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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s630740745
|
p02273
|
u358919705
| 2,000
| 131,072
|
Wrong Answer
| 20
| 7,636
| 679
|
Write a program which reads an integer _n_ and draws a Koch curve based on recursive calles of depth _n_. The Koch curve is well known as a kind of You should start (0, 0), (100, 0) as the first segment.
|
from math import sin, cos, pi
def koch(px, py, qx, qy, n):
if n == 0:
return True
sx = (2 * px + qx) / 3
sy = (2 * py + qy) / 3
tx = (px + 2 * qx) / 3
ty = (py + 2 * qy) / 3
ux = sx + (tx - sx) * cos(pi / 3) - (ty - sy) * sin(pi / 3)
uy = sy + (tx - sx) * sin(pi / 3) - (ty - sy) * cos(pi / 3)
koch(px, py, sx, sy, n - 1)
print("{:.8f} {:.8f}".format(sx, sy))
koch(sx, sy, ux, uy, n - 1)
print("{:.8f} {:.8f}".format(ux, uy))
koch(ux, uy, tx, ty, n - 1)
print("{:.8f} {:.8f}".format(tx, ty))
koch(tx, ty, qx, qy, n - 1)
print("{:.8f} {:.8f}".format(0, 0))
koch(0, 0, 100, 0, 2)
print("{:.8f} {:.8f}".format(100, 0))
|
s096063314
|
Accepted
| 40
| 7,976
| 696
|
from math import sin, cos, pi
def koch(px, py, qx, qy, n):
if n == 0:
return True
sx = (2 * px + qx) / 3
sy = (2 * py + qy) / 3
tx = (px + 2 * qx) / 3
ty = (py + 2 * qy) / 3
ux = sx + (tx - sx) * cos(pi / 3) - (ty - sy) * sin(pi / 3)
uy = sy + (tx - sx) * sin(pi / 3) + (ty - sy) * cos(pi / 3)
koch(px, py, sx, sy, n - 1)
print("{:.8f} {:.8f}".format(sx, sy))
koch(sx, sy, ux, uy, n - 1)
print("{:.8f} {:.8f}".format(ux, uy))
koch(ux, uy, tx, ty, n - 1)
print("{:.8f} {:.8f}".format(tx, ty))
koch(tx, ty, qx, qy, n - 1)
n = int(input())
print("{:.8f} {:.8f}".format(0, 0))
koch(0, 0, 100, 0, n)
print("{:.8f} {:.8f}".format(100, 0))
|
s017219674
|
p04043
|
u552502395
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 123
|
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.
|
L = list(map(int, input().split()))
L.sort()
if L[0] == 5 and L[1]==5 and L[2]==7:
print("Yes");
else:
print("No");
|
s504223148
|
Accepted
| 18
| 2,940
| 124
|
L = list(map(int, input().split()))
L.sort()
if L[0] == 5 and L[1]==5 and L[2]==7:
print("YES");
else:
print("NO");
|
s543550692
|
p02408
|
u874395007
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 409
|
Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.
|
N = int(input())
cards = []
for i in range(N):
line = input().split()
a = line[0]
num = int(line[1])
cards.append((a, num))
tarinai_list = []
for a in ['S', 'H', 'C', 'D']:
for num in range(1, 14):
for card in cards:
if a == card[0] and num == card[1]:
break
tarinai_list.append((a, num))
for l in tarinai_list:
print(' '.join(map(str, l)))
|
s346468231
|
Accepted
| 20
| 5,616
| 468
|
suits = ['S', 'H', 'C', 'D']
table = [[False] * 14 for i in range(4)]
N = int(input())
for _i in range(N):
mark, num = input().split()
num = int(num)
if mark == 'S':
table[0][num] = True
if mark == 'H':
table[1][num] = True
if mark == 'C':
table[2][num] = True
if mark == 'D':
table[3][num] = True
for i in range(4):
for j in range(1, 14):
if not table[i][j]:
print(f'{suits[i]} {j}')
|
s002544504
|
p02613
|
u223215060
| 2,000
| 1,048,576
|
Wrong Answer
| 164
| 9,212
| 323
|
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())
a = 0
b = 0
c = 0
d = 0
for i in range(n):
s = str(input())
if s == 'AC' :
a += 1
elif s == 'WA':
b += 1
elif s == 'TLE':
c += 1
else:
d += 1
print('AC ร {}'.format(a))
print('WA ร {}'.format(b))
print('TLE ร {}'.format(c))
print('RE ร {}'.format(d))
|
s150371612
|
Accepted
| 159
| 9,220
| 319
|
n = int(input())
a = 0
b = 0
c = 0
d = 0
for i in range(n):
s = str(input())
if s == 'AC' :
a += 1
elif s == 'WA':
b += 1
elif s == 'TLE':
c += 1
else:
d += 1
print('AC x {}'.format(a))
print('WA x {}'.format(b))
print('TLE x {}'.format(c))
print('RE x {}'.format(d))
|
s624914821
|
p02612
|
u688375653
| 2,000
| 1,048,576
|
Wrong Answer
| 35
| 9,164
| 216
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
def input_int():
return map(int, input().split())
def one_int():
return int(input())
def one_str():
return input()
def many_int():
return list(map(int, input().split()))
a=one_int()
print(a%1000)
|
s918721163
|
Accepted
| 30
| 9,168
| 257
|
def input_int():
return map(int, input().split())
def one_int():
return int(input())
def one_str():
return input()
def many_int():
return list(map(int, input().split()))
a=one_int()
if a%1000>0:
print(1000-a%1000)
else:
print(0)
|
s029074857
|
p04043
|
u054412614
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 100
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a,b,c=map(int,input().split())
if a+b+c==17:
if a==7 or b==7 or c==7:
print("YES")
print("NO")
|
s716340605
|
Accepted
| 17
| 3,060
| 206
|
a,b,c=map(int,input().split())
if a+b+c==17:
if a==5 and b==5:
print("YES")
elif a==5 and c==5:
print("YES")
elif b==5 and c==5:
print("YES")
else:
print("NO")
else:
print("NO")
|
s295197457
|
p03214
|
u374171306
| 2,525
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 289
|
Niwango-kun is an employee of Dwango Co., Ltd. One day, he is asked to generate a thumbnail from a video a user submitted. To generate a thumbnail, he needs to select a frame of the video according to the following procedure: * Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video. * Select t-th frame whose representation a_t is nearest to the average of all frame representations. * If there are multiple such frames, select the frame with the smallest index. Find the index t of the frame he should select to generate a thumbnail.
|
N = int(input())
a = list(map(int,input().split()))
sum = 0
for i in a:
sum += i
avg = sum/N
result = 0
tmp = 0
for i in range(len(a)):
ans = avg - a[i]
ans = ans if ans < 0 else ans * -1
if i == 0:
result = i
tmp = ans
elif tmp > ans:
result = i
print(result)
|
s036552831
|
Accepted
| 17
| 3,064
| 282
|
N = int(input())
a = list(map(int,input().split()))
sum = sum(a)
avg = sum/N
result = 0
tmp = 0
for i in range(len(a)):
ans = avg - a[i]
ans = ans if ans > 0 else ans * -1
if i == 0:
result = i
tmp = ans
elif tmp > ans:
result = i
tmp = ans
print(result)
|
s941393182
|
p03386
|
u433380437
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,196
| 213
|
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())
A=[]
for i in range(a,min(a+k+1,b+1)):
A.append(i)
for j in range(b,max(b-k,a),-1):
A.append(j)
A.sort(reverse=False)
B=list(set(A))
for p in range(len(B)):
print(B[p])
|
s813008554
|
Accepted
| 26
| 9,196
| 218
|
a,b,k = map(int,input().split())
A=[]
B=[]
for i in range(a,min(a+k,b+1)):
A.append(i)
for j in range(b,max(b-k,a-1),-1):
A.append(j)
B=list(set(A))
B.sort(reverse=False)
for p in range(len(B)):
print(B[p])
|
s028786285
|
p03673
|
u368796742
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 26,016
| 117
|
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
n = int(input())
l = list(map(int,input().split()))
ans = []
for i in l:
ans.append(i)
ans = ans[::-1]
print(ans)
|
s893002989
|
Accepted
| 176
| 26,180
| 256
|
n = int(input())
l = list(map(int,input().split()))
if n % 2 == 0:
ans = [l[i] for i in range(n-1,-1,-2)] + [l[i] for i in range(0,n-1,2)]
print(*ans)
else:
ans = [l[i] for i in range(n-1,-1,-2)] + [l[i] for i in range(1,n-1,2)]
print(*ans)
|
s452476073
|
p03449
|
u224050758
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 720
|
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel?
|
N = int(input())
HIGH = [int(x) for x in input().split()]
LOW = [int(x) for x in input().split()]
def get_sum(diff_list):
max_abs_value = max(diff_list, key=abs)
max_abs_index = diff_list.index(max_abs_value)
corrects = [HIGH[i] if i <= max_abs_index else LOW[i] for i, _ in enumerate(diff_list)]
print(corrects)
return sum(corrects)
def main():
first = HIGH.pop(0)
last = LOW.pop()
diff_list = []
high = low = 0
for i in range(N - 1):
high = high + HIGH[i]
low = low + LOW[i]
diff_list.append(high - low)
sum = get_sum(diff_list) if len(diff_list) != 0 else 0
result = first + sum + last
print(result)
if __name__ == "__main__":
main()
|
s021480213
|
Accepted
| 17
| 3,064
| 439
|
N = int(input())
HIGH = [int(x) for x in input().split()]
LOW = [int(x) for x in input().split()]
def main():
first = HIGH.pop(0)
last = LOW.pop()
mid_len = len(HIGH)
mid_sum = 0
for i in range(mid_len)[::-1]: # reversed
mid_sum = max(mid_sum, sum(HIGH))
HIGH[i] = LOW[i]
mid_sum = max(mid_sum, sum(LOW))
result = first + mid_sum + last
print(result)
if __name__ == "__main__":
main()
|
s084613857
|
p03679
|
u396391104
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 117
|
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
x,a,b=map(int,input().split())
if a<b:
print("delicious")
elif (b-a)<=x:
print("safe")
else:
print("dangerous")
|
s528166273
|
Accepted
| 17
| 3,060
| 118
|
x,a,b=map(int,input().split())
if a>=b:
print("delicious")
elif (b-a)<=x:
print("safe")
else:
print("dangerous")
|
s214125881
|
p02389
|
u681232780
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,580
| 74
|
Write a program which calculates the area and perimeter of a given rectangle.
|
a,b=map(int,input().split())
print('้ข็ฉ',a*b,'ๅจใฎ้ทใ',2*a+2*b)
|
s984654957
|
Accepted
| 20
| 5,580
| 49
|
a,b=map(int,input().split())
print(a*b,2*a+2*b)
|
s229107576
|
p02614
|
u396472025
| 1,000
| 1,048,576
|
Wrong Answer
| 29
| 8,988
| 8
|
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
print(0)
|
s753061871
|
Accepted
| 65
| 9,064
| 426
|
H, W, K = map(int, input().split())
C = [list(input()) for _ in range(H)]
ans = 0
for h_ptn in range(2 ** H):
for w_ptn in range(2 ** W):
cnt = 0
for h in range(H):
for w in range(W):
if (h_ptn >> h) & 1 == 0 and (w_ptn >> w) & 1 == 0:
if C[h][w] == '#':
cnt += 1
if cnt == K:
ans += 1
print(ans)
|
s886539732
|
p02697
|
u069125420
| 2,000
| 1,048,576
|
Wrong Answer
| 74
| 9,216
| 79
|
You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given.
|
N, M = map(int, input().split())
for i in range(0, M):
print(i + 1, N - i)
|
s428959221
|
Accepted
| 74
| 9,188
| 330
|
N, M = map(int, input().split())
M_f = M * 2 + 1
if M % 2 != 0:
for i in range(0, M // 2):
print(i + 1, (M) - i)
for i in range(0, M // 2 + 1):
print(M + 1 + i, M_f - i)
else:
for i in range(0, M // 2):
print(i + 1, (M + 1) - i)
for i in range(0, M // 2):
print(M + 2 + i, M_f - i)
|
s170492670
|
p03579
|
u830592648
| 2,000
| 262,144
|
Wrong Answer
| 2,105
| 107,600
| 495
|
Rng has a connected undirected graph with N vertices. Currently, there are M edges in the graph, and the i-th edge connects Vertices A_i and B_i. Rng will add new edges to the graph by repeating the following operation: * Operation: Choose u and v (u \neq v) such that Vertex v can be reached by traversing exactly three edges from Vertex u, and add an edge connecting Vertices u and v. It is not allowed to add an edge if there is already an edge connecting Vertices u and v. Find the maximum possible number of edges that can be added.
|
import sys
sys.setrecursionlimit(10**6)
n,m=map(int,input().split())
s=[[]for i in range(n+1)]
c=[0]*(n+1)
for i in range(m):
a,b=map(int,input().split())
s[a].append(b)
s[b].append(a)
def dfs(v,t):
c[v]=t
print('start :'+str(c))
for i in s[v]:
if c[i]==t:
return False
if c[i]==0:
dfs(i,-t)
# return False
else:
return True
if dfs(1,1):
q=c.count(1)
print((n-q)*q-m)
else:
print((n*(n-1)//2)-m)
|
s216070048
|
Accepted
| 541
| 31,156
| 491
|
import sys
sys.setrecursionlimit(10**6)
n,m=map(int,input().split())
s=[[]for i in range(n+1)]
c=[0]*(n+1)
for i in range(m):
a,b=map(int,input().split())
s[a].append(b)
s[b].append(a)
def dfs(v,t):
c[v]=t
# print('start :'+str(c))
for i in s[v]:
if c[i]==t:
return False
if c[i]==0 and not dfs(i,-t):
return False
else:
return True
if dfs(1,1):
q=c.count(1)
print((n-q)*q-m)
else:
print((n*(n-1)//2)-m)
|
s193988837
|
p02663
|
u921352252
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 9,404
| 300
|
In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?
|
import time
import datetime
from datetime import timedelta
H1,M1,H2,M2,K=map(int,input().split())
delta1=timedelta(minutes=M1,hours=H1)
delta2=timedelta(minutes=M2,hours=H2)
delta=delta2-delta1
delta_minutes=delta.total_seconds()/60
if delta_minutes>30:
print(delta_minutes-30)
else:
print(0)
|
s340040007
|
Accepted
| 22
| 9,400
| 293
|
import datetime
from datetime import timedelta
H1,M1,H2,M2,K=map(int,input().split())
delta1=timedelta(minutes=M1,hours=H1)
delta2=timedelta(minutes=M2,hours=H2)
delta=delta2-delta1
delta_minutes=delta.total_seconds()/60
if delta_minutes>=K:
print(round(delta_minutes-K))
else:
print(0)
|
s365171171
|
p03472
|
u648212584
| 2,000
| 262,144
|
Wrong Answer
| 542
| 32,880
| 341
|
You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, โฆ, 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 โค i โค 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 โค i โค N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?
|
N,M = map(int,input().split())
X = [list(map(int,input().split())) for _ in range(N)]
X = sorted(X,key=lambda X:-X[0])
a = X[0][0]
X = sorted(X,key=lambda X:-X[1])
count = 0
for i in range(N):
if a <= X[i][1]:
M -= X[i][1]
count += 1
if M <= 0:
break
print(M,count)
if M > 0:
count += -((-M)//a)
else:
count += 1
print(count)
|
s759119760
|
Accepted
| 550
| 32,880
| 306
|
N,M = map(int,input().split())
X = [list(map(int,input().split())) for _ in range(N)]
X = sorted(X,key=lambda X:-X[0])
a = X[0][0]
X = sorted(X,key=lambda X:-X[1])
i = 0
count = 0
while M > 0:
if i < N and X[i][1] > a:
M -= X[i][1]
i += 1
else:
count += -(-M//a)
break
count += 1
print(count)
|
s997279731
|
p02742
|
u368579103
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 134
|
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
h,w = list(map(int,input().split()))
if h % 2 == 0:
ans = h / 2 * w
else:
ans = ((h + 1) / 2 * w) - (int(w / 2) * 1)
print(ans)
|
s047152859
|
Accepted
| 17
| 3,060
| 171
|
h,w = list(map(int,input().split()))
if h % 2 == 0:
ans = h / 2 * w
elif h == 1 or w == 1:
ans = 1
else:
ans = ((h + 1) / 2 * w) - (int(w / 2) * 1)
print(int(ans))
|
s027749541
|
p03637
|
u625963200
| 2,000
| 262,144
|
Wrong Answer
| 63
| 14,252
| 218
|
We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 โค i โค N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective.
|
n=int(input())
A=list(map(int,input().split()))
cnt1,cnt2,cnt3=0,0,0
for a in A:
if a%4==0:
cnt1+=1
elif a%2==0:
cnt2+=1
cnt3+=cnt2//2
print('Yes') if cnt3>=n//2 else print('No')
|
s900401669
|
Accepted
| 62
| 15,020
| 192
|
n=int(input())
A=list(map(int,input().split()))
cnt1,cnt2,cnt3=0,0,0
for a in A:
if a%4==0:
cnt1+=1
elif a%2==0:
cnt2+=1
cnt1+=cnt2//2
print('Yes') if cnt1>=n//2 else print('No')
|
s498758327
|
p03155
|
u568789901
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 66
|
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board. The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns. How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
|
N=int(input())
H=int(input())
W=int(input())
print((N//H)*(N//W))
|
s499067353
|
Accepted
| 17
| 2,940
| 68
|
N=int(input())
H=int(input())
W=int(input())
print((N-H+1)*(N-W+1))
|
s965304572
|
p03711
|
u023127434
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,028
| 195
|
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 โค x < y โค 12), determine whether they belong to the same group.
|
x, y = map(int, input().split())
a = [1, 3, 5, 7, 8, 10, 12]
b = [4, 6, 9, 11]
c = [2]
if (x in a and y in a) or (x in b and y in b) or (x in c and y in c):
print("YES")
else:
print("NO")
|
s266835306
|
Accepted
| 28
| 9,192
| 195
|
a = [1, 3, 5, 7, 8, 10, 12]
b = [4, 6, 9, 11]
c = [2]
x, y = map(int, input().split())
if (x in a and y in a) or (x in b and y in b) or (x in c and y in c):
print("Yes")
else:
print("No")
|
s708871206
|
p03351
|
u559313689
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,128
| 164
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a, b, c, d = map(int, input().split())
x=abs(b-a)
y=abs(c-b)
z=abs(c-a)
if z <= d:
print('Yes')
elif y <= d and z <= d:
print('Yes')
else:
print('No')
|
s570656686
|
Accepted
| 28
| 9,148
| 164
|
a, b, c, d = map(int, input().split())
x=abs(b-a)
y=abs(c-b)
z=abs(c-a)
if z <= d:
print('Yes')
elif y <= d and x <= d:
print('Yes')
else:
print('No')
|
s960631414
|
p02670
|
u490195279
| 2,000
| 1,048,576
|
Time Limit Exceeded
| 2,206
| 38,116
| 1,325
|
Tonight, in your favourite cinema they are giving the movie Joker and all seats are occupied. In the cinema there are N rows with N seats each, forming an N\times N square. We denote with 1, 2,\dots, N the viewers in the first row (from left to right); with N+1, \dots, 2N the viewers in the second row (from left to right); and so on until the last row, whose viewers are denoted by N^2-N+1,\dots, N^2. At the end of the movie, the viewers go out of the cinema in a certain order: the i-th viewer leaving her seat is the one denoted by the number P_i. The viewer P_{i+1} waits until viewer P_i has left the cinema before leaving her seat. To exit from the cinema, a viewer must move from seat to seat until she exits the square of seats (any side of the square is a valid exit). A viewer can move from a seat to one of its 4 adjacent seats (same row or same column). While leaving the cinema, it might be that a certain viewer x goes through a seat **currently** occupied by viewer y; in that case viewer y will hate viewer x forever. Each viewer chooses the way that minimizes the number of viewers that will hate her forever. Compute the number of pairs of viewers (x, y) such that y will hate x forever.
|
N=int(input())
S=list(map(int,input().split()))
count=0
P=[0]*(N*N+1)
for j in range(N*N):
P[j+1]=S.index(j+1)
for i in range(N*N):
player=S[i]
hayasa=P[player]
que=[[] for _ in range(N)]
d=[-1]*(N*N+1)
que[0].append(player)
d[player]=0
j=0
finish=0
while True:
for kouho in que[j]:
if kouho%N==0:
finish=1
break
if kouho%N==1:
finish=1
break
if kouho+N>N*N:
finish=1
break
if kouho<N:
finish=1
break
if d[kouho+N]==-1:
if P[kouho+N]>hayasa:
que[j+1].append(kouho+N)
d[kouho+N]=0
else:
que[j].append(kouho+N)
d[kouho+N]=0
if d[kouho-N]==-1:
if P[kouho-N]>hayasa:
que[j+1].append(kouho-N)
d[kouho-N]=0
else:
que[j].append(kouho-N)
d[kouho-N]=0
if d[kouho+1]==-1:
if P[kouho+1]>hayasa:
que[j+1].append(kouho+1)
d[kouho+1]=0
else:
que[j].append(kouho+1)
d[kouho+1]=0
if d[kouho-1]==-1:
if P[kouho-1]>hayasa:
que[j+1].append(kouho-1)
d[kouho-1]=0
else:
que[j].append(kouho-1)
d[kouho-1]=0
if finish==1:
count+=j
|
s936294108
|
Accepted
| 1,498
| 118,800
| 841
|
from numba import njit
import numpy as np
@njit
def main(N, P):
lista = [1, -1, N, -N]
d = np.zeros((N, N), dtype=np.int32)
d = d.ravel()
sit = np.ones_like(d)
stack = np.empty_like(d)
count = 0
for i in range(N*N):
d[i] = min(i % N, (N-1-i) % N, i//N, N-1-i//N)
for player in P:
sit[player-1] = 0
count += d[player-1]
stack[0] = player
num = 0
while num >= 0:
p = stack[num]
num -= 1
a = d[p-1]+sit[p-1]
for i in lista:
if p+i < 1 or p + i > N*N:
continue
if d[p+i-1] > a:
d[p+i-1] = a
num += 1
stack[num] = p+i
print(count)
N = int(input())
P = np.array(input().split(), np.int32)
main(N, P)
|
s566157781
|
p03095
|
u426108351
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 3,316
| 144
|
You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order.
|
s = list(input().split())
import collections
s = collections.Counter(s)
s = s.most_common()
ans = 1
for i in s:
ans *= i[1]
print(ans-1)
|
s226582274
|
Accepted
| 27
| 4,140
| 161
|
n = input()
s = list(input())
import collections
s = collections.Counter(s)
s = s.most_common()
ans = 1
for i in s:
ans *= i[1]+1
print((ans-1)%1000000007)
|
s891847550
|
p03997
|
u037430802
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 67
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2)
|
s854985139
|
Accepted
| 18
| 2,940
| 72
|
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2))
|
s798271302
|
p02389
|
u998185318
| 1,000
| 131,072
|
Wrong Answer
| 40
| 6,516
| 113
|
Write a program which calculates the area and perimeter of a given rectangle.
|
import re
numbers = input()
lines = re.split(" ", numbers)
a = int(lines[0])
b = int(lines[1])
print(a * b / 2)
|
s207792382
|
Accepted
| 20
| 5,592
| 149
|
numbers = input()
lines = numbers.split(" ")
a = int(lines[0])
b = int(lines[1])
area = str(a * b)
line = str(2 * (a + b))
print(area + " " + line)
|
s009140122
|
p03151
|
u286955577
| 2,000
| 1,048,576
|
Wrong Answer
| 1,233
| 34,656
| 469
|
A university student, Takahashi, has to take N examinations and pass all of them. Currently, his _readiness_ for the i-th examination is A_{i}, and according to his investigation, it is known that he needs readiness of at least B_{i} in order to pass the i-th examination. Takahashi thinks that he may not be able to pass all the examinations, and he has decided to ask a magician, Aoki, to change the readiness for as few examinations as possible so that he can pass all of them, while not changing the total readiness. For Takahashi, find the minimum possible number of indices i such that A_i and C_i are different, for a sequence C_1, C_2, ..., C_{N} that satisfies the following conditions: * The sum of the sequence A_1, A_2, ..., A_{N} and the sum of the sequence C_1, C_2, ..., C_{N} are equal. * For every i, B_i \leq C_i holds. If such a sequence C_1, C_2, ..., C_{N} cannot be constructed, print -1.
|
import numpy as np
def solve(A, B):
diff = np.array(A) - np.array(B)
shortage = list(filter(lambda d: d < 0, diff))
remainder = sorted(list(filter(lambda d: d > 0, diff)), reverse=True)
short_sum = sum(shortage)
counter = 0
for r in remainder:
if short_sum >= 0: return len(shortage) + counter
short_sum += r
counter += 1
return -1
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
print(solve(A, B))
|
s500345376
|
Accepted
| 115
| 18,484
| 505
|
import operator
def solve(A, B):
diff = list(map(operator.sub, A, B))
shortage = list(filter(lambda d: d < 0, diff))
remainder = list(filter(lambda d: d > 0, diff))
remainder.sort(reverse=True)
short_sum = sum(shortage)
if short_sum == 0: return 0
counter = len(shortage)
for r in remainder:
short_sum += r
counter += 1
if short_sum >= 0: return counter
return -1
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
print(solve(A, B))
|
s032772711
|
p03478
|
u955907183
| 2,000
| 262,144
|
Wrong Answer
| 51
| 3,636
| 290
|
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())
allsum = 0
for i in range(1,n+1):
l = list(str(i))
li = [int(s) for s in l]
tmpsum = sum(li)
print (tmpsum)
#for s in range(0,len(l)):
# print(s)
if ((tmpsum >= a) and (tmpsum <= b)):
print("ok")
allsum = allsum + i
print(allsum)
|
s811471706
|
Accepted
| 37
| 3,064
| 257
|
n, a, b = map(int, input().split())
allsum = 0
for i in range(1,n+1):
l = list(str(i))
li = [int(s) for s in l]
tmpsum = sum(li)
#for s in range(0,len(l)):
# print(s)
if ((tmpsum >= a) and (tmpsum <= b)):
allsum = allsum + i
print(allsum)
|
s480192564
|
p03623
|
u934868410
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b = map(int,input().split())
if abs(x-a) > abs(x-b):
print('A')
else:
print('B')
|
s676978470
|
Accepted
| 17
| 2,940
| 88
|
x,a,b = map(int,input().split())
if abs(x-a) < abs(x-b):
print('A')
else:
print('B')
|
s447849524
|
p02389
|
u369093003
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,576
| 44
|
Write a program which calculates the area and perimeter of a given rectangle.
|
a,b=map(int,input().split(" "))
print(a*b)
|
s085403033
|
Accepted
| 20
| 5,580
| 52
|
a,b=map(int,input().split(" "))
print(a*b,2*a+2*b)
|
s576702310
|
p03377
|
u074445770
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 85
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
a,b,x=map(int,input().split())
if 0<=x-a<=b:
print("Yes")
else:
print("No")
|
s959055775
|
Accepted
| 17
| 2,940
| 85
|
a,b,x=map(int,input().split())
if 0<=x-a<=b:
print("YES")
else:
print("NO")
|
s845679213
|
p02394
|
u229478139
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 284
|
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
l = list(map(int, input('Please input int: ').split()))
W = l[0]
H = l[1]
x = l[2]
y = l[3]
r = l[4]
if (0 <= x <= W) and (0 <= y <= H):
if (0 <= x - r) and (x + r <= W) and (0 <= y - r) and (y + r <= H):
print('Yes')
else:
print('No')
else:
print('No')
|
s672238061
|
Accepted
| 20
| 5,596
| 132
|
w, h, x, y, r = list(map(int, input().split()))
if (r <= x <= w - r) and (r <= y <= h - r):
print('Yes')
else:
print('No')
|
s917451019
|
p04044
|
u927870520
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 145
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1โฆiโฆmin(n,m)), such that s_j = t_j for all indices j(1โฆj<i), and s_i<t_i. * s_i = t_i for all integers i(1โฆiโฆmin(n,m)), and n<m.
|
n,l=list(map(int,input().split()))
S=[list(input()) for i in range(n)]
[S[i].sort() for i in range(len(S))]
S.sort()
ans=''.join(S[0])
print(ans)
|
s972996116
|
Accepted
| 17
| 3,060
| 99
|
n,l=list(map(int,input().split()))
S=[input() for i in range(n)]
S.sort()
ans=''.join(S)
print(ans)
|
s915783814
|
p02743
|
u614875193
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 68
|
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
a,b,c=map(int,input().split())
print('NYoe s'[a**.5+b**.5<c**.5::2])
|
s717304015
|
Accepted
| 17
| 2,940
| 115
|
a,b,c=map(int,input().split())
if a*a+b*b+c*c-2*(a*b+b*c+c*a)>0 and c-a-b>0:
print('Yes')
else:
print('No')
|
s334579465
|
p03160
|
u208497224
| 2,000
| 1,048,576
|
Wrong Answer
| 639
| 22,724
| 389
|
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.
|
import numpy as np
n = int(input())
h_list = list(map(int, input().split()))
dp = [0] * n
dp[1] = np.abs(h_list[1] - h_list[0])
print(dp)
for ashiba_index in range(2, n):
dp[ashiba_index] = min(dp[ashiba_index-1]+np.abs(h_list[ashiba_index] - h_list[ashiba_index-1]),
dp[ashiba_index-2] + np.abs(h_list[ashiba_index]-h_list[ashiba_index-2]))
print(dp[n-1])
|
s192934990
|
Accepted
| 631
| 22,760
| 379
|
import numpy as np
n = int(input())
h_list = list(map(int, input().split()))
dp = [0] * n
dp[1] = np.abs(h_list[1] - h_list[0])
for ashiba_index in range(2, n):
dp[ashiba_index] = min(dp[ashiba_index-1]+np.abs(h_list[ashiba_index] - h_list[ashiba_index-1]),
dp[ashiba_index-2] + np.abs(h_list[ashiba_index]-h_list[ashiba_index-2]))
print(dp[n-1])
|
s431705160
|
p03377
|
u333731247
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 57
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A,B,X=map(int,input().split())
if A<=X:
print('Yes')
|
s663786024
|
Accepted
| 17
| 3,060
| 122
|
A,B,X=map(int,input().split())
if A<=X and A+B>=X:
print('YES')
elif A>X:
print('NO')
elif A+B<X:
print('NO')
|
s962201608
|
p03228
|
u008173855
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 237
|
In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.
|
a, b, k = map(int,input().split())
cnt = 0
while True:
if cnt==k:
break
if a%2!=0:
a-=1
b+=a/2
a = a/2
cnt+=1
if cnt==k:
break
if b%2!=0:
b-=1
a+=b/2
b = b/2
cnt+=1
|
s389506845
|
Accepted
| 17
| 3,060
| 259
|
a, b, k = map(int,input().split())
cnt = 0
while True:
if cnt==k:
break
if a%2!=0:
a-=1
b+=a/2
a = a/2
cnt+=1
if cnt==k:
break
if b%2!=0:
b-=1
a+=b/2
b = b/2
cnt+=1
print(int(a),int(b))
|
s472919869
|
p02928
|
u727801592
| 2,000
| 1,048,576
|
Wrong Answer
| 376
| 3,188
| 241
|
We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.
|
n,k=map(int, input().split())
a=list(map(int, input().split()))
ans=0
for i in range(n-1):
for j in range(i+1,n):
if a[i]>a[j]:
ans+=1
print(10**9+7)
x=(ans*k*(k+1))//2
y=(n*(n-1))//2
z=(k*(k-1))//2
print((x+(y-ans)*z)%(10**9+7))
|
s948844482
|
Accepted
| 794
| 3,188
| 251
|
n,k=map(int, input().split())
a=list(map(int, input().split()))
ans=0
rev=0
for i in range(n-1):
for j in range(i+1,n):
if a[i]>a[j]:
ans+=1
if a[i]<a[j]:
rev+=1
x=(ans*k*(k+1))//2
y=(rev*k*(k-1))//2
print((x+y)%(10**9+7))
|
s073581385
|
p02243
|
u426534722
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,672
| 605
|
For a given weighted graph $G = (V, E)$, find the shortest path from a source to each vertex. For each vertex $u$, print the total weight of edges on the shortest path from vertex $0$ to $u$.
|
import sys
readline = sys.stdin.readline
from heapq import heapify, heappush, heappop
INF = float("inf")
def MAIN():
n = int(input())
G = [[i, INF] for i in range(n)]
G[0][1] = 0
m = {}
for _ in range(n):
A = list(map(int, readline().split()))
m[A[0]] = {}
for i in range(2, len(A), 2):
m[A[0]][A[i]] = A[i + 1]
dp = [(0, 0)]
while dp:
cost, u = heappop(dp)
for v, c in m[u].items():
if G[v][1] > G[u][1] + c:
G[v][1] = G[u][1] + c
heappush(dp, (G[v][1], v))
print(*G)
MAIN()
|
s277957998
|
Accepted
| 340
| 40,036
| 652
|
import sys
readline = sys.stdin.readline
from heapq import heapify, heappush, heappop
INF = float("inf")
def MAIN():
n = int(input())
adj = [[]] * n
for _ in range(0, n):
U = list(map(int, input().split()))
adj[U[0]] = zip(U[2::2], U[3::2])
def dijkstra():
d = [INF] * n
d[0] = 0
hq = [(0, 0)]
while hq:
u = heappop(hq)[1]
for v, c in adj[u]:
alt = d[u] + c
if d[v] > alt:
d[v] = alt
heappush(hq, (alt, v))
print("\n".join(f"{i} {c}" for i, c in enumerate(d)))
dijkstra()
MAIN()
|
s394145298
|
p04011
|
u886274153
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 222
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
N, K, X, Y = [int(input()) for j in range(4)]
def price(N, K, X, Y):
if N < K:
return N * X
else:
print("exceed", K, X, K * X + (N-K) * Y)
return N * X + (N-K) * Y
print(price(N, K, X, Y))
|
s786411994
|
Accepted
| 17
| 2,940
| 173
|
N, K, X, Y = [int(input()) for i in range(4)]
def price(N, K, X, Y):
if N < K:
return N * X
else:
return K * X + (N-K) * Y
print(price(N, K, X, Y))
|
s918114014
|
p03854
|
u911276694
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,444
| 686
|
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`.
|
def extract_6(x):
strtemp=' '
if(len(x)>=6):
temp=[0]*6
for i in range(6):
temp[i]=x[len(x)-6+i]
else:
temp='false'
strtemp=''.join(temp)
print(strtemp)
return(strtemp)
def extract_5(x):
strtemp=' '
if(len(x)>=5):
temp=[0]*5
for i in range(5):
temp[i]=x[len(x)-5+i]
else:
temp='false'
strtemp=''.join(temp)
print(strtemp)
return(strtemp)
S=input()
flag=1
while(flag==1):
if(extract_6(S)=='dreamer'):
S=S.rstrip('dreamer')
elif(extract_6(S)=='eraser'):
S=S.rstrip('eraser')
elif(extract_5(S)=='dream'):
S=S.rstrip('dream')
elif(extract_5(S)=='erase'):
S=S.rstrip('erase')
else:
flag=0
if len(S)!=0:
print('NO')
else:
print('YES')
print(S)
|
s771456075
|
Accepted
| 159
| 3,316
| 768
|
def extract_6(x):
strtemp=' '
if(len(x)>=6):
temp=[0]*6
for i in range(6):
temp[i]=x[len(x)-6+i]
else:
temp='false'
strtemp=''.join(temp)
return(strtemp)
def extract_5(x):
strtemp=' '
if(len(x)>=5):
temp=[0]*5
for i in range(5):
temp[i]=x[len(x)-5+i]
else:
temp='false'
strtemp=''.join(temp)
return(strtemp)
def extract_7(x):
strtemp=' '
if(len(x)>=7):
temp=[0]*7
for i in range(7):
temp[i]=x[len(x)-7+i]
else:
temp='false'
strtemp=''.join(temp)
return(strtemp)
S=input()
flag=1
while(flag==1):
if(extract_7(S)=='dreamer'):
S=S[:-7]
elif(extract_6(S)=='eraser'):
S=S[:-6]
elif(extract_5(S)=='dream'):
S=S[:-5]
elif(extract_5(S)=='erase'):
S=S[:-5]
else:
flag=0
if len(S)!=0:
print('NO')
else:
print('YES')
|
s529588901
|
p03049
|
u269318751
| 2,000
| 1,048,576
|
Wrong Answer
| 33
| 3,700
| 360
|
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.
|
#python 3.5.2
N = int(input())
s_list = [input() for i in range(N)]
cnt = sum([1 for s in s_list if "AB" in s])
a_end, b_start, a_b = 0, 0, 0
for s in s_list:
if s[0] == "B" and s[-1] == "A":
a_b += 1
elif s[0] == "B":
b_start += 1
elif s[-1] == "A":
a_end += 1
print(a_end, b_start, a_b)
print(min(a_end, b_start)+a_b+cnt)
|
s374461523
|
Accepted
| 34
| 3,700
| 523
|
#python 3.5.2
N = int(input())
s_list = [input() for i in range(N)]
cnt = sum([s.count("AB") for s in s_list])
a_end, b_start, a_b = 0, 0, 0
for s in s_list:
if s[0] == "B" and s[-1] == "A":
a_b += 1
elif s[0] == "B":
b_start += 1
elif s[-1] == "A":
a_end += 1
if a_b == 0:
cnt += min(a_end, b_start)
else:
cnt += a_b - 1
if a_end >= 1:
cnt += 1
a_end -= 1
if b_start >= 1:
cnt += 1
b_start -= 1
cnt += min(a_end, b_start)
print(cnt)
|
s089978473
|
p03693
|
u735069283
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 105
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r,g,b=map(int,input().split())
result = (r*100+g*10+b)%4
if result==0:
print('yes')
else:
print('no')
|
s838649651
|
Accepted
| 17
| 2,940
| 79
|
r,g,b=map(int,input().split())
print('YES' if (r*100+g*10+b)%4 == 0 else 'NO')
|
s700323501
|
p02831
|
u517099533
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 841
|
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.
|
a, b = list(map(int, input().split()))
def prime(n):
a = []
while n % 2 == 0:
a.append(2)
n //= 2
f = 3
while f * f <= n:
if n % f == 0:
a.append(f)
n //= f
else:
f += 2
if n != 1:
a.append(n)
return a
a = sorted(prime(a))
b = sorted(prime(b))
print(a, b)
c = set()
def cal(a, b):
c = set()
for i in range(len(a)):
c.add(a[i])
for i in range(len(b)):
c.add(b[i])
ans = []
e = 1
for i in range(len(c)):
d =c.pop()
if a.count(d) >= b.count(d):
for j in range(a.count(d)):
ans.append(d)
else:
for j in range(b.count(d)):
ans.append(d)
for i in range(len(ans)):
e *= ans[i]
return e
print(cal(a, b))
|
s849085132
|
Accepted
| 17
| 3,064
| 829
|
a, b = list(map(int, input().split()))
def prime(n):
a = []
while n % 2 == 0:
a.append(2)
n //= 2
f = 3
while f * f <= n:
if n % f == 0:
a.append(f)
n //= f
else:
f += 2
if n != 1:
a.append(n)
return a
a = sorted(prime(a))
b = sorted(prime(b))
c = set()
def cal(a, b):
c = set()
for i in range(len(a)):
c.add(a[i])
for i in range(len(b)):
c.add(b[i])
ans = []
e = 1
for i in range(len(c)):
d =c.pop()
if a.count(d) >= b.count(d):
for j in range(a.count(d)):
ans.append(d)
else:
for j in range(b.count(d)):
ans.append(d)
for i in range(len(ans)):
e *= ans[i]
return e
print(cal(a, b))
|
s358770222
|
p04029
|
u633450100
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 34
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N = int(input())
print(N*(N+1)/2)
|
s283460653
|
Accepted
| 17
| 2,940
| 34
|
N = int(input())
print(N*(N+1)//2)
|
s741726103
|
p03973
|
u619819312
| 2,000
| 262,144
|
Wrong Answer
| 268
| 7,084
| 271
|
N people are waiting in a single line in front of the Takahashi Store. The cash on hand of the i-th person from the front of the line is a positive integer A_i. Mr. Takahashi, the shop owner, has decided on the following scheme: He picks a product, sets a positive integer P indicating its price, and shows this product to customers in order, starting from the front of the line. This step is repeated as described below. At each step, when a product is shown to a customer, if price P is equal to or less than the cash held by that customer at the time, the customer buys the product and Mr. Takahashi ends the current step. That is, the cash held by the first customer in line with cash equal to or greater than P decreases by P, and the next step begins. Mr. Takahashi can set the value of positive integer P independently at each step. He would like to sell as many products as possible. However, if a customer were to end up with 0 cash on hand after a purchase, that person would not have the fare to go home. Customers not being able to go home would be a problem for Mr. Takahashi, so he does not want anyone to end up with 0 cash. Help out Mr. Takahashi by writing a program that determines the maximum number of products he can sell, when the initial cash in possession of each customer is given.
|
n=int(input())
a=[int(input())for i in range(n)]
c=0
b=1
d=0
while c<n:
if a[c]==b:
b+=1
c+=1
else:
if a[c]%b==0:
d+=max((a[c]-2-b)//b+1,a[c]//b-1)
c+=1
else:
d+=a[c]//b
c+=1
print(d)
|
s105469461
|
Accepted
| 237
| 7,084
| 311
|
n=int(input())
a=[int(input())for i in range(n)]
c=0
b=1
d=0
while c<n:
if a[c]==b:
b+=1
c+=1
else:
if a[c]%b==0:
d+=((a[c]-2-b)//b+1if b!=1 else a[c]-1)
c+=1
b+=(1 if b==1 else 0)
else:
d+=a[c]//b
c+=1
print(d)
|
s630636725
|
p03548
|
u169501420
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 233
|
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?
|
# -*- coding: utf-8 -*-
[X, Y, Z] = [int(i) for i in input().split()]
personal_width = Y + Z
chair_width = X - 2 * Z
number = chair_width // personal_width
amari = X % personal_width
if amari >= X:
number += 1
print(number)
|
s928115938
|
Accepted
| 17
| 3,060
| 229
|
# -*- coding: utf-8 -*-
[X, Y, Z] = [int(i) for i in input().split()]
personal_width = Y + Z
chair_width = X - Z
number = chair_width // personal_width
amari = X % personal_width
if amari >= X:
number += 1
print(number)
|
s908386243
|
p02842
|
u677121387
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 91
|
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
N = int(input())
X = int(N//1.08)
if (X*1.08)//1 == N:
print(X)
else:
print(":(")
|
s911858907
|
Accepted
| 17
| 2,940
| 132
|
N = int(input())
X = int(N//1.08)
if (X*1.08)//1 == N:
print(X)
elif ((X+1)*1.08)//1 == N:
print(X+1)
else:
print(":(")
|
s808741121
|
p03353
|
u672475305
| 2,000
| 1,048,576
|
Wrong Answer
| 1,766
| 4,340
| 207
|
You are given a string s. Among the **different** substrings of s, print the K-th lexicographically smallest one. A substring of s is a string obtained by taking out a non-empty contiguous part in s. For example, if s = `ababc`, `a`, `bab` and `ababc` are substrings of s, while `ac`, `z` and an empty string are not. Also, we say that substrings are different when they are different as strings. Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
|
s = input()
k = int(input())
dct = []
for i in range(len(s)):
for j in range(k+1):
word = s[i:i+j]
if word not in dct:
dct.append(word)
dct.sort()
for d in dct:
print(d)
|
s625049104
|
Accepted
| 33
| 3,188
| 331
|
s = input()
k = int(input())
lst = sorted(list(set(s)))
tank = []
for tg in lst:
for i in range(len(s)):
if s[i]==tg:
for j in range(k):
if i+j<=len(s) and (s[i:i+j+1] not in tank):
tank.append(s[i:i+j+1])
if len(tank) >= k:
break
tank.sort()
print(tank[k-1])
|
s940631038
|
p02612
|
u337949146
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,152
| 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)
|
s970017768
|
Accepted
| 27
| 8,940
| 36
|
print((1000-int(input())%1000)%1000)
|
s327325882
|
p02277
|
u917432951
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,704
| 1,057
|
Let's arrange a deck of cards. Your task is to sort totally n cards. A card consists of a part of a suit (S, H, C or D) and an number. Write a program which sorts such cards based on the following pseudocode: Partition(A, p, r) 1 x = A[r] 2 i = p-1 3 for j = p to r-1 4 do if A[j] <= x 5 then i = i+1 6 exchange A[i] and A[j] 7 exchange A[i+1] and A[r] 8 return i+1 Quicksort(A, p, r) 1 if p < r 2 then q = Partition(A, p, r) 3 run Quicksort(A, p, q-1) 4 run Quicksort(A, q+1, r) Here, A is an array which represents a deck of cards and comparison operations are performed based on the numbers. Your program should also report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
def partition(A,p,r):
x = A[r]
i = p-1
for j in range(p,r):
if A[j][1] <= x[1]:
i = i+1
(A[i],A[j]) = (A[j],A[i])
(A[i+1],A[r]) = (A[r],A[i+1])
return i+1
def quicksort(A,p,r):
if p < r:
q = partition(A,p,r)
quicksort(A,p,q-1)
quicksort(A,q+1,r)
def isStable(A,n):
for i in range(1,n-1):
if A[i-1][1] == A[i][1] and A[i-1][2] > A[i][2]:
return False
return True
if __name__ == '__main__':
n = (int)(input())
A = []
for i in range(0,n):
tmp = list(input().split())
A.append((tmp[0],(int)(tmp[1]),i))
quicksort(A,0,n-1)
print("Stable" if isStable(A,n) else "Not Stable" )
for x in A:
print(x[0],x[1])
|
s406779650
|
Accepted
| 1,080
| 22,516
| 1,057
|
def partition(A,p,r):
x = A[r]
i = p-1
for j in range(p,r):
if A[j][1] <= x[1]:
i = i+1
(A[i],A[j]) = (A[j],A[i])
(A[i+1],A[r]) = (A[r],A[i+1])
return i+1
def quicksort(A,p,r):
if p < r:
q = partition(A,p,r)
quicksort(A,p,q-1)
quicksort(A,q+1,r)
def isStable(A,n):
for i in range(1,n-1):
if A[i-1][1] == A[i][1] and A[i-1][2] > A[i][2]:
return False
return True
if __name__ == '__main__':
n = (int)(input())
A = []
for i in range(0,n):
tmp = list(input().split())
A.append((tmp[0],(int)(tmp[1]),i))
quicksort(A,0,n-1)
print("Stable" if isStable(A,n) else "Not stable" )
for x in A:
print(x[0],x[1])
|
s498922794
|
p03478
|
u702018889
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,116
| 148
|
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())
ans=0
for i in range(1,a+1):
k=list(str(i))
m=list(map(int,k))
if a<=sum(m)<=b:
ans+=i
print(ans)
|
s467497399
|
Accepted
| 41
| 9,120
| 148
|
a,b,c=map(int,input().split())
ans=0
for i in range(1,a+1):
n=list(str(i))
m=list(map(int,n))
if b<=sum(m)<=c:
ans+=i
print(ans)
|
s865870138
|
p03971
|
u261886891
| 2,000
| 262,144
|
Wrong Answer
| 109
| 4,016
| 400
|
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
N, A, B = list(map(int, input().split()))
S = input()
cnt1 = 0
cnt2 = 0
for x in S:
if x == 'a':
if cnt1 < A+B:
print("YES")
cnt1 += 1
else:
print("NO")
elif x == 'b':
if cnt1 < A+B and cnt2 < B:
print("YES")
cnt1 += 1
cnt2 += 1
else:
print("NO")
else:
print("NO")
|
s025445341
|
Accepted
| 101
| 4,244
| 400
|
N, A, B = list(map(int, input().split()))
S = input()
cnt1 = 0
cnt2 = 0
for x in S:
if x == 'a':
if cnt1 < A+B:
print("Yes")
cnt1 += 1
else:
print("No")
elif x == 'b':
if cnt1 < A+B and cnt2 < B:
print("Yes")
cnt1 += 1
cnt2 += 1
else:
print("No")
else:
print("No")
|
s659498389
|
p03574
|
u989345508
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,444
| 447
|
You are given an H ร W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
h,w=map(int,input().split())
s=[input()+"." if i!=h else "."*(w+1) for i in range(h+1)]
def count_8(i,j):
global h,w,s
cnt=s[i-1][j-1]=="#"+s[i-1][j+1]=="#"+s[i+1][j-1]=="#"+s[i+1][j+1]=="#" \
+s[i-1][j]=="#"+s[i][j-1]=="#"+s[i+1][j]=="#"+s[i][j+1]=="#"
return cnt
for i in range(h):
for j in range(w):
if s[i][j]=="#":
print("#",end="")
else:
print(count_8(i,j),end="")
print()
|
s716463461
|
Accepted
| 21
| 3,064
| 472
|
h,w=map(int,input().split())
s=[input()+"." if i!=h else "."*(w+1) for i in range(h+1)]
def count_8(i,j):
global h,w,s
cnt=str((s[i-1][j-1]=="#")+(s[i-1][j+1]=="#")+(s[i+1][j-1]=="#")+(s[i+1][j+1]=="#") \
+(s[i-1][j]=="#")+(s[i][j-1]=="#")+(s[i+1][j]=="#")+(s[i][j+1]=="#"))
return cnt
for i in range(h):
s_sub=""
for j in range(w):
if s[i][j]=="#":
s_sub+="#"
else:
s_sub+=count_8(i,j)
print(s_sub)
|
s284709733
|
p04011
|
u693694535
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 83
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
N=int(input())
K=int(input())
X=int(input())
Y=int(input())
Q=N*X+Y*(N-K)
print(Q)
|
s104380784
|
Accepted
| 17
| 2,940
| 117
|
N=int(input())
K=int(input())
X=int(input())
Y=int(input())
if N<=K:
Q=N*X
else:
Q=K*X+(N-K)*Y
print(int(Q))
|
s550726446
|
p02421
|
u144068724
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,576
| 343
|
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each. Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
|
n = int(input())
t_point = 0
h_point = 0
for i in range(n):
card = input().split()
print(card)
taro = card[0]
card.sort()
card.reverse()
print(card)
if card[0] == card[1]:
t_point += 1
h_point += 1
elif card[0] == taro:
t_point += 3
else:
h_point += 3
print(t_point,h_point)
|
s575313410
|
Accepted
| 30
| 7,568
| 311
|
n = int(input())
t_point = 0
h_point = 0
for i in range(n):
card = input().split()
taro = card[0]
card.sort()
card.reverse()
if card[0] == card[1]:
t_point += 1
h_point += 1
elif card[0] == taro:
t_point += 3
else:
h_point += 3
print(t_point,h_point)
|
s752189975
|
p02796
|
u509150616
| 2,000
| 1,048,576
|
Wrong Answer
| 2,207
| 43,516
| 589
|
In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep.
|
import numpy as np
def solve(XL, N):
XL.sort(key = lambda x:x[0])
for i in range(N):
if XL[i][0] == None:
pass
else:
right = XL[i][0] + XL[i][1]
for j in range(i+1, N):
if XL[j][0] == None:
pass
elif XL[j][0] - XL[j][1] < right:
XL[j][0] = None
count = 0
for i in range(N):
if XL[i][0] != 0:
count += 1
return count
N = int(input())
XL = []
for i in range(N):
x, l = map(int, input().split(' '))
XL.append([x, l])
print(solve(XL, N))
|
s480693947
|
Accepted
| 320
| 25,428
| 329
|
def solve(XL, N):
XL.sort(key=lambda x:x[1])
count = N
for i in range(1, N):
if XL[i][0] < XL[i-1][1]:
XL[i][1] = XL[i-1][1]
count -= 1
return count
N = int(input())
XL = []
for i in range(N):
x, l = map(int, input().split(' '))
XL.append([x-l, x+l])
print(solve(XL, N))
|
s800632920
|
p03610
|
u865413330
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,316
| 112
|
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s = list(input().split())
result = ""
for i in range(len(s)):
if i % 2 != 0:
result += s[i]
print(s)
|
s015344358
|
Accepted
| 45
| 3,956
| 114
|
s = list(input())
result = ""
for i in range(len(s)):
if (i+1) % 2 != 0:
result += s[i]
print(result)
|
s302678392
|
p02841
|
u263339477
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 79
|
In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month.
|
a, b=map(int, input().split())
c, d=map(int, input().split())
0 if a==c else 1
|
s865005484
|
Accepted
| 18
| 2,940
| 86
|
a, b=map(int, input().split())
c, d=map(int, input().split())
print(0 if a==c else 1)
|
s992752886
|
p03719
|
u604655161
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 197
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
def ABC_61_A():
A,B,C = map(int, input().split())
if C>=A and C<=B:
print('YES')
else:
print('NO')
if __name__ == '__main__':
ABC_61_A()
|
s875643493
|
Accepted
| 17
| 2,940
| 197
|
def ABC_61_A():
A,B,C = map(int, input().split())
if C>=A and C<=B:
print('Yes')
else:
print('No')
if __name__ == '__main__':
ABC_61_A()
|
s421703070
|
p04043
|
u102242691
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 159
|
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 = list(map(int,input().split()))
if a.sort() == [5,5,7]:
print("YES")
else:
print("NO")
|
s910000240
|
Accepted
| 17
| 2,940
| 171
|
a = list(map(int,input().split()))
a.sort()
#print(a)
if a == [5,5,7]:
print("YES")
else:
print("NO")
|
s620055745
|
p03473
|
u636775911
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 45
|
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?
|
#coding:utf-8
n=int(input())
n-=48
print(n)
|
s764091390
|
Accepted
| 17
| 2,940
| 46
|
#coding:utf-8
n=int(input())
n=48-n
print(n)
|
s683469339
|
p03555
|
u875769753
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,040
| 144
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
C1 = input()
C2 = input()
ans = 'YES'
if C1[0] != C2[2]:
ans = 'NO'
if C1[1] != C2[1]:
ans = 'NO'
if C1[2] != C2[1]:
ans = 'NO'
print(ans)
|
s107729958
|
Accepted
| 33
| 8,932
| 144
|
C1 = input()
C2 = input()
ans = 'YES'
if C1[0] != C2[2]:
ans = 'NO'
if C1[1] != C2[1]:
ans = 'NO'
if C1[2] != C2[0]:
ans = 'NO'
print(ans)
|
s192978028
|
p02842
|
u556225812
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 90
|
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
N = int(input())
x = N // 1.08
M = x * 1.08 // 1
if N == M:
print(x)
else:
print(':(')
|
s387296831
|
Accepted
| 17
| 2,940
| 117
|
import math
N = int(input())*100
x = math.ceil(N / 108)
M = x*108
if M//100 == N//100:
print(x)
else:
print(':(')
|
s388574511
|
p00205
|
u847467233
| 1,000
| 131,072
|
Wrong Answer
| 30
| 5,596
| 434
|
ไปฒ่ฏใ 5 ไบบ็ตใงใใใใใใใใใใจใซใชใใพใใใใใใใใใจใฏใใฐใผใใใงใญใใใผใจใใ 3ใคใฎๆใใใใใฐใผใจใใงใญใฎๅ่ฒ ใชใใฐใผใใๅใกใใปใใงใญใใ่ฒ ใใใใใงใญใจใใผใชใใใใงใญใใๅใกใใปใใผใใ่ฒ ใใใใใผใจใฐใผใชใใใผใใๅใกใใปใฐใผใใ่ฒ ใใใจใใใซใผใซใงใใๅ
จๅกใๅใๆใใพใใฏใฐใผใใใงใญใใใผๅ
จใฆใๅบใๅ ดๅใฏใใใใใใจใชใใพใใ 5 ไบบใฎใใใใใใฎๆใๅ
ฅๅใจใใใใใใใฎไบบใฎๅๆใๅบๅใใใใญใฐใฉใ ใไฝๆใใฆใใ ใใใใใใใใใฎๆใฏใใฐใผใฏ 1ใใใงใญใฏ 2ใใใผใฏ 3 ใฎๆฐๅญใง่กจใใพใใๅๆใฏใๅใกใใ 1ใใ่ฒ ใใใ 2ใใใใใใใ 3 ใฎๆฐๅญใง่กจใใๅ
ฅๅ้ ใซๅพใฃใฆๅบๅใใพใใ
|
# AOJ 0205: Rock, Paper, Scissors
# Python3 2018.6.25 bal4u
h = [0]*5
while True:
try:
for i in range(5): h[i] = int(input())
except: break
print(type(h), h)
if len(set(h)) == 1 or len(set(h)) == 3:
print(*[3]*5, sep='\n')
continue
ans = [1]*5
for i in range(5):
if h[i] == 1: ans[i] += (h.count(3) > 0)
elif h[i] == 2: ans[i] += (h.count(1) > 0)
else: ans[i] += (h.count(2) > 0)
print(*ans, sep='\n')
|
s404357430
|
Accepted
| 30
| 5,596
| 415
|
# AOJ 0205: Rock, Paper, Scissors
# Python3 2018.6.25 bal4u
h = [0]*5
while True:
try:
for i in range(5): h[i] = int(input())
except: break
if len(set(h)) == 1 or len(set(h)) == 3:
print(*[3]*5, sep='\n')
continue
ans = [1]*5
for i in range(5):
if h[i] == 1: ans[i] += (h.count(3) > 0)
elif h[i] == 2: ans[i] += (h.count(1) > 0)
else: ans[i] += (h.count(2) > 0)
print(*ans, sep='\n')
|
s033927637
|
p03555
|
u306142032
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
c = input()
c_r = input()
if reversed(c) == c_r:
print("Yes")
else:
print("No")
|
s931305783
|
Accepted
| 17
| 2,940
| 90
|
c = input()
c_r = input()
c = c[::-1]
if c == c_r:
print("YES")
else:
print("NO")
|
s365093698
|
p02399
|
u828399801
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,648
| 151
|
Write a program which reads two integers a and b, and calculates the following values: * a รท b: d (in integer) * remainder of a รท b: r (in integer) * a รท b: f (in real number)
|
splited = input().split(" ")
f = int(splited[0])/int(splited[1])
r = int(splited[0])%int(splited[1])
d = int(splited[0])//int(splited[1])
print(d,r,f)
|
s070731138
|
Accepted
| 30
| 7,644
| 171
|
splited = input().split(" ")
f = int(splited[0])/int(splited[1])
r = int(splited[0])%int(splited[1])
d = int(splited[0])//int(splited[1])
print(d, r, "{0:.5f}".format(f))
|
s888470459
|
p02401
|
u535719732
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,608
| 260
|
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.
|
ans = 0
while True:
num = list(map(str,input().split()))
if(num[1] == "?"): break
a = int(num[0])
b = int(num[2])
if(num[1] == "+"): ans=a+b
elif(num[1] == "-"): ans=a-b
elif(num[1] == "*"): ans=a*b
else: ans=a/b
print(ans)
|
s340797392
|
Accepted
| 20
| 5,596
| 286
|
ans = 0
while True:
num = list(map(str,input().split()))
if(num[1] == "?"): break
a = int(num[0])
b = int(num[2])
op = num[1]
if(op == "+"): ans=a+b
elif(op == "-"): ans=a-b
elif(op == "*"): ans=a*b
elif(op == "/"): ans=a/b
print("%d" %ans)
|
s590129856
|
p03862
|
u994521204
| 2,000
| 262,144
|
Wrong Answer
| 191
| 20,788
| 516
|
There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies. Snuke can perform the following operation any number of times: * Choose a box containing at least one candy, and eat one of the candies in the chosen box. His objective is as follows: * Any two neighboring boxes contain at most x candies in total. Find the minimum number of operations required to achieve the objective.
|
n, x = map(int, input().split())
A = list(map(int, input().split()))
cnt = [0] * (n - 1)
ans1 = 0
for i in range(n - 1):
cnt[i] = A[i] + A[i + 1]
cnt[i] -= x
cnt[i] = max(cnt[i], 0)
temp_cnt = cnt[::]
print(cnt)
for i in range(n - 1):
if temp_cnt[i] == 0:
continue
if i < n - 2:
ans1 += temp_cnt[i]
temp_cnt[i + 1] -= min(A[i + 1], temp_cnt[i])
temp_cnt[i] = 0
temp_cnt[i + 1] = max(temp_cnt[i + 1], 0)
# print(ans1)
ans1 += temp_cnt[-1]
print(ans1)
|
s752001377
|
Accepted
| 192
| 19,908
| 518
|
n, x = map(int, input().split())
A = list(map(int, input().split()))
cnt = [0] * (n - 1)
ans1 = 0
for i in range(n - 1):
cnt[i] = A[i] + A[i + 1]
cnt[i] -= x
cnt[i] = max(cnt[i], 0)
temp_cnt = cnt[::]
# print(cnt)
for i in range(n - 1):
if temp_cnt[i] == 0:
continue
if i < n - 2:
ans1 += temp_cnt[i]
temp_cnt[i + 1] -= min(A[i + 1], temp_cnt[i])
temp_cnt[i] = 0
temp_cnt[i + 1] = max(temp_cnt[i + 1], 0)
# print(ans1)
ans1 += temp_cnt[-1]
print(ans1)
|
s914848154
|
p03455
|
u652150585
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 98
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
num=list(map(int,input().split()))
if num[0]*num[1]%2==0:
print("even")
else:
print("odd")
|
s130597870
|
Accepted
| 17
| 2,940
| 98
|
num=list(map(int,input().split()))
if num[0]*num[1]%2==0:
print("Even")
else:
print("Odd")
|
s782947332
|
p03469
|
u902917675
| 2,000
| 262,144
|
Wrong Answer
| 26
| 8,996
| 34
|
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.
|
x = input()
print("2018" + x[6:])
|
s209941608
|
Accepted
| 27
| 8,980
| 34
|
x = input()
print("2018" + x[4:])
|
s388984764
|
p03139
|
u102960641
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 63
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
n,a,b = map(int, input().split())
print(min(a,b), min(a+b-n,n))
|
s147547952
|
Accepted
| 17
| 2,940
| 71
|
n,a,b = map(int, input().split())
print(min(a,b), max(min(a+b-n,n),0))
|
s039968824
|
p03699
|
u747602774
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 145
|
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
N=int(input())
li=[]
for n in range(N):
li.append(int(input()))
print(li)
li.sort()
S=sum(li)
while S%10==0 and li!=[]:
S-=li.pop(0)
print(S)
|
s029359531
|
Accepted
| 17
| 3,060
| 211
|
N=int(input())
li=[]
for n in range(N):
li.append(int(input()))
li.sort()
S=sum(li)
while S%10==0 and li!=[]:
if li[0]%10==0:
li.pop(0)
else:
S-=li.pop(0)
if li==[]:
print(0)
else:
print(S)
|
s975958420
|
p02389
|
u279483260
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,632
| 164
|
Write a program which calculates the area and perimeter of a given rectangle.
|
def calc(a, b):
area = a * b
circumference = (a * 2) + (b * 2)
return area, circumference
a, b = map(int, input('a, b???').split())
print(calc(a, b))
|
s448454035
|
Accepted
| 20
| 5,588
| 129
|
# -*- coding: utf-8 -*-
a, b = map(int, input().split())
area = a * b
circle = (a + b) * 2
print('{} {}'.format(area, circle))
|
s476606548
|
p02274
|
u196653484
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 655
|
For a given sequence $A = \\{a_0, a_1, ... a_{n-1}\\}$, the number of pairs $(i, j)$ where $a_i > a_j$ and $i < j$, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the following program: bubbleSort(A) cnt = 0 // the number of inversions for i = 0 to A.length-1 for j = A.length-1 downto i+1 if A[j] < A[j-1] swap(A[j], A[j-1]) cnt++ return cnt For the given sequence $A$, print the number of inversions of $A$. Note that you should not use the above program, which brings Time Limit Exceeded.
|
count=0
def merge(A, left, mid, right):
global count
L=A[left:mid]+[2**32]
R=A[mid:right]+[2**32]
i=0
j=0
for k in range(left,right):
if L[i] <= R[j]:
A[k] = L[i]
i += 1
else:
A[k] = R[j]
j += 1
count+=1
def merge_sort(A, left, right):
if left+1 < right:
mid=(left+right)//2
merge_sort(A, left, mid)
merge_sort(A, mid, right)
merge(A, left, mid, right)
if __name__ == "__main__":
n=int(input())
A=list(map(int,input().split()))
merge_sort(A,0,n)
print(*A)
print(count)
|
s086971547
|
Accepted
| 2,070
| 27,480
| 732
|
count=0
def merge(A, left, mid, right):
global count
h=[]
L=A[left:mid]+[2**32]
R=A[mid:right]+[2**32]
i=0
j=0
for k in range(left,right):
if L[i] <= R[j]:
A[k] = L[i]
h.append(L[i])
i += 1
else:
A[k] = R[j]
h.append(R[j])
j += 1
if i < len(L)-1:
count += len(L) - 1 - i
def merge_sort(A, left, right):
if left+1 < right:
mid=(left+right)//2
merge_sort(A, left, mid)
merge_sort(A, mid, right)
merge(A, left, mid, right)
if __name__ == "__main__":
n=int(input())
A=list(map(int,input().split()))
merge_sort(A,0,n)
print(count)
|
s467526187
|
p03047
|
u163449343
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 49
|
Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi. How many ways are there to choose K consecutive integers?
|
n,k = map(int, input().split())
print((n-2)*k//k)
|
s584844110
|
Accepted
| 17
| 3,060
| 146
|
n,k = map(int, input().split())
ans = 0
num = [i for i in range(1, k+1)]
while num[-1] <= n:
num = [i+1 for i in num]
ans += 1
print(ans)
|
s687550744
|
p03815
|
u576432509
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 79
|
Snuke has decided to play with a six-sided die. Each of its six sides shows an integer 1 through 6, and two numbers on opposite sides always add up to 7. Snuke will first put the die on the table with an arbitrary side facing upward, then repeatedly perform the following operation: * Operation: Rotate the die 90ยฐ toward one of the following directions: left, right, front (the die will come closer) and back (the die will go farther). Then, obtain y points where y is the number written in the side facing upward. For example, let us consider the situation where the side showing 1 faces upward, the near side shows 5 and the right side shows 4, as illustrated in the figure. If the die is rotated toward the right as shown in the figure, the side showing 3 will face upward. Besides, the side showing 4 will face upward if the die is rotated toward the left, the side showing 2 will face upward if the die is rotated toward the front, and the side showing 5 will face upward if the die is rotated toward the back. Find the minimum number of operation Snuke needs to perform in order to score at least x points in total.
|
x=int(input())
xi=x//11
if xi==0:
print(2*xi)
else:
print(2*xi+1)
|
s820286143
|
Accepted
| 17
| 2,940
| 102
|
x=int(input())
n=x//11
if x%11==0:
print(2*n)
elif x%11<=6:
print(2*n+1)
else:
print(2*n+2)
|
s265693889
|
p02280
|
u805716376
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,624
| 928
|
A rooted binary tree is a tree with a root node in which every node has at most two children. Your task is to write a program which reads a rooted binary tree _T_ and prints the following information for each node _u_ of _T_ : * node ID of _u_ * parent of _u_ * sibling of _u_ * the number of children of _u_ * depth of _u_ * height of _u_ * node type (root, internal node or leaf) If two nodes have the same parent, they are **siblings**. Here, if _u_ and _v_ have the same parent, we say _u_ is a sibling of _v_ (vice versa). The height of a node in a tree is the number of edges on the longest simple downward path from the node to a leaf. Here, the given binary tree consists of _n_ nodes and evey node has a unique ID from 0 to _n_ -1.
|
n = int(input())
deg = [0]*n
child, parent = {}, {}
sib = [None]*n
for i in range(n):
a = list(map(int, input().split()))
if a[1] != -1:
child[i] = a[1:]
sib[a[1]] = a[2]
sib[a[2]] = a[1]
deg[i] = 2
for j in child[i]:
parent[j] = i
else:
child[i] = []
root = (set(child)-set(parent)).pop()
depth = [None]*n
depth[root] = 0
hei = [None]*n
parent[root] = -1
sib[root] = -1
print(child)
def dfs(s):
if len(child[s]) == 0:
hei[s] = 0
return
for i in child[s]:
depth[i] = depth[s] + 1
dfs(i)
hei[s] = max(hei[i] for i in child[s]) + 1
dfs(root)
for i in range(n):
node_type = 'root' if parent[i] == -1 else 'leaf' if len(child[i]) == 0 else 'internal node'
print(f'node {i}: parent = {parent[i]}, sibling = {sib[i]}, degree = {deg[i]}, depth = {depth[i]}, height = {hei[i]}, {node_type}')
|
s605531279
|
Accepted
| 20
| 5,644
| 1,228
|
n = int(input())
deg = [0]*n
child, parent = {}, {}
sib = [None]*n
for i in range(n):
a = list(map(int, input().split()))
if a[1] != -1:
if a[2] == -1:
child[a[0]] = [a[1]]
sib[a[1]] = a[2]
deg[a[0]] = 1
else:
child[a[0]] = a[1:]
sib[a[1]] = a[2]
sib[a[2]] = a[1]
deg[a[0]] = 2
for j in child[a[0]]:
parent[j] = a[0]
elif a[2] != -1:
child[a[0]] = [a[2]]
sib[a[2]] = a[1]
deg[a[0]] = 1
for j in child[a[0]]:
parent[j] = a[0]
else:
child[a[0]] = []
root = (set(child)-set(parent)).pop()
depth = [None]*n
depth[root] = 0
hei = [None]*n
parent[root] = -1
sib[root] = -1
def dfs(s):
if len(child[s]) == 0:
hei[s] = 0
return
for i in child[s]:
depth[i] = depth[s] + 1
dfs(i)
hei[s] = max(hei[i] for i in child[s]) + 1
dfs(root)
for i in range(n):
node_type = 'root' if parent[i] == -1 else 'leaf' if len(child[i]) == 0 else 'internal node'
print(f'node {i}: parent = {parent[i]}, sibling = {sib[i]}, degree = {deg[i]}, depth = {depth[i]}, height = {hei[i]}, {node_type}')
|
s426800742
|
p01086
|
u037441960
| 8,000
| 262,144
|
Wrong Answer
| 20
| 5,604
| 453
|
A _Short Phrase_ (aka. Tanku) is a fixed verse, inspired by Japanese poetry Tanka and Haiku. It is a sequence of words, each consisting of lowercase letters 'a' to 'z', and must satisfy the following condition: > (The Condition for a Short Phrase) > The sequence of words can be divided into five sections such that the total > number of the letters in the word(s) of the first section is five, that of > the second is seven, and those of the rest are five, seven, and seven, > respectively. The following is an example of a Short Phrase. > > do the best and enjoy today at acm icpc In this example, the sequence of the nine words can be divided into five sections (1) "do" and "the", (2) "best" and "and", (3) "enjoy", (4) "today" and "at", and (5) "acm" and "icpc" such that they have 5, 7, 5, 7, and 7 letters in this order, respectively. This surely satisfies the condition of a Short Phrase. Now, _Short Phrase Parnassus_ published by your company has received a lot of contributions. By an unfortunate accident, however, some irrelevant texts seem to be added at beginnings and ends of contributed Short Phrases. Your mission is to write a program that finds the Short Phrase from a sequence of words that may have an irrelevant prefix and/or a suffix.
|
while True :
n = int(input())
if(n == 0) :
break
else :
A = [5, 7, 5, 7, 7]
ans = 0
w = [len(input()) for i in range(n)]
for i in range(n) :
k = 0
s = 0
for j in range(i, n) :
s += w[j]
if(s == A[k]) :
s = 0
k += 1
if(k == 5) :
ans = i + 1
break
elif(s > A[k]) :
break
if(ans != 0) :
break
print(ans)
|
s281421007
|
Accepted
| 30
| 5,616
| 773
|
while True :
n = int(input())
if(n == 0) :
break
else :
S = [len(input()) for i in range(n)]
T = [5, 7, 5, 7, 7]
ans = 0
for i in range(n) :
s = 0
w = 0
for j in range(i, n) :
s += S[j]
if(s == T[w]) :
s = 0
w += 1
if(w == 5) :
ans = i + 1
break
else :
pass
elif(s > T[w]) :
break
else :
pass
if(ans != 0) :
break
else :
pass
print(ans)
|
s353404115
|
p03474
|
u697696097
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 305
|
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
import sys
a,b=map(int,input().split())
ss=input().strip()
if len(ss) != (a+b+1):
print("No")
sys.exit()
#a 3 123-345
nums=[0,1,2,3,4,5,6,7,8,9]
for i in range(len(ss)):
if i==a:
if i!="-":
print("No")
sys.exit()
elif i not in nums:
print("No")
sys.exit()
print("Yes")
|
s831457473
|
Accepted
| 17
| 3,064
| 333
|
import sys
a,b=map(int,input().split())
ss=input().strip()
if len(ss) != (a+b+1):
print("No")
sys.exit()
#a 3 123-345
nums=["0","1","2","3","4","5","6","7","8","9"]
for i in range(len(ss)):
if i==a:
if ss[i]!="-":
print("No")
sys.exit()
elif ss[i] not in nums:
print("No")
sys.exit()
print("Yes")
|
s139539292
|
p03470
|
u496975476
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 186
|
An _X -layered kagami mochi_ (X โฅ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
#! /usr/bin/env python3
n = int(input())
d = list(map(int, input().split(" ")))
result = 1
d.sort()
for i in range(1, len(d)):
if d[i-1] < d[i]:
result += 1
print(result)
|
s605919328
|
Accepted
| 17
| 3,060
| 204
|
#! /usr/bin/env python3
n = int(input())
d = []
result = 1
for _ in range(0, n):
d.append(int(input()))
d.sort()
for i in range(1, len(d)):
if d[i-1] < d[i]:
result += 1
print(result)
|
s149768989
|
p04029
|
u231647664
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 241
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
a = str(input())
b = list(a)
c = len(b)
strs = []
for i in range(c):
if b[i] == 'B':
if i == 0 or len(strs) == 0 or b[i-1] == 'B':
continue
else:
strs = strs[:-1]
else:
strs.append(b[i])
print(''.join(strs))
|
s028637446
|
Accepted
| 18
| 2,940
| 55
|
a = int(input())
l = range(1, a+1)
b = sum(l)
print(b)
|
s337580613
|
p02678
|
u623283955
| 2,000
| 1,048,576
|
Wrong Answer
| 832
| 67,424
| 375
|
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
N,M=map(int,input().split())
l=[set() for i in range(N)]
for i in range(M):
a,b=map(int,input().split())
l[a-1].add(b-1)
l[b-1].add(a-1)
ope=[1]*N
ope[0]=0
s={0}
ans=[-1]*N
while len(s):
news=set()
for j in s:
for k in l[j]:
if ope[k]:
news.add(k)
ans[k]=j
ope[k]=0
s=news
print('yes')
for i in range(1,N):
print(ans[i]+1)
|
s931176400
|
Accepted
| 780
| 67,612
| 375
|
N,M=map(int,input().split())
l=[set() for i in range(N)]
for i in range(M):
a,b=map(int,input().split())
l[a-1].add(b-1)
l[b-1].add(a-1)
ope=[1]*N
ope[0]=0
s={0}
ans=[-1]*N
while len(s):
news=set()
for j in s:
for k in l[j]:
if ope[k]:
news.add(k)
ans[k]=j
ope[k]=0
s=news
print('Yes')
for i in range(1,N):
print(ans[i]+1)
|
s952905734
|
p03493
|
u320325426
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 25
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
print(input().count("0"))
|
s689431015
|
Accepted
| 17
| 2,940
| 25
|
print(input().count("1"))
|
s302469794
|
p03992
|
u258009780
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 39
|
This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s.
|
d = input()
print(d[0:3] + " " + d[4:])
|
s354331477
|
Accepted
| 17
| 2,940
| 33
|
d=input()
print(d[:4]+" "+d[4:])
|
s332858447
|
p03944
|
u141574039
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,064
| 507
|
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 โฆ i โฆ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 โฆ i โฆ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
W,H,N=map(int,input().split())
S=[0]*N; p=0
for i in range(N):
x,y,a=map(int,input().split())
S[i]=[x,y,a]
print("x,y,a",S)
e=101;w=-1;n=101;s=-1
for i in range(N):
if S[i][2]==1:
w=max(w,S[i][0])
elif S[i][2]==2:
e=min(e,S[i][0])
elif S[i][2]==3:
s=max(s,S[i][1])
elif S[i][2]==4:
n=min(n,S[i][1])
print("w,e,s,n",w,e,s,n)
for i in range(1,H+1):
for j in range(1,W+1):
if (j<=w and w>0) or (j>=e and e<101) or (i>=n and n<101) or (i<=s and s>0):
p=p+1
print(W*H-p)
|
s270459990
|
Accepted
| 20
| 3,064
| 507
|
W,H,N=map(int,input().split())
S=[0]*N; p=0
for i in range(N):
x,y,a=map(int,input().split())
S[i]=[x,y,a]
#print("x,y,a",S)
e=101;w=-1;n=101;s=-1
for i in range(N):
if S[i][2]==1:
w=max(w,S[i][0])
elif S[i][2]==2:
e=min(e,S[i][0])
elif S[i][2]==3:
s=max(s,S[i][1])
elif S[i][2]==4:
n=min(n,S[i][1])
#print("w,e,s,n",w,e,s,n)
for i in range(1,H+1):
for j in range(1,W+1):
if (j<=w and w>0) or (j>e and e<101) or (i>n and n<101) or (i<=s and s>0):
p=p+1
print(W*H-p)
|
s372822356
|
p02283
|
u564398841
| 2,000
| 131,072
|
Wrong Answer
| 20
| 7,740
| 1,055
|
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue. Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property: * Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$. The following figure shows an example of the binary search tree. For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk. A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields _left_ , _right_ , and _p_ that point to the nodes corresponding to its left child, its right child, and its parent, respectively. To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree. 1 insert(T, z) 2 y = NIL // parent of x 3 x = 'the root of T' 4 while x โ NIL 5 y = x // set the parent 6 if z.key < x.key 7 x = x.left // move to the left child 8 else 9 x = x.right // move to the right child 10 z.p = y 11 12 if y == NIL // T is empty 13 'the root of T' = z 14 else if z.key < y.key 15 y.left = z // z is the left child of y 16 else 17 y.right = z // z is the right child of y Write a program which performs the following operations to a binary search tree $T$. * insert $k$: Insert a node containing $k$ as key into $T$. * print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively. You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
|
treeRoot = {'left': None, 'right': None, 'p': None, 'val': None}
root = None
def insert(val):
global root
z = {'left': None, 'right': None, 'p': None, 'val': val}
x = root
y = None
while x:
y = x
if z['val'] < x['val']:
x = x['left']
else:
x = x['right']
z['p'] = y
if y is None:
root = z
else:
if z['val'] < y['val']:
y['left'] = z
else:
y['right'] = z
def inorder(root):
if root is None:
return
inorder(root['left'])
print(root['val'])
inorder(root['right'])
def preorder(root):
if root is None:
return
print(root['val'])
preorder(root['left'])
preorder(root['right'])
M = int(input())
for _ in range(M):
line = input()
if line == 'print':
inorder(root)
preorder(root)
break
inst, k = line.split()
k = int(k)
if inst == 'insert':
insert(k)
elif inst == 'find':
pass
elif inst == 'delete':
pass
|
s833969901
|
Accepted
| 9,150
| 153,872
| 1,500
|
class Node:
def __init__(self, val=None):
self.l = None
self.r = None
self.p = None
self.v = val
def insert(self, val):
if self.v is None:
self.v = val
return
n = Node(val)
cur_node = self
prev_node = None
while cur_node:
prev_node = cur_node
if val < prev_node.v:
cur_node = prev_node.l
else:
cur_node = prev_node.r
if prev_node is None:
self.v = n.v
else:
if val < prev_node.v:
prev_node.l = n
else:
prev_node.r = n
def __str__(self):
return str(self.v)
def preorder(self):
if self.v:
print(' {0}'.format(self.v), end='')
if self.l:
self.l.preorder()
if self.r:
self.r.preorder()
def inorder(self):
if self.l:
self.l.inorder()
if self.v:
print(' {0}'.format(self.v), end='')
if self.r:
self.r.inorder()
M = int(input())
tree = Node()
for _ in range(M):
line = input().strip()
if line == 'print':
tree.inorder()
print('')
tree.preorder()
print('')
else:
inst, k = line.split()
k = int(k)
if inst == 'insert':
tree.insert(k)
elif inst == 'find':
pass
elif inst == 'delete':
pass
|
s200032350
|
p02557
|
u299869545
| 2,000
| 1,048,576
|
Wrong Answer
| 883
| 52,196
| 1,454
|
Given are two sequences A and B, both of length N. A and B are each sorted in the ascending order. Check if it is possible to reorder the terms of B so that for each i (1 \leq i \leq N) A_i \neq B_i holds, and if it is possible, output any of the reorderings that achieve it.
|
from heapq import heappush, heappop
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
all_freq = {}
cnt_b = [0] * 220000
for num in b:cnt_b[num] += 1
for num in a:
if num not in all_freq:
all_freq[num] = 0
all_freq[num] += 1
for num in b:
if num not in all_freq:
all_freq[num] = 0
all_freq[num] += 1
heap = []
for num, freq in all_freq.items():
if freq > n:
print("No")
exit()
if cnt_b[num] > 0:
heappush(heap, (-freq, num))
ans = []
def get_max():
(freq, num) = heappop(heap)
freq = -freq
while all_freq[num] != freq:
(freq, num) = heappop(heap)
freq = -freq
return num
for num in a:
most_freq_num = get_max()
if most_freq_num != num:
ans.append(most_freq_num)
all_freq[most_freq_num] -= 1
cnt_b[most_freq_num] -= 1
if cnt_b[most_freq_num] > 0:
heappush(heap, (-all_freq[most_freq_num], most_freq_num))
else:
second_freq_num = get_max()
heappush(heap, (-all_freq[most_freq_num], most_freq_num))
ans.append(second_freq_num)
all_freq[second_freq_num] -= 1
cnt_b[second_freq_num] -= 1
if cnt_b[second_freq_num] > 0:
heappush(heap, (-all_freq[second_freq_num], second_freq_num))
if cnt_b[num] == 0:continue
all_freq[num] -= 1
heappush(heap, (-all_freq[num], num))
print(*ans)
|
s791635895
|
Accepted
| 873
| 52,224
| 1,467
|
from heapq import heappush, heappop
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
all_freq = {}
cnt_b = [0] * 220000
for num in b:cnt_b[num] += 1
for num in a:
if num not in all_freq:
all_freq[num] = 0
all_freq[num] += 1
for num in b:
if num not in all_freq:
all_freq[num] = 0
all_freq[num] += 1
heap = []
for num, freq in all_freq.items():
if freq > n:
print("No")
exit()
if cnt_b[num] > 0:
heappush(heap, (-freq, num))
ans = []
def get_max():
(freq, num) = heappop(heap)
freq = -freq
while all_freq[num] != freq:
(freq, num) = heappop(heap)
freq = -freq
return num
for num in a:
most_freq_num = get_max()
if most_freq_num != num:
ans.append(most_freq_num)
all_freq[most_freq_num] -= 1
cnt_b[most_freq_num] -= 1
if cnt_b[most_freq_num] > 0:
heappush(heap, (-all_freq[most_freq_num], most_freq_num))
else:
second_freq_num = get_max()
heappush(heap, (-all_freq[most_freq_num], most_freq_num))
ans.append(second_freq_num)
all_freq[second_freq_num] -= 1
cnt_b[second_freq_num] -= 1
if cnt_b[second_freq_num] > 0:
heappush(heap, (-all_freq[second_freq_num], second_freq_num))
all_freq[num] -= 1
if cnt_b[num] == 0:continue
heappush(heap, (-all_freq[num], num))
print("Yes")
print(*ans)
|
s632503206
|
p03493
|
u212328220
| 2,000
| 262,144
|
Wrong Answer
| 56
| 5,224
| 141
|
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.
|
import random
s1 = random.randint(0,1)
s2 = random.randint(0,1)
s3 = random.randint(0,1)
masu_list = (s1, s2, s3)
print(masu_list.count(1))
|
s218179277
|
Accepted
| 20
| 3,316
| 63
|
import random
masu_list = input()
print(masu_list.count('1'))
|
s822606087
|
p03097
|
u353797797
| 2,000
| 1,048,576
|
Wrong Answer
| 714
| 11,440
| 1,209
|
You are given integers N,\ A and B. Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists. * P_0=A * P_{2^N-1}=B * For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit.
|
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def main():
def parity_onebit(x):
return bin(x).count("1") & 1
def make_route(s, g, state):
if state.count(-1) == 1:
return [s, g]
sep_dig = bin(s ^ g)[::-1].find("1")
s_bit = (s >> sep_dig) & 1
Lstate = state[:]
Rstate = state[:]
Lstate[sep_dig] = s_bit
Rstate[sep_dig] = 1 - s_bit
for i in range(16):
ng = 0
i_copy = i
for sk in Lstate[::-1]:
ng <<= 1
if sk == -1:
ng += i_copy & 1
i_copy >>= 1
else:
ng += sk
ns = ng ^ (1 << sep_dig)
if ng != s and parity_onebit(ng) != parity_onebit(s) and ns != g:
break
res = make_route(s, ng, Lstate)
state[sep_dig] = 1 - s_bit
res += make_route(ns, g, Rstate)
return res
n, s, g = map(int, input().split())
if parity_onebit(s) == parity_onebit(g):
print("NO")
exit()
ans=make_route(s, g, [-1] * n)
print(*ans)
main()
|
s093485760
|
Accepted
| 744
| 14,888
| 1,033
|
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def bit(xx):return [format(x,"b") for x in xx]
popcnt=lambda x:bin(x).count("1")
def solve(xx,a,b):
if len(xx)==2:return [a,b]
k=(a^b).bit_length()-1
aa=[]
bb=[]
for x in xx:
if (x>>k&1)==(a>>k&1):aa.append(x)
else:bb.append(x)
i=0
mid=aa[i]
while mid==a or (mid^1<<k)==b or popcnt(a)&1==popcnt(mid)&1:
i+=1
mid=aa[i]
#print(bit(xx),a,b,k,bit(aa),bit(bb),format(mid,"b"))
return solve(aa,a,mid)+solve(bb,mid^1<<k,b)
def main():
n,a,b=MI()
if (popcnt(a)&1)^(popcnt(b)&1)==0:
print("NO")
exit()
print("YES")
print(*solve(list(range(1<<n)),a,b))
main()
|
s675998429
|
p03493
|
u923010184
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 184
|
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] = map(str,input().split())
b = 0
if a[0] == 1:
c = b + 1
else:
c = b
if a[1] == 1:
d = c + 1
else:
d = c
if a[2] == 1:
e = d + 1
else:
e = d
print(e)
|
s189391288
|
Accepted
| 17
| 2,940
| 25
|
print(input().count("1"))
|
s203719246
|
p03457
|
u256106029
| 2,000
| 262,144
|
Wrong Answer
| 381
| 27,380
| 775
|
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())
trip = []
for i in range(n):
array = list(map(int, input().strip().split()))
trip.append(array)
def func(num: int, query):
for j in range(len(query)):
if (query[j][0] % 2 == 0 and not (query[j][1] + query[j][2]) % 2 == 0) or \
(not query[j][0] % 2 == 0 and (query[j][1] + query[j][2]) % 2 == 0) or \
query[j][1] + query[j][2] > query[j][0]:
return print('No')
elif (query[j][0] % 2 == 0 and (query[j][1] + query[j][2]) % 2 == 0) or \
(not query[j][0] % 2 == 0 and not (query[j][1] + query[j][2]) % 2 == 0) and \
query[j][1] + query[j][2] <= query[j][0] and \
j + 1 == num:
return print('YES')
func(num=n, query=trip)
|
s520052290
|
Accepted
| 448
| 27,324
| 561
|
n = int(input())
travel = [[0, 0, 0]]
for i in range(n):
array = list(map(int, input().strip().split()))
travel.append(array)
def func(num, query):
for j in range(num):
time_temp = query[j+1][0] - query[j][0]
x_temp = abs(query[j+1][1] - query[j][1])
y_temp = abs(query[j+1][2] - query[j][2])
if not (x_temp + y_temp) % 2 == time_temp % 2:
print('No')
return
if x_temp + y_temp > time_temp:
print('No')
return
print('Yes')
func(num=n, query=travel)
|
s532995884
|
p03416
|
u507116804
| 2,000
| 262,144
|
Wrong Answer
| 62
| 2,940
| 154
|
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
a,b=map(int, input().split())
n=b-a
k=0
for i in range(n+1):
A=str(a)
if A[0]==A[4] and A[1]==A[3]:
k+=1
else:
k+=0
a+=1
|
s554563052
|
Accepted
| 71
| 2,940
| 164
|
a,b=map(int, input().split())
n=b-a
k=0
for i in range(n+1):
A=str(a)
if A[0]==A[-1] and A[1]==A[-2]:
k+=1
else:
k+=0
a+=1
print(k)
|
s493235128
|
p01852
|
u546285759
| 5,000
| 262,144
|
Wrong Answer
| 20
| 7,520
| 61
|
่่ฅฟใใใฏๆใฎๆฐใๅขใใใใๆธใใใใใงใใ๏ผ ่่ฅฟใใใฎๅใซใฏ n ๅใฎใพใใใ
ใใใใ๏ผ ่่ฅฟใใใฏๆใๆใฃใฆใพใใใ
ใใฎๅๆฐใๆฐใใใใจใใฆใใ๏ผ ่่ฅฟใใใฎๆใๅใใๅฝขใฏๆใใฆใใใๆใใฆใใชใใใฎ 2 ใคใใ็กใ๏ผ ่่ฅฟใใใฏ [2 ้ฒๆฐ](http://www.asahi-net.or.jp/~ax2s-kmtn/ref/bdh.html)ใ็่งฃใใฆใใ๏ผ ่่ฅฟใใใฏๅๆใซ 2 ้ฒๆฐใฎๆกใๅฏพๅฟใใใฆๆฐใๆฐใใใใจใๅบๆฅใ๏ผ ่่ฅฟใใใฏๅฏพๆฐใ็่งฃใใฆใใชใ๏ผ ่่ฅฟใใใฎใใใใซใพใใใ
ใใๆฐใไธใใใฎใซๅฟ
่ฆใชๆใฎๆฌๆฐใฎๆๅฐๅคใๆฑใใ๏ผ
|
n = int(input())
print(1 if n == 0 else len(str(bin(n))[2:]))
|
s071612278
|
Accepted
| 40
| 7,708
| 70
|
n = int(input())
print(0 * (n == 0) + len(str(bin(n))[2:]) * (n != 0))
|
s394479923
|
p03379
|
u909643606
| 2,000
| 262,144
|
Wrong Answer
| 126
| 25,472
| 202
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
n=int(input())
a=[int(i) for i in input().split()]
enu=max(a)
k=(enu)/2
sa=float("inf")
for i in range(n):
if abs(k-a[i])<sa and a[i]!=enu:
sa=abs(k-a[i])
aru=a[i]
print(enu,aru)
|
s423180173
|
Accepted
| 304
| 25,348
| 303
|
n=int(input())
x=[int(i) for i in input().split()]
nn=n//2
y=x[:]
x.sort()
low_median=x[nn-1]
high_median=x[nn]
[print(high_median) if y[i]<high_median else print(low_median) for i in range(n) ]
|
s968619933
|
p03456
|
u163421511
| 2,000
| 262,144
|
Wrong Answer
| 31
| 9,324
| 115
|
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 = map(int, input().split())
num = int(A+B)
result = "Yes" if num**0.5 == int(num**0.5) else "No"
print(result)
|
s886142996
|
Accepted
| 31
| 9,436
| 103
|
A, B = input().split()
num = (int(A+B))**0.5
result = "Yes" if num.is_integer() else "No"
print(result)
|
s625339515
|
p02419
|
u177808190
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,556
| 254
|
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
|
import sys
c = 0
for num, line in enumerate(sys.stdin):
if line == 'END_OF_TEXT':
break
if num == 0:
hoge = line
continue
for word in line.strip().split():
if word.lower() == hoge:
c += 1
print (c)
|
s739470020
|
Accepted
| 20
| 5,556
| 254
|
c = 0
hoge = ''
while True:
sent = input()
if sent.strip() == 'END_OF_TEXT':
break
if hoge:
for word in sent.split():
if word.lower() == hoge:
c += 1
else:
hoge = sent.strip()
print (c)
|
s145420541
|
p02613
|
u135197221
| 2,000
| 1,048,576
|
Wrong Answer
| 140
| 16,504
| 169
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
from collections import Counter
n = int(input())
words = Counter([input() for i in range(n)])
for i in ("AC", "WA", "TLE", "RE"):
print(f"{i} X {words.get(i, 0)}")
|
s257021435
|
Accepted
| 142
| 16,396
| 169
|
from collections import Counter
n = int(input())
words = Counter([input() for i in range(n)])
for i in ("AC", "WA", "TLE", "RE"):
print(f"{i} x {words.get(i, 0)}")
|
s884730484
|
p03455
|
u497326082
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 93
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b = map(int,input().split())
if (a + b) % 2 == 1:
print("Odd")
else:
print("Even")
|
s891766894
|
Accepted
| 17
| 2,940
| 86
|
a,b = map(int,input().split())
if a*b % 2 == 1:
print("Odd")
else:
print("Even")
|
s387027129
|
p03456
|
u951601135
| 2,000
| 262,144
|
Wrong Answer
| 54
| 3,060
| 154
|
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 = map(str,input().split( ))
#print(a+b)
count=0
for i in range(1,int(a+b)):
if(i*i == int(a+b)):
print('Yes')
break
if(count==0):print('No')
|
s585786840
|
Accepted
| 18
| 3,060
| 59
|
print("No"if int("".join(input().split()))**.5%1 else"Yes")
|
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