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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s433273055
|
p02412
|
u713218261
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,724
| 307
|
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
while True:
(n, x) = [int(i) for i in input().split()]
if n == x == 0:
break
for a in range(1, n + 1):
for b in range(2, n + 1):
for c in range(3, n + 1):
count = 0
if (a + b + c) == x:
count += 1
print(count)
|
s310156226
|
Accepted
| 500
| 6,724
| 357
|
while True:
(n, x) = [int(i) for i in input().split()]
if n == x == 0:
break
count = 0
for a in range(1, n+1):
for b in range(a+1, n+1):
for c in range(b+1, n+1):
if (a + b + c) == x:
count += 1
elif (a + b + c) > x:
break
print(count)
|
s435052348
|
p00005
|
u847467233
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,620
| 304
|
Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b.
|
# AOJ 0005 GCD and LCM
# Python3 2018.6.9 bal4u
def lcm(a, b):
return a/gcd(a, b)*b
def gcd(a, b):
while b != 0:
r = a % b
a = b
b = r
return a
while True:
try:
a = list(map(int, input().split()))
print(gcd(a[0], a[1]), lcm(a[0], a[1]))
except EOFError:
break
|
s469508341
|
Accepted
| 20
| 5,604
| 309
|
# AOJ 0005 GCD and LCM
# Python3 2018.6.9 bal4u
def lcm(a, b):
return a // gcd(a, b) * b
def gcd(a, b):
while b != 0:
r = a % b
a = b
b = r
return a
while True:
try:
a = list(map(int, input().split()))
print(gcd(a[0], a[1]), lcm(a[0], a[1]))
except EOFError:
break
|
s187218672
|
p02972
|
u464244643
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 8,228
| 441
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
import sys
sys.setrecursionlimit(10 ** 6)
def input():
return sys.stdin.readline()[:-1]
def main():
N = int(input())
a = [-1]+list(map(int, input().split()))
b = [0] * (N+1)
for i in reversed(range(1,N+1)):
total = 0
for j in range(i,N+1):
if i%j==0:
total += b[j]
if a[i] != (total%2):
b[i] = 1
print(*b[1:])
if __name__ == "__main__":
main()
|
s081236137
|
Accepted
| 377
| 14,124
| 548
|
import sys
sys.setrecursionlimit(10 ** 6)
def input():
return sys.stdin.readline()[:-1]
def main():
N = int(input())
a = [-1]+list(map(int, input().split()))
b = [0] * (N+1)
for i in reversed(range(1,N+1)):
total = 0
for j in range(i,N+1,i):
total += b[j]
if a[i] != (total%2):
b[i] = 1
ans = []
cnt = 0
for i, n in enumerate(b):
if n!=0:
ans.append(i)
cnt += 1
print(cnt)
print(*ans)
if __name__ == "__main__":
main()
|
s289575855
|
p03469
|
u075303794
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 42
|
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.replace('8/', '9/')
print(s)
|
s106566191
|
Accepted
| 17
| 2,940
| 42
|
S=str(input())
print(S.replace('7/','8/'))
|
s727639277
|
p03457
|
u377834804
| 2,000
| 262,144
|
Wrong Answer
| 269
| 26,988
| 321
|
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.
|
N = int(input())
TXY = [list(map(int, input().split())) for _ in range(N)]
pre_t, pre_x, pre_y = 0, 0, 0
for t, x, y in TXY:
diff_t = t - pre_t
diff_pos = abs(x-pre_x) + abs(y-pre_y)
if diff_t > diff_pos or diff_pos-diff_t % 2 == 1:
print('No')
exit()
else:
pre_t, pre_x, pre_y = t, x, y
print('Yes')
|
s906551144
|
Accepted
| 270
| 26,864
| 324
|
N = int(input())
TXY = [list(map(int, input().split())) for _ in range(N)]
pre_t, pre_x, pre_y = 0, 0, 0
for t, x, y in TXY:
diff_t = t - pre_t
diff_pos = abs(x-pre_x) + abs(y-pre_y)
if diff_t < diff_pos or (diff_t-diff_pos) % 2 == 1:
print('No')
exit()
else:
pre_t, pre_x, pre_y = t, x, y
print('Yes')
|
s066785650
|
p03854
|
u607930911
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,188
| 199
|
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()
judge = "YES"
s = s.replace("eraser", "0")
s = s.replace("erase", "0")
s = s.replace("dreamer", "0")
s = s.replace("dream", "0")
if not s.isdigit():
judge = "NO"
print(s)
print(judge)
|
s135241110
|
Accepted
| 20
| 3,188
| 190
|
s = input()
judge = "YES"
s = s.replace("eraser", "0")
s = s.replace("erase", "0")
s = s.replace("dreamer", "0")
s = s.replace("dream", "0")
if not s.isdigit():
judge = "NO"
print(judge)
|
s851731734
|
p02613
|
u736443076
| 2,000
| 1,048,576
|
Wrong Answer
| 144
| 16,112
| 199
|
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)]
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'))))
|
s392057996
|
Accepted
| 136
| 16,184
| 125
|
n = int(input())
s = [input() for i in range(n)]
ans = ['AC', 'WA', 'TLE', 'RE']
for i in ans:
print(i, 'x', s.count(i))
|
s205283444
|
p03737
|
u977370660
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 72
|
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
|
a,b,c = input().split()
s1 = a[0]
s2 = b[0]
s3 = c[0]
print(s1+s2+s3)
|
s006048650
|
Accepted
| 17
| 2,940
| 83
|
N = input().upper()
N = N.split()
print("{}{}{}".format(N[0][0],N[1][0],N[2][0]))
|
s135557790
|
p03434
|
u375870553
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 317
|
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.
|
def main():
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
sw = 0
alice = 0
bob = 0
for i in a:
if sw == 0:
alice += i
sw = 1
else:
bob += i
sw = 0
print(alice-bob)
if __name__ == '__main__':
main()
|
s019384100
|
Accepted
| 17
| 3,060
| 330
|
def main():
n = int(input())
a = [int(i) for i in input().split()]
a.sort(reverse=True)
sw = 0
alice = 0
bob = 0
for i in a:
if sw == 0:
alice += i
sw = 1
else:
bob += i
sw = 0
print(alice-bob)
if __name__ == '__main__':
main()
|
s356422258
|
p02613
|
u598684283
| 2,000
| 1,048,576
|
Wrong Answer
| 145
| 9,224
| 324
|
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 x " + str(ac))
print("WA x " + str(wa))
print("TLE x " + str(tle))
print("RE x " + str(re))
|
s182256314
|
Accepted
| 144
| 9,044
| 324
|
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 " + str(ac))
print("WA x " + str(wa))
print("TLE x " + str(tle))
print("RE x " + str(re))
|
s718196038
|
p03861
|
u518042385
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 2,940
| 149
|
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?
|
x,y,z=map(int,input().split())
t1=0
t2=0
for i in range(1,x):
if i%z==0:
t1+=1
for i in range(1,y):
if i%z==0:
t2+=1
print(t2-t1)
|
s775541408
|
Accepted
| 17
| 2,940
| 88
|
a,s,d=map(int,input().split())
if a%d==0:
print(s//d-a//d+1)
else:
print(s//d-a//d)
|
s466715091
|
p02613
|
u465423770
| 2,000
| 1,048,576
|
Wrong Answer
| 145
| 9,204
| 264
|
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.
|
num = int(input())
count = {"AC":0,"WA":0,"TLE":0,"RE":0}
for _ in range(num):
s = input()
count[s] += 1
print("AC x {}".format(count["AC"]))
print("WA x {}".format(count["WA"]))
print("TLE x {}".format(count["TLE"]))
print("RE {}".format(count["RE"]))
|
s464682318
|
Accepted
| 146
| 9,212
| 266
|
num = int(input())
count = {"AC":0,"WA":0,"TLE":0,"RE":0}
for _ in range(num):
s = input()
count[s] += 1
print("AC x {}".format(count["AC"]))
print("WA x {}".format(count["WA"]))
print("TLE x {}".format(count["TLE"]))
print("RE x {}".format(count["RE"]))
|
s379080153
|
p03448
|
u995861601
| 2,000
| 262,144
|
Wrong Answer
| 50
| 3,064
| 222
|
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())
pattern = 0
for i in range(c+1):
for j in range(b+1):
for k in range(a+1):
if 500*a - 100*b - 50*c == x:
pattern += 1
print(pattern)
|
s217865645
|
Accepted
| 51
| 3,060
| 222
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
pattern = 0
for i in range(c+1):
for j in range(b+1):
for k in range(a+1):
if 500*k + 100*j + 50*i == x:
pattern += 1
print(pattern)
|
s821358558
|
p02407
|
u187646742
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,608
| 85
|
Write a program which reads a sequence and prints it in the reverse order.
|
a = list(map(int, input().split()))
a.sort(reverse=True)
print(" ".join(map(str, a)))
|
s846198358
|
Accepted
| 40
| 7,756
| 122
|
_ = input()
a = list(map(int, input().split()))
a = list(a[i] for i in range(len(a)-1,-1,-1))
print(" ".join(map(str, a)))
|
s066403644
|
p03436
|
u327248573
| 2,000
| 262,144
|
Wrong Answer
| 32
| 3,956
| 740
|
We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j). Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player repeatedly moves Kenus up, down, left or right by one square. The game is completed when Kenus reaches square (H, W) passing only white squares. Before Snuke starts the game, he can change the color of some of the white squares to black. However, he cannot change the color of square (1, 1) and (H, W). Also, changes of color must all be carried out before the beginning of the game. When the game is completed, Snuke's score will be the number of times he changed the color of a square before the beginning of the game. Find the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed, that is, Kenus can never reach square (H, W) regardless of how Snuke changes the color of the squares. The color of the squares are given to you as characters s_{i, j}. If square (i, j) is initially painted by white, s_{i, j} is `.`; if square (i, j) is initially painted by black, s_{i, j} is `#`.
|
from collections import deque
from itertools import chain
H, W = map(int, input().split(' '))
Maze = [list(input()) for _ in range(H)]
White = list(chain.from_iterable(Maze)).count('.')
Maze[0][0] = 1
q = deque([])
current = [0, 0]
q.append(current)
dir_x = [1, -1, 0, 0]
dir_y = [0, 0, 1, -1]
while len(q) > 0:
now_y, now_x = q.popleft()
print(now_y, now_x)
if now_y == H-1 and now_x == W-1 : break
for (dx, dy) in zip(dir_x, dir_y):
if 0 <= now_y + dy <= H-1 and 0 <= now_x + dx <= W-1 and Maze[now_y + dy][now_x + dx] == '.':
Maze[now_y + dy][now_x + dx] = Maze[now_y][now_x] + 1
q.append([now_y + dy, now_x + dx])
if Maze[H-1][W-1] == '.': print(-1)
else: print(White - Maze[H-1][W-1])
|
s843218868
|
Accepted
| 28
| 3,316
| 716
|
from collections import deque
from itertools import chain
H, W = map(int, input().split(' '))
Maze = [list(input()) for _ in range(H)]
White = list(chain.from_iterable(Maze)).count('.')
Maze[0][0] = 1
q = deque([])
current = [0, 0]
q.append(current)
dir_x = [1, -1, 0, 0]
dir_y = [0, 0, 1, -1]
while len(q) > 0:
now_y, now_x = q.popleft()
if now_y == H-1 and now_x == W-1 : break
for (dx, dy) in zip(dir_x, dir_y):
if 0 <= now_y + dy <= H-1 and 0 <= now_x + dx <= W-1 and Maze[now_y + dy][now_x + dx] == '.':
Maze[now_y + dy][now_x + dx] = Maze[now_y][now_x] + 1
q.append([now_y + dy, now_x + dx])
if Maze[H-1][W-1] == '.': print(-1)
else: print(White - Maze[H-1][W-1])
|
s288642320
|
p03359
|
u806855121
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 95
|
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())
ans = a - 1
if b > a:
print(ans+1)
else:
print(ans)
|
s245146425
|
Accepted
| 17
| 2,940
| 80
|
a, b = map(int, input().split())
if b >= a:
print(a)
else:
print(a-1)
|
s096703893
|
p03853
|
u652656291
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 117
|
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
h,w = map(int,input().split())
l = list(map(str,input().split()))
for i in range(len(l)):
print(l[i])
print(l[i])
|
s391484721
|
Accepted
| 17
| 3,060
| 112
|
h, w = list(map(int, input().split()))
c = [input() for i in range(h)]
for i in range(h*2):
print(c[i // 2])
|
s099884099
|
p03160
|
u715107458
| 2,000
| 1,048,576
|
Wrong Answer
| 50
| 14,476
| 432
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
from functools import lru_cache
N = int(input())
heights = [int(h) for h in input().split()]
@lru_cache(None)
def dfs(i):
if i == len(heights) - 1:
return 0
if i >= len(heights):
return float("inf")
return min(abs(heights[i] - heights[i + 1]) + dfs(i + 1) if i + 1 < len(heights) else float("inf"),
abs(heights[i] - heights[i + 2]) + dfs(i + 2) if i + 2 < len(heights) else float("inf"))
|
s780255115
|
Accepted
| 137
| 14,976
| 849
|
import sys
from functools import lru_cache
sys.setrecursionlimit(20000000)
N = int(input())
heights = [int(h) for h in input().split()]
# heights = [10, 30, 40, 20]
# heights = [10, 10]
# heights = [30, 10, 60, 10, 60, 50]
@lru_cache(None)
def dfs(i):
if i == len(heights) - 1:
return 0
return min(abs(heights[i] - heights[i + 1]) + dfs(i + 1) if i + 1 < len(heights) else float("inf"),
abs(heights[i] - heights[i + 2]) + dfs(i + 2) if i + 2 < len(heights) else float("inf"))
def helper():
dp = [float("inf")] * len(heights)
dp[0] = 0
for i in range(1, len(heights)):
dp[i] = min(dp[i], dp[i - 1] + abs(heights[i] - heights[i - 1]))
if i > 1:
dp[i] = min(dp[i], dp[i - 2] + abs(heights[i] - heights[i - 2]))
return dp[-1]
# print(dfs(0))
print(helper())
|
s305601350
|
p03471
|
u117193815
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 3,060
| 220
|
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=map(int, input().split())
a=10000
b=5000
c=1000
for i in range(n+1):
for j in range(n-i):
for k in range(n-i-j):
if (i*a)+(j*b)+(k*c)==y:
print(i,j,k)
exit()
print(-1,-1,-1)
|
s608795299
|
Accepted
| 782
| 3,060
| 184
|
n,y=map(int, input().split())
a=-1
b=-1
c=-1
for i in range(n+1):
for j in range(n-i+1):
k=n-i-j
if (i*10000)+(j*5000)+(k*1000)==y:
a=i
b=j
c=k
print(a,b,c)
|
s337371739
|
p03478
|
u524765246
| 2,000
| 262,144
|
Wrong Answer
| 33
| 3,064
| 390
|
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).
|
lst = list(map(int, input().split()))
ans = 0
tot = 0
cnt = 0
for i in range(1,lst[0]+1):
flg = True
val = i
while flg:
cnt += 1
tot = tot + val%10
val = val//10
if val==0:
if ( lst[1] <= tot and
tot <= lst[2] ):
ans = ans + i
tot = 0
flg = False
print("Sum: " + str(ans))
|
s570567792
|
Accepted
| 34
| 3,064
| 379
|
lst = list(map(int, input().split()))
ans = 0
tot = 0
cnt = 0
for i in range(1,lst[0]+1):
flg = True
val = i
while flg:
cnt += 1
tot = tot + val%10
val = val//10
if val==0:
if ( lst[1] <= tot and
tot <= lst[2] ):
ans = ans + i
tot = 0
flg = False
print(ans)
|
s843203510
|
p03854
|
u143492911
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,188
| 146
|
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().replace("eraser",",").replace("eraser",",").replace("dreamer",",").replace("dream",",").replace(",","")
print("YES" if s=="" else "NO")
|
s488050412
|
Accepted
| 23
| 6,516
| 99
|
import re
s=input()
print("YES") if re.match("^(dream|dreamer|erase|eraser)+$",s) else print("NO")
|
s789296413
|
p03563
|
u209951743
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 128
|
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.
|
N = int(input())
K = int(input())
S = 1
for i in range(N):
if S < K:
S = S*2
else:
S = S + K
print(S)
|
s732375729
|
Accepted
| 17
| 2,940
| 57
|
R = int(input())
G = int(input())
A =G * 2 - R
print(A)
|
s199527831
|
p03556
|
u763881112
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,108
| 36,824
| 98
|
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
import numpy as np
n=int(input())
for i in range(10**9):
if(i*i>n):
print((i-1)**2)
|
s438010668
|
Accepted
| 17
| 3,060
| 32
|
print(int(int(input())**0.5)**2)
|
s142545992
|
p02612
|
u809819902
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,140
| 24
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
print(int(input())%1000)
|
s217177993
|
Accepted
| 29
| 9,156
| 53
|
n=int(input())
print(0 if n%1000==0 else 1000-n%1000)
|
s318546299
|
p03998
|
u135521563
| 2,000
| 262,144
|
Wrong Answer
| 24
| 3,316
| 527
|
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.
|
a = input()
b = input()
c = input()
a_pos = 0
b_pos = 0
c_pos = 0
now = 'a'
while True:
print(a_pos, b_pos, c_pos, now)
if now == 'a':
if len(a) < a_pos + 1:
print('A')
break
now = a[a_pos]
a_pos += 1
elif now == 'b':
if len(b) < b_pos + 1:
print('B')
break
now = b[b_pos]
b_pos += 1
elif now == 'c':
if len(c) < c_pos + 1:
print('C')
break
now = c[c_pos]
c_pos += 1
|
s525955205
|
Accepted
| 23
| 3,064
| 491
|
a = input()
b = input()
c = input()
a_pos = 0
b_pos = 0
c_pos = 0
now = 'a'
while True:
if now == 'a':
if len(a) < a_pos + 1:
print('A')
break
now = a[a_pos]
a_pos += 1
elif now == 'b':
if len(b) < b_pos + 1:
print('B')
break
now = b[b_pos]
b_pos += 1
elif now == 'c':
if len(c) < c_pos + 1:
print('C')
break
now = c[c_pos]
c_pos += 1
|
s599651206
|
p03719
|
u153729035
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 65
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
A,B,C=list(map(int,input().split()))
print(['NO','YES'][A<=C<=B])
|
s269536126
|
Accepted
| 17
| 2,940
| 65
|
A,B,C=list(map(int,input().split()))
print(['No','Yes'][A<=C<=B])
|
s373560531
|
p02613
|
u954153335
| 2,000
| 1,048,576
|
Wrong Answer
| 156
| 16,616
| 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.
|
import collections
n=int(input())
data=[]
for i in range(n):
num=input()
data.append(num)
data=collections.Counter(data)
print("AC × {}".format(data["AC"]))
print("WA × {}".format(data["WA"]))
print("TLE × {}".format(data["TLE"]))
print("RE × {}".format(data["RE"]))
|
s461168501
|
Accepted
| 159
| 16,620
| 274
|
import collections
n=int(input())
data=[]
for i in range(n):
num=input()
data.append(num)
data=collections.Counter(data)
print("AC x {}".format(data["AC"]))
print("WA x {}".format(data["WA"]))
print("TLE x {}".format(data["TLE"]))
print("RE x {}".format(data["RE"]))
|
s217238147
|
p03826
|
u060793972
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 52
|
There are two rectangles. The lengths of the vertical sides of the first rectangle are A, and the lengths of the horizontal sides of the first rectangle are B. The lengths of the vertical sides of the second rectangle are C, and the lengths of the horizontal sides of the second rectangle are D. Print the area of the rectangle with the larger area. If the two rectangles have equal areas, print that area.
|
a,b,c,d=map(int,input().split())
print(min(a*b,c*d))
|
s636534461
|
Accepted
| 17
| 2,940
| 52
|
a,b,c,d=map(int,input().split())
print(max(a*b,c*d))
|
s486634704
|
p03386
|
u773246942
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,316
| 164
|
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())
C = []
for i in range(A, A+K):
C.append(i)
D = []
for i in range(B, B-K, -1) :
D.append(i)
E = set(C) | set(D)
print(E)
|
s252736684
|
Accepted
| 17
| 3,060
| 295
|
A, B, K =map(int,input().split())
if B-A <= K:
for i in range(A, B+1):
print(i)
else:
C = []
for i in range(A, A+K):
C.append(i)
D = []
for i in range(B, B-K, -1) :
D.append(i)
E = set(C) | set(D)
F = sorted(E)
for i in range(len(F)):
print(F[i])
|
s640978200
|
p03548
|
u580362735
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 62
|
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 + 2*Z))//(Y+Z))
|
s804043424
|
Accepted
| 17
| 2,940
| 65
|
X,Y,Z = map(int,input().split())
print((X - (Y + 2*Z))//(Y+Z)+1)
|
s006365623
|
p03409
|
u419686324
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 4,596
| 777
|
On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i). A red point and a blue point can form a _friendly pair_ when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point. At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.
|
N = int(input())
R = [tuple([int(x) for x in input().split()]) for _ in range(N)]
B = [tuple([int(x) for x in input().split()]) for _ in range(N)]
xl = [set() for _ in range(2 * N + 1)]
yl = [set() for _ in range(2 * N + 1)]
for b in B:
x, y = b
for i in range(x):
xl[i].add(b)
for j in range(y):
yl[j].add(b)
cands = []
for r in R:
x, y = r
s = xl[x] and yl[y]
cands.append([r, len(s), s])
def f(cands, used, n):
if not cands:
return n
ret = 0
hd, tl = cands[0], cands[1:]
r, _, s = hd
for t in s - used:
u = used.union({t})
ret = max(ret, f(tl, u, n + 1))
ret = max(ret, f(tl, used, n))
return ret
import operator
cands.sort(key=operator.itemgetter(1))
print(f(cands, set(), 0))
|
s280731441
|
Accepted
| 19
| 3,188
| 384
|
N = int(input())
R = [[int(x) for x in input().split()] for _ in range(N)]
B = [[int(x) for x in input().split()] for _ in range(N)]
import operator
R.sort(key=operator.itemgetter(1), reverse=True)
B.sort()
ans = 0
for b in B:
xb, yb = b
for r in R:
xr, yr = r
if xr < xb and yr < yb:
ans += 1
R.remove(r)
break
print(ans)
|
s022894239
|
p02692
|
u961674365
| 2,000
| 1,048,576
|
Wrong Answer
| 313
| 17,328
| 5,248
|
There is a game that involves three variables, denoted A, B, and C. As the game progresses, there will be N events where you are asked to make a choice. Each of these choices is represented by a string s_i. If s_i is `AB`, you must add 1 to A or B then subtract 1 from the other; if s_i is `AC`, you must add 1 to A or C then subtract 1 from the other; if s_i is `BC`, you must add 1 to B or C then subtract 1 from the other. After each choice, none of A, B, and C should be negative. Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.
|
n,a,b,c = map(int,input().split())
ss=['' for _ in range(n)]
ans=['' for _ in range(n)]
wa=a+b+c
for i in range(n):
s=input()
ss[i]=s
i=0
if wa==0:
print('No')
exit()
elif wa==1:
for s in ss:
if s=='AB':
if c:
print('No')
exit()
elif b:
a+=1
b-=1
ans[i]='A'
elif a:
b+=1
a-=1
ans[i]='B'
elif s=='AC':
if b:
print('No')
exit()
elif a:
c+=1
a-=1
ans[i]='C'
elif c:
a+=1
c-=1
ans[i]='A'
elif s=='BC':
if a:
print('No')
exit()
elif b:
c+=1
b-=1
ans[i]='C'
elif c:
b+=1
c-=1
ans[i]='B'
i+=1
elif wa==2:
for s in ss:
if s=='AB':
if a==0 and b==0:
print('No')
exit()
elif a==0:
a+=1
b-=1
ans[i]='A'
elif b==0:
b+=1
a-=1
ans[i]='B'
elif a==1 and b==1:
if i==n-1:
ans[i]='A'
else:
if ss[i+1]=='AC' or ss[i+1]=='AB':
a+=1
b-=1
ans[i]='A'
elif ss[i+1]=='BC':
b+=1
a-=1
ans[i]='B'
elif s=='AC':
if a==0 and c==0:
print('No')
exit()
elif a==0:
a+=1
c-=1
ans[i]='A'
elif c==0:
c+=1
a-=1
ans[i]='C'
elif a==1 and c==1:
if i==n-1:
ans[i]='A'
else:
if ss[i+1]=='AC' or ss[i+1]=='AB':
a+=1
c-=1
ans[i]='A'
elif ss[i+1]=='BC':
c+=1
a-=1
ans[i]='C'
elif s=='BC':
if b==0 and c==0:
print('No')
exit()
elif b==0:
b+=1
c-=1
ans[i]='B'
elif c==0:
c+=1
b-=1
ans[i]='C'
elif b==1 and c==1:
if i==n-1:
ans[i]='B'
else:
if ss[i+1]=='BC' or ss[i+1]=='AB':
b+=1
c-=1
ans[i]='B'
elif ss[i+1]=='AC':
c+=1
b-=1
ans[i]='C'
i+=1
elif wa>2:
for i in range(n):
s=ss[i]
if s=='AB':
if a==0 and b==0:
print('No')
exit()
elif a==0:
a+=1
b-=1
ans[i]='A'
elif b==0:
b+=1
a-=1
ans[i]='B'
else:
if i==n-1:
ans[i]='A'
a+=1
b-=1
else:
if b>=a:
a+=1
b-=1
ans[i]='A'
else:
b+=1
a-=1
ans[i]='B'
elif s=='AC':
if a==0 and c==0:
print('No')
exit()
elif a==0:
a+=1
c-=1
ans[i]='A'
elif c==0:
c+=1
a-=1
ans[i]='C'
else:
if i==n-1:
ans[i]='A'
a+=1
c-=1
else:
if c>=a:
a+=1
c-=1
ans[i]='A'
else:
c+=1
a-=1
ans[i]='C'
elif s=='BC':
if b==0 and c==0:
print('No')
exit()
elif b==0:
b+=1
c-=1
ans[i]='B'
elif c==0:
c+=1
b-=1
ans[i]='C'
else:
if i==n-1:
ans[i]='B'
b+=1
c-=1
else:
if c>=b:
b+=1
c-=1
ans[i]='B'
else:
c+=1
b-=1
ans[i]='C'
print(i,a,b,c)
print('Yes')
for x in ans:
print(x)
|
s091019027
|
Accepted
| 225
| 16,988
| 5,249
|
n,a,b,c = map(int,input().split())
ss=['' for _ in range(n)]
ans=['' for _ in range(n)]
wa=a+b+c
for i in range(n):
s=input()
ss[i]=s
i=0
if wa==0:
print('No')
exit()
elif wa==1:
for s in ss:
if s=='AB':
if c:
print('No')
exit()
elif b:
a+=1
b-=1
ans[i]='A'
elif a:
b+=1
a-=1
ans[i]='B'
elif s=='AC':
if b:
print('No')
exit()
elif a:
c+=1
a-=1
ans[i]='C'
elif c:
a+=1
c-=1
ans[i]='A'
elif s=='BC':
if a:
print('No')
exit()
elif b:
c+=1
b-=1
ans[i]='C'
elif c:
b+=1
c-=1
ans[i]='B'
i+=1
elif wa==2:
for s in ss:
if s=='AB':
if a==0 and b==0:
print('No')
exit()
elif a==0:
a+=1
b-=1
ans[i]='A'
elif b==0:
b+=1
a-=1
ans[i]='B'
elif a==1 and b==1:
if i==n-1:
ans[i]='A'
else:
if ss[i+1]=='AC' or ss[i+1]=='AB':
a+=1
b-=1
ans[i]='A'
elif ss[i+1]=='BC':
b+=1
a-=1
ans[i]='B'
elif s=='AC':
if a==0 and c==0:
print('No')
exit()
elif a==0:
a+=1
c-=1
ans[i]='A'
elif c==0:
c+=1
a-=1
ans[i]='C'
elif a==1 and c==1:
if i==n-1:
ans[i]='A'
else:
if ss[i+1]=='AC' or ss[i+1]=='AB':
a+=1
c-=1
ans[i]='A'
elif ss[i+1]=='BC':
c+=1
a-=1
ans[i]='C'
elif s=='BC':
if b==0 and c==0:
print('No')
exit()
elif b==0:
b+=1
c-=1
ans[i]='B'
elif c==0:
c+=1
b-=1
ans[i]='C'
elif b==1 and c==1:
if i==n-1:
ans[i]='B'
else:
if ss[i+1]=='BC' or ss[i+1]=='AB':
b+=1
c-=1
ans[i]='B'
elif ss[i+1]=='AC':
c+=1
b-=1
ans[i]='C'
i+=1
elif wa>2:
for i in range(n):
s=ss[i]
if s=='AB':
if a==0 and b==0:
print('No')
exit()
elif a==0:
a+=1
b-=1
ans[i]='A'
elif b==0:
b+=1
a-=1
ans[i]='B'
else:
if i==n-1:
ans[i]='A'
a+=1
b-=1
else:
if b>=a:
a+=1
b-=1
ans[i]='A'
else:
b+=1
a-=1
ans[i]='B'
elif s=='AC':
if a==0 and c==0:
print('No')
exit()
elif a==0:
a+=1
c-=1
ans[i]='A'
elif c==0:
c+=1
a-=1
ans[i]='C'
else:
if i==n-1:
ans[i]='A'
a+=1
c-=1
else:
if c>=a:
a+=1
c-=1
ans[i]='A'
else:
c+=1
a-=1
ans[i]='C'
elif s=='BC':
if b==0 and c==0:
print('No')
exit()
elif b==0:
b+=1
c-=1
ans[i]='B'
elif c==0:
c+=1
b-=1
ans[i]='C'
else:
if i==n-1:
ans[i]='B'
b+=1
c-=1
else:
if c>=b:
b+=1
c-=1
ans[i]='B'
else:
c+=1
b-=1
ans[i]='C'
#print(i,a,b,c)
print('Yes')
for x in ans:
print(x)
|
s609726561
|
p03605
|
u705418271
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,080
| 72
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
n=str(input())
if n[0]==9 or n[1]==9:
print("Yes")
else:
print("No")
|
s894063718
|
Accepted
| 31
| 9,080
| 57
|
n=input()
if "9" in n:
print("Yes")
else:
print("No")
|
s702768042
|
p03476
|
u102126195
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 7,960
| 663
|
We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime. You are given Q queries. In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.
|
PN = [i for i in range(100001)]
i = 2
print(PN[1:20])
while True:
if i >= len(PN):
break
j = i * 2
while True:
if j >= len(PN):
break
PN[j] = 0
j += i
while True:
i += 1
if i >= len(PN):
break
if PN[i] != 0:
break
PN[1] = 0
PN_2 = []
for i in range(100001):
if PN[i] != 0 and PN[int((PN[i] + 1) / 2)] != 0:
PN_2.append(PN[i])
else:
PN_2.append(0)
Q = int(input())
lr = []
for i in range(Q):
cnt = 0
l, r = map(int, input().split())
for j in range(l, r + 1):
if PN_2[j] != 0:
cnt += 1
print(cnt)
|
s302808556
|
Accepted
| 969
| 9,068
| 727
|
PN = [i for i in range(100001)]
i = 2
while True:
if i >= len(PN):
break
j = i * 2
while True:
if j >= len(PN):
break
PN[j] = 0
j += i
while True:
i += 1
if i >= len(PN):
break
if PN[i] != 0:
break
PN[1] = 0
PN_2 = [0]
DPN = [0]
for i in range(1, 100001):
if PN[i] != 0 and PN[int((PN[i] + 1) / 2)] != 0:
PN_2.append(PN[i])
DPN.append(DPN[i - 1] + 1)
else:
PN_2.append(0)
DPN.append(DPN[i - 1])
#print(PN[1:30])
#print(PN_2[1:30])
#print(DPN[1:30])
Q = int(input())
lr = []
for i in range(Q):
cnt = 0
l, r = map(int, input().split())
print(DPN[r] - DPN[l - 1])
|
s656547169
|
p03730
|
u906017074
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 119
|
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 = input().split()
ans = 'NO'
if a[len(a)-1] == b[0]:
if b[len(b)-1 == c[0]]:
ans = 'YES'
print(ans)
|
s058851497
|
Accepted
| 17
| 2,940
| 235
|
a, b, c = [int(i) for i in input().split()]
ans='NO'
i = 1
f = a % b
while(1):
tmp = a * i
mod = tmp % b
if mod == c:
ans = 'YES'
break
if i != 1 and mod == f:
break
i += 1
print(ans)
|
s996615869
|
p00002
|
u506705885
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,372
| 189
|
Write a program which computes the digit number of sum of two integers a and b.
|
answers=[]
while True:
try:
nums=input().split()
answers.append(len(nums[0])+len(nums[1]))
except:
break
for i in range(len(answers)):
print(answers[i])
|
s792266806
|
Accepted
| 20
| 7,656
| 199
|
answers=[]
while True:
try:
nums=input().split()
answers.append(len(str(int(nums[0])+int(nums[1]))))
except :
break
for i in range(len(answers)):
print(answers[i])
|
s059173884
|
p03494
|
u750651325
| 2,000
| 262,144
|
Wrong Answer
| 26
| 8,996
| 166
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
A = list(map(int, input().split()))
a = 0
count = 0
N = [0]*len(A)
for i in range(len(A)):
if A[i] % 2 == 0:
N[i] += 1
print(min(N))
|
s973312912
|
Accepted
| 31
| 9,084
| 203
|
a = int(input())
A = list(map(int, input().split()))
N = [0]*len(A)
for i in range(a):
n = A[i]
while n % 2 == 0:
if n % 2 == 0:
n /= 2
N[i] += 1
print(min(N))
|
s363979159
|
p03433
|
u939552576
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 72
|
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())
print('YES' if N % 500 <= A else 'NO')
|
s181797789
|
Accepted
| 17
| 2,940
| 72
|
N = int(input())
A = int(input())
print('Yes' if N % 500 <= A else 'No')
|
s966829146
|
p03485
|
u341736906
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 89
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a,b = map(int,input().split())
if (a+b) % 2==0:
print((a+b)/2)
else:
print((a+b+1)/2)
|
s373667792
|
Accepted
| 17
| 2,940
| 73
|
from math import ceil
a,b = map(int,input().split())
print(ceil((a+b)/2))
|
s992819015
|
p03129
|
u657221245
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 119
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
a = list(map(int, input().split()))
b = - ( -a[0] // 2 )
if b >= a[1]:
print('yes')
else:
print('no')
|
s815107519
|
Accepted
| 17
| 2,940
| 119
|
a = list(map(int, input().split()))
b = - ( -a[0] // 2 )
if b >= a[1]:
print('YES')
else:
print('NO')
|
s507318326
|
p03493
|
u278379520
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 43
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
a=list(input().split())
print(a.count('1'))
|
s873363162
|
Accepted
| 17
| 2,940
| 91
|
a=input()
b=0
if a[0]=='1':
b+=1
if a[1]=='1':
b+=1
if a[2]=='1':
b+=1
print(b)
|
s879201961
|
p03737
|
u820351940
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 53
|
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
|
"".join(map(lambda x: x[0].upper(), input().split()))
|
s340178019
|
Accepted
| 17
| 2,940
| 60
|
print("".join(map(lambda x: x[0].upper(), input().split())))
|
s127002180
|
p04043
|
u873715358
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 81
|
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 = input().split().sort()
if a == [5, 5, 7]:
print('YES')
else:
print('NO')
|
s260305469
|
Accepted
| 17
| 2,940
| 100
|
a = [int(n) for n in input().split()]
a.sort()
if a == [5, 5, 7]:
print('YES')
else:
print('NO')
|
s124947540
|
p03712
|
u137228327
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,456
| 254
|
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.
|
from collections import deque
H,W = map(int,input().split())
st = [str(input()) for _ in range(H)]
print(st)
st = deque(st)
for i in range(H):
st[i] = '#'+st[i]+'#'
st.appendleft('#'*(W+2))
st.append('#'*(W+2))
for i in range(H+2):
print(st[i])
|
s113401206
|
Accepted
| 28
| 9,180
| 136
|
H,W = map(int,input().split())
st = ['#' + str(input()) +'#' for _ in range(H)]
st = ['#'*(W+2)] + st + ['#'*(W+2)]
print(*st,sep='\n')
|
s512260119
|
p02401
|
u215335591
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,612
| 294
|
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:
x = [i for i in input().split()]
if x[1] == '?':
break
a = int(x[0])
b = int(x[2])
if x[1] == '+':
print (a + b)
elif x[1] == '-':
print (a - b)
elif x[1] == '*':
print (a * b)
elif x[1] == '/':
print (a / b)
|
s457278313
|
Accepted
| 20
| 5,604
| 295
|
while True:
x = [i for i in input().split()]
if x[1] == '?':
break
a = int(x[0])
b = int(x[2])
if x[1] == '+':
print (a + b)
elif x[1] == '-':
print (a - b)
elif x[1] == '*':
print (a * b)
elif x[1] == '/':
print (a // b)
|
s811446917
|
p03415
|
u994988729
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 48
|
We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}. Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.
|
print([input()[0],input()[1],input()[2]],sep="")
|
s348173777
|
Accepted
| 18
| 2,940
| 50
|
print("".join([input()[0],input()[1],input()[2]]))
|
s489206792
|
p02831
|
u056830573
| 2,000
| 1,048,576
|
Wrong Answer
| 51
| 3,064
| 1,081
|
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
number_of_guests = input().split()
A = int(number_of_guests[0])
B = int(number_of_guests[1])
A_divisors = []
x = 2
while True:
if (A % x == 0):
A /= x
A_divisors.append(x)
else:
x += 1
if (A <= 1):
break
y = 2
B_divisors = []
while True:
if (B % y == 0):
B /= y
B_divisors.append(y)
else:
y += 1
if (B <= 1):
break
A_only = [x for x in A_divisors if (x not in B_divisors)]
B_only = [y for y in A_divisors if (y not in A_divisors)]
common_divisors = []
A_divisors = [3, 3, 2]
B_divisors = [3, 3, 4, 2, 3]
for z in (A_divisors + B_divisors):
if z in A_divisors and z in B_divisors:
if (A_divisors.count(z) >= B_divisors.count(z)):
common_divisors.append(z**A_divisors.count(z))
else:
common_divisors.append(z**B_divisors.count(z))
A_divisors = list(filter(lambda x: x != z, A_divisors))
B_divisors = list(filter(lambda x: x != z, B_divisors))
result = 1
for x in A_only + B_only + common_divisors:
result *= x
print(x)
|
s099878125
|
Accepted
| 51
| 3,064
| 1,034
|
number_of_guests = input().split()
A = int(number_of_guests[0])
B = int(number_of_guests[1])
A_divisors = []
x = 2
while True:
if (A % x == 0):
A /= x
A_divisors.append(x)
else:
x += 1
if (A <= 1):
break
y = 2
B_divisors = []
while True:
if (B % y == 0):
B /= y
B_divisors.append(y)
else:
y += 1
if (B <= 1):
break
A_only = [x for x in A_divisors if (x not in B_divisors)]
B_only = [y for y in B_divisors if (y not in A_divisors)]
common_divisors = []
for z in (A_divisors + B_divisors):
if z in A_divisors and z in B_divisors:
if (A_divisors.count(z) >= B_divisors.count(z)):
common_divisors.append(z**A_divisors.count(z))
else:
common_divisors.append(z**B_divisors.count(z))
A_divisors = list(filter(lambda x: x != z, A_divisors))
B_divisors = list(filter(lambda x: x != z, B_divisors))
result = 1
for x in A_only + B_only + common_divisors:
result *= x
print(result)
|
s651916007
|
p00003
|
u804558166
| 1,000
| 131,072
|
Wrong Answer
| 50
| 5,596
| 193
|
Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so.
|
n = int(input())
for i in range(n):
x, y, z =map(int, input().split())
if x*x + y*y == z*z or y*y + z*z == x*x or x*x + z*z == y*y :
print("Yes")
else:
print("No")
|
s255965765
|
Accepted
| 30
| 5,592
| 193
|
n = int(input())
for i in range(n):
x, y, z =map(int, input().split())
if x*x + y*y == z*z or y*y + z*z == x*x or x*x + z*z == y*y :
print("YES")
else:
print("NO")
|
s818523528
|
p02747
|
u568576853
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 123
|
A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.
|
S=input()
while len(S)>=2:
if S[:2]=='hi':
S=S[2:]
else:
break
if len(S)==0:
print('YES')
else:
print('NO')
|
s973378993
|
Accepted
| 17
| 2,940
| 124
|
S=input()
while len(S)>=2:
if S[0:2]=='hi':
S=S[2:]
else:
break
if len(S)==0:
print('Yes')
else:
print('No')
|
s693851097
|
p02255
|
u162598098
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,404
| 197
|
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.
|
def insertionSort(lis,leng):
for i in range(leng):
v = lis[i]
j = i - 1
while j >= 0 and lis[j] > v:
lis[j+1] = lis[j]
j = j-1
lis[j+1]=v
|
s376649947
|
Accepted
| 30
| 8,000
| 348
|
def insertionSort(lis,leng):
for i in range(leng):
v = lis[i]
j = i - 1
while j >= 0 and lis[j] > v:
lis[j+1] = lis[j]
j = j-1
lis[j+1]=v
print(*lis)
if __name__ == "__main__":
lengg=int(input())
liss=list(map(int, input().split()))
insertionSort(liss,lengg)
|
s777393455
|
p03635
|
u851704997
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 55
|
In _K-city_ , there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city?
|
s = input()
x = len(s) - 2
print(s[0] + str(x) + s[-1])
|
s000298012
|
Accepted
| 17
| 2,940
| 58
|
n,m = map(int,input().split())
print(str((n - 1)*(m - 1)))
|
s281360950
|
p03606
|
u072717685
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 131
|
Joisino is working as a receptionist at a theater. The theater has 100000 seats, numbered from 1 to 100000. According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive). How many people are sitting at the theater now?
|
n = int()
seats = [list(map(int,input().split())) for _ in range(n)]
nop = [s[-1] - s[0] + 1 for s in seats]
r = sum(nop)
print(r)
|
s364473891
|
Accepted
| 20
| 3,188
| 141
|
n = int(input())
seats = [list(map(int, input().split())) for _ in range(n)]
nop = [s[-1] - s[0] + 1 for s in seats]
r = sum(nop)
print(r)
|
s075021074
|
p03037
|
u881141729
| 2,000
| 1,048,576
|
Wrong Answer
| 589
| 3,064
| 386
|
We have N ID cards, and there are M gates. We can pass the i-th gate if we have one of the following ID cards: the L_i- th, (L_i+1)-th, ..., and R_i-th ID cards. How many of the ID cards allow us to pass all the gates alone?
|
while True:
try:
n, m = map(int, input().split())
start = 1
end = n
for i in range(m):
a, b = map(int, input().split())
if a > end or b < start:
print(0)
break
start = max(start, a)
end = min(end, b)
else:
print(end-start+1)
except:
break
|
s986104769
|
Accepted
| 331
| 3,064
| 456
|
while True:
try:
n, m = map(int, input().split())
start = 1
end = n
for i in range(m):
a, b = map(int, input().split())
if a > end or b < start:
print(0)
for j in range(m-i-1):
input()
break
start = max(start, a)
end = min(end, b)
else:
print(end-start+1)
except:
break
|
s408917115
|
p03369
|
u048945791
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 37
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
s = input()
print(s.count("o") * 100)
|
s140104617
|
Accepted
| 17
| 2,940
| 43
|
s = input()
print(700 + s.count("o") * 100)
|
s103397239
|
p03854
|
u021916304
| 2,000
| 262,144
|
Wrong Answer
| 26
| 3,804
| 159
|
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`.
|
import re
s = input()
re.sub('eraser','0',s)
re.sub('erase','0',s)
re.sub('dreamer','0',s)
re.sub('dream','0',s)
print('YES' if set(s) == set('0') else 'NO')
|
s447491806
|
Accepted
| 23
| 3,656
| 215
|
import re
s = input()
s = re.sub('eraser','0',s)
#print(s)
s = re.sub('erase','0',s)
#print(s)
s = re.sub('dreamer','0',s)
#print(s)
s = re.sub('dream','0',s)
#print(s)
print('YES' if set(s) == set('0') else 'NO')
|
s731968094
|
p02747
|
u223904637
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 103
|
A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.
|
a='hi'
b=''
p=[]
for i in range(6):
b+=a
p.append(a)
s=input()
print('Yes' if s in p else 'No')
|
s631886643
|
Accepted
| 18
| 3,188
| 104
|
a='hi'
b=''
p=[]
for i in range(6):
b+=a
p.append(b)
s=input()
print('Yes' if s in p else 'No')
|
s569337669
|
p03816
|
u072717685
| 2,000
| 262,144
|
Wrong Answer
| 61
| 22,144
| 306
|
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written. He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept. Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
|
import sys
from collections import Counter
read = sys.stdin.read
readlines = sys.stdin.readlines
def main():
n, *a = map(int, read().split())
ac = Counter(a)
acl = ac.values()
r = len(acl)
t = (sum(ac.keys()) - r)&1
r -= t ^ 1
print(r)
if __name__ == '__main__':
main()
|
s834406205
|
Accepted
| 55
| 20,572
| 204
|
import sys
read = sys.stdin.read
readlines = sys.stdin.readlines
def main():
n, *a = map(int, read().split())
r = len(set(a))
r -= (n - r)&1
print(r)
if __name__ == '__main__':
main()
|
s621817436
|
p03049
|
u539517139
| 2,000
| 1,048,576
|
Wrong Answer
| 37
| 3,064
| 238
|
Snuke has N strings. The i-th string is s_i. Let us concatenate these strings into one string after arranging them in some order. Find the maximum possible number of occurrences of `AB` in the resulting string.
|
n=int(input())
a=0
b=0
ab=0
x=0
for i in range(n):
s=input()
l=len(s)-1
if s[0]=='B':
if s[l]=='A':
ab+=1
else:
b+=1
elif s[l]=='A':
a+=1
x+=s.count('AB')
if ab>0:
x+=(a>0)+(b>0)
x+=max(0,ab-1)
print(x)
|
s791823686
|
Accepted
| 36
| 3,064
| 279
|
n=int(input())
a=0
b=0
ab=0
x=0
for i in range(n):
s=input()
l=len(s)-1
if s[0]=='B':
if s[l]=='A':
ab+=1
else:
b+=1
elif s[l]=='A':
a+=1
x+=s.count('AB')
if ab>0:
x+=ab-1
x+=(a>0)+(b>0)
a-=1;b-=1
a=max(0,a);b=max(0,b)
x+=min(a,b)
print(x)
|
s066982958
|
p03591
|
u066337396
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 69
|
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`.
|
if input().startswith('YAKi'):
print('Yes')
else:
print('No')
|
s536577928
|
Accepted
| 17
| 2,940
| 69
|
if input().startswith('YAKI'):
print('Yes')
else:
print('No')
|
s312500637
|
p03760
|
u440975163
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,032
| 142
|
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
o = list(str(input()))
x = list(str(input()))
ln = []
for i in o:
for j in x:
ln.append(i)
ln.append(j)
print(''.join(ln))
|
s671653183
|
Accepted
| 28
| 8,952
| 159
|
o = list(str(input()))
x = list(str(input()))
ln = []
for i in range(len(o)):
ln.append(o[i])
if len(x) > i:
ln.append(x[i])
print(''.join(ln))
|
s931662071
|
p03457
|
u404561212
| 2,000
| 262,144
|
Wrong Answer
| 429
| 11,752
| 570
|
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.
|
N = int(input())
t = []
x = []
y = []
for i in range(N):
a_list = list(map(int, input().split()))
t.append(a_list[0])
x.append(a_list[1])
y.append(a_list[2])
for n in range(N):
if n == 0:
offset_x = abs(x[n])
offset_y = abs(y[n])
else:
offset_x = abs (x[n] - x[n-1])
offset_y = abs (y[n] - y[n-1])
offset_total = offset_x + offset_y
time_offset = 1 if n==0 else t[n] - t[n-1]
if offset_total != time_offset:
print("No")
break
else:
print("Yes")
|
s459552232
|
Accepted
| 433
| 11,872
| 1,175
|
def create_decrement_list(num):
decrement_list = []
while(num >= 0):
decrement_list.append(num)
num -=2
return decrement_list
def check_route(num_list, offset):
is_path_possible = False
for num in num_list:
if offset == num:
is_path_possible = True
break
return is_path_possible
def route_checker(N, list_t, list_x, list_y):
for n in range(N):
if n == 0:
offset_x = abs(list_x[n])
offset_y = abs(list_y[n])
else:
offset_x = abs (list_x[n] - list_x[n-1])
offset_y = abs (list_y[n] - list_y[n-1])
offset_total = offset_x + offset_y
time_offset = list_t[n] if n==0 else list_t[n] - list_t[n-1]
time_offset_list = create_decrement_list(time_offset)
if check_route(time_offset_list, offset_total)==False:
print("No")
break
else:
print("Yes")
N = int(input())
t = []
x = []
y = []
for i in range(N):
a_list = list(map(int, input().split()))
t.append(a_list[0])
x.append(a_list[1])
y.append(a_list[2])
route_checker(N, t, x, y)
|
s537194451
|
p00024
|
u553148578
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,628
| 75
|
Ignoring the air resistance, velocity of a freely falling object $v$ after $t$ seconds and its drop $y$ in $t$ seconds are represented by the following formulas: $ v = 9.8 t $ $ y = 4.9 t^2 $ A person is trying to drop down a glass ball and check whether it will crack. Your task is to write a program to help this experiment. You are given the minimum velocity to crack the ball. Your program should print the lowest possible floor of a building to crack the ball. The height of the $N$ floor of the building is defined by $5 \times N - 5$.
|
import math
print(math.ceil((math.ceil(4.9*(float(input())/9.8)**2)+5)/5))
|
s944249273
|
Accepted
| 20
| 5,640
| 105
|
import math
while 1:
try: print(math.ceil((math.ceil(4.9*(float(input())/9.8)**2)+5)/5))
except: break
|
s675076705
|
p03854
|
u309120194
| 2,000
| 262,144
|
Wrong Answer
| 55
| 9,240
| 358
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
S = input()
while True:
if S[-5:] == 'erase': S = S[:-5]
else:
if S[-6:] == 'eraser': S = S[:-6]
else:
if S[-5:] == 'dream': S = S[:-5]
else:
if S[-7:] == 'dreamer': S = S[:-7]
else: break
if len(S) == 0: print('Yes')
else: print('No')
|
s319460102
|
Accepted
| 61
| 9,164
| 324
|
S = input()
S = S[::-1]
words = {'maerd', 'remaerd', 'esare', 'resare'}
while len(S) > 0:
orig = S
for w in words:
if S.startswith(w):
S = S[len(w):]
break
if orig == S:
break
if len(S) == 0: print('YES')
else: print('NO')
|
s341670864
|
p03598
|
u244466744
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,088
| 180
|
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
N = int(input())
K = int(input())
x = list(map(int, input().split()))
sum = 0
for i in range(N):
if x[i] >= K - x[i]:
sum += K - x[i]
else:
sum += x[i]
print(sum)
|
s921187955
|
Accepted
| 25
| 9,004
| 190
|
N = int(input())
K = int(input())
x = list(map(int, input().split()))
sum = 0
for i in range(N):
if x[i] >= K - x[i]:
sum += 2 * (K - x[i])
else:
sum += 2 * x[i]
print(sum)
|
s207895226
|
p03545
|
u381282312
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,052
| 309
|
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.
|
n=list(map(str,input()))
cnt = len(n) - 1
for i in range(2**cnt):
op = ['-'] * cnt
for j in range(cnt):
if((i>>j) & 1):
op[cnt - 1 - j] = '+'
formula = ''
for p_n, p_o in zip(n, op+['']):
formula += (p_n + p_o)
if eval(formula) == 7:
print(formula + '==7')
break
|
s838064967
|
Accepted
| 28
| 8,964
| 356
|
n=list(map(str,input()))
cnt = len(n) - 1
for i in range(2**cnt):
op = ['-'] * cnt
for j in range(cnt):
if((i>>j) & 1):
op[cnt - 1 - j] = '+'
formula = ''
for p_n, p_o in zip(n, op+['']):
# n->p_n: 1
# op+['']->p_o: +, +, +, ''
formula += (p_n + p_o)
if eval(formula) == 7:
print(formula + '=7')
break
|
s250939132
|
p03795
|
u616522759
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 53
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
N = int(input())
x = 800 * N
y = N // 15
print(x - y)
|
s232372548
|
Accepted
| 17
| 2,940
| 59
|
N = int(input())
x = 800 * N
y = N // 15 * 200
print(x - y)
|
s187918772
|
p03456
|
u616413858
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 92
|
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.
|
joint = input().replace(" ", "")
num = int(joint)
root = num**(1/2)
print(root.is_integer())
|
s516526414
|
Accepted
| 17
| 2,940
| 152
|
joint = input().replace(" ", "")
num = int(joint)
root = num**(1/2)
# print(root)
print("Yes" if root.is_integer() else "No")
|
s842017649
|
p03478
|
u815666840
| 2,000
| 262,144
|
Wrong Answer
| 35
| 3,060
| 147
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n,a,b = map(int,input().split())
num = 0
for i in range(n+1):
if a <= sum(list(map(int,list(str(i))))) <= b:
num = num + 1
print(num)
|
s145385429
|
Accepted
| 37
| 3,060
| 148
|
n,a,b = map(int,input().split())
num = 0
for i in range(n+1):
if a <= sum(list(map(int,list(str(i))))) <= b:
num = num + i
print(num)
|
s310003157
|
p03472
|
u589381719
| 2,000
| 262,144
|
Wrong Answer
| 393
| 11,640
| 361
|
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?
|
from collections import deque
N,H = map(int,input().split())
A=[]
B=[]
d=deque()
max_axe = 0
max_throw = 0
ans=0
for i in range(N):
a,b = map(int,input().split())
A.append(a)
B.append(b)
A.sort(reverse=True)
B.sort(reverse=True)
for i in B:
if i>=A[0]:
H-=i
ans+=1
else:
break
ans+=int(-(-H/A[0]))
print(ans)
|
s806824620
|
Accepted
| 397
| 11,580
| 390
|
from collections import deque
N,H = map(int,input().split())
A=[]
B=[]
d=deque()
max_axe = 0
max_throw = 0
ans=0
for i in range(N):
a,b = map(int,input().split())
A.append(a)
B.append(b)
A.sort(reverse=True)
B.sort(reverse=True)
for i in B:
if i>=A[0]:
H-=i
ans+=1
if H<=0:break
else:
break
if H>0:ans+=int(-(-H//A[0]))
print(ans)
|
s095603166
|
p03779
|
u140251125
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 85
|
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
# input
X = int(input())
i = 1
while 2 * X <= i * (i - 1):
i += 1
print(i)
|
s178052777
|
Accepted
| 30
| 2,940
| 84
|
# input
X = int(input())
i = 1
while 2 * X > i * (i + 1):
i += 1
print(i)
|
s621993290
|
p03836
|
u059210959
| 2,000
| 262,144
|
Wrong Answer
| 40
| 5,968
| 598
|
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.
|
# encoding:utf-8
import copy
import random
import bisect
import fractions
import math
import sys
mod = 10**9+7
sys.setrecursionlimit(mod)
sx,sy,gx,gy = map(int,input().split())
ans = []
# go
way1 = ["U"] * (gy-sy) + ["R"] * (gx-sx)
#return
way2 = ["D"] * (gx-sx) + ["L"] * (gy-sy)
way3 = ["L"] + ["U"] * (gy-sy+1) + ["R"] * (gx-sx+1) + ["D"]
way4 = ["R"] + ["D"] * (gx-sx+1) + ["L"] * (gy-sy+1) + ["U"]
ans += way1 + way2 + way3 + way4
print("".join(ans))
|
s479515533
|
Accepted
| 40
| 5,964
| 616
|
# encoding:utf-8
import copy
import random
import bisect
import fractions
import math
import sys
mod = 10**9+7
sys.setrecursionlimit(mod)
sx,sy,gx,gy = map(int,input().split())
ans = []
# go
way1 = ["U"] * (gy-sy) + ["R"] * (gx-sx)
#return
way2 = ["D"] * (gy-sy) + ["L"] * (gx-sx)
way3 = ["L"] + ["U"] * (gy-sy+1) + ["R"] * (gx-sx+1) + ["D"]
way4 = ["R"] + ["D"] * (gy-sy+1) + ["L"] * (gx-sx+1) + ["U"]
ans += way1 + way2 + way3 + way4
print("".join(ans))
# print(len(ans))
|
s312948289
|
p02388
|
u316584871
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,560
| 31
|
Write a program which calculates the cube of a given integer x.
|
s = float(input())
print(s**3)
|
s678403235
|
Accepted
| 20
| 5,576
| 29
|
s = int(input())
print(s**3)
|
s412434885
|
p02255
|
u387731924
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 287
|
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())
nums = []
nums = input().split()
def insertationSort(A,N):
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 = j - 1
A[j+1] = v
print(' '.join(A))
insertationSort(nums,n)
|
s480226412
|
Accepted
| 20
| 5,600
| 437
|
n = int(input())
nums = []
nums = input().split()
intNums = list(map(int,nums))
def insertationSort(nums,intNums):
print(' '.join(nums))
for i in range(1,len(nums)):
v = intNums[i]
j = i - 1
while j>=0 and intNums[j]>v:
intNums[j+1] = intNums[j]
j = j - 1
intNums[j+1] = v
nums = list(map(str,intNums))
print(' '.join(nums))
insertationSort(nums,intNums)
|
s861327258
|
p02833
|
u667024514
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 123
|
For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
|
n = int(input())
if n % 2 == 1:
print(0)
exit()
ans = 0
for i in range(1,50):
ans += (n // (5 ** i))
print(ans)
|
s470646868
|
Accepted
| 17
| 2,940
| 132
|
n = int(input())
if n % 2 == 1:
print(0)
exit()
n = n//2
ans = 0
for i in range(1,50):
ans += (n // (5 ** i))
print(ans)
|
s556249304
|
p02412
|
u957680575
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,600
| 254
|
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
while True:
n,x=map(int,input().split())
if n==0 and x==0:
break
sum=0
for i in range(x//3,n+1):
if (x-i)%2==0:
sum+=((x-i)/2)-1
else:
sum+=(x-i-1)/2
print(sum)
|
s418941927
|
Accepted
| 30
| 7,516
| 233
|
while True:
n,x=map(int,input().split())
if n==0 and x==0:
break
sum=0
for i in range(x//3+1,n+1):
for j in range((x-i)//2+1,i):
if 0<x-i-j<j :
sum+=1
print(int(sum))
|
s260955991
|
p03607
|
u798129018
| 2,000
| 262,144
|
Wrong Answer
| 197
| 11,884
| 98
|
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 = set()
for i in range(N):
ai = int(input())
a.add(ai)
print(len(list(a)))
|
s038211027
|
Accepted
| 186
| 17,884
| 238
|
n = int(input())
d = {}
for i in range(n):
a = input()
if a not in d:
d[a] = 1
else:
d[a] = d[a] + 1
ans = 0
for val in d.values():
if val % 2 == 1:
ans += 1
else:
continue
print(ans)
|
s871197016
|
p02390
|
u636711749
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,720
| 82
|
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
S =int(input())
h = S // 3600
m = S % 3600 // 60
s = S // 60
print(h,m,s, sep=':')
|
s335484720
|
Accepted
| 40
| 6,724
| 84
|
S = int(input())
h = S // 3600
m = S % 3600 // 60
s = S % 60
print(h, m, s, sep=':')
|
s954141546
|
p03610
|
u332331919
| 2,000
| 262,144
|
Wrong Answer
| 47
| 9,368
| 225
|
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
#072_b
s = input()
a = []
if s.islower() and 1 <= len(s) and len(s) <= 10 ** 5:
b=""
for n in range(len(s)):
if n % 2 != 0:
a.append(s[n])
for m in range(len(a)):
b=b+a[m]
print(b)
|
s481165470
|
Accepted
| 49
| 9,388
| 225
|
#072_b
s = input()
a = []
if s.islower() and 1 <= len(s) and len(s) <= 10 ** 5:
b=""
for n in range(len(s)):
if n % 2 == 0:
a.append(s[n])
for m in range(len(a)):
b=b+a[m]
print(b)
|
s365243237
|
p02261
|
u535719732
| 1,000
| 131,072
|
Wrong Answer
| 30
| 5,604
| 614
|
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
n = int(input())
a = list(map(str,input().split()))
data1 = a[:]
data2 = a[:]
def bubble_sort(data):
for i in range(len(data)):
for j in range(len(data)-1,i,-1):
if(data[j][1] < data[j-1][1]):
data[j],data[j-1] = data[j-1],data[j]
return(data)
def selection_sort(data):
for i in range(len(data)):
minj = i
for j in range(i,len(data)):
if(data[j][1] < data[minj][1]):
minj = j
data[i],data[minj] = data[minj],data[i]
return(data)
data1 = bubble_sort(data1)
data2 = selection_sort(data2)
print(*data1)
print("Stable")
print(*data2)
print("Stable" if data1 == data2 else "NotStable")
|
s863110627
|
Accepted
| 20
| 5,616
| 615
|
n = int(input())
a = list(map(str,input().split()))
data1 = a[:]
data2 = a[:]
def bubble_sort(data):
for i in range(len(data)):
for j in range(len(data)-1,i,-1):
if(data[j][1] < data[j-1][1]):
data[j],data[j-1] = data[j-1],data[j]
return(data)
def selection_sort(data):
for i in range(len(data)):
minj = i
for j in range(i,len(data)):
if(data[j][1] < data[minj][1]):
minj = j
data[i],data[minj] = data[minj],data[i]
return(data)
data1 = bubble_sort(data1)
data2 = selection_sort(data2)
print(*data1)
print("Stable")
print(*data2)
print("Stable" if data1 == data2 else "Not stable")
|
s394287283
|
p03997
|
u276785896
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 76
|
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(1/2 * (a + b) * h)
|
s434152309
|
Accepted
| 17
| 2,940
| 68
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h//2)
|
s791932470
|
p03779
|
u067986021
| 2,000
| 262,144
|
Wrong Answer
| 159
| 4,408
| 220
|
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
X = int(input())
i = 1
x = X
while(True):
if x == i:
print(i)
break
elif x >= (i * 2 + 1):
x = x - i
i += 1
elif x < i:
break
else:
i += 1
print(i, x)
|
s887124060
|
Accepted
| 27
| 3,060
| 79
|
X = int(input())
i = 1
x = 1
while(x <= X):
x += i
i += 1
print(i - 1)
|
s008143701
|
p03068
|
u829356061
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 153
|
You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.
|
A = int(input())
B = input()
C = int(input())
mark = B[C-1]
for i in range(A):
if B[i] != mark:
B = B.replace(B[i], '*', 1)
print(1)
print(B)
|
s736320080
|
Accepted
| 18
| 3,068
| 145
|
A = int(input())
B = input()
C = int(input())
mark = B[C-1]
for i in range(A):
if B[i] != mark:
B = B.replace(B[i], '*', 1)
print(B)
|
s354324717
|
p02646
|
u587518324
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 9,220
| 506
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
#!/usr/bin/env python
def run(a, v, b, w, t):
if a == b:
return 'Yes'
elif b < a:
if a - (v * t) <= b - (w * t):
return 'Yes'
else:
return 'No'
else:
if b + (w * t) <= a + (v * t):
return 'Yes'
else:
return 'No'
def main():
A, V = list(map(int, input().split()))
B, W = list(map(int, input().split()))
T = int(input())
print(run(A, V, B, W, T))
if __name__ == '__main__':
main()
|
s951678619
|
Accepted
| 25
| 9,200
| 594
|
#!/usr/bin/env python
def run(a, vt, b, wt):
if a < b:
if b + wt <= a + vt:
return 'YES'
else:
return 'NO'
else:
if a - vt <= b - wt:
return 'YES'
else:
return 'NO'
def main():
A, V = list(map(int, input().split()))
B, W = list(map(int, input().split()))
T = int(input())
print(run(A, V*T, B, W*T))
if __name__ == '__main__':
main()
|
s915348025
|
p03050
|
u010090035
| 2,000
| 1,048,576
|
Wrong Answer
| 144
| 3,936
| 412
|
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())
div=[]
for i in range(1,int((n**0.5)//1)+1):
if(n%i==0):
div.append(i)
div.append(n//i)
div=sorted(list(set(div)))
ans=[0]
#i//m + i%m
#m*x + x = i
#x = i/(m+1)
for divi in div:
m=divi-1
if(n//divi < m):
ans.append(m)
print(ans)
print(sum(ans))
|
s918179685
|
Accepted
| 156
| 3,936
| 401
|
n=int(input())
div=[]
for i in range(1,int((n**0.5)//1)+1):
if(n%i==0):
div.append(i)
div.append(n//i)
div=sorted(list(set(div)))
ans=[0]
#i//m + i%m
#m*x + x = i
#x = i/(m+1)
for divi in div:
m=divi-1
if(n//divi < m):
ans.append(m)
print(sum(ans))
|
s522352349
|
p02567
|
u368796742
| 5,000
| 1,048,576
|
Wrong Answer
| 2,075
| 31,536
| 1,895
|
You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. The type of i-th query is represented by T_i. * T_i=1: You are given two integers X_i,V_i. Replace the value of A_{X_i} with V_i. * T_i=2: You are given two integers L_i,R_i. Calculate the maximum value among A_{L_i},A_{L_i+1},\cdots,A_{R_i}. * T_i=3: You are given two integers X_i,V_i. Calculate the minimum j such that X_i \leq j \leq N, V_i \leq A_j. If there is no such j, answer j=N+1 instead.
|
class segtree:
## define what you want to do ,(min, max)
sta = -1
func = max
def __init__(self,n):
self.size = n
self.tree = [self.sta]*(2*n)
def build(self, list):
for i,x in enumerate(list,self.size):
self.tree[i] = x
for i in range(self.size-1,0,-1):
self.tree[i] = self.func(self.tree[i<<1],self.tree[i<<1 | 1])
def set(self,i,x):
i += self.size
self.tree[i] = x
while i > 1:
i >>= 1
self.tree[i] = self.func(self.tree[i<<1],self.tree[i<<1 | 1])
## take the value of [l,r)
def get(self,l,r):
l += self.size
r += self.size
res = self.sta
while l < r:
if l & 1:
res = self.func(self.tree[l],res)
l += 1
if r & 1:
res = self.func(self.tree[r-1],res)
l >>= 1
r >>= 1
return res
def max_right(self, l, x):
if l == self.size:
return l
l += self.size
res = self.sta
check = True
while check or (l & -l) != l:
check = False
while l%2 == 0:
l >>= 1
if x < self.func(res,self.tree[l]):
while l < self.size:
l <<= 1
if not x < self.func(res,self.tree[l]):
res = self.func(res,self.tree[l])
l += 1
return l - self.size
res = self.func(res,self.tree[l])
l += 1
return self.size
n,q = map(int,input().split())
a = list(map(int,input().split()))
seg = segtree(n)
seg.build(a)
for _ in range(q):
t,x,v = map(int,input().split())
if t == 1:
seg.set(x-1,v)
elif t == 2:
print(seg.get(x-1,v))
else:
print(seg.max_right(x-1,v)+1)
|
s626843221
|
Accepted
| 2,164
| 31,388
| 2,906
|
class SegTree:
""" define what you want to do with 0 index, ex) size = tree_size, func = min or max, sta = default_value """
def __init__(self,size,func,sta):
self.n = size
self.size = 1 << size.bit_length()
self.func = func
self.sta = sta
self.tree = [sta]*(2*self.size)
def build(self, list):
""" set list and update tree"""
for i,x in enumerate(list,self.size):
self.tree[i] = x
for i in range(self.size-1,0,-1):
self.tree[i] = self.func(self.tree[i<<1],self.tree[i<<1 | 1])
def set(self,i,x):
i += self.size
self.tree[i] = x
while i > 1:
i >>= 1
self.tree[i] = self.func(self.tree[i<<1],self.tree[i<<1 | 1])
def get(self,l,r):
""" take the value of [l r) with func (min or max)"""
l += self.size
r += self.size
res = self.sta
while l < r:
if l & 1:
res = self.func(self.tree[l],res)
l += 1
if r & 1:
res = self.func(self.tree[r-1],res)
l >>= 1
r >>= 1
return res
def max_right(self, l, x):
if l == self.n:
return l
l += self.size
res = self.sta
check = True
while check or (l & -l) != l:
check = False
while l%2 == 0:
l >>= 1
if not self.func(res,self.tree[l]) < x:
while l < self.size:
l <<= 1
if self.func(res,self.tree[l]) < x:
res = self.func(res,self.tree[l])
l += 1
return l - self.size
res = self.func(res,self.tree[l])
l += 1
return self.n
def min_left(self, r, x):
if r == 0:
return 0
r += self.size
res = self.sta
check = True
while check and (r & -r) != r:
check = False
r -= 1
while (r > 1 and r%2):
r >>= 1
if not self.func(res, self.tree[r]) < x:
while r < self.size:
r = 2*r + 1
if self.func(res, self.tree[r]) < x:
res = self.func(res, self.tree[r])
r -= 1
return r + 1 - self.size
res = self.func(self.tree[r],res)
return 0
n,q = map(int,input().split())
a = list(map(int,input().split()))
seg = SegTree(n,max,-1)
seg.build(a)
for _ in range(q):
t,x,v = map(int,input().split())
if t == 1:
seg.set(x-1,v)
elif t == 2:
print(seg.get(x-1,v))
else:
print(seg.max_right(x-1,v)+1)
|
s365964394
|
p03544
|
u103902792
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 130
|
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())
ans = [2,1]
if n == 1:
print(2)
exit()
for _ in range(n-2):
ans.append(ans[-1]+ans[-2])
print(ans[-1])
|
s382331237
|
Accepted
| 17
| 2,940
| 133
|
n = int(input())
ans = [2,1]
if n == 1:
print(1)
exit()
for _ in range(n-1):
ans.append(ans[-1]+ans[-2])
print(ans[-1])
|
s483027338
|
p03693
|
u075520262
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 181
|
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?
|
a,b,c=map(int, input().split())
d = int(a+b+c) % 4
if d != 0:
print('NO')
else:
print('YES')
|
s791569769
|
Accepted
| 17
| 2,940
| 196
|
a,b,c=map(int, input().split())
d = int(str(a)+str(b)+str(c)) % 4
if d != 0:
print('NO')
else:
print('YES')
|
s336913655
|
p03993
|
u664481257
| 2,000
| 262,144
|
Wrong Answer
| 99
| 14,516
| 843
|
There are N rabbits, numbered 1 through N. The i-th (1≤i≤N) rabbit likes rabbit a_i. Note that no rabbit can like itself, that is, a_i≠i. For a pair of rabbits i and j (i＜j), we call the pair (i，j) a _friendly pair_ if the following condition is met. * Rabbit i likes rabbit j and rabbit j likes rabbit i. Calculate the number of the friendly pairs.
|
# -*- coding: utf-8 -*-
# !/usr/bin/env python
# vim: set fileencoding=utf-8 :
"""
#
# Author: Noname
# URL: https://github.com/pettan0818
# License: MIT License
# Created: 2016-09-28
#
# Usage
#
"""
import sys
def input_single_line():
return input()
def input_two_line():
"""Receive Two Lined Inputs.
Like this.
N
1 2 3 4 5
"""
sys.stdin.readline()
target = sys.stdin.readline()
target = target.rstrip("\n")
target = target.split(" ")
return target
def search_lovers(target_list: list) -> None:
"""Search Simple."""
lovers = []
for i in target_list:
if target_list[i-1] == i-1:
lovers.append(i-1)
print(len(lovers))
if __name__ == "__main__":
import doctest
doctest.testmod()
target = input_two_line()
|
s607305419
|
Accepted
| 121
| 17,908
| 1,171
|
# -*- coding: utf-8 -*-
# !/usr/bin/env python
# vim: set fileencoding=utf-8 :
"""
#
# Author: Noname
# URL: https://github.com/pettan0818
# License: MIT License
# Created: 2016-09-28
#
# Usage
#
"""
import sys
def input_single_line():
return input()
def input_two_line():
"""Receive Two Lined Inputs.
Like this.
N
1 2 3 4 5
"""
sys.stdin.readline()
target = sys.stdin.readline()
target = target.rstrip("\n")
target = target.split(" ")
return target
def search_lovers(target_list: list) -> None:
"""Search Simple.
>>> target_list = [2, 3, 1]
>>> search_lovers(target_list)
0
>>> target_list = [2, 1, 4, 3]
>>> search_lovers(target_list)
2
>>> target_list = [5, 5, 5, 5, 1]
>>> search_lovers(target_list)
1
"""
target_list = [int(i) - 1 for i in target_list]
lovers = []
ans = 0
for i, n in enumerate(target_list):
if target_list[n] == i:
ans = ans + 1
print(int(ans/2))
if __name__ == "__main__":
import doctest
doctest.testmod()
target = input_two_line()
search_lovers(target)
|
s843029651
|
p03993
|
u393971002
| 2,000
| 262,144
|
Wrong Answer
| 146
| 14,008
| 180
|
There are N rabbits, numbered 1 through N. The i-th (1≤i≤N) rabbit likes rabbit a_i. Note that no rabbit can like itself, that is, a_i≠i. For a pair of rabbits i and j (i＜j), we call the pair (i，j) a _friendly pair_ if the following condition is met. * Rabbit i likes rabbit j and rabbit j likes rabbit i. Calculate the number of the friendly pairs.
|
N = int(input())
a = list(map(int, input().split()))
a = [i - 1 for i in a]
count = 0
for i in range(N):
print(a[i])
if i == a[a[i]]:
count += 1
print(count // 2)
|
s007828243
|
Accepted
| 85
| 14,008
| 164
|
N = int(input())
a = list(map(int, input().split()))
a = [i - 1 for i in a]
count = 0
for i in range(N):
if i == a[a[i]]:
count += 1
print(count // 2)
|
s443317291
|
p03657
|
u616188005
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 165
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a,b = map(int,input().split())
if a%3==0:
print("Possible")
elif b%3==0:
print("Possible")
elif a+b%3==0:
print("Possible")
else:
print("Impossible")
|
s548160336
|
Accepted
| 17
| 2,940
| 167
|
a,b = map(int,input().split())
if a%3==0:
print("Possible")
elif b%3==0:
print("Possible")
elif (a+b)%3==0:
print("Possible")
else:
print("Impossible")
|
s860829045
|
p03408
|
u951601135
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 354
|
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
N=int(input())
s=[input() for i in range(N)]
M=int(input())
l=[input() for i in range(M)]
list_str=list(dict.fromkeys(s+l))
s_l=[]
for i in range(len(list_str)):
s_l.append(s.count(list_str[i])-l.count(list_str[i]))
print(all([i < 0 for i in s_l]))
if(all([i < 0 for i in s_l])):
t='a'
while(t==any(s_l)):
t+='a'
print(t)
else:print(max(s_l))
|
s870087455
|
Accepted
| 18
| 3,064
| 314
|
N=int(input())
s=[input() for i in range(N)]
M=int(input())
l=[input() for i in range(M)]
list_str=list(dict.fromkeys(s+l))
s_l=[]
for i in range(len(list_str)):
s_l.append(s.count(list_str[i])-l.count(list_str[i]))
if(all([i < 0 for i in s_l])):
print(0)
else:print(max(s_l))
|
s409769705
|
p04039
|
u729133443
| 2,000
| 262,144
|
Wrong Answer
| 20
| 2,940
| 71
|
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.
|
n,k,*a=open(0).read().split()
while set(a)&{n}:n=str(int(n)+1)
print(n)
|
s924124047
|
Accepted
| 137
| 2,940
| 70
|
n,a=open(0)
n=int(n.split()[0])
while set(a)&set(str(n)):n+=1
print(n)
|
s163264106
|
p03456
|
u127856129
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 108
|
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()
c=(int(a+b))
for i in range(350):
if c==i*i:
print("Yes")
else:
print("No")
|
s240821418
|
Accepted
| 17
| 2,940
| 107
|
from math import sqrt
a,b=input().split()
c=sqrt(int(a+b))
if c==int(c):
print("Yes")
else:
print("No")
|
s228125423
|
p03408
|
u103902792
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 250
|
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
n = int(input())
d = {}
for _ in range(n):
s = input()
if s not in d:
d[s] = 0
d[s] += 1
m =int(input())
for _ in range(m):
s = input()
if s not in d:
d[s] = 0
d[s] -= 1
ans = 0
for key in d:
ans += max(d[key], 0)
print(ans)
|
s690395570
|
Accepted
| 17
| 3,060
| 253
|
n = int(input())
d = {}
for _ in range(n):
s = input()
if s not in d:
d[s] = 0
d[s] += 1
m =int(input())
for _ in range(m):
s = input()
if s not in d:
d[s] = 0
d[s] -= 1
ans = 0
for key in d:
ans = max(d[key], ans)
print(ans)
|
s696501168
|
p03161
|
u336011173
| 2,000
| 1,048,576
|
Wrong Answer
| 2,206
| 38,524
| 398
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
import numpy as np
N, K = map(int, input().split())
hs = np.array(list(map(int, input().split())))
i = 0
total_cost = 0
a = np.zeros(N, int)
a[0] = 0
a[1] = abs(hs[1] - hs[0])
for i in range(2, N):
bmin = float('inf')
for j in range(K):
if i-j+1 >= 0:
a[i] = min(a[i], a[max(i-(j+1),0)] + abs(hs[i] - hs[max(i-(j+1),0)]))
else:
break
print(a[-1])
|
s252370260
|
Accepted
| 796
| 38,696
| 304
|
import numpy as np
N, K = map(int, input().split())
hs = np.array(list(map(int, input().split())))
i = 0
total_cost = 0
a = np.zeros(N, dtype=np.int64)
a[0] = 0
a[1] = abs(hs[1] - hs[0])
for i in range(2, N):
k = min(K,i)
a[i] = np.min(a[i-k:i] + np.abs(hs[i] - hs[i-k:i]))
print(a[-1])
|
s541752747
|
p03407
|
u389910364
| 2,000
| 262,144
|
Wrong Answer
| 23
| 3,572
| 835
|
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
import functools
import os
INF = float('inf')
def inp():
return int(input())
def inpf():
return float(input())
def inps():
return input()
def inl():
return list(map(int, input().split()))
def inlf():
return list(map(float, input().split()))
def inls():
return input().split()
def debug(fn):
if not os.getenv('LOCAL'):
return fn
@functools.wraps(fn)
def wrapper(*args, **kwargs):
print('DEBUG: {}({}) -> '.format(
fn.__name__,
', '.join(
list(map(str, args)) +
['{}={}'.format(k, str(v)) for k, v in kwargs.items()]
)
), end='')
ret = fn(*args, **kwargs)
print(ret)
return ret
return wrapper
a, b, c = inl()
if a + b == c:
print('Yes')
else:
print('No')
|
s914008968
|
Accepted
| 22
| 3,572
| 836
|
import functools
import os
INF = float('inf')
def inp():
return int(input())
def inpf():
return float(input())
def inps():
return input()
def inl():
return list(map(int, input().split()))
def inlf():
return list(map(float, input().split()))
def inls():
return input().split()
def debug(fn):
if not os.getenv('LOCAL'):
return fn
@functools.wraps(fn)
def wrapper(*args, **kwargs):
print('DEBUG: {}({}) -> '.format(
fn.__name__,
', '.join(
list(map(str, args)) +
['{}={}'.format(k, str(v)) for k, v in kwargs.items()]
)
), end='')
ret = fn(*args, **kwargs)
print(ret)
return ret
return wrapper
a, b, c = inl()
if a + b >= c:
print('Yes')
else:
print('No')
|
s001510024
|
p03380
|
u325956328
| 2,000
| 262,144
|
Wrong Answer
| 249
| 24,140
| 599
|
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
from scipy.misc import comb
#from scipy.special import comb
n = int(input())
A = sorted(list(map(int, input().split())))
ans = 0
out = []
if n % 2 == 1:
middle = A[n // 2]
last = A[-1]
print(last, middle)
# print(ans)
elif n == 2:
print(*A[::-1])
else:
cand = [A[n // 2 - 1], A[n // 2]]
last = A[-1]
for i in range(2):
middle = cand[i]
comb_last_middle = comb(last, middle)
if ans < comb_last_middle:
out.append([last, middle])
ans = comb_last_middle
print(last, middle)
|
s695861194
|
Accepted
| 157
| 38,388
| 553
|
import numpy as np
n = int(input())
a = list(map(int, input().split()))
a.sort()
def get_nearest_value(arr, num):
idx = np.abs(np.asarray(arr) - num).argmin()
return arr[idx]
first = a[-1]
second = get_nearest_value(a, first / 2)
print(first, second)
|
s990780840
|
p02397
|
u300641790
| 1,000
| 131,072
|
Wrong Answer
| 60
| 7,476
| 123
|
Write a program which reads two integers x and y, and prints them in ascending order.
|
while 1:
x,y = [int(i) for i in input().split()]
if (x == 0 and y == 0):
break
print(str(x)+" "+str(y))
|
s204665480
|
Accepted
| 60
| 7,480
| 140
|
while 1:
l = list(map(int,input().split()))
if (l[0] == 0 and l[1] == 0):
break
l.sort()
print(" ".join(map(str,l)))
|
s898215114
|
p03719
|
u859897687
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 77
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a,b,c=map(int,input().split())
if a<=c<=b:
print("YES")
else:
print("NO")
|
s738902677
|
Accepted
| 17
| 2,940
| 77
|
a,b,c=map(int,input().split())
if a<=c<=b:
print("Yes")
else:
print("No")
|
s305751581
|
p02865
|
u217303170
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,060
| 30
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n=int(input())
print((n+1)//2)
|
s995789898
|
Accepted
| 29
| 9,080
| 68
|
n = int(input())
if n%2==0:
print(n//2-1)
else:
print(n//2)
|
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