted | 17 | 2,940 | 38 | print('A'+'RB'[int(input())<1200]+'C') |
s882619423 | p02420 | u283452598 | 1,000 | 131,072 | Wrong Answer | 20 | 7,504 | 261 | Your task is to shuffle a deck of n cards, each of which is marked by a alphabetical letter. A single shuffle action takes out h cards from the bottom of the deck and moves them to the top of the deck. The deck of cards is represented by a string as follows. abcdeefab The first character and the last character correspond to the card located at the bottom of the deck and the card on the top of the deck respectively. For example, a shuffle with h = 4 to the above deck, moves the first 4 characters "abcd" to the end of the remaining characters "eefab", and generates the following deck: eefababcd You can repeat such shuffle operations. Write a program which reads a deck (a string) and a sequence of h, and prints the final state (a string). | moto=""
while True:
x = input()
if x == "-":
print(moto)
break
try:
num = int(x)
cop = moto[:num]
moto = moto[num:]
moto = moto+cop
except:
if moto:
print(moto)
moto = x | s805530872 | Accepted | 20 | 7,592 | 171 | while True:
s = input()
if s == '-':
break
m = int(input())
for mi in range(m):
h = int(input())
s = s[h:] + s[:h]
print(s) |
s950927309 | p03387 | u363825867 | 2,000 | 262,144 | Wrong Answer | 28 | 9,032 | 174 | You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations. | A, B, C = sorted(list(map(int, input().split())))
print(A, B, C)
r = C-B
B += r
A += r
print(A, B, C)
s = C-A
if s % 2 == 0:
s = s//2
else:
s = (s//2) + 2
print(r+s)
| s521678172 | Accepted | 28 | 9,112 | 144 | A, B, C = sorted(list(map(int, input().split())))
r = C-B
B += r
A += r
s = C-A
if s % 2 == 0:
s = s//2
else:
s = (s//2) + 2
print(r+s)
|
s672121391 | p03471 | u145415974 | 2,000 | 262,144 | Wrong Answer | 2,104 | 3,060 | 286 | The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough. | N,Y = (int(x) for x in input().split())
for a in range(N+1):
for b in range(N+1):
for c in range(N+1):
if Y==10000*a+5000*b+1000*c and N==a+b+c:
ans = [a,b,c]
break
else:
ans = [-1,-1,-1]
print(*ans)
| s729974132 | Accepted | 779 | 3,060 | 212 | N,Y = map(int, input().split())
ans = [-1,-1,-1]
for a in range(N+1):
for b in range(N+1-a):
c = N-a-b
if Y == 10000*a+5000*b+1000*c:
ans = [a,b,c]
break
print(*ans)
|
s351535308 | p02743 | u995861601 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 115 | Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold? | import math
a, b, c = map(int, input().split())
c -= a+b
if c > math.sqrt(a*b):
print("Yes")
else:
print("No") | s791075045 | Accepted | 19 | 3,064 | 109 | a, b, c = map(int, input().split())
c -= a+b
if c >= 0 and a*b*4 < c**2:
print("Yes")
else:
print("No") |
s034055640 | p03110 | u813286880 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 236 | Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total? | N = int(input())
x = []
u = []
for _ in range(N):
inp = input().split(' ')
if inp[1] == 'JPN':
x.append(int(inp[0]))
else:
x.append(float(inp[0]))
u.append(inp[1])
sum = 0
for i in range(N):
sum += x[i]
print(sum)
| s122738114 | Accepted | 20 | 2,940 | 223 | N = int(input())
x = []
for _ in range(N):
inp = input().split(' ')
if inp[1] == 'JPY':
x.append(int(inp[0]))
else:
x.append(float(inp[0]) * 380000.0)
sum = 0
for i in range(N):
sum += x[i]
print(sum)
|
s341959394 | p03998 | u706414019 | 2,000 | 262,144 | Wrong Answer | 31 | 8,944 | 323 | Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game. | Sa = list(input())
Sb = list(input())
Sc = list(input())
s = Sa.pop(0)
while True:
if Sa==[]or Sb==[] or Sc ==[]:
break
if s == 'a':
s = Sa.pop(0)
elif s == 'b':
s = Sb.pop(0)
else:
s = Sc.pop(0)
if Sa ==[]:
print('A')
elif Sb == []:
print('B')
else:
print('C')
| s151341155 | Accepted | 29 | 8,848 | 379 | Sa = list(input())
Sb = list(input())
Sc = list(input())
s = Sa.pop(0)
while True:
if s == 'a':
if Sa ==[]:
print('A')
break
s = Sa.pop(0)
elif s == 'b':
if Sb ==[]:
print('B')
break
s = Sb.pop(0)
else:
if Sc ==[]:
print('C')
break
s = Sc.pop(0)
|
s010826015 | p03494 | u516361212 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 150 | There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform. | n = input()
list_a = list(map(int,input().split()))
ans = 0
while all(a % 2 == 0 for a in list_a):
list_a = [a / 2 for a in list_a]
ans += 1 | s188230687 | Accepted | 19 | 3,060 | 162 | n = input()
list_a = list(map(int,input().split()))
ans = 0
while all(a % 2 == 0 for a in list_a):
list_a = [a / 2 for a in list_a]
ans += 1
print(ans) |
s429726976 | p00137 | u546285759 | 1,000 | 131,072 | Wrong Answer | 30 | 7,688 | 176 | 古典的な乱数生成方法の一つである平方採中法のプログラムを作成します。平方採中法は、フォンノイマンによって 1940 年代半ばに提案された方法です。 平方採中法は、生成する乱数の桁数を n としたとき、初期値 s の2乗を計算し、その数値を 2n 桁の数値とみて、（下の例のように 2 乗した桁数が足りないときは、0 を補います。）その中央にある n 個の数字を最初の乱数とします。次にこの乱数を 2 乗して、同じ様に、中央にある n 個の数字をとって、次の乱数とします。例えば、123 を初期値とすると 1232 = 00015129 → 0151 1512 = 00022801 → 0228 2282 = 00051984 → 0519 5192 = 00269361 → 2693 26932 = 07252249 → 2522 の様になります。この方法を用いて、初期値 s（10000未満の正の整数）を入力とし、n = 4 の場合の乱数を 10 個生成し出力するプログラムを作成してください。 | n = int(input())
for i in range(n):
s = int(input())
print("Case {}".format(i+1))
for _ in range(10):
s = int("{:08d}".format(s ** 2)[2:7])
print(s) | s577556772 | Accepted | 30 | 7,524 | 177 | n = int(input())
for i in range(n):
s = int(input())
print("Case {}:".format(i+1))
for _ in range(10):
s = int("{:08d}".format(s ** 2)[2:6])
print(s) |
s359498597 | p02265 | u848218390 | 1,000 | 131,072 | Wrong Answer | 20 | 5,604 | 564 | Your task is to implement a double linked list. Write a program which performs the following operations: * insert x: insert an element with key x into the front of the list. * delete x: delete the first element which has the key of x from the list. If there is not such element, you need not do anything. * deleteFirst: delete the first element from the list. * deleteLast: delete the last element from the list. | n = int(input("command number:"))
L = []
for i in range(n):
com = input()
if com == "deleteFirst":
del L[0]
elif com == "deleteLast":
del L[len(L)-1]
else:
c = com.split()
if c[0] == "insert":
L.insert(0, int(c[1]))
elif c[0] == "delete":
for i in range(len(L)):
if L[i] == int(c[1]):
del L[i]
break
print(L)
for i in range(len(L)):
if i == len(L)-1:
print(L[i])
else:
print(L[i], end=' ')
| s344197370 | Accepted | 4,950 | 45,884 | 450 | from collections import deque
n = int(input())
L = deque()
for i in range(n):
com = input()
if com[0] == 'i':
C = com.split()
L.appendleft(int(C[1]))
else:
if com[6] == 'F':
L.popleft()
elif com[6] == 'L':
L.pop()
else:
C = com.split()
x = int(C[1])
try:
L.remove(x)
except:
pass
print(*L)
|
s650061550 | p03997 | u089142196 | 2,000 | 262,144 | Wrong Answer | 19 | 2,940 | 62 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | a=int(input())
b=int(input())
h=int(input())
print((a+b)*h/2) | s023105330 | Accepted | 16 | 2,940 | 63 | a=int(input())
b=int(input())
h=int(input())
print((a+b)*h//2) |
s542937689 | p03730 | u445511055 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 463 | We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`. | # -*- coding: utf-8 -*-
def main():
"""Function."""
a, b, c = map(int, input().split())
check = set()
temp = a
flag = 0
while temp % b not in check:
print(check)
print(temp)
if c in check:
flag = 1
break
else:
check.add(temp % b)
temp = temp + a
if flag == 1:
print("YES")
else:
print("NO")
if __name__ == "__main__":
main()
| s622632905 | Accepted | 18 | 3,060 | 422 | # -*- coding: utf-8 -*-
def main():
"""Function."""
a, b, c = map(int, input().split())
check = set()
temp = a
flag = 0
while temp % b not in check:
if c in check:
flag = 1
break
else:
check.add(temp % b)
temp = temp + a
if flag == 1:
print("YES")
else:
print("NO")
if __name__ == "__main__":
main()
|
s268657294 | p03679 | u580697892 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 83 | 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())
print("delicious" if a+b <= x else "dangerous") | s158925634 | Accepted | 17 | 2,940 | 158 | # coding: utf-8
X, A, B = map(int, input().split())
if A - B >= 0:
print("delicious")
elif A + X - B >= 0:
print("safe")
else:
print("dangerous")
|
s947557310 | p03214 | u408375121 | 2,525 | 1,048,576 | Wrong Answer | 17 | 3,060 | 163 | Niwango-kun is an employee of Dwango Co., Ltd. One day, he is asked to generate a thumbnail from a video a user submitted. To generate a thumbnail, he needs to select a frame of the video according to the following procedure: * Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video. * Select t-th frame whose representation a_t is nearest to the average of all frame representations. * If there are multiple such frames, select the frame with the smallest index. Find the index t of the frame he should select to generate a thumbnail. | n = int(input())
a = list(map(int, input().split()))
mean = sum(a) / n
INF = 10**5
flg = INF
for i in range(n):
if abs(mean - a[i]) < flg:
idx = i
print(idx) | s736970747 | Accepted | 17 | 3,060 | 191 | n = int(input())
a = list(map(int, input().split()))
mean = sum(a) / n
INF = 10**5
flg = INF
for i in range(n):
if abs(mean - a[i]) < flg:
idx = i
flg = abs(mean - a[i])
print(idx)
|
s937622471 | p03455 | u530169312 | 2,000 | 262,144 | Wrong Answer | 30 | 9,104 | 182 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. |
a,b=map(int,input().split())
c = a*b
if c / 2 ==0:
print('Even')
else:
print('Odd') | s661706977 | Accepted | 24 | 8,920 | 182 |
a,b=map(int,input().split())
c = a*b
if c % 2 ==0:
print('Even')
else:
print('Odd') |
s810045787 | p03545 | u351241495 | 2,000 | 262,144 | Wrong Answer | 19 | 3,444 | 592 | 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. | s=input()
listA=[0,0,0,0]
i=0
for x in s:
listA[i]=int(x)
i+=1
a=listA[0]
b=listA[1]
c=listA[2]
d=listA[3]
if(a+b+c+d==7):
print("%d+%d+%d+%d==7"% (a,b,c,d))
elif(a+b+c-d==7):
print("%d+%d+%d-%d==7"% (a,b,c,d))
elif(a+b-c+d==7):
print("%d+%d-%d+%d==7" % (a, b, c, d))
elif(a+b-c-d==7):
print("%d+%d-%d-%d==7" % (a, b, c, d))
elif(a-b+c+d==7):
print("%d-%d+%d+%d==7" % (a, b, c, d))
elif(a-b-c+d==7):
print("%d-%d-%d+%d==7" % (a, b, c, d))
elif(a-b-c-d==7):
print("%d-%d-%d-%d==7" % (a, b, c, d))
elif(a-b+c-d==7):
print("%d-%d+%d-%d==7" % (a, b, c, d)) | s355859736 | Accepted | 19 | 3,064 | 584 | s=input()
listA=[0,0,0,0]
i=0
for x in s:
listA[i]=int(x)
i+=1
a=listA[0]
b=listA[1]
c=listA[2]
d=listA[3]
if(a+b+c+d==7):
print("%d+%d+%d+%d=7"% (a,b,c,d))
elif(a+b+c-d==7):
print("%d+%d+%d-%d=7"% (a,b,c,d))
elif(a+b-c+d==7):
print("%d+%d-%d+%d=7" % (a, b, c, d))
elif(a+b-c-d==7):
print("%d+%d-%d-%d=7" % (a, b, c, d))
elif(a-b+c+d==7):
print("%d-%d+%d+%d=7" % (a, b, c, d))
elif(a-b-c+d==7):
print("%d-%d-%d+%d=7" % (a, b, c, d))
elif(a-b-c-d==7):
print("%d-%d-%d-%d=7" % (a, b, c, d))
elif(a-b+c-d==7):
print("%d-%d+%d-%d=7" % (a, b, c, d)) |
s267017347 | p03636 | u633548583 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 51 | 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=str(input())
a=len(s[1:-1])
print(s[0]+'a'+s[-1]) | s229001292 | Accepted | 17 | 2,940 | 37 | a,*b,c=input()
print(a+str(len(b))+c) |
s920095666 | p03377 | u816116805 | 2,000 | 262,144 | Wrong Answer | 20 | 3,316 | 90 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | a,b,x = map(int,input().split())
if a <= x <= a+b:
print("Yes")
else:
print("No")
| s914528780 | Accepted | 17 | 2,940 | 90 | a,b,x = map(int,input().split())
if a <= x <= a+b:
print("YES")
else:
print("NO")
|
s143879592 | p02393 | u917722865 | 1,000 | 131,072 | Time Limit Exceeded | 9,990 | 5,584 | 354 | Write a program which reads three integers, and prints them in ascending order. | abc_s = input()
abc_a = abc_s.split(" ")
abc = [int(x) for x in abc_a]
sorted = False
while sorted == False:
sorted == True
if abc[0] > abc[1]:
sorted == False
tmp = abc[0]
abc[0] = abc[1]
abc[1] = tmp
if abc[1] > abc[2]:
sorted == False
tmp = abc[1]
abc[1] = abc[2]
abc[2] = tmp
print("{0} {1} {2}".format(abc))
| s685497509 | Accepted | 20 | 5,600 | 363 | abc_s = input()
abc_a = abc_s.split(" ")
abc = [int(x) for x in abc_a]
sorted = False
while sorted == False:
sorted = True
if abc[0] > abc[1]:
sorted = False
tmp = abc[0]
abc[0] = abc[1]
abc[1] = tmp
if abc[1] > abc[2]:
sorted = False
tmp = abc[1]
abc[1] = abc[2]
abc[2] = tmp
print("{0[0]} {0[1]} {0[2]}".format(abc))
|
s578394132 | p03694 | u767995501 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 52 | It is only six months until Christmas, and AtCoDeer the reindeer is now planning his travel to deliver gifts. There are N houses along _TopCoDeer street_. The i-th house is located at coordinate a_i. He has decided to deliver gifts to all these houses. Find the minimum distance to be traveled when AtCoDeer can start and end his travel at any positions. | l=sorted(map(int,input().split()))
print(l[-1]-l[0]) | s885405855 | Accepted | 20 | 3,064 | 60 | input()
l=sorted(map(int,input().split()))
print(l[-1]-l[0]) |
s394335538 | p03693 | u940780117 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 94 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4? | R,G,B =map(str,input().split())
RGB=R+G+B
if int(RGB) %4 ==0:
print('Yes')
else:
print('No') | s847883205 | Accepted | 17 | 2,940 | 126 | cards=list(map(int,input().split()))
num=cards[0]*100+cards[1]*10+cards[2]
if num%4==0:
print('YES')
else:
print('NO') |
s717436195 | p03694 | u898143674 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 59 | It is only six months until Christmas, and AtCoDeer the reindeer is now planning his travel to deliver gifts. There are N houses along _TopCoDeer street_. The i-th house is located at coordinate a_i. He has decided to deliver gifts to all these houses. Find the minimum distance to be traveled when AtCoDeer can start and end his travel at any positions. | a = list(map(int, input().split()))
print(max(a) - min(a))
| s002927467 | Accepted | 17 | 2,940 | 67 | input()
a = list(map(int, input().split()))
print(max(a) - min(a))
|
s167861068 | p02257 | u339728776 | 1,000 | 131,072 | Wrong Answer | 20 | 5,588 | 288 | A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list. | x = int(input())
count = 0
for i in range(0, x):
a = int(input())
if a % 2 == 0:
count += 1
else:
c =int( a / 2)
for j in range ( 3, c ):
if a % j == 0:
count += 1
break;
print(x-count)
| s052890023 | Accepted | 2,240 | 5,604 | 262 | x = int(input())
count = 0
for i in range(0, x):
a = int(input())
for j in range ( 2, a ):
c = int(a)
if a % j == 0:
count += 1
break;
if j * j > c:
break;
print(x-count)
|
s612535553 | p04043 | u384793271 | 2,000 | 262,144 | Wrong Answer | 27 | 9,020 | 121 | Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order. | abc = list(map(int, input().split()))
if abc.count(5) == 2 and abc.count(7) == 1:
print('Yes')
else:
print('No') | s983835427 | Accepted | 25 | 8,964 | 114 | s = list(map(int, input().split()))
if s.count(5) == 2 and s.count(7) == 1:
print('YES')
else:
print('NO') |
s496274837 | p03607 | u803647747 | 2,000 | 262,144 | Wrong Answer | 211 | 16,268 | 325 | You are playing the following game with Joisino. * Initially, you have a blank sheet of paper. * Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times. * Then, you are asked a question: How many numbers are written on the sheet now? The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game? | N = int(input())
a_list = [int(input())for _ in range(N)]
count_dct = {}
for input_num in a_list:
if input_num in count_dct.keys():
count_dct[input_num] += 1
else:
count_dct[input_num] = 1
counter = 0
for i, j in count_dct.items():
if j%2 == 1:
counter +=1
else:
pass | s334653885 | Accepted | 216 | 16,268 | 345 | N = int(input())
a_list = [int(input())for _ in range(N)]
count_dct = {}
for input_num in a_list:
if input_num in count_dct.keys():
count_dct[input_num] += 1
else:
count_dct[input_num] = 1
counter = 0
for i, j in count_dct.items():
if j%2 == 1:
counter +=1
else:
pass
print(counter) |
s072821591 | p03455 | u972795791 | 2,000 | 262,144 | Wrong Answer | 30 | 8,904 | 89 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a,b=input().split()
a,b=map(int,(a,b))
if a*b%2==0:
print("EVEN")
else :
print("0dd") | s901592068 | Accepted | 27 | 8,956 | 89 | a,b=input().split()
a,b=map(int,(a,b))
if a*b%2==0:
print("Even")
else :
print("Odd") |
s615061573 | p02850 | u865119809 | 2,000 | 1,048,576 | Wrong Answer | 506 | 30,000 | 369 | Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different. Among the colorings satisfying the condition above, construct one that uses the minimum number of colors. | n = int(input())
graph = [[] for i in range(n)]
for i in range(1,n):
a,b = map(int,input().split())
graph[a-1].append([b,i])
ans = [0]*(n-1)
def col(now,color):
c = 1
for [b,j] in graph[now-1]:
if c == color:
c += 1
ans[j-1] = c
col(b,c)
c += 1
print(max(ans))
for i in range(1,n):
print(ans[i-1])
| s579959098 | Accepted | 588 | 70,996 | 420 | import sys
sys.setrecursionlimit(10 ** 7)
n = int(input())
graph = [[] for i in range(n)]
for i in range(1,n):
a,b = map(int,input().split())
graph[a-1].append([b,i])
ans = [0]*(n-1)
def col(now,color):
c = 1
for [b,j] in graph[now-1]:
if c == color:
c += 1
ans[j-1] = c
col(b,c)
c += 1
col(1,0)
print(max(ans))
for i in range(1,n):
print(ans[i-1])
|
s079283157 | p03338 | u811169796 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,068 | 155 | You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position. | n = int(input())
s = list(input())
ans = 0
for i in range(1,n-1):
sa = set(s[:i])
sb = set(s[i+1:])
ans = max(ans,len(sa&sb))
print(ans)
| s719490989 | Accepted | 18 | 3,064 | 147 | n = int(input())
s = list(input())
ans = 0
for i in range(1,n-1):
sa = set(s[:i])
sb = set(s[i:])
ans = max(ans,len(sa&sb))
print(ans) |
s387332872 | p03997 | u894440853 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 73 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | a, b, h = map(int, [input() for _ in range(3)])
print(round((a + b) * h)) | s612712078 | Accepted | 17 | 2,940 | 80 | a, b, h = map(int, [input() for _ in range(3)])
print(round(((a + b) * h) / 2))
|
s425612623 | p03563 | u484856305 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 42 | Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it. | a=int(input())
b=int(input())
print(b-2*a) | s537268580 | Accepted | 17 | 2,940 | 42 | r=int(input())
g=int(input())
print(2*g-r) |
s383616650 | p03385 | u127856129 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 123 | You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`. | a=input()
if a=="abc" and a=="acb" and a=="bac" and a=="bca" and a=="cab" and a=="cba":
print="Yes"
else:
print="No"
| s494994729 | Accepted | 21 | 3,060 | 121 | a=input()
if a=="abc" or a=="acb" or a=="bac" or a=="bca" or a=="cab" or a=="cba":
print("Yes")
else:
print("No")
|
s094345106 | p04029 | u822179469 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 85 | 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())
total = 0
for i in range(n):
total = total + i*i
print(total) | s688828796 | Accepted | 17 | 2,940 | 85 | n = int(input())
total = 0
for i in range(n+1):
total = total + i
print(total) |
s532835880 | p03623 | u392319141 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 70 | Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|. | x, a, b = map(int, input().split())
print(min(abs(x - a), abs(x - b))) | s242577286 | Accepted | 17 | 2,940 | 75 | x, a, b = map(int, input().split())
print('AB'[abs(x - a) > abs(x - b)::2]) |
s531080847 | p03719 | u604262137 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 133 | You are given three integers A, B and C. Determine whether C is not less than A and not greater than B. | table = input().split()
A,B,C = int(table[0]),int(table[1]),int(table[2])
if A <= C and C<= B:
print('yes')
else:
print('no') | s917350854 | Accepted | 17 | 2,940 | 133 | table = input().split()
A,B,C = int(table[0]),int(table[1]),int(table[2])
if A <= C and C<= B:
print('Yes')
else:
print('No') |
s175868019 | p03090 | u668785999 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,356 | 266 | 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):
for i in range(1,N):
for j in range(i+1,N+1):
if(i+j != N+1):
print(*[i,j])
else:
for i in range(1,N):
for j in range(i+1,N+1):
if(i+j != N):
print(*[i,j]) | s621545903 | Accepted | 28 | 9,264 | 291 | N = int(input())
print(N*(N-1)//2 - N//2)
if(N%2 == 0):
for i in range(1,N):
for j in range(i+1,N+1):
if(i+j != N+1):
print(*[i,j])
else:
for i in range(1,N):
for j in range(i+1,N+1):
if(i+j != N):
print(*[i,j]) |
s645596716 | p04012 | u848647227 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 168 | Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful. | a = list(input())
br = []
for ar in a:
br.append(a.count(ar))
x = 0
for b in br:
if b % 2 == 1:
x += 1
if x == 0:
print("YES")
else:
print("NO") | s882686349 | Accepted | 17 | 2,940 | 168 | a = list(input())
br = []
for ar in a:
br.append(a.count(ar))
x = 0
for b in br:
if b % 2 == 1:
x += 1
if x == 0:
print("Yes")
else:
print("No") |
s573791126 | p02259 | u456917014 | 1,000 | 131,072 | Wrong Answer | 20 | 5,600 | 248 | Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode. | N=int(input())
A=list(map(int,input().split()))[::-1]
flag=1
count=0
while flag:
flag=0
for i in range(N-1):
if A[i]<A[i+1]:
A[i],A[i+1]=A[i+1],A[i]
flag=1
count+=1
print(A[::-1],count,sep='\n')
| s760393048 | Accepted | 20 | 5,612 | 247 | N=int(input())
A=list(map(int,input().split()))[::-1]
flag=1
count=0
while flag:
flag=0
for i in range(N-1):
if A[i]<A[i+1]:
A[i],A[i+1]=A[i+1],A[i]
flag=1
count+=1
print(*A[::-1])
print(count)
|
s169913417 | p04039 | u497046426 | 2,000 | 262,144 | Wrong Answer | 75 | 3,064 | 251 | Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K. She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change). However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money. Find the amount of money that she will hand to the cashier. | numbers = {str(i) for i in range(10)}
N, K = map(int, input().split())
*D, = map(str, input().split())
D = set(D)
available = numbers - D
ans = 0
for i in range(N, 100000):
nums = set(str(i))
if nums < available:
print(i)
break | s965720108 | Accepted | 65 | 3,064 | 252 | numbers = {str(i) for i in range(10)}
N, K = map(int, input().split())
*D, = map(str, input().split())
D = set(D)
available = numbers - D
ans = 0
for i in range(N, 100000):
nums = set(str(i))
if nums <= available:
print(i)
break |
s764911014 | p02694 | u392058721 | 2,000 | 1,048,576 | Wrong Answer | 21 | 9,156 | 104 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time? | import math
x = int(input())
d = 100
a = 0
while x >= d:
d += math.floor(d*0.01)
a += 1
print(a) | s235341659 | Accepted | 22 | 9,160 | 104 | import math
x = int(input())
d = 100
a = 0
while x > d:
d += math.floor(d*0.01)
a += 1
print(a) |
s344475480 | p02408 | u986199217 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 321 | Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond. | Suit = ["S","H","C","D"]
deck = []
for i in range(4):
for n in range(1,14):
tmp = (Suit[i]+" "+str(n))
deck.append(tmp)
u = int(input())
for _ in range(u):
s = str(input())
if s in deck:
deck.remove(s)
for y in range(len(deck)):
print('{} {}'.format(deck[y][0],deck[y][2]))
| s657384672 | Accepted | 20 | 5,596 | 317 | Suit = ["S","H","C","D"]
deck = []
for i in range(4):
for n in range(1,14):
tmp = (Suit[i]+" "+str(n))
deck.append(tmp)
#print (deck)
u = int(input())
for _ in range(u):
s = str(input())
if s in deck:
deck.remove(s)
for y in range(len(deck)):
print('{}'.format(deck[y]))
|
s935773329 | p03472 | u500297289 | 2,000 | 262,144 | Wrong Answer | 357 | 11,256 | 345 | You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total? | N, H = map(int, input().split())
A = []
B = []
for _ in range(N):
a, b = map(int, input().split())
A.append(a)
B.append(b)
a = max(A)
B.sort(reverse=True)
ans = 0
for b in B:
if b < a:
break
H -= b
ans += 1
if H <= 0:
print(ans)
exit()
import math
ans += math.floor(H / a)
print(int(ans))
| s539559871 | Accepted | 351 | 11,276 | 344 | N, H = map(int, input().split())
A = []
B = []
for _ in range(N):
a, b = map(int, input().split())
A.append(a)
B.append(b)
a = max(A)
B.sort(reverse=True)
ans = 0
for b in B:
if b < a:
break
H -= b
ans += 1
if H <= 0:
print(ans)
exit()
import math
ans += math.ceil(H / a)
print(int(ans))
|
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