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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s213551943
|
p03730
|
u984664611
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 746
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
a, b, c = map(int, input().split())
flag = False
a_mod = a % b
if a_mod == 0 and c == 0:
flag = True
elif a_mod == 0:
pass
elif c % a_mod == 0:
flag = True
else:
# c % ((a%b*n)%b) == 0
for i in range(1, b-1):
ia_mod = (i*a_mod) % b
if ia_mod == 0:
continue
elif c % ia_mod == 0:
flag = True
break
if flag:
print('Yes')
else:
print('No')
|
s357461401
|
Accepted
| 17
| 3,060
| 746
|
a, b, c = map(int, input().split())
flag = False
a_mod = a % b
if a_mod == 0 and c == 0:
flag = True
elif a_mod == 0:
pass
elif c % a_mod == 0:
flag = True
else:
# c % ((a%b*n)%b) == 0
for i in range(1, b-1):
ia_mod = (i*a_mod) % b
if ia_mod == 0:
continue
elif c % ia_mod == 0:
flag = True
break
if flag:
print('YES')
else:
print('NO')
|
s425386126
|
p02645
|
u730807152
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,020
| 30
|
When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him.
|
a = str(input())
print(a[3:])
|
s457037840
|
Accepted
| 23
| 9,088
| 26
|
a = input()
print(a[0:3])
|
s863321400
|
p02399
|
u369313788
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,712
| 97
|
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
a, b = [int(i) for i in input().split()]
d = a // b
r = a % b
f = a / b
print((d),(r),float(f))
|
s928231224
|
Accepted
| 30
| 7,708
| 114
|
a, b = [int(i) for i in input().split()]
d = a // b
r = a % b
f = a / b
print("{0} {1} {2:.5f}".format(d, r, f))
|
s677300546
|
p03997
|
u059210959
| 2,000
| 262,144
|
Wrong Answer
| 50
| 6,092
| 347
|
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.
|
# encoding:utf-8
import copy
import random
import bisect
import fractions
import math
import sys
mod = 10**9+7
sys.setrecursionlimit(mod)
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2)
|
s516418002
|
Accepted
| 54
| 6,604
| 348
|
# encoding:utf-8
import copy
import random
import bisect
import fractions
import math
import sys
mod = 10**9+7
sys.setrecursionlimit(mod)
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h//2)
|
s713510595
|
p03469
|
u077852398
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,012
| 36
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
a = input()
a.replace('2017','2018')
|
s645823252
|
Accepted
| 32
| 8,980
| 43
|
a = input()
print(a.replace('2017','2018'))
|
s679801005
|
p03644
|
u816070625
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 85
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
N=int(input())
i=0
while 2**i<N:
i+=1
if 2**i==N:
print(i)
else:
print(i-1)
|
s831812123
|
Accepted
| 17
| 2,940
| 93
|
N=int(input())
i=0
while 2**i<N:
i+=1
if 2**i==N:
print(2**i)
else:
print(2**(i-1))
|
s260967876
|
p03712
|
u190525112
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 221
|
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
x = list(map(int, input().split()))
a=[input() for i in range(x[0])]
for i in range(x[1]):
print('#',end="")
print('')
for i in a:
print('#{0}#'.format(i))
for i in range(x[1]):
print('#',end="")
print('')
|
s336169787
|
Accepted
| 18
| 3,060
| 224
|
x = list(map(int, input().split()))
a=[input() for i in range(x[0])]
for i in range(x[1]+2):
print('#',end="")
print('')
for i in a:
print('#{0}#'.format(i))
for i in range(x[1]+2):
print('#',end="")
print('')
|
s084407776
|
p03067
|
u548308904
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 231
|
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.
|
inp = [int(f) for f in input().split()]
if inp[0] > inp[2]:
if inp[1] < inp[2]:
print('Yes')
else:
print('No')
elif inp[0] > inp[2]:
if inp[1] < inp[2]:
print('Yes')
else:
print('No')
|
s972661455
|
Accepted
| 17
| 3,060
| 231
|
inp = [int(f) for f in input().split()]
if inp[0] > inp[2]:
if inp[1] < inp[2]:
print('Yes')
else:
print('No')
elif inp[0] < inp[2]:
if inp[1] > inp[2]:
print('Yes')
else:
print('No')
|
s561324352
|
p02397
|
u602702913
| 1,000
| 131,072
|
Wrong Answer
| 60
| 5,616
| 125
|
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 x<y:
print(x,y)
elif y<x:
print(y,x)
elif x==y==0:
break
|
s480636179
|
Accepted
| 50
| 5,608
| 204
|
while True:
x,y=map(int,input().split())
if x<y:
print(x,y)
continue
elif y<x:
print(y,x)
continue
elif x and y !=0:
print(x,y)
continue
elif x==y==0:
break
|
s930206806
|
p03943
|
u689835643
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 138
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
x, y, z = map(int, input().split())
i = x + y + z
if i - max(x, y, z) < max(x, y, z) or i % 2 != 0:
print('NO')
else:
print('YES')
|
s954740923
|
Accepted
| 17
| 2,940
| 141
|
x, y, z = map(int, input().split())
i = x + y + z
if i - max(x, y, z) == max(x, y, z) and i % 2 == 0:
print('Yes')
else:
print('No')
|
s615715023
|
p03548
|
u178432859
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
x,y,z = map(int, input().split())
print(x//(y+z+1))
|
s342697790
|
Accepted
| 17
| 2,940
| 100
|
x,y,z = map(int, input().split())
if x%(y+z) >= z:
print(x//(y+z))
else:
print((x//(y+z))-1)
|
s453842794
|
p02842
|
u606090886
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 38
|
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
n = int(input())
print((n/108*100)//1)
|
s337992287
|
Accepted
| 17
| 2,940
| 181
|
n = int(input())
flag = False
x = int(n / 108 * 100)
for i in range(x-5,x+5):
if int(i*1.08) == n:
flag = True
x = i
if flag:
print(x)
else:
print(":(")
|
s095296932
|
p03836
|
u733337827
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 176
|
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx, sy, tx, ty = map(int, input().split())
dx = tx-sx
dy = ty-sy
s = 'R'*dx + 'U'*dy
s += 'L'*dx + 'D'*dy
s += 'D' + 'R'*(dx+1) + 'U'*dy
s += 'U' + 'L'*(dx+1) + 'D'*dy
print(s)
|
s664520381
|
Accepted
| 17
| 3,064
| 197
|
sx, sy, tx, ty = map(int, input().split())
dx = tx-sx
dy = ty-sy
s = 'U'*dy + 'R'*dx
s += 'D'*dy + 'L'*dx
s += 'L' + 'U'*(dy+1) + 'R'*(dx+1) + 'D'
s += 'R' + 'D'*(dy+1) + 'L'*(dx+1) + 'U'
print(s)
|
s018708989
|
p02972
|
u821432765
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 12,004
| 259
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
n=int(input())
a=list(map(int, input().split()))
eo=[]
ans=[]
for i in range(n):
k=eo[:i+1:n-i]
if len(k) <= 1: eo.append(a[-(i+1)])
else: eo.append((sum(eo)+a[n-i-1])%2)
if eo[-1]==1: ans.append(i+1)
eo=eo[::-1]
print(eo.count(1))
print(*ans)
|
s315422834
|
Accepted
| 476
| 17,192
| 453
|
from math import ceil
N = int(input())
D = ceil(N/2)
A = [0] + [int(i) for i in input().split()]
boxes = [0] * (N+1) # 1-indexed
for i in range(D, N+1):
boxes[i] = A[i]
for i in range(N-D, 0, -1):
c = 0
for j in range(i+i, N+1, i):
if boxes[j]:
c ^= 1
if c==A[i]:
boxes[i]=0
else:
boxes[i]=1
res = set()
for i in range(1, N+1):
if boxes[i]:
res.add(i)
print(len(res))
print(*res)
|
s538983044
|
p03563
|
u391328897
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 49
|
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.
|
r = int(input())
g = int(input())
print((g-r)/2)
|
s295426363
|
Accepted
| 17
| 2,940
| 49
|
r = int(input())
g = int(input())
print(2*g - r)
|
s869584752
|
p02742
|
u613292673
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 109
|
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
H, W = list(map(int, input().split()))
n = -(-H // 2)
if (W % 2) == 0:
print(n*W)
else:
print(n*W-1)
|
s794627347
|
Accepted
| 17
| 2,940
| 101
|
H, W = list(map(int, input().split()))
if H == 1 or W ==1:
print(1)
else:
print(-(-H*W // 2))
|
s608574513
|
p03623
|
u073139376
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 78
|
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(['B', 'A'][abs(x - a) > abs(x - b)])
|
s823107838
|
Accepted
| 17
| 2,940
| 79
|
x, a, b = map(int, input().split())
print(['A', 'B'][abs(x - a) > abs(x - b)])
|
s241385168
|
p03457
|
u731368968
| 2,000
| 262,144
|
Wrong Answer
| 410
| 3,064
| 365
|
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.
|
def intime(t, dx, dy):
x, y = map(abs, (dx, dy))
if t < 2 and x == 0 and y:
return False
return True if dx + dy < t else False
N = int(input())
bx, by = 0, 0
ans = True
for i in range(N):
t, x, y = map(int, input().split())
if not intime(t, x - bx, y - by):
ans = False
bt, bx, by = t, x, y
print('Yes' if ans else 'No')
|
s291103014
|
Accepted
| 384
| 3,064
| 246
|
n = int(input())
bt,bx,by=0,0,0
ans=True
for i in range(n):
t, x, y = map(int, input().split())
dt=t-bt
l=abs(x-bx)+abs(y-by)
if dt<l:
ans = False
if (dt-l)%2:ans=False
bt,bx,by=t,x,y
print('Yes' if ans else 'No')
|
s871686256
|
p03997
|
u316386814
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 86
|
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())
ans = (a + b) * h / 2
print(ans)
|
s861118114
|
Accepted
| 17
| 2,940
| 87
|
a = int(input())
b = int(input())
h = int(input())
ans = (a + b) * h // 2
print(ans)
|
s235209702
|
p03371
|
u155828990
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 160
|
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
a,b,c,x,y=map(int,input().split())
a=0
c*=2
if(x>y):
a=min((a*x+b*y),((x-y)*a+y*c),c*max(x,y))
else:
a=min((a*x+b*y),((y-x)*a+y*c),c*max(x,y))
print(a)
|
s055105236
|
Accepted
| 19
| 3,060
| 159
|
a,b,c,x,y=map(int,input().split())
s=0
c*=2
if(x>y):
s=min((a*x+b*y),((x-y)*a+y*c),c*max(x,y))
else:
s=min((a*x+b*y),((y-x)*b+x*c),c*max(x,y))
print(s)
|
s641559534
|
p03379
|
u666198201
| 2,000
| 262,144
|
Wrong Answer
| 428
| 25,556
| 273
|
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())
A=list(map(int,input().split()))
B=[]
if N==2:
print(A[1])
print(A[0])
exit(0)
for i in range(N):
B.append(A[i])
B.sort()
print(A)
print(B)
for i in range(N):
if A[i]<B[N//2+1]:
print(B[N//2])
else:
print(B[N//2-1])
|
s821751434
|
Accepted
| 367
| 25,220
| 254
|
N=int(input())
A=list(map(int,input().split()))
B=[]
if N==2:
print(A[1])
print(A[0])
exit(0)
for i in range(N):
B.append(A[i])
B.sort()
for i in range(N):
if A[i]<B[N//2]:
print(B[N//2])
else:
print(B[N//2-1])
|
s613531412
|
p03486
|
u857330600
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 173
|
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s=sorted(str(input()))
t=sorted(str(input()))
t.reverse()
print(s,t)
l=[s,t]
lis=sorted(l)
if s==t:
print('No')
else:
if l==lis:
print('Yes')
else:
print('No')
|
s597542416
|
Accepted
| 17
| 2,940
| 162
|
s=sorted(str(input()))
t=sorted(str(input()))
t.reverse()
l=[s,t]
lis=sorted(l)
if s==t:
print('No')
else:
if l==lis:
print('Yes')
else:
print('No')
|
s725730232
|
p02613
|
u247066121
| 2,000
| 1,048,576
|
Wrong Answer
| 155
| 16,328
| 386
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
S = [input() for i in range(N)]
count_AC, count_WA, count_TLE, count_RE = 0, 0, 0, 0
for s in S:
if s == 'AC':
count_AC+=1
elif s == 'WA':
count_WA+=1
elif s == 'TLE':
count_TLE+=1
else:
count_RE+=1
print('AC × '+str(count_AC))
print('WA × '+str(count_WA))
print('TLE × '+str(count_TLE))
print('RE × '+str(count_RE))
|
s783669932
|
Accepted
| 149
| 16,192
| 382
|
N = int(input())
S = [input() for i in range(N)]
count_AC, count_WA, count_TLE, count_RE = 0, 0, 0, 0
for s in S:
if s == 'AC':
count_AC+=1
elif s == 'WA':
count_WA+=1
elif s == 'TLE':
count_TLE+=1
else:
count_RE+=1
print('AC x '+str(count_AC))
print('WA x '+str(count_WA))
print('TLE x '+str(count_TLE))
print('RE x '+str(count_RE))
|
s618404238
|
p03721
|
u112007848
| 2,000
| 262,144
|
Wrong Answer
| 219
| 27,284
| 190
|
There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3.
|
n, k = map(int, input().split(" "))
a = [list(map(int, input().split(" "))) for i in range(n)]
count = 0
for x, y in a:
count += y
if k >= y:
print(x)
break
else:
print("None")
|
s300312185
|
Accepted
| 396
| 28,404
| 202
|
n, k = map(int, input().split(" "))
a = sorted([list(map(int, input().split(" "))) for i in range(n)])
count = 0
for x, y in a:
count += y
if k <= count:
print(x)
break
else:
print("None")
|
s042155610
|
p03846
|
u013408661
| 2,000
| 262,144
|
Wrong Answer
| 55
| 14,008
| 311
|
There are N people, conveniently numbered 1 through N. They were standing in a row yesterday, but now they are unsure of the order in which they were standing. However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person. According to their reports, the difference above for person i is A_i. Based on these reports, find the number of the possible orders in which they were standing. Since it can be extremely large, print the answer modulo 10^9+7. Note that the reports may be incorrect and thus there may be no consistent order. In such a case, print 0.
|
n=int(input())
a=list(map(int,input().split()))
stack=[0]*(10**5+1)
p=10**9+7
for i in a:
stack[i]+=1
if n%2==0:
for j in range(1,n+1,2):
if stack[j]!=1:
print(0)
exit()
print(pow(2,n//2,p))
exit()
for j in range(0,n+1,2):
if stack[j]!=1:
print(0)
exit()
print(pow(2,n//2,p))
|
s018697770
|
Accepted
| 57
| 14,008
| 365
|
n=int(input())
a=list(map(int,input().split()))
stack=[0]*(10**5+1)
p=10**9+7
for i in a:
stack[i]+=1
if n%2==0:
for j in range(1,n+1,2):
if stack[j]!=2:
print(0)
exit()
print(pow(2,n//2,p))
exit()
for j in range(0,n+1,2):
if j==0 and stack[j]==1:
continue
if j!=0 and stack[j]!=2:
print(0)
exit()
print(pow(2,n//2,p))
exit()
|
s078941371
|
p02865
|
u434282696
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 51
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n = int(input());print(n//2 if n%2==0 else n//2 -1)
|
s135645112
|
Accepted
| 17
| 2,940
| 51
|
n = int(input());print(n//2 if n%2==1 else n//2 -1)
|
s969160730
|
p03719
|
u410903849
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,096
| 155
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
A, B, C = map(int, input().split())
if C >= A and C <= B:
print('YES')
else:
print('NO')
|
s819852638
|
Accepted
| 27
| 9,164
| 97
|
A, B, C = map(int, input().split())
if C >= A and C <= B:
print('Yes')
else:
print('No')
|
s854377268
|
p03469
|
u516242950
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 90
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
date = input().split('/')
date[0] == '2018'
print(date[0] + '/' + date[1] + '/' + date[2])
|
s157669366
|
Accepted
| 17
| 2,940
| 90
|
date = input().split('/')
date[0] = '2018'
print(date[0] + '/' + date[1] + '/' + date[2])
|
s561096528
|
p03416
|
u870998297
| 2,000
| 262,144
|
Wrong Answer
| 56
| 2,940
| 155
|
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
A, B = [int(i) for i in input().split()]
count = 0
for i in range(A, B+1):
str_A = str(A)
if str_A == str_A[::-1]:
count += 0
print(count)
|
s496840417
|
Accepted
| 52
| 2,940
| 155
|
A, B = [int(i) for i in input().split()]
count = 0
for i in range(A, B+1):
str_i = str(i)
if str_i == str_i[::-1]:
count += 1
print(count)
|
s548892522
|
p03997
|
u881472317
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 74
|
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)
|
s366193756
|
Accepted
| 17
| 2,940
| 79
|
a = int(input())
b = int(input())
h = int(input())
print(int((a + b) * h / 2))
|
s281348354
|
p00005
|
u877201735
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,624
| 218
|
Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b.
|
def gcd(a, b):
if (a % b) == 0:
return b
return gcd(b, a%b)
while True:
try:
nums = list(map(int, input().split()))
print(gcd(nums[0], nums[1]))
except EOFError:
break
|
s191881532
|
Accepted
| 20
| 7,628
| 287
|
def gcd(a, b):
if (a % b) == 0:
return b
return gcd(b, a%b)
def lcd(a, b):
return a * b // gcd(a, b)
while True:
try:
nums = list(map(int, input().split()))
print(gcd(nums[0], nums[1]), lcd(nums[0], nums[1]))
except EOFError:
break
|
s140394836
|
p03478
|
u905582793
| 2,000
| 262,144
|
Wrong Answer
| 40
| 3,060
| 193
|
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 sm(x):
xls=list(str(x))
a=0
for i in range(len(xls)):
a+=int(xls[i])
return a
n,a,b=map(int,input().split())
ans=0
for i in range(1,n+1):
if a<=sm(i)<=b:
ans+=1
print(ans)
|
s144972359
|
Accepted
| 37
| 3,060
| 193
|
def sm(x):
xls=list(str(x))
a=0
for i in range(len(xls)):
a+=int(xls[i])
return a
n,a,b=map(int,input().split())
ans=0
for i in range(1,n+1):
if a<=sm(i)<=b:
ans+=i
print(ans)
|
s959690123
|
p03574
|
u411858517
| 2,000
| 262,144
|
Wrong Answer
| 32
| 3,572
| 613
|
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
W,H=map(int,input().split())
arr = [input() for j in range(W)]
ans = [['0' for i in range(H)] for j in range(W)]
for i in range(0, W):
for j in range(0, H):
bomb_num = 0
if arr[i][j] == '#':
ans[i][j] = '#'
break
for k in (-1, 0, 1):
for l in (-1, 0, 1):
if (-1<i+k<W) and (-1<j+l<H):
if arr[i+k][j+l] == '#':
bomb_num += 1
ans[i][j] = str(bomb_num)
for i in range(W):
for j in range(H):
print(ans[i][j], end = '')
print('\n')
|
s408749925
|
Accepted
| 28
| 3,064
| 567
|
W, H=map(int,input().split())
arr = [input() for j in range(W)]
ans = ['' for i in range(W)]
for i in range(0, W):
for j in range(0, H):
bomb_num = 0
if arr[i][j] == '#':
ans[i] += '#'
else:
for k in (-1, 0, 1):
for l in (-1, 0, 1):
if (-1<i+k<W) and (-1<j+l<H):
if arr[i+k][j+l] == '#':
bomb_num += 1
ans[i] += str(bomb_num)
for i in range(W):
print(ans[i])
|
s056769285
|
p03054
|
u543954314
| 2,000
| 1,048,576
|
Wrong Answer
| 44
| 3,896
| 472
|
We have a rectangular grid of squares with H horizontal rows and W vertical columns. Let (i,j) denote the square at the i-th row from the top and the j-th column from the left. On this grid, there is a piece, which is initially placed at square (s_r,s_c). Takahashi and Aoki will play a game, where each player has a string of length N. Takahashi's string is S, and Aoki's string is T. S and T both consist of four kinds of letters: `L`, `R`, `U` and `D`. The game consists of N steps. The i-th step proceeds as follows: * First, Takahashi performs a move. He either moves the piece in the direction of S_i, or does not move the piece. * Second, Aoki performs a move. He either moves the piece in the direction of T_i, or does not move the piece. Here, to move the piece in the direction of `L`, `R`, `U` and `D`, is to move the piece from square (r,c) to square (r,c-1), (r,c+1), (r-1,c) and (r+1,c), respectively. If the destination square does not exist, the piece is removed from the grid, and the game ends, even if less than N steps are done. Takahashi wants to remove the piece from the grid in one of the N steps. Aoki, on the other hand, wants to finish the N steps with the piece remaining on the grid. Determine if the piece will remain on the grid at the end of the game when both players play optimally.
|
from collections import Counter
h,w,n = map(int, input().split())
sx,sy = map(int, input().split())
sd = Counter(input())
td = Counter(input())
wb,hb = 0,0
a1 = (sd["L"]+1,w-sd["R"])
a2 = (sx-td["L"],sx+td["R"])
if (a1[0] <= a1[1] and (a1[0]<=a2[1] or a1[1]<=a2[0])) or a2[0]>a2[1]:
wb = 1
b1 = (sd["U"]+1,w-sd["D"])
b2 = (sy-td["U"],sy+td["D"])
if (b1[0] <= b1[1] and (b1[0]<=b2[1] or b1[1]<=b2[0])) or b2[0]>b2[1]:
hb = 1
if wb&hb:
print("YES")
else:
print("NO")
|
s239973174
|
Accepted
| 216
| 3,888
| 630
|
h,w,n = map(int,input().split())
sr,sc = map(int,input().split())
t = input()
s = input()
hei = [1,h]
wid = [1,w]
for i in range(n-1,-1,-1):
if s[i] == "L":
wid[1] = min(wid[1]+1,w)
elif s[i] == "R":
wid[0] = max(wid[0]-1,1)
elif s[i] == "U":
hei[1] = min(hei[1]+1,h)
elif s[i] == "D":
hei[0] = max(hei[0]-1,1)
if t[i] == "L":
wid[0] += 1
elif t[i] == "R":
wid[1] -= 1
elif t[i] == "U":
hei[0] += 1
elif t[i] == "D":
hei[1] -= 1
if wid[0]>wid[1] or hei[0]>hei[1]:
print("NO")
exit(0)
if hei[0] <= sr <= hei[1] and wid[0] <= sc <= wid[1]:
print("YES")
else:
print("NO")
|
s575266742
|
p03565
|
u625963200
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 404
|
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
S=list(input())
T=list(input())
if len(S)<len(T):
print('UNRESTORABLE')
else:
for i in range(0,len(S)-len(T)+1):
flag=False
for j in range(i,len(T)):
if S[j]=='?' and S[j]==T[j-i]:
flag=True
else:
flag=False
if flag:
S[i:]=T
for j in range(len(S)):
if S[j]=='?':
S[j]='a'
print(*S,sep='')
exit()
print('UNRESTORABLE')
|
s669713063
|
Accepted
| 17
| 3,064
| 314
|
S=list(input())
T=list(input())
ans=''
for i in range(len(S)-len(T)+1):
for j in range(i,i+len(T)):
if S[j]!=T[j-i] and S[j]!='?':
break
else:
ans=S[:i]+T+S[i+len(T):]
if ans=='':
print('UNRESTORABLE')
else:
for i in range(len(ans)):
if ans[i]=='?':
ans[i]='a'
print(*ans,sep='')
|
s494771220
|
p03486
|
u777028980
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 250
|
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.
|
hoge=list(input())
huga=list(input())
hoge.sort()
huga.sort()
flag=0
for i in range(min(len(hoge),len(huga))):
if(hoge[i]>huga[i]):
flag=1
if(flag==0):
if(len(hoge)>len(huga)):
flag=1
if(flag==0):
print("Yes")
else:
print("No")
|
s980114106
|
Accepted
| 17
| 3,064
| 326
|
hoge=list(input())
huga=list(input())
hoge.sort()
huga.sort(reverse=True)
flag=0
for i in range(min(len(hoge),len(huga))):
if(hoge[i]>huga[i]):
flag=1
break
elif(hoge[i]<huga[i]):
flag=-1
break
if(flag==0):
if(len(hoge)>=len(huga)):
flag=1
if(flag==1):
print("No")
else:
print("Yes")
|
s151224467
|
p03385
|
u897302879
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 50
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
print('Yes' if sorted(input()) == 'abc' else 'No')
|
s565286090
|
Accepted
| 19
| 3,060
| 59
|
print('Yes' if ''.join(sorted(input())) == 'abc' else 'No')
|
s301794859
|
p03050
|
u140137089
| 2,000
| 1,048,576
|
Wrong Answer
| 498
| 3,316
| 150
|
Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied: * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds. Find all favorite numbers and print the sum of those.
|
N = int(input())
tmp = N ** 0.5
sum = 0
k =1
while k < tmp:
m = N // k -1
if N % k == 0 and m > tmp-1 and k < m:
sum += m
k += 1
|
s966012110
|
Accepted
| 489
| 3,060
| 164
|
N = int(input())
tmp = N ** 0.5
sum = 0
k =1
while k < tmp:
m = N // k -1
if N % k == 0 and m > tmp-1 and k < m:
sum += m
k += 1
print(sum)
|
s076369152
|
p03997
|
u321804710
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 75
|
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())
s = (a + b)*h/2
print(s)
|
s865858198
|
Accepted
| 17
| 2,940
| 84
|
a = int(input())
b = int(input())
h = int(input())
s = int((a + b) * h / 2)
print(s)
|
s683807206
|
p03605
|
u052332717
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 94
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
s = input()
if s[0:1] == '9' or s[1:1] == '9':
print('Yes')
else:
print('No')
|
s498774635
|
Accepted
| 18
| 2,940
| 111
|
n = list(input())
ans = 0
for s in n:
if s == '9':
print('Yes')
break
else:
print('No')
|
s799745183
|
p03836
|
u339199690
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 147
|
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())
res = ""
res += "U" * (ty - sy) + "R" * (tx - sx)
res += "D" * (ty - sy) + "L" * (tx - sx)
print(res)
|
s051145290
|
Accepted
| 17
| 3,064
| 269
|
sx, sy, tx, ty = map(int, input().split())
res = ""
res += "U" * (ty - sy) + "R" * (tx - sx)
res += "D" * (ty - sy) + "L" * (tx - sx)
res += "L" + "U" * (ty - sy + 1) + "R" * (tx - sx + 1) + "D"
res += "R" + "D" * (ty - sy + 1) + "L" * (tx - sx + 1) + "U"
print(res)
|
s669197101
|
p02612
|
u629607744
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,200
| 269
|
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.
|
def i():
return int(input())
def i2():
return map(int,input().split())
def s():
return str(input())
def l():
return list(input())
def intl():
return list(int(k) for k in input().split())
n = i()
print(n%1000)
if n%1000 == 0:
print(0)
else:
print( 1000 - n%1000)
|
s953555269
|
Accepted
| 28
| 9,176
| 255
|
def i():
return int(input())
def i2():
return map(int,input().split())
def s():
return str(input())
def l():
return list(input())
def intl():
return list(int(k) for k in input().split())
n = i()
if n%1000 == 0:
print(0)
else:
print( 1000 - n%1000)
|
s887281546
|
p02614
|
u092061507
| 1,000
| 1,048,576
|
Wrong Answer
| 50
| 9,212
| 1,860
|
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
H, W, K = map(int, input().split())
c = [input() for i in range(H)]
"""
H = 6
W = 6
K = 3
c = ['...#.#',
'...###',
'#.....',
'...#..',
'...###',
'...##.']
"""
ret = 0
for i in range(2**H):
for j in range(2**W):
num = 0
mask_h = str(bin(i))[2:].zfill(H)
mask_w = str(bin(j))[2:].zfill(W)
for k in range(H):
if mask_h[k] == '0': continue
for l in range(W):
if mask_w[l] == '0': continue
if c[k][l] == '#':
num += 1
if num == K:
ret += 1
print('num',num)
print (ret)
|
s823035260
|
Accepted
| 49
| 9,220
| 1,834
|
H, W, K = map(int, input().split())
c = [input() for i in range(H)]
"""
H = 6
W = 6
K = 3
c = ['...#.#',
'...###',
'#.....',
'...#..',
'...###',
'...##.']
"""
ret = 0
for i in range(2**H):
for j in range(2**W):
num = 0
mask_h = str(bin(i))[2:].zfill(H)
mask_w = str(bin(j))[2:].zfill(W)
for k in range(H):
if mask_h[k] == '0': continue
for l in range(W):
if mask_w[l] == '0': continue
if c[k][l] == '#':
num += 1
if num == K:
ret += 1
print (ret)
|
s738299468
|
p03471
|
u883792993
| 2,000
| 262,144
|
Wrong Answer
| 848
| 3,060
| 247
|
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 = list(map(int, input().split()))
answer = "-1 -1 -1"
for a in range(0,N+1):
for b in range(0,N-a+1):
c = N-a-b
sum = a*10000 + b*5000 + c*1000
if answer == sum:
answer = "%s %s %s"%(a,b,c)
print(answer)
|
s938162517
|
Accepted
| 855
| 2,940
| 243
|
N,Y = list(map(int, input().split()))
answer = "-1 -1 -1"
for a in range(0,N+1):
for b in range(0,N-a+1):
c = N-a-b
sum = a*10000 + b*5000 + c*1000
if Y == sum:
answer = "%s %s %s"%(a,b,c)
print(answer)
|
s649043223
|
p02613
|
u323045245
| 2,000
| 1,048,576
|
Wrong Answer
| 149
| 9,204
| 278
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n=int(input())
ac=0
wa=0
tle=0
re=0
for i in range(n):
tmp = input()
if tmp == "AC":
ac+=1
elif tmp == "WA":
wa+=1
elif tmp == "TLE":
tle+=1
else:
re+=1
print("AC ×",ac)
print("WA ×",wa)
print("TLE ×",tle)
print("RE ×",re)
|
s329660783
|
Accepted
| 146
| 9,188
| 274
|
n=int(input())
ac=0
wa=0
tle=0
re=0
for i in range(n):
tmp = input()
if tmp == "AC":
ac+=1
elif tmp == "WA":
wa+=1
elif tmp == "TLE":
tle+=1
else:
re+=1
print("AC x",ac)
print("WA x",wa)
print("TLE x",tle)
print("RE x",re)
|
s160022982
|
p03455
|
u970308980
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 64
|
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())
print('Even' if a%b==0 else 'Odd')
|
s415059376
|
Accepted
| 17
| 2,940
| 89
|
a, b=map(int, input().split())
if (a * b) % 2 == 0:
print('Even')
else:
print('Odd')
|
s904656809
|
p02690
|
u746206084
| 2,000
| 1,048,576
|
Time Limit Exceeded
| 2,205
| 9,224
| 473
|
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
|
x=int(input())
s={0}
n=1
stop=0
while(1):
N=n**5
for i in s:
if N-i==x:
if i>0:
sig=int(i//abs(i))
else:
sig=1
print(n,sig*int(pow(abs(i),1/5)))
stop=1
elif N-i==-x:
if i>0:
sig=int(i//abs(i))
else:
sig=1
print(n,sig*int(pow(abs(i),1/5)))
stop=1
if stop==1:
break
s|={N,-N}
|
s008222038
|
Accepted
| 22
| 9,412
| 545
|
x=int(input())
s={0}
n=1
stop=0
while(1):
N=n**5
for i in s:
if N-i==x:
if i>0:
sig=1
elif i<0:
sig=-1
else:
sig=0
print(n,int(sig*pow(abs(i),1/5)))
stop=1
elif i-N==x:
if i>0:
sig=1
elif i<0:
sig=-1
else:
sig=0
print(int(sig*pow(abs(i),1/5)),n)
stop=1
if stop==1:
break
s|={N,-N}
n+=1
|
s831247249
|
p02389
|
u797636303
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,580
| 48
|
Write a program which calculates the area and perimeter of a given rectangle.
|
a,b = map(int, input().split())
c=a*b
d=2*(a+b)
|
s109665599
|
Accepted
| 20
| 5,584
| 59
|
a,b = map(int, input().split())
c=a*b
d=2*(a+b)
print(c,d)
|
s829367862
|
p03672
|
u499381410
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 122
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
s = input()
cnt = len(s)
while True:
if s[:len(s) // 2] == s[len(s) // 2:]:
break
s = s[:-2]
cnt -= 2
|
s822238942
|
Accepted
| 17
| 3,060
| 205
|
s = input()
if len(s) % 2:
s = s[:-1]
else:
s = s[:-2]
cnt = len(s)
while True:
if s[:len(s) // 2] == s[len(s) // 2:]:
break
s = s[:-2]
cnt -= 2
print(cnt)
|
s664322494
|
p02612
|
u304590784
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,156
| 158
|
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())
ans = N
for i in range(1,10):
ans -= 1000
if ans >= 1000:
continue
elif 1000 > ans >= 0:
print(ans)
break
|
s564905878
|
Accepted
| 26
| 9,156
| 97
|
N = int(input())
if N % 1000 != 0:
ans = 1000 - (N % 1000)
print(ans)
else:
print(0)
|
s674299165
|
p02578
|
u668078904
| 2,000
| 1,048,576
|
Wrong Answer
| 131
| 36,448
| 202
|
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
N=int(input())
a=input().split()
A=list(map(int,a))
print(A)
tmp=0
count=0
for i in range(N):
if(i==0):
tmp=A[0]
if(tmp<=A[i]):
#print(A[i])
tmp=A[i]
else:
count+=1
print(count)
|
s543218416
|
Accepted
| 134
| 32,372
| 227
|
N=int(input())
a=input().split()
A=list(map(int,a))
#print(A)
tmp=0
spring_high=0
for i in range(N):
if(i==0):
tmp=A[0]
if(tmp<=A[i]):
#print(A[i])
tmp=A[i]
else:
spring_high+=tmp-A[i]
print(spring_high)
|
s136341511
|
p03486
|
u630657312
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 132
|
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s = list(input())
t = list(input())
s.sort()
t.sort()
s = ''.join(s)
t = ''.join(t)
if s < t:
print('Yes')
else:
print('No')
|
s907272627
|
Accepted
| 17
| 3,060
| 144
|
s = list(input())
t = list(input())
s.sort()
t.sort()
t.reverse()
s = ''.join(s)
t = ''.join(t)
if s < t:
print('Yes')
else:
print('No')
|
s630885204
|
p02742
|
u203669169
| 2,000
| 1,048,576
|
Wrong Answer
| 50
| 5,496
| 570
|
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
from fractions import gcd
from collections import Counter, deque, defaultdict
from heapq import heappush, heappop, heappushpop, heapify, heapreplace, merge
from bisect import bisect_left, bisect_right, bisect, insort_left, insort_right, insort
from itertools import accumulate, product, permutations, combinations
N = [int(_) for _ in input().split()]
H, W = N[0], N[1]
def solve(H, W):
cnt = 0
for i in range(H):
if i % 2 == 0:
cnt += int(W / 2) + 1
else:
cnt += W - (int(W / 2) + 1)
return cnt
|
s527199740
|
Accepted
| 17
| 3,060
| 431
|
from math import floor
N = [int(_) for _ in input().split()]
H, W = N[0], N[1]
def solve(H, W):
acnt = floor(W / 2) + 1
bcnt = W - acnt
if H == 1 or W == 1:
return 1
else:
return floor((H * W + 1) / 2)
# if H % 2 == 0:
# cnt = acnt * floor(H/2) + bcnt * floor(H/2)
# else:
# cnt = acnt * (floor(H/2) + 1) + bcnt * floor(H/2)
# return cnt
print(solve(H, W))
|
s096065905
|
p04029
|
u655622461
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
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())
candy = N * (N+1) / 2
print(candy)
|
s907301575
|
Accepted
| 17
| 2,940
| 58
|
N = int(input())
candy = N * (N + 1) / 2
print(int(candy))
|
s557717408
|
p02276
|
u643021750
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,604
| 471
|
Quick sort is based on the Divide-and-conquer approach. In QuickSort(A, p, r), first, a procedure Partition(A, p, r) divides an array A[p..r] into two subarrays A[p..q-1] and A[q+1..r] such that each element of A[p..q-1] is less than or equal to A[q], which is, inturn, less than or equal to each element of A[q+1..r]. It also computes the index q. In the conquer processes, the two subarrays A[p..q-1] and A[q+1..r] are sorted by recursive calls of QuickSort(A, p, q-1) and QuickSort(A, q+1, r). Your task is to read a sequence A and perform the Partition based on the following pseudocode: Partition(A, p, r) 1 x = A[r] 2 i = p-1 3 for j = p to r-1 4 do if A[j] <= x 5 then i = i+1 6 exchange A[i] and A[j] 7 exchange A[i+1] and A[r] 8 return i+1 Note that, in this algorithm, Partition always selects an element A[r] as a pivot element around which to partition the array A[p..r].
|
def partition(p, r):
x = A[r]
i = p - 1
for j in range(p, r):
if A[j] <= x:
i += 1
A_tmp = A[i]
A[i] = A[j]
A[j] = A_tmp
t = A[i+1]
A[i+1] = A[r]
A[r] = t
n = int(input())
A = [int(_) for _ in input().split()]
q = partition(0, n-1)
for i in range(n):
if i:
print(" ")
if i == q:
print("[")
print(A[i])
if i == q:
print("]")
print("\n")
|
s884553305
|
Accepted
| 300
| 16,364
| 519
|
def partition(p, r):
x = A[r]
i = p - 1
for j in range(p, r):
if A[j] <= x:
i += 1
A_tmp = A[i]
A[i] = A[j]
A[j] = A_tmp
t = A[i+1]
A[i+1] = A[r]
A[r] = t
return i+1
n = int(input())
A = [int(_) for _ in input().split()]
q = partition(0, n-1)
for i in range(n):
if i:
print(" ", end="")
if i == q:
print("[", end="")
print(A[i], end="")
if i == q:
print("]", end="")
print("")
|
s920250144
|
p03455
|
u350179603
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,316
| 85
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b = map(int, input().split())
if (a*b)%2==0:
print("Odd")
else:
print("Even")
|
s704326966
|
Accepted
| 17
| 2,940
| 85
|
a,b = map(int, input().split())
if (a*b)%2==0:
print("Even")
else:
print("Odd")
|
s330440216
|
p03671
|
u533679935
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 133
|
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.
|
a,b,c=map(int,input().split())
if a+b < c+b or b+c <c+b:
print(c+b)
elif c+b < a+b or b+c <a+b:
print(a+b)
else:
print(a+c)
|
s907421639
|
Accepted
| 17
| 3,064
| 206
|
x,y,z=map(int,input().split())
if (x+y) <= (x+z) and (x+y)<= (y+z):
print(x+y)
elif (x+z) <= (x+y) and (x+z) <= (y+z):
print(x+z)
elif (y+z) <= (x+y) and (y+z) <= (x+z):
print(y+z)
else:
print(x+z)
|
s525222318
|
p03377
|
u693378622
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 78
|
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())
print("Yes" if a<=x and x<=a+b else "No" )
|
s498198474
|
Accepted
| 17
| 2,940
| 78
|
a, b, x = map(int, input().split())
print("YES" if a<=x and x<=a+b else "NO" )
|
s518304887
|
p03456
|
u443996531
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 106
|
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()
ab = int(a+b)
ans = "No"
i=3
while i*i<=ab:
if i*i == ab:
ans="Yes"
print(ans)
|
s075286127
|
Accepted
| 18
| 2,940
| 113
|
a,b = input().split()
ab = int(a+b)
ans = "No"
i=3
while i*i<=ab:
if i*i == ab:
ans="Yes"
i+=1
print(ans)
|
s686847510
|
p03910
|
u867378674
| 2,000
| 262,144
|
Wrong Answer
| 1,518
| 3,572
| 285
|
The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems. The score allocated to the i-th (1≦i≦N) problem is i points. Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve. As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him. Determine the set of problems that should be solved.
|
n = int(input())
def calc(n):
mx = 1
while True:
if n <= mx * (mx + 1) // 2:
break
mx += 1
return mx
ans = []
while n > 0:
x = calc(n)
ans.append(x)
n -= x
if x > n:
ans.append(n)
break
print(*ans, sep="\n")
|
s952499937
|
Accepted
| 22
| 3,572
| 233
|
N = int(input())
x = 1
for _ in range(100000):
if (x + 2) * (x + 1) // 2 > N:
break
x += 1
H = [i for i in range(x, 0, -1)]
dif = N - sum(H)
for i in range(dif):
H[(i) % len(H)] += 1
print(*sorted(H), sep="\n")
|
s057089559
|
p03456
|
u496018856
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,192
| 204
|
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.
|
# -*- coding: utf-8 -*-
import math
a, b = map(str, input().split())
c=a+b
c=int(c)
if(isinstance(math.sqrt(c),int)):
print("Yes")
else:
print("No")
|
s027778521
|
Accepted
| 17
| 2,940
| 201
|
# -*- coding: utf-8 -*-
import math
a, b = map(str, input().split())
c=a+b
c=int(c)
if(math.sqrt(c).is_integer()):
print("Yes")
else:
print("No")
|
s540978036
|
p02416
|
u776758454
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,696
| 196
|
Write a program which reads an integer and prints sum of its digits.
|
def main():
while True:
number_string = input()
if number_string == '0': break
print(list(number_string))
print(sum(map(int, list(number_string))))
main()
|
s107612400
|
Accepted
| 20
| 7,752
| 161
|
def main():
while True:
number_string = input()
if number_string == '0': break
print(sum(map(int, list(number_string))))
main()
|
s944377106
|
p03139
|
u824425238
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 658
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
# coding: utf-8
import sys
import math
def main(s):
n = s[0]
a = s[1]
b = s[2]
if not(n.isdigit()) or int(n) <= 0 or int(n) > 100:
return -1
if not(a.isdigit()) or int(n) < 0 or int(a) > int(n):
return -1
if not(b.isdigit()) or int(b) < 0 or int(b) > int(n):
return -1
n = int(n)
a = int(a)
b = int(b)
reta = a
retb = (n - a) - b
if retb > 0 :
retb = 0
else:
retb = abs(retb)
print('{reta} {retb}'.format(reta=reta,retb=retb))
return 0
if __name__ == '__main__':
args = sys.argv
if len(args) > 1:
s = args[1:]
main(s)
|
s663463278
|
Accepted
| 17
| 3,060
| 263
|
# coding: utf-8
import sys
import math
args = input().split()
n = int(args[0])
a = int(args[1])
b = int(args[2])
reta = a
retb = (n - a) - b
if retb > 0:
retb = 0
else:
retb = abs(retb)
print('{reta} {retb}'.format(reta=min(a,b),retb=retb))
sys.exit(0)
|
s002884639
|
p02612
|
u550146131
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,136
| 32
|
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())
M=N%1000
print(M)
|
s714356435
|
Accepted
| 30
| 9,148
| 74
|
N=int(input())
if N%1000==0:
print("0")
else:
print(1000-(N%1000))
|
s670311097
|
p03455
|
u143903328
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 87
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int,input().split())
if a * b % 2:
print('odd')
else:
print('even')
|
s293486840
|
Accepted
| 17
| 2,940
| 87
|
a, b = map(int,input().split())
if a * b % 2:
print('Odd')
else:
print('Even')
|
s337556900
|
p03997
|
u075595666
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 445
|
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 = input()
b = input()
c = input()
a = a.replace("a","")
b = b.replace("b","")
c = c.replace("c","")
turn = 'a'
for i in range(500):
if turn == 'a':
if int(len(a)) == 0:
print("A")
break
else:
turn = str(a[0])
a = a[1:]
if turn == 'b':
if int(len(b)) == 0:
print('B')
break
else:
turn = str(b[0])
b = b[1:]
if turn == 'c':
if int(len(c)) == 0:
print('C')
break
else:
turn = str(c[0])
c = c[1:]
|
s446113671
|
Accepted
| 17
| 2,940
| 74
|
a = int(input())
b = int(input())
h = int(input())
print(int((a + b)*h/2))
|
s637432820
|
p02401
|
u369313788
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,660
| 252
|
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while True:
a, op, b = input().split()
a = int(a)
b = int(b)
if op == '?':
break
if op == '+':
print(a+b)
if op == '-':
print(a-b)
if op == '*':
print(a*b)
if op == '/':
print(a/b)
|
s452831712
|
Accepted
| 20
| 7,656
| 253
|
while True:
a, op, b = input().split()
a = int(a)
b = int(b)
if op == '?':
break
if op == '+':
print(a+b)
if op == '-':
print(a-b)
if op == '*':
print(a*b)
if op == '/':
print(a//b)
|
s973746159
|
p03574
|
u301624971
| 2,000
| 262,144
|
Wrong Answer
| 73
| 3,872
| 873
|
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
import itertools
H, W = map(int, input().split())
S = []
ans = []
for _ in range(H):
S.append(list(input()))
ans.append(['
op =[]
for i in itertools.product([-1, 0, 1], repeat=2):
op.append(i)
for i, s1 in enumerate(S):
for j, s2 in enumerate(s1):
if(s2 == '.'):
counter = 0
for o in op:
print(i,j,o)
if(i == 0 and o[0] == -1):
continue
elif(i == H-1 and o[0] == 1):
continue
elif(j == 0 and o[1] == -1):
continue
elif(j == W-1 and o[1] == 1):
continue
else:
if(S[i + o[0]][j + o[1]] == "#"):
counter += 1
ans[i][j] = str(counter)
for a in ans:
print(''.join(a))
|
s375685673
|
Accepted
| 26
| 3,316
| 1,314
|
def myAnswer(H:int,W:int,S:list) -> None:
XY = [[-1,-1],[-1,0],[-1,1],[0,-1],[0,1],[1,-1],[1,0],[1,1]]
index = []
ans = [[0]*W for _ in range(H)]
for i in range(H):
for j in range(W):
if(S[i][j] == "#"):
ans[i][j] = "#"
index.append([i,j])
for i,j in index:
for x,y in XY:
if(0 > (j + x) or (j + x) >= W or 0 > (i + y) or (i + y) >= H):
continue
elif(ans[y+i][x+j]== "#"):
continue
else:
ans[y + i][x + j] += 1
for a in ans:
string = ""
for s in a:
string += str(s)
print(string)
def modelAnswer(H:int,W:int,S:list) -> int:
dx = [1,0,-1,0,1,-1,-1,1]
dy = [0,1,0,-1,1,1,-1,-1]
for i in range(H):
for j in range(W):
if(S[i][j] == "#"):continue
num = 0
for d in range(8):
ni = i + dy[d]
nj = j + dx[d]
if(ni < 0 or H <= ni):continue
if(nj < 0 or W <= nj):continue
if(S[ni][nj] == "#"): num+=1
S[i][j] = str(num )
for s in S:
ans = ""
for a in s:
ans += a
print(ans)
def main():
H,W = map(int,input().split())
S = [list(input()) for _ in range(H)]
myAnswer(H,W,S[:])
if __name__ == '__main__':
main()
|
s715752388
|
p03623
|
u485435834
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 218
|
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|.
|
location_list = list(map(int, input().split()))
ax_length = abs(location_list[0] - location_list[1])
bx_length = abs(location_list[0] - location_list[2])
if ax_length > bx_length:
print('A')
else:
print('B')
|
s953979639
|
Accepted
| 17
| 2,940
| 218
|
location_list = list(map(int, input().split()))
ax_length = abs(location_list[0] - location_list[1])
bx_length = abs(location_list[0] - location_list[2])
if ax_length < bx_length:
print('A')
else:
print('B')
|
s675461350
|
p03352
|
u048945791
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 152
|
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
x = int(input())
maxInt = 1
i = 2
while i * i <= x:
exp = i * i
while exp <= x:
exp *= i
else:
maxInt = max(maxInt, exp)
i += 1
print(maxInt)
|
s239758895
|
Accepted
| 17
| 3,064
| 157
|
x = int(input())
maxInt = 1
i = 2
while i * i <= x:
exp = i * i
while exp <= x:
exp *= i
else:
maxInt = max(maxInt, exp // i)
i += 1
print(maxInt)
|
s981035880
|
p03563
|
u076764813
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 59
|
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.
|
r = float(input())
g = float(input())
t = 2*g -r
print(t)
|
s328530054
|
Accepted
| 17
| 2,940
| 55
|
r = int(input())
g = int(input())
t = 2*g -r
print(t)
|
s283715339
|
p03024
|
u960653324
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 146
|
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
S = list(str(input()))
if S.count("o") >= 8:
print("Yes")
else:
if 15 - len(S) + S.count("o") >= 8:
print("Yes")
else:
print("No")
|
s273348510
|
Accepted
| 17
| 2,940
| 146
|
S = list(str(input()))
if S.count("o") >= 8:
print("YES")
else:
if 15 - len(S) + S.count("o") >= 8:
print("YES")
else:
print("NO")
|
s811746654
|
p03796
|
u221841077
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 3,464
| 107
|
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
N = int(input())
pow = 1
for i in range(1,N+1):
pow = pow*i
num = 10**9+7
print(pow%num)
print(pow)
|
s626951258
|
Accepted
| 35
| 2,940
| 100
|
N = int(input())
pow = 1
num = 10**9+7
for i in range(1,N+1):
pow = pow*i%num
print(pow%num)
|
s139117633
|
p03854
|
u975966195
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,188
| 339
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
s = input()[::-1]
words = [w[::-1] for w in ['dream', 'dreamer', 'erase', 'eraser']]
can = 'Yes'
while True:
match_any = False
for w in words:
if s.startswith(w):
s = s.replace(w, '', 1)
match_any = True
break
if not match_any:
can = 'No'
break
break
print(can)
|
s211071977
|
Accepted
| 82
| 3,188
| 326
|
s = input()[::-1]
words = [w[::-1] for w in ['dream', 'dreamer', 'erase', 'eraser']]
can = 'YES'
while s:
match_any = False
for w in words:
if s.startswith(w):
s = s.replace(w, '', 1)
match_any = True
break
if not match_any:
can = 'NO'
break
print(can)
|
s109279114
|
p03379
|
u063052907
| 2,000
| 262,144
|
Wrong Answer
| 315
| 25,472
| 200
|
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())
lst_x = list(map(int, input().split()))
lst_x.sort()
b1 = lst_x[n//2 - 1]
b2 = lst_x[n//2]
for x in lst_x:
if x <= b1:
ans = b2
else:
ans = b1
print(ans)
|
s614509774
|
Accepted
| 288
| 26,016
| 168
|
n = int(input())
lst_x = list(map(int, input().split()))
lst = lst_x[:]
lst.sort()
b1 = lst[n//2 - 1]
b2 = lst[n//2]
for x in lst_x:
print(b2 if x <= b1 else b1)
|
s263126798
|
p02422
|
u195186080
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,628
| 353
|
Write a program which performs a sequence of commands to a given string $str$. The command is one of: * print a b: print from the a-th character to the b-th character of $str$ * reverse a b: reverse from the a-th character to the b-th character of $str$ * replace a b p: replace from the a-th character to the b-th character of $str$ with p Note that the indices of $str$ start with 0.
|
sen = input()
n = int(input())
for i in range(n):
cmd = input().split()
sta = int(cmd[1])
end = int(cmd[2])
if len(cmd) == 4:
sen = sen[0:sta] + cmd[3] + sen[end+1:]
elif cmd[0] == 'reverse':
rev = sen[end+1:sta:-1]
sen = sen[0:sta] + rev +sen[end+1:]
elif cmd[0] == 'print':
print(sen[sta:end+1])
|
s260170609
|
Accepted
| 20
| 7,700
| 427
|
sen = input()
n = int(input())
for i in range(n):
cmd = input().split()
sta = int(cmd[1])
end = int(cmd[2])
if cmd[0] == 'replace':
sen = sen[0:sta] + cmd[3] + sen[end+1:]
elif cmd[0] == 'reverse':
rev = sen[sta:end+1]
rev = rev[::-1]
# print(rev)
sen = sen[0:sta] + rev +sen[end+1:]
elif cmd[0] == 'print':
print(sen[sta:end+1])
|
s100734347
|
p03433
|
u420200689
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 336
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
# -*- coding: utf-8 -*-
while True:
N = int(input("N = "))
A = int(input("A = "))
if (N >= 1 and N <= 10000) and (A >= 0 and A <= 1000):
break
if (N / 500) > 0:
x = N % 500
if A >= x:
print('Yes')
else:
print('No')
else:
if A >= N:
print('yes')
else:
print('No')
|
s378943684
|
Accepted
| 17
| 3,060
| 323
|
# -*- coding: utf-8 -*-
while True:
N = int(input())
A = int(input())
if (N >= 1 and N <= 10000) and (A >= 0 and A <= 1000):
break
if (N / 500) > 0:
x = N % 500
if A >= x:
print('Yes')
else:
print('No')
else:
if A >= N:
print('yes')
else:
print('No')
|
s119478616
|
p03447
|
u422267382
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 114
|
You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?
|
X,A,B=[int(input()) for i in range(3)]
x=0
XX=X-A
while XX>0:
XX-=B
print(XX)
x+=1
print(X-A-B*(x-1))
|
s810102681
|
Accepted
| 17
| 3,064
| 101
|
X,A,B=[int(input()) for i in range(3)]
x=0
XX=X-A
while XX>=0:
XX-=B
x+=1
print(X-A-B*(x-1))
|
s198912904
|
p03469
|
u750389519
| 2,000
| 262,144
|
Wrong Answer
| 25
| 8,944
| 25
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
S=input()
S[4],print('8')
|
s621178352
|
Accepted
| 27
| 8,968
| 29
|
S=input()
print("2018"+S[4:])
|
s974076564
|
p02613
|
u561294476
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,168
| 194
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
S = [i for i in str(input())]
print("AC x " + str(S.count('AC')))
print("WA x " + str(S.count('WA')))
print("TLE x " + str(S.count('TLE')))
print("RE x " + str(S.count('RE')))
|
s551658334
|
Accepted
| 158
| 16,272
| 221
|
N = int(input())
S = ["00"]
for i in range(N):
S.append(str(input()))
print("AC x " + str(S.count('AC')))
print("WA x " + str(S.count('WA')))
print("TLE x " + str(S.count('TLE')))
print("RE x " + str(S.count('RE')))
|
s710227066
|
p04030
|
u023229441
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 115
|
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 i in s:
if i==0 or i==1:
ans+=i
else:
if len(ans)!=0:
del ans[-1]
print(ans)
|
s542798634
|
Accepted
| 17
| 2,940
| 122
|
s=input()
ans=""
for i in s:
if i=="0" or i=="1":
ans+=i
else:
if len(ans)!=0:
ans=ans[:-1]
print(ans)
|
s805899670
|
p03360
|
u627417051
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 131
|
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?
|
A, B, C = input().split()
K = int(input())
A = int(A)
B = int(B)
C = int(C)
X = (A, B, C)
M = max(X)
print(M ** K + A + B + C -M)
|
s461313672
|
Accepted
| 17
| 3,060
| 137
|
A, B, C = input().split()
K = int(input())
A = int(A)
B = int(B)
C = int(C)
X = (A, B, C)
M = max(X)
print(M * (2 ** K) + A + B + C -M)
|
s315262089
|
p03385
|
u606033239
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 86
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
S = input()
print(["YES","NO"][S.count("a")>=2 or S.count("b")>=2 or S.count("c")>=2])
|
s631825100
|
Accepted
| 18
| 2,940
| 81
|
S = sorted(input())
if S == ["a","b","c"]:
print("Yes")
else:
print("No")
|
s508174367
|
p03644
|
u676091297
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 107
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n=int(input())
cnt=1
for i in range(0, 100):
cnt*=2
if cnt >= n:
print(cnt/2)
break
|
s227582708
|
Accepted
| 18
| 2,940
| 131
|
import math
n=int(input())
cnt=1
for i in range(0, 100):
cnt*=2
if cnt > n:
print(math.floor(cnt/2))
break
|
s688759711
|
p03150
|
u724732842
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 496
|
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
s = list(input())
ss = ['k','e','y','e','n','c','e']
p = 6
a = 0
b = 0
c = 0
for i in range(len(ss)):
if s[i] == ss[i]:
continue
else:
a = i
break
else:
print('YES')
exit()
for j in range(len(s)-1,-1,-1):
if p >= 0 :
if (s[j] == ss[p]):
print('ヌッ',c)
p -= 1
c += 1
else:
b = c
break
else:
print('YES')
exit()
if a+b == 7:
print('YES')
else:
print('NO')
|
s789184169
|
Accepted
| 18
| 3,060
| 359
|
S = list(input())
s = ['k','e','y','e','n','c','e']
a = 0
for i in range(len(s)):
if S[i] == s[i]:
continue
else:
a = i
break
else:
print('YES')
exit()
ss = s[a:]
for i in range(len(ss)):
if ss[i] == S[len(S)-len(ss)+i]:
continue
else:
print('NO')
exit()
else:
print('YES')
|
s854139644
|
p03854
|
u936378263
| 2,000
| 262,144
|
Wrong Answer
| 74
| 3,956
| 268
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
words = ["dream", "dreamer", "erase", "eraser"]
s = input()
dp = [0] * (len(s)+1)
dp[0] = 1
for i in range(len(s)+1):
if dp[i]!=1: continue
for word in words:
if word in s[i:i+len(word)]:
dp[i+len(word)] = 1
print("Yes" if dp[len(s)] else "No")
|
s711191538
|
Accepted
| 76
| 3,956
| 268
|
words = ["dream", "dreamer", "erase", "eraser"]
s = input()
dp = [0] * (len(s)+1)
dp[0] = 1
for i in range(len(s)+1):
if dp[i]!=1: continue
for word in words:
if word in s[i:i+len(word)]:
dp[i+len(word)] = 1
print("YES" if dp[len(s)] else "NO")
|
s021837385
|
p03719
|
u052066895
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 74
|
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("YES" if A<=C and C<=B else "NO")
|
s988170624
|
Accepted
| 17
| 2,940
| 74
|
A,B,C = map(int, input().split())
print("Yes" if A<=C and C<=B else "No")
|
s920034227
|
p02972
|
u480300350
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 18,884
| 728
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
n = int(input())
A = list(map(int, input().split()))
ans = 0
B = [0] * n
ansList = []
def makeList(num):
start = num + 1
mul = n // start
return [start * j - 1 for j in range(1, mul+1)]
def makeRes(indices):
flag = 0
for elm in indices:
if B[elm] == 1:
flag = (flag + 1) % 2
return flag
for i in range(n-1, -1, -1):
if i + 1 > n / 2:
if A[i] == 1:
B[i] = 1
ans += 1
ansList.insert(0, i)
else:
indices = makeList(i)
res = makeRes(indices)
# print(res)
if A[i] != res:
B[i] = 1
ans += 1
ansList.insert(0, i)
print(ans)
print(*ansList)
|
s238132547
|
Accepted
| 662
| 23,988
| 999
|
from collections import deque
def main():
n = int(input())
A = list(map(int, input().split()))
ans = 0
B = [0] * n
# ansList = []
ansList = deque()
def makeList(num):
start = num + 1
mul = n // start
return [start * j - 1 for j in range(1, mul+1)]
def makeRes(indices):
flag = 0
for elm in indices:
if B[elm] == 1:
flag = (flag + 1) % 2
return flag
for i in range(n-1, -1, -1):
if i + 1 > n / 2:
if A[i] == 1:
B[i] = 1
ans += 1
ansList.appendleft(i+1)
else:
indices = makeList(i)
res = makeRes(indices)
# print(res)
if A[i] != res:
B[i] = 1
ans += 1
ansList.appendleft(i+1)
print(ans)
if ans != 0:
print(*(list(ansList)))
if __name__ == "__main__":
main()
|
s476441377
|
p03448
|
u409064224
| 2,000
| 262,144
|
Wrong Answer
| 52
| 3,060
| 219
|
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())
res = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if (500*a)+(100*b)+(50*c) == x:
res += 1
else:
print(res)
|
s293867756
|
Accepted
| 51
| 3,060
| 213
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
res = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if (500*i)+(100*j)+(50*k) == x:
res += 1
print(res)
|
s075526001
|
p03386
|
u039623862
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 229
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a, b, k = map(int, input().split())
l = set()
for x in range(a, a + k):
if x > b:
break
l.add(x)
for x in range(b - k, b + 1):
if x < a:
continue
l.add(x)
l = list(l)
l.sort()
print(*l, sep='\n')
|
s443583857
|
Accepted
| 17
| 3,060
| 176
|
a, b, k = map(int, input().split())
u = set()
for i in range(k):
if a+i <= b:
u.add(a + i)
if b-i >= a:
u.add(b - i)
print(*sorted(list(u)), sep='\n')
|
s298066621
|
p00031
|
u358919705
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,604
| 134
|
祖母が天秤を使っています。天秤は、二つの皿の両方に同じ目方のものを載せると釣合い、そうでない場合には、重い方に傾きます。10 個の分銅の重さは、軽い順に 1g, 2g, 4g, 8g, 16g, 32g, 64g, 128g, 256g, 512g です。 祖母は、「1kg くらいまでグラム単位で量れるのよ。」と言います。「じゃあ、試しに、ここにあるジュースの重さを量ってよ」と言ってみると、祖母は左の皿にジュースを、右の皿に 8g と64g と128g の分銅を載せて釣合わせてから、「分銅の目方の合計は 200g だから、ジュースの目方は 200g ね。どう、正しいでしょう？」と答えました。 左の皿に載せる品物の重さを与えるので、天秤で与えられた重みの品物と釣合わせるときに、右の皿に載せる分銅を軽い順に出力するプログラムを作成して下さい。ただし、量るべき品物の重さは、すべての分銅の重さの合計 (=1023g) 以下とします。
|
while True:
try:
w = int(input())
except:
break
print(*[2 ** i if w // 2 ** i else '' for i in range(10)])
|
s359327727
|
Accepted
| 30
| 7,636
| 205
|
while True:
try:
w = int(input())
except:
break
ans = [2 ** i if w // 2 ** i % 2 else 0 for i in range(10)]
while ans.count(0):
del ans[ans.index(0)]
print(*ans)
|
s178913781
|
p03997
|
u594859393
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 56
|
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 _ in range(3)]
print(a*b*h//2)
|
s385356183
|
Accepted
| 17
| 2,940
| 58
|
a,b,h = [int(input()) for _ in range(3)]
print((a+b)*h//2)
|
s145282118
|
p03162
|
u591524911
| 2,000
| 1,048,576
|
Wrong Answer
| 2,109
| 54,084
| 418
|
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
N= int(input())
import numpy as np
arr = np.array([input().split() for _ in range(N)], dtype=np.int8)
dp = np.zeros_like(arr)
for i, arr_ in enumerate(arr):
for j in range(len(arr_)):
if i == 0:
dp[i][j] = arr_[j]
continue
for k in range(len(arr_)):
if j != k:
dp[i][j] = max(dp[i-1][k] + arr_[j], dp[i][j])
print(np.max(dp[-1]))
#print(dp)
|
s695994509
|
Accepted
| 596
| 53,492
| 305
|
N= int(input())
arr = [list(map(int,input().split())) for _ in range(N)]
dp = [ list([0,0,0]) for i in range(N)]
for i,l in enumerate(arr):
if i == 0:
dp[0] = l
continue
dp[i] = [max(dp[i-1][1:])+l[0], max(dp[i-1][0],dp[i-1][2])+l[1],max(dp[i-1][:2])+l[2]]
print(max(dp[-1]))
#print(dp)
|
s209541069
|
p03698
|
u168416324
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 131
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s=list(input())
li=[]
non=False
for c in s:
if c in li:
non=True
break
if non:
print("yes")
else:
print("no")
|
s970431942
|
Accepted
| 17
| 2,940
| 146
|
s=list(input())
li=[]
non=False
for c in s:
if c in li:
non=True
break
li.append(c)
if non:
print("no")
else:
print("yes")
|
s069265713
|
p02678
|
u920977317
| 2,000
| 1,048,576
|
Wrong Answer
| 2,207
| 47,036
| 1,026
|
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
def main():
RoomNum,PathNum=map(int,input().split())
Link=[]
for _ in range(PathNum):
array=input().split()
Link.append(list(map(int,array)))
#Link.append(list(map(int,[array[1],array[0]])))
Link.sort()
room_que=deque([1])
sign=[]
room_num=[]
while len(room_que)>0 and len(Link)>0:
current_room=room_que.popleft()
for link in Link[:]:
if (current_room in link):
room_que.append(link[link.index(current_room)^1])
if (link[link.index(current_room)^1] in room_num)==False:
sign.append(current_room)
room_num.append(link[link.index(current_room)^1])
Link.remove(link)
if len(sign)!=RoomNum-1:
print("No")
else:
RS=zip(room_num,sign)
RS=sorted(RS)
room_num,sign=zip(*RS)
print("yes")
for i in sign:
print(i)
if __name__=="__main__":
main()
|
s448039854
|
Accepted
| 653
| 34,972
| 883
|
from collections import deque
def main():
RoomNum,PathNum=map(int,input().split())
G=[[] for _ in range(RoomNum)]
for _ in range(PathNum):
a,b=map(int,input().split())
G[a-1].append(b-1)
G[b-1].append(a-1)
dist=[-1]*RoomNum
que=deque([])
dist[0]=0
que.append(0)
while len(que)>0:
v=que.popleft()
for i in G[v]:
if dist[i]!=-1:
continue
dist[i]=v
que.append(i)
if (-1 in dist):
print("No")
else:
print("Yes")
for i in dist[1:]:
print(i+1)
if __name__=="__main__":
main()
|
s740795718
|
p03377
|
u263824932
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 110
|
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 X<A:
print('No')
elif X>A+B:
print('No')
else:
print('Yes')
|
s011319135
|
Accepted
| 17
| 2,940
| 94
|
A,B,X=map(int,input().split())
if X>=A and A+B>=X:
print('YES')
else:
print('NO')
|
s759238968
|
p03544
|
u131464432
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,100
| 117
|
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
N = int(input())
L = [0]*(N+1)
L[0] = 2
L[1] = 1
for i in range(2,N+1):
L[i] = L[i-1] + L[i-2]
print(L)
print(L[N])
|
s594265718
|
Accepted
| 31
| 9,152
| 108
|
N = int(input())
L = [0]*(N+1)
L[0] = 2
L[1] = 1
for i in range(2,N+1):
L[i] = L[i-1] + L[i-2]
print(L[N])
|
s428422082
|
p03434
|
u907549201
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 138
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
input()
li = list(map(int, input().split()))
li.sort(reverse=True)
alice = [i for index,i in enumerate(li) if index % 2 == 0]
print(alice)
|
s923545467
|
Accepted
| 20
| 3,060
| 167
|
input()
li = list(map(int, input().split()))
li.sort(reverse=True)
alice = sum([i for index,i in enumerate(li) if index % 2 == 0])
ans = alice * 2 - sum(li)
print(ans)
|
s020367818
|
p02748
|
u736470924
| 2,000
| 1,048,576
|
Wrong Answer
| 2,229
| 2,008,476
| 715
|
You are visiting a large electronics store to buy a refrigerator and a microwave. The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen. You have M discount tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time. You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required.
|
import itertools
A, B, M = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
m = {}
for _ in range(M):
x, y, z = map(int, input().split())
key = str(x) + "&" + str(y)
if key in m.keys():
if m[key] < z:
m[key] = z
else:
m[key] = z
A_B = list(itertools.product(range(A), range(B)))
A_B_sum = list(map(lambda x: x[0] + x[1], (itertools.product(a, b))))
for i in range(len(A_B)):
# print(str(A_B[i][0]) + "&" + str(A_B[i][1]))
if str(A_B[i][0]) + "&" + str(A_B[i][1]) in m.keys():
print(str(A_B[i][0]) + "&" + str(A_B[i][1]))
A_B_sum[i] -= m[str(A_B[i][0]) + "&" + str(A_B[i][1])]
print(min(A_B_sum))
|
s675828128
|
Accepted
| 418
| 18,612
| 271
|
A, B, M = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
min_a = min(a)
min_b = min(b)
ans = min_a + min_b
for _ in range(M):
i, j, c = map(int, input().split())
ans = min(ans, a[i - 1] + b[j - 1] - c)
print(ans)
|
s816192244
|
p02796
|
u536600145
| 2,000
| 1,048,576
|
Wrong Answer
| 348
| 11,048
| 227
|
In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep.
|
N = int(input())
left = []
right = []
for i in range(N):
x, dis = map(int, input().split(" "))
left.append(x-dis)
right.append(x+dis)
cnt = 1
for i in range(N-1):
if right[i] <= left[i+1]:
cnt += 1
|
s173292683
|
Accepted
| 464
| 22,104
| 305
|
N = int(input())
robots = []
for _ in range(N):
x, dis = map(int, input().split())
robots.append([x-dis,x+dis])
robots = sorted(robots, key=lambda x: x[1])
cnt = 0
now = -float("inf")
for left, right in robots:
if left >= now:
cnt += 1
now = right
print(cnt)
|
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